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H: Why does 3-to-5 volt with transistor not work the way it should? A quick google for "3v to 5v transistor" gives me dozens of circuit diagrams that look like option A below with a NPN transistor and a pull-up resistor. If I build the diagram in iCircult, it works as expected. However, when I wire it up on a breadboard, it doesn't work quite the way it should. Without any power to the base, I get Vout = 4.95v. With power to the base I get Vout = 0v so it is inverting my wave. A few experiments later, I solve my problem using a PNP transistor and a pull-down resistor (option B). In both cases I am taking Vout from the Collector side of the transistor. It's a pretty simple circuit, you'd think it would just work. I double and triple checked the connections. I have also posted a photo of the breadboard below. I always seem to have weird problems like this when working with transistors, so now I am sort of on a quest to understand why option A doesn't work like it should for me? AI: Both circuits invert your input signal. The second one will not work properly. If you want a non-inverting level shifter you can use this circuit: simulate this circuit – Schematic created using CircuitLab When the input is close to 3.3V, the transistor is off and R2 pulls the output up to 5V. When the input is close to ground, the transistor is saturated and the output is equal to the input plus maybe 50mV.
H: Analog Multiplexer In what ways can an analog multiplexer i.c fail? I want to use this in our system ,so i need the complete details of it,so that even if the mux fails,the output should connect to one of the input signals. Is this possible? AI: It's likely not possible. The most common failure modes for ICs in my experience are for a pin to fail open-circuit, or short-circuited to either power or ground. For an IC to fail with one I/O pin connected to another one is very unlikely. To get the behavior you want, you'd have to design an ancillary circuit that detects a failure condition (it would have to detect all possible failure conditions) and activates a second multiplexer to select input from the fail-over input instead of from the first mux. And then you'd have to worry about failure modes of that mux...and so on.
H: Working of a current mirror as an active load Ordinary resistor as load in a transistor amplifier has a certain short coming…we cant have voltage and current gain simultaneously ….. In this circuit a current mirror is actings as an active load in place of RL …I wonder how this current mirror able to give voltage as well as current gain at same time?? AI: With a load resistor, the current through the resistor obeys ohm's law and varies with the voltage. However, an active current mirror allows a fixed current, even with large variations in voltage. This basically makes the amplifier work more effectively by presenting a much higher impedance. This high impedance increases the gain.
H: Dipole antenna VS. coil for RF transmission or recieving Correct me if I am wrong, but all RF light waves are created via a changing electric and magnetic field. But when looking at all kinds of radios and RF circuits, this light wave is always made with a simple wire antenna. Why are coils not used instead? Wouldn't a coil of the same length create a stronger magnetic field thereby creating a more efficient 'antenna' that produces RF waves? I know that there has to be a reason why coils are not used, but I cannot find an explanation. It makes more sense to me to have a coil creating a fluxuating electric/ magnetic field rather than a piece of wire. Thanks in advance! AI: Take a look at this picture of a dipole antenna: - Both electric AND magnetic fields are produced by the antenna. Magnetic fields are produced by straight wires with current flowing in them i.e. the wire doesn't need to be wound into a coil to produce magnetism. Now, the physics (well, a bit of it): - Space has an impedance of approximately 377 ohms and an antenna has to produce an E field in the right proportion to the H (magnetic) field to maximize convsersion of electrical energy flowing to the antenna into EM power. The ratio of E field to H field is 377 ohms so trying to produce a bigger H field than is necessary is a waste of time because the impedance will be wrong. See this wiki article for extra reading.
H: Is this true: "At over 700 Hz, current simply flows over your body"? I have heard this in a movie clip. I was just curious to know, is this really true? Because the one thing that I do know, is that at a high voltage, the current decreases, so that it doesn't harm a human body. AI: Skin Depth The human body does have a "skin effect" but it's not as thin as you might think. Electric currents are confined to the outside of a conducting body, but humans are not very conductive, so the fields penetrate quite deep. The best example that comes to mind is 2.45 GHz - we all know that a microwave oven cooks about 2 or 3 cm into a piece of meat - this penetration depth is closely related to the skin depth. The primary reason that you don't feel high frequency current is that the nerves and cells can't respond to anything above ?about? 1 kHz. I've discussed this in a previous answer, more about the safety aspects than the skin effect itself, but it might help. Nerve effects are the primary cause of injury due to electricity, mainly the heart of course. If the frequency is high enough that it can't influence the nerves, then all you have to worry about is the heating effect. For a potentially lethal 100 V at 20 mA, only 2 W is dissipated in the body, which is insignificant compared to the 200 W of normal body heat (though it will be concentrated at the entry and exit points). So at high frequencies you can carry a much higher current than would be lethal at low frequencies, possibly without pain or injury. High voltage and lower current It's not true that the current is lower at high voltage. In fact, a higher voltage will usually cause a larger current to flow, than a low voltage. High voltage overhead transmission lines might be 400 kV but they also carry hundreds of amps. When it comes to human safety, higher voltage are almost always more dangerous.
H: What's this substance looking like glass cloth in an electrolytic cap? This epic image of a blown electrolytic cap is from this question. It looks like the cap is made of something that looks like glass fiber cloth inside. What is that material that the inside of the cap is made of? AI: It's paper or something much like paper (Edit: see below). Used to separate the aluminum foil layers. It is normally soaked in (very polar) electrolyte so it acts as a conductor connected to the cathode. The actual dielectric layer is a very thin aluminum oxide coating on the anode plate. See this description from Nichicon entitled General Description of Aluminum Electrolytic Capacitors. The plates are coiled up inside the can with the paper separating them, in order to get a lot of area and to use both sides of most of the plates. C = \$\frac{\epsilon_r \epsilon_0 A}{t}\$ where t is the thickness of the oxide layer, A is the total area of the plates and \$\epsilon_r\$ and \$ \epsilon_0\$ are the relative permittivity of the oxide and the permittivity of free space respectively. Edit: Low-ESR capacitors tend to use other separators than paper because the lower resistance electrolyte is more corrosive. Glass cloth is ones such material. From US4321076: Low ESR capacitors are preferably made using a woven glass cloth or Celgard 3500 microporous polypropylene film, manufactured by Celanese Corp., as these are not attacked by the electrolyte. Other materials which can be used include porous woven polyolefin cloth, porous perfluoroethylene cloth, polyolefin spun bonded fiber paper, and polypropylene screening. (That's the simple description, in reality both plates have oxide layers, but the anode layer is thicker on a polarized aluminum electrolytic capacitor).
