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H: What is the plastic keying filler called for IDC/Dupont connectors? Some male IDC and Dupont connectors have a pin missing, with the corresponding female connector having that hole blocked up with some plastic, so that the cable can only be mated one way. I can however only see for sale female IDC and Dupont connectors that have all their holes unblocked. I am having no luck figuring out how you are supposed to block off the keying holes yourself. Is there a particular plastic piece you insert to achieve this, or is it just a matter of cramming the hole with whatever you can find and perhaps melting it in place? The professional ones look either moulded or as if a plastic pin has been pushed into the hole which would be a nice touch, but I have no idea what this plastic pin may be called and all my guesses (key, block, filler, etc.) are coming up blank. Is it possible to buy these keying pieces for both IDC and Dupont connectors, and if so what are they called? Here is an example pic from Google showing a keying pin on a Dupont connector: And another showing a blue keying pin on an IDC connector: AI: It may depend on the manufacturer/vendor, but Digikey seems to sort them as "Keying Plugs" or "Keying Pins". See below for an image of what they look like, this one is for a TE Connectivity plug.
H: Does grounding to earth make a circuit (allowing electricity to flow)? This question is mostly extending on a question that has been asked here: Single terminal of voltage source attached to earth ground I'm learning the basics of electronics and this question has always bothered me. The post above explains that a single terminal of a voltage source, when connected to ground (earth), will have no current flowing, as there is no circuit. It then goes on to talk about how capacitance might play a small factor. My main question is this: Why then would you get electrocuted if you touched a live wire while grounded, since it is not a proper circuit? Is it different with AC and DC? Also, when the single terminal of a flyback transformer (usually the anode cap on a CRT monitor) is placed near the ground, it creates a bright arc to the ground. How does that happen if it's not a complete circuit? In high voltage projects, most people connect the live wire from the flyback transformer straight to ground to complete the circuit. AI: Why then would you get electrocuted if you touched a live wire while grounded, since it is not a proper circuit? Because it IS a proper circuit. If the return wire of the high voltage source is grounded then, touching the live wire will close a circuit where your body (also grounded as per the quote above) is the "load" and a current will flow. If the high voltage source were floating i.e. it didn't make a connection to ground then you might feel a little tingle when first touching the live wire but this is due to discharging the small amount of capacitance that the floating voltage source has (with respect to ground). Also, when the single terminal of a flyback transformer (usually the anode cap on a CRT monitor) is placed near the ground, it creates a bright arc to the ground. How does that happen if it's not a complete circuit? Usually, the high voltage winding of a CRT flyback transformer is grounded (or partially grounded via a resistor) and therefore the arc is completing a circuit.
H: Learning how to choose integrated DC to DC boost converter I am designing a microprocessor driven wristwatch with a VFD tubes 2 digits display. These are the three major requirements: VFDs operate at 15V uC operates at 3V I have to use as less space as possible My plan is to use a 3V battery and a dc to dc step-up converter to get the 15V when I need them (that is, when I decide I want to display the time on the two VFDs). Now I have to select a suitable integrated circuit, which fits my very special needs, and this is something I've never done. So, these are the specs I have for the booster: Vin 3V Vout 15V Iout 300mA Based on this parameters, I've identified the following component: LT3495 Do you think it would be suitable for what I need? Does a designer normally search for a new component this way, or normally are more specifications, more details needed? AI: 300 mA out at 15 volts is a power of 4.5 watts and, with the inefficiency of the converter you will need to supply about 5 watts from the 3 volt battery (or 1.67 amps). However, the output resistance of a Duracell CR17345 is about 0.25 ohms and taking this amount of current instantly drops the battery terminal voltage from 3 volts to about 2.6 volts. You are now on the slippery slope of it never working - the voltage drops and your DC-DC converter demands even more current to get 5 watts it needs to sustain 15 volts on the output at 300 mA. See how the battery terminal voltage has dropped to about 2.6 volts with a load current of just 1 amp after only 6 minutes - given the output resistance is 0.25 ohms, a 1 amp load will immediately cause a voltage drop to 2.75 volts. The LT3495 requires an input voltage of at least 2.5 volts so after a few tens or maybe hundreds of milli seconds it will shut down then burst back to life (as the load is removed due to the shut-down) then continue repeating this and flashing the display on and off cyclically. You need to find a display technology that isn't so hungry or, have a better battery.
H: What is the truth about 1.5 V "lithium" cells? At least one manufacturer out there is marketing "Lithium" cells in familiar AA and AAA sizes, as direct replacements for those standard 1.5 V sizes, boasting the typically better than alkaline longevity you'd expect from lithium. But we all know the range of lithium technology cell voltage is expected to be 3 V for single use cells, up to a max of around 4.2 for li-Ion variations of rechargeable at max charge. All my attempts to research what the truth is (short of buying and cutting one open) have resulted in little more than manufacturers hype. Can anyone shed light on what is going on with these? A stretch to think perhaps they actually have embedded buck converters under the hood? Or has a genuine 1.5 V lithium technology actually been invented? I've included one manufacturer's photo as a reference AI: Lithium batteries come in many different chemistries, and it is the chemistry that governs the voltage. The most common chemistries are on the order of 3-4V, but there are chemistries which have a 1.5V terminal voltage. The wiki page for Lithium batteries has a list of many different chemistries and their voltages. A Lithium anode with an Iron Disulphide cathode (\$\mathrm{Li-FeS_2}\$) is one such example of a 1.5V terminal voltage, and is the chemistry used in the AA replacement batteries as per the datasheet link on the Wiki page, and in @pjc50's answer.
H: Intuitive way to see pole/zero directly from schematic The image below is the electrical configuration and the transfer function of a type II compensator using OTA. By looking at the transfer function, we can easily see that the circuit has two poles (one pole is at origin) and one zero. However, I am wondering if there is an intuitive way to see the poles, zero directly from the schematic. In other words, why do these pole and zero exist? I think we can base on poles/zeroes definition to get some insight. Poles and Zeros of a transfer function are the frequencies for which the value of the denominator and numerator of transfer function becomes zero respectively. From this definition, I can see why there is an origin pole here. At DC (zero frequency), two capacitors C1 and C3 act as open circuit so the equivalent impedance of the two parallel branches is infinity. This leads to infinite output voltage. Is there an intuitive way to see the other poles and zero here? (from poles/zeroes definition or something else) AI: First get an RLC nomograph and transparent or translucent paper. Since adding log graphs of Z(f) becomes like linear superposition, you can then add up each node remembering that Av is impedance ratio on inverting side and 1+Av for non-Inverting side. Pole zeros that combine multiple resistors and capacitors can be intuitively ignored at f near 0 and ∞ as open and short. (R1/(R2+C1))//C2 is can be a lead/lag circuit. These are used in PLL filters and SMPS filters to have two ranges where the R ratios dominate the band no phase shift) This occurs when Zc is too low or too high compared to R to affect a change, which you can see on the nomograph. Just identify which ratios you are using change colors for negative feedback where the slope inverts ( negative impedance conversion). With practise you ad more parts and find the resonant frequency of RC circuits from phase cancellation where the -Zc = +Zc making band reject or band pass filters. Q or -3dB BW is then the reactive/real impedance change or visa versa depending if series or shunt using negative impedance Caps ( -ve feedback) or inductors.
H: Iphone Lightning Charger - How would one go about it? Hello, im completely new to EE, but im doing my best to learn. For a starter, i want to make a device that can charge an iphone with a battery pack. Ive read up a little, and i think this would work, for the power itself, im not sure however. Can anyone try to explain why it would, or would not work, and/or how i would go about making such a device. I hope this question isnt to wide, but i could try to narrow it down a little if needed. EDIT: I noticed the R1 component should actually be on 6 Ohms, so pretend its there. Thank you :) AI: This approach has numerous problems -- there's no active voltage regulation as a primary point. Resistors do not drop a fixed voltage, rather it's dependent upon the current flowing through them. In some instances you will have a much higher voltage than the phone will likely tolerate, which may cause permanent damage. Your options are: A linear regulator. Linear regulators take a higher input voltage, and burn the difference between the input and output voltage off as heat. They provide extremely clean and stable outputs, but are extremely inefficient and require upwards of 2 volts higher than the output to remain stable. A mechanical equivalent is dragging a brake on a motor to keep its speed regulated. A switching regulator. Switching regulators use inductors as "flywheels" to transform voltages at high efficiency. A mechanical example here is to strike a heavy spinning flywheel with a fast, lightweight hammer whenever its speed drops. The "ping" of the hammer adds a bit of energy back to the flywheel, which keeps its speed in the right area. These circuits "ping" the inductor several thousand to several millions of times per second. The transfer of energy here is very efficient. There are premade switchmode modules all over eBay for pennies. I'd highly suggest repurposing an existing device to suit your needs, since that will satisfy about 30% of EE anyway.
H: SMPS with output adjustable from 0-100% - what topology? I need a power supply with an output adjustable from 0 to 12Vcc-10A. I'm not sure if I make a 12Vcc fixed supply and then put it trought a PWM switch with a filter to adjust the amplitude, or I try to make a supply with 110v/220vac input to 0-12Vcc (flyback or forward). I am afraid a single SMPS won't be able to generate an output which I can adjust from 0 to 100%. AI: First, for 10A output current, flyback would be totally the wrong choice due to high losses on output windings and high flyback spikes to the primary. Two-switch flyback may be a good choice but still the output winding is the issue for 10A current. Go for either two-switch forward, half bridge or even LLC half bridge. Designing a variable output SMPS is a challenging task: The first problem comes from generating supply voltage for the controller block: The auxiliary winding concept in conventional isolated converters is totally useless for variable SMPS, because the output voltage of this winding will be load- or output voltage-dependent. So you'll need a separate supply. Actually, a self-oscillating (also known as "Ringing Choke") converter with, say, 15V/200mA output will do the job. But, as you might guess, cost and placement issues... Still it's the best solution. The another problem is generating 0V. IIRC, some articles about this tells that a negative reference is needed. But I'm not sure. Anyway, if your minimum output voltage requirement was 2.5VDC then simple TL431 and divider network with a linear pot would simplify the design. And the last problem is light load or even unloaded state. For forward and half-bridge, you need an output inductor. It's easy to get 10A output from a 33uH toroidal inductor with a switching frequency of 100kHz. But what if the output current goes down to 10mA (Suppose you want to test an LED with your supply, for example)? You'll need either an output inductor with high inductance (say, 1mH) or a heavy dummy load to waste some of output power for unloaded state which decreases efficiency. LLC half-bridge does not need an output inductor but even for this, you still need to waste at least %5 of max output power on a dummy load. Of course you can get the converter working with 12VDC/10mA output even with %1 dummy load, but if you decrease the output voltage then the converter will enter discontinuous mode and you'll see that the output voltage is fluctuating. Rather than output voltage, you'll also need to regulate output current (for current limiting or output shorting). This will need an extra comparator which brings extra care for designing output and feedback blocks (feedback stabilization).
H: AC powered microcontroller issue Context: I got this fancy freezer from Liebherr a while ago and it stopped working. The compressor would fail to run continuosly. I have narrowed down the issue to the control board (behind the front panel). The micro-controller would constantly attempt to turn the triac ON, but fail to maintain the ON state. The triac is responsible for switching on the compressor. Suspect: The following power supply circuit (might be wrong though... multilayer PCB): simulate this circuit – Schematic created using CircuitLab The C1 is X2 MPF capacitor. R1 is a 0.5W resistor. C2 is an electrolytic (already replaced). The two diodes next to the ceramic capacitor are two schottky diodes. If I attach a regulated 3V power supply across the capacitor (the polarized one), the micro controller is happily chooching. The current draw from the external power supply does not seem to be high. but as soon as I switch off the external feed, the leds on the display dim and the micro-controller fails to keep the triac in the on state. So the question is: how do I troubleshoot this further? Picture: Another Detail: The X2 Metalised polyester foil capacitor is rated 680nF but my multimeter reads 104nF. Could that be the issue? Update Failed Component: C1. Thanks to Simon B AI: It looks like a relatively standard capacitive dropper circuit. C1 provides the current limiting. R1 prevents a surge on power-on. The two diodes, plus the two pairs of diodes (why pairs?) form a bridge rectifier, with C2 providing the smoothing. With the correct load applied, there should be an appropriate voltage on the + and - terminals. Without the load, the voltage could rise to the point where C2 goes bang. You should be able to test each of the components in turn with a multimeter. If C1 is failing, it may not be letting through enough current. Beware, there would normally be a high value resistor across C1 in similar circuits. Without that resistor, if the circuit has been plugged in recently, there could be a substantial voltage across C1.
H: Relay closes without load connected across its terminals but doesn't work when it's loaded why would something like that happen? all the relay has to do is to close its contacts, what difference would the load do? AI: In theory it shouldn't, but perhaps your relay drive circuit is operating at the margins and is being thrown off a bit by interference of some sort when the load is present. Looking at things, you have a PIC micro, which if the IO pin is putting 5V into R3, delivers 500 microamps into the base of Q2. The DC gain of a 2N2222, let's optimistically call it 100, would yield 50mA current through the collector. LED2, powered by the 12V battery, would want nearly 50mA of current by itself ((12V Battery - 1.8V LED voltage drop) / 220 ohms = ~48mA). Perhaps the relay is only barely getting enough juice in this setup? Try change R3 to something lower, like 1K or something, and increase R5 to at least 1K if not more. Or even take out LED2 entirely to see what happens. And check the specs on the relay coil - how much current does it need to operate? You could just test it with the battery and a meter, too.
H: Annoying problems getting started with logic circuits (SN74LS) I'm trying to get started making logic circuits but I'm not making much progress even with the most basic. I have a SN7408 which is an AND gate however it doesn't matter what I put over the inputs (1A and 1B) I get continuity from Vcc to 1Y which is rather irksome. I have been playing around with a few USB power sources 5v from a 230v mains adapter. 5v from my macbook and 5v from a recently purchased phone powerbank. I'm guessing I'm burning out the chips. How can I stop doing that? Any help gratefully appreciated! Cheers, Andrew - AI: To expand on Peter's explanation, your two inputs are currently either switched to +5V, or open. The problem here is that both +5 and "open" are seen as logic "one", and so the AND gate output will always be a "one". You can fix this by putting strong pulldown resistors (220 Ohm) on the inputs, but with TTL it is much better to use pull-up resistors and switch to ground. The pullups can be 1K to 10K or so. With CMOS logic the resistors can be much larger, but they are still necessary.
H: Converting -3.7V to -32V (both negative), Is it a boost or a buck? I need to get create a split supply at +/-32V, but all I have to work with are LiPo batteries. I was thinking something like the block diagram below. The positive part is simple, but the negative part is not so straight-forward. simulate this circuit – Schematic created using CircuitLab I tired googling this, but there is not much literature on dc-dc conversion with negative voltage. Main questions I have and need help with are: When you convert -3.7V to -32V, is that considered a boost or buck? The output voltage is lower, but the absolute value is greater. Judging by the lack of information, it seems this is not a typical problem, but is it practical? Can the negative "boost" be done using the same positive boost IC? Ideally, I'd like to use the same chip if possible. Any recommendation on reading material or schematics? Or any other method of getting +/-32V? AI: Going from -3.7V to -32V to is a boost, a negative boost. To make things more less confusing, here's are application notes Designing a negative boost converter from a standard positive buck converter and Positive Buck Regulator Makes Negative Boost DC/DC Converter. (The regulators in these app notes are intended to be positive bucks, but repurposed as negative boosts.) For a simple inductive boost converter, it's advisable not to have a boost factor greater than 6. This applies to both positive and negative boost. p.s. We’re situated roughly in the same neck of the woods, interestingly.
H: STM32F072 - Working with HAL and FatFS I am trying to make small data logger project with STM32F072 as an opportunity to get more familiar with STM's HAL. With CubeMX I generated HAL code that has SPI and also includes FatFS. Right now I am having issue gluing SPI low level drivers to FatFS. Can someone please point me in the right direction or post some examples on how to glue HAL SPI functions to FatFS? Right now I see that there is following declaration generated by HAL, but I can't seem to find any actual implementation that works. Diskio_drvTypeDef USER_Driver = { USER_initialize, USER_status, USER_read, #if _USE_WRITE USER_write, #endif /* _USE_WRITE == 1 / #if _USE_IOCTL == 1 USER_ioctl, #endif / _USE_IOCTL == 1 */ }; Any examples or documentation on how to implement this would be very much appreciated. AI: I've finally manage to implement low-level drivers and they seem to work nicely. I'm sharing source code of my implementation in case somebody else gets stuck at the same part that I was.
H: Is electric charge a state variable since derivative of charge flow is defined as current? \$ i=\frac{dq}{dt} \$ Is 'q' a state variable in general? AI: Yes, if we understand the same thing by state variable. If we are trying to do a time-step by step simulation of a circuit, then we might proceed as follows a) using the following initial conditions for each step source voltages and currents the ground voltage (=0) the voltage across any capacitor (as it depends on the capacitor charge) the current through any inductor (as it depends on the inductor flux) b) now solve for all the remaining voltages and currents c) now update the capacitor voltages due to the time integrated current that flows into them, and update the inductor currents due to the time integrated voltage across them, these become the initial conditions for the next step A related notion is that charge in a capacitor, and current in an inductor, stores energy.
H: Regulated AC power supply? I've noticed mains voltage tends to be 120V +/- 10% for 108-132VAC. How would you go about building a regulated AC-to-AC power supply which works off of mains and outputs 120VAC with a much lower (less than 1%) variation? The output must be a sine wave at 120Vrms. Edit: For those asking, the output would be fairly low, 1.5A - 2A, 2.25A max. Circuit to be powered: Expected input: 0V-5.5KVrms sine wave at 60Hz Simulated output: +/- 0-100KV DC. The objective here would be to connect components in the order: mains --> power conditioner --> variac --> 1:46 center-tapped transformer insulated for high voltage --> Cockcroft Walton Multiplier outlined above. To achieve a variable 0-100KV DC (effectively 0-200KV DC with both channels factored in) power supply. I am not certain the power conditioning stage is the way to go but assumed it would be since it is much easier to stabilize low voltage (120Vrms) than it is to stabilize high voltage (5.52KVrms.) Relatively minor ripple of 5% will grow into +/- 10% by the end of the CW stage at full power, yielding around 20KV off of the expected value - not an insignificant amount. AI: The classic solution adopted by the TC community is a Variac, but they can tend to get quite heavy for kW throughput. If you only want to cope with a small range of input variation, say 10% = +/- 12v, then you might consider a small Variac, followed by a 120v->12v transformer, the secondary of which is put in series with your mains supply. Let's say you had a 1A variac and a 120VA transformer. The transformer could deliver 10A, so could trim a 1200 watt supply, using 120 watt components. There's no free lunch of course, the increase in power throughput goes together with the decrease in range. If you wanted to be clever, you could custom wind the 12v transformer to give you (say) 0.5v taps, and electronically switch them with (say) back to back MOSFETs, depends what your level of skill and requirement for automatic operation was. A binary tapped 12v transformer and some manual switches would make a cheap half-way house.
