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H: The importance of knowing that charges only occur in nature as integral multiples
So I read the following about electric charge from Fundamentals of Electric Circuits (fifth edition) by Charles K. Alexander and Matthew N.O. Sadiku :
"According to experimental observations, the only charges that occur in nature are integral multiples of the electronic charge e = - 1.602 X 10^(-19) C"
Now I understand what is meant by the "the only charges that occur in nature are integral multiples " part - it means that the charges only exist as 1e, 2e or -1e; it cannot exist as 1/2 e, 0.8654e or πe etc.
What I don't understand is: what am I supposed to deduce by knowing that charge only exists as integral multiples? What would happen if the charges existed, for example, as 0.5e?
AI: The simplest thing to deduce is that charge, when observed under the conditions of the experiment, only comes in 'packets' of 1e. The simplest way to ensure that is that there are 'particles', each having an integral charge.
This is the simplest deduction to make. From here, it would be total nonsense to suppose that charge comes in smaller packets, like e/2, or e/3.
That's why it comes as a surprise to most people when learning about quarks, the components of nuclear particles, that they do have a charge in units of e/3. You don't observe quarks under the conditions of Millikan's oil drop experiment. In fact you don't observe quarks directly at all, but that's a different question. |
H: PIC microcontroller voltage drop
It is my first project with a PIC microcontroller, so I am sorry if my question seems too basic.
I have programmed the PIC as a frequency counter with two inputs: one input for a PLL, and another input for a clock (I am not using an external oscillator... This clock is used as a windows of 1 second so that the PIC can count on that 1 second how much pulses does the PLL make).
However, the big problem is the following: Each part works well alone with 5 volts of output, but when I connect the clock to the microcontroller's input, the output voltage of the clock drops from 5 volts to less than 1 volt. With less than 1 volt, the PIC is not able to recognize my 1 second windows. Why does this happend? How can I avoid or correct this problem?
Schematics of PIC and PLL:
Schematics of clock: The output is Q1 of 4018. That output goes directly to the input of the PIC (the input of the pic is where that square signal of the first image goes) Evcerything is fed with 5 volts
AI: As mentioned in the comments above - if your microcontroller pin isn't set as an input you will see this problem, as you're effectively shorting it to ground.
Another possibility is that your supply is browning out. Have you got decoupling caps on the VCC pins on each chip? |
H: Parallel resistor algorithm
I have build a circuit that functions as a variable load. It consists of 4 resistors in parallel - let's call them R1,R2,R3,R4 - each with an activation switch (implemeted with a transistor). R1 has the lowest value, the other resistors have value: Rn = R1*2n-1.
I'm currently trying to find an algorithm that calculates the nearest resistance that is above a desired value, Rdes. Does anybody have an idea of how this could be implemented, without looping through all the possible values?
AI: Does anybody have an idea of how this could be implemented, without
looping through all the possible values?
It's called successive approximation and is used quite effectively in determining the digital value of an analogue signal. So, with 4 resistors you would make 4 decisions in order to decide which of the 15 values above zero was most appropriate: -
To make this work like an ADC, feed the unknown resistor with a constant current and do the same for the controllable resistance. Then feed both signals to a comparator that "makes the decision".
Alternatively, buy a cheap ADC and throw away your transistor and resistor network in favour of a much more stable R-2R network and inbuilt algorithm. |
H: Oscilloscope probe compabitibility
I bought a very old Philips oscilloscope some time ago (see my post. This is a Philips PM 3253). Although this Philips really has a lot of features, I cannot really use it well, since I cannot do good readings below 1 MHz because of the bad screen (focus problems, lines stay very long on the screen, or the signals are 'clipped').
Now I saw another one at a local 2nd hand site: Metrix OX8050
I was wondering if the probes between these two scopes (or actually in general in case this one is already sold) are interchangeable.
AI: Basic probes generally are interchangeable
Basic probes means probes with BNC connectors. Specialist probes, with extra pins and built-in amplifiers (such as differential probes or current probes) are generally not compatible with other brands of scope.
One thing to watch for is the input capacitance of the scope. Capacitance can vary between typically 10 and 15pF. Decent quality probes are adjustable, but not all of them can adjust far enough for all scopes. If you don't get them adjusted right, high frequencies will be distorted, and square waves may be slightly rounded or may ring. |
H: how to connect i2c peripherals to digispark Attiny85
I am new to micro controllers and I have problem with understanding and connecting I2C devices to Attiny85 digiSpark micro controller.
According to this picture SDA and SCL are on the same pin and I am confused how can I connect any i2c device to this controller. Please can somebody help me in order to achieve this?
To be more specific, I have oled display like this one and I want to connect it to digispark.
If somebody can help me I will appreciate that! Thanks
AI: I'm not sure why the linked image shows SCL and SDA on the same pin.
If you look at the data sheet for the Attiny on page 60 you can see that SDA is on the PBO pin (D0 on the linked picture). |
H: How does this CMOS OTA with common-mode feedback work?
In class I learned about the circuit shown below. It is basically a simple CMOS OTA with a common-mode feedback realized by a resistive divider. But I am not quite sure how it works.
Lets say that I have a VDD of 1.8V and I want my common-mode level ideally at 900mV, so exactly half of VDD. Therefore the voltage vcmref is 900mV. If the voltage Vx is at 900mV, I have zero difference between Vx and vcmref, therefore the amplifier will output zero Volts. But now M6 sees zero volts at its gate, meaning it experiences a huge overdrive. How does this make sense? Shouldn't M6 be off if the common-mode voltage equals vcmref?
Furthermore: If I want to measure the open-loop gain of this circuit, I need to break the feedback loop somewhere. Let's say I break it after the amplifier. Is it correct to connect the gate of M6 to ground then, since it would be floating otherwise?
simulate this circuit – Schematic created using CircuitLab
AI: I have zero difference between Vx and vcmref, therefore the amplifier will output zero Volts.
And there is your misconception. The output of the opamp will not be 0 V, it will be a voltage that will result in Vx = Vcmref.
Yes but Vx = Vcmref so...
Note that the opamp has a very high gain. So if the opamp needs to output say 1 V, divided by a gain of 1000 that is only 1 mV which is "close enough" to assume Vx = Vcmref. You can make the opamp's gain 1000000 if you like an then Vx - Vcmref would be 1 uV.
In the ideal "everything totally balanced" situation the drain currents of M3, M4 and M5 will be identical. Also M6 must then have the same drain current (think what happens if the drain current of M6 is slightly higher or lower) so M6 must have the same Vgs as M5.
And that will be the voltage coming out of the opamp: Vdd - Vgs5 in the ideal "everything totally balanced" situation.
In practice things will not be totally balanced but that does not matter much, the gain of the opamp will still make Vx - Vcmref = very small and that's all we need.
For open loop you cannot just break the loop like that because then M6 will have an Id = 0 and the common mode will be way off. You have somehow to make sure M6 has the same Id as M5.
In practice it will be very difficult to actually measure open loop gain as it will be very high so noise and disturbances are amplified a lot.
In the simulator I generally use a trick that makes closes the loop for DC but allows to plot the open loop gain for AC.
Here's how I would modify the circuit to simulate that:
simulate this circuit – Schematic created using CircuitLab
I added from Vx an RC lowpass filter, 10 Mohm and 1 F (yes one Farad !), then an AC = 1 voltage source (note that DC = 0). For DC this does "nothing" in the simulator, it behaves the same as your original schematic. Ergo the DC operating point will be the same as in the closed loop situation, meaning the CMM feedback loop still works !!! :-)
However for AC (anything above a few Hz) the loop is broken as the resistor and 1 F capacitor attenuate everything. Simulate this in an AC simulation and plot the magnitude and phase at Vx. That is the open loopgain. |
H: Why is this opto isolator input turning off so slowly?
I feed a nice square wave into Q1/2, and on U40/1 I see a full turn off take almost a millisecond! What is happening, and can I fix it (ideally by not adding components)?
AI: There is some capacitance associated with the input. When the transistor turns off, this capacitance must be discharged through the LED of opto. As the data sheet suggests, at an input of about 1.5 volts the LED will turn off, and the output will switch. As you have determined, this takes about 3 usec, and you should compare this to the output delay. (Hint: they are pretty close to the same.)
Once the voltage drops below about 1.5 volts, the exponential nature of LED voltage/current means that the LED will start drawing less current, and rate of voltage drop will be reduced. At very low voltages the current will become very low indeed. This will lead to a long "tail" on the input voltage, as you see.
Try adding a 1k resistor from the input to ground, and see if that doesn't speed things up. There is no need to do this for an operational circuit, since it only cares about the voltage above about 1.5 volts, but try it anyways just to see what happens. |
H: How to do Non contact USB wiretapping?
I am relatively new to electronics but came up with this project, not for nefarious purposes but simply as a cool project. Simply put I want to try and read the data traveling through a USB cable with a non contact device (like a digital clamp meter but for dc). I have looked at Hall effect sensors as an option but I'm not sure they would be sensitive/ fast enough (since speed is the main issue). The constant 5 volt power line could be used as a baseline for detecting signals in the communications line. Are there any other alternatives to a Hall effect sensor in this scenario?
AI: A clamp-on detector which does not break out the individual wires won't work. USB is differential. Signal is carried on two wires, one labelled + and the other -. The sum of the currents in the two signal wires is zero, so the net magnetic field is also zero (at any distance from the pair). |
H: what is the main differences between the capacitors?
I have read many documents on these terminologies. I find almost same working for these capacitors.
Please make me clear what is the main difference between bypass capacitor, decoupling capacitor & smoothing capacitor?
Thanks
AI: They are all capacitors with different usage. I'll give you some of my own experience.
Simply, bypass capacitor is used to bypass AC signal, since only AC voltage can go through a capacitor.
Decoupling capacitor is typically used between two electrical components, in order to bring up the signal to a certain level(in a complicate circuits, not all the devices will get the AC transition fast or accurate enough.)
smoothing capacitor is normally used in a power supply when converting AC to DC.
My answer may not be 100% right. But there are plenty of articles talking about these capacitors, you can easily look up online, google is your best friend. |
H: Can AC floating ground be used as a common ground to a DC circuit?
I'm designing a system that uses a power supply to convert 120V AC current from the wall socket to 12V DC current. I'm looking at a Jameco SP-240-12.
Originally I thought pins 4 and 5 output negative DC current, meaning I would have to ground my circuit elsewhere. After seeing the datasheet though, I realized that DC- are grounding points.
But that got me thinking, from my understanding no current runs through the floating ground normally and it is just there as a safety feature. Is it possible that I could have used pin 3 and grounded DC current on an AC floating ground?
Edit: Originally I was unclear, I was asking if I can ground my DC devices to pin 3, even though it gets connected to the "frame ground" of an AC plug, like shown in the image below:
AI: Terminal Pin No. Assignment (from OP's screenshot):
AC Live
AC Neutral
Frame Ground (not "floating ground").
DC output -V
DC output -V
DC output +V
DC output +V
simulate this circuit – Schematic created using CircuitLab
Figure 1. Three output configurations are possible: floating, grounded negative or grounded positive.
... no current runs through the floating ground normally and it is just there as a safety feature.
It's not "floating ground" it's frame or chassis ground. This must be connected to mains earth as the power supply has a metal chassis. Failure to do this correctly could result in a live case. (e.g., If one of the heatsink insulation kits breaks down.)
You have the option of letting the output float with respect to ground or tie either positive or negative to ground. In most cases we tie negative to ground.
Figure 2. Datasheet entry for grounding continuity shows that it is specified at < 100 mΩ to chassis but that the test result was 8 mΩ. (PE stands for "protective earth".)
I was asking if I can ground my DC devices to pin 3, even though it gets connected to the "frame ground" of an AC plug, like shown in the image ...
No. There is no internal connection between FG and -V so there would be no current return path to your power supply. Connect your return path to -V and link the second -V to FG if required. |
H: The reason for multimeters to have a high input impedance
My question applies not only to multimeters, but all devices which require a low or a high impedance at the input or output.
Based on my understanding, a device would require a high input impedance, so that the input signal does not get affected by it. In other words, if the circuit was like shown below, then at point A, the effective input signal would be like an output at the voltage divider, hence Rinput / (Rinput + R_source), but since R_input is large, this value will likely become close to unity.
simulate this circuit – Schematic created using CircuitLab
Am I correct in understanding this? Also, is it the same reason multimeters have a high impedance? I.e., so the measurement is not affected and is correct?
AI: It's not that multimeters have a high impedance - just voltmeters do.
When measuring voltage, you connect the voltmeter (a multimeter in voltmeter mode) to the nodes between which you want to measure: it is a parallel connection. In order to measure correctly, you need the impedance of the voltmeter to be as high as possible so that very little current goes through it, so that it doesn't disturb the circuit (the current that was going through the, for example, resistance that you are measuring voltage on will continue to be almost exactly the same, thus the voltage drop will as well).
When measuring current, on the other hand, you connect the ammeter in series. This time, to prevent the device from disturbing the measuring, its impedance should be as low as possible, so that the circuit keeps behaving the same way (no voltage drops in the ammeter and the current flowing in that branch of the circuit stays the same). |
H: 240VAC to 240VDC
When a 240 Vrms mains voltage is rectified with a bridge rectifier and reservoir capacitor, the output voltage becomes 340 VDC, on average. Then, I would put a buck converter that oscillates at 20 kHz (say) and step it down back to 240V. But how would you regulate that 340V down at 240V AND eliminate (all or as much as possible) that resultant high-frequency ripple?
What components would you add?
Edit: This is all wrong. Instead of trying to use a buck converter, I would build a Full Bridge converter with PFC.
AI: When a 240 Vrms mains voltage is rectified with a bridge rectifier and reservoir capacitor, the output voltage becomes 340 VDC, on average.
Correct, if smoothing capacitor added.
Then, I would put a buck converter that oscillates at 20 kHz (say) and step it down back to 240V. But how would you step that 340V down to 240V AND eliminate (all or as much as possible) that resultant high-frequency ripple?
You wouldn't. You'd use a 240 V AC to 240 V DC switched mode power supply and do it in one step.
Figure 1. Switched mode power supply block diagram.
What components would you add? And what should be the base voltage to the switching transistor?
You generally wouldn't use a BJT transistor. I suspect you've picked up some terminology somewhere but don't have enough understanding at present to do this project. Pick something at low voltages first.
The 240V sine wave gives 240 Vrms, but the peak-to-peak voltage is actually about 340V.
Almost right. \$240 \sqrt 2 = 340 \; V_{peak} \$. The peak to peak voltage of the AC is double that. |
H: 120v DC to 5v DC conversion
For our e-Kart project we need to power a raspberry pi 3 (5V 2A) of a battery of 120V.
We are having trouble finding a good solution as these high voltages seem to high for a conventional (self made) SMPS and its not AC voltage that needs conversion.
Someone suggested using a normal USB wall charger directly. Would this be a possibility? And if not, why?
Any suggestions are welcome, as of now it seems that an extra battery might be needed..
Kind regards!
EDIT:
We were able to produce 5V using a Sony wall charger!
However, as Transistor noted below, it seems that this wont work with all charger types, the one that shipped with the Raspberry Pi (notably bigger than the Sony one) did not work.
The Sony one was rated 100-240VAC as was the RPi one.
AI: Someone suggested using a normal USB wall charger directly. Would this be a possibility?
It is a possibility but a little unlikely.
Figure 1. SMPS diagram.
Most SMPS use the configuration shown in Figure 1. At first glance it would appear that since the PSU has a rectifier anyway that you should be able to connect DC either before or after the rectifier. The problem is the DC voltage: even a "universal" 100 - 240 V AC PSU running at lowest input voltage of 100 V AC will have a peak of 100√2 = 140 V on the DC bus. Your 120 V battery might be just too low to work.
If you can find one that will work down to 85 V AC (-> 120 V peak) you might get by. Some laptop chargers will do this and you could consider this as a first stage voltage reduction into, say, a 19 V to 5 V converter of your own design.
As suggested in the comments, you could design your own. 120 V DC is in the dangerous category, however, so a pre-fabbed solution would probably be better. |
H: How do I generate ~50 microamps of DC across dynamic resistance?
How do I generate ~50 microamps of steady direct current across a load (tissue sample) with dynamic resistance that varies between 2kohm and 20kohm with good visibility into circuit operating conditions and safeguards against elevated current and voltage?
Ideally the power source will be an over the counter battery.
Update: I completely missed the existence of constant current ICs, but see the answers for additional info on monitoring and alternatives.
AI: A brief review of the specifications leads to:
Constant current of 50 µA into 20 kΩ (worst case) gives required voltage \$ V = IR = 50µ \times 20k = 1000 mV = 1 \; V \$. This should be easy to do from a 9 V battery.
A circuit that can provide ~50 microamps of steady direct current to a 1 square inch electrode applied to heterogeneous tissues with good visibility into circuit operating conditions and safeguards against elevated current.
I think you can satisfy both requirements by using a 50 µA constant current source which is limited to some maximum voltage at the terminals.
The constant current source may eliminate the need to read the current. Since it is constant we need only read the voltage to calculate the resistance from \$ R = \frac {V}{I} = \frac {V}{50µ} \$.
Since 1 V is adequate to drive the required current an upper limit of 1.2 V seems reasonable to give you some headroom.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Very sketchy outline of required system.
I1 is the current source. It needs to have the upper voltage control built in.
VM1 is shown for reference. The voltage across your tissue sample would appear here.
A simple micro such as an Arduino or any of the others would provide a simple means of reading, scaling, averaging, etc., the readings and outputting to a display or logger. |
H: Converting pulse into volt?
I am working on the lab which is allowed to ask other people for hints.
Below picture is part that I don't understand.
In the image, program called "Audacity" is used to capture that waveform.
My question is, is there a way to convert each Y-axis values of each pulse in signal in audacity into volts? (since the description in the photo made an example with volts V)
For example, first pulse in first signal in the picture, I can assume that it spans from around -0.1 to +0.8 by just looking at it. How can I convert this value into volt, V?
AI: It is not clear what you want. However, the text says the vertical span of the display is 4 volts. Thus 1.0 on the vertical axis is 2 volts and -1.0 is -2 volts. Knowing that, you can convert any other y-axis value to volts (just multiply by 2). Thus a span of =0.1 to +0.8 represents a voltage span of -0.2 to 1.6 volts. |
H: Charge battery with AC
EDIT:
Question 1: what type of current will be created when passing magnets past a coil? AC or DC?
Question 2: What is the best way of storing the electricity generated with a contactless dynamo?
I'm working on a contactless dynamo project for my bicycle, (similar to this project: http://www.instructables.com/id/Contactless-dynamo-powering-bike-safety-lights/) except my concept includes being able to turn on/off the lights at will instead of them receiving power directly from the coil as shown on the link I just shared. I'm attaching some photos for reference.
I was successfully able to get a simple LED to flicker with 6 small neodymium magnets and one 12V ~80 Ohm coil. From what the internet and some friends have mentioned, I would need to convert the AC being induced by the magnets passing by the coil into a battery for a more complex bike lighting system.
My ideas for the bike lighting system are to power a front-facing headlight, a rear-facing brake light, increase luminescence of rear-facing brakelight when I apply pressure to handbrakes, a blinker system sensitive to turns and toggle-able with hand switches. Similar to how a lighting system would work on a scooter or motorcycle.
Any insights into this would be much appreciated!
AI: Two to four diodes and a capacitor will get you a long ways, without the complications of battery charging (which is picky to do well, depending on the type/chemistry, some being more difficult than others.) The concept you are after at that level is not too complicated - a rectifier (converts AC to DC) and a filter (smooths out the pulsing DC.) You'll want those anyway for a system with a battery, but with a small LED current and a fairly large capacitor, you may find that your application can be adequately covered with just the capacitor.
You'll also typically get more output if you shift your generator position towards the rim of the wheel, so the magnets go past the coil faster. More magnets will also help, and there's more room for them out there.
Mind you, it's far more convenient with current technology to simply use a rechargeable battery powered light, in practical terms for most riders - you really need an odd set of circumstances to make the "on bike generator" actually make sense in the era of rechargeable lithium-ion batteries and LED lights. But it can be fun to tinker with.
simulate this circuit – Schematic created using CircuitLab |
H: LVDS Displays - Swapped EVEN and ODD Channels?
A friend of mine gave me 2 LCD display driver boards (VGA/HDMI to LVDS). One has RTD2553V chip (from Realtek) and the other has RTD2660H (I cannot share the datasheets because they are not public). The main difference between them is the latter one accepts HDMI input. He said that he's currently using the first one and wants to use the second one due to that difference, so he asked me to check if it is possible without modifying LVDS display connection.
Anyway, I checked the connections between chip and LVDS output socket (2x15 vertical header /w 2.54mm pitch). And I noticed that EVEN and ODD channels are swapped. For example, pin7 is TXE3- in the 1st board but it is TXO3- in the 2nd board. Likewise, pin29 is TXO1+ in the 1st board but it is TXE1+ in the 2nd board. And it is true for all of channel outputs (i.e. if a pin is EVEN on the 1st board, the same pin is ODD on the 2nd board), only VDD and GND connections are matching.