H: Is there a correct way to document RoHS compliance? I work for a small electrical engineering company. Our products are not presently RoHS, but we're planning to switch over. As best I can tell, this is a matter of ensuring that our individual components are RoHS, and using lead-free solder. But what documents do we need to keep on hand? Do we need a file of letters from each component manufacturer? Or is it enough to glance at Digikey and see a green leaf next to the component name? AI: There is a great booklet here that details manufacturer's obligations. Here is an extract of the latest obligations: - The booklet is called: - RESTRICTION OF HAZARDOUS SUBSTANCES (RoHS) REGULATIONS 2012 See also this UK government website
H: Aux and Microphone to same input I have a loudspeaker for my car and I've added an aux cord to the system so that I can play music over it. There is also a microphone attached to it and what I've done is add a 3 way switch between the two so I can either use the microphone or the aux cable. I don't really like this setup and I want to be able to use both at the same time but I'm a software engineer, not electrical, so I'm not sure if tying the voltage of the aux and microphone to the same input is safe. Here is a diagram with what I currently have: simulate this circuit – Schematic created using CircuitLab And I want to remove the switch: simulate this circuit My roommate suggested adding diode to ensure the signal only goes one way but said that would lose about .7 volts. I don't like that as the input isn't that strong already and I have to manually amplify music in audacity so that you can hear it. While I'm asking a question, does anyone know how I would go about amplifying these signals through hardware because the microphone you really have to scream into and the aux is generally pretty soft compared to the tones/sirens it can generate. Unfortunately I don't have the schematics to the circuit board which is probably needed for something like that. Here are some pictures of the circuit board/system: Circuit http://zyphox.com/misc/pa1.jpg Front Panel http://zyphox.com/misc/pa2.jpg AI: In this case, there is only one microphone input configuration - the one that is hardwired to the unit. If you can measure the peak to peak voltage of the microphone, you can work out how different it is to the aux input (1VPP). Play a loud sine tone (low freq) into the microphone (as loud an input as is likely to be shouted into it), then measure it with the AC setting on a multimeter. You might be able to use a simple potential divider to match the signal levels and then two series resistors to merge them together. Failing that, rebuffer them with an opamp. Using series resistors will help to reduce one driving the other. If this were DC, diodes would work but as Techydude mentioned - music is AC, and using didoes will absolutely kill the signal. Does the mic have a push-to-talk button on it? If not, and the mic is constantly picking up it might end up picking up feedback from the aux input. EDIT: Measuring AC volts with a multimeter will give you RMS. You need to convert it to peak to peak to make a real comparison. Google "RMS to peak calculator".
H: What makes an arbitrary waveform generator arbitrary? Why are some signal or waveform generators called "Arbitrary Waveform Generators"? Am I misunderstanding the word arbitrary? This question is not necessarily about the word arbitrary but more about arbitrary signals. How is a arbitrary waveform generator different from a plain old waveform generator? AI: An AWG is more like a soundcard on steroids, while your POWG contains circuitry for generation of a classical set of waveforms. The lines are a tiny bit blurry since there are DDS waveform generators that only do a standard configurable set. Basically for the classical waveform generators and the clunky pushbottons for waveforms (square, sawtooth, sine etc.) you activated a distinct circuit that was responsible for the type of waveform. In the modern ones, you use DDS (direct digital synthesizers) that are highspeed DACs disciplined by an accurate reference clock. Since they are more or less fed with a waveform (lots of them can take .wav or even .mp3 files and play them back at desired frequencies) the waveforms they can produce are called arbitrary.
H: How to convert 10K Pot raw output ADC data to voltage and resistance from arduino uno? I have a simple and basic question. I am using Arduino uno to read the output from a 10K potentiometer. I understand that Uno uses a 10 bit ADC, therefore it can display outputs from 0 as 0V to 1024 as 5V. Question: I am getting these values from the pot and displayed on the serial monitor of arduino IDE, while I vary the resistance of the 10K Pot. How am I supposed to calculate or know the output voltage and resistance corresponding to the values from 0 to 1024? Another question is, when I vary the POT's knob to a maximum point i.e. 5V, the output displayed is 1007 or 1008. Isn't it supposed to output 1024 for 5V? AI: A converted value of 1024 would correspond to 5v - or to whatever voltage you put on the chip's Vref pin. But with 10 bits, the maximum number the converter can deliver is 1023. It's a small difference, but keep it in mind. To find the voltage corresponding to a the converted value: V = (value / 1024) * Vref. Thus a reading of 1008 corresponds to 4.92v, assuming 5.00 volts at Vref.
H: Why PNP transistor might be opened even though (almost) nothing is connected to the base? I have a simple printed circuit board with simple circuit on it. Very beginner. No SMD. As a part of this circuit I have transistor BC557. If I don't solder the base pin of this transistor and test the circuit then transistor is closed (as expected) but when I solder it to a path on the board that has nothing else soldered to it (there were two resistors but I desoldered them) then the transistor is magically open now. Transistor operates on voltages around 16V (on emitter) and on very small currents. The only load between collector and the mass are in parralel: 100k resistor and MOSFET gate-source. Can printed circuit board be leaky in a way? Somehow allowing some current to flow from the base that's not connected to any other component? Adding 1uF capacitor between base path and emitter path on PCB doesn't seem to help. I noticed that if base is left unsoldered and I touch it with a screwdriver I'm touching with a finger then the transistor opens sometimes, even if I don't touch anything else in the circuit at the same time. Can I somehow make it less sensitive? This behavior didn't occur when I had same circuit powered by the same battery and tested with the same load but on breadboard. The circuit I'm trying to build is that one: http://serwis.avt.pl/manuals/AVT746.pdf And it worked on lower voltages (less than 9V). I tried putting 100k resistor between emitter and +16V but it doesn't seem to help either, transistor still is open when base is soldered to unconnected path (but it caused MOSFET to overheat, even though the sorce-drain current is just 2A). I also swapped the transistor for another one (same) to rule out damaged element. BTW This circuit cuts the battery power to the receiver on the right after a delay roughly controlled by C2. It turns it back on after pressing the switch. Cool thing about this one is that it can turn on an off large currents (in theory up to 30A) and work with voltages from 6 to 16V. I replaced switch with transoptor (+480 Ohm resistor in series) so I can control this circuit with another one. AI: Could be leaky, could be pickup of AC hum from the mains or RF from AM radio etc. as the transistor will act as a detector. To tell the difference, put a 1uF ceramic (low leakage) capacitor from base to emitter and see if the issue disappears. In any case, you can reduce the sensitivity simply by connecting a relatively high-value resistor from emitter to base. Something in the range 10K-100K will likely work for you. Edit: One thing that can cause a great deal of leakage is the use of acid-core (plumber's) solder. If this is accidentally used, clean the board with warm soap and water, preferably followed by isopropanol alcohol. I have seen mains circuits burn up as a result of this. As an aside, the solder used in plumbing was often 50:50 PbSn, so it would not solder as nicely (and will require much higher temperatures) compared to good electronic grade 63:37 solder.