H: Optical Communication Two Sources One Receiver I'm working on a project that requires me to wirelessly transmit analog information over a short range (few meters at maximum) to a receiving end from two transmitting ends. The tricky part is I can only use one receiver and I have to build these transmitter-receiver parts from scratch, meaning no microcontrollers or ready-made modules (like those 433MHz RF transceivers that I see commonly used). I want to use IR or LASER LEDs for this purpose but am unsure how that would turn out. What kind of output would I observe from a photodiode if I were to shine two LASER beams on it simultaneously? For instance, if I were to drive the two LASER diodes with PWM and have a significant difference between their frequencies (5kHz and 100kHz for example), would I be able to filter the two transmitted signals using basic filters with corresponding center frequencies on the receiving end? I don't need extreme precision, as long as I can compare the amplitudes of my original signals on the receiving end I'm good. I realize that with this kind of set-up the best method would be time division multiplexing but I'm unsure how to achieve that with analog circuitry. Would you guys suggest any other method for this type of wireless communication? I don't think I have enough experience with electronics to build my own RF transceiver but if you have any tips on that it would also be appreciated. Thank you! AI: I think you are almost there with the pwm idea. Keep the pwm modulation at the same frequency and then take the output of this at each transmitter and use it to modulate a much higher carrier frequency, different in each transmitter. Filter the two carriers at the receiver and you end up with two PWM modulated signals. As long as you are not saturating the photodiode the received signal should be the sum of the two incoming signals.
H: Resistor dissipating 0.1W in heatshrink tubing I have made a LED lamp by soldering together a 3mm red LED and a 1K resistor form ebay. I then wrapped it in heatshrink, to insulate it and make it look nice. But now I've done some maths and I realized that the resistor will have to dissipate 0.1W (12V power supply -> 10V/1KOhm=10mA -> 10V*10mA=0.1W) and I am worried that it might run a bit too hot and malfunction (cause a fire). Am I right to worry or will it be fine? It has been on for about an hour now and it is only a little warm on the outside (probably not even 30°C), but I don't know what thermal resistance it has. AI: We could get into some maths here about the thermal resistance of heat shrink tubing but you are worrying unnecessarily. If the outside is 30C the inside is not going to more than 50C.
H: What are the reasons that the emitter is heavily doped in an NPN BJT? Is doping the emitter more than the collector crucial? Would current not flow at all if both regions were equally doped? Why? Is there a way to explain this in a simple fashion? AI: The emitter, as the name suggests, emits carriers. In case of an NPN transistor those will be electrons. A higher doping level means more of these carriers will be generated (per volume and time) compared to a region with less doping. These "shallow doped" areas are the base and the collector. A BJT works by the carriers from the emitter "overwhelming" the amount of (opposite) carriers (in case of an NPN: holes) in the base region. Also, the base region is narrow. This makes the chance that a carrier originating from the emitter recombines in the base, small. If that chance was large (due to a highly doped base) then the base current (to fill up all those used-up holes) would be much larger ! So: the Base needs to have a lower doping level (compared to the emitter) so that the beta (current amplification) will be large, which is what we want. The collector could be doped more and the beta could then still be high. Remember that beta is related to the doping ratios of emitter and base. In an NPN under non-saturated operation the collector is at a high voltage so all the carrier electrons will be "sucked out" and travel to the positive supply. The electrons originating from the collector therefore cannot influence what happens at the emitter-base junction. However a low doping level on the collector does increase the size of the depletion region and this increases the maximum collector-base voltage. Also it is much easier to manufacture a BJT with a shallow doped collector, then the base and emitter regions are doped "on top of" the doping of the collector. Also the collector region is where the Vcb voltage drop occurs and power is dissipated. The heat resulting of this has to be taken away and having the collector directly sit on the metal case of the transistor helps with this.
H: How much will the output voltage of a transformer drop as I add loads? I have a transformer that says 6V 600mA 3.6VA, AC current (but the voltage measured is actually 8.65Vac). I will add a 4 diode rectifier WITHOUT capacitor (I say without in order to simplify the calculus). I want to connect LEDs (200mA per LED). So, I should be able to connect 3 LEDs. I want to know how to calculate the series resistor for each LED. But I don't know the REAL output voltage of the transformer. Will the voltage drop after I add each LED (this is what I think it will happen) or only after I reach the maximum power consumption? AI: With a transformer that small, the load regulation will be very poor, maybe 20% (that is, loaded voltage = 80% of unloaded voltage), or could even be worse. The voltage will drop steadily as you add more load. Your best bet is to measure it under load. You can do this safely by measuring the voltage off load, then compute the dropper resistors for that voltage. Attach the LEDs. The transformer voltage will drop under load, so having the wrong values will be safe. Take new measurements under load, rinse and repeat.
H: I designed a constant speed PWM motor driver using an op-amp and mosfet. Will it work? Can you point out some problems? I'm currently working driving two motors using some sort of PWM or switching such that they run at constant speed, regardless of load. I want to control this entire circuit using one Arduino PWM output pin.I believe that the mosfet I'm using is a logic-level one intended to be driven at 4.5V. The two motors combined shouldn't draw more than 30A stall. Here's what I have so far: simulate this circuit – Schematic created using CircuitLab The two diodes are for protection. The resistor divider is simply to account for the fact that the Arduino can only output 5V. My logic is as follows: If the voltage drop across the motor is too low, then the op-amp turns the mosfet on, allowing current to flow. If the voltage across the motors is too high, then the mosfet turns off. My concern is that, rather than oscillating, the mosfet will just end up in a stable state, thus dispersing a lot of heat. Will this happen? If so, how can I make the circuit oscillate to avoid this problem? Thanks! AI: Bad idea to use opamp to drive a gate. Much better to find a gate driver. Otherwise your system may overheat, suffer noises on input, etc. You will reduce the frequency to deduce heat, and the motor will stop working or will burn.. Also if you wand analov driver, you should better start with BJT instead MOSFET. Then indeed opamp may work (not sure about the specific opamp). Actually with MOSFET it works sometimes too, but for you will be easier to work with BJT, it's much more linear. For analog driver you will need much more advanced control loop. But frankly i a little skeptical about describing it here. Google analog servo drive. And last, but not least : for a brushed motor to keep the speed constant (or almost constant) just control it's voltage. Use some kind of variable voltage converter with current limit, and that's it.
H: Why is there a short between my VCC and GND in this circuit? I am doing a project where I am using a FPGA board which connects to a circuit I designed. It has a quad Comparator (LM339N), a few resistors and an LDR; two distance sensors (GP2Y0A51SK0F) are connected to it by wires, and the circuit gets it's voltage from the FPGA board (DE0-Nano, 4.80V). The comparators are also powered by the 4.80V from the board. This is the schematic: As you can see, the output of each distance sensor (this sensor only has 3 pins, I couldn't make the fourth hidden) goes to a comparator. The point was to turn the distance sensor into a proximity sensor and output '0' when it sees something at a certain distance. The FPGA then receives that '0' and acts. Somewhat the same with the LDR. I can confirm everything worked when I used two different voltages supply when I tested this - one being 5V that goes into the distance sensor and one 3.3V that goes to the comparators. Both 3.3V and 5V came from a DC power supply. Now, I'm using the DE0-Nano to supply power, the only source being one pin of 4.8V and one pin of GND. As you can see from the schematic, everything has one single power supply. My problem - things are not working with the distance sensors; Trying to measure voltage at the output of each of the sensors I just get 0V, so obviously the comparators output nothing as well. I'm using a multimeter to try and check what the problem is, and when I tried to check for shorts (with a mode called continuity, I think), one of the following may happen (or together): A short between VCC and GND - but there is no physical connection between them, I check that with resistance mode. A short between the VCC and GND pins of the LM339N 1 and 2 will never happen at the same time. A short between one of the outputs of the comparators (U1 or U2) and VCC. When this happens, The 3K pull-up resistors are also a short. Like I said, there is no physical connection between VCC and GND so I assume that the short the multimeter is seeing is due to pins having the same potential? Is it possible that the pull-up resistors are too small and that is why this problem happens? Is it possible that I fried the distance sensors? Thank you for your help, I have been testing this for days and trying to figure out what the problem is. Edit - here are a few screen shots from scope: This is the power pin: This is the power pin, zoomed: This is the sensor's output: The output stays like this no matter what, so it doesn't really seem like it is working. maybe the amplitude is increasing but I don't think it matters (it could also be the other sensor and I don't remember). Those screen shots are from one sensor only - the other shows different results, but when checking their output, none seems to actually respond. AI: You have fallen into a classic XY problem1 trap. However, thanks for briefly mentioning the original issue, as that makes the overall situation easier to understand. You have a genuine problem ("X"): things are not working with the distance sensors. That is not enough detail to help you with the actual problem, but I'll come back to that. You are having problems interpreting the results from an (unfortunately inappropriate) multimeter test, and that is what you were asking about ("Y"). However "Y" is the wrong issue on which to be spending your time, because getting an answer to that, won't help you with solving the original problem "X". As confirmation, below is the answer to your question "Y" (about the apparent shorts you measured) but it doesn't help resolve the real problem "X" (about the sensors, when powered by the DE0-Nano): I'm using a multimeter to try and check what the problem is, and when I tried to check for shorts (with a mode called continuity, I think), one of the following may happen [...] Using a typical multimeter in that way on a board with ICs, will usually lead to lots of false positive apparent shorts displayed on the meter. This is due to the ESD protection diodes (and other ESD protection structures) inside most ICs, which can conduct during that "continuity" testing. So stop doing down this rabbit hole and back-up a few steps :-) Here is what I suggest: Go back to your earlier configuration which you say worked i.e. external separate power supply for the 3.3V and 5V sections. I hope (and expect) this will work again. This would confirm your hardware has no "short circuits". I suspect that either additional output capacitance and/or a higher current capability of that separate DC power supply (compared to the DE0-Nano's power output), is why that configuration worked - see next point. Note that some Sharp distance sensors have unusual power requirements. Specifically they pulse their IR LEDs at relatively high currents, for brief periods. Due to those high currents, they are well-known for needing lots of decoupling capacitance close to the sensors, especially if the power supply is weak, or has lots of inductance (e.g. long / thin wires etc.). Your sensor's datasheet mentions average supply current (12mA to 22mA) but not a maximum current! Therefore use an oscilloscope at the sensor's power pin to view the waveform, and you may find it drops below the minimum acceptable power voltage for the sensor (e.g. in brief dips, when the internal IR LED is pulsed) due to the factors I mentioned above. I expect decoupling capacitors will be required as a minimum. That may be part or all of the reason why "things are not working with the distance sensors" when they are powered from the DE0-Nano. If you are still having problems with the sensors only when they are powered from the DE0-Nano, then focus on that and ask for more troubleshooting help. Show some photos of your setup, explain any constraints, use short power wires, expect to be asked to provide scope screen captures, and you will likely also need to provide info on the DE0-Nano and what power it can supply. See these pages for examples of info about those Sharp sensors and the need for additional power supply decoupling near the sensor; some suggest also adding a low-pass filter on the output signal itself, although I doubt that output signal ripple is your current problem: How to improve Sharp GP2Dxxx sensors Testing GP2D12 Sensor: Oscilloscope Traces, Capacitors, and Distance Sensor Results Sharp distance sensors and some other relevant Electrical Engineering Stack Exchange questions: Can't get consistent readings from Sharp IR range detector Where to put stabilising capacitor? Daisy chain (parallel) of IR sensors Sharp Infrared sensor-Filtering supply ripple Where to place a capacitor to smoothen IR sensor reading? How do I properly use Sharp GP2Y0A21YK0F sensor? Can't get consistent readings from Sharp IR range detector Sharp IR distance sensor outputting consistently high voltage 1 There are several slightly different definitions of the XY problem so I linked to a Google search above, for readers who want to learn more. The terminology I use is: "X [is] the underlying problem and Y the exposed question or request" which is adapted from here.
H: Using NPN transistors as on/off switches for a LED circuit I have a very basic LED circuit: simulate this circuit – Schematic created using CircuitLab All D# are 3Vx0.02A LED. In order for this circuit to work, we need a 25Ω resistor for each LED quartet: Total LEDs current = 0.02A*4 = 0.08A Voltage drop = 5V - 3V = 2V Resistor value = 2V/0.08A = 25Ω I didn't have a 25Ω resistor, so I put a 27Ω resistor instead. Now, I want to add an Arduino in order be able to power on and off each LED quartet programmatically. I don't want to overload the Arduino, so I thought it would be ok to have an external power supply which will give the LEDs all the current they need to work, without worrying about burning the Arduino itself. I thought I could either use transistors or MOSFETs as logical on/off switches, but I chose transistor because they are way less expensive than MOSFETs. I thought about getting a 2N3904, because it supports up to 200mA and the circuit requires 0.08A per LED quartet. So far I know the circuit should look something like this: simulate this circuit I'm definitely not an electronic expert, I assumed it should be like that because I looked a site which talked about Arduino controlled LEDs via transistor used as on/off switches. It didn't explain a thing, it just had the circuit so what I want to ask is: why the 10kΩ resistor? Is the value right? Do I need a 1/4W 10kΩ resistor or a 1/2W one? The 27Ω resistor should remain 27Ω or should be changed when a transistor is introduced? AI: Your basic concept is right, but there are some issues: Paralleling 4 LEDs is not a great idea. It would be better to give each LED its own resistor. You say the LEDs drop 3 V. Figure about 200 mV for the saturated transistor, which leaves 1.8 V across the resistor. (1.8 V)/(20 mA) = 90 Ω. The common value of 100 Ω would work fine. You have to consider the gain of the transistor in determining its base resistor. As you say, the transistor will pass 80 mA when on. Let's say we want to run the transistor at 20:1 collector/base current, since it can be counted on to have more gain than that. That means you want about 4 mA base current. If the digital output goes to 5 V when high, you have about 4.3 V across the resistor. That comes out to (4.3 V)/(4 mA) = 1.08 kΩ. We have some slop in the calculations so 1.1 kΩ should work fine. In any case, 10 kΩ is likely too high. Working backwards, that provides 430 µA base current, which means you are relying on the transistor to have a gain of at least 186. That is well above the guaranteed minimum, so 10 kΩ is clearly too much resistance.
H: Is it possible to store charge in a capacitor with lower voltage to another capacitor with higher voltage? As in the figure above, C1 is charged to V1 volts while C2 is charged up V2 volts which is higher than V1. Is there any circuit that can store all the charge in the C1 to C2? After the process, C1 will be equal to or close to GND and C2 will be equal to or close to V2 + V1 volts. Is it possible? AI: You need some sort of switching circuit. Examples are a capacitive charge pump, and a boost converter. Here is the basics of a boost converter: When the switch is turned on, the input voltage is applied to the inductor. That causes the current thru the inductor to ramp up linearly. You open the switch at some time before the current reaches the saturation level of the inductor. Now the output voltage minus the input voltage is applied to the inductor in reverse. This causes the current to ramp down linearly. However, during this time current is transferred from the input to the output. When the current gets to zero, the diode will block current trying to flow in reverse thru the inductor back to the input. If you keep turning the switch on and off, you will keep transferring energy from the input to the output.
H: Simple Transistor Circuit Question I know this question title is not the best, but it describes the question well. I am confused with one thing. Why does this circuit work: simulate this circuit – Schematic created using CircuitLab By work I mean turn the 12V relay on. And this doesn't. simulate this circuit I have tested it out on a breadboard. The first one is able to run the Relay and the second one can't. Why is that so? What difference does it make if the load is before or after the transistor. PS: I am using BD435 transistor. And a motor instead of relay. AI: The 2nd circuit doesn't work because to turn on the transistor you need a base emitter voltage of 0.7 volts and that means, at best the emitter is at about 4.3 volts and that would be the voltage on the relay. Think about the first circuit and why it works - it's easy to forward bias the base emitter junction because the emitter is at ground. With the emitter connected to the relay, the moment you apply the base voltage the emitter must rise with that base voltage but be 0.7 volts lower. This only partially turns on the transistor. Or if you expected the emitter to be 12 volts in the 2nd example then the base would need to be 12.7 volts. The bottom line is that the forward biased base emitter junction IS a diode and behaves like one.
H: Do capacitors conduct electricity while charging? I was reading a schematic for a timer circuit, and it made use of a capacitor and a transistor. The description below the schematic explained that the current through the base of the transistor was not enough to switch it on until the capacitor was charged. I am quite confused as to the behavior of capacitors during charging. I have read countless descriptions of the device, and I understand that it stores energy in the electromagnetic field surrounding its component plates. I still do not understand their behavior, though. I have come up with two hypothesis as to why the timer circuit is designed like this: The capacitor conducts electricity only while charging. While it is charging, the current is directed through the capacitor, ignoring the transistor and going back into the voltage source. After it has charged, the current no longer flows through the capacitor and flows through the transistor instead. The capacitor conducts electricity only while charged. While it is charging, the circuit is open and electricity flows through neither the capacitor nor the transistor, all of it ending up in the electromagnetic field of the capacitor until it is charged, when the capacitor is able to conduct electricity. Then, the current flows through both the capacitor and the transistor. Of these two hypothesis based on what I have read, I am leaning towards #1. I would like to know if a capacitor conducts electricity while charging, and if one or neither of these are right, and which one if so. Thank you in advance - I just learned how a transistor works, and now I am struggling with capacitors. :) Additional information: this question is asked with the perspective of someone who only works with direct current. Here is a link to the diagram (I had to dig through my browser history): http://www.instructables.com/community/Timer-Circuit/ AI: For DC. It is not correct to say that the capacitor conducts current. Current is the movement of charge. Charge moves through the resistor and is stored in the capacitor like a reservoir, as it fills up the pressure (Voltage) increases. When the capacitor voltage reaches the base turn on level the current flows into the transistor turning it on. In practice a current does move on the opposite side of the cap. This is to balance the charge that is building on the positive plate of the capacitor. It gives the appearance that the capacitor is "conducting" but the current does not flow through the capacitor because the dielectric between the plates is an insulator.