Without further ado, I'm finalizing this verbiage with the actual question. Can it be used? Will that difference cause any problems?
AI: It should not be used just like the first, I think.
For full compatibility, you need to swap EVEN / ODD channels.
And for this you will have to edit the flash firmware on the board.
Datasheet
Try to connect and submit a test image, as you will not be able to "kill" the display in any case. |
H: incorrect voltage applied, dev board still worked
I took this guy from here - http://rlx.sk/en/accelerometer-gyro/3582-vibration-sensor-module-801s-er-sen801svs-.html and without noticing I have applied 3V3 instead of the advertised 5V to it. It also seemed to me it has worked fine. Can anyone tell me why it has still worked or if it makes any difference in this case ?
AI: The module contains an extremely simple circuit:
The sensor is probably a mechanical one (metal ball or spring) so that works at almost any voltage. The opamp is an LM393 which works form 2 V supply and more.
So there's no reason why it would not work at 3.3 V.
Note however that I personally would not use this circuit as it has issues in its design. There's a reference circuit in the datasheet of the RZ801 which will work much better. |
H: MOSFET with Integrated Driver for Boost Converter
simulate this circuit – Schematic created using CircuitLab
Trying to build a boost converter using a PIC microcontroller with minimum number of external components. Was hoping to find an NMOS with integrated driver (dashed rectangle above) in a small surface mount package (e.g. SOT-23-5) to use as the switch and drive it directly from the PIC output pin. Searched for 'gate drivers' and 'power switches' but couldn't find any suitable part.
Edit:
This is a commonly used configuration (driver+MOS) and was under the impression that more devices in a single package would be available. Why are devices like this seemingly unobtainable?
AI: Is it possible to find an NMOS with integrated driver
No (bold statement!), I think this is a rare beast and there are a few reasons that spring to mind: -
To properly turn on the MOSFET (without damage) you need a supply rail of a few volts (maybe 5 volts) to no more than 20 volts and this instantly restricts application usage of such a device.
The low end supply voltage (if a tad too low) might indeed cause the MOSFET to fail because it doesn't turn on adequately enough so, an extra pin for under-voltage lock-out (UVLO) would likely be needed and, if you look at a lot of integrated switchers, these are quite common. Not turning on enough would generate too much heat in many applications and burn the MOSFET.
The device would need to protect against a supply voltage greater than 20 volts else the gate source region could become damaged by over-voltage and, not surprisingly, many integrated switchers do have an internal linear voltage regulator and this usually means a Vcc pin to hang a power supply decoupler on. Can you see that you are beginning to adopt the fuller functionality of a proper integrated switcher?
To drive the MOSFET in light load applications requires a peak drive current to the gate-source parasitic capacitance of a few mA i.e. just enough current to adequately activate the gate without wasting power. In higher load power applications this has to rise to several hundred mA to avoid switching losses being too great. Now, you need another pin that sets this current appropriately for the application.
The majority of engineers that require a boost converter will not try and use a microprocessor as the "brain" because they recognize that trying to control duty cycle under varying load and input voltage scenarios is best left to an integrated switching device.
There may be other good reasons that people can think of. |
H: SDHC over SPI: OCR Register Reading 0x000000FF, nBusy bit 0 even after IN_IDLE_STATE clears to zero
Has anyone else enoucntered this problem before?
I'm using an STM32F105VC, SPI port 1 at a speed of either ~100kHz, ~200kHz and ~800kHz.
I've tried using a 16GB Sandisk microSDHC and a 2GB Transcend microSDSC both providing the same results.
I have the Host Capacity Support (HCS) bit set when I send ACMD41, which loops until the idle state bit in R1 response is no longer set (usually after 3 repeats; seems too fast maybe?).
I have tried with the SS toggling between each Tx/Rx transaction and with it held low for the whole init procedure, made no difference. Stuff bits as 0's or 1's made no difference. Sending around 1s worth of ACMD41s before checking the OCR made no difference.
If you think you can help give me a shout for more info on my setup. Or if you can think of any obvious errors I may have made then it would all be greatly appreciated!
AI: Apologies, got tunnel vision on this bug. The issue was actually a typo in the code; sending CMD8 instead of CMD58 (Both of which return 5 byte responses). Hopefully this doesn't happen to anyone else but gonna leave this here just in case! |
H: Testing an oscilloscope
This question is a bit related to an earlier question: Question
With my current scope (Philips PM 3253, 20 MHz 2 channels), I cannot measure the quality of edges from square waves from the test signal, because if I put the time to less than 1 ms, I either don't see anything or a straight line.
I assume this is a problem of the scope (it has a quite bad screen, the beam stays 'forever' or is fading away fastly, so all the time I have to press the 'reset 'beam'' button.
Is there something I do wrong? I already know I should use the 10:1 setting on the proble, but still have the same problems.
So I'm in the market for a new (old analog secondhand oscilloscope) and was wondering how to best do a test similar to this?
My idea: use a square wave (which I guess never is 100% square). I put the time to e.g. 10 us or whatever and I would see square waves which connect the 0 and 5V (or + or - volts) together with a 'rather' steep line. And I want to measure the time from 0 to 5V or vice versa. This should be something in us I assume.
AI: I guess you have a problem with triggering.
Follow following 3 steps:
Select correct trigger source. You probably have a choice between CH1, CH2, LINE, EXT. Select CH1 if the signal is picked up at channel1.
Select correct trigger mode. You probably have a choice between AUTO, NORM, and a few other modes. If you select NORM the beam will only be visible if trigger is triggered (see 3.). If you use AUTO, you should always see the beam. So select AUTO.
If you now see the signal but it is not standing still trigger is not tiggering; probably because the trigger level is beyond the signal range (i.e. too low or too high). So select an appropriate trigger level |
H: Digital voltage/current controller for DIY bench power supply
I'm trying to design a portable rechargeable bench power supply. I have a 12V sealed lead-acid battery (WP7.2-12) which I can use as the main power source. These batteries are cheap and easy to source, and have reasonable power output specifications.
I would like to be able to digitally control the output voltage and current using a microcontroller.
I'm seeking advice for a voltage/current output controller which would be reasonably easy to interface with a MCU. It needs to be digitally controlled (i.e. serial, SPI, I2C, 8-bit bus, etc), and should be able to provide two modes of operation:
Voltage output with selectable voltage and maximum current
(0-12V, 0-10A)
Constant current output with selectable current output (0-10A)
(Maybe these features need to be implemented as two separate circuits, using relays to connect the output to one or the other...)
Would an LED driver such as those referenced here be a good route for this type of controller, or would it be best to roll-my-own circuitry for this? Any pointers/tips appreciated.
Obviously there are other components (i.e. overcurrent protection, battery charge controller, deep-discharge prevention, user interface, etc.) that must be implemented as well, but for this question I'm simply seeking a way to control the output voltage/current per the above.
Thanks!
AI: Find a suitable voltage regulator that can supply up to 12 volts at 10 amps and has a feedback (FB) pin that is used for setting the output voltage via a potential divider.
If your input voltage is only 12 volt then use a pre-boost regulator to deliver maybe 15 volts to the regulator described in (1).
If you consider that maybe 11.5 volts is enough at the output then you don't need a pre-regulator if your main regulator has enough grunt and low drop-out to do the job (but you won't quite get 12 volts of course).
Two more things to do and the first revolves around using a digital pot to control the voltage fed back to the FB pin. Alternatively inject a current into the FB node that can "con" the regulator into assuming a diufferent voltage output. Here's the first idea: -
This was taken from figure 8 in this ADI document.
The second idea is to use this type of injection using a current output DAC: -
Or like this: -
And the second thing to do is have a small resistor in series with the output of the regulator (but before the feedback resistors) that can be used to measure load current. You can use a circuit like this that takes the high-side measurement and references it to 0 volts that can be read by an ADC: -
The above taken from here.
Or you go the whole-hog and get yourself a step-Down controller with Digital Power System Management like the LTC3886: - |
H: How to determine the nominal current of an LED?
I salvaged the LED disk from a defunct LED light bulb. (The problem was an arced transformer, the LEDs are intact.) It has several unmarked white SMD LEDs soldered on it: 5 lines parallel, with 3 LEDs in series on each line.
I want to use this disk in a strobe. My aim is to get the maximum brightness out of it without damaging the LEDs. Is there any way I can determine (vaguely) its nominal current rating, rather than guessing it?
AI: It has several unmarked white SMD LEDs ...
... without marking you cannot get a datasheet and without datasheet there you don't have a chance of finding out the real current.
However there is a possibility of getting a rough impression of the current needed by the LEDs:
Attach a current source and a voltage meter to the LEDs (the current source may be a voltage source with a resistor in series).
Slowly increase the current from 0A until the product of current and voltage is the power you estimate (the nominal power of the bulb multiplied by the efficiency of the transformer; maybe 5W for a 7W bulb). |
H: Is current calculation in copper wire for AC simple or not?
I'm trying to learn electronics by myself, and I'm starting with simple current calculations. Let's assume we have a straight wire of length L connecting a source and a sink. I know the voltage (V), and for finding the current I need to apply Ohm's law:
$$I = \frac{V}{R}$$
This formula works fine for a DC source, and it's an easy one. What happens when one has an AC system? I saw that the formula changes to:
$$I = \frac{V}{Z}$$
where Z is the impedance, which is a complex number. This formula is a new universe for me because of this complexity. Or at least that's how it looks when trying to learn it.
From this website I saw that one could calculate the impedance at different frequencies, but AC also has to do with magnetic fields as well, right? There are two issues I don't understand yet:
Why is the impedance a complex number and what does the imaginary number suppose to reveal?
Is the medium around that wire changing the current calculations due to the magnetic field? If so, how does it get into OHM's law?
I know it might sound stupid to most of you, but for me is not easy to understand.
AI: It's got nothing to do with magnetic fields in a simple situation (skin effect does come into this later but for now ignore it and just learn the basics of impedances). Walk first, then run.
If the load is a resistor then the load impedance = R (or Z = R). So you get a sinewave current with a sinewave voltage and the two waveforms are in sync: -
However, in AC circuits there are capacitors and inductors and these numerically are represented by complex numbers. Simple reason: the voltage current relationship is at 90 degrees. See this for a capacitor: -
And for an inductor: -
So, if you have any inclination about complex numbers this should make sense. If you are a bit rusty on complex numbers then you probably need to do some more research on the topic.
Pictures taken from here and this might be a useful learning resource. |
H: pulse response of opamp integrator
I was simulating the below opamp integrator circuit with pulse current.As this is an inverting integrator I thought it would integrate the input current and give me a triangular waveform.
But after simulation in PSpice I got the following result.
Can anyone explain why the current flowing through capacitor jumping to 0 to 2u instead -1u to 1u.
Also why Vout is almost constant at 15v.
From my understanding, if dc current flows through capacitor then the voltage across through capacitor will be ramp.
AI: Assuming this was an ideal op amp your question is answered by the initial condition at time zero for your simulation. If the current in the capacitor was zero at time zero, and the voltage at the output was zero at time zero, the current and voltage waveforms would be as you describe.
Since this is in the real world, and the initial conditions are I = -1u, the op amp starts out saturated because C1 is charged to a large negative voltage.
To get the desired result, simulate a switch across C1 with the switch closed, and open the switch a short time after your current starts. If you vary the start time delay, the triangle wave will have a differing DC component. |
H: LC oscillator versus Kirchhoff's law for meshes
Analyzing the behavior of an LRC-circuit I got stuck on a seeming contradiction between the vector diagram of the circuit and Kirchhoff's law for meshes.
Please consider the circuit shown below and follow the description:
In its initial state, at t=0, the capacitor is fully loaded, so its voltage is equal to that of the DC voltage source (for example 5V). No current is flowing, so there is no voltage drop at the resistor and no induction voltage at the inductance.
At t=1s, the DC voltage is switched to zero Volts. This results in a closed circuit of L1, R1 and C1 with no source, thus a passive circuit. However, the circuit contains electrical energy which is stored in the electric field in the capacitor. Therefore, a transient process will start at t=1s. During this process, the capacitor is discharged and the energy is dissipated in the form of heat into the environment by the resistor. At the end of the process, the capacitor will be discharged and no current will flow anymore.
Now for the question: As the vector diagram below shows, the voltages at the inductance, capacitor and resistor cannot add up to zero, except each of these voltages is zero itself. (UR is perpendicular to both UL and UC). But according to Kirchhoff's law for meshes, the voltages have to add up to Zero during the transient process (closed mesh of L, R, C). So we have a seeming contradiction here. What's the resolution to that? (Or: Where is the error within my understanding?)
I know that the vector diagram is not exact, because what I consider here, is a transient process with decreasing amplitudes while the vector diagram represents stationary behavior with constant amplitudes. However, that shouldn't make a difference, because the phase angles of the voltages are correctly represented, aren't they?
AI: I know that the vector diagram is not exact, because what I consider
here, is a transient process with decreasing amplitudes while the
vector diagram represents stationary behavior with constant
amplitudes.
Not just inexact but wrong.
However, that shouldn't make a difference, because the phase angles of
the voltages are correctly represented, aren't they?
You CANNOT use a phasor diagram for this - phasor diagrams are reserved for steady state AC analyses (with a constant excitation frequency). For instance, if you apply a DC level of 12 volts across an inductor, current rises at a rate of V/L - where is the 90 degrees phase shift here? No, please don't use phasor diagrams for transient analyses. |
H: Beginner question about CPU clock
So I just started learning AVR and I am currently working with the ATMega328P. I have a question regarding the CPU clock. Currently I am using the 8MHz internal clock. Now I know that F_CPU tells the compiler what my clock speed is. The problem here is that the CLKDIV8 fuse is programmed by default, so I was wondering what value would I enter for F_CPU? Will it be 8000000 or 1000000.
I would be glad if someone could clear this confusion.
AI: 1Mhz, until/unless you change the CLKDIV8 fuse, then it's 8MHz |
H: Beginner PLC: Configure Ethernet/IP Connection
I am really new to PLCs programming, connecting, etc. I received a 1756-L72 with Logix5572 Processor and an EthernetIP card. My goal here is to connect it to my laptop along with another device (a Bar Code Reader). Both devices are connected to an unmanaged ethernet switch and one of the port to my laptop.
Now the problem is that all my Rockwell Softwares are on a VM. Therefore, I need to connect all those devices to the VM on my laptop and it does not seem to work.
My ethernet IP on the VM is 192.168.198.128 while my PLC is 192.168.0.205. I know i have to get the 3 first equal but I don't know how to change my PLC IP address. I tried following this website steps but it did not work.
I know it might not be clear, so I'm available to answer any questions you guys have.
Thanks,
luisarcher
AI: This site is really for electrical / electronics design questions so your question isn't a great fit. However ...
On VMWare, and I presume most other VM software, there is an option on the VM settings to configure your network connection as NAT or bridged. You have two choices:
NAT
Configure the IP address and subnet mask you want on the host machine.
Set the VMWare to NAT. It will then pick up the host address.
Bridged
Leave the host PC as is.
Configure the IP address and subnet mask on the VM.
Set the VMWare to Bridged.
It's often much easier to use a router with DHCP enabled. Configure it with the browser interface to the PLC subnet (using a unique IP address) and then let the PC obtain its address by DHCP. In this case you would use NAT for the VM settings.
Mr. PLC and PLC Talk are great sites for help with industrial controls. |
H: Is 0603 vs 0402 assembly mass production cost optimization still relevant?
Historically, choosing a design avoiding 0402 and finer pitch components was advantageous for mass production cost savings. The yields were improved and accuracy requirements for pick and place machines were reduced. This allowed vendors to choose from a larger number of manufacturing facilities and identify cost saving opportunities.
Does that kind of thinking still hold any water given that 0402s and BGAs have been popular for a long time? Are there still factories that specialize in low cost larger pitch manufacturing?
Just to be clear, I'm only talking about production volumes in the millions.
AI: Many assembly houses these days do 0402 with the same machines they do anything else, possibly using a different needle, though I'd suspect they'd be using the 0402 capable needle for 0603 and 0805 as well.
Is it still true you'll have more choices when you don't go below 0603? Sure. Most likely. There's a lot of cheap assembly houses that are cheap because they use the old equipment of the others, probably everywhere around the world. Some assembly houses may do down to 0201, but not be too happy to do the smaller stuff, because it requires extra operator attention.
However, when you go into volume, the cost of the more advanced assembly house will likely not weigh up to the cost difference for smaller circuit boards, more efficient systems and/or lower component cost. And some times using 0402 or even 0201 offers better per-component performance as well, such as lower parasitic effects.
Obviously if a 5000 unit reel of 0603 capacitors costs $15 and the 10000 unit reel of 0402 of the same value cost $20, that'll add up when you're making 10's of thousands with 10 each, but not really do much at all below using a reel per month.
Because boards are now almost always made to 5mil/5mil standard, the board won't likely be much more expensive if you make anything more compact with tiny components, but at high volumes the board space savings will start weighing as well. If a panel costs $100 and with 0603 the panel can fit 20 PCBs, but with 0402 it can fit 25 PCBs, that usually saves much more in volume than any extra cost you have at assembly in high quantities.
In all, if you want to be fully sure you'd need to do a cost estimation, including an RFQ to a few assembly houses that tickle your fancy. All the assembly/full-service houses I use are always ready to pick up the phone or answer an e-mail with questions about comparative costs. And more often than not I find the cost increase of something "unwise" 10 years ago falls into the less than a few percent now. And the same will happen later to stuff we think expensive now, so, really, you need to regularly keep asking them if things have changed if you want to be the best designer you can be.
Summarising: The only reason I don't do at least 0402 in my designs is if it's a hobby thing for me or others, where I want to be as quick as possible with replacing components, or I want others to be able to use my design as well, as I am not even noticing significant cost increase up to 160mmx160mm boards at 10 units. Over average past orders. |
H: Op. Amp. read variation when Earth is connected with Signal GND
I've made this simple circuit to read the line voltage (the 1 pin is conected in a uC by a resistor):
In the image NEUTRO is Mains Neutral AC, FASE is Mains Line AC and TENSÃO is the signal to the uC (microcontroller).
The circuit worked very wel until a connecting with the PC's serial connector has been made with de eqp. This connection had increased the signal readed (1 pin), making a measure error by 30Vrms in the uC.
When i was probing and investigating what would be the problem, i noticed that in the PC everything is connected with earth! USB shield and GND pin, GND of audio conector, GND signal tracks in the motherboard, shield and GND of the serial conector, everything.
I've tested 3 PCs and all of them is the same.
Searching in the internet i have found conflicted opinions about the union between Signal GND and Earth.
So:
What is better to stop the circuit functioning variation? Make the union once of all?
If this represent's users risks why in the PC everithing is Earthed?
And finally what kind of problem this connection could cause? Could expose the circuit to line noise for example?
Any other thing to concern?
Thank you!
AI: What are you doing with your resistor values? You need matching to obtain minimal common mode errors (such as when neutral becomes connected to a different point or ground): -
Study the above. R1 and R2 and R3 = R4 and R5 and R6. Even C1 = C2 and importantly R7 and R8 are equal.
If you got rid of your R12 (s/c) and made R1 and R9 each 16k4 (twice 8k2) then it would work better. You also need to ensure that your R11 and R14 are rated for the incoming supply voltage maximum peak voltage. That is why you see three resistors in series in the reference design I showed. |
H: How to make precise holes in Eagle
I need to create 3 circles in Layer 20 in eagle, of witch one should be in the center of my board and have a diameter of 10 mm and the other 2 should have a diameter of 4,8 mm. The two circles should be symmetrical spaced from the circle in the middle (the one that is 10 mm) and the distance from center of either one of the circles to the one in the center should be 40 mm, also all the centers of the circles should be on (x= z, y=0).
Thanks!
Mihai
AI: If you right click on the part / via or any object in the eagle pcb router, you can access a "properties window", you can here enter the exact coordinate of the object.
You just have to do the math and then enter the position. I'll let you do the math part, I'm pretty lazy... |
H: Rectifying diode for boost converter in nixie clock
I've been working on this schematic to create a nixie clock. But I don't understand the need for recifying diode D5 in the schematic. Wouldn't a standard diode work? Does it have to do with the "back emf" produced by the inductor?
AI: D5 is part of a high frequency boost circuit formed around the MOSFET Q7 and the inductor L1. together they dump energy into capacitor C4 and they will use high switching frequencies possibly circa 100 kHz. Generic boost converter: -
Your boost converter featuring D5: -
This means that you must use a fast diode or the extended reverse recovery time will significantly reduce your high voltage rail and/or efficiency. Reverse recovery time means that when a diode goes from forward to reverse bias, current will flow against the grain for a short period of time and for a 1N4007 (for instance) this can be as long as 30 us. 30 us would severely disrupt the working of your boost regulator.
Fast diodes are in the low tens of nano seconds with a diode like the BAS16 being about 5 ns (for example). Nixie tubes are cool BTW! |
H: KICAD PCB coil design related
My almost completed pcb design in KICAD requires a coil which needs to be accurate in dimension (coil is optimized using a optimization calculation algorithm developed in our lab which needs to be realized as is). If I draw using gui tools in KICAD, it is hard to make it accurate in terms of coordinates even with lowest possible grid size. I am looking for something similar to Allegro PCB editor where we can draw traces using instructions in KICAD. I believe it is possible since, I saw a python console in Pcbnew tool. Can somebody help with any information or tips that can at least draw lines with coordinates provided in instructions?