H: Please explain FFT Frequency Spectrum and different filters It seems that I am not able to find the right key-words in order to get the answers from google but how can I recognize different filters based on their FFT Frequency Spectrum and how do different filter look like in their FFT Frequency Spectrum? I understand that the FFT represents a signal as a spectrum of its frequencies but I don't really know how to interpret a Frequency Spectrum plot of a FFT. I don't know how different filters would look like such as high-pass, low-pass and band-pass filters. Could somebody explain and/or link me to some resources that could explain that to me? Please understand that I don't need a deep mathematical explanation I just need to be able to interpret and distinguish different FFT FS plots. Thank you! For example: AI: I will try and explain things as simply as I can. The frequency spectrum shows how the filter looks in the frequency domain.That filter is the one which is multiplied (in the frequency domain) with your input signal so you can tell what type of filter your filter is by looking at the frequencies you filter stops e.g a filter with all zeros will stop everything and one with all ones will let all frequencies pass. For example in you image of the plot of \$H_1[k]\$, look at the frequencies greater than 0 (ignore the negative part of the graph for now - it is just a reflection of the positive side), from the graph you can see that low frequencies in the filter are stopped and higher frequencies are let through, therefore that is a high pass filter. In the second image (again ignoring the negative parts), you can see that very low frequencies and very high frequencies are let through and only a small band of frequencies is stopped.That filter is therefore a band-stop filter.
H: Controlling an electric water heater I'm trying to control the temperature of water heated electrically. On devices sold commercially I see that temperature is not being controlled, and the heating elements are either turned on or off (based on the feedback from some thermostats), probably because it is expected that the hot water will be mixed with cold water afterwards. For a project I need to control the temperature of the water by controlling a heating element, using a power around 5KW - 10KW, and I guess I must do that by switching the power using some kind of SCR or Triac. What would be the right approach to do this? Any help that leads me in the right direction will be very helpful. Thanks in advance. AI: The controller of a water heater usually has a high hysteresis, keeping the temperature within the range of a few °C. If you need more precision, you can build something on your own. It is simple to switch the power of your heater on or off. Maybe, you have a look at this application note about zero crossing switches. These devices contain an opto-coupler to insulate your circuit from the mains and a control unit which switches the AC mains during its next zero crossing. The benefits are less EMI and less power dissipation, and of course you do not have to do much on the mains side. On the other side, this devices switches the current for entire half-waves only, so if you want a "duty-cycle" of 1%, you need to switch one half-wave of 100, which results in a period of 1 second (for 50Hz mains). But this is not such a problem in your case. The document also contains an example circuit for temperature control, which is a good starting point BUT a water heater contains some volume of water, and it takes some time until the heat from the heater spreads into the volume. This can lead to heavy oscillations of your temperature. The built-in thermostat is so slow that this does not happen. For a well-regulated temperature you will need some kind of PID-controller which controls the heater based on current and previous temperature measurements. While a PID can be designed in analog electronics, a microcontroller can be used as well, and it may be preferred by you, as you have a high reputation at stackoverflow... Implementing a PID in software is easy, the hard part is finding parameters for which it regulates your temperature as fast and precise as possible, without overshooting / oscillations. Another point: Depending on the size of your water reservoir, the water may form levels of different temperature. Maybe, you think about a way to circulate the water inside the reservoir for a more homogenous / faster temperature distribution. By the way: You say 5-10KW, which is quite much and usually not taken from a single phase, but from a three-phase system, and you need to switch the three phases.
H: How serial resistors actually reduce EMI? I am recently working with a GSM based system, and there was this advice in the datasheet of the GSM module: 22 Ω resistors should be connected in series between the module and the SIM card so as to suppress the EMI spurious transmission and enhance the ESD protection. I tried to do a little search and I have found a document, PCB Design Guidelines For Reduced EMI, it has similar statement in it, but no explanation. Put a 50 –100 Ω resistor in series with every output pin, and 35 –50 Ω resistor on every input pin. An other part says: (Series Termination, Transmission Lines) Series resistance is an inexpensive solution to termination and ringing problems, and is the preferred method for microcomputer-based systems where minimizing the differential-mode noise is also a concern. One more possibly related part: Impedance Matching at Inputs and again, the series resistance is the most likely solution. Resistance placed at the driver increases the output impedance, as seen by the trace and the input pin, thus matching the high impedance of the input I have found something in this document as well, Understanding Radiated EMI it says: Add series resistance? May help - Less current (good and bad current) flows through high impedance - May reduce EMI by reducing currents flowing outside IC All in all, I need a little clarification about the topic, so my question is: How serial resistors actually reduce EMI and what is the principle? AI: Well one it slows down the rise time of your signal which reduces the high frequency content. So that could help you if you didn't need your edges to be that fast. It also lowers the current flowing through the trace and back through its return path which would lower the strength of the field created around it (and radiated out). I would add that I've often seen this added lazily, or perhaps by the "emi group" after the design is done. It's fine if you didn't need the edge rates but if you're trying to go fast slowing down your edges or destroying your eye may not be what you want.