H: RC integrator circuit - can a high RC product really by approximately equal to Vin / R? This question is very much related to Why do R and C have to be small for differentiator circuit asked earlier. On page 26 of Art Of Electronics, 2nd Edition, the book makes the statement that if we manage to keep \$V\$ much less than \$V_{in}\$, by keeping the product \$RC\$ large, then. $$ C {dV \over dt} $$ is approximately equal to $$ V_{in} \over R $$ The original equation, prior to simplification, is $$ I = C {dV \over dt} = { { V_{in} - V } \over R } $$ Why is this so? Or, Why must we manage a relation of \$V\$ and \$V_{in}\$ by juggling \$R\$ and \$C\$. I know \$RC\$ is the time constant. I suspect that if you make it long enough, \$V\$ can always be made to be less than \$V_{in}\$ given some \$dV \over dt\$. If that is so, I see the simplification above. However if I expand the original equation I am not seeing it mathematically. $$ I = C {dV \over dt} = { { V_{in} \over R } - { V \over R } } $$ Yeah, if you make \$R\$ large enough, \$V \over R\$ will be insignificant, but won't \$V_{in} \over R\$ be getting close to insignificant as well? Is it always the case the \$V\$ will be less than \$V_{in}\$. If so, I see the simplification above. AI: A large time constant (a large value of RC), means that V will never grow large relative to Vin (It will eventually of course, but we're approximating). Another way of putting it, is that the current through the resistor is relatively constant because V isn't changing, and this is accomplished by making R or C (or both) large. Imagine a resistor charging an "infinite" capacitor, the voltage on the capacitor would be constant. Similarly, if the resistor is infinite, voltage doesn't change either. Now obviously, neither of these cases is useful, V needs to change at least a bit to be helpful, but hopefully that helps you see where the approximation comes from.
H: Xbee series 2 drops off network without apparent cause - diagnostic suggestions? I've built a meter reader, based around an ATMega 328P and an XBee series 2. The initial protoype built with an Adafruit trinket pro (3.3V) and it worked like a charm. It was stable for 10ish days, before it ran out of battery. To improve battery performance, I moved it to a breadboard-duino. This is where the problem starts. I start it with a newly charged battery back (4*1.5V hybrid NiMH) and it's fine for ~7ish days, as the battery pack voltage slowly drops from 5.7V down to ~5.1V. Some time around this point (and it varies by >0.1V) the XBee dissociates from the network. I know this because 1) Power cycling the xbee only restores connectivity 2) The XBee association LED activates, rather than blinking and 3) The 328P can be regularly seen trying to put the XBee to sleep. However, if I restart the xbee (or entire circuit, ether way) the issue repeats itself - but in a timeframe of a day. This implies to me it's related somehow to the battery decay, as that's the only thing that doesn't get reset. However: Power output is regulated to 3.3V by a TS2950CT 3.3V regulator Measured circuit voltage is 3.27V, nonfluctuating on my meter The trinket pro worked all the way down to a supply voltage of 3.6V There's a 100uF capacitor between the +3.3V and ground for smoothing After the blocking diode, supply voltage is still 4.69V Drop out voltage on the regulator is listed at 450mV, worst case, with a 100mA load. This is, at worst, half that. All the components are rated at 3V, except for... The only dodgy thing I've done is (accidentally) used a 16Mhz ATMega. I've got an 8Mhz crystal in the post, and I'm going to replace that ASAP. But it's Christmas, and if I'm lucky it'll be here by new year. So, what I'm asking people is - what else can I check? What could it be? Where could I start looking? Oh, and, relevant seasonal greetings. They'll be merry if I can get this thing to work! AI: The only thing I can think of is that fast current spikes are pulling your supply voltage down again in fast spikes. Try putting a 300nF ceramic in parallel with your 100uF smoothing capacitor. This is much better at dealing with fast spikes than the 100uF. Do you have decoupling capacitors on all your chip power supplies again this may help with this problem.
H: What is shunt field resistance in a DC Generator? Is the "shunt field resistance" the resistance of the coils in a shunt generator? Googling https://www.google.com/search?q=%22shunt+field+resistance%22&oq=%22shunt+field+resistance%22&sourceid=chrome&ie=UTF-8 didn't give any helpful results. AI: Yes, a great explanation in the NI link of your search. The total field resistance is made up of the field coil pair (in series) and one (occasionally 2) limit resistor. In old DC car generators there was typically 1 resistor, which set low current and when shorted out, high current output. Controlled by a voltage sensing relay and an overcurrent relay.
H: Phasor analysis of AC circuits: Which is correct: impedance triangle or phasors as resistors? Phasor analysis is convenient with AC circuits when looking for simple quantities like impedance. After converting capacitive and inductive quantities to reactances, I then treat them like resistors. It's a convenient way to make AC circuit analysis relatively simple. However, I found out today that I may have been doing it wrong, and I need help understanding why. For example, the circuit below: To find the impedance in this circuit, I would first find XL, then Xc: \begin{equation} {X_L}=2{\pi}fL=52.779{\quad}{\quad}{X_C}=-{\frac{1}{2{\pi}fC}}=-16.579 \end{equation} Then I would treat the reactances as resistances, in this case using the simple parallel resistor formula: \begin{equation} Z = \frac{1}{\frac{1}{R}+\frac{1}{X_L}+\frac{1}{X_C}} \end{equation} The result I get this way is Z = 24.75&ohm; In studying for the FE exam, I've come across a method which relies on the impedance triangle: Through some trig on the relations on this diagram, the formula for impedance in a parallel AC circuit with reactive components becomes: \begin{equation} Z = \frac{1}{ \sqrt{\left(\frac{1}{R}\right)^2+\left(\frac{1}{X_L}+\frac{1}{X_C}\right)^2} } \end{equation} The result I get this way is Z = 24.17&ohm; As you can see, they both produce a very similar result, but not the same. One is an approximation, and I have a suspicion based on the impedance triangle that the latter method is correct. Can someone please elaborate as to why that is or isn't the case? Thank you. AI: Oops. They are the same. I was using a calculator and was failing to insert the imaginary 'i' (or 'j', depending on your preference) in the reactances when using the first method. After doing so and extracting the magnitude of the complex result, the answers turn out identically.
H: Software Defined Radio hardware Most software defined radio peripherals involve a FPGA. In designing a SDR peripheral, I would like to know whether its possible to design a GNU Radio Compatible SDR hardware peripheral without any FPGAs. i.e. Directly sending the output from the ADCs to a PC via USB. If possible, I think My interest to eliminate FPGAs (at some other cost) is because FPGA soldering, design is a difficult process that is not properly explained by most vendors. If there is a way I could eliminate them from my design it will be great. My interest is to design something similar to this. In the above image I found from TI, is it possible that we directly connect the output of the ADC to the to the PC via some sort of a USB connection? Perhaps, transmission is possible this way as well. I would also like to know the exact purpose(s) of using FPGAs in SDR peripheral designs. Im sure there may be many. But I would like to know the most imperative ones. UPDATE: As suggested by Neil_UK in the answers section, if the limitation is not being able to transmit the raw ADC samples via a USB interface to the PC due to slowness in USB speed, What would be the highest practical bandwidth that we could have in a system that directly connects the ADC to a USB? perhaps a USB 3.0? is it only the bandwidth that will be limited by this decision? What other fundamental features would go missing if SDRs were desiged this way? (provided its not required to operate on signals e.g. demodulate them) AI: Note: we're going to oversimplify SDR hardware for illustration in this answer ADC --> PC is (one of) the functions of the FPGA The FPGA serves different functions in different SDR designs, but one of its primary purposes is to "just connect the ADC to the PC". I think you don't fully appreciate the process of moving data through the Universal Serial Bus (USB) and the process of capturing the radio signal. SDR's typically require at least two ADC streams (quadrature capture after down conversion) and usually have multiple quadrature channels to do advanced things like MIMO, RADAR/Beam-forming, etc... The FPGA is required to multiplex the digital data streams coming off the various ADC's, to format the data in a manner compatible with USB (and ultimately gnuRadio), and to receive control information from the PC/gnuRadio and effect the various changes to the ADC's and down-conversion components. Without the FPGA you would have to implement these functions with other hardware and your design would end up much more complex, rather than the simplicity you seek. SDR designs have evolved to their current state and they are much lower cost/simpler today than they have been in years past. Examples Some common SDR products from Ettus and Pervices Devices illustrate the central glue-logic nature of the FPGA in these architectures. Note the placement of the FPGA between the high-speed analog converters and the relevant external data interfaces.
H: What difference of marked relays on scheme? I have two relays that marked as red and orange. What is the difference between them? And what relays (SMD) I should use to create pcb? Thanks for answer. AI: According to this web site it is a "Step relay Mechanical interlock relay". As far as the contact for K2 goes, I would guess its a 'make before break'.
H: Low pass filter not working correctly? I have made low pass filter both on Breadboard and in Proteus.Both of them have different problems. I have kept voltage gain at "3". In breadboard,voice output in the earphone is lower than input.In input I have a microphone When I blow air by mouth, It give me rough voice hearing by earphone. In case of simulation in Proteus,When I applied above 333 Hz freq, output in speaker is still beep. I cant understanding what is this happening.. Please help me...I am beginner. AI: 1) You are using a single supply. Analyze the circuit for positive inputs. This requires negative outputs. Where will the - voltage come from? 2) You do not specify your op amp. Most op amps are completely incapable of driving 8 ohms, which the usual nominal speaker impedance. You need a power stage for the output - subject, of course, to fixing part 1). 3) Using 500 ohms for feedback and 150 ohms input is too low. For instance, you need to check to see if your AC source is happy driving 150 ohms. A good function generator will drive 50 ohms, but many sources aren't happy doing that. Use 10k and 30k. This will also let you use a smaller capacitor.
H: How to measure charging current for TP4056 module? TP4056 module is connected to usb port on my PC resistor is default 1.2k... When I put multimetar on output of module I get about 120-130 mA... I am not sure why is that, should I get 1A? From datasheet it say for resitor 1.2k chargin current is 1A. Maybe I doing something wrong, anyone know what is problem? AI: When I put multimetar on output of module I get about 120-130 mA That is the expected "trickle charge" current. should I get 1A? Not necessarily. You only expect to see 1A (assuming RPROG=1.2kΩ) under specific circumstances e.g. battery voltage >2.9V (VTRIKL), and that is after a short period of rising charge current, up to a maximum of 1A - see the TP4056 datasheet. If a battery is not attached or <2.9V then you would only measure a maximum of the trickle charge current, as you reported.
H: Can a transistor have negative Q point? I'm already having trouble understanding Q-point of a transistor. I have been assigned to draw Q point as follows: Calculate and draw a graph of Q point , VCE(cutoff) , Ic(sat) if IB = 100uA, 200uA, 300uA, 400uA, 500uA, 600uA and β = 150 Note: The values in the diagram will be replaced if mentioned in the question. The problem is that when base current is 400(micro Amp) it gives -3.2 V VCE and negative values for other proceeding base current values. How should I plot it in the graph? Before plotting the graph I want to know if it is even possible to have negative VCE. If yes then how will current flow towards ground as it's at high potential now? Please let me know solution asap. Thank You. AI: As a rule of thumb, you should never have a negative \$V_{CE}\$ in your DC biasing. If you do, the emitter acts as a collector and the collector acts as an emitter. (They're not designed for that, so it won't work very well.) In your circuit, there's no way to get a negative \$V_{CE}\$. The emitter is connected to the lowest voltage in the circuit (ground). What actually happens when you increase \$I_B\$ is that the collector voltage drops until it reaches \$V_{CEsat}\$, which is about 0.2 volts. Then it doesn't drop anymore, regardless of how much current you pump into the base. In other words, the collector current is limited by the resistor. You have 10 volts and 220 ohms, so you'll never have a collector current higher than about 45 milliamps.
H: Purpose of the diodes in DRV8301 motor driver circuit This schematic is used to drive a motor with DRV8301. But the datasheet of DRV8301 doesn't show diodes. What is their purpose? VDD is 14.8V (16Vmax). MOSFETS are BSC016N04LS Schematic with diodes: The datasheet of DRV8301: AI: The exact purpose of those Zener diodes cannot be inferred from that schematic alone, without further informations such as MOSFETs part numbers. As others have already said, it is most probably a protection device acting as a clamp. One problem that those diodes could be intended to prevent, and which has not been mentioned yet, is \$V_{gs}\$ spikes due to quick transitions of \$V_{ds}\$. This is often indicated in literature as high dv/dt rates. There are fairly big intrinsic capacitances due to the MOSFET structure between source, drain and gate. In particular, \$C_{dg}\$ and \$C_{gs}\$ are relevant for the problem at hand. If the device is OFF and \$V_{ds}\$ experiences a rapidly varying transient, some of this transient can be coupled to the gate, giving rise to \$V_{gs}\$ spikes that can destroy the device or make it turn ON unexpectedly for a short time. For further details see this application note from International Rectifier (AN-936): The Do’s and Don’ts of Using MOS-Gated Transistors. Some excerpts from section 3 (emphasis mine): Excessive voltage will punch through the gate-source oxide layer and result in permanent damage. This seems obvious enough, but it is not so obvious that transient gate-to-source overvoltages can be generated that are quite unrelated to, and well in excess of, the amplitude of the applied drive signal. The problem is illustrated by reference to Figure 2. (...) If we assume that the impedance, Z, of the drive source is high, then any positive-going change of voltage applied across the drain and source terminals (caused, for example, by the switching of another device in the circuit) will be reflected as a positive- going voltage transient across the source and the drain terminals, in the approximate ratio of: $$ \frac{1}{1+ \frac{C_{gs}}{C_{dg}}} $$ The above ratio is typically about 1 to 6. This means that a change of drain-to-source voltage of 300V, for example, could produce a voltage transient approaching 50V between the gate and source terminals. In practice this “aiming” voltage will not appear on the gate if the dv/dt is positive because the MOS-gated device goes in conduction at approximately Vgs = 4V, thereby clamping the dv/dt at the expense of a current transient and increased power dissipation. However, a negative-going dv/dt will not be clamped. This calculation is based upon the worst case assumption that the transient impedance of the drive circuit is high by comparison with the gate-to-source capacitance of the device. This situation can, in fact, be quite easily approximated if the gate drive circuit contains inductance—for example the leakage inductance of an isolating drive transformer. This inductance exhibits a high impedance for short transients, and effectively decouples the gate from its drive circuit for the duration of the transient. The negative-going gate-to-source voltage transient produced under the above circumstances may exceed the gate voltage rating of the device, causing permanent damage. It is, of course, true that since the applied drain transient results in a voltage at the gate which tends to turn the device ON, the overall effect is to an extent self-limiting so far as the gate voltage transient is concerned. Whether this self-limiting action will prevent the voltage transient at the gate from exceeding the gate-source voltage rating of the device depends upon the impedance of the external circuit. Spurious turn-on is of itself undesirable, of course, though in practical terms one may grudgingly be able to accept this circuit operating imperfection, provided the safe operating area of the device is not violated. It should be remembered that a collapse of voltage across the device (i.e., a negative-going dv/dt) will produce a transient negative voltage spike across the gate-source terminals. In this case, of course, there will be no tendency for the device to turn ON, and hence no tendency for the effect to be self-limiting. A zener diode connected to clamp positive transients will automatically clamp negative-going transients, limiting them to the forward conduction voltage drop of the zener. Since you ruled out some other reasons given in other answers (ESD protection, protection from spikes coming from the gate driver), I think that the most likely reason is to prevent high dv/dt-induced spikes to damage the device, especially because the load (motor) is highly inductive and could well generate transients with steep edges.
H: Verification of this CMOS realisation I have to make the CMOS-equivalent of this function: \$A'*B'+C'\$. I made this CMOS-circuit, but I'm not sure if it's correct: simulate this circuit – Schematic created using CircuitLab The course I'm following isn't very clear about this subject. AI: If the ′ means that the signal is inverted then it is fine.
H: Can't get simple electromagnet to work I'm trying to make a very simple electromagnet. I found a inductor, took it's copper wire and wrapped it around a nail. When I connect my bench power supply and crank it up to 20v, but I can't even pick up tiny metal pieces, it just seems like there's no magnetism. I also noticed that there is no current flowing and I can't figure out why. Is my connection or choice for materials wrong? I also took super thin copper wire from a DC motor and wrapped it around many times a smaller nail, but same effect - no magnetism. I would expect it ether become magnetic, or at least short circuit, but I get nether.. Any suggestions on what am I doing wrong? Thanks! AI: The wire is insulated with a thin layer of enamel. You have to remove it before you can get current to flow through your coil. Use fine sandpaper to clean the ends of the wire, or use a lighter flame to burn the enamel off.
H: What determines a laser's strength? What specification of a laser will cause a laser to be more "powerful" over another more specifically in cutting capabilities? If multiple, is there one that increases the strength more than the other specs? For example does wattage affect it more so than the wavelength? Sorry if I am completely off on this one, pretty new to electronics and still have some misconceptions. AI: The best laser for cutting depends on three factors: Wavelength. This will impact the absorption into the target, The vast majority of high powered cutting lasers are between 1 um (He-Ne @1.5 um) and 10 um (Co2 @10.6 um) Power. The gas Lasers are easy to excite and produce large powers (up to kW range), though Fiber Lasers (very small initial beam diameter) are catching on quickly in the 100 W range. Lensing. To cut you want very small Laser diameter with minimal kerf and it's difficult to make lenses that will focus very large powers due to any attenuation causing lens heat damage. Great info on Co2 lenses here. There is a great graphic here that shows the wavelengths and achievable powers.