AI: You can use the Python console like this:
import pcbnew
from pcbnew import wxPointMM
board = pcbnew.GetBoard()
tracks = board.GetTracks()
new_track = pcbnew.TRACK(board)
new_track.SetStart(wxPointMM(STARTX,STARTY))
new_track.SetEnd(wxPointMM(ENDX,ENDY))
tracks.Append(new_track)
Alternatively, you can click on a track in pcbnew and type 'E' (or right-click and select Properties)
This will bring up the Track and Via Properties window where you can choose the start/end positions to finer than the user grid. |
H: Application of the Sampling Theorem to a System
I have the following system:
$$G(s)=\frac{1}{10s^{2}-5s+2}$$
that I want to discretize. How do I apply the sampling theorem to a system so I can choose the correct sampling period?
Furthermore if I use a wrong sampling frequency how can I demonstrate that there is something wrong with the system? I know that we won't be able to recover the right analog data if something like that happens but I don't understand how I can actually "witness" it.
AI: How do I apply the sampling theorem to a system so I can choose the
correct sampling period?
You can't apply a sampling regime to a transfer function of a system because all the TF contains are placements for poles and zeroes and it contains no knowledge about the input to the system in the frequency domain.
You could make an argument that the TF is basically a low pass filter and that beyond a certain input frequency the output will be very small but there's nothing in your question that allows this judgement call to be made.
Furthermore if I use a wrong sampling frequency how can I demonstrate
that there is something wrong with the system?
Engineers decide beforehand what the input frequency limits are and apply anti-alias filters to prevent significant aliasing errors so, if something comes along that is larger than expected in normal situations and might cause an aliasing error then you could compare the RMS of the anti aliasing filter output with the raw signal and make a decision based on this. |
H: Designing an LED Driver
I'm a beginner in electronics, and started the 2nd chapter of AoE, regarding transistors. I've decided to design my own LED driver, as mentioned on page 76 (3rd ed.)
simulate this circuit – Schematic created using CircuitLab
Some specs:
V1 = +3.3 [V], V2_LOW = 0 [V] , and V2_HIGH = +3.3 [V].
The current (I1) through D1 should be 5 [mA].
Using a green LED (for example), the voltage drop V_D1 is 2 [V] at 5 [mA].
BETA of Q1 is at least 25.
Given the above, I need to figure out the values of R1 and R2. Here's my attempt:
To make sure that I1 = 5 [mA] when there is a signal, I'll need to find an appropriate R1:
V_D1 + V_R1 = V1, so the voltage drop across R1 (V_R1) is 3.3 - 2 = 1.3 [V].
V_R1 = I1*R1, so R1 should be 1.3[V]/5[mA] = 2.6 [k].
Now, I need to find R2 such that there is saturation when there is a signal:
I1 < BETA*I2 = BETA*(V2/R2), so R2 should be no more than BETA*V2/I1 = 16.5 [k].
In conclusion, R1 = 2.6 [k] and R2 = 16.5 [k].
Is my solution correct? If not, where and why is it faulty?
AI: You have made a small mistake when computing the \$I_2\$. You say \$I_2 = V_2/R_2\$, which is not correct.
You should say \$I_2 = (V_2-V_b)/R_2\$, since \$V_b\$ is not equal to 0.
Then, you can say \$V_b = V_{be}\$ and assuming you will have the transistor in saturation \$V_{be}\$ is actually known and is equal typically to 0.7 V.
Moreover, there is also a mistake in the collector side calculations. More particulary you have thought the voltage drop across \$R_1\$ wrong.
From Kirchoff's law you can write \$V_1 = V_{D1}+I_1 \cdot R_1 + V_{ce}\$. In this equation you are missing \$V_{ce}\$. If you assume the transistor will be in saturation it is typical to assume \$V_{{ce}_{sat}} = 0.2V\$ (this value is given at the datasheet of the transistor). Now you can calculate \$R_1\$! Note this is the complete formula. The way you did it is also correct, since \$V_{{ce}_{sat}}\$ is quite small. You just make some approximation by ignoring it. |
H: EEE-INST-002 Wire Derating - Single Wire vs. Bundle/Multi-Conductor
We have an aerospace application and are having trouble defining EEE-INST-002's current derating guidelines in regards to wire.
The issue that I have isn't about stranded versus solid. MIL-W-22759 does seem to classify tinned, copper strands as one wire so long as the outer diameter of said wire is of its declared gauge. Similarly, we don't have any cause for concern when it comes to multi-conductor, shielded, jacketed cable in the sense of, say, Ethernet or something similar.
What I'm having trouble understanding is what exactly does NASA consider to be a "wire bundle," and when should we use the derating guideline for bundled wires/multi-conductor cables?
Consider this portion of our application:
We need a power input of something like 85A @ 70V. INST-002 would dictate that we need 8x 12AWG wires since, according to the derating table, 12AWG wire is rated at 25A, and 4x would cover the positive bus and 4x the negative (with some margin to spare).
But wait! We just tie-wrapped these wires inside of our box, making this a harness (bundle) of 8x wires. So....if these are now part of a bundle, did our derating of each 12AWG wire in this bundle suddenly drop to 11.5A according to INST-002? Why?
I found an article titled "Derating Standards and Thermal Modelling Tools
for Space Harness Designs" which boils down to "why in the world do multi-conductor cable ratings vary so widely between different space agencies?" But I was not able to glean any sort of reasonable answer from it regarding my own application.
Any thoughts? And more importantly, are my boss's fears that we suddenly need to double the current rating for every cable in our unit unfounded, or at least able to be allayed?
AI: Imho the 25A is for a single instance of the cable in an open situation . As soon as you group 4 well 8 together into a box the cooling is reduced so the current capacity has to be reduced.
This explanation is the same as cables laid open in a loft or attic or wall with an air space. But as soon as the cables are covered or surrounded by insulation the current capacity is reduced. |
H: Transistor as switch and voltage drop
I've started studying basic electronics, and in this video I noticed he had a similar example as in my textbook: https://youtu.be/sTu3LwpF6XI?t=3m28s
Above is a screenshot if you don't bother to click on the link.
So, in essence.. when the switch is off, the LED to the right on. This means, there's a voltage drop across the second resistor and a voltage drop across the LED.
Now, by turning the switch on, we'll have voltage U across both first two paths. The first resistor should make sure not too much current is flowing through the base and thus, turning the transistor on. Now when the transistor is on, the voltage will be lowered to 0 after the second resistor.
Why the LED turns off at this point is because there's no (or at least not enough) current flowing through it (the potential is zero so there can't be). This should be due to the resistance of the LED itself. Or is there anything else going on here?
AI: Your understanding is pretty much correct.
When the transistor is turned on it is driven into saturation and becomes a very low resistance in parallel with the LED. This pulls the voltage down towards zero - maybe 0.2 V and this isn't enough to light the LED.
Figure 1. LED I versus V curves.
If you pick the colour of your LED and have a look at the graph you can see that at 0.2 V virtually no current will go through the LED.
Your example circuit is using the transistor to "shunt" the current away from the LED. This is a bit wasteful of energy as it draws more current when the LED is off than when it's on. The more usual arrangement is to put the LED in series with the resistor. |
H: safety relays and PLCS
I am having trouble trying to get my head around using safety relays in PLC controlled machinery. I want to install a PLC into a machine that needs upgrading, it is straight forward enough, well i thought so until i ran into two safety relays im just not sure how or if i should tie them in to the PLC. The relays are controlling 4 estops and 12 guard door sensors, which i thought the PLC would be capable of doing. So how do i keep the two seperate or do they get hardwired to each other? The net seems to be very scarce on this info
AI: It is very important for safety, regulatory and legal reasons that you get this right.
Basics
First you must assume that your PLC could fail with an output on, off, or switching due to software or hardware issues.
You should, of course, put normal safety interlocks into your code but you do not rely on these for personnel safety.
You then hardwire (* see note) the safety circuits so that even if the PLC goes berserk the e-stop will always render the machine safe.
Monitoring
Your safety relay will have auxiliary contacts one of which can drive an input to the PLC so that the PLC knows the status of the safety circuit.
Safety wiring
Safety circuits now almost all use dual channels and monitoring. E-stops have two contacts that are forced apart (not sprung apart) when the e-stop button is pressed. Similarly guard door switches have two contacts and again the act of opening the door will force the contacts apart. If one channel fails to open on e-stop the other will make the system safe but the monitoring will prevent a reset. This forces the user to investigate the problem.
See my answer to Symbol or marking on safety relay for an in-depth of how a safety relay works.
The situation has changed rapidly over the last few years with safety rated logic controllers. These are beyond the scope of this answer.
The relays are controlling 4 estops and 12 guard door sensors ...
No. The e-stops and guard door switches are controlling the safety relay or, if you prefer, the safety relay is monitoring the switches. The safety relay is controlling the load to ensure energy (electrical, pneumatic, hydraulic, etc.) is de-energised when required.
Related: my answers to the questions below may be helpful.
safety relays and PLCS
Symbol or marking on safety relay
Self Checking Circuit
Specifying relay for breaking injection molding control circuit |
H: Is it legit to use DC source with voltage regulator to create dual power supply for op-amp?
My op-amp needs +12V and -12V voltage supply. I have only one DC source which can output 24V (I can use simple voltage divider to get +12V). I also have one voltage regulator. Is my approach above correct?
AI: If you're a little bit creative, and you don't mind the ground connection being non-standard, you can do this:
simulate this circuit – Schematic created using CircuitLab
The reason that the ground is non-standard on this is because a split supply is usually grounded at the center-tap (0V). If your 24V supply is floating (not grounded), then you can still do this, but if the supply's negative output is permanently grounded, then you have to remember that what the 12-0-12 circuit calls its reference is actually 12V above ground. That might make a difference for any outside-world connections you may have.
Also, this circuit as drawn requires that the negative side of the load draw at least a little bit more current than the positive side. That's because the regulator can only work in one direction. Or if the positive side is always loaded more, then you can "turn it upside down", so to speak, and use a negative regulator instead.
That said though, you're using this for op-amps. They don't actually care about the center tap anyway. All they actually see are the extremes that their power pins are connected to, and signals wiggling in relation to that. The split rails are only specified as a matter of convenience. So if all you need is a signal reference, and not an actual power rail, then you can do this:
simulate this circuit
Use a spare op-amp channel to buffer the voltage divider into something "stiff". This allows the divider itself to use large values to save power. You can use any value that you happen to have on hand; the only requirements to make it work are R1 = R2 and you're satisfied with their current draw.
You'll also notice that I've grounded the op-amp's output. Yes, that's legal, provided that it's the only ground point. (1) Again, if your 24V supply's output is already grounded, then you have to use what it gives you and not add your own.
(1) In fact, I helped to repair my brother's bass guitar amp a few years ago, and the power amp itself actually did this! What would normally be the output went to ground instead, and the speaker was driven between the input power transformer's center tap and ground. Don't know why they would have done it that way, but it worked just fine. |
H: 7 position switch, only one position active at a time?
I am wiring a 6 channel camera switcher (http://www.brandmotion.com/360-vision/blind-spot/6-way-video-switcher-9002-6118.html)
And in addition to the automatic triggers (there are 6 trigger wires, when one receives 12v, it switches to that camera) I want to add a manual switch. I've looked and looked for a 7PST (I think thats the term) ideally in a push button format (7 individual buttons) but I cannot find anything.
So my question is either 1) do these switches exist? or 2) How can I wire 7 individual switches so that only 1 circuit/switch can be putting out 12v at a single point in time?
AI: You are wanting a push button selector switch that works like the old style car radio channel selector buttons. These are available as a series of push button switches that have an interlock mechanism such that when any one button is pressed it causes any previously set button to pop out. Finding the exact correct thing for your application could be a challenge though.
You may be better off designing a thing with a small micro controller that accepts the necessary switch selector input signals and in turn drives outputs to some electronic signal analogue mux/switches that manage the video signal routing. Then you can have the software control the behavior as needed and the switches can be very simple momentary push button tact buttons. |
H: Capacitor plates; smooth or rough?
I am building an aluminium capacitor for fun and it got me thinking. Is it better for the plates on a capacitor to be rough or smooth?
My thinking is that it is better for the surface to be rough as it increases the surface area? I think that is why the carbon nano-tube super-capacitors are so good?
Or is there a trade off for roughness? i.e. increases one property while decreasing another property.
Apologies if this is simple. I am new to electronics.
Thanks.
AI: A rough surface, if it were rough enough to make a difference, would result in field lines converging on the high points (the parts closer to the other plate) of the surface. Which is another way of saying, higher electric field strength near those points. Which would result in a lower applied voltage causing dielectric breakdown near those high points, and a lower WV rating for the capacitor. |
H: What would "push to open" mean in the context of this switch?
I'm not an electrician, but I have a basic understanding of circuits
I had bought a switch that was labeled NC, and when it was unpressed, the light connected to the switch turned on
When it was pressed, it turned off the light
I'm looking at this product right now
https://edwards-signals.com/index.cfm?pid=69&level=48
and it says "push to open" and "push to close"
does "push to open" mean the same as NC?
Logically it seems to make sense that way, but the different terminology is tripping me up, and I just wanted to confirm before making a large purchase
AI: A switch is able to carry current when it is closed --- that is, when the two contacts are touching.
If you put the switch in series with a light bulb, for example, the switch needs to be closed to allow current to flow through the light bulb and light it up.
NC stands for normally closed. That means the normal state of the switch is closed, and it will be opened (breaking the circuit) when it is activated (for example, by pushing a button).
So yes, "push to open" is the same as NC. FWIW, "push to open" is not standard EE jargon --- possibly it's more common in some specific sub-field like the industrial controls that seem to be the market for the website you linked to. |
H: Why modern cellphones do not have an externally visible antenna?
Old phones had either a fixed antenna inside a housing which was externally visible, or a retractable antenna, which was visible when used. I'm assuming this was needed to obtain a certain total length.
In modern cellphones all the antennas are inside the case. The height dimension of older cellphones was similar to current modern cellphones (smartphones). Why now externally visible antennas are not needed?
AI: Short answer: external antennas are unattractive to the user. The consumer drives the market, and these days, consumers want thinner, lighter, faster, etc. This is actually a huge problem for antennas because the physics are fundamentally limiting.
Most antennas in cell phones are some sort of variant of a PIFA (planar inverted-F antenna). Antenna-Theory has some cool articles on PIFAs and the now-famous iPhone 4 antenna. PIFAs are fairly pattern invariant, and provide a decent impedance match. They have low gain but if you do a link budget calculation, you'll see that your noise floor is like -110 dBm (I think... I forget but somewhere in that range).
The miracle that makes phones work is digital communications. Advanced modulation, extremely robust protocols, MIMO, etc have all contributed to ensuring communication even when the antenna on the phone itself is a terrible radiator (-5 dBi gain). If you do the link budget calculation to reach the nearest tower, you'd see that it works, so long as your link budget factors in all of the communication properties. That is, you have to consider what signal level the radio (cell phone) needs to demodulate the digital data and get information. Beyond analog gain (antenna, amplifiers, etc), you have digital gain through things like error-control coding (i.e. coding gain) and robust modulation techniques like OFDM (the standard for LTE and the future). Multiple access techniques (CDMA, TDMA), and MIMO also contribute to this. If you look at the "generation standards" like 3G and 4G, you'll see where on the protocol level things get super complicated. All of this goes into making cell networks work.
There's been a huge hiring thrust in the mobile industries to hire antenna engineers to design functioning antennas for thin phones. One of the toughest challenges facing folks now is that cell phones need many antennas that all couple and communicate on the same bands. So you have to be able to integrate multiple radios and antennas on a small platform and make sure everything works. When you don't do it right, you get the iPhone 4/4S... :-) |
H: Power Supply (220vac to 5v dc/2A)
I'm designing a power supply for a smart switch project which requires a 5v dc/2A supply for usb charger and 12V dc for relay operation. is there any way i could do that? my input is 220ac, which will be rectified to 310(approx) dc. any suggestion will be much appreciated. Thanks
AI: I highly recommend to use external AC/DC adaptor and connects to your project. It has many benefits.
(1) AC/DC has CE marking for both EMI and safety so you do not need to worry on that. It expensive to conducts CE certification.
(2) Protects yourself from lawyer who may raise legal cases under poorly designed AC/DC build in project.
(3) Many more benefits if you think about it more.
You need to implement sufficient EMI filtering and protection on your project from external 5V 2A PSU which is quite easy.
Good lucks |
H: Maximum current draw from 12v sealed lead acid battery?
I am currently building a "battery box", for camping, and kayaking, etc. It has a few different features on it, such as a 150W inverter, a 48W spotlight, 12V accessory outlet, and strobe lights. The inverter and the spotlight seem to draw quite a bit of power, 7 Amps and 3 Amps respectively. (correct me if that doesn't sound right.)
This will all be powered by an ExpertPower 12v 8ah sealed lead acid battery, however I do not know if it can handle the load that I will put on it. I tried looking around, but most people were using this battery for some sort of home security system.
I found a datasheet on the bottom of their website that I think would tell me that (http://www.expertpower.us/exp1280), but I cannot interpret it because of its complexity. If someone could also show me the process of calculating this, that would be great, as I am very eager to learn.
AI: I think you will be disappointed with that battery.
Lead acid batteries are best on low rate discharge. Most these days are rated at 20hrs. That battery is rated 8Ah, so will deliver that capacity when discharged over a 20hr period, at 400mA. At higher currents, the capacity will be less. Here are a few lines taken from the discharge capacity table in the data sheet, for constant current discharge, down to a cell voltage of 1.75v (more of that later!)
current period capacity
0.4A 20Hr 8.0Ah
4.8A 1Hr 4.8Ah
16.5A 10min 2.8Ah
so there's quite a capacity penalty to high rates of discharge.
A 150W inverter will take around 15A (assuming 85% efficiency) to deliver full power, 7A is only around half maximum load.
The lifetime of a lead acid battery, before it wears out, is strongly related to its depth of discharge. That battery rates 260 cycles at 100% DOD, ie to 1.75v. You can double that lifetime if you only discharge to 50%, and x5 if you go to 30%, that is, stop discharge at a higher voltage. Depending on how you want to use it, weight and capacity may be more important than cycle lifetime to you.
It's not clear whether you have a vehicle with you, or everything is carried in your kayak. If you have the capacity to carry more weight, then a cheap auto battery is by far the best bang for buck for the sort of currents you are talking about.
Both nickel and lithium chemistries stand up to high rates of discharge better than lead, and lithium in particular gives a much better power and energy per unit weight. |
H: How overvoltaging makes graphic processing unit operate faster?
My teacher told me that we are trying to make transistors smaller so that they require less electricity to operate, hence make them faster (because changing form 1 signal to 0 or vice versa need less energy, or so he said). Meanwhile, cryptocurrency miners and gamers overvoltage their GPUs for better performances. Why are 2 points contradict each others? What's happening under the hood when we overvoltage GPUs for better performance?
P/s: I had some thought whether i should ask this in SuperUser or ElectricalEngineering before making up my mind and ask this question here.
AI: Very generally speaking -- the clock-speed of the device will directly drive it's performance (operations per second). Chips are designed (and extensively verified) for correct operation at some frequency determined by the manufacturer. Increasing the clock speed (overclocking) will increase performance, but potentially at the risk of instability / incorrect operation as it is running outside of manufacturer parameters. Often, this is not an issue for most users as they happen to never exercise the path in the chip that fails at the excess clock-speed, or they happened to get lucky with the specific die they got, and it can in fact operate stably at a higher frequency (process variation from die to die is a very real thing).
Increasing the voltage of a device on its own won't cause it to operate "faster" in the sense that you'll see more operations per second. Increasing it in conjunction with an overclock may increase the stability of said overclock by allowing the transistors to operate faster (shorter rise/fall times, if I remember correctly) -- at the expense of increased heat dissipation, and potential damage to the chip if the voltage is raised too high (modern chips often hover close to 1V these days); even 100-200mV can kill the chip / cause premature death (electromigration). By increasing transistor performance, you're allowing them to perform "better" to support the increased demands from the increased clock rate (shortened setup and hold times).
Summary -- most people over-voltage their devices to support an overclock.
Extra: In some occasions, increasing the voltage very slightly can help with stability at stock clocks, but you've arguably been sold a marginal device at that point. I have a habit (with honestly no real scientific backing) of bumping my DRAM voltage up by 50mV when I have 4 sticks installed -- for my small sample size, it seems to help stability wise supporting overclocking of the CPU (while keeping the stock DRAM frequency). |
H: Multi-emitter transistor: does the transistor turn on per-emitter or as a whole?
In a multi-emitter transistor (like in TTL ICs), does the transistor switch on and conduct through the emitters independently or together? That is, if there's a base current through emitter \$E_1\$, does this result in collector current through just \$E_1\$ or through both \$E_1\$ and \$E_2\$? For example, suppose \$V_{BE_1}\$ is 0.8V but \$V_{BE_2}\$ is 0.1V, so only the first is enough to switch the transistor on.