H: Why would MOSFET transistor heat up? I had a circuit on PCB. I used solder with acid-core to solder it. This combined with my mess soldering caused various shorts between components. One of those components was MOSFET transistor BUZ11 which due to shorts heated up to the point of slight discoloring of radiator and giving a little smoke before I shut off the power. I wonder what could cause this heat up. Current that was fed through circuit was about 2A. Well within range of this component. Voltage was 15V. There were no oscilators, no inductors, just two capacitors. So the question is: How to heat up MOSFET transistor without passing through it (drain-source) more current than it can handle or applying higher voltage than it can handle. AI: If you look at the data sheet https://www.fairchildsemi.com/datasheets/BU/BUZ11.pdf you'll see the thermal resistance to ambient, which for a non-heatsunk TO-220 (such as you are using) is in the vicinity of 75 deg/W, if the transistor is dissipating 2 watts the chip will get to ~175 C (2 x 75 + 25 ambient), which is more than enough to do damage. If you were pulling 2 amps, a 1-volt drop on the transistor will produce this much power. If you look at the Output Characteristics (fig 5) this suggests a Vgs on the order of 3 volts. As ijmilburn has pointed out, you need more than this to turn the FET fully on.
H: Phasor diagrams for RLC series circuit? So, here's the circuit: How can I make phasor diagrams for the circuit in case when $$I_C<I_L$$ and $$I_C>I_L$$ Edit: What made me confused is the current conditions, it's actually voltage $$U_C<U_L$$ and $$U_C>U_L$$ AI: What @EMFields said is correct, i think you are getting confused between the current through each element and the values of each element. Because they are in series the current through each element will be equal. But the Value of each element will affect that very current.First convert the inductance and capacitance into reactances, and find the overall impedance. $$X_l=\omega L$$ $$X_c = \frac{1}{\omega C}$$ $$Z = R+j(X_l-Xc)$$ The sign of the imaginary part will tell you if the circuit is predominantly inductive or predominantly capacitive . And hence if the current will lag the voltage or lead the voltage respectively.
H: Current Shunt Vs Current Clamp Meter - 15% difference Which one is right? I have a variable AC power supply. 0-12V 0-1800A 60hz. I have always been told to trust the current shunts and not current clamps when measuring current. The current clamps are supposed to accurate to less then 2%. Our current clamps have always been about 5-15% different then the current shunts though. I want to know why, and which one I should believe. My boss wants me to prove to him that the shunts are more accurate than the clamps but I don't know where to begin. All items are calibrated annually. Fluke 353 (1.5% accurate) Ideal 61-746 (1.7% accurate) Current shunt 500A 50mV (10:1 ratio) (Empro) Current shunt 3000A 50mV (60:1 ratio) (Ram Meters) Current shunt 2000A 50mV (40:1 ratio) (Empro) (EDIT: Added) mV meters for the shunts are fluke 45 and Fluke 289 both read the same I adjusted the power supply based on the current shunts. 500A shunt - first Iteration Shunt Fluke Ideal Target mV Current 50 5.5 55 48 49 13% 11% 100 10.5 105 91 92 13% 12% 200 20.1 201 178 182 11% 9% 300 30.1 301 263 269 13% 11% 3000A shunt - first Iteration Shunt Fluke Ideal Target mV Current 100 1.716 102.96 98.3 100.3 5% 3% 200 3.345 200.7 190.7 194.7 5% 3% 300 5.023 301.38 284.8 292.1 6% 3% 400 6.695 401.7 379.4 389.1 6% 3% 500 8.356 501.36 471 484 6% 3% 500A shunt - second Iteration Shunt Fluke Ideal Target mV Current 50 5.125 51.25 45.3 46.5 12% 9% 100 10.304 103.04 91.9 93.5 11% 9% 200 20.3 203 181.2 185.1 11% 9% 300 30.225 302.25 269.3 275.8 11% 9% 400 40.21 402.1 357 366.2 11% 9% 2000A shunt - first Iteration (Fluke 45) Shunt Fluke Ideal Target mV Current 50 1.21 48.4 37.17 38 23% 21% 100 2.44 97.6 84.7 86.4 13% 11% 200 5.02 200.8 181.9 186.3 9% 7% 300 6.31 252.4 230.5 235.8 9% 7% 400 10.09 403.6 360.2 372.2 11% 8% 500 11.31 452.4 405 415 10% 8% 2000A shunt - second Iteration (Fluke 45) Shunt Fluke Ideal Target mV Current 50 1.26 50.4 29.33 29.5 42% 41% 100 2.48 99.2 65.7 65.5 34% 34% 200 5.14 205.6 141.3 144.1 31% 30% 300 7.46 298.4 207 211.5 31% 29% 400 9.89 395.6 284.4 290.6 28% 27% 500 12.5 500 355.3 362.6 29% 27% 2000A shunt - third Iteration (Fluke 289) Shunt Fluke Ideal Target mV Current 50 1.21 48.4 37.2 37.4 23% 23% 100 2.51 100.4 76.6 77.3 24% 23% 200 5.14 205.6 165.5 168.8 20% 18% 300 7.58 303.2 252.6 257.8 17% 15% 400 10.11 404.4 344.6 351.7 15% 13% 500 12.49 499.6 418 426 16% 15% EDIT: After I calculated the error rates it looks my 500A is off more than the 3kA. I am going to try it with a 3rd shunt tomorrow. EDIT2: 200A 50mV Shunt Fluke 353 Clamp Meter Fluke 45 AI: If you do the math, the Fluke 353 and Ideal 61-746 are within 2.2% error (STD 0.4%). This is well within the level of accuracy of the machines given (1.5% and 1.7%) with the Ideal always larger than the Fluke. For me, this correlation means they are the most accurate. The current shunt is a manganin resistor. 100µΩ, 25W. Online references state accuracies of ±0.25% (which is down from your 2%). This should be the most accurate according to workplace lore. If you look at the errors for both the 500A and 3000A, they start high (10%, 3%, 2.5%) and go low (<0.5%). This does make sense because it is a resistor and even though temperature coefficient is 0.00001, it will be most accurate at rated values. Difference between 500A current shunt and Fluke/Ideal ranges from 9-13% lower (3000A from 3-6%). This tells me there is a systemic error. The Fluke 45 and Fluke 289 give the same answer. I'd look at how there are connected to the current shunt. We are just dealing with 50mV. Thicker, shorter wires possibly? (Not even sure how it is connected.) Your problem is you have three answers and you don't know which is correct. Two agree, but the most accurate (in principle) is always higher. You need another reference which can be verified some way. I always try to go back to basics. I'd go borrow a calorimeter and boil some water. Edit... From Calibrating DC Current Shunts: Techniques and Uncertainties The five error sources inherent to current shunts are: 1) Connection 2) Temperature 3) Frequency 4) Drift 5) Thermal emf And... Most modern metrology-grade shunt manufacturers are aware of these problems and have attempted to design them out of their products. Figure 5 shows a metering type shunt highly susceptible to current and potential connection errors I substitute your image because his was similar. The resistance of shunt Manganin rises about 20 ppm (0.002 %) per degree C around lab ambient. Applying current causes self-heating, which changes resistance. This change is not linear. Some shunts rise to a maximum resistance at a certain current / temperature level, then fall as temperature continues to rise. How fast do you make your measurements? A shunt must stabilize at each temperature / current level. The thermal mass of a shunt includes its resistance element, its end blocks, the current cable lugs and connection hardware, and the cables themselves. At higher current levels, a shunt may require more than an hour to reach thermal equilibrium. This is when measurement should begin. Not that they say it's not significant for 50/60 Hz. But you could try your measurements with shielded twisted pair. Figure 8 shows ac coupling between current and potential circuits. Coupling can be reduced by connecting current leads in line with the shunt and routing potential leads together into a shielded, twisted pair extending at right angles from the shunt (green, not red). Would this account for 9-13% error always in the same direction? The jury is out, but I'd say yes. It does give you things you can try. I'd pour over that report and you should be able to easily prove that the clamp meters are the most accurate.