H: slew rate of a mosfet Is there way to calculate slew rate of following circuit? simulate this circuit – Schematic created using CircuitLab Datasheet for Transistor Datasheet for Diode I chose the PMOS and Diode arbitrarily from what I could choose from Circuit Lab. If VTRIG goes from 5V to 0V, how can I calculate the slew rate (V/us)? AI: A rough approximate is to use the RdsOn for -5V. It is guaranteed to be 0.3Ω but that is for -10V @25'C with 7.2A Pulse, width = 300 μs; duty cycle = 2 %. your situation is different and vague; Initial conditions : Vcap (unknown ) Cap part number (unknown ) and trace inductance (unknown ) cap ESR or Dissipation Factor (unknown ) Diode ESR 1N4148 0.1W ~ 10ohm ( assumed to be larger than 1nF Cap ESR which is expected to have ESRC value=T < 0.1us for ceramic The RdsOn starts at a high value due to Vdes =5V and Vgs=-5V and not being a "logic level" gate controlled FET requires a certain amount of calculations from the datasheet below With the slew rate being dV/dt=Ic/C for the cap and Ic = (V+-Vcap)/(RdsOn+ESR(diode)) it becomes highly nonlinear. But with initial conditions of ESR diode=10, and ESR or RdsOn of FET=10Ω then dropping to 0.34 Ω as Vds drops below 3V. Ic=C dV/dt=1e-9 (5V)/(10Ω+10Ω) and C= 1e-9F for Vds=3~5V slew rate dV/dt= Ic/C= 5V/20Ω*1e9= = 0.25V/ns Low confidence in results due to inadequate info.
H: Creating AND gate with transistors I'm trying to create an AND gate with transistors. But I've stumbled on a problem. I've also connected a LED to out. When I turn A on the LED is off. When I turn B on the LED is on??????. When I turn A and B on the LED is on Why is it when B is on and A is off the LED turns a little bit on? And is it possible to turn the output fully off when B is on and A is off. AI: Because you are using BJTs, the LED is likely turning on due to B's base current. In detail: current flows through B's 10k resistor, through the base-emitter junction, and through the LED. To "fix" this, the easiest way would be to use MOSFETs (how logic gates are typically made); current does not flow into a MOSFET gate (in steadystate).
H: Best way to get 5.5KVrms/7777VAC? What's the best way to get a 5.5KVrms (7,777V peak) full-wave AC signal at 60Hz strong enough to drive two 8-stage CW multipliers at 150pF per capacitor? Image for reference: Not a duplicate of "how to make a 120VAC power supply." AI: A neon sign transformer driven from a lower than rated voltage. Unfortunately, two microwave oven transformers (MOTs) back to back in anti-series driven at the rated voltage won't quite give you the voltage you need. MOTs are so agressively designed that there is no scope for increasing the input voltage a little. However, if you increase the input frequency, then the input voltage can be increased pro-rata, a 50% uplift in both could get you where you want to be.
H: Difference between difference between bc547 and 2n2222 What is the difference between bc547 and 2n2222A transistor. I have use bc547 for a project (Water level indicator). Can i use 2n2222A instead of bc547? How each transistors are selected for each project? AI: There is not much difference when it comes to electrical properties of the two transistors. You can have a look at the datasheets : P2N2222A BC546/547/548/549/550 The two major differences that I find are: the power dissipation (a measure of power that transistor can provide): 2N2222A - 625mW BC547 - 500mW collector current (output current in common emitter config): 2N2222A - 600mA BC547 - 100mA So your application defines your selection in such cases. If you need more current amplification for same biasing voltages you might want to go for 2A2222N. Read the datasheets carefully and make your choice.
H: Simple resistor circuit I'm stuck with this simple problem (please be aware that this is my first non trivial circuit!): simulate this circuit – Schematic created using CircuitLab Due to Kirchoff law (I chose clockwise as positive) I've written: $$-i_1+i_0+i_2=0$$ $$i_4-i_5-i_0=0 $$ $$i_1-i_3-i_4=0$$ $$i_3-i_2+i_5=0$$ So I go on with $$i_1-i_2=i_4-i_5$$ $$i_1=i_3+i_4$$ Then, for tensions: $$V_1+V_3+V_2=0$$ $$V_4+V_5+V_3=0$$ $$V_0+V_1+V_4=0$$ I obtain the system of equations: $$i_3(R_1+R_2+R_3)+R_1 i_4+ R_2 i_5=0$$ $$i_3 R_3+ i_4R_4 + i_5 R_5=0$$ $$V_0+R_1 (i_3+i_4)+ R_4 i_4 =0$$ I've solved the problem, but trying some simulators I realize that the result is correct if all the resistors are of the same value, and completely wrong otherwise. What's the problem? Thanks a lot for your precious help! AI: I got the error. If I suppose current flows down through R3, then the second equation for tensions should follow the sign convention. It should be $$V_4+V_5-V_3=0$$, instead of $$V_4+V_5+V_3=0$$
H: Cycloconverter , MOC3021 problems? i am doing a cycloconverter single phase design and i have a problem , i use microcontroller to control the firing angle ,i have used zero crossing voltage (741) to trigger INTERRUPT port of MCU , i have controled two positive waves well but two negative waves dont work in reality . my question is " i want to use directly MOC3021 to control the two negative waves with load resistor 100ohm ?" And i will use MOC3021 inside triac to control negative waves instead of 2 SCR negative direction but it doesnt work.Can the MOC3021 drive directly load 100ohm Vin=6VAC [![enter image description here][2]][2] VIN=6VAC Rload=100ohn Thanks guys! AI: You need to exceed certain AC voltages on the switching side for this to work properly. Once triggered the triac will have an on-state voltage. For the MOC3021 this is typically 1.4 volts but can be as high as 3 volts. However, the AC waveform needs to exceed this by some margin to get the device to trigger. I think, if it's working on positive cycles, you are close to making it work on negative cycles too. Maybe try lowering the load resistor a bit and see if this begins to happen. BTW what is your diode forward current? Your question doesn't specify this.
H: Long range Arduino to Arduino communication // electronics beginner here I have been using VirtualWire with 433mhz modules to have an arduino successfully send data to an other one. Now, for telemetry purposes I need this one way communication to be operated 20km away. I do not want sophisticated stuff on that , the ideal way for me would be to amplify the signal and keep using 433mhz. I do not need a high bit rate : 2Kb/s are way enough . I can even go below if needed. Any ideas of an electrical setup that would send binary data that far ? Is it possible to keep using the light and great VirtualWire library or am I doing wrong ? AI: Your current set up probably works ok at 100m between units and you want it to work at 200 times this distance (20 km). In a perfect universe (in free space) you will require 40 thousand times more transmit power so, if you are using 1 mW transmit power you will be required to increase this to at least 40 watts. Here's the Friis transmission equation and this relates to free-space: - What it basically says is that for the same antennas and transmitting at the same frequency, the power received relates to the reciprocal of distance squared. It works with ratios too. If 1 mW gets you 100 metres, to achieve the same received power at twice this distance you need to transmit 4 mW. At ten times the distance you need to transmit 100 mW. At 100 times the distance the power required is 10 watts etc etc. so 40 watts is by no means a silly number. Given that the path between transmitter and receiver will likely be littered with a plethora of other 433 MHz users, to obtain the same performance you are likely to need to transmit over 100 watts. So, you use an illegal amount of brute force or a legal amount of sophistication with a little bit of brute force. Of course, some sophistication can be achieved by restricting the data rate sent. This is because the receiver could then be designed to operate with a much smaller bandwidth and not have to fight against the noise. Here's a generally accepted formula for the amount of power needed by a receiver at ambient temperatures: - Power required in dBm is -154dBm + 10log\$_{10}\$(data rate) dBm So at 1 bit per second, your receiver could be designed to have a really tight bandwidth (thus excluding noise) and it would need a power of -154 dBm. At 1000 bits per second that receive power rises to -124 dBm and at 10,000 bits per second it will be -112 dBm. Given that your current receiver might be capable of receiving 10 kbps, redesigning it with a much restricted band width (to say 100 bps) will mean the receive power could be 20 times lower. Thus the 100 watts previously mentioned, with some sophistication (i.e. redesigning the receiver), might now mean a transmit power of 20 dB lower at 5 watts. Goin back to the Friis transmission equation, if instead of 433 MHz you transmitted using 96.8 MHz (4.5 times the wavelength) the received power would be 20 times higher thus you could reduce transmit power by a further factor of twenty. Using antennas with gain (directionality) is another way of focussing the power onto a far-distant receiver although at 433 MHz and the curvature of the earth (at 20 km) this is probably not a great option unless you use repeaters. So, sophistication over brute force usually works and is what I advise.
H: Behavior of Art of Electronics highpass filter depending on resistance value While reading The Art of Electronics (3rd Ed.), I was stumped by a particular assertion in section 1.7.1C, "Blocking capacitor": For instance, every stereo audio amplifier has all its inputs capacitively coupled, because it doesn't know what dc level its input signals might be riding on. In such a coupling application you always pick R and C so that all frequencies of interest ... are passed without loss ... That determines the product RC ... You've got the product, but you still have to choose individual values for R and C. You do this by noticing that the input signal sees a load equal to R at signal frequencies ... so you choose R to be a reasonable load, i.e., not so small that it's hard to drive, and not so large that the circuit is prone to signal pickup from other circuits in the box. What do the authors mean with a small resistance being "hard to drive"? Also, why does a high-value resistance affect signal pickup? I can only guess, but my hypothesis is that signal pickup may pop up because the cutoff frequency is lowered when R is increased, and lower frequencies pass through. A low resistance may become hard to drive because of wasted power? The schematic: simulate this circuit – Schematic created using CircuitLab AI: You should read that sentence in the context of what was explained in the introductory chapters about signal sources and voltage dividers. As can be shown the signal source will see the amplifier input as a simple resistance (R) in the passband. Once you set the bandwidth (i.e. the RC product) you are left with one degree of freedom, i.e. choosing the R value. The R value is chosen as a trade-off between a very large (ideally infinite) resistance (good to avoid attenuation, but bad for noise pickup) and a very low (ideally zero) resistance (no noise pickup, but no signal transfer either!). To be more explicit, in the passband the circuit is like this: simulate this circuit – Schematic created using CircuitLab $$ V_{in} = V_s \cdot \frac{R}{R_s + R} $$ So, if R is much bigger than Rs, then Vin will be almost equal to Vs (no attenuation). But making R bigger will increase the input impedance of the amplifier, thus increasing the chance for noise pickup (noise sources can couple capacitively to the input terminal, and they appear as high impedance sources, so we want their (noise) signal to be attenuated, hence low R).
H: Why isn't the Thevenin equivalent impedance calculated as a function of frequency? When I look for calculations of Thevenin equivalent circuits, the equivalent impedance, \$Z_{th}\$, is always calculated with the frequency at which the source operates in all the examples I can find. Now, I understand that we use the equivalent circuit to be able to plug in any circuit we want between the points of the circuit for which we calculated the equivalent, but those calculations restrict \$Z_{th}\$ to a certain frequency. Why can't I work out \$Z_{th}\$ as a function of frequency? AI: The impedance of capacitors and inductors is frequency dependent. You can express Zth as a function of frequency but to calculate it for a particular example you need to know the frequency.
H: What's a good book title for Op-Amp design? I already have OpAmps for everyone but like many electronics texts, it describes what they are,how they work, what they do, and how to use them as components in circuits(monolithic op-amp ICs) but doesn't go in depth about the internals of an individual op-amp and how to actually design an op-amp oneself! That's what I'm looking for: A book that describes in detail the op-amp microelectronic circuit design down to the transistor level! In particular because I'm seeking to design one and first test it with discrete transistors. AI: There are several books that deal with the design of op amps. A notable classic is P. R. Gray, P. J. Hurst, S. H. Lewis, R. G. Meyer, Analysis and Design of Analog Integrated Circuits. 5th ed., Wiley, 2009 (Amazon link). Another one I like, quite recent, is S. Franco, Analog Circuit Design: Discrete & Integrated, McGraw-Hill Education, 2014 (Amazon link).
H: Can 1.5v Lithium AA batteries be recharged? I hope this is the best place for this question. I have some Energizer lithium AA batteries (1.5v), that work very nicely in electronics that have heavy constant usage, such as my Apple Magic Trackpad. Some time back though, I found myself with a solar power unit from an outside patio umbrella (it had built-in lights), and noticed that the batteries inside were simply 3 generic AA cell nimh batteries. Just for kicks, I decided to see what would happen if I tried recharging some of my AA Lithium batteries. They charge very, very slowly of course, but it definitely works. The batteries get up to a voltage (as read by a voltmeter) of 1.6v, but I've noticed that they don't hold their charge anywhere near as long as they originally did. It's a rough guess but I'd say I probably get about 20% of the original lifespan of the battery. Still feels kinda cool to be using the sun's energy rather than throwing them batteries out each time. So I guess my question is in a few parts... Are there any downsides to recharging these batteries in this way? Why is the lifespan so drastically reduced? Is there a way to increase the lifespan after recharging? Is there anything else I should know with regard to recharging batteries like this? AI: Lithium Primary batteries are not meant to be recharge. Can you physically recharge them? As you noticed, yes, but it's not a good idea. They can explode from this. To be honest, if they could be recharged, Energizer would market them like that. If they could be recharged and life expanded, they would have found a way, considering how many billions they have in funds. If you want to recharge lithium batteries, get standard lithium secondary cells. In fact, you "measuring it" at 1.6V means its DEAD: A “good” battery will generally have an Open Circuit Voltage (OCV) >1.74 volts. Any battery with an OCV <1.70 (after it has been allowed to recover) is completely discharged. Although an alkaline battery may read “good” at 1.6 volts, this reading on a LiFeS2 battery indicates the product has been discharged. The reason the lifespan has shorten is because of the chemical reaction converting the chemicals that make up the cell as the battery drains. Simply charging it does not undue this basic chemical change. 3. NO. 4. DON'T.
H: Do Twisted Nematic LCD need refresh (polarity reversal)? The Wikipedia entry for the history of the LCD during the 1970s and 80s, states that twisted nematic (TN) field effect LCDs were available in the 1970s. Later, in 1983, super-twisted nematic (STN) structure for passive matrix addressed LCDs was invented. From Passive and Active-matrix, it is stated that STN requires constant refreshing: STN LCDs have to be continuously refreshed by alternating pulsed voltages of one polarity during one frame and pulses of opposite polarity during the next frame. whereas, apparently, the TN appears not to require refreshing, see the description of Twisted Nematics. Without wishing to quote the entire description, there is no mention of refreshing, or field reversal: A voltage of about 1 V is required to make the crystal align itself with the field, and no current passes through the crystal itself. Thus the electrical power required for that action is very low. So, assuming that the Wikipedia entry has not omitted something, with TN LCD was the refreshing mechanism required or not? As an aside, if it was not, why then was it required for STN? AI: All LCDs have to be driven with AC. When you use DC instead, you get a ghost image which fades away rapidly. The difference between TN and STN is the amount of twist each LC molecule does, nothing more. You can have an active matrix with TN and STN, but TN has nearly died out, what's advertized as "TN panel" is STN in fact. For low-temperature applications, TN is better than STN and may be still available.
H: Would a potentiometer/resistor still draw the same current This is a very basic question but here it goes: For example we have a potentiometer going to an LED. the current of the output of the poteniometer is lowered, but wouldn't the current of the wire connecting to the inputs of the potentiometer still have the same current? Like this poorly drawn circuit Would the red/black wire still have the same current as without/without the resistor or potentiometer? Or was I hiding under a rock this whole time and missed out on some basic law/physics. AI: Yes. Current is the same at all nodes in a series circuit. The Wires in that circuit are just very very low resistance resistors. There is n Amps going through the red wire and black wire, the same that is going through the 1kΩ Potentiometer. What they divide is the voltage. The Voltage across the whole circuit is the sum of the voltage across each node. So some tiny fraction of a volt across the red wire, some tiny fraction of a volt across the black wire, and the majority across the potentiometer.
H: Can I Hook a Raspberry Pi Zero to a Laptop Display Via HDMI to TTL and TTL to LVDS? Dilemma: I have a project in which I am trying to use a Raspberry Pi Zero to display static content through a laptop screen. I am trying to make my unit as thin as possible and use as little power as possible. Previous searches have found solutions that would work, but are less than ideal for what I am trying to do. I would also like to mention that I am no electrical engineer and I struggle at reading datasheets and making sure I won't blow anything up, which is why I need your help. An example of a previous solution: Some solutions mention using an LCD controller from eBay that had input and output in the form of HDMI, VGA, DVI and LVDS. However, these boards require something like 9V minimum, where I am trying to limit my build to a small 5V DC power supply. They are also massive in size and more importantly, height, which is also something I am trying to avoid. Goal: My goal is to be able to power the Pi from a 5V 2.4A DC wall adapter, and power the screen through the micro USB port (micro USB > USB A Hub > micro USB cable) so that only one small external power cable is required. Eventually I would like to see if battery power is an option, but as far as this question goes, I am concerned only about the 5V power supply. My idea is to use an HDMI cable out from the Pi that goes to a 40-pin TTL breakout board that is typically used to power small TTL TFT displays and takes a micro USB in for providing power. From that board, I would take the TTL and run it through an LVDS converter board to try and output something that can communicate to the screen that I have available from an old laptop. The screen datasheet can be found at: http://www.datasheetspdf.com/datasheet/B156XW02-V0.html I ran out of links, sorry I'm new... And the laptop screen model is: B156XW02 V.0 This is an LED backlit 15.6 inch screen that uses a single 40-pin LVDS connector to the motherboard. Some Info/My Attempt: I figure that from the datasheet about the screen, the LCD has a max Voltage of 3.6, and a max current of 400 mA (0.4 A) and the backlight has a max Voltage of 3.4 and typical forward current (no max listed) of 20 mA (Seems really low?). It also lists the LED power consumption at 4.4 Watts max, which leads me to believe if the Volts are at 3.4 max, the current would be 1.29 amps max? It seems to me that the max Volts required is 3.6 and max current required is .4 A or 1.29 A (depending on what is correct)? Since the Pi can output 5V and .5 A from the micro USB out (after the hub), or 5V and 2A from a special PiHat, I would say this is feasible. My question to you fine folks is: Will my idea work? If not, where does this breakdown, and can I add anything to make this work? AI: For a start - the only useful figure you can get from the screen datasheet so far is its power consumption, which is a nice figure you can use to know what to expect in terms of battery life etc. Other than that, nothing of interest there unless you plan to interface with I2C of the panel (which you likely won't). I vote for the "get a HDMI-to-LVDS board" way. Those boards do the task you want without TTL in between (possibly, having TTL-to-LVDS on the board, but any possible problems with connecting two modules together are already sorted out). I encourage you to do just that - they're cheap enough. What's with the 9V then? Those boards mainly use 5V and lower inside, having a regulator of some sorts to produce the 5V/lower for the display logic. You can remove that regulator, power the 5V parts from the Pi and, if anything on the board actually uses 9V (say, backlight driver), you can then use a boost converter, which won't consume much energy since the bulk of current should be taken by the 5V line you'll connect directly to the Pi without conversion losses. As for the height concerns - you can desolder unnecessary connectors, remove some capacitors and replace some with the ones laying flat. Inductors are kinda harder to minify but they're likely to belong to the power regulator parts and if you're going the "remove regulator" route, you can remove them. The connectors can be replaced with angled ones (which lowers height) or bypassed altogether by soldering to the board (think power-menu-button board which typically comes with those HDMI-to-LVDS boards and that you won't need to remove). For the size concern - you can find a board without the audio amplifier (which is often populated on those boards), or maybe HDMI-only. In some circumstances, boards can even be cut (provided you can see there are no traces that are important for you - or you can jumper those with small wires). Oh, and the link you provided to "TTL-to-LVDS converter" is not really a converter - it's a PCB with two connectors soldered together, there are no active components to make the conversion. The actual converter is somewhere lower in "Suggested items", I guess.