I looked at this question, and it seems to imply that if the transistor conducts, both emitters get equal current, but I'm not sure I'm interpreting that correctly.
(My question isn't about the operation of TTL, but about multi-emitter transistors in general.)
AI: if there's a base current through emitter E1, does this result in
collector current through just E1 or through both E1 and E2?
Just think what would happen in an AND gate if both emitters received current when only one was taken low: -
Imagine input A is low and input B is high. Input A takes current because it is low and if that caused input B to fall to 0 volts then the signal driving input B is being corrupted by that action. |
H: Can a rocker switch with two positions be an SPDT?
I have rocker switches that look very similar to this:
In my current understanding, a 2-positions rocker is meant to be an SPST, and a 3-positions rocker (see below) is meant to be an SPDT:
From this source, I find the following image showing an SPST rocker switch:
I observe here that it has only two pins. The 3-pins switch they have is actually an SPDT looking like so:
On this source, they show how to wire a rocker switch, like so:
I have a few questions coming from this:
Why do they have the "on" label on the same side as the "power". Isn't that wrong? From the "internal construction" image (left), I would say it is the other way round.
Is this 3-pins rocker an SPST switch? Why does it have 3-pins, whereas the second image (from the top of the post) only needs two for the (supposedly) same functionality? I guess it has to do with the ground, but I don't understand it.
Still in the 3-pins rocker, what prevents me from using the "power" and "ground" pins as accessories and put the power on the "accessory" pin in order to use the rocker as an SPDT switch? Would that be a correct SPDT switch?
AI: SPST stands for Single Pole, Single Throw. SPDT stands for Single Pole, Double Throw. The 'Pole' is the number of circuits that can be controlled by the switch at any time. So a SP switch can only power one circuit at a time, a DP switch can do 2. The 'Throw' is how many 'on' positions it has (I am 90% confident on this, feel free to correct me).
So: An SPST switch you will only get an On/Off whereas if you had an SPDT switch, you could get a switch with 2 On positions and only have one circuit powered at any time, so you would try and find an 'On/On' SPDT switch.
Bonus answer: If you still wanted to use a switch for this, you could search for an 'On/Off/On' SPDT switch which will only power up 2 circuits (one at a time) but the middle position will be the 'Off' position, so gives you that 'break before make' feel when switching between your circuits.
Also, rocker switches can be SPST or SPDT. A quick search on any component site will give you all the options and descriptions you need.
Hope this helps
EDIT
1 - It isn't really 'wrong' as it doesn't really matter which side you use as ground and which you use as power if you are switching between the 2. Chances are, this switch you happened to get the diagram from has some form of indication (such as a 'red' part of the button to show ON) that means you would connect the power to this side.
2 - A 3 pin rocker is usually always going to be an SPDT switch, unless one of the pins is disconnected completely. Please see my previous answer for an explanation on how these switches work and you should get an understanding of it. And if you just want an on/off switch, all of these could do it, depending on how you wire it.
3 - Nothing stops you from using the middle pin as power and using either side to connect to different circuits. As I said in reply to your first question, you most likely happened to get a switch diagram of one with indicators on it, whereas you can buy blank looking ones too. Again, read my description and you can see that it doesn't actually matter, however, most of these datasheets and examples will have to show you something, they can't give an example and say 'it doesn't matter what you wire where' can they?
Again, I hope this has answered your (revised) question. |
H: A problem with charging
I am beginner in this field. I made a charger with 7805, 10 uF, 100 uF, 100 nF capacitors and 9 V non-chargeable + 3 V rechargeable batteries. The output current is 0.6 A (I used battery instead of solar panel in picture).
And the phone charges but it is hanging on previous percentage. What's the problem? Can non-chargeable battery charge rechargeable battery?
AI: Your poor little 9V battery has probably given up its last electron.
9V batteries don't contain much energy, and you are throwing away a lot of that in the 7805.
If you draw 0.6A through it, you will be dropping 4 volts at 0.6A. That's 2.4Watts being wasted. That's nearly as much as actually makes it to your phone. You are throwing away half of what the battery can provide.
9V batteries are rated at maybe 200mAh (200mAh at 9V gives 1.8Wh.)
Your phone battery is probably rated for 2000 mAh (2000mAh at 4.5 V gives 9Wh.)
But you are wasting half of the 1.8Wh from the battery just heating the 7805. And, you can't get all of the energy of the battery because the 7805 won't work after the battery voltage drops below about 7 or 8 Volts. The 7508 needs 2 to 3 Volts of difference between input and output to work.
So, you are wasting a lot of energy in the 7805, and most of what is left you can't get out of the battery.
Solution:
Don't use 9Volt radio batteries to charge a phone. Use something with more capacity, like AA cells.
Don't use a linear regulator like the 7805. Use a switching regulator. They are more efficient by far.
Use a couple of AA cells in parallel and a boost converter to make 5V out of the 1.5V from the AA cells.
You could use the 7805 solution if you have something beefier to power it. You could use it to charge your phone from a car battery, for example. You'd need a big heat sink to cool it, though. From 13V to 5V is 8 Volts difference. At .6A, that'd be at least 5Watts of power wasted as heat. |
H: Powering LED strips with 5v power supply - safety question
I am making an LED strip installation. I will be using 15 meters of APA102 LEDs. They require constant voltage, I believe. https://cdn-shop.adafruit.com/datasheets/APA102.pdf
I calculated that I need 23.4A at 5v, or around 120w. Adding 20% for safety, I figure I need a 150W 5v supply.
I will be buying a power supply from Digikey: https://www.digikey.ca/product-detail/en/tdk-lambda-americas-inc/LS150-5/285-1817-ND/1918828.
I will be running large wires from the power supply to each of the strips (ie in parallel), while running the LED strips' data lines in series. The Arduino will control the strips, and that will be powered by the supply as well (that should be fine, right?).
I will encase the power supply and Arduino in an aluminium encasement. This will hang from the ceiling outdoors, and be subject to low temperatures in winter (Canada). I am figuring on a temperature range of -10C to 30C. I will be testing it indoor at first, of course.
My confusion is with safety of the power supply. I am relatively new to electronics, and the high current and power makes me a bit nervous, so I want to verify things first.
I think that I need to hook up the live AC wire to L on the power supply, the neutral AC to N, and the ground AC to GND. I then need to hook up the Arduino and LEDs Vcc to V+, and Arduino/LED GND to V- (correct?).
The installation instructions from the power supply http://www.us.tdk-lambda.com/ftp/manuals/ls_manual.pdf states "This power supply has a possibility for hazardous voltage to appear at output terminal depending on the type of failure. The outputs of these products must be earthed in the end equipment to maintain SELV. If the outputs are not earthed, they must be considered hazardous and must not be made user accessible."
Would you be able to explain what this means I need to do? I have read quite a bit about this, and have an idea of what it means, but I want to be absolutely certain.
I think it means that I need to, in this situation, hook up V- to the AC GND. Do I have this correct?
Also, do I need to hook up the chassis of both the power supply and the encasement to GND?
Thank you for you help.
EDIT:
Thanks for your answers. I found this series of articles that made things much more clear for me. https://www.acromag.com/white-paper-electrical-ground-rules
Also, I am still wondering how to verify that the PE and Neutral on the AC in my house are actually independent, up to the mains board. This is the best I have found: https://soundforums.net/features/192052-basic-receptacle-testing
AI: I think the way you want to do the connections certainly makes sense. It is as I would to them as well.
And as you said, for safety reasons and to fulfill the requirement of the manual, you should connect the V- wire to the protective earth (FG) terminal. The point is, if something goes wrong and there is current flowing out of the V- terminal, it will find a low impedance route to the earth through this connection and the earth wire coming from your home's electrical installation and not through the person that potentially comes in contact with the circuit.
That is also the reason, why you should hook up the chassis of both the power supply and the encasement to the earth. However, I make the educated guess that the chassis of the power supply already makes contact with the FG terminal (you could easily check this out with a multimeter) and so no special care should be taken there. The connection of the chassis of your encasement to the earth could be done either with an extra wire or with the mounting screws/holes -if any- that you will use to mount the power supply, or with both ways.
All these of course under the condition that the PE terminal of your house's installation is indeed wired to earth.
One last point: If you indeed indeed to hang the whole thing from the ceiling, pay attention that this method of mounting is NOT recommended, as you can see in Chapter 5 of the manual you have linked. The reason is, as they say, that in this case the heat from the electronics will be trapped and not be able to leave the enclosure. That would mean probably that the output power will be reduced and/or the power supply could shut-down due to high temperatures or worse be damaged. And you are anyway quite at the limit if you see the derating curves, so I wouldn't do it like you intend to.
Lastly, I assume you have already considered this, but it is also imperative that your enclosure has air vent holes. |
H: Why laptop reboots after electrostatic discharge
So why does this happens when my laptop is plugged to the charger? Typical charger scheme looks like this:
What is the path for discharge current? Low voltage part seems to be isolated from power lines.
Some forums says that the reason the laptop was not connected to the grounded outlet but how can grounding helps?
AI: Actually a typical modern "isolated" charger schematic looks probably more like this: -
Note the 2.2nF capacitor connecting output terminals back to the rectified non-isolated supply. It is present to cut-down on EMC on the output and will of course be a pretty good path for a discharge of current to the AC wiring and hence earth.
What is the path for dicharge current? Low voltage part seems to be
isolated from power lines.
Not on a modern power supply - static discharges will fly through the aforementioned capacitor. |
H: Snubber Circuit working
How does snubber circuit work in phase control circuit of Single Phase AC Motor? I have tried to know the causes but I was confused with RC circuit function with this phase control circuit.
AI: Very simply, the snubber provides a load to dissipate energy from rapid changes in voltage associated with reactive components. In your diagram, consider what happens when the triac experiences a voltage spike. The triac itself has capacitance between the line and the gate, and the spike may provide enough current through this capacitance to turn it "on."
The capacitor provides a high current path through the resistor to ground, preventing the spike voltage from rising quickly and providing a resistive load to dissipate the energy. At lower frequencies (like the motor frequency) the capacitor impedance is high and the current is low. |
H: Intermediate frequency
I'm a newbie to communication systems and circuitry. I was going through superhet receivers while trying to undertand AM/FM/DAB receiver architecture. For FM, IF is fixed to 10.7 MHz and AM 455 KHz. Once the carrier wave is shifted to this IF frequency, it is amplified and filtered. In DAB, it is mentioned that IF is 0 (Source). So, is it possible to set IF frequency as 0?
AI: The 10.7 and 455 kHz IF frequencies are more "historic" choices for superheterodyne receivers. The choices for these frequencies might have to do with availability of components (filters etc.) and/or frequency of the LO signal.
To receive for example a 100 MHz FM station, when IF = 10.7 MHz the LO can be either 89.3 MHz or 110.7 MHz. If an IF of 1 MHz was chosen then the LO frequency would need to be much closer to the frequency you want to receive introducing all kinds of problems.
The main problem is that the LO signal can find its way back into the receiver (LO feedthrough) so the radio would be receiving its own LO oscillator. Also the signal you want to receive should have a bandwidth of less than the IF frequency. For an FM receiver a 1 MHz IF would affect sound quality as it cannot properly handle the a 1 MHz frequency deviation of the FM signal.
Yes but then when IF = 0 Hz the LO is at the same frequency as the signal you want to receive !
Yes that is correct ! But in more modern receivers which are mainly implemented on a chip, not using discrete components, the issues caused by the LO being at the same frequency as the RX signal can be solved using a more complex architecture (Quadrature LO, LO supressing mixers, digital DC calibration etc.)
Modern receivers like for DAB and the ones in your (smart)phone are almost without exception Zero-IF or same thing, different name: Direct conversion receivers. See this lecture for more information.
These type of receivers are easier to implement on-chip. The issues like DC-offset and LO suppression can be handled by calibration and digital signal processing.
There are also (on-chip) AM and FM receivers using zero-IF so that superheterodyne architecture with 10.7 MHz IF is not strictly needed.
But in "classical" designs there was less choice to do things differently. |
H: Modbus c terminal
I have to communicate with a device. There are A,B,C Terminals.
A and B are standard. But What is The C Terminal? Is it ground?
AI: "C" is usually ground in RS485 configurations that need ground: -
And, given that MODBUS uses (or can use) RS485: -
I suspect "C" means ground for MODBUS too. I don't think the "C" designator is used in MODBUS TCP. |
H: How to Estimate Power to Ground Current?
I need to automate a power to ground test for a device which outputs 12V / 30A.
There is a .4 ohm resistive load connected to the device which it is driving, however if I remove that load and go straight to ground, the current will be much larger.
I = V/0 = Infinity
Of course the resistance of the conductor will not be exactly zero but I'm assuming the current will be very high, probably high enough to ruin my multimeter! I'm trying to rate a relay to switch the load.
Do I just measure the resistance through the relay and the associated wires in order to gauge the inrush current requirement or is there a better way to do it?
(The problem with the former method is that I need to measure a relay I don't have.)
AI: Recognizing that your driver will never have a zero ohm source impedance, there will be an expected upper limit that it can reach in a true short circuit condition. You should be able to estimate this value. Then by applying a low resistance current shunt in series with the "short", you can measure the near short circuit condition. For example, a shunt of 0.01 ohms would allow you to measure up to 1200 amps.
You will need to include all resistances associated with the short circuit apparatus in your final measurement results. This includes connector resistances, wire resistance, relay contact resistances, etc. In order to reduce the relay contact resistance, you may wish to consider an industrial control "contactor" which is essentially a heavy duty relay designed for high current applications. |
H: Problem with MAX756 Boost converter
I'm using MAX756 as a DC-DC boost converter to provide 3.3V from of a single 1.5 AA battery. I've implemented the circuit on page 1 of datasheet with 1N5819 instead of 1N5817 and 144uF caps between Vin and Gnd instead of 150uF. Also Vin comes from bench power supply.
I've tested three 22uH inductors with different current capabilities (all above the required 1.2A due to datasheet (page 6).
~SHDN and 3/~5 is connected to out. LBO is left open.
The problem is the Vout is 3.3V in no-load. But as soon as I insert the 220Ohm resistors, Vout drops and in Iout = 60mA, the Vout drops below 2V.
Here is the picture of the setup: lower rail in Vin and upper rail is Vout
I've shown the other two inductors that I tested.
Anyone knows where the problem might be?
Thanks in advance.
P.S.
Glenn W9IQ says:
Take a bench supply capable of 2 amps output and set it to 7 volts.
Put a 4.7 ohm, 10 watt non inductive resistor in series with your
inductor and put this circuit across the supply. Monitor the inductor
voltage with your scope. You should see a nice exponential, asymptotic
curve that heads toward zero when power is applied. This will happen
within 1 us of applying power. If the beginning of the curve has a
different slope, your inductor is in saturation at less then 1.5 amps.
But your problem is more likely the ESR of your inductor and
capacitor. Use the right parts - it makes a difference! –
I've paralleled 5 , 2 Watts 22 Ohms resistor (so 4.4 Ohm in theory and 4.7 due to DMM).
Here is the circuit:
And here is the scope's result immediately after applying the power :
AI: {Added diagrams of SinglePoint VDD and SinglePoint GND}
Circuit works with no load and with 220 ohm (12mA) Iload, with 3.3 volts output.
Consider inductive bounce on the breadboard. Assume 2" of loop (in Gnd or elsewhere), or 50nH. Assume 0.1Amp in the IC's internal FET; the IC timing circuits have a minimum ON time, and thus some minimum current; we'll use 0.1amp.
Those internal FETs can be very fast, so assume 10 nanosecond to turn OFF or ON.
Result? Vbounce = L * dI/dT = 50nH * 0.1a /10nS = 0.5 volt bounce.
Nothing is well-controlled when GND or the GND rail is bouncing around with 0.5 volt.
Suggestion: build your circuit atop a sheet of copper (one-side copperclad).
Here is one physical setup that will generate this 0.5 volt upset
simulate this circuit – Schematic created using CircuitLab
By the way, this is not the output swing, but inductive upset within (inside) the GND wiring between the IC pins. (for this example) |
H: Need help to calculate a proper fuse for a 19v transformer
I've successfully installed my Ring Door Bell Pro yesterday.
Ring Door Bell Pro integrates with an existing door chime. I used a voltmeter and confirmed the transformer's output voltage is 19 V AC.
Due to the Ring Door Bell Pro size, my installation at the entrance door exposes two wires going to the unit. The wires are safe; however, someone can pull them hard and they can short. I tested and when I short the wires, they spark.
I am thinking to add a fuse right after the transformer. this way, if I am not at home, and any short circuit happens by an intruder, there is no risk of fire in my basement where the transformer is.
What is the proper fuse model, type, number ar amp for such application? It would be nice if I can buy it from HomeDepot
AI: The Buss Fuses "Fuseology" article will help you select a fuse.
Fuses are more complicated than you might expect.
You will probably find that the doorbell transformer has an internal fuse. |
H: Why are there no 256-bit or 512-bit microprocessors?
In an 8-bit microprocessor its data bus consists of 8 data lines. In a 16-bit microprocessor its data bus consists of 16 data lines and so on.
Why is there neither a 256-bit microprocessor nor a 512-bit microprocessor? Why don't they simply increase the number of the data lines and create a 256-bit microprocessor or a 512-bit microprocessor?
What is the obstacle that prevents creating a 256-bit microprocessor or a 512-bit microprocessor?
AI: Think about it. What exactly do you envision a "256 bit" processor being? What makes the bit-ness of a processor in the first place?
I think if no further qualifications are made, the bit-ness of a processor refers to its ALU width. This is the width of the binary number that it can handle natively in a single operation. A "32 bit" processor can therefore operate directly on values up to 32 bits wide in single instructions. Your 256 bit processor would therefore contain a very large ALU capable of adding, subtracting, ORing, ANDing, etc, 256 bit numbers in single operations. Why do you want that? What problem makes the large and expensive ALU worth having and paying for, even for those cases where the processor is only counting 100 iterations of a loop and the like?
The point is, you have to pay for the wide ALU whether you then use it a lot or only a small fraction of its capabilities. To justify a 256 bit ALU, you'd have to find an important enough problem that can really benefit from manipulating 256 bit words in single instructions. While you can probably contrive a few examples, there aren't enough of such problems that make the manufacturers feel they will ever get a return on the significant investment required to produce such a chip. If it there are niche but important (well-funded) problems that can really benefit from a wide ALU, then we would see very expensive highly targeted processors for that application. Their price, however, would prevent wide usage outside the narrow application that it was designed for. For example, if 256 bits made certain cryptography applications possible for the military, specialized 256 bit processors costing 100s to 1000s of dollars each would probably emerge. You wouldn't put one of these in a toaster, a power supply, or even a car though.
I should also be clear that the wide ALU doesn't just make the ALU more expensive, but other parts of the chip too. A 256 bit wide ALU also means there have to be 256 bit wide data paths. That alone would take a lot of silicon area. That data has to come from somewhere and go somewhere, so there would need to be registers, cache, other memory, etc, for the wide ALU to be used effectively.
Another point is that you can do any width arithmetic on any width processor. You can add a 32 bit memory word into another 32 bit memory word on a PIC 18 in 8 instructions, whereas you could do it on the same architecture scaled to 32 bits in only 2 instructions. The point is that a narrow ALU doesn't keep you from performing wide computations, only that the wide computations will take longer. It is therefore a question of speed, not capability. If you look at the spectrum of applications that need to use particular width numbers, you will see very very few require 256 bit words. The expense of accelerating just those few applications with hardware that won't help the others just isn't worth it and doesn't make a good investment for product development. |
H: 20 Amp automotive thermatic fan PWM controller
I have an older car which tends to over heat in the summer, so a 20 Amp electric fan was added in front of the radiator. Currently its driven by a on / off relay.
I want to build an Arduino PWM controller but I am not confident switching that much current especially in an automotive environment.
Been testing the VNP35n07 MOSFET but I dont think I am doing it correctly. When tested in the on state there is a noticeable difference in fan speed (noticed via noise) and the draw is about 1 amp less. Understand there will be lose from the MOSFET, but should it be that high?
The wiring for the test is as follows:
VNP35n07
Pin 1: +5.5V
Pin 2: Fan's ground wire
Pin 3: Ground
My intention is to connect the Arduino PWM signal directly to pin 1.
Any feedback / comments welcome.
AI: You will need to have a higher input voltage to the FET to get it to the lowest RDS-ON. Right from the data sheet that you linked you can see the difference between having an input at 5V versus 10V. |
H: How do you resize a part in PSpice?
I have a number of parts that I want to re-size so as to fit on paper. How would you resize a part?
AI: Ah you can't really re-size a part unless you open up the part itself and move some pins around and redraw it to be smaller. I assume you were wanting to scale it instead of doing all that. What you could do is go under options, schematic page properties, and then choose a larger page size. |
H: Driving PWM lights with parallel darlington arrays
I am building some service light arrays, which I want to operate using the Arduino's PWM outputs.