H: Increasing Frequency of a Signal I was wondering if it would be possible to increase the frequency of a signal by a fixed amount. Basically, I have a negative wire soldered to a speaker and a positive wire that very quickly breaks/unbreaks the circuit. As expected, this causes the speaker to pulse each time the positive wire breaks and unbreaks. However, the frequency is very low and I was wondering if it would be possible to have a circuit that "adds some frequency" to the current frequency so that I would get a continuous beeping instead of some pulsing with noticeable space in between. The circuit would have to be pretty simple since I'm making it myself. Is this feasible or should I just be happy with what I have? AI: A mixer might be what you're looking for. A mixer multiplies two signals together. \$x_{out}(t) = x_1(t) \times x_2(t)\$ if \$x_1\$ and \$x_2\$ are two input signals. From the multiplicative property of the Fourier transform, this means that in frequency space, the output contains sum and difference frequencies of the input. Say \$x_1(t)\$ is the signa produced by you moving your wire around. And \$x_2(t)\$ is a sine wave at some frequency \$f_0\$. Then if you put these two signals into a mixer, the output will be your \$x_1\$ signal shifted up in frequency by \$f_0\$. There is also a shifted version of the negative-frequency part of your input signal, but that might not be important or it might be what you actually want, depending what you're trying to do. You can also see that if \$x_1(t)\$ is a square wave that's sometimes 1 V and sometimes 0 V, and you multiply it by a sine wave, you get out a pulsed sine wave, or a series of beeps, if \$f_0\$ is an audio frequency.
H: How can one verify if the FCC/CE certification mark on an electronic component or product is genuine? Some companies can be unscrupulous and simply paste a label to fake FCC CE. How can one be sure the label is real? Is there a website for checking the validity of the FCC CE mark? AI: Search the FCC database by their stated FCC ID code: http://transition.fcc.gov/oet/ea/fccid/ The first 3 or 5 characters is the Grantee Code (assigned to a company). There isn't an equivalent for CE.
H: 2 NPN Transistor Square Wave Oscillator? I have built the circuit found here. I substituted with 22K resistors at R2 and R3, and substituted C1 and C2 with .022uF capacitors. These are all the components I have lying around. Everything else is the same. Even the transistor part number. I realize that these substitutions will affect the frequency of the square wave. However, scrolling through every possible time interval I don't see a square wave. I see a sawtooth wave. I don't quite understand how this circuit works, so it was my intention to build it, then dissect exactly how it works, but I can't seem to even get it working correctly. Current Waveform: I'm unable to dissect the circuit and see exactly how it works, therefore I can't determine why I'm getting a sawtooth wave instead of a square wave. I'm sure that it has something to do with the substitutions but the reason this effects the waveform is my interest. My time interval is at 20uS. The period of the wave is very nearly 18uS. That makes the frequency of the wave 55.555kHz? That can't be right, can it? It also seems that what I have done by accident could be used on purpose to integrate a square wave into a sawtooth wave? AI: The reason for the sawtooth shape at the collectors of the transistors is caused by the B-E junction of the transistors and C1 / C2. You can insert resistors in series with the base of each transistor - I'd start with 10k and see what happens. Be sure that the resistors are in series directly at the base of each transistor. That should give you a much more square-looking waveform at the transistor collectors. The reason the sawtooth waveform happens is that the E-B junction of the transistors looks like a very-low impedance when the capacitors are charging via the 1k collector resistors. Think of it this way: the capacitors are almost in parallel with the collector E-C leads. There is a 0.6V region when the capacitor is beginning to charge that is high-impedance but as soon as the E-B voltage reaches 0.6V, the impedance drops to a very low value.
H: LTSpice capacitor ciruit behaviour doubt? I have basic knowledge of electronics and i have been using LTSpice to clear my concepts lately. I also have breadboard, multimeter etc. The first circuit i tested is a simple capacitor resistor circuit... Below is the second circuit.... What confuses me is how in the first circuit the current through R1 resistor increases again. Is this some sort of resonating circuit? Why does this phenomenon occur.. I am a novice to this field. Please explain. AI: Look at the current value. It is very small. Even less current change. This is just an error. The capacitor is charged and has no rezoansa. Make ".tran 1 uic" and you will see the charging current of the capacitor through a resistor. Bordodynov.