H: What relay do i need to control 220v 16amp motor (water pump)? I am new to Arduino and Digital electronics, I have purchased an Arduino UNO, and want to control my water pump from it. The water pump works on 220V and 15A. can you guide me on this spec? what relays do i need etc? AI: You must be careful when choosing the relay. Firstly, it must be appropriate for the your circuit power supply. For Ex. if your power supply is 5V DC, then you have to choose one which its supply voltage is 5V DC. Secondly,its contact current must be appropriate for the your load current (water pump). My sugggest is that you should choose min. 20A and over one in your circuit. And I have attached an circuit about your question, you can connect a pump instead of the ac bulb. simulate this circuit – Schematic created using CircuitLab
H: Complex Numbers, Phasor Notation, and Eulers Identity On Page 30 in the Art Of Electronics (2nd Edition). The two images below describe the conservation of Sinusoidal signals to a complex representation and back. I am perfectly ok with the first image. The next image has me stumped. In particular, I am stumped at the description of \$V(t)\$ and \$I(t)\$ For \$V(t)\$, I understand why they say, \$V(t) = \mathcal{Re}(\mathbf{V}e^{jwt})\$. I do not understand the next line of the simplification, \$V(t) = \mathcal{Re}(\mathbf{V})\operatorname{cos}(wt) - \mathcal{Im}(\mathbf{V})\operatorname{sin}(wt)\$. \$e^{jwt}\$ should simplify to \$\operatorname{cos}(wt) + j\operatorname{sin}(wt)\$. I don't understand how \$\mathbf{V}\$ is distributed to the second equation. I guess \$I(t)\$ would follow similarly. Can someone help with the factoring? AI: I'll just zero in on the last bit. The authors are using bold, capitalized V and I to indicate that they are complex values. Not merely real values. That's all it is. So, assume \$\mathbf{V}=5 j\$. Then: $$ \begin{align} V\left(t\right)&=\operatorname{Re}\left(\mathbf{V}\cdot e^{j \omega t}\right) \\ &= \operatorname{Re}\left(\mathbf{V}\cdot \left[\operatorname{cos}\left(\omega t\right)+i\operatorname{sin}\left(\omega t\right)\right]\right) \\ &= \operatorname{Re}\left(5 i\cdot \left[\operatorname{cos}\left(\omega t\right)+i\operatorname{sin}\left(\omega t\right)\right]\right) \\ &= \operatorname{Re}\left( 5 i\cdot\operatorname{cos}\left(\omega t\right)+5 i\cdot i\operatorname{sin}\left(\omega t\right)\right) \\ &=\operatorname{Re}\left( 5 i\cdot\operatorname{cos}\left(\omega t\right)-5 \operatorname{sin}\left(\omega t\right)\right) \\ &= -5 \operatorname{sin}\left(\omega t\right) \end{align} $$ But perhaps \$\mathbf{V}=3-4 j\$? This just provides phase information where the polar magnitude is \$5\:\textrm{V}\$. But the result is arrived at in the same fashion as above. It's just that you now have sine and cosine terms mixed into the result.
H: How does Gunn diode generate oscilations? The Gunn diode can produce spontaneous oscillation if biased to the -ve resistance region of operation. How ? What exactly makes it tick ? AI: It is possible because Gunn diodes are usually made of compound semiconductors (usually N type) therefore there are two energetic valleys of different value. See the picture below. What it means is that if you add strong electric field, electrons will climb up the valley and have higher effective mass so they won't be as fast as the ones in the lower energetic valley. Electrons will start to pile up there at the higher valley forming a domain of electric field which will travel across the structure bringing diode into negative resistance region. As this domain reaches the end of a structure, it dissolves and new domains can be formed.
H: Ethernet LEDs won't turn off when i unplug the Ethernet cable I'm working on a custom board equipped with a Ka-Ro TX28 module and the LAN9500A USB to Ethernet controller without EEPROM for the network interface. Operating system is Linux (Yocto). The LAN9500A is connected to a RJ45 jack with integrated Magnetics and LEDs. I can see the LAN9500A is detected and correct driver is loaded..., Everything works fine when a cable is plugged... The only problem is that when I unplug the Ethernet cable from the RJ45 socket, Linux does not detect it, and the status LEDs stay ON or may be blinking... However, if I again plug the network cable, link goes down then up again and it works okay... Does anyone have any ideas? AI: Not enough information. What you describe is not normal, and can be a cause of various issues: issues with software configuring chip working in wrong mode (e.g. external PHY mode); issues with hardware like you using wrong transfomer (e.g. non-symmetrical one), slightly wrong values for RX/TX pull-ups. Please refer to reference design to ensure you have correct circuit, and parts you use are correct for the chip. Reviewing application notes would also may help. If you still have issues, detail your question with circuit diagram and software settings, or apply directly to the Microchip support.
H: DAC controlled buck regulator I am trying to learn about buck converters. So far I think I have a grasp of the main idea of the basic topology: simulate this circuit – Schematic created using CircuitLab From what I read and based from examples, the SW is driven with PWM signal from an MCU, so the output is based on the signal duty cycle. Now I am trying to reverse-engineer a power module using a buck converter. I found out that the MCU does not provide any PWM signal, but using DAC features. Is it possible to create an adjustable buck regulator using DAC signal as control instead of PMW? If so, how does it work? If there are missing point from this question, please ask and I will try to improve this question. AI: Is it possible to create an adjustable buck regulator using DAC signal as control instead of PMW? PWM is the fundamental mechanism inside a buck regulator that allows it to have such high power transfer efficiencies. If you wish to replace that mechanism with some form of linear voltage mechanism then, the regulator is no longer a buck regulator but more likely a linear regulator with lower power transfer efficiencies. However, if you wish to use a DAC output to feed a circuit that converts the DAC output to PWM that is OK. Such a circuit exists i.e. an LTC6992: - I am trying to learn about a buck converter, and so far I think I have grasp the main idea of the basic topology The basic circuit that converts power you have probably grasped but you haven't acquired all the detail to make a decent buck regulator. So, whether you used an MCU IO pin or a DAC/LTC6992 to control the switch, you still need a feedback loop to stabilize the buck regulator as shown below: - The trouble with simplified images you find on-line, is that they give you less than half of the picture - you need a switch (MOSFET), inductor and capacitor but you also need: - Something to generate PWM Something to drive the MOSFET Something to measure the output voltage and compare it with a known reference Something to convert this to PWM to feed to the MOSFET driver
H: Can I use an opto-isolator or opto-coupler instead of a relay to operate AC source? I am designing light control board using a micro controller. As I want to make my circuit board smaller, I am thinking of using an opto-isolator or opto-coupler instead of using relays. Is it OK to replace a relay with an opto-isolator" or opto-coupler? Can an opto-isolator work on more current? AI: If your load is AC AND low power then some opto-isolators will work such as the MOC3021. It has a triac output capable of driving a low power load in the order of a few mA but, more likely you would want to drive a more powerful device such as this: - The external triac would typically be a BT136 and there are a pile of google images here that show this type of configuration. The BT136 comes in a T0-220 package that isn't that small but maybe some space saving can be achieved. A lower power BT131-600 could handle 100 watt loads but RMS current is limited to below 1 amp. It is available in a T0-92 package. You could also consider using a solid-state-relay although a lot of them will contain pretty much the above Triac circuit or back-to-back MOSFETs: - (source: bristolwatch.com)
H: What effect does Permanent Magnet have on an air-core coil? Given an air-core coil of known inductance and a given frequency of applied voltage, you can predict the coil reactance, the current, and the voltage/current phase relationship. What effect, if any, will a permanent magnet placed at one end of the coil have on the predicted properties of the coil? Additionally, if the reactance for the given frequency is altered by the magnet, can the applied frequency be changed such that a new reactance (in the presence of the magnet) can be matched to the original reactance without the magnet. The former paragraph is at the core of what I'm trying to get at, because I want to know if the magnet will have any kind of transformer effect (parasitic or otherwise) on the oscillating coil. Will the presence of a magnet simply change the inductance of the coil, or will it cause other behaviors (similar to those seen when two coils share mutual inductance)? Or is there something else altogether I am missing? The arrangement of coil and magnet is end-to-end as the field polarities would be aligned along their axes like this: coil <--> magnet _//////_ [N \| S] The arrangement would also allow for the magnet to slide into the coil, thereby creating a sort of magnet-core coil. AI: The reactance of the coil will change and there are two opposing mechanisms at play: - The permeability of the magnet will tend to increase inductance The conductivity of the magnet will cause eddy currents that tend to decrease inductance To what extent that either increases or decreases the coil's inductance is guesswork and will vary at different frequencies. I've seen situations where reactance/inductance doesn't change at all. Another effect is that the eddy currents in the magnetic material will cause the magnet to take power from the AC circuit i.e. losses will increase. At high frequencies the magnet will act like a plate of a capacitor and thus there will be a resonance effect. Again, without detail of the experiment this is hard to quantify.
H: How to choose correct value of inductor in a circuit I have a PCB where SIM900 is used. As SIM900 needs at least 2amps of current during transmission burst so I have used LM2576 for its power. Below is its schematic: In the circuit, I am using 200uH inductor. During a call, SIM900 stops suddenly and inductor is blown. This happens for the first time so I replaced it and again I tried and the same thing happened. Inductor starts heating up and then it get blown. What value of inductor should I use with this circuit.? AI: It sounds like the inductor is not rated for the current. You want 2 A average out, so the inductor must be rated for more than that. Purchasing a component from a local market without a datasheet is no way to source parts. The LM2567 is a rather old part requiring high inductance. I've only looked at the first page of the datasheet, but it says right up front that it only runs at 52 kHz fixed frequency. That requires a large inductor. You need to actually read the datasheet. Right on page 1 it shows a Schottky diode. Your 1N4004 is totally inappropriate here. It's long reverse recovery time is probably significantly stressing the LM2576. That may actually be the real cause of the failure. The inductor appears to be the problem after the LM2567 output fails shorted to the input. Especially when you don't really understand how something works, you need to follow the directions carefully.
H: Equivalent circuit It seems at first sight of the first circuit that all the resistors of resistance r are connected in parallel and i have thus simplified it to the second circuit. But then on numbering the nodes and redrawing the circuit i got the third circuit. I got the same equivalent resistance for all the three circuits which i think is a coincidence because i do not not know why the second circuit is equivalent to the first one (i drew that way because the in the first circuit it seems that all the resistors are in parallel). I basically have two questions:Is the first circuit equivalent to the second circuit or the third circuit or both? Why is the second circuit equivalent to the first one if it is so? AI: Yes they are all the same. 5 resistors R in parallel. The actual equivalent circuit would be the following... simulate this circuit – Schematic created using CircuitLab Regarding "Why is the second circuit equivalent to the first one if it is so?": All the nodes you marked as separate, are all connected by "wires", so they are all at same potential, and are indeed the same point (assuming the "wires" are ideal with no resistance, as it is in 99.99% of schematics). Here's an example about this: simulate this circuit All this means you can actually draw schematics in any way you want, you can draw wires 1 m long, you can draw them curved, in the shape of a dog, it doesn't matter. Of course you should try to make your designs simple and easy to read, or it will be difficult for anyone to understand it.
H: Is there a shorter or L-shaped version of PowerPole connector? if not, is there an engineering reason why not? I've not worked with this connector in many years. I'm working with a robotics team and these connectors are very long and bulky and take up [more, think] space [than they need to ]. I'm trying to find a shorter or L-shaped version. Have found none, wondering if I missed it or if there is an engineering reason why they can't be made that way. AI: They're built that way because they are optimized for 1. Power handling and 2. Cost. The pins are sheet metal construction (folded from a flat sheet) and typically tin-plated copper. As such cannot rely on a solid core to improve conduction and better materials (e.g. gold) to reduce resistance. All they have to work with is surface area and since the diameter of the pin must remain small enough to retain strength (it's a rolled up sheet) yet large enough to be manipulated by conventional forming equipment, the only degree of freedom left is the length... ergo the longer pins.
H: Determining current draw of arduino I am trying to determine the correct batteries required for my Arduino (Atmega 328) project. I have components which have both voltage and current ratings. I am adding up all the maximum current specifications from the components to determine the ideal mAH. I am currently stuck on the Atmega328 chip. I read the datasheet, but I cant seem to find the specific current draw of the chip. My understanding is that the operating voltage is ~5V, and the current draw of the chip would be ~40mA. But if I am using 6 pins (3 analog, 3 digital), assuming they all output 40mA at the same time (which is unlikely as the 3 analog pins take in x,y,z values from an accelerometer), would it be safe to assume the current draw of the chip would be 240 mA? (this sounds wrong) AI: The datasheet of a microcontroller tells you the current it uses internally. Only you can determine how much additional current there might be going into the power pin or out the ground pin due to other pins sinking or sourcing current. From what you say, the micro itself can take up to 40 mA. I don't know that particular micro, but that's certainly plausible. The datasheet is also telling you the maximum that output pins are allowed to source. This is of no use in a power calculation. You use this information during the design to make sure this limit is not exceeded. The actual current that the external circuit draws from a output pin is dependent on the external circuit. Consider the limiting case where a pin is left unconnected. Clearly no current is going out that pin. For example, let's say you're lighting a LED directly from a digital output. The processor runs on 5 V, and you have a green LED to ground with 300 Ω resistor in series. The current going out of the pin when high will be about 10 mA. The fact that the output could have sourced up to 25 mA, for example, is irrelevant.
H: 0 or 1 as desired output after certain threshold I'm a high school student, making a project which uses LDR, as we know resistance across the terminals varies according to the amount of light falling on it, but in my project I just need two definite states, either high or low as output, i.e after certain threshold the output should high/low. Problem is that, I can do this with an Arduino but I want to keep it as simple as possible, so I'm looking solution of this problem in form of any combination of logic gates available in market. I'm not still in college and don't have anyone around me capable of helping me with this matter, so I'm here looking forward to you guys. Thank You AI: First you need to turn the variable resistance of the LDR into a variable voltage, then you compare that voltage to a threshold. The first is done with a simple resistor divider, with the LDR being one of the two resistor. The second is done with a comparator. Many opamps can work like comparators if used open loop. Here is a example circuit: When more light shines on the LDR, its resistance goes down. It and R2 form a resistor divider that makes a fraction of the power voltage at the junction between the two resistors. The fraction is dependent on both resistors. When the top one is lowered, the output voltage goes up. R2 should be about the resistance the LDR has at the light threshold you want to switch at. That gives you most sensitivity right around that light level. R3 sets the light threshold. All it does is produce a voltage that is a fraction of the supply voltage, depending on how the knob is turned. IC1 compares the two voltages. Its output goes high when the + input is higher than the - input, and low when + is lower than -. One subtlety with this circuit is that it will probably oscillate or be unstable when the light is right at the threshold. In theory, the two opamp inputs are equal then. There is always some noise, so sometimes one will be more positive than the other, then in the next instant it will be reversed. It's actually informative to experiment with this circuit as it is and to observe the behavior right at the threshold. To get snap action, add a big resistance from OUT to the + input. If the output is just at the point where it goes high, that higher output will make the plus input just a little higher, which makes OUT more solidly high, etc. Note that now going down requires a little lower threshold. A resistor from OUT to + adds hysetersis, which means there are actually two thresholds, a slightly high one for going up, and a slightly low one for going down. 1 MΩ should work for the hysteresis feedback resistance in this case.
H: How do you find the equivalent resistance for the following question? In the given circuit, i need to find the current flowing in the circuit when the switch is closed and for this i need to calculate the equivalent resistance between the terminals of the cell. How would you calculate the equivalent resistance between the terminals of the cell? I calculated the equivalent resistance to be by taking 6 ohms and 3 ohms resistors in parallel and the equivalent resistance of these two in series with the 10 ohms resistor. The equivalent resistance between the terminals of the cell comes out to be 2 ohms and the current hence comes out to be 5/6 A which is not present in the options. (The options are-:2 A, 1 A, 3 A and 0 A.) AI: When the switch is shorted there's no current through the 6 Ohm and the 3 Ohm resistors so the current value would be 10/10=1A.
H: What does this bus signal representation mean? Often, when looking at digital system signal plots over time there are these graphs with two parallel lines that often "swap" place. I am wondering what precisely they represent? I noticed they usually are used for buses, but for the rest I'm clueless. It's probably quite trivial. This is an example of the graphs I mean, in this case from a discussion about how CPU's execute instructions: I've tried googling, but I couldn't find an explanation. AI: The parallel lines indicate a multi-bit bus, and represent any possible combinations of low/high values. This bus could be two bits, or 64 bits, or whatever the design calls for. As shown here, the grey regions indicate that the data value is unknown or indeterminate. In this case, immediately after the positive clock transition, the address and data are unknown. Soon, after that, the address data becomes stable and known. Some time after that the data is known and can be read. edit: As Peter says, the grey regions can also mean "don't care".