This lighting is going to be powered by 5V OR 12V with a nominal current of 700mA per array. There will be 4 arrays total.
Elsewhere in my design I am using ULN2803s; I am thinking of again using this chip. There is obviously a problem though; as the datasheet notes, the maximum collector current is 500mA. However, since the ULN2803 happens to have 8 darlington arrays, I was thinking of putting the darlingtons into parallel pairs, giving me 4 pairs with a max current of an amp. I do know that one of the tradeoffs for the darlington's high current gain is a slow turnoff time, but I am reasonably sure that the accuracy of the PWM is not a very important consideration.
There are a few things on which I am hoping this question can answer for me:
Does anyone know the total power dissipation that this IC can handle? I can see that the maximum Vce is 50V and the maximum collector current is 500mA, so is it safe for me to assume that the maximum power dissipation for the entire chip is 200W or thereabouts?
I know there is going to be an issue with mismatched transistors leading to current hogging, as noted in this wiki article. I know it has something to do with the fact that the voltage drop across the darlington pairs will be different between transistors because of minute manufacturing differences, leading to different collector currents in each pair and worst case to a failure in one of the parallel pairs. I also have found that it can be solved by putting a small resistor on the emitter or collector of each darlington in the parallel pair. I'm not exactly sure which as it seems there is some conflicting information. Additionally, I'm not sure how to calculate this resistance given the datasheet and my parameters. Could anyone shed some light on the exact theory behind this, and how I would go about calculating this resistance?
Alternatively, is there another chip or collection of discrete components that you would recommend I use? I would prefer an array, if possible, to minimize assembly costs.
AI: Not sure how you calculated 200W, but the ULN2803 is not capable of anywhere near that. We can work it out from the datasheets thermal resistance figures though - the datasheet gives a figure for the SOIC package of 73.14°C/W. Assuming a maximum operating temperature of 125°C and maximum ambient of 70°C, this gives us (125 - 70) / 73.14 = 0.75W.
To calculate how much power is dissipated by the chip at a certain current, we look at the saturation voltage for the outputs. If we select the 350mA Ic value, we see that it can be up to 1.6V. So the power dissipated will be 0.35 * 1.6 = 0.56W, which is pretty close to the package limit. At 500mA continuous we will need some heatsinking to meet a 70°C max ambient rating.
You don't say if the 700mA is for each string of LEDs or for the total of 4 arrays. Either way though, paralleling the ULN2803s is not a good idea. The darlingtons have a high saturation voltage and bipolar transistors are not good in parallel due to the possibilites of thermal runaway. EDIT - the above is for separate ICs, if you are using pairs from the same ULN2803 then (as m.Alin points out) the datasheet says this is okay (as the transistors are well matched and thermally linked) Still, the overall current level (I'm assuming 4 * 700mA = 2.8A) is too high for the ULN2803.
A better idea would be to use some MOSFETs - you can get logic level MOSFETs that can be driven directly from the Arduinos PWM outputs, and will have a very low Rds and hence low power dissipation. Something like this has an Rds of only 25mΩ at 4.5V gate drive, and can handle up to 6.7A drain current. At 700mA the dissipation will only be I^2 * R = 0.7^2 * 0.025 = 12.25mW.
EDIT - about using discrete bipolar transistors in parallel:
Because of their thermal characteristics (gain and leakage rise with temp) and variable gain, some form of control is needed when paralleling bipolar transistors. An emitter resistor to provide negative feedback is one scheme that is commonly used:
When a transistor heats up and draws more current, the voltage across the emitter resistor rises, stealing voltage from Vbe and keeping things from running away. You would size the emitter resistor according to the maximum current you want. Also, having all transistors mounted on the same heatsink is a good idea.
In general though, unless there is a very good reason for using bipolars, use MOSFETs.
To demonstrate what the emitter resistor does, have a look at this circuit:
The input voltage (SIG) is 1V.
Both transistors are of a similar type, but we will sweep one's gain from 50 to 500. There are no emitter resistors (set to 1 nanoOhm so the effect is irrelevant)
This is a very crude representation of what might happen when one gets hotter than the other (it's difficult to simulate thermal runaway effects in SPICE, and I didn't have enough time to create/find an appropriate model to simulate it transiently)
Anyway, here is the simulation:
We can see the collector currents are quite different (to be expected) and if uncontrolled the transistor drawing the higher current may go into thermal runaway, as more current -> more power disspated -> higher junction temperature -> higher leakage/gain -> more current -> repeats...
Now if we add a 10Ω emitter resistor to each transistor and simulate again:
We get a much different result, the collector currents are within a few mA of each other. The emitter resistor adds negative feedback and limits the collector current. It does this because the higher the collector current, the higher the current through the emitter resistor and hence the voltage drop across the resistor increases. With the fixed base voltage, this "steals" voltage from the transistor V(base-emitter), which reduces the base current. The transistor collector current can only go so high.
We can calculate the emitter resistor pretty easily. Say we want a maximum collector current of 100mA, and the base-ground voltage is a maximum of 1V. The base-emitter drop is maybe ~0.7V, so there is 1V - 0.7V = 0.3V left for the resistor. So:
0.3V / 0.1A = 3Ω
The above is simplified, but should give you the idea. Thermal effects will alter various parameters, Vbe changes with temp/current, etc. Ultimately you just want to make sure that the process of thermal runaway cannot start, so limiting the gain in some way is necessary. Since the transistor gain is finite the lower the base resistor, the less of a "hard" limiting effect it has (however this is not too important for many applications as long as it stops runaway) |
H: 18f4550 comm clarification and advice
Please excuse my ignorance. I need your advise and clarification on something. I'm using the 18f4550 and I'm a little curious about the communication protocols. I am not using the USB features. It has 1-EUSART, and 1 (SPI/I2C) comm. My question is, the EUSART and SPI seem to share pin26 (on the 40pdip). I need to use both. Is it normal for 2 comm protocols to share a pin and just toggle the config of that pin throughout the program?
Are the SPI devices on that line going to be confused at the UART data they see on it, and vice versa? How likely is it they could react to some signal from the other protocol? Or is it good practice never to dual-use such a pin and spec a chip with 2 separate protocol pin allocations?
Thank you.
AI: It looks like the UART transmit line and the SPI data out line share a pin on that part. That happens sometimes. Some of the newer PICs have alternate pins for some of the peripherals, or have the remappable pin feature. The 18F4550 is a rather old part.
Trying to share the pin sounds like a lot more trouble than it's worth. Fortunately, SPI is very easy to do in firmware if you're the master. If this PIC needs to be a SPI slave, then you probably should use a different PIC. Otherwise, just do the master function in firmware. SPI is very simple, so that is easy to do. Since it's synchronous and you own the clock, it doesn't matter if the processor occasionally goes off to a interrupt routine or something. |
H: What difference does it make to an appliance whether the supply is sinusoidal, or any other shape?
Just what the title states.
I understand some appliance/s may be damaged if the supply is any shape other than sinusoidal. E.g. sawtooth, square ...
Why is this the case? How does the shape of the supply signal matter to an appliance?
AI: Many factors are associated with supply waveform like power and heat generation,charge accumulation, effective time period etc.A circuit is mostly designed keeping in mind, a type of input signal. Changes in those parameters may make circuit behave abnormally and thus sending it into an undefined state and even causing damages.
Let us understand with some examples:-
1 . In AC motor feeding a sawtooth or square waveform wont help since the magnetic field will be discontinuous with time.Also suppose you switched on a DC motor with sawtooth waveform, which usually needs PWM(consider a duty cycle of 40%) in rectangular waveform, results will be disappointing.
2 . If you are able to operate this device with different basic waveforms then please let me know!!!
3 . A simple LED circuit
At square waveform your led will glow forever (Vmax=100V,Vmin=0V,R2=R1=1K,C1=1u) however with sinusoidal waveform its hard to imagine the previous result. |
H: How to go from a development board to a production board?
Ok, I am a software guy and new to electronics. My product needs a small computer and currently I have developed everything on development boards like raspberry pi and such. As the development board does not have all the hardware that I need, I have added the missing hardware through USB and GPIO's but obviously the production board can not be like that.
So my question is how do I go to the next step to come up with the production version of my hardware? In other words, what would a hardware specialist do to turn its development board into a production ready product? These are what comes to my mind
1> Get the schematic of current development boards like the pi (or any publicy available schematic that is close to my basic hardware requirement like the cpu type and such) and try to find some one to add the missing pieces to it and re-organize the position of the ports (like usb port and such) to your desire and make it the production product. This is possible because CAD software like Eagle are very powerful and adding a couple of more hardware chips (say like a temperature sensor) and changing the locations of ports is fast to do.
2> Are you stupid? The design of such board is very complex, specially a computer like pi and you have to hire a board design consulting company (or whatever they are called) to design this production ready board for you. If so, how should I talk to? US or China?
3> If your are building a production level hardware, you need to have the design team in-house and you need to hire the right ppl with such expertise to do so. This is not something that you want to contract out. you must have it internally as practice has shown. If it was a simple board you could have done yourself, but a computer needs a lot of work (although there are many commonality among them) and you have to start from scratch as there are so many details involved.
Oh, my production size is not big. I need 5,000 units every 6 month. The main thing is the a good reliable final board design with my requirements to get it to a PCB manufacturer.
As you see, I am sort of lost in this hardware manufacturing space and your insight and personal experience will be very valuable to me.
Many Thanks!
AI: Development to 50,000 units every six months? I wish all my projects went like that :) If you don't have the experience there's no reason you can't hire a consulting company to make the board for you. It won't be cheap but they'll get the job done. It's a little riskier too if you don't know the guys you're hiring enough to trust them with the design.
50,000 units is not a small run so if you're really going to do that you should have no trouble finding a manufacturer here in the US or over seas who would work with you. Keep in mind you'll need the cash to buy your parts, and order your boards upfront.
So I'll go through each approach for you:
1.) Do it yourself
Making a Schematic
Start with the reference schematics you have, then find yourself a tool you like. I'm an Orcad guy, I've used Mentor and many others. Just pick one you're comfortable with and you can afford (Eagle is cheap I understand). If you're lucky you can get your reference board schematics in a format that you can modify. If not you'll have to create parts in your schematic tool. Creating parts basically involves looking up each parts datasheet to get it's pin out and then creating a symbol with pin names and numbers to match. Then you can use those symbols in your schematic and connect them up the way they need to be. That's the simple version, oh and double and triple check that your schematic symbols match the pins on the datasheet.
Here's some links to schematic tools
Orcad: http://www.cadence.com/products/orcad/pages/default.aspx
ViewDraw: http://www.mentor.com/products/pcb-system-design/design-creation/dxdesigner/
Altium: http://www.altium.com/
Eagle: http://www.cadsoft.de/?language=de&lang=en
Free: http://www.sunstone.com/PCB123.aspx
Layout Your PCB
Now you have a schematic that's a big step, from here you could give that schematic to a contractor and ask him to do the layout for you (that's the drawing of the actual traces on the board). You could also opt to do it yourself, it's both easier and harder than the schematics. Drawing connections and placing parts isn't too hard, but knowing where to put things, how many layers, how to route traces correctly for things like cross talk and emissions, and especially how to do the decoupling correctly takes a little know how. If you're committed to it and you review the reference schematics for each of your pieces you can make it through. Oh and you'll spend a lot of time looking at datasheets, and drawing footprints if the standard ones don't work. If you ever took a CAD class in school it's pretty much like that. Ask questions here if you go this route.
Here's some links to layout tools, there are certainly others
Orcad Layout: http://www.cadence.com/products/orcad/orcad_pcb_designer/pages/default.aspx
Allegro Layout: http://www.cadence.com/products/pcb/pages/default.aspx
Mentor Pads: http://www.mentor.com/products/pcb-system-design/layout-routing/
Altium: http://www.altium.com/
Eagle: http://www.cadsoft.de/?language=de&lang=en
Free: http://www.sunstone.com/PCB123.aspx
Decoupling, SI, and power design
Decoupling, Signal integrity and power design are huge areas and too detailed for this post. However if you're going to get into pcb design you should know them. I could write posts on top of posts about it :) I'd at least check out these two guys and get their books, or at least browse around their websites:
Dr Howard Johnson http://www.sigcon.com/
Eric Boggaton http://www.bethesignal.com/bogatin/index.php
Both of them are pretty nice guys and will answers questions if you ask them, you can also join the SI-List over at http://www.freelists.org/archive/si-list It's a great place to ask questions.
That may be more than you're ready to do so there levels of how involved you can get and how involved you need to be on this front. For your design I'd suggest following the app notes and reference design and keeping all your caps as close as possible.
From ok to better here's some ways you can handle signal/power integrity:
Ignore it (NO!!! :)
Just use a bunch of the highest value smallest size caps
you can get and keep them close to your chips Design your own
decoupling cap system in pspice, and then wing it on the placement of
them in layout
Use an excel calculator like the one Altera provides
for it's tools http://www.altera.com/literature/ug/pdn_tool_stxiv.zip
(pretty useful if you have no other tools)
Design your cap system in spice, and then use a full simulator
I've done all of those depending on where I was and what I can afford. When I can get it I love to use Sigrity to do both SI and PI analysis http://www.sigrity.com/ They're actually owned by Cadence now. No affiliation here I just really like their tools.
You can also hire guys to do it for you, I've only ever used http://www.teraspeed.com/ for that but I know there are others. It's not cheap though!
Generating Files To Send To Board House
Once you finish your layout you'll need to quadruple check it because you're about to pay actual money for bare boards. At this point you can generate cad files, either Gerbers or ODB++ files. You send these files to a board house to get a quote. Pricing is based on complexity and how impatient you are. You should probably order a small number, ask them for say 10 or best value that should give a good place to start. ( I should point out that there are some board houses that offer their own free software tools if you want to go that route, it restricts you to them but hey it's free).
You should review these gerber files too not just generate them I've always used the free GC-Prevue from http://www.graphicode.com/GC-Prevue. There's also a nice commercial tool out there that some of my cad guys love called Blueprint http://www.downstreamtech.com/support-viewers.php. There's others too but I always like to look at the final design on a projector and pick out problems. I'll also print the top and bottom layers out in hi-res on a laser printer and make sure the parts fit the footprints I made. If I'm feeling particularly obsessive I might print all the layers on transparencies and look them over. Really, really obsessive I might send the top and bottom layers out as a two layer board just to see how things fit together.
Order Your Proto Parts
At this point you should be ordering parts for your proto-run so they arrive when your boards do. If you don't think you can handle soldering yourself you'll need to pick an assembly house to do you run for you. I can think of a few that handle small runs and they should be easy to find. You'll need to send them your gerbers ahead of time so they can make a solder stencil for you board. Then send them the parts kit, and ship them the bare boards when they come in.
Bare PCB Production
There are a lot of good board houses out there:
Cheaper ones like PCB Express ( the guys with the free software) http://www.pcbexpress.com/
I also use Advanced Circuits in Colarado a lot for my hobby projects, and some fast proto types as well http://www.4pcb.com/ They have an assembly service too that I've never used.
For my US production PCBs I use DDI http://www.ddiglobal.com/ now via systems http://www.viasystems.com/ or Vermont Circuits http://www.vtcircuits.com/
PCB Assembly Services
For small to medium US assembly services I use IMS in NH http://www.imscorp-us.com/ They'll do 10 boards for me or 10,000 and their quality is great. I've used them for years. For crazy big runs I'd use a Flextronics or someone like that but that's a whole different league, and not what you're looking for. There are plenty of others, probably even near you. There's a family owned place by me called Edmond Marks that does good work. http://www.edmondmarks.com/ and Advanced who I mentioned before likes to call me and tell me about their assembly options as well.
Over Seas
So most of my China production experience is with million unit plus volumes so that's not as helpful to you, but let me tell you it's a whole different experience :) I do know that people like IMS can help you take something over there if you get a little bit of volume so that's what I'd suggest. My advice to you would be pick a US partner who has the ability to outsource to a Mexico or China plant if you need it. You may not find as much of a cost advantage as you might think for your board though. Especially if you don't have a lot of hand operations.
Done!
If all that goes ok you'll have protos back that you can play with, and you'll have a good time finding all the things you did wrong that you need to fix for your next proto run.
Compliance and Testing
I should mention also that no matter what you do if you're going to sell these you'll also need to do FCC compliance testing (or other countries if you're selling internationally). In addition there are environmental regs like RoHS and REACH that apply both here and internationally. Don't sell 50,000 units with doing compliance testing, fines are a b*.
Here's some links just to the wiki pages for those:
http://en.wikipedia.org/wiki/Restriction_of_Hazardous_Substances_Directive
http://en.wikipedia.org/wiki/Title_47_CFR_Part_15
http://en.wikipedia.org/wiki/Registration,_Evaluation,_Authorisation_and_Restriction_of_Chemicals
Typically I pick a compliance lab that's near me. Now that happens to be NTS http://www.nts.com/, but I've also use TUV http://www.tuv.com/global/en/index.html, met labs http://www.metlabs.com/ and even UL http://www.ul.com/ themselves once or twice. I've also used small independent places. They can all help you but I like to pick someplace close by so I can sneak in when I need to.
You may also want to do UL safety testing to ensure your product is safe, in which case any UL lab mentioned above can help you. My guess is you'd be under UL 60950 which is for telecom products.
2.) Use Consultants
Listen everyone here started out at one point with no idea how do to a schematic or layout a board. If we can learn you can to. That said if you can afford it there's nothing wrong with have a consultant do it for you. Just remember that no one loves your product like you do so stay on top of them. I don't consider the PI boards to be very complex but it's not exactly a beginner board. Personally I would stick to the US or Canada for my first attempt. However if you really are going to order 20,000-50,000 I know there are small China (prob US too) manufacturer who would take your design, do the work and then manf for you just to get the business. I've worked with guys like that before, but just keep in mind it's not that hard for them to copy your design... :) Happens all the time.
Also distance, time shift, and language barrier can be difficult but not impossible to over come. One nice thing about this is if you have a day job you can work at night on your project with your guys overseas. (I've certainly never done that before...)
These are the only guys I've ever done a product with, there are countless others but here are some examples who did well by me:
Dave Durfee at https://www.baycomp.com/
Bernie Gobeil at http://www.signalcraft.com/
Hector at Encadis http://www.encadis.com/
3.) Build Your Own Team
Well listen if you can do it hire the right people, I've done pretty well walking into small places that are a mess on the hardware side and fixing lots of issues. Having the right people with the right knowledge (maybe the right tools if you're lucky). That's really invaluable. But that shouldn't scare you into riding out into the unknown by yourself. This would definitely be the safest route, but hey if we all took the safe route what fun would that be.
You could also consider outsourcing and building your team in another country. I find that this is full of pitfalls though. You really need to know what you're doing yourself in order to manage this, it's hard to outsource effectively if you don't have the expertise in house to know what's going on.
Finishing up
Some last words of advice from a guy who's made a lot of products :P If you really have a channel to move 50,000 units then great. If that's just speculation though don't stick your neck out buying a big order to keep your prices down. Find a way to make it work where you're only making say a 100 and you can still sell them without losing money.
Lastly if your pi project is epic enough to sell 50k units consider doing a kickstarter (www.kickstarter.com) project and seeing if you can pre-sell any. They have a new requirement that you have a working proto and demonstrate what you would do with the money, but many a cool project have been given life there.
Good Luck, and ask us questions as you go. |
H: Easier-to-machine alternative to pogo pins?
My project has a rectangular 5 x 6 grid, 1mm pitch pin header. Unfortunately, it needs to be cycled by the user over 300 times!
We've looked at pogo pins, because of the high cycle count, but the three moving parts per pin and the tiny size of the precision spring makes pogo pins understandably costly to manufacture!
What are some alternative high-cycle, <= 1mm pitch pin array shapes we can consider? Some ideas:
Micro "flat springs" -- like this, but each .6 mm long:
Contacts independently seated on a springy / cushiony foam.
"Grid of triangles" where each female contact is a set of three soft metal balls forming a triangle whose area is < 1mm sq, and where each male contact is a larget soft metal ball that contacts the center of the triangle -- hopefully touching all three corners of the corresponding female triangle pad.
Some kind of liquid or gel conductive medium between pad array connectors, with some way to insulate adjacent lines.
AI: Check out this conductive material, it's a silicon layer with tons of tiny wires in it. They use it in scope probes, testers, and other things. I wanted a tiny bit for myself but they wouldn't just sell me a square :)
http://www.shinpoly.com/products/interconnectors/maf01.shtml |
H: Directional pickup of human ear?
Given the following:
x, y, z of omnidirectional source source
x, y, z of center of listener's head
pitch, roll, yaw of head
What is a simple fuction that will return the decibel attenuation and the delay of the left and right ears' perception of this source?
AI: The accurate function is the "head related transfer function"
This varies between people. The data can be measured and turned into an impulse response. There are some archives of these. But that is not simple: if you use real data, it means you have a function with a large number of coefficients.
HRTF's have been approximated in software. E.g. video games which place sounds relative to the player, who is wearing headphones.