H: Find the power at the output of a random input + noise? Problems on Linear Systems with Random Inputs In this problem I don't know how to start. the answer might be long. so it would be really helpful if you help me with the concept and the idea how to start AI: Point A is quite straightforward. To achieve \$\text{SNR}=1000\equiv30\text{dB}\$ you will need the output signal power to be 1000 times higher the output noise power. The input noise is white, meaning its PSD is flat. The value is given, i.e. \$10^{-3}\frac{W}{Hz}\$. (They should really provide measurement units in my opinion.) \$H(\omega)\$ is an ideal low pass filter, the cutoff frequency is \$f_c=\frac{\omega_c}{2\pi}=159Hz\$. The output noise power is thus: $$ P_n^{out}=\int_{-\infty}^{+\infty} S_n^{in}(f)\cdot\|H{f}\|^2df=2\int_0^{f_c}10^{-3}\cdot4df= 1.27W $$ Yeah you've got a HUGE input noise. Finally: $$ P_s^{out}=1000P_n^{out}=1.27kW $$ Now to part B. You are given the input signal power spectral density. You can calculate the output powers (signal and noise) independently because \$S_{XX}(\omega)\$ is independent from \$M(t)\$. The output signal power is now: $$ P_s^{out}=\int_{-\infty}^{+\infty}S_{XX}(\omega)\|H{\omega}\|^2d\omega= 2\int_0^{10^3}\frac{c}{1+\omega^2}\cdot4d\omega=8c\left[tan^{-1}(\omega)\right]_0^{10^3}\approx c\cdot 12.55W $$ Finally: $$ c = \frac{1.27kW}{12.55W}\approx 101 $$ Please note that since \$S_{XX}(\omega)\$ is a second order low pass function with the poles in \$\omega=1\$ you could have probably used some sort of approximation. In this case, for the signal, the value of the filter cutoff frequency does not matter much: the signal PSD is so low up there that if the cutoff frequency happened to be several times higher the signal power would have been remained the same. The noise power instead is flat, so the cutoff frequency matters a lot. I pinpointed this fact because you will be in this situation very often.
H: Close loop circuit vs an open one in regards to an electrical bicycle light system From what I know about DC current, for it to flow you need 2 points with different voltage (while AC doesn't flow from point to point), this why DC batteries discharge overtime, when not used, because they are surrounded with a 0V environment. I am building a light system for my electrical bicycle and have 2 options. I can use my 12V source as a close loop or I can just connect my light system (one end) to my bicycle metal frame and use the bicycle as 0V. My questions is what will be the difference? AI: In general, it would not work unless you attached the negative of you battery to you bicycle frame as well in which case you could avoid the need of an extra wire. Though, since electricity flows through the least resistant path and I'd have to guess that a proper copper wire would have less resistance than a bike frame (you should not trust me on this one, just use a multimeter and check this one out yourself), I'd probably avoid using the frame as a wire but if you want to place your battery quite far away from your "light system" connecting your batteries negative terminal to the frame might save you some wire and could make your project look a tad bit better without the wire hanging around.
H: Question on DC analysis of BJT - Common Base I'm trying to solve a question from my book. The answer for the question is given but I keep getting a different answer. I don't know if I'm doing it the wrong way or the book is wrong. This is the question: This is my working: The answer is apparently 0.921mA but I get a different answer. The question also asks for voltage gain and current gain. I did not attempt those parts yet since I got stuck on the first part. AI: $$ I_E= \frac{(V_{EE}-V_{EB})}{R_E}=0,93 \text{mA} $$ $$ I_E=I_B+I_C=\frac{I_C}{\beta}+I_C$$ Therefore, $$ I_C=I_E \times \frac{\beta}{(1+\beta)}=0.921 \text{mA} $$
H: Do If else have priority in verilog? I have some query about the priority of if else in verilog. For example. If (a) b else if c d else if e f else g At here, those a,b,c...g are all searched ? How do those have priorities in if~else in verilog? AI: An if-else construct infers a priority routing network. Following example is taken directly from a verilog book : if (m==n) r = a + b + c ; else if (m > n ) r = a - b ; else r = c + 1 ; The circuit is as shown : As you can see, the outermost if is given a priority in the sense that it is routed first if a match is found, bypassing others. Ref : Fpga prototyping by verilog
H: How can I objectively compare TCXO performance? I would like to evaluate an alternate TCXO used in a GPS receiver circuit. Our current TCXO works good but I'm looking to test a lower cost alternate. The alternate part is Epson TG - 5035CJ / CG / CE (datasheet). Import specifications like voltage, frequency tolerance, aging..., current consumption, and package size are acceptable according to our GNSS chipset manufacturer's recommendation. I've done some basic GPS testing including Time To First Fix (TTFF) and Top 4 satellite CNo number comparison, as well as overall GPS performance. All of the tests I've done so far have been at room temperature. I have access to a temp chamber, GNSS simulator, Vector network signal generator, and a decent Tektronix Oscilloscope. Our network analyzer is currently out of shop. I do not have a test jig but I'm hoping to somehow objectively measure performance differences between our existing and new part. EDIT What specific tests can be done on both TCXOs in-circuit or out-of-circuit with the equipment I have available? Is it possible to measure jitter or are there any other performance characteristics that would be useful to measure? AI: I am going to make some assumptions here about the clocks in your measurement equipment being much better than both tcxos, otherwise it will be hard to tell which of those is causing deviations. Just for a simple illustration, say you have two unknown oscillators that should be 10MHz, but one is really off my much. You use one for your frequency counters reference, and measure the other. In one case the counter says 9.9MHz, in the other 10.1MHz. Not much gained here. Always keep this and similar issues in mind. For a more in depth look, google up some guidelines about oscillator measurements, applications notes AN10007 and AN10033 from sitime seem to contain some useful tips for measurement setups. Also informing yourself about allan variance might be useful. You always need to keep in mind what the requirements for your product are. Measuring one to deviate by 2.1ppm and the other by 1.8ppm doesn't mean much if your application is fine with anything under 10ppm. For long term stability tests, you need something to compare to. Either you have a high end ocxo in one of your frequency counters, or you pick your best tcxo and discipline it by gps. Both should have very good mid-term stabilities, so you can then run your frequency counter with those as references, pick some high enough gate time, and collect the data of your frequency counter for long enough so you are confident that it will be meaningful for your application, then compare those for the new and old part. For short term stability (which blurs over to jitter) you need to decide which kind of instability you are interested in, and there are tons of ways to measure jitter and similar, that all depend on your equipment abilities. You could configure a spectrum analyzer for a rather long sweep time around the frequency you are interested in and compare the results of the tcxos. Phase noise will be visible as side bands here (if your analyzers LO is good enough). If your scope is good enough you can use its abilities to measure a certain amount of clock cycles and build a histogram out of that and compare both (some scopes have even a feature for histograms). Sometimes just using the infinite persistance of a scope can tell you something about the jitter. Additionally it might make sense to compare the output waveforms and if they are usable or at the edge case of what your circuit can handle. A lot of cheapos on the market have rather weird clipped sine waves. So since the sole responsibility of an oscillator is to generate a certain waveform (usually sine) with a precise frequency, there is not much more to measure here: Deviation from that intended waveform, as well as deviation from the frequency. Of course, since you mentioned it, if your product requirements are such that it has to work within certain temperatures, make use of the temperature chamber, but only after you are confident in your measurement setup on the bench, and your abilities to read the data acquired.