H: Can a AC generator be powered by D.C.? I have a very specific need. I need to avoid a VFD due to its square waves - my research requires exact sinwave frequencies of electrical current. With that said I have a Tesla turbine which powers a AC Generator. The tesla turbine is powered via heavy water - For reasons I can't go into the power is collected, transformed a number of times and then fed back into the water apparatus to 'push' down on the water which then goes back into the tesla turbine. I need a way to generate AC from the turbine, convert it into D.C.for storage, the the D.C. Needs to be able to be fed back into a reliable AC sinewave for reuse. Probably by a AC gen... What I was thinking is to have the AC generator which is being driven by the turbine ->feed to a battery stack, have that power stored as D.C.- Then have the D.C. Which is stored to Power a AC generator ... The catch - I need the AC frequency to be modular - as in I need to be able to change the final AC frequency at will... A VFD produces square waves - so I thought the second AC gen would be ideal. Any thoughts, problems I'm going to have and solutions? AI: Drive the second AC generator with a variable speed dc drive they are manufactured for propulsion systems these days. This will give you variable frequency.
H: What is the role of these inductors? My understanding is that inductors in a dc circuit act like a wire and don't do anything. So what are the ones labeled FLxx in this schematic for? Could they be replaced with a simple wire and have no affect on the circuit function? AI: First, from the way they are specified, those parts are more likely ferrite beads than inductors. For more information about ferrite beads, see this old question: Regarding ferrite bead However, the function of the ferrite bead and the inductor is somewhat similar. In this circuit, they are most likely (it would be more clear if you included the whole circuit and not just a piece) filtering the power supply nets. They are there to ensure that the current flowing through those wires is mainly DC. The chips that are powered by these nets will have variable current requirement over time. In some circuits, the current requirement can change very quickly. These inductors ensure that when that happens, the extra current (or sudden drop in current) is supplied by the bypass capacitors (for example C42 and C46) rather than by the up-stream power supply. This reduces time varying currents in the supply lines coming from wherever they are coming from, which is likely to reduce radiation (EMI) produced by this device. If the supply is noisy (and the noise is in the correct frequency band), these ferrites could also help prevent that noise affecting the down-stream components. If you replace these inductors (or ferrite beads) with simple wires, the most likely impact would be increased radiated emissions from your system, or possibly interference between different subcircuits in your system (which we can't predict without seeing more of your circuit).
H: How does this push button work? I'm on a quest for wireless coffee. I've succesfully hacked my Nespresso with a NodeMCU that will now let me wirelessly make a coffee. That is, when the machine is already on. I need to figure out a way to turn it on programatically also, but the button for turning on is different. After opening up my Nespresso Lattisima+ (model EN520) I found this push button - a kind of button that I don't know. How do I find out how it works (i.e. which pins to solder my relay onto in order to switch it on programmatically) without killing myself? I see a F5 and SW1 next to some pins, but on the back of the printboard I can see that there are pins "hidden", meaning big white block as well. Thx for helping me stay alive and curious :) AI: Bär (Baer without the umlaut) is now a brand of Johnson Electric. Check their website: http://www.johnsonelectric.com/en/features/energy-saving-push-button-switches-for-coffee-machines Seems these "Tippmatic" switches have some timer built-in.
H: LTC4002 Li-Ion Battery Charger... How do I connect the Comp pin? I am attempting to use the LTC4002 Li-Ion switch-mode battery charger in my project. I'm fairly green with electronics, so some of the conventions used in data sheets probably evade me. On page 8 of the Linear Technology-supplied data sheet (link below), I see the following description for pin 1: COMP (Pin 1/Pin 1): Compensation, Soft-Start and Shut- down Control Pin. The COMP pin is the control signal of the inner loop of the current mode PWM. Charging begins when the COMP pin reaches 800mV. The recommended compensation components are a 0.47μF (or larger) capacitor and a 2.2k series resistor. A 100μA current into the compensation capacitor also sets the soft-start slew rate. Pulling the COMP pin below 360mV will shut down the charger. Ok. Nice description. But when I look at their suggested schematic (shown on both page 1 and page 14), it shows Pin 1 (the Comp pin) connected to a 0.47μF cap and then to a 2.2K ohm resistor, then on to ground. From the description of pin 1, it seems they are implying that I need to be supplying the 800mV, like I need to supply the 100μA of current. But how would I do that when they show the pin tied to ground through a cap and a resistor? Right now, my board is otherwise complete and ready for fab, but I am loathe to write that check until I'm sure the device will work. I have the parts, but they are all extremely tiny fine-pitch surface-mount parts that are difficult to bread-board without breakout boards... So: What will the behavior of this part be if laid out like their schematic? Can I leave it as it is? Does the description simply provide some details in case I want more granular control over the charging process? Any assistance will be hugely appreciated! Thanks! The data sheet can be found here AI: The 100µA current source on the COMP pin is internal to the part -- see the block diagram on page 9 of the datasheet. This means that the soft-start time is controlled by the value of the capacitor in the RC network connected to the pin -- there's no need to inject an external current into COMP for such a basic function.
H: Phase-Shifting Commutation I posted a question a while ago about a circuit to realize a voltage controlled phase. I wasn't able to give a full discussion about why I was interested in the circuit at the time but I am in a position to do so now. I have filed a provisional patent for "phase shifting commutation" with this abstract: The apparatus employs magnetoresistive sensors (28) to generate a sinusoid having a time-varying frequency commensurate with the angular rate of change of a motor’s rotating shaft (40). The resultant signal is then fed to a mixture of fixed- and voltage-controlled phase shifting circuits (22, 64) (32) whose outputs are then amplified and fed to independently wired stator phases (56). The apparatus also includes a rotary transformer (24, 36, 46, 60) whose current is generated by amplifying a frequency-shifted version of this shaft sinusoid. The rotary transformer (24, 36, 46, 60) in turn supplies current to the rotor system’s electromagnet to generate a controllable magnetic on the rotor system despite the rotor system’s motion. My former employer is investigating this methodology and has the right to use the technology but I am in control of the IP. I am building a mini dynamometer chassis with BLDC motors and I want to build the circuitry to achieve this commutation technique so that I have something to demonstrate. A voltage controlled phase shifting circuit seems entirely achievable as there are many available off the shelf that operate with center frequencies in the GHz range. Also consider that this technology is key to phased-array systems. So I am wondering if anyone can help me build this phase-shifting circuit or can recommend some solution that I don't know about. Off-the-shelf phase shifting circuit Detailed discussion of phase shifting circuits Previous thread of discussion AI: In the analog domain, i think like a voltage controlled phase shift would necessitate a voltage controlled impedance of some kind. There are all kinds of funky ways this could be done for reactive components like with varactors (diodes whose reverse transfer capacitance has a high voltage dependence), saturable inductors (a varying DC voltage across the windings will induce a current which lowers the small signal inductance), etc. Voltage controlled resistances are realized often for things like automatic gain control circuits for fixed output amplitude from variable input amplitudes, or sine oscillator stabilization. These could be furnished with JFETs or optocouplers or blasting voltage controlled heat at RTD's or all kinds of other funky ways. Now whether you need orthogonal control of phase and amplitude or not and how precise your open loop phase shift vs voltage characteristics need to be would dictate the direction of the design in addition to other practical matters like power, size, ranges, etc.
H: What does "open drain when programmed for the wait function" mean (in micro controller spec language)? So I am reading the Zilog Product specification for the Z8440/1/2/4, Z84C40/1/2/3/4 serial input/output controller (here it is http://www.zilog.com/docs/z80/ps0183.pdf). And there is one sentence I dont understand the meaning of: Here is the schematic with the leg marked: I can understand, for example, TxDA, TxDB. Transmit Data (outputs, active High) which to me means that the pin will have voltage (it will be pulled high), when the controller will starts sending data to the output channel. But what does it mean by "open drain when programmed for Wait function"? What is this drain and how to open it? AI: Read this small piece of article from Wikipedia about what open drain is and why it has right to exist. Programmed means that there's some internal register in the chip which you can write with, let's say, 0 or 1 - and in one state output will have logic 0 or logic 1, at another state of this config register it will have this open drain output, which is very useful to connect to another outputs for logical ANDing of the signals, eliminating usage of a number of AND gates in order to get output signal which is AND function of all input signals. Note that regular logic outputs can not be connected together, open drain/open collector can. In this particular case, machine based on Z80 chipset may have several sources of wait signals - for example DMA, video processor, machine core core logic - and you just connect all these open drain outputs together, connect pull-up resistor to the resulting line, and have AND function of everything connected. If any of signals goes low (or all go low), resulting signal will be low. In terms of current flow, if open drain/open collector output is deactivated (to be simple - output transistor is turned off) there's minimal current flowing through it. If it is active, then current is flowing through pull-up resistor and collector-emitter junction, and this current is limited by this pull-up resistor. Look at the page 110 of the manual, for Write register 1 diagram. It seems this Z8440 specification does not detail how its programming works exactly, I am afraid you will have to read whole Z80 family manual to find it out, or find out another type of document which would focus on application/programming rather than circuit design and chip internals.
H: Trimpot potentiometer rotational life Vs load life I'm dealing with an application which makes use of an adjustable linear voltage regulator (LT3080). A trimpot potentiometer is used in place of the set resistance Rset to regulate the output voltage. A constant set current of 10 uA flows into the Rset and gives the output voltage following the equation Vout = Rset*Iset The linear regulator is powered by a 12V DC input source. I would like the output voltage to vary between 5V DV and 12V DC (minus the voltage dropout). Therefore I would split Rset in the series of two resistors: a 500 KOhm fixed resistor and a variable trimpot of 1 MOhm. In this way the minimum Rset will be 500 KOhm and therefore the minimum output voltage will be 5V. The maximum Rset will be 1.5 MOhm and the output voltage will go at maximum allowed. I was keen in choosing a high accuracy (0.1%) fixed Rset resistor and a good accuracy (10%) trimpot. I'm now concerned about the life of the trimpot. The datasheet of the trimpot I chose (BOURNS 3362P-1-105LF) states that rotational life is 200 cycles. I'm aware that cermet potentiometers like this one have limited rotational life. On the other hand, there is one value I am most concerned about: load life. Here again, the datasheet states 1,000 hours 0.5 watt @ 70 °C. Since I am burning a meaningless amount of power in the potentiometer, can I expect longer life? And the most important question here: What will happen after these values have been passed? Will the potentiometer be like a shortcut or and open circuit? Or there will just be derating from the set resistor value? The system is expected to work in standard room conditions, at an ambient temperature of +25C, non condensing. simulate this circuit – Schematic created using CircuitLab AI: The two main failure modes of a trimpot are ... a) the track wearing away, from excess mechanical use, leading to a high resistance and noisy connection track to wiper b) contaminants interrupting the contact between track and wiper, leading to a high resistance and noisy connection track to wiper Most use, yours included, can probably ignore the first problem. The second problem has two main components, Large particles, like dust, and reactive contaminant gases, like sulphur and nitrogen compounds. These will generally react faster when hot. As you are going to be running your trimpot cool, you might like to hope that you would get more than 1000 hours out of it. The connection you've shown is particularly nasty in adjustable voltage regulators, as the loss of track contact gives high resistance and therefore high output voltage (the nastier type of error). You can reduce the amount of damage such a fault can do by connecting the wiper to the unconnected top of the track. That way, if the wiper loses contact, the resistance only goes to 1.5M, not to infinity. A capacitor ADJ-> GND (up to 1uF permitted by the regulator data sheet) will reduce the size of the positive excursion that any momentary losses of wiper to track contact cause as you adjust the setting. It's worth bearing in mind that a trimpot is going to be orders of magnitude less reliable than a soldered connection to a resistor, and possibly an order of magnitude less reliable than a large potentiometer, so try to avoid them if you can. Ideally, in this sort of application, you would not use a trimpot. If you want to set the voltage carefully to a setting and leave it there, then it's worth soldering in a resistor, measuring the voltage, calculating a new value, rinse and repeat. If you want to adjust the value often, then get a 'real' 1/4" shaft potentiometer, or an IC 'digipot', that's programmable from a microcontroller.
H: Checking array input value Verilog How can I check the value of array input in Verilog? I tried this code but, I got an error "op_code is not a constant" module mutlti_proccessor(data_out, flags, data_1, data_2, op_code); input [7:0]data_1; input [7:0]data_2; input [3:0]op_code; output [7:0]data_out; output [4:0]flags; /* flags[0]:ZF "Zero Flag", flags[1]:SF "Sign Flag", flags[2]:CF "Carry Flag", flags[3]:OF "Overflow Flag", flags[4]:COF "Compare Flag" / wire carry; / Set falgs to zero / assign {flags[0], flags[1], flags[2], flags[3], flags[4]} = {1'b0, 1'b0, 1'b0, 1'b0, 1'b0 }; if(op_code==00000) Ripple_Adder RA1(data_out, carry, data_1, data_2, 0); endmodule Where Ripple_ADD is another module that sums up two 8-bit numbers.Note: there is no clock. Thanks .... AI: Your problem lies in the line of code: if(op_code==00000) Ripple_Adder RA1(data_out, carry, data_1, data_2, 0); You don't seem to have understood that Verilog is a "Hardware Description Language", not a procedural programming language. It does not execute code line by line, but rather is synthesized into hardware based on what you are describing. You are trying to use an if statement incorrectly. What your code is describing is that if the op_code is 0, then create some hardware called Ripple_Adder. However, what happens if op_code is not 0? Does it need to infer the adder? Should it delete the adder? Simple answer, it can't - the hardware is fixed at synthesis, you can't add or remove bits of it when running. This is why it is complaining about op_code not being a constant - it doesn't know whether or not Ripple_Adder should be included or not. The solution, is to always include the hardware, and then infer a multiplexer to select whether or not the output of it is used. Something like this: wire [7:0] ripple_data_out; //Output data for the ripple adder Ripple_Adder RA1(ripple_data_out, carry, data_1, data_2, 0); //Infer the ripple adder always @ begin case (op_code) 5'b00000: data_out = ripple_data_out; //If the op_code matches, connect ripple_data_out to data_out through the mux. //... Add more cases here for other op_code values ... default: data_out = 8'b0; endcase end
H: What difference does the clock make in synchronous vs asynchronous communication? I am new to embedded programming and have very little knowledge in how digital/electronic system works, but I have been given a task to bring up USART communication between two chips. For that, I have started learning from the basics of serial communication etc etc. What I don't understand in serial protocol is that what is the difference between synchronous and asynchronous communication? Let us say I have two devices d1 and d2. They both are aware of their baud rates. d1 sends data at a baudrate of 9600 and d2 receives data at the same rate. Then my question is, how the clock plays role in synchronous communication? Means with baud-rate I think we have all the information of the communication channel. With this confusion I am not able to clearly understand the exact difference between synchronous and a synchronous communication. AI: With asynchronous data transmission, when there is no payload data to be sent, the data line becomes idle and therefore the receiver waits for a transition that marks the beginning of new incoming data. In this respect there is no definite phase relationship between data previously received and the new data arriving. This is why it is called "asynchronous" With synchronous data, either a clock is permanently present as a seperate signal or, the clock is embedded into the data (as per Manchester encoding or scrambling) so that the receiver is always aware when real payload data could be present. When no payload data is present the clock is still present. Manchester encoded data: - how the clock plays role in synchronous communication? Whether you extract the clock from the data (or have a dedicated clock line) you need a clock; that is fundamental to any data transmission. Asynchronous transmission re-creates a clock internally based on the agreed baud rate and the first transition of data following an idle period. From this point on until the end of the data block it generates a clock internally for the whole transmission. So the clock plays a vital role in both asynchronous and synchronous data communications. with baud-rate I think we have all the information of the communication channel Not quite - knowing baud rate at both ends is useful but it doesn't tell you when bits actually change state in the data - this "synchronization" to the data is performed by the idle-to-start bit transmission in asynchronous data and is everpresent in synchronous data communication.
H: Can a 50 Hz, 220 VAC transformer work on 40 Hz, 180VAC? I am designing an AC to DC converter. My power source is a single phase synchronous alternator that has a voltage range of 170~260 VAC and a frequency of 40~60 Hz. Can i use a transformer designed on 50 Hz and 220 VAC to work on my alternator specs mentioned above? AI: Can a 50 Hz, 220 VAC transformer work on 40 Hz, 180VAC? Yes it probably can - the initial worry is saturation problems due to operating at the lower frequency but, with the voltage dropping to 180 V at 40 Hz, this produces virtually the same magnetization current as 220 V at 50 Hz. Strictly speaking, if the transformer is nominally rated for 220 V at 50 Hz, you should run it at a nominal voltage of 176 V at 40 Hz.
H: ESD protection for raspberry pi I am a software engineer . I am working on a project based on raspberry pi and sensors, I am using sensors to get data and I am using an ADS1115 ADC to convert sensor analog data to digital. I am concerned about the ESD for raspberry pi so i have used a metal case for this project. Also, I am using a MEANWELL Net 35-b power supply . Question: I read on a blog that the simplest way to have ESD protection is to use a metal case and connect the metal case to the ground of the power supply also ground of raspberry pi and power supply are connected. Is this the correct way to protect circuit and raspberry pi? AI: ESD is misunderstood by many so beware ! You can indeed use a metal box to enclose your Rpi but it might not be needed. What causes an ESD discharge ? It is cased by someone (who is somewhat electrically charged) touching the conductive parts of an electronic device. This charge wants to equalize with your surroundings and tries to flow away through anything conductive you touch. You might have noticed that you sometimes can get a small electric shock if you touch anything electrically conductive. But if you touch it again, there's no shock. That first shock was an ESD discharge event. If your Rpi is somewhere where no-one can or will touch it, no case is needed. To prevent someone touching the sensitive parts of an Rpi, you do not need a metal case, a plastic case will provide the same protection against touching. All the sensitive chips on a Rpi board have ESD protection build-in and this is usually enough protection even against a charged human touching any part of the Rpi's board. Although I am very well aware of ESD, what damage it can cause and what needs to be done to protect a sensitive chip against that (I design on-chip circuits for a living), I know that I do not need to take special measures to protect a device like an Rpi against ESD. Just placing it in a plastic case is more than enough protection. Even if you would ESD discharge through one of the connectors for example, the Rpi can handle that.