Wikipedia:
http://en.wikipedia.org/wiki/Head-related_transfer_function
Intro to HRTFs slides:
http://www.umiacs.umd.edu/~ramani/cmsc828d_audio/HRTF_INTRO.pdf
SLAB: NASA's Open Source Spatial Audio Renderer (claims to have a "listener HRTF (Head-Related Transfer Function) database"; maybe something can be borrowed from this):
http://ti.arc.nasa.gov/opensource/projects/slab-spatial-audio-renderer/ |
H: Measuring distance at angle
I have a question, I want to measure a distance from an object of unknown shape and material, and I want to measure it under different angles, I want to measure a distance to just one specific point of object and accuracy has to be good.
Can someone please point me to right direction?
AI: A standard laser rangefinder (I have a Bosch DLE70) can give you 1.5mm precision from 5cm to many meters (mine works up to 70 meters) and works in daylight and for every texture and object I've tried.
Also look at this video |
H: Why do many electronics operate on 5 and 3.3 volts?
Many modern appliances use a 5V power connection, but internally they work with 3.3V. Wouldn't it be easier to have 5V everywhere?
Examples are many USB devices or routers (they use 5V for power but 3.3V for serial communication.)
AI: 5 V became much used in early logic families, and especially TTL. While TTL is very much passé now everybody still talks about "TTL levels". (I even hear UART decribed as "TTL bus", which is a misnomer: it's a logic level communication channel, but may well be a different voltage than 5 V.) In TTL 5 V was a good choice for the setpoints of the BJTs and for a high noise immunity.
The 5 V level was retained when technology switched to HCMOS (High-Speed CMOS), with 74HC as the best-known family; 74HCxx ICs can operate at 5 V, but the 74HCT is TTL-compatible for its input levels as well. That compatibility may be required in mixed technology circuits, and that's the reason why 5 V won't be completely abandoned soon.
But HCMOS doesn't need the 5 V like TTL's bipolar transistors did. A lower voltage means lower power consumption: an HCMOS IC at 3.3 V will typically consume 50 % or less power than the same circuit at 5 V. So you create a microcontroller which internally runs at 3.3 V to save power, but has 5 V I/Os. (The I/O may also be 5 V-tolerant; then it works at the 3.3 V levels, but won't be damaged by 5 V on its inputs. Next to compatibility 5 V also offers a better noise immunity.
And it goes further. I've worked with ARM7TDMI controllers (NXP LPC2100) with a core running on 1.8 V, with 3.3 V I/Os. The lower voltage is an extra power saving (only 13 % of a 5 V controller), and lower EMI as well. The drawback is that you need two voltage regulators.
So that's the trend: internally ever lower voltages for lower power consumption and EMI, and externally a higher voltage for better noise immunity and connectivity. |
H: Can I use 16V power adapter for scanners that need 12V?
Possible Duplicate:
Choosing power supply, how to get the voltage and current ratings?
Will they be damaged? The two HP deskjet 2400 and 4300C scanners need 12V, 1.25A as I read. I only have a 16V, 3.36A power adapter with the connector they require.
Will I break them?
AI: No, you must use a 12V adapter rated for >= 1.25A. |
H: Microphone replacement
I have a device containing an active mic, an amplifier, and an AD converter. I would like to swap the microphone with another completely different mic. I have no tech specs on the microphone that I'm replacing. Would it be a good idea measure the voltage output of the original mic in response to some sine waves at various frequencies and then wind a simple toroidal transformer to get the output of the new mic to look similar?
...or would this be totally dumb?
AI: As someone who designs pro-audio equipment for a living I can confidently say that this would be totally dumb. Here's why:
No tech specs on the old mic.
No tech specs on the new mic.
No tech specs on the device that you want to swap mics on.
No mention on what performance levels you want.
Hand winding transformers for audio is difficult at best.
Measuring the voltage at various frequencies is only part of the equation.
Given your interest in VST Plugins, I am going to assume that you want pro-audio level of performance. And based on your previous questions, I am going to assume that you are fairly new at electrical engineering (that's not a criticism, as we all have to start somewhere and that is after all why you are here).
Mic pre-amp circuits are difficult, even for seasoned EE's. Doing a pro-audio level circuit (or even a consumer grade circuit) is difficult without good specs and the test equipment to verify that it is working correctly.
Using a transformer for this is not at all advised. Yes, that is how they used to do it "back in the day" but designing and building a transformer that works well across the entire audio frequency band is difficult. I know many really good pro-audio EE's, but maybe only one of them could design this transformer. You could buy a reasonable Jensen transformer that is designed for this, but it will cost you a minimum of US$100/each and probably won't have the required turns ratio.
If you want to do this as a learning process, and I highly approve of doing things to just learn stuff, this is still the wrong thing to do. Modifying gear like this, when you don't have a complete understanding of what is going on (no specs, etc.) is just super frustrating. You would be much better off just buying the exact thing that you need, and picking a different audio project to work on. Otherwise you run the risk of damaging an otherwise good piece of gear, not getting the device you really want, and getting frustrated and abandoning your EE ambitions. And that's never a good thing. |
H: What is wrong with this MSP430 code?
I am using the MSP-EXP430F5438 evaluation board, which has pushbuttons connected to P2.6 and P2.7 and LEDs connected to P1.0 and P1.1.
I want to make it such that P2.6 sets the LEDs on and P2.7 resets them. To do that, I wrote the following code:
#include "msp430x54x.h"
int main(void)
{
WDTCTL = WDTPW + WDTHOLD; // Stop watchdog timer
P1DIR |= BIT0 + BIT1; // Set P1.0 and P1.1 to output direction
P2DIR = 0x00; // Set the P2 to input direction
P1OUT = 0x00; // Blank all LEDs
while(1)
{
if(P2IN & 0x40) // When P2.6 is pushed
P1OUT = 0x03;
if(P2IN & 0x80) // When P2.7 is pushed
P1OUT = 0x00;
__delay_cycles(100000);
}
}
But it doesn't work properly; sometimes it works erratically (the LEDs turn on when P2.6 is pressed but they don't turn off when P2.7 is pressed), sometimes it doesn't work at all.
Any ideas?
AI: There are no pullups on P2.6 and P2.7.
The MSP has internal pullup / pulldowns but you have to set them up yourself.
So as part of your initialization, add this
P2REN |= (1<<6) | (1<<7); // turn on pullups
P2OUT |= (1<<6) | (1<<7); // set them to pull up
Then the inputs won't float randomly when the switches aren't pressed. |
H: MOSFET or bipolar to drive a solenoid valve?
I have to drive a 12V solenoid valve. In the datasheet it's written: Power: 8 W. So the current drawn by the solenoid is 0.666 A.
Usually, to drive small relay (about 10 or 20 mA) I use a bipolar transistor but with this kind of current is it possible? Or do I have to use a MOSFET because my bipolar will be burning?
AI: I'm afraid I don't agree with Matt's conclusions.
You can find BJTs with low saturation voltages. A quick search at Digikey turned up this one. At 1 A the saturation voltage is typically 45 mV, that's equivalent to a FET with an \$R_{DS(ON)}\$ of 45 mΩ. Not bad at all for a BJT.
And even the 200 mV Matt used as a typical value is absolutely no problem: at 0.67 A that's 130 mW, and the "lot of wasted energy" after a year of continuous operation is 1.17 kWh. For the NSS60601MZ4 that's 0.26 kWh. Not what I call "a lot".
The problem lies elsewhere. The minimum \$h_{FE}\$ of 120 at 1 A is specified for a \$V_{CE}\$ of 2 V, i.e. the transistor is not in saturation. The 45 mV is for a base current of 100 mA and 1 A collector current. Now that's what I call a waste: the 100 mA doesn't flow through the load at all. If you control the transistor from a microcontroller you'll have the problem that a microcontroller can't source that much current. And, finally, the datasheet notes that the 100 mA is pulsed, so it may not be able to do this continuously (though I couldn't find a maximum value for base current).
That's the problem with BJTs, especially for currents higher than about 0.5 A. And here the MOSFET shines. Drive current: near zero. \$R_{DS(ON)}\$: what you want to pay for, 1 mΩ is possible. You'll have to look at the minimum gate voltage; for driving by a microcontroller you want a logic level gate FET.
So, if FETs are so great, why do we still use BJTs? Cost. A BJT may be 50 % cheaper than a FET. It's that simple really.
edit
Scott mentions a Darlington to overcome the driving current problem with the BJT. I failed to mention it, and I should have. Not because it's a solution, but because it's not! A Darlington has a (much) higher saturation voltage than a common BJT; I've seen values up to 4 V maximum. But even 1.5 V would be much if you have a low voltage supply like 12 V: the reduced voltage across the solenoid means that only about 75 % of the required 8 W is available, and that may be too little to activate the solenoid. Also, the transistor will dissipate 1 W, and a Darlington is more expensive than a common BJT, so the cost advantage doesn't count either. |
H: Resettable latching switches
I don't know if self-contained units like this exist, but I'd like to think they do, and I just don't know what they're called, who makes them, or where to find them.
Please bear with me, for a moment. Imagine a lamp, a simple On/Off latching switch, and a battery connected in series. The lamp is powered when the latching switch is in its on position.
Is there another switch-type or device that can be added to the circuit, that if pressed whilst the simple switch is on, opens/breaks the circuit until the simple switch is off. The new device is effectively a latching normally closed switch, but special because it can reset its own state to closed, if the simple switch changes position to off, and current no longer flows.
Ideally, I'd like something with these characteristics:
12V 0.5A
10-30mm diameter or width
panel mountable
LED indicator (not required, but would be nice!)
ideally looks like a latching switch but resets mechanically.
Any help appreciated.
AI: The function you're looking for is very common in, for example, a sump-pump controller. If the sump starts to overflow (because the pump has failed), an alarm contact closes, and it usually sounds a very loud buzzer. There's a momentary-action switch called "Alarm Silence" that will shut off the buzzer (so that you can fix the problem in peace), but then it will re-arm itself once the original alarm contact opens.
It is easily implemented with a single SPDT relay, as shown below. S1 is the alarm contact, S2 is the "silence" pushbutton, and the relay K1 keeps the load (buzzer) switched off until S1 opens again and the relay drops out. "Power Source" and "Load" can be pretty much anything, as long as the coil of K1 is compatible with the power source. |
H: Square wave / Sine wave is more audible
Consider you have 50 Hz both square wave as well as sine wave put on to a same speaker at different time. Which signal is more audible and what is the reason behind that ?
We know that by taking the Fourier series of the square wave we get multiple periodic sinusoidal wave, each having the frequency as the multiple of the fundamental frequency. So which will be more audible, whether a sine wave having single frequency or the square wave?
AI: The square wave will be more audible. 50 Hz is fairly low and most speakers will not reproduce that frequency very well. Since a sine wave will have only 50 Hz there may not be that much audio reaching the human ear, and even then the human ear will not respond very efficiently to it. A square wave, on the other hand, will have lots of harmonics that the speaker will reproduce very well and the harmonic frequencies will be spread right through the ideal frequency range for human hearing (300Hz to 3 KHz). |
H: Replace UM 66 IC
I'm trying to build a circuit of a Doorbell, but it's very hard to find ICs from local stores.
I didn't find UM 66. So, someone told me to look for:
3481 series
HT3820
4822
AM993-9A
Neither of them was found, also.
So my question is:
What other ICs could I possibly use in this circuit?
Also, is there a specific site or place where I can search for some ICs and directly find equivalents/similars or it requires knowledge and experience?
AI: For old/obsolete ICs like this, you can sometimes find them on eBay.
Here is one listing from China for the UM66, and another one from France.
Note that there are some unscrupulous semiconductor vendors on eBay so be careful, I would always buy from somewhere like Digikey, Mouser, Farnell, etc if you can (I had a quick look but couldn't see anything particularly suitable)
You could quite easily make a similar circuit using a small micro, a 555 timer or two, using discrete transistors, etc. It depends on your level of knowledge and whether you actually want to design a circuit or simply build an existing one. There are many doorbell circuits/kits out there that do have parts you can find - Velleman make lots of kits, here is a doorbell/tone generator kit that may be of interest. |
H: Purpose of these two transistors
Hi does anybody know what's the purpose of the Q1 and Q2 transistors,
sounds to me like they are constant current sources. Am I wrong?
Could somebody explain me the purpose of those Q1 and Q2 transistor
configuration there?
Note, This is a simple preamp circuit.
AI: They act like a diode to bias the output transistor bases. They are like a Vbe multiplier (also known as a rubber diode - see here) without the multiplication (i.e R1 0Ω, R2 infinite in the diagram below), so the drop is ~0.6V.
In a class AB or class B amp they are often used to provide a trimmable ~2*Vbe drop to bias the output transistors:
Note the input is at the upper side, so one ~2*Vbe multiplier (~1.4V, depending on desired bias)) can be used. In your circuit it's in the middle, so you need 2 drops each of around 0.6V. |
H: What do I need for a basic RF circuit?
I'm looking for any resource that can help me build a very very basic RC circuit. I'm talking as simple as using RF to turn on/off an LED. The purpose of this circuit is for research and education. If anyone knows where I can get parts for simple RF circuits or even a tutorial or schematic that would be really appreciated.
From what I've gathered, (and I'm an absolute noob to this) I will need a transmitter and a receiver... is that correct?
Honestly guys, I know you are all probably annoyed with this question and trust me, I wouldn't bother you all if I could find my own way into RF for beginners but really I'm having the toughest time finding good resources for RF circuit building. All I'm looking for is something to get me started in the right direction and then I can go from there.
I've googled about a thousand things to try and find info on this but all I end up getting are a bunch of RC planes/boats sites. Nothing for just regular basic do-it-yourself circuit building.
Please no infrared. I'm only interested in radio control. Thanks!
AI: If you want complete DIY, which is probably not you best starting point, then something like this is about as simple as you can get. Receiver at left. Transmitter at right.
That is from this wonderful page of RF ideas Homebrew RF Circuit Design Ideas These tend to be circuits only with no notes so a bit of a leap for a beginner.
with the actual circuit diagram here
A wireless doorbell may do what you want.
"ASK" RF data modules abound - finding them amongst the 1000 at a time Alibaba adverisers is the trick.
Here is one at $US6 with free shipping Uses nRF24L01 transceiver IC from Nordic.
From Sparkfun - uses same IC - $20.
Description: The nRF24L01 module is the latest in RF modules from SparkFun. This module uses the 2.4GHz transceiver from Nordic Semiconductor, the nRF24L01+. This transceiver IC operates in the 2.4GHz band and has many new features! Take all the coolness of the nRF2401A and add some extra pipelines, buffers, and an auto-retransmit feature - very nice!
You can DIY it with the IC, burt here's what's on board (+ a power supply)
There are many low power short range low cost wireless data modules available.
A large number of idea starters here look for small unhoused modules.
A bit fancy but would do what you want - but also somewhat complex to use and at GBP13.50 a bit expensive Pololu Wixel
Similar
433 MHz ISM band transceivers. $15US/1.here
Transmitter $3 free shipping.
Receivers will be ~ similar here
panStamp here
Open source
panStamps are small wireless modules programmable from the Arduino IDE. Each module contains an Atmega328p MCU and a Texas Instruments CC1101 RF interface, providing the necessary connectivity and processing power to create autonomous low-power wireless motes. Everything contained into a compact DIP-24 module.
Store $18.55 basic unit.
Units like this rare widely available at modest cost. This site only sells them in volume (probably). |
H: Interfacing microcontroller with 1.8V logic
I am in the process of selecting a microcontroller to interface (~30 signal lines) to an FPGA that operates at 1.8V voltage levels.
I have found NXP's Cortex M0's to operate down to 1.8V which is great, but the bulk of Cortex M3 chips out there have a minimum operating voltage of 2V.
Should I restrict myself to the M0's? Or should I just use level translators? Are there Cortex M3 implementations that have an I/O voltage of 1.8v?
Note that I do have other voltage levels available on the board, but i just need the micro's I/O voltage to be compatible with 1.8V logic.
AI: NXP has 74 different Cortex-M0 controllers, so if the M0's performance is good enough (they run at 50 MHz maximum), then you'll probably find a type which has the right features for your application.
If you want more options or need a higher performance you could go for the Cortex-M3 combined with a GTL2000.
The GTL2000 offers you 22 fast channels which you can use unidirectional or bidirectional, and as you can see in the diagram you can interface with several voltage levels with a single device. The GTL2000 can work with levels between 1 V and 5 V on either side. This is a cost-effective solution if you have many lines between the controller and FPGA, but there are also versions with less transceivers if you can do with less. The GTL2010 is a 10-channel version. |
H: Getting 3.3 (V) from a 5(V) rail - Emitter Follower, or something else?
I am using a Bluetooth module (RN-41 SM) that is spec'd to run at 3.3(V).
I can operate this module at 5(V) however, due to another part of the circuit, I need to run the micro controller at 5(V)
The issue I have is that the only rail I have available is a 5(V) rail. My idea was to use a 3.3(V) zener diode and buffer it through an emitter follower, however when I tried this, it failed on me. I have to supply at max 100 mA to the circuit.
I am using a garden variety transistor (2N3904).
I think my issue may have been related to biasing the transistor properly. Could anyone guide me on how to bias a circuit like this or suggest any other method I could use to supply the 3.3 (V) rail.
Thank you for your time.
AI: Why not an LDO (Low Drop-Out) regulator? They're easy to use: capacitor on the input, and one on the output, and have much better performance than any zener circuit, load regulation-wise and line regulation-wise.
The TC1269 (just a random pick) has a 2.5 % output voltage accuracy, can supply 300 mA, and has dropout voltage of maximum 160 mV at 100 mA. Dissipation will be less than a shunt regulator as well.
The TC1269 is only 31 cents in 1s at Digikey. A zener + transistor may cost as much.
Digikey lists 3600 types which can supply at least 100 mA.
If you really want a voltage follower then remember that the output voltage is about 0.7 V ( the base-emitter junction voltage drop) lower than the base voltage, so you'll want a 4 V zener to get 3.3 V out. |
H: Need chargeable momentary 15 watt regulated voltage source
I'm working on a RF telemetry side-project that requires an amplifier IC to be run for about 5 seconds bursts between charges. The system is portable and runs off of a 3.3V battery that can supply about 100 mA. The IC runs at 10V and draws 1.5 Amps, so 15 Watts. I can't simply use a boost converter on the device's battery because it can't supply nearly that much power. So I need to store the necessary charge and then dispense it at a regulated voltage.
The output voltage of my charge storage system should be well regulated at 10 volts, and I need it to be able to supply a constant 1.5 Amps for at least 5 seconds, after which I will turn it off and recharge it.
The goals here are to minimize size and amount of time it takes to charge the thing. I assume the solution is either going to be some sort of battery or capacitor with some circuitry to regulate the voltage output.
I don't know very much about practical charge storage and power regulation, but if someone knows a good general approach I can look into the details.
edit: Assuming money is no object, anyone interested in elaborating on using a super-cap?
AI: 15 W \$\times\$ 5 s = 75 J. If you want to store that in a capacitor you'll need a capacitance of
\$ C = \dfrac{2 \times 75 J}{(10 V)^2} = 1.5 F \$
There are supercaps with even higher capacitances, but types which can supply a 1.5 A current are between expensive and Damn Expensive™.
A battery will be a better choice. For instance a LiPo followed by a boost converter, to have a nicely regulated 10 V, even at varying battery voltage. Use another boost converter to charge the battery between load bursts; I would shut-down the charger while you're transmitting.
If the 3.3 V can supply 100 mA that's 330 mW. Suppose the two boost converters have an efficiency of 85 % then to get the required 75 J output you'll need 105 J input. At 330 mW that will take 315 seconds, that's 5 minutes and 15 seconds. So you can transmit a 5 second burst once every 5 minutes and 20 seconds. If you need a higher frequency you'll have to find another power source.
edit
I almost forgot: the battery has a less than 100 % efficiency too. You can't get every joule you put into it out again. So in practice the charging time will have to be longer than the 5 minutes.
I would use two LiPo cells in series. Then you'll have 7.4 V nominal, 8.4 V fully charged. That's relevant, because the closer the voltage to the 10 V the less current will be required from the battery. If you draw 1.5 A at 10 V a 3.7 V cell would have to supply 4.8 A if we take again the 85 % for the booster. Two cells in series will only have to supply 2.4 A.
edit 2
Dave's ultracap seems to be a good alternative, but he should mention that the 3.3 V to 2.5 V regulator also has to be a switching regulator, unless you're satisfied with a 75 % efficiency there. If we use eBay as a price reference you can buy a set of two 240 mAh LiPo's for 4.5 dollar, free shipping. That means they only will be discharged for 1 % during the 5 s burst. So while 14 dollar for the cap isn't bad, and definitely something to remember, the LiPo's are only a third of that.
It's also worth noting that stepping down from 3.3 V to 2.5 V and then up to 10 V is less efficient than up from 3.3 V to 7.4 V, and then to 10 V. That's because a switcher's efficiency is related to both the in-out voltage difference and their ratio. |
H: Fixing/Replacing Conductor Ribbon Cable
I tried to fix a keyboard that was having some intermittent issues by cleaning the conductors on the ribbon cable. I used an emery-board to rub off the rust and patina on the pins at the end of the ribbon, but it turns out that the pins are really soft metal and I ended up rubbing off the conductors themselves.