H: PIC microcontroller DAC output incorrect Using a PIC16F1825 microcontroller I was seeing incorrect values being output on the DACOUT pin. The setup stipulated using the fraction 8/32 * VREF+ where VREF+ was an external 2.5V reference. The output should have been 0.625 but was twice that value. (have answer myself) AI: The answer is two fold: Firstly - pull up resistors are enabled by default on all input ports when WPUEN# in OPTION_REG is cleared. This will affect analog inputs like VREF+ and may affect analog outputs like DACOUT. I'm not certain on the last point as I have not tested that assertion - it seems wrong on the face of it though. This is true for the PIC16F1825 but may not be true for other PIC micros in other ranges. Secondly - the DACOUT and VREF+ pins are, on this micro, shared with the PGD and PGC pins. If a PICKIT3 is connected, that device has 4k7 pull down resistors on both these pins. If using VREF+ or DACOUT, make sure pullups are disabled and the PICKIT3 is disconnected.
H: Control Engineering: Model parameter estimation for a motor which will be under various loads I am creating a closed loop controller for a motor which is part of a robotic arm. I give it a required angle, the controller calculates a control input, the motor moves, a pot feedbacks the angle, repeat, and the arm ends up at the required angle. I am using state variable feedback, with integral action, and in order to calculate the control gains required for stability and speed I need a mathematical model of my system. To get this model I am using statistical model identification techniques. I am inputting a random control signal into the motor, and recording this and the angle of the arm at each sample time. I'll then have sample input and output data of my system, and I can use a mathematical optimization technique to find the model parameters that best fit this data. My Question When under actual operation, the robot arm will be lifting a variety of weights. Does this mean the model I found with no load (intrinsic load) is invalid? How do I get a mathematical model of this system when the load on the motor can change? AI: Usually control loops are tuned under around the operational conditions. But the PID loops are usually robust enough to take care of some deviations (especially with the right tuned integral part. The Proportional only won't handle it.). If you want to increase the robustness, use several nested loops (for position control, for example, you might want to implement an internal velocity loop. For the velocity loop - internal torque/current loop). The idea is, to make the outer position loop not to care about the underlying physical system (well, to some extent), it is done with the inner loop. The inner loop will (given it is velocity loop) will take velocity reference as input from the position PID controller, assuming the velocity loop is perfect (or modelled with some transfer function for better results). The same can be done with the velocity loop, by implementing the inner torque-regulating loop. So only the innermost loop will have to take in consideration the actual physical system parameters. The parameters for the outer loop plants will be the artificial ones, derived from the inner ones. simulate this circuit – Schematic created using CircuitLab
H: 64k x 8 RAM - what exactly does it mean? I'm looking at this microcontroller because it has the protocols that I want and a really interesting price. My doubt is related with this field: RAM Size 64K x 8. What exactly does it mean? I know what it means RAM but 8 blocks. How it works? AI: It means that the RAM is organised in a structure which is 64K Words wide. The x8 then means that each word is 8 bits. So 32Kx16 would be a RAM which is 32K Words wide, at 16bits per word.
H: how long will an 11w bulb last on 3 x 500mAh 3.7v cells I have this question that I can't seem to figure out. I have an 11 Watt bulb and 3 500mAh 3.7V Lithium Polymer Cells. And the question is how long the bulb will last. Can anyone help me with the calculations? AI: Looking at only power available and drain: 500mAh * 3 * 3.7V = 5.55 Watt Hours. 5.55 WH / 11w = .504 hours.
H: Load sensing switch I made the circuit here to automatically switch on my workshop dust extractor when I use a power tool. It works fine when used with a drill press or router. However with a Dremel 4000 it only switches on when the Dremel is at its highest speed. I imagine this is because the Dremel has an electronic speed adjustment. Is it possible to modify the circuit to work with the Dremel or should I look at using a different approach - perhaps using a current sensing transformer? AI: It is not functioning for the Dremel tool because the Dremel is too small of a load. For a larger load, the four diodes in the schematic allow about 1-2v of the AC mains into the transformer, which is wired up "backwards" intentionally, to boost this to >5v or so, so the solid-state relay can turn on. But the Dremel is such a small load, that insufficient electricity is making it through the transformer to "turn on" the solid-state relay. Component values could be changed to allow this tool to trigger the switch, such as by changing the transformer or diodes used, but this will provide too much voltage to the relay when using larger loads, and could damage it. Additional design work could be done to mitigate over-voltage, but then it's no longer a simple and elegant solution and was basically redone from scratch. As an easy fix, plug a work lamp and the Dremel into this together, and it should work as intended with no modifications.
H: ESP8266 "rom" boot code is overwritten when upgrade firmware? You can flash the firmware using ESPTOOL with a command like ./esptool.py write_flash 0x00000 my_app.elf-0x00000.bin 0x40000 my_app.elf-0x40000.bin this will overwrite the SPI FLASH chip with a new contents. But if you overwrite 0x0 region is the bootloader that allows flash the chip through UART lost? or is it a real rom mask code? AI: The bootloader and the basic peripheral library is in ROM on the ESP8266 chip itself. What you write to using esptool is the external flash chip. Due to the way the content of this flash chip is loaded into RAM (either 'permanently' or only cached) this ROM image is split into segments. (There can be more segments than these two, for instance for data.) Side note: this blogpost by me has an overview of the various ESP8266 modules.
H: How can I calculate time constant and charge time of RRC circuit? This circuit is part of my time delay circuit. I want to calculate exact value of delay time. Please help me. simulate this circuit – Schematic created using CircuitLab AI: There are two time constants here. One is when the input is driven and the other when it is open. Probably the input is driven high, then left to float low. When the input is driven, you assume it has 0 impedance to ground. Note that this impedance is parallel to R1. What is R1//0? When the input is open-circuit, then it has infinite impdance to ground. The effective resistance seen at the top of C1 to ground should be obvious in that case.