H: Single step debug and timer's counter value I am debugging with cortex m7, i found that when I halt the CPU, the timer's counter value still upward increases. After reading it's reference maunal, I found a register that can stop timer's clock feed when the debugger halts the CPU. However, even I configure this register, I still got counter value increases for about 15 in disassembly single step(timer clock is same as cpu's). Is this the same for all microcontroller? So I can only see the right behavior in simulator? Thank you! AI: This sort of detail varies considerably between microcontrollers. This specifically has to do with how exactly debugging and single stepping is implemented. Manufacturers don't want to add a lot of hardware just for debugging, because that burdens the case where a large company wants to buy 100 k pieces at a time for a production run. As a result, all kinds of clever schemes exist for re-using existing hardware. One common scheme is to have the part execute special code to help implement the debugging interface. To the processor, this is just code, so counters and the like keep running. The switch to stop the counter during single stepping in your case probably only means the peripherals are shut off when the chip is idle, not when the debug code is running and communicating with the debugger. It may be in your case that this debug code takes 15 cycles per single step.
H: What is needed to switch a relay via logic output (e.g. from Raspberry Pi)? I'm working on a project with a raspberry pi involving some sensor and a relay. Actually I'm Using a board like this one: So I connect GND,5V, and a GPIO to the IN pin to control the relay state: ON/OFF. Now I'm starting to move from the prototype stage.. I'm working on create the PCB, and I'm wondering if I can drive the Relay, with a "naked" version like this: is it possible? Why all the other components are required on the board, figure 1? resistor, transistor etc.. I'm expecting to connect: D to the source and one between A or B to the destination C to GND E to a GPIO ..correct? Maybe I'm using "rocket to kill an ant".. I'm using the relay for control a low voltage (between 12V and 24V DC)..is there any more appropriate component or integrated circuit that you can propose? AI: The transistor is used to drive the coil that activates the relay. The diode is a flyback diode which is used to discharge the coil when switching the relay off, otherwise the transistor will be damaged. You need to design those components onto the board if you want to use this relay "naked". Basically, you are right but you need a transistor to drive the relay because the microcontroller can't deliver the required ~90mA at 5V. It looks a bit overdesigned if you only want to switch 24 VDC. It depends on many parameters (current, voltage, switching frequency) which relay you can use.
H: Operational Amplifier Circuit I am given this circuit The voltage at each inverting ends should be 0, since the non-inverting ends are grounded. However, given that VIN is positive, there's a current flowing between each of the non-inverting ends (through the capacitor and resistor: hence to me it's equivalent to saying that the voltage through a series capacitor-resistor circuit is 0 even though there's current through both components). How can this be the case? AI: For reference and to protect against future edits, here is the circuit we are discussing: This is simple to analyze if you break it into its two distinct parts. You left out component designators in the schematic, so it will be difficult to talk about. You'll just have to guess which ones I mean. Next time, draw the schematic properly. The IC1 is just a classic integrator. Due to the feedback, the opamp does what it needs to so that the - input is kept at 0 V. This means the current flowing thru R1 is directly proportional to Vin. The only place that current can go is thru C1. By the nature of what a capacitor does, the voltage across C1 is the integral of the current thru it. The IC2 is just a classic inverting amplifier. Ignore the V1 input for now. The gain from the output of IC1 is simply -R3/R2. Note that IC1 actually inverts while integrating. IC2 inverts the signal again, so Vout is the positive integral of Vin. V1 is just another input added to that from IC1. The gain from V1 to Vout is -R3/R4. This is simply added to the previous result. In that sense, you can think of IC2 as a summing amplifier.
H: No current in this circuit? I had to find the resistance of this circuit between A and B. My teacher said that since the blank wires have zero resistance, the potential difference across the ends of the blank wire is zero(according to v = ir) and thus 1,4 are equipotent and 2,5 are equipotent and the circuit is thus simplified.The equivalent resistance comes out to be 2r/5. MY QUESTION:If the points 1,4 and 2,5 are equi-potent how can current flow through the circuit. MY REASONING: The potential difference across each branch of a parallel circuit is same and in this case it is 0 and hence no current can flow through the circuit. AI: "If the points 1,4 and 2,5 are equi-potent how can current flow through the horizontal circuit." No current flows "through the circuit", current flows through parts of that circuit. "The potential difference across each branch of a parallel circuit is same and in this case it is 0 and hence no current can flow through the circuit." The circuit as shown is not fully parallel, only partially. Did you try to rearrange the circuit to make the parallel parts stand out? If you do it will become much easier to understand.
H: What are the key differences between Tantalum and Tantalum Polymer capacitors? I've noticed that most component retailers split these types, and am wondering what comparative differences there are between them. I know that "regular" tantalum caps usually use MnO2 as an electrolyte, whereas polymer ones use (generally unspecified) polymers instead. What are the main performance differences between them? Is there a key factor that should decide which type is used in a specific design scenario? AI: Conventional MnO2 tantalum capacitors are the ideal choice for applications with requirements for high temperatures (currently up to 175C), high voltage (up to 50V) and established reliability. The technology offers highest CV in a small package. It’s possible to use them up to 80% of rated voltage, but in low impedance circuits further derating needs to be applied - see 7]. Tantalum Polymer Capacitors are the best choice for consumer applications with low ESR requirements such as DC/DC converters in notebooks, PDA, telecom and other applications. The parts can be used up to 80% of rated voltage. Manufacturer’s specifications of lead-free reflow process capability, temperature range ratings, leakage current and appropriate storage/handling in accordance to the MSL level should be verified for specific application needs. NbO OxiCapTM Capacitors offer the safest, available alternative among the various capacitor technologies with good cost-versus-performance value. The parts can be used up to 80% of rated voltage and are compatible with lead-free reflow requirements. The excellent steady state reliability makes these parts a favorite choice not only for consumer applications but also for high end, automotive, computer and professional designs. Appropriate temperature derating needs to be applied for temperatures over 85°C. http://www.avx.com/docs/techinfo/New_Tantalum_Technologies.pdf
H: Why do electrical systems have neutral, live and ground when electronic systems have only VCC and ground? Wouldn't it be simpler if electrical systems take away the neutral and just have live and ground similar to electronic systems? AI: Wouldn't it be simpler if electrical systems take away the neutral and just have live and ground similar to electronic systems? Ground is localized to a house or installation. Ground is not provided by the electricity companies because that would be unsafe - how could a safe ground be sourced from an electrical generator hundreds of miles away? How could you detect if current was flowing through some unfortunate person (in the process of being electrocuted) if one of the power feeding wires is actually ground - how could you tell that the current was flowing through that person without a neutral wire? How would you distinguish it from normal load current? That's the whole purpose of residual current devices (UK name) or ground fault current interrupters (USA name): -
H: A question on the differences between a bench power supply and a non-bench power supply in terms of current voltage regulation? Before asking my question I would like to briefly clarify what I understand by a bench power supply opeartion: Let's have a bench power supply which is set to 10V 1A CC. As far as I understand, this means upto a 1A current the power supply can supply 10V CV accross the load. So if I hook up a 10 Ohm resistor as a load to the terminals of this power supply I will measure 10V accross it and V/R = 1A current through it.(I'm assuming a special resistor which can handle enough power for the sake of this example) But let's say if I hook up a 5 Ohm resistor instead. In this case the current will not increase to 2A since it is set to 1A CC. The current regulator inside the power supply will not let the load to sink more than 1A current. Which means the current will remain the same as 1A. But this time since V = I*R, the power supply's voltage will accrdingly drop to 5V. This is what understand in case of bench power supplies. Correct me if I'm wrong. Question: Most of the power supplies (SMPS or linear such as 9V adapters ect.) in the market have two ratings always mentioned on them: Rated voltage output and max current. Are they operate the same way as the bench power supplies? Let me ask the question by an example. If we have a power supply such as the following adapter: https://www.alibaba.com/product-detail/10v-1a-power-supply-adapter-10w_60310531013.html?s=p 12V and 1A ratings written on it. If it were a bench supply with CC, I would say connecting a 6 Ohm load would sink only 1A and would drop supply's output voltage to 6V. But is that also what happens with most of the power supplies(not bench ones) in the market? I mean the same scneario happens as in bench supplies? Or it depends? How can we know it without trying? Here are two examples of non-bench power supplies(typical ones used most commonly): https://www.alibaba.com/product-detail/10v-1a-power-supply-adapter-10w_60310531013.html?s=p http://uk.rs-online.com/web/p/embedded-switch-mode-power-supplies-smps/0413655/ Basically I'm asking what happens when we lower the load for a typical non-bench power supply and try to exceed its max current? Would there be a current and voltage regulation as in bench power supllies where CC activates or what would happen? AI: This is what I (and most EEs) would call a bench power supply or lab supply. The currents and voltages written near the output terminals are the maximum ratings. These are the maximum currents and voltages this device can deliver. Since this is a lab supply, it is posible to adjust the maxima for those values through the user interface. Then depending on the load the supply will regulate the voltage (the load does not draw more current than what you have set) or it will regulate the current (the load draws more current than what you have set so the voltage is lowered as to not exceed that current). Devices like this: and this: Are what I call power adapters. They simply deliver a certain voltage. They can be used up to the current indicated. You cannot expect these devices to go into current regulation (lowering the voltage to maintain that maximum current) !!! Depending on the power adapter, it might be overloaded and damaged or (better) switch off. Or the fuse might blow. Power adapter are simply not suitable for current regulation.
H: Microphone supply circuit comparison? I'm going to connect an electret mic (WM-61A) to my computer's "mic in", and am very confused on the reasoning behind a couple different diagrams/schematics that could possibly be used to do this with. The first one is from a thread on Head-Fi and seems to be the standard that everyone uses there: This is good to me, but then when I did a search for a couple different options, I found this: There's also another schematic using an AD654 op-amp specifically for phantom power. The one above claims to provide a "Phantom Ground"; what exactly is this compared to the first schematic and compared to a virtual ground? Could any of these be used for my purpose? Please and thank you for any guidance given! AI: The extra circuits you have researched do not apply if all you are trying to do is interface an electret microphone to a computer mic-in port. Those extra circuits produce a mid-rail voltage (such as 4.5 volts when using a 9 volt battery) and are useful when developing an op-amp amplifier circuit and all you have is a single power supply such as a 9 volt battery. The mid-rail voltage can be regarded as ground for any audio inputs or outputs. It is sometimes called "phantom" or "virtual" ground but I prefer the term "midrail generator" because it says what it does. You don't need this for what you are wanting to do.
H: Are these two circuits equal and what does the arrow symbolize? I'm not sure what the arrow means, anyway V+ and V- were given as 5V and -5V respectively and Vin changes from -6V to 6V (so maybe an AC voltage source would've been appropriate). AI: Given the information provided, I think the circuits are correct. The arrows designate the currents through D1 and D2. For clarity you should also include a ground, most likely between the 2 5VDC power supplies. Something like this: simulate this circuit – Schematic created using CircuitLab Edit: fixed negative terminal of V3 to be connected to ground per Naz' suggestion.
H: Lithium Battery Protection Circuit - Why are there two MOSFETs in series, reversed? I was studying a battery protection chip and reference circuit (below) commonly used in cell phone Li-ion batteries, and am confused by the two MOSFETs in series on the negative terminal EB-. According to this question, I now understand that MOSFETs can conduct in either direction S-D or D-S. My questions are: 1. Why are there TWO MOSFETs in this circuit? Why not just one? 2. If they conduct in either direction, why are FET1 and FET2 installed with opposite polarities? How does this benefit the circuit? AI: It's for two reasons. Well, actually just for one, but with two factors. A MOSFET can conduct in both directions when turned on, as it truly is just a resistive channel that is opened or closed. (Just like a tap, it's open with a tiny resistance, closed with huge resistance or a small gradation in between.) But, a MOSFET also has what is called a body diode, which is indicated by the little arrow. That body diode always conducts when it is forward biased. It looks a bit like this: simulate this circuit – Schematic created using CircuitLab (weird text label aside to make the image look less bombastic) This is inside all MOSFETs, due to their internal construction, so it's not an option. Some MOSFETs are manufactured specially so that the diode becomes more useful to certain applications, but there's always a diode there. As pointed to in comments; the Body-Diode is a consequence of the substrate connection. I remember seeing a rare one or two MOSFET types with that connection on a separate pin, but they'd be hard to find. (And you would likely want to connect the pin normally anyway, for current capability) This means, that if you use only one in a current path that can conduct in two ways, one way will always conduct with approximately one diode's voltage drop. Sometimes you want that, sometimes you don't. When you don't you connect two MOSFETs in reverse, and the total picture becomes this: simulate this circuit When the one body diode conducts, the other blocks and vice-versa. Now in the case of a battery protection, both MOSFETs are connected with their gate to an independent I/O pin, because when the battery is empty, it is allowed to be charged and when it's full it is allowed to be discharged. So the chip only turns on the MOSFET whose diode blocks the allowed directions, and if the battery is at one extreme of its use-case, its body diode will at least allow current in the other direction, even if the over or under voltage situation persist a while after current starts flowing. Whether or not this might cause problems with MOSFET heating when a battery behaves super weird is a separate point and has up to now be proven to be not an issue. Usually the body diode only conducts a fraction of a second before the over/under voltage disappears and both MOSFETs turn back on. The diodes shown in the schematic might have indicated this fact (my eyes glossed over them initially), but it's equally likely they intend for you to place better diodes to support higher safe discharge currents from a full battery or charge currents into an empty one.
H: Why is it said that when voltage source connected in series would only increase the voltage and not the current? By adding cells in series, a larger-voltage bat- tery can be made, whereas adding cells in parallel results in a battery with a higher current-output capacity. This is what a book that I am reading says and hence what I am understanding is Voltage is added when Voltage sources are connected in series and Current s added when voltage sources are connected in parallel. Considering the ohm's law, 1)for V1=10v and V2=10v, R = 1K, if I keep only keep V1 as voltage source, V across R would be 10v and I(current) would be 10mA. 2)Now if I connect V1 and V2 in series , V = 20V and I would be 20mA. Did I miss something or the current actually doubled up? I even simulated the curcuit in the Multisim and i faced with a similiar result. Any help in understanding the concept would be helpful. AI: Maximum limits is the key here not ohms law so.... Forget ohms law. Putting batteries in series has the maximum current limited by the weakest battery. We're talking about a single battery being able to supply x amps with two in series only able to the same maximum.
H: Disadvantages of dimming LEDs via current regulation as opposed to via pulse frequency? I bought an LED stripe from näve (very similar to this model), but found that it created flicker in video recordings. The flicker got less visible with lower frame rates, but even at 24 fps it was there. I found that the frequency of the flicker was certainly not 50/60 Hz, but I believe the controller varied it to set the brightness (as I tuned it brighter or darker, the number of flicker "bands" caught on video changed) - and as I moved towards the maximum brightness, the flicker got less perceptible in the video recording. I was able to fix this by connecting the stripe to another controller. As far as I can tell, this controller does not pulse the light, but rather adjusts the current. It causes no flicker at any frame rate (24/25/30/50/60Hz), and no bands at any shutter speed (up to 1/4000 s). If I go over a certain current, the LEDs just stop getting brighter, and the controller emits a faint humming noise - but no part of it seems to overheat. I am wondering - why would LED bands not use this approach? Is there any detriment to me using it? AI: This answer contradicts the assumption that constant current degrades performance for colour rendering. As far as the LED is concerned I assert there is no compromise using Constant current whereas PWM may cause "blue shift" from phosphor photon "quenching " OR insufficient dwell time ( two distinctly different phosphors and time sensitivities (Too long or too short. ) Phosphors for LEDs start with those typically used in traditional TV CRT's and Fluorescent tubes where used was UV to excite the phosphor for a fairly long time so that flicker at 60Hz would be less noticeable, thus the time constant of decay was in the 1~10 ms range. Although LEDs use Blue as a substrate so phosphors in the blue range are not needed. Why is an LED white? Think of White LEDs with massive thick transparent BLUE emitting substrate and a hundred atoms thick coated of phosphor that is critically thick to vary the colour from cool (thinner) to warm ( thicker). Like some FL tubes they may be tri-phosphor for broader spectral and higher CRI quality levels which also affect other factors. The decay time < shutter time causes the dark bands in the image. The same phosphors were used in the Blue LED's with a very thin layer of phosphor which converts 10 to 20% of the Blue light into a smooth spectrum of yellow,green,red using yttrium aluminum garnet (YAG) phosphor (doped with cerium). The decay times are slower but the Red content was weaker, so mostly used in "Cool" white LED's. This decay time is affected by PWM choices from 1k to 10kHz, where <=1Khz may be noticeable on camera but > 1kHz might not and depends on both the PWM and the LED phosphor and CCT,CRI values. Now with higher efficacy (>120LPW) White LED's they have added more dopants to improve CRI as well with more red emissions. Phosphor selection will continue to evolve with quality indicators like CCT/CQS that also affect cost. They include oxynitride phosphors is the MSi2O2N2:Eu2+ (where M = Ca2+, Sr2+, Ba2+) compositions whose emission ranges from 575 to 675 nm. The phosphors are selected for cost, efficacy, thermal stability and longevity more so than response time, so PWM rates may have to increase for camera work or use filtered PWM. Filtered PWM has some losses associated with the real conductance or resistance losses in Chokes and Capacitors (ESR), but this is the most effective way to eliminate PWM flicker and may be added to any controller. This is the same as the LC low pass filter (LPF) used in Class-E PWM audio power amplifiers. For Linear voltage controlled current sources or sinks, there is of course the linear loss across the regulator. This only becomes a disadvantage if there is a large voltage drop in the regulator. Typically LED brightness range is limited by the quality or ESR of the power LED and thus the voltage range required from say 10% to 100% brightness. In very high Efficacy LEDs this can be from 2.8V to 2.95V and in lesser efficacy LEDs, this can vary from 2.8 to 3.2V and worse. Thus this range can be 5% to 12% (150mV/3V=5%, 400mV/3.2V~=12%). This implies the input supply voltage for the string must be buck,boost or series regulated to create this voltage drop to dim the LEDs and its pass current * voltage drop of 0 to 5% or 0 to 12%V * unknown I [Amp] implies the power loss Vdrop*I=P of this regulator, which can be quite good <5% if well designed to match the string voltage +/-5 to 10%. This requires good DC-DC design principles with a desired response speed for modulated light with current sense and fast current regulated intensity. Thus the PWM ripple current amplitude and time interval and the LED epiwafer phosphor decay times all affect the dark bands found in time aperture captured images. To prevent flicker the current ripple must be reduced, as the sliding shutter capture method may alias or interfere wit the PWM dead time from phosphor decay, when PWM is used. A line filter may be used to match the current peaks used with sufficient attenuation of the PWM frequency being 20~30dB or more depending on the camera image quality desired on grey scale background.. Linear or filtered PWM is the ideal way to control LED's to prevent the issues associated with diode saturation, efficacy loss, colour shift, but adds a little cost with the LC filter depending on quality and depth (-dB@1kHz) of the filter., (similar to an AC line filter cost) There are also phosphors designed for LED photoflash that require a minimum of 50ms duration for optimal CRI, that may also reduce flicker effect with PWM. http://www1.eere.energy.gov/buildings/publications/pdfs/ssl/chowdhury_phosphors-panel_sandiego2014.pdf
H: Art Of Electronics Second Edition - Math Question - Complex Number Current Calculation How does the author obtain the expression I is approximately equal to .059sin(wt) In the beginning of the book, 30 pages back, the author stated that V was equal to Asin(wt) (if you ignore phase), why is V now equal to Acos(wt)? More importantly though, How does I = V/Z bring in sin(wt) when V nor Z had sin in it? Is sin(wt) equal to j where there is a 90 degree phase shift only? Thanks for any help. This is a real great site and hopefully I will get through Chapter 1 of this book. Which should be downhill from there (I think). I also don't calculate .059. I get .029 at best assuming w=23.14f AI: First, the author mentions that, "The phase of the voltage is arbitrary," so he's defining the source voltage waveform as a cosine wave. The current function I(t) will be referenced to this function. Next, since \$V(t) = Acos(\omega t)\$ and \$Z = -j/\omega C\$, then $$I(t) = V/Z = A \omega Ccos(\omega t) / -j = j A \omega C cos(\omega t) $$ Since a multiplication by \$ j \$ represents a 90 degree phase shift, we yield $$I(t) = A \omega C sin(\omega t) $$ which is what the book describes as well. Now for the 0.059 coefficient: just check your math. You're correct in \$ \omega = 2 \pi f\$ where \$f = 60 Hz \$, thus $$ A \omega C = 156 (2 \pi 60) 1e ^{-6} = 0.059$$
H: why is this oscillator not working? Falstad link I've taken this oscillator from an introduction book. I've more or less understood the explanation in this book for this circuit, but in practice, it doesn't work: the LED should be flashing on and off each second. AI: Since you're simulating it your transistors are perfectly matched. You actually need mismatched transistors (betas maybe 10% different) for oscillators like this to work.