The ribbon had a little bit of slack so, I trimmed it to where the first break in the conductors were so that I could insert the remainder into the slot. Unfortunately there was not enough metal left exposed to make the connection (<1mm on some of the pins).
I thought about different ways I might be able to expose more of the metal but considering how fragile it already was, I figured that scraping or rubbing the plastic covering was out of the question. I considered using a chemical to dissolve some of the plastic, but I have no idea (or way of finding out) what to use or moreover, how to control it precisely enough. I figured that using a soldering iron to soften the plastic and push it away would work because I could more easily control the heat (more or less).
Unfortunately, the metal seems to have been extremely fragile (maybe it’s solder‽). Not only did the plastic warp a little bit and not soften easily to be pushed out of the way, but one of the lines even got a hole in it right through the metal.
Now I’m left with a ribbon cable (as seen below) and wondering how I can fix it. Is there a way to fix it as is or perhaps attach an extension?
AI: I think this is the type of membrane keyboard where the metal is extremely thin. It's often not copper foil as you might expect, rather I think it's printed silver paste. Basically, it has no structural integrity of its own, relying on the plastic for that. If you rub away the surface of the metal, you quickly discover that the surface is all there is to it.
If your ribbon cable consists of two layers of plastic, with the tacks printed onto the inside surface of one, then you may have some success. Try to peel the two layers apart, and cut the non-track layer a few mm shorter.
Alternatively, you might try repairing the broken tracks with some wire glue.
Follow the instructions on Repair the Keyboard Membrane.: |
H: How is this design for a basic PCB pick and place?
I have zero experience with any serious mechanical design, but out of necessity and curiosity, I'm attempting to build a pick and place machine (for my hobby projects as well as low-volume PCB production) -- but a very basic version of it, customized to my own typical applications.
Specs: I'm trying to build the system with:
Cost < US $100 (excluding the vacuum pickup, microscope, etc.)
Board/panel area: Approx. one square foot (not important)
Speed of about 1 part picked and placed within 5 seconds (not important).
The "probe" (see figure below) is intended to be a vacuum pickup (as well as a miniature USB digital microscope attached)
Resolution/step-size of 0.3 mm or less (my smallest part footprints are 1206 resistors and 3 mm QFNs).
Accuracy and repeatability not too important since I have visual/magnified supervision of the process via the USB microscope.
My 1st draft of a very barebones structure, so far including 3 steppers, 3 rods threaded, a USB microscope, and a vacuum pickup:
Operation:
On my PC, for each part to be placed, I store (X, Y) coordinates for its corresponding tape reel as well as coordinates for target position on the PCB.
Y-axis motor/rod/pickup moves to tape reel and picks up part, then moves along Y-axis to target position's Y-coordinate on PCB.
X-axis motor/rod/PCB moves along X-axis so as to allow X-coordinate alignment too.
Z-axis motor/rod/part descends to PCB to place part, then rises.
Repeat until completion.
I supervise any mis-alignments or part misses, etc. via the digital microscope viewed on my PC monitor.
If any adjustments need to be made during any of this, I can just manually pause and adjust the position/action using the computer.
Here are my questions:
Is the mechanical setup drawn above too simple to accomplish the movement? Based on my reading of some literature and watching some videos of pick and places, the systems look much more intricate in build form, and also only either the PCB or the vacuum pickup moves, not both -- whereas in mine, I have one moving along the X-axis and the other along Y-axis (so as to simplify the stage/build).
What will be some key determinants you can think of that will make possible the resolution of 0.25 mm or better? I presume a good choice of stepper/motor (e.g., steps/revolution) is a start.
I see there is one laughably major flaw: Rotation of any of the three rods will cause the PCB or the vacuum pickup, or the picked-up part, respectively, to be rotated along with the rod! Any simple modification to solve this?
AI: Many home made pick and place machines are very similar to CNC milling machines, and this is where you should take your inspiration from.
The machine consists of three linear axes, each of which consists of:
some kind of linear bearing or rail to allow the axis to slide freely.
some kind of motor to actuate the movement.
These two parts will probably make up the bulk of the cost of your machine. Your budget is extremely tight; you're looking at less than $20 per axis! I'm tempted to say that this is impossible, but I hate naysayers, and I love a challenge.
As you've already pointed out, your design is flawed because there's nothing really to prevent rotation of the parts on the threaded rods. It's also missing the important rotary axis which is needed to rotate the parts to the correct orientation before placement. Some designs get around this by placing some of the parts, then asking the operator to rotate the PCB 90º, then placing more parts, etc. You might want to take this option.
Your real problem is the budget, and you're going to have to work very hard to either make many of the parts yourself (those that you can make) or find those parts cheaply somehow (perhaps from broken down machines). One place you look is in old printers. They contain quite nice linear rails which you can salvage, including a fast motor and encoder strip.
Motors: There are two types of motor you can choose from:
Servo Motors. You'll basically be making these up yourself. They consist of a DC motor, electronics to drive the motor, a sensor to measure the position of the motor, and a controller which calculates how much power to apply to the motor to get it to the correct position quickly and accurately.
Stepper Motors. This type of motor doesn't spin freely, rather it can be commanded to move one step at a time. You don't need a position sensor, but you do need to keep track of exactly how many steps you've made in each direction to know exactly where you are, and how far you need to go to get to your next position.
I would recommend the stepper motor approach. Most small CNC machines use these. You should also try to find a driver which supports some microstepping. Not only does this increase your resolution, but it also helps overcome resonance at certain speeds. If you want fast motion, then you'll need acceleration. If you're accelerating, then you'll likely hit the motor's resonant speed and miss steps.
Resolution: High resolution is not that difficult to achieve. For example, if you're using a stepper motor with 200 steps per revolution, driving an M8 threaded rod (which has a 1.25mm pitch) then you can expect each step to produce 1.25mm/200 = 0.00625mm of movement. However, that doesn't mean that your machine is accurate to 0.00625mm. Thread non-linearity, backlash, step drift, and other factors will conspire to increase your error.
Software: Writing the software for this kind of machine isn't that difficult, but it all takes time. Why not check out The Open PNP Project. Their software is already full of features.
Complexity: Unfortunately, as with all robotics projects, you start off with grand goals of simplicity. You can often get simple things working quickly, but you eventually discover that you do need quite a lot of complexity to get things working well, reliably and for a long time. There is no particular problem having the PCB move on one axis, and the head move on another axis. One might think that the moving PCB will shake of the components, but this is unlikely to be a problem. The components are usually very light (unless you're placing large connectors or very large ICs) and they're stuck in a blob of solder paste. I often clumsily manhandle by PCBs into the reflow oven, and I've never seen a part slide out of place. However, if you have a lot of parts to place, then you're moving quite a large table around, and you'll need longer rails and a stiffer table.
Pick up: This is going to be another expensive part, unless you want to suck on a tube to pick up each part. Vacuum pumps can be surprisingly expensive (if your budget is only $100) and you'll also need a valve. You may also need to make a removable pick head so that you can pick parts of different sizes. Small parts need a small tube (obviously) but big parts need a larger tube because they're heavier, and need more surface area for the vacuum to operate over. |
H: Leaving a RC servo motor idle for long term?
I've been searching for awhile and can't find an answer so hopefully someone here can help me :)
I just want to know if leaving a servo motor idle, powered but no input, is bad for it? I'm planning on leaving a servo motor powered indefinitely and only sending it a signal to move a bit maybe once a day.
I was trying to setup something with a relay so my microcontroller can control the power to the servo but it's a little bit of a pain, not that it can't be done.
So, should I definitely not leave my servo motor powered for days/weeks?
Thanks!
AI: This is just like any other motor really, if you're running it inside spec it should be fine. Its lifetime isn't indefinite though. I've done a couple of days with no problems. |
H: interfacing MMA7361L with PIC ADC
I'm trying to interface an mma7361L accelerometer (on breakout board with caps) to the analog inputs of an 18f46k20. According to the PIC datasheet, the Vmax input voltage is Vdd + 0.3 which means I have to get the signal between 0V - 3.6V.
I only need to measure the Z-axis and from what I can tell it produces 850mV/g but I don't know if that's centered at an already offset voltage, or at zero. Does anyone here have experience with this chip? If so, what kind of signal conditioning do you recommend if any?
Many thanks.
AI: I didn't used the device. But I can answer one part of your question.
When you take a look at page 3 of the datasheet under Output Signal you see a value for Zero-g thats typically 1.65 volts for VDD = 3.3V. Even more detail is given in footnote 5 on the same page:
The device can measure both + and – acceleration. With no input
acceleration the output is at midsupply. For positive acceleration the
output will increase above VDD/2. For negative acceleration, the
output will decrease below VDD/2.
So yes, the output signals is centered. |
H: Watchdog Signature in Flash memory or RAM?
We used to use Watchdog signatures in our control applications.
Basically we used to allocate a reserved memory in RAM(uninitialized) and used to store the watchdog signatures.
On power on, the system checks for the watchdog signature and based on that we used to find out whether it is watchdog reset or normal reset that has happened.
My question is, which memory is better for storing the watchdog signatures whether it is RAM or Flash?
AI: Usage of memory storage signatures to determine if a watch dog timer is what has caused your last reset is a scheme that should be avoided at all costs. That scheme is fraught with problems and does not provide a fool proof scheme.
Here are some key factors to consider here that lead to this being a problematic approach. (There are others as well so analyze carefully).
1) Signatures stored into RAM may or may not be legitimate after a power cycle reset becasue RAM can maintain its content for some period of time after a power cycle.
2) A watch dog reset is likely caused by a spurious error in the hardware electronics some place. How can you be certain that this has not compromised the storage of the signatures?
3) The storage of signatures into FLASH or other non-volatile memory is problematic due to the fact that the watch dog timeout / reset could occur during the time that such signatures are being stored.
It is far superior to setup your system hardware with logic that captures the "reset reason" into special flop-flops or register bits that the software can poll in its startup sequence to determine if the restart was due to a watch dog timeout. If your system processor does not have this built right into the chip then consider substituting another processor that does include this capability. In cases where the watch dog is implemented outside the controller look into adding a simple external circuit that can provide the "reset reason" to the processor through an I/O pin. |
H: Reviewing an AC dimmer - trying to understand
I'm apology if the title is kinda generic but I found a (on a AC) which use IC and a triac and there is some points I don't really get.
Image from this application note:
Datasheet used as reference.
About the PTA0 pin :
How can the pin accept 120v since the range of the ADC is VSS to VDD (Page 161).
The maximum absolute voltage for input is VSS-0.3 to VDD+0.3 which is far away from 120v, (Page 149)
What about -120v? The PTA0 shouldn't have a diode to be a voltage above 0v?
Why using resistors for the zero-crossing on the PTA0 pin? If it is to limit the current I tought input pin handled them.
Z1 shouldn't do a shortcut when it reach the breakdown voltage from C1 to Ground? (C1 -> D2 -> Z1) (I assume the top graph is the positive side)
I'm trying to know why it need 0.5W, I don't think it may be the IC (25mA maximum for PTA* pins, but I don't find the power consuption overall (50mA could be the overall consuptions (including PTA* output pin I hope) ?)
Also, there is an exemple using a generic opto coupler where they connect the anode to a 100k which then go to the AC :
I know LED work with current instead of voltage... but i'm kinda surprise they work at 120v. Is it really possible or this is only true for opto coupler?
Should -120v destroy the led since it is much more higher than the regular 5v reverse input voltage?
Thanks
AI: I)
It doesn't. Whether it's low or high, the resistors act as a pull-up while also providing the needed signal. The current, for a supposed logic low would be \$115\sqrt{2}/940k\Omega = 173 \mu A\$
As Some Hardware Guy said, it's floating so the reference is 0V because its supply is mono-alternance and done through a capacitive impedance (reactance) of \$1\mu F\ = 2.65k\Omega \$ (60Hz) or\$ 3.18k\Omega\$ (50Hz).
They're there for not actually putting the pin to 160V, your fear at #1.
II)
C1 has the mentioned reactance, acting as an "active" resistor, limiting current through Z1.
I couldn't find the 0.5W you're talking about.
The rest:
The LED will still have ~2V foreard voltage drop and the maximum current will be: \$ I_{max}=\frac{115\sqrt{2}-2}{100k\Omega} = 1.62mA\$
There won't be -120V since its reference will be the IC's supply, therefore the bridge will clamp the voltage.
You're welcome, but you should really try those simple circuits that have proven to be good enough, instead of choosing this. |
H: Emulate many-pole, dual throw switch with transistors
How can I use transistors and other discrete components to make a circuit that functions like a DPDT (double pole double throw) switch, and can be extended for any number of poles required? For examples assume an input voltage of 12VDC, required current of 200mA per pole and use general transistors and standard values where possible.
Note: it must act as double throw; i.e. power goes to two separate outputs, not one output or floating. Please include some form of schematic.
AI: You can implement this with semiconductors provided the power supply to the circuit stays on all the while you want to have the transistors conducting current. This problem is also much easier to solve if you design for the current flow through the switches to always be the same direction.
Let's say that you design for 12V coming into the C (common) terminal of each "switch" and the NC (normally closed) and NO (normally open) "connections" will always be used to gate power to a ground referenced load from the incoming 12V.
For each pole you will need two FETs driven from a pair of control signals that are logically inverted from each other. You drive one FET ON while the other is OFF and flip the states when the "switch" changes state. In this 12V design you may find it easiest to use a P-FET since turning on the FET works nicely by pulling the FET gate to GND with the control signal. (If you use N-FETs it is necessary to provide sufficient gate drive above the 12V level signals that you are trying to switch). The two FET drive signals can derive from a single control signal by adding an inverter within the path to one of the FET gate drives.
Example Schematic - Note that you have to select the appropriate P-FET per your requirements. |
H: Fabricating control panels
Electronic projects that aren't in a box are fragile. I have a project completing soon that has a reasonably complex control layout. It's going to need some manner of control panel that includes plenty of writing and has different sections.
I can think of a few options:
Get a panel engraved
Print the panel on a color printer and laminate it.
Option #1 is the most professional but can be expensive. Therefor I am exploring option #2. Once the project works, perhaps dropped the cash on option #1 will be a consideration.
I'm wondering what software other board members use for this sort of thing. I've used a nice drafting package called Draft IT from CADlogic to draw the panel. Now I realize I need to color it. I don't think the free version (or any version, for that matter) of the software supports color. This means if I produce a PNG file using DraftIT, color it with some other software, and then need to make a change to the original drawing, I'll have to redo all my work.
Here's the latest cut of the control panel. It isn't quite ready - I haven't marked where holes need to be drilled. I'm working to ensure the control layout is ergonomic and that the controls actually fit. Like icebergs, many controls are bigger underneath...
So we see a lot of text in the upper left. It's a table that defines the behavior of the rotary switches represented by the large circles at the bottom. The seven smaller circles represent 10mm LEDs. The three small rectangles represent toggle switches. The remaining medium sized circle is a speed control knob. Overall dimensions are 10" x 5.25".
So as you see, I have a lot going on here. I need the text to be readable; I would like the main power switch denoted by a read box. I'd like the column of LEDs and the toggle beneath also within a colored box as they are related.
How can I make this into reality, people?
AI: Color print on photo paper covered with a transparent sheet is the way I used in the past for a device I made.
I used INKSCAPE to draw it. |
H: Low voltage oscillator for joule thief
I have a transformer with 2 input leads and 3 output leads. I hooked up an LED to 2 of the output and put a dead AA battery on the input and the LED blinked for a second. I have no data on the transformer, but I am wondering if it is possible (and if so what is the schematic or instructions on how to do it) to put a low voltage oscillator before the transformer. I know how to work with 12 volts with a power MOSFET and a 555 timer, but my question is if this is possible with low voltage.
Any information is great.
AI: If you don't know anything about the transformer then just give it a try and see what you get - a blocking oscillator can be made with just about any transformer, so the chances are pretty good it will work first time. There are a couple of variations of the "Joule Thief" (blocking oscillator) circuit, here is the most basic:
The transformer does not have to be the hand wound toroid shown, and the transistor can be any general purpose NPN bipolar. Note the windings must be connected in opposite polarity (dot convention) You may have to vary the resistor a bit to achieve desired operation (frequency should increase with resistance)
As it happens, I wrote a blog entry a while back on the blocking oscillator circuit used in a buzzer, which was itself part of a cheap "mole repeller". It discusses the operation in a little more detail and shows an alternative topology, so may be of some use to you (you want the bit under the heading "what frequency does the buzzer run at?")
EDIT - to clarify about the connections and some caution:
Since you have a centre tapped secondary, you should be able to use this assuming it's a standard centre tap - the centre tap would the topmost connection to V+ in the circuit above, then one lead to the resistor, and the other to the transistor collector.
In all cases, make sure you have a load (e.g. your LED) connected before you apply power - otherwise you may get pretty high voltages developing at the transistor side, which will damage it. Also, as Acme Fixer points out wisely, if it's a mains transformer you will get high voltage developing at the primary winding, so make sure these are insulated well. The currents will be low from a much stepped up and half dead battery, but could still give a nasty zap so treat with caution. |
H: What limits the modulation of digital signals?
I am trying to understand why there are different speed on different frequency bands. I see that e.g. ZigBee can do 250kb/s for both 2.4GHz and 868/915MHz, but WiFi are also 2.4GHz with much larger speeds, why are they different? How come 2.4GHz is so slow for ZigBee and so fast for WiFi when they operate in the same frequency band?
And is it possible to obtain larger speeds on lower frequency bands just by modulating the signal the same as WiFi does? Is modulation impaired by lower frequencies or is it only dependent on how good the modulating hardware is?
Thank you very much in advance!
AI: The maximum useful datarate is a fraction of the carrier frequency. However, the difference between ZigBee and WiFi has nothing to do with that. 2.4 GHz is so high, that it is not even close to the limiting factor for data rate.
Generally, there are two tradeoffs with higher data rate over the same frequency carrier: RF bandwidth and power.
A perfect sine wave, like a unmodulated carrier, has a bandwidth of zero. As soon as you start messing with that, the bandwidth becomes non-zero. All modulation schemes by definition must modulate the carrier in some way, so all cause the bandwidth of the transmitted signal to be non-zero. This means that signal will use up some finite amount of the RF spectrum. There are legal and practical reasons why you want to keep the amount of RF spectrum space to a minimum. Different modulation schemes have different characteristics, but one way or the other you're going to pay for high data rate with a high spectrum width, or bandwidth.
The other issue is power. This is the main reason ZigBee is much lower data rate than WiFi. I won't go into all the reasons why since it seems off topic to the question and the answer could be a whole book, but it takes more power to transmit a higher data rate at the same signal to noise ratio as a lower one. In some sense, you can trade off data rate with signal to noise ratio at the same power. There are also power and dollar costs in the electronics that have to handle the higher data rate independent of the RF power. Since ZigBee is intended for small low power battery operated devices, a tradeoff was made between data rate, power requirements, distance, and usefulness. Because battery operation was so important, the data rate was made lower. WiFi, on the other hand, had little constraint on the power required for the electronics. Its intended use also demands higher data rate, so a different tradeoff was made. |
H: How do I use an Arduino to switch another IC's pin (reset) to ground?
I'm a releative newbie. I'm using an Arduino to control a Cowlacious audio board that uses the ISD1700 chipcorder series of chips. Periodically, I need to pull the reset pin on the chipcorder to ground to reset it to it's original state. It seems to me that with the low currents involved I should be able to do this directly, but I'm not having luck. Using an Arduino pin set to digital out and leaving it HIGH except when switching doesn't work (I assume the chipcorder doesn't like the voltage when it's high). How can I get this to work?
P.S. I know I could use a transistor as a switch, or some sort of relay, but do I need to do that?
AI: The ISD1700 reset pin is active low with an interal 600K pullup resistor, so pulling the pin low with an output pin of the Arduino should work I would guess.
Some things to double check:
Are the ground pins of the Arduino and the Cowlacious audio board connected?
Did you configure the pin on the Arduino to be an output? (pinMode command)
What is the voltage of the reset pin when the Arduino pulls it low? If it is zero, then that part is working. What is the voltage of the pin when the arduino sets it high? If the Cowlicous board and the Arduino have different HIGH voltages then this might cause a problem. In this case you could try leaving the pin as in input except when you pull the pin low. e.g.
pinMode(X, INPUT);
....
digitalWrite(X, LOW);
pinMode(X, OUTPUT);
delay(100);
pinMode(X, INPUT);
or like you say, use a FET to switch it. |
H: emi standards/best practices for PC cases
According to an EE friend, the FCC specifies EMI tolerances (max outputs) for PC components, including cases. However, here it says that there are standards for a number of components but not cases.
1) Am I understanding correctly that there are no requirements for cases?
2) According to the EE in question constructiong a PC case from anything but (mostly) metal will result in excessive EMI radiation for my purposes (audio) leading to stray signals. Furthermore he says that ventilation holes (number, diameter, and position) are calculated based on stray em wavelength. Is this the case?
3) If all of the above are the case, is there perhaps some lining I could apply to non-metallic enclosures to control stray EMI?