H: Seemingly excess trace length reason I was looking at the traces of a computer's mother board and noticed that some traces appear excessively long and zig-zaggy(Really couldn't think of a better term). I am wondering why the designer would have added all the extra turns. Is this just a way to fill the empty spaces? Or just this have a purpose such as increasing inductance/capacitance or for EMI? I am aware that there are probably different rules for different things, but is there a semi-generic reason for this? I can't figure out what the little "bumps" in the trace would be necessary for. It should be noted that the traces in the picture connect 2 RAM sticks to the main processor. AI: This is an example of length matching (which can be necessary for both differential pairs and certain interface bus types). This is done so that a set of signals that are launched together arrive within a short time of each other (how much they can be off depends on the timing error budget available); note that no amount of length matching will guarantee perfect timing across the matched group due to various effects (even manufacturing tolerances have an effect here). Although the signals in question are highly likely to be impedance controlled that is a separate subject; the length match is specifically for timing; impedance controls are to maintain signal integrity. It is certainly true that no amount of length matching will be of use if there is a failure to properly implement SI where it may be required.
H: Can I use opamp to convert 5V DC into 10V DC? Can I use an opamp as given below to convert 5V to 10V DC? If yes then why do we go for BOOST converters? If both methods are suitable then which is better method and why? AI: Yes, that would seem to work. As drawn your op-amp has a gain of 2, so with an input voltage of 5V, the output is 10V. 10V is within the supply voltage of -15V to +15V, so no problem there. However there is a bit of confusing terminology in your question: most people would not talk of this op-amp "converting" a voltage in this way. Usually "converting X volts to Y volts" means specifically converting one voltage source intended to be used as a power supply to another voltage source. An op-amp might work for that purpose, but more common would be to go for a linear regulator like LM7812 or LM317. These are able to provide a more stable output voltage, come in packages able to dissipate more heat, and often have thermal overload protection and other such nice features. To your other question: why use boost converters? In a linear regulator, input current equals output current, but input voltage is higher than output voltage. Considering that power is the product of voltage and current, it's also the case for a linear regulator than input power is greater than output power. This excess power is converted to heat, which means it gets hot, and isn't very efficient for a battery powered device, say. If you require a large amount of current the heat may be excessive. Boost converters on the other hand are not linear circuits, and can in theory convert voltages with 100% efficiency. In practice, with real components, efficiency around 70% to 90% is achievable.
H: Choose Vcc for MOSFET driving a 48v Motor The circuit above is used to drive a 48v motor as shown. I am using: IRF3205 MOSFETs (datasheet) IR2110 High and Low side driver (datasheet) 22uF bootstrap capacitor for PWM frequency of 30 KHz. I managed to find online and here on the website the meaning of bootstrap circuitry and how it provides the extra voltage to overcome the high side MOSFET source voltage. The problem I am facing is the choice of a suitable Vcc value. I am running the motor on 4 sealed Lead-acid 12V 12Ah batteries in series for a total of 48v. The data sheet of the IR2110 shows a maximum value for Vcc =25 V and I see this makes sense as the maximum voltage that the IRF3205's low side Gate can take= 20V with respect to the Source. If I choose Vcc= 25V, which I think will already blow up the low side MOSFET's gate, then on the high side I will get total of 50V minus the voltage drop across the bootstrap diode, since the capacitors will be charged with 25V and we have Vcc=25V giving a total of 50V. This is not sufficient to turn on the Gate of the high side MOSFET since its Source will have 48V so Vgs=50-48=2V How can I achieve a Vgs>=8V with the restriction of Vcc given that I can't use an external supply other than my four 12v sealed Lead acid batteries that are in series? AI: You've got a few options to obtain a suitable voltage for driving MOSFET gates: Take +12V from the first lead-acid battery in your series-connected bank of lead-acid batteries. Even a drained lead-acid battery will supply sufficient voltage for gate driving. Use a step-up converter to make higher voltage from +5V (the VDD). With a capacitive charge pump, you can 2x the voltage. With an inductive boost, you aren't limited to only 2x. Use a step-down regulator (linear or buck) to create the suitable gate driving voltage from +48V. Either of these methods can be used to create VCC in the range of voltages that you require. source: block diagram is from p.4 of the IR2110 datasheet see also: related thread about bootstrap gate driving
H: Can I store all channels from the signal coming from a digital antenna? Whatever signal comes out of a digital antenna carries many basic broadcast channels (ABC, FOX etc.) Once this signal reaches the TV, some extraction circuit extracts (or tunes to) the channel I want to watch. If all this is true, then, why can I store the raw signal coming from the antenna on a device, and extract the desired channel(s) later as needed? AI: You can, and this is what a software defined radio does. The only requirement is that your sample rate in the recording is sufficiently high to capture the entire range of frequencies you wish to record, and that the bits per sample provide sufficient dynamic range. It should be noted that a "digital antenna" actually outputs an analog signal. In fact it's not digital in any sense, except that it's designed to receive the frequencies used for digital television. The signal does not become digital until after it's demodulated by your TV. Depending on regional details of course, the digital TV broadcast band goes from approximately 470 to 700 MHz. That means the band is 700 - 470 = 230 MHz wide, meaning a sample rate at least twice that is required: 460 MHz. Considering the typical computer sound card (and MP3s, CDs, etc) typically has a sample rate of 0.048 MHz, recording the entire TV broadcast band would require some specialized equipment, and the recording would be relatively large.
H: Triggering a Thyristor I am trying to model a thyristor DC switch but I am having problems with properly turning "ON" the SCR that is used as the switch. I am using a 2N1595 SCR and I've looked at its datasheet and it seems like the parameters I have for my gate current source and gate resistance should trigger the SCR. However, it seems the thyristor continues blocking the current in the circuit even after the gate current gets applied. How can I fix this? This is the circuit design on ORCAD Capture: This is the current/time graph of R6: This is the current/time graph of R5: AI: It looks like your voltage source is backwards for the gate drive to the SCR, the gate voltage for an SCR needs to positive, you are giving it a negative voltage. Here is a good reference for SCR's
H: Smith chart plot from simulation is opposite from experiment? I did a simulation of an antenna and replicated the antenna in an experiment however the simulation is coming out to be roughly the opposite. From the experiment I was expecting this plot to spiral from 1420 to 650 clockwise, however it's counterclockwise. I don't have a background in electrical engineering and I was wondering, what could flip the pattern like this? Could it be that my dielectric for the simulation isn't chosen correctly or could it be that the dimensions of the simulation are incorrect? Any help would be greatly appreciated! EDIT: the measurement was done with a vector network analyzer connected to the antenna with a SMA connector cable on one port. It's a two port system but the other port has a 50 ohm terminator. AI: Always refer to clock for increasing frequency and for standing wave impedances on simple antennae , it should be clockwise increasing f over the band of interest. So it is showing as it should with many standing wave resonance points closest to the centre thru an octave in frequency. Except your impedance is mismatched and circling around and only achieves perfect Gamma=1 near 0.7GHz. (for a real measurement), this assumes your cables are calibrated with a fixed R to give a perfect dot at center, which may not be true if using cheap coax instead of semi-rigid copper tubing for the test.
H: Cat 5 cable 100MHz bandwidth meaning Category 5 cable on Wikipedia: The bandwidth of category 5 and 5e is the same (100 MHz) and the physical cable construction is the same, and the reality is that most Cat 5 cables meet Cat 5e specifications, though it is not tested or certified as such. So what does cable bandwidth mean? is it correct to say that cat 5 cable frequency response is almost flat in a region with 100Mhz width and 3-dB attenuation? Something like this with B=100Mhz: If correct, what is f0 frequency for cat 5 cable? And how the cat5 max speed, i.e. 100Mbps, is related to 100MHz bandwidth? If we want to send digital signal with some simple line code such as NRZ, what will be the duration of a bit (bit time) in milliseconds? AI: Basically the quote you have shown was written by someone who either didn't understand what they were talking about, or oversimplified it. The bandwidth of the cable is a result of the resistance of the cable and the fact that it is capacitive. These act like an R-C low pass filter limiting the bandwidth. Additionally the distributed inductance and capacitance of the cable are frequency dependent so have a more complex affect on the bandwidth. The "insertion loss" which is a measure of the gain of the cable is dependant on both frequency, but also on length. The longer the cable, the more lossy it is. Here is one example of the insertion loss of a CAT5e cable, this for a 100m length: Image Source Here we see that at 100MHz, the loss is actually more than 20dB - a lot more than the 3dB (power gain) or 6dB (voltage gain) point that would be used to specify the bandwidth of a first order low-pass filter as the quote implies. Further up the Wikipedia page more accurately explains where the 100MHz figure comes from: The specification for category 5 cable was defined in ANSI/TIA/EIA-568-A, with clarification in TSB-95. These documents specify performance characteristics and test requirements for frequencies up to 100 MHz Basically, the requirements in terms of insertion loss that must be met for a cable to be classified as Cat5 are only specified up to 100MHz. Beyond this point the specifications of the cable are undefined by the standard - though manufacturers may well provide data at the higher frequencies. In terms of how the Ethernet speed relates to frequency, that is a touch more complex. The raw data is encoded using various schemes which are described in brief in the wiki page here. For 100BASE-TX, several encoding schemes are used which result in a maximum fundamental frequency of 31.25MHz through the cable itself.
H: Uses of a silicon semiconductor and why its resistance at different temperatures We conducted an experiment in which we found the variation in the resistance of a fairly pure silicon sample between temperatures of about 400K to 600K, and we found a value for the energy gap of silicon of our sample. We were comparing the resistance variation with the model: \$ R=R_0e^{\frac{E_g}{2k_BT}} \$ Firstly, does anyone know what the name of this model is? I have spent several hours trying to find out about this model (unfortunately it is not possible for me to ask my practical demonstartor) to no avail. This model also predicts that this relationship will only be followed when the temperature of the semiconductor is sufficiently high- within the 'intrinsic region'. However all of my searches about the 'intrinsic region' of a semiconductor simply come back with intrinsic and extrinsic semiconductors. Nothing about the 'intrinsic region'. Also, I am trying to figure out why such information may be important. I know silicon as a semiconductor has many applications for example in detecting the temperature, however I have not come across any examples of devices that would use the properties of silicon at such a high temperature. AI: It's the energy-band model. The idea is materials have separated conduction and valence bands in which electrons may reside. Eg is the size of the gap between these bands. In metals, the gap is zero, the bands overlap. In insulators, the gap is so huge that it is impossible for electrons to reach the valence band (temperature >>1000K needed). Semiconductors are inbetween, the gap is a few eV (Si: 1eV@300K). In think this "intrinsic region" should just mean "temperature-depending conduction" in contrast to the conduction introduced by doping atoms. The latter change the energy niveaus of the bands and the size of the band gap. When the temperature is sufficiently low, there are other effects which govern the resistance measured instead (meaning: it's still there but you cannot measure it independently.)
H: Deciding on lithium battery chemistry, lithium ion (protected) or Lithium Iron Phosphate (no protection) Safety is critical I am designing a product requiring high power and energy storage capability. I have narrowed down the selection to two (most cost effective) systems: a 3s system of protected 18650 lithium ion cells (11.1V), or a 4s system of unprotected 26650 lithium iron phosphate cells (12.8V) Both systems would be charged with a stand alone battery charging IC, with a thermister safety mechanism. I am also planning on including a fuse that will disable the system with excessive current draw (10A) in the case of the unprotected LiFePO4 cells. Which system would offer a better safety profile for the use with a consumer product? AI: Let's start with a very important point: With Lithium based batteries a thermistor is a nice last line of defence, but you seem to be making it the centre point of your thoughts. If your thermistor triggers on a Lithium based battery, purely from charging, you might want to hook that up to an airhorn so you know to start running. Do not get me wrong: I fully support having a thermistor in everything that's to do with bucket-loads of energy. But with Lithium based batteries, that's not the be-all and end-all of safety it was with NiCd/NiMH. Hyperbole aside, if a Lithium based battery is hot enough to trigger a charging chip's thermal limits, your batteries are already being destroyed. If you want a safe consumer product, regardless the type of lithium, you need the appropriate charging scheme. If you want it to be a quality product, just assembling a pack from protected cells is going to get you into trouble. While individual cells will likely not get damaged during its operational life, the operational life will probably be limited. When an individual cell goes into protection it will open up the chain, where the other batteries are still "active". There's a number of scenarios when a manufacturer uses single-cell-spec'ed MOSFETs in the protection that will not turn out great for your batteries or their service life. So, if you want to be taking yourself seriously, you need to build the pack properly, with the appropriate protection board. If you want to be actually safe, with a separate pack wired into electronics, this protection needs to be inside the pack to prevent damaged wires from causing short-currents. This putting either choice on level ground, since protection boards can be had for your voltages for either chemistry, you need to weigh the advantages and disadvantages of each chemistry against your individual needs. LiFePO4 will be much safer from self-combustion, even when hard-shorted, but not holy! They are still a bear you should not be intentionally poking. Although there's videos out there showing the pneumatic insertion of a nine-inch nail into well built, branded LiFePO4 cells with no fallout. LiFePO4 also is much more resistant to formation of so-called dendrites on its lithium electrode, causing it to be less sensitive to short term or low-grade over-voltage. On the other hand, LiFePO4 in most cases has no more than 3C discharge allowance and is usually advised at 0.5C charge currents. Of course, in Lithium-CobaltOxide based batteries ("Li-Ion") the very high discharge types will always still have significantly reduced life at these currents. LiFePO4 is more expensive (at the moment) and has a lower storage density (for now) and identifying the right manufacturer in the more affordable cells can still be difficult. The fact that destroying such a cell is harder to begin with makes that even harder when samples are sent. That is, the grey area is much thinner, and if you test samples it's that grey area where your data points live. However, the reason I stock shitloads of quality LiFePO4 and that they are used in high-end automotive, is that they have clear advantages, even if you need a much higher capacity to get the same high currents; They have a lower self-discharge rate They are a little less affected by residing in high or low charge states They can be safely discharged below freezing (a major point for automotive) They contain a somewhat reasonable amount of remaining energy below freezing, 45~80% for LiFePO4, versus 0~35% for most LiIon. Where the exact number depends on the temperature. They have a relatively level discharge curve for the majority of the energy content. Minor: Their voltage levels are compatible with many ARM type controllers. All that said, I see you mention balancing nowhere. This also is less of a task with LiFePO4, but some plan for keeping the cells balanced should still be considered. Whether you bottom-balance, top-balance, full-balance, or active-balance that's your choice. For either chemistry. Or not balance at all, but again, usable life will be reduced noticeably.
H: Fluorescent tubes in series? Is it possible to use fluorescent tubes wired in series, provided that the ballast can supply a high enough voltage and power? Problem description F28T5 are too long (4ft) for me, so instead I plan to use pairs of F14T5 side by side, which are 2ft long. The problem is this would require more ballasts. Typically, one ballast can power one or two lamps, like so But I would like to use the ballast like this My theory is that 2 x F14T5 would be electrically equivalent to one F28T5, if wired in series. The ballast will provide the appropriate current, which would be almost the same. Is this correct? or would the plasma have very different characteristics with an intermediate electrode? Starting issues? Another potential problem is starting the lamp. My understanding is that the plasma in the tube must first be ionized, typically by heating a filament at both ends. But with lamps in series, the middle two filaments would be cold since they are not connected to the ballast. Would this prevent the lamp from starting? or does it only need to be heated from one end? (Note that with two lamps in series, each lamp has one heated filament. But with three or more, the middle lamps will be cold. Would it also work with three or more lamps?) Would using a different type of ballast (instant start) that uses a high voltage to induce breakdown avoid this problem? AI: The lights are gas discharge tubes, so have a struck voltage usually between 60-80 V. So on 110 V mains you won't have enough to keep them alight in series, but at 220-240 V you would have enough voltage. If you have a simple inductor/choke limiter, and glow starters this may be your solution: With electronic ballasts it will depend on the connection complexity they support, but there are plenty that support 2, 3 and 4 tubes with heater activation and no starters. Here's a pointer to a 2 tube electronic ballast.

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