Many thanks in advance
Joe
AI: Am I understanding correctly that there are no requirements for cases?
Correct. The requirement is for the complete product. Be aware that the requirement is not just for your product to pass the test once in a type test, but for every unit in use to meet the requirements.
According to the EE in question constructiong a PC case from anything but (mostly) metal will result in excessive EMI radiation for my purposes (audio) leading to stray signals. Furthermore he says that ventilation holes (number, diameter, and position) are calculated based on stray em wavelength. Is this the case?
That's basically correct. It is possible to design a circuit that can pass the requirements with no shielding provided by the case; but doing so requires either a fairly simple product or extreme care (and probably trade-offs in functionality) when designing it. Most digital products you see depend on the shielding provided by their cases to meet the requirements.
It's also correct that the ventilation holes in your case should be designed in part to maintain the shielding properties of your case. Each opening in the case has the potential to act as a slot antenna, which will radiate efficiently at wavelengths above 4x or so the longest dimension of the opening.
Also, although each opening might not radiate efficiently, an array of holes in the case can produce constructive interference, leading to substantial radiation at certain frequencies, which may be what your friend was thinking of when he mentioned the number of holes affecting the radiation.
If all of the above are the case, is there perhaps some lining I could apply to non-metallic enclosures to control stray EMI?
Yes. You could either build a shielding enclosure around your circuit and then install the whole shield into the externally-visible case, or you could apply a metallic coating to the inside of a plastic or other non-metallic case. For example, Laird Technologies offers Eccoshield ES conductive lacquer coating. I've never priced these materials, but I suspect that they will cost more than simply using a metal case.
One trick with a paint-on coating is that typically you want the shield to be connected to the circuit ground or an external ground by some means, so you'll need to have some kind of reliable contact to the shield. |
H: Use 3.5V power LED connection on motherboard to control a 12V lighting accessory in a computer case?
Background for the problem:
I rebuilt my computer in a new case where the components fit a bit better. Everything is great, and I have no issues with it. However, because of the way they engineered the front panel LEDs, I don't have an LED indicator of computer sleep status. The power light is powered off their control board, which is powered from the PSU directly, so I don't have a sleep indicator.
Now, I could easily just get an LED and extension cable for a simple power LED installation, but I got to thinking that I'd like to try something a bit more elaborate. That is, instead of a single 3.5V LED, I want to use a 12V light source (such as a LED spotlight or cold cathode) and install that to have something that could indicate sleep/power status with a bit more style. I have a mesh grill on the front of my case, so I figure something that illuminates under the mesh would be cool looking and noticeable but also not be disruptive in a dark room (like if I'm sleeping). There's some hurdles to this, though. For one, I obviously can't power one of these straight from the motherboard connector. It doesn't provide enough voltage. For another, the PSU shuts off my case lighting when in sleep mode.
Proposed solution:
So, I'd like to make something to do what I want. I'm a programmer but I have tinkered with electronics kits as a child, so I'm familiar with some (very) basics, and have assembled circuits before (at least, from pre-written designs). It has been a while though and my efforts in designing my own analog circuitry are best not to be spoken of, so I hope I can do something simple, here. :)
What I figure I would have to do is build a simple circuit that uses the power output from the power LED connection on the motherboard to power a switch (of some sort) that would provide power to the 12V light source from a battery, which could be recharged from the PSU. I believe the 3.5V connection on the motherboard just provides straight power to a LED (if one is connected). But instead of a LED, I want to make it so that while the computer is powered on, it would close the circuit for the 12V light so it would illuminate, while simultaneously charging the battery. When in standby mode, it would light the LED (my motherboard causes it to blink) by drawing power from the now-charged battery.
Question:
I think that a simple relay switch would probably be adequate for controlling the 12V circuit, but they're noisy. :) Is there a solid-state alternative? Since I'm dealing with two power sources, I'm not sure if it's possible for me to isolate them easily in a single circuit without physical separation (as a relay switch would provide). As I said, I only tinkered with electronics kits as a child. I didn't take electrical engineering in college, as I got into computer programming instead. But if it is and is simple enough to implement, I'd prefer it.
Also, I'm sure that dealing with a rechargeable battery brings with it its own set of caveats and things to worry about (like worries of overcharging). Is there anything else that I'd have to take into account?
AI: The motherboard connection expect a LED.
Just use an optocoupler or an optotriac to drive next stages. I had once a 250W halogen lamp connected to the hard disk activity led via phototriac/triac :-) |
H: How do I choose my optocoupler to drive a solenoid with a MOSFET?
I need to drive a solenoid with a MOSFET. The solenoid is an 8W solenoid and it will be connected to a +12V voltage.
And this drive command have to be isolated with an optocoupler but I don't know how to choose it.
What characteristics do I need to watch in the datasheet? What is important? Are there some good brands and some bad ones?
AI: A common optocoupler has a current output: you connect the output transistor to Vcc and the emitter will source the current. How much depends on the CTR, or Current Transfer Ratio. That's not very much, and is usually expressed as a percentage. For instance a CTR of 30 % means that you need 10 mA input to get 3 mA output. Use those 3 mA to drive the base of a BJT. You'll want a Darlington to drive get more than 100 mA collector current.
But a Darlington has a high saturation voltage, and may take away too much from the solenoid's power supply voltage. A MOSFET may be better. But MOSFETs are voltage driven, not current driven like BJTs. So you have to convert the optocoupler's output current into a voltage. Nothing more easy: add a resistor between gate and ground, and the current through it will cause a voltage drop, which will turn on the FET.
The nice thing is that you can choose the voltage just by picking the right resistor value. For instance, our 3 mA will cause a 4.5 V gate voltage across a 1.5 kΩ resistor. You may be tempted to choose the resistor value rather high, but that's not necessarily a good idea. The optocoupler has a leakage current when off (called "dark current") and that will also cause a gate voltage. You'll have to make sure that won't get high enough to activate the FET. If the dark current is 10 µA ( a rather high value) then the 1.5 kΩ resistor will show 15 mV on the FET's gate, and that will be low enough not to switch it on. The 4.5 V from the 3 mA will be enough if you pick a logic level gate FET.
The LTV817 is a low-cost optocoupler which is perfect for this: minimum 50 % CTR, a dark current of only 100 nA, and a maximum collector-emitter voltage of 35 V.
Since the LTV817 has such a low dark current the value of R1 can be increased to 15 kΩ. Then 300 µA is enough to get the 4.5 V gate voltage, and the dark current will only cause a 1.5 V voltage across the resistor. At a 50 % CTR you'll only need 600 µA input current. Use 2 mA to have some margin.
For the FET there are plenty of options. The FDC855, for instance, will give you enough current at 4.5 V gate voltage, giving a negligible on-resistance of 36 mΩ: the voltage drop is only 24 mV, and the power dissipation 16 mW (that's 0.2 % of the solenoid's power).
Edit: Selecting the right FET
Like I said there are lots of FETs suitable for your application. I often refer to the FDC855 because it has a good balance between cost and features. For cost the rule is; the lower \$R_{DS(ON)}\$, the more expensive you FET. Yours has only to switch 0.67 A, that's average, and then an extremely low \$R_{DS(ON)}\$ (you can get them down to 1 mΩ) is not really necessary.
You found the PMF290XN cheap (though at Digikey it's only 25 % cheaper than the FDC855, not 80 %). It has a somewhat higher \$R_{DS(ON)}\$ of 350 mΩ, but that's still no problem. Voltage drop is 240 mV, and power dissipation 160 mW. That's more than the FDC855, but still OK.
The higher \$R_{DS(ON)}\$ also puts a limit to the current. For the PMF290XN that's 1 A, which isn't great, but enough for the application. The 2 A you read in the datasheet is pulsed (a single 10 µs pulse). Don't read it as 2 A continuous would be allowed, the 1 A is Absolute Maximum Rating. The (pulsed) higher currents just show where the graph is heading to.
Have a look at figures 6 and 7 also. Figure 6 shows that 3 V is sufficient for 1.5 A drain current, so more than enough for your 0.67 A. Figure 7 shows that you need 3.5 V for an \$R_{DS(ON)}\$ of 350 mΩ at 0.67 A. |
H: Single-phase to 3-phase conversion
I have seen some boxes (with some capacitor circuits inside) which do single-phase power supply to 3-phase power supply conversion. The problem with them is that they cause the load motor to heat up, because the phase difference between the legs is 90 degrees instead of 120 degrees.
Is it possible to design a circuit that produces the third phase with exact 120 degree phase difference using only the original single phase?
Is it possible to use the available phase legs equally to produce the third phase, so that loading on a single phase is avoided?
How would this circuit look?
Note:
The motors in question are 3.7 kW-5.6 kW induction motors.
AI: I'm understanding this question to mean that you're trying to run a three-phase motor off a single-phase line. If you're trying to run the motor directly off the AC line, the phase angles involved will make it difficult to get the motor started, which is part of the reason three-phase exists in the first place. Single-phase motors usually have motor start caps for just that reason. That sounds like what you're describing.
The simple answer to your question is that to get three-phase AC from single-phase AC, you need to rectify the single-phase AC line into DC, then run the DC back through an inverter to get controlled three-phase AC. There are other electronic approaches, but they're less common in my (limited) experience. There are also mechanical approaches, which may be more convenient if you have the parts.
I'd suggest using a drive to operate your three-phase motor. Typical variable-frequency three-phase drives are exactly what I described above: a rectifier, followed by an inverter. I can't speak as to what's on the market in a given power class, but larger three-phase drives typically have terminals for the three-phase AC line input, the DC bus, and the three-phase motor output. If you have those terminals, you have two options.
One is to run single-phase AC through the three-phase input of the drive. If the voltages are correct, the drive should operate fine. The caveat is that you'll have to derate the drive somewhat. The input diodes are spec'd assuming that the drive's constant-power load will be distributed among three legs of the rectifier. If you distribute that same load over just two legs, those diodes will get hotter. The internal bus capacitors will also get hotter, because they'll see more ripple current without the third phase. Check with the drive manufacturer for the derating info.
If your drive has DC bus terminals, your other option is to skip it's internal rectifier and use an external one. Rectify the single-phase AC, then use that DC as the input to the drive. This will let you avoid derating the drive. My company makes something exactly for that purpose, though its power range may be larger than is cost-effective for your application. You'd have to price both options out to find out for sure. Read this for more details. |
H: How does a TFT panel work?
I have to use a TFT panel but I have never used any before and I really can't get how it works and how I can interface it with a microcontroller.
I found INT057ATFT which seems to have an integrated TFT controller so I guess I just interface it with the 8080 interface of my microcontroller.
But I have been advised to use a TFT like the TM056KDH01 and this one I really don't get how I can interface it with a microcontroller. On the pin descriptions I don't even see the RGB pins.
Could you explain me how these screens work and how to use and interface them?
For the microcontroller I will choose it depending on the screen I take so I don't have made my choice yet. I will take something which is possible to interface with the screen.
Another thing is: is there any advantages using one or the other type?
After some research I found out that it would not be easy to interface a tft panel with an RGB digital interface with the microcontroller. Then I thought about using a tft controller and I found the SSD1963 which seems great because it can drive tft with a resolution up to 800/480 but it needs 3.3V supply AND 1.2V supply.
Does anyone know about another controller with the same caracteristic but with only the 3.3V supply?
AI: The INT057ATFT uses an SSD1961 controller, whose I/O can be configured in Motorola 6800 mode or Intel 8080 mode. So if your microcontroller has an 8080 type interface you can connect it without problems. But also a microcontroller without external address or data bus can talk to the LCD controller, of course.
The left side connections are the only ones you should look at. The ones on the right are the controller's business; you don't have to worry about those. That's very different from the TM056KDH01 you refer to in a comment. That's an LCD controller which needs continuous data, combined with vertical and horizontal sync. You can compare it somewhat with video + sync: if you stop supplying a video signal you won't have a picture. Only here you have 18 digital lines instead of one analog. The SSD1961's microcontroller interface runs asynchronous of the LCD refresh. You just have to fill the buffer.
There are a couple of control lines, like D/C (Data/Control) select, R/W and enable, which will be familiar for users who have worked with text LCD modules with a 44780 controller. The SSD1961 datasheet has a list of commands you can use to configure the LCD. You supply the commands via the 18-bit data bus. That allows to send a complete pixel data of 3 x 6 bit (RGB) simultaneously. An 8080 microcontroller only has an 8 bit bus, and you can configure the device to use only 8-bit; the unused data lines should remain floating. In 8-bit mode you need less I/O pins on you microcontroller, but each pixel will need three successive writes. |
H: Protection of ADC inputs
I am using an ADC MCP3424.
After breaking one of the ADC inputs, I was saw this thread:
ADC input protection?
I also read Microchip app note TB3013. Figure 3 in particular:
The ADC's analog inputs need to be clamped at ± 0.3V of the supply rails.
The Vf of a schottky like BAT85 depends on the current flowing through it. So for a Vf < 0.3V, we would need a series resistance of 10k, to keep the current less than 1mA (assuming the Opamp Supply is +15V/-5V). Now the problem is wouldn't the 10k resistance slow down the charging time of the sampling capacitor and reduce my 18-bit accuracy? I am measuring slow moving DC voltages.
Is there any solution for the problem ?
AI: Putting a 10K resistor on the input will not change things to any great extent. Yes, technically it will slow down charging the sampling cap. It will act as a low pass filter where the cutoff frequency is 1/(2*pi*R*C). That works out to a cutoff frequency of 4.97 MHz. So any frequency above 4.97 MHz will be significantly attenuated.
This ADC has several modes, but the quickest mode it has runs as 240 samples per second. Meaning that it can handle frequencies less than 120 Hz. The RC filter caused by the input cap and the 10K resistor has a frequency response that is still 4 orders of magnitude higher!
In short, that 10k resistor on the input isn't going to do anything negative for you. |
H: How do I protect power out connector from having power in connected to it?
I'm building an adapter for rotary encoder that reads input of encoder and sends pulses by RS485. I'm using two RJ-45 connectors to provide power both for adapter and encoder. Both connectors have the same pinout. The right side connector receives 12V, theres linear regulator on board that powers RS485 transmitter/receiver chips and encoder.
I want to make some protection mechanism that would prevent failure when user connects 12V cable to output (left side) connector. Indication LED about wrong connection would be bonus.
All I was able to come up is a mosfet Q1 that has it's gate tied to +12V and 5V zener diode in series with resistor and "wrong connection" led.
Could someone advice me on more elegant/better way to do it?
AI: You're MOSFET doesn't seem to be doing anything useful. Is it connected to the shield pins of the RJ45? I would just get rid of it.
You really need the blocking diode on the output side of the 7805, not the input side. 78xx-series regulators really do not like to be back-driven. To compensate for the diode's forward voltage drop in normal operation, you could put a similar diode in series with the 7805's ground pin.
That, plus your "wrong connection indicator" circuit, should be all you need. |
H: How to improve torque and RPM of a DC motor?
I have an FA-130 motor (DC) with permanent magnet, my power source is a 2 AA batteries (rechargeable) so a total of 2.4v.
Assume that all cases will start from the same specification, theoretically, what would happen if I do the following?
Case 1: Increase/Decrease strength of permanent magnets. What would happen to torque and RPM? Why?
Case 2: Increase/Decrease size of magnet wires. What would happen to torque, power consumption and RPM? Why?
Case 3: Increase/Decrease the size of armature. What would happen to torque, power consumption and RPM? Why?
Case 4: Increase/Decrease the number of turns (coil). What would happen to torque, power consumption and RPM? Why?
In general, how can I increase the torque and RPM of this motor given a constant voltage?
Please, explain it as if you're talking to a 6-year old kid, I'm not knowledgeable in this field but I want to know the concept.
AI: I'm going to assume that this 6 year old has at least a little background in physics. I'm going to start off by answering why each result will occur with a lot of math to describe the physics behind it all. Then I will answer each case individually with the math providing the reasoning behind each result. I will wrap up by answering your "in general" question.
Why?
The answer to all of your "Why?" questions is: Physics! Specifically Lorentz's law and Faraday's law. From here:
The torque of the motor is determined by the equation:
$$\tau = K_t \cdot I~~~~~~~~~~(N \cdot m)$$
Where:
\$\tau = \text{torque}\$
\$K_t = \text{torque constant}\$
\$I = \text{motor current}\$
The torque constant, \$K_t\$, is one of the main motor parameters that describe the specific motor based on the various parameters of its design such as magnetic strength, number of wire turns, armature length, etc. as you've mentioned. Its value is given in torque per amp and is calculated as:
$$K_t = 2 \cdot B \cdot N \cdot l \cdot r~~~~~~~~~~(N \cdot m / A)$$
Where:
\$B = \text{strength of magnetic field in Teslas}\$
\$N = \text{number of loops of wire in the magnetic field}\$
\$l = \text{length of magnetic field acting on wire}\$
\$r = \text{radius of motor armature}\$
The Back-EMF voltage is determined by:
$$V = K_e \cdot \omega~~~~~~~~~~(volts)$$
Where:
\$V = \text{Back-EMF voltage}\$
\$K_e = \text{voltage constant}\$
\$\omega = \text{angular velocity}\$
Angular velocity is the speed of the motor in radians per second (rad/sec) which can be converted from RPM:
$$\text{rad/sec} = \text{RPM}\times\dfrac{\pi}{30}$$
\$K_e\$ is the second main motor parameter. Funnily enough, \$K_e\$ is calculated using the same formula as \$K_t\$ but is given in different units:
$$K_e = 2 \cdot B \cdot N \cdot l \cdot r~~~~~~~~~~(volts/rad/sec)$$
Why does \$K_e = K_t\$? Because of the physical law of Conservation of Energy. Which basically states that the electrical power put into the motor needs to equal the mechanical power got out of the motor. Assuming 100% efficiency:
\$P_{in} = P_{out}\$
\$V \cdot I = \tau \cdot \omega\$
Substituting the equations from above we get:
\$(K_e \cdot \omega) \cdot I = (K_t \cdot I) \cdot \omega\$
\$K_e = K_t\$
Cases
I'm going to assume that each parameter is being changed in isolation.
Case 1: Magnetic field strength is directly proportional to the torque constant, \$K_t\$. So as magnetic field strength is increased or decreased, the torque, \$\tau\$, will increase or decrease proportionally. Which makes sense because the stronger the magnetic field, the stronger the "push" on the armature.
Magnetic field strength is also directly proportional to the voltage constant, \$K_e\$. However \$K_e\$ is inversely proportional to the angular velocity:
$$\omega = \dfrac{V}{K_e}$$
So, as the magnetic field increases, the speed will decrease. This again makes sense because the stronger the magnetic field, the stronger the "push" on the armature so it will resist a change in speed.
Because power out is equal to torque times angular velocity, and power in equals power out (again, assuming 100% efficiency), we get:
$$P_{in} = \tau \cdot \omega$$
So any change to torque or speed will be directly proportional to the power required to drive the motor.
Case 2: (A bit more math here that I didn't explicitly go over above) Going back to Lorentz's law we see that:
$$\tau = 2 \cdot F \cdot r = 2 (I \cdot B \cdot N \cdot l) r$$
Therefore:
$$F = I \cdot B \cdot N \cdot l$$
Thanks to Newton we have:
$$F = m \cdot g$$
So...
$$\tau = 2 \cdot m \cdot g \cdot r$$
If you keep the length of the wire the same but increase its gauge, the mass will increase. As can be seen above, mass is directly proportional to torque just like magnetic field strength so the same result applies.
Case 3: The radius of the armature, \$r\$ in our equations above, is again directly proportional to our motor constants. So, once again, we have the same results as we increase and decrease its length.
Starting to see a pattern here?
Case 4: The number of turns of our wire, \$N\$ in our equations above, is also directly proportional to our motor constants. So, as usual, we have the same results as we increase and decrease the number of turns.
In general
If it isn't obvious by now, torque and speed are inversely proportional:
There is a trade-off to be made in terms of power input to the motor (voltage and current) and power output from the motor (torque and speed):
$$V \cdot I = \tau \cdot \omega$$
If you want to keep the voltage constant, you can only increase current. Increasing current will only increase torque (and the total power being supplied to the system):
$$\tau = K_t \cdot I$$
In order to increase speed, you need to increase voltage:
$$\omega = \dfrac{V}{K_e}$$
If you want to keep the input power constant, then you need to modify one of the physical motor parameters to change the motor constants. |
H: Laptop charger with broken ground prong
I have an old house (1927) that has been completely renovated, but all the plugs in the living room are 2 pronged. I have a charger that had the ground prong broken off. I want to know if it's bad on my laptop or if it's a fire danger to my house. One I'm willing to risk, one I'm not...
AI: No, the lack of a safety ground connection is not a danger either to your laptop or to your house ... unless your laptop was going to catch fire anyway for other reasons, but in that case, the ground connection woudn't have made any difference.
But considering your house's wiring generally, for additional peace of mind, you might consider installing ground-fault circuit breakers and/or arc-fault circuit breakers in your breaker box ... assuming, of course, that you have a breaker box and not a fusebox to begin with. However, beware of nuisance trips with these breakers, which may be an indication that the wiring is in generally poor shape and needs to be upgraded. |
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