H: 4/5 band resistor - extra black band I have two resistors which need to be replaced. I tried to determine the value, but it can't make sense of these black bands: For reference the bands are: brown - silver - green - brown - black brown - silver - red - brown - black Since a silver line can only be part of the multiplier, that means the black line on the right has to be the tolerance. Black can not be used for a tolerance from what I've read. Is this black line a tolerance value or something else? AI: I don't think I've ever seen a resistor value with a leading zero (usually those super-low values are marked with the value in numerals), but that's what those appear to be. 0.012\$\Omega\$ 1% and 0.015\$\Omega\$ 1% are what I see (reading from right to left, as Olin suggests, but I think the silver is a multiplier and the brown is the tolerance). Here's a 0.015\$\Omega\$ 1% resistor (courtesy Digikey):
H: Scoring a PCB across traces I'm wondering if it's possible to score a PCB (such that I can easily break off the scored section) while still being able to have traces going across where the score would be. I have a rectangular board, and I'd like to have a little edge connector tab on the side of it for temporary use, with the ability to just snap it off at a certain point. Is this possible? I'm picturing something like a v-score on one side, while the traces go along the other, but I have next to no experience with PCBs so this is all conjecture. AI: Yes, this is possible. It's commonly done to provide a temporary test fixture. In the attached example the traces are on internal layers connecting the temporary board connector used for testing. After testing the connector is removed and the traces are broken.
H: Parallel networks and references I have a very simple question regarding parallel RLC networks (or any similar circuit). When given a problem where the reference voltages and current directions are allocated 'strangely', is my intuition correct to make all voltage references positive on one side (top) and the current directions consistent (down)? No equations or dependencies are given. AI: You can (in general) select the polarity of current sources and voltage sources any way you want. Then, when you solve the circuit (via nodal or mesh analysis) if you guessed wrong on any of the sources the voltage or current value will come out negative. This tells you the actual direction is opposite from what you picked. Good practice is to choose initial polarities to be consistent (as you suggested) and then solve the circuit!
H: How do MCUs and RAM wire up for programmatic access? I am new to electronics and microcontroller programming and need either confirmation that my understanding on a few items is correct, or I am in desperate need of correction on them! I have a C-based server that requires about 2GB RAM to function properly. I am interested in seeing if there's any way to flash this program to an ARM-based MCU, such as the Raspberry Pi or the STM32. If my understanding of MCU componentry is correct, the Raspberry Pi simply cannot handle this RAM demand, because its a Broadcom/ARM chip affixed to an SoC with fixed memory ICs. However, it is also my understanding that non-SoCs such as the STM32 can be wired up to any kind of memory chip. I found these 4GB memory cards, and am wondering if it is possible to place one of these on a PCB alongside an ARM-based MCU, thereby allowing the ARM chip to actually run a very large program on it. So I ask: is this possible to do? If not, why? How are they incompatible with each other? If so, how are they typically wired up? Does an engineer look at the datasheets for both chips and figure out the wiring/IO between them? I guess at the end of the day, I'm wondering why there aren't more MCUs out there with crazy big memory; and I'm thinking it's either because most embedded/MCU programs don't need that much memory, or its because - for some reason - they can't. AI: The answer is economy. The optimal processes for CPU and RAM are different, so when both are on the same chip some compromise must be found. For a small chip (not that much RAM) this does not hurt so much, less than adding an external RAM and using a lot of IO pins to interface to it. For a large RAM (currently the switchover seems to be at ~ 100Kb) it makes more sense to put the CPU and RAM on different chips. Making large RAMs is a very specialized high-volume buisiness, so better use what is there in the market than developm a new chip. The RAM you link to is PC-style DDR3 dynamic RAM module. Such a module has a specific interface which is not compatible with the (old) standard address-lines + data-lines + control bus found on most MCU's. I think you would need a glue chip (bridge) to make such a connection. And yes, engineers read the specs and make the wiring (including anyb glue chips needed, power supply things, careful timing etc.) This might sound simple but it is not.
H: Superposition on a circuit with just one source I'm trying to analyse this circuit (state variable filter): All the places on the web I've found solve it first using superposition, creating an equation of the form \$ V_2 = kV_1 + kV_4 + kV_3 \$. However, I can't understand how they use superposition in this circuit. There is only one source so how can it work? How can they get to that equation? AI: Superposition is actually used. For a start assume V3 and V4 are zero then V2 = -V1*R4/R1 Then assume V1 and V4 are zero then V2 = \$\dfrac{V3\cdot R2}{R2+R3}\cdot (1+\dfrac{R4}{R1\|\|R5})\$ Then assume V1 and V3 are zero and solve V2 in terms of V4 Finally, add all the different versions of V2 together as per superposition
H: With regard to GPS signals, what is the significance of I-parts and Q-parts of Complex signal? What is the significance of I-part & Q-part magnitudes and imbalances in the screenshot? They are listed next to the red Complex signal diagram. Imbalance of I-part: 4.7% Imbalance of Q-part: 25.8% Magnitude of I-part: 56.5% Magnitude of Q-part: 55.7% I am running uBlox uCenter and looking at the Extended Hardware Status message. Is there an ideal percentage of I and Q parts? AI: For direct conversion, you demodulate down to zero and use two ADCs, which means the system will have separate DC offsets and gain factors for the I and Q channels. For demodulation, these need to be compensated for (in a superheterodyne system, this problem doesn't exist, because there is only one ADC, and the DC offset of the system is outside the analyzed signal). The DC offset can be determined by a low-pass-filter -- simple averaging over some time is usually enough, and works for everything that isn't a CW signal near the LO frequency, just subtract it, and the I/Q constellation is properly centered. That is usually a fairly large error, so the compensation is required in order to get a usable baseband signal. There are other imperfections in signal reconstruction as well, the most obvious ones in the constellation diagram are gain difference between I and Q channels (stretching in I or Q direction) phase offset between I and Q channels not exactly 90 degrees (shearing) The phase offset seems to be ignored in that application, because it's usually small and doesn't really matter for GPS reception.
H: Very odd open-drain output design I ran across a open-drain design and am baffled by the resistor between the nFET source and ground. There are only two purposes that I see it could serve. 1) It's series source termination, matching the impedance of a transmission line attached to the drain. However if this is the case, R would be better placed on the drain terminal. 2) Protection against a ground short in the event of an oxide breakdown shorting gate to source. Again, even in this case R would be better placed at the gate of the nFET. Am I missing anything? The drain is indeed connected to a transmission line, and is pulled up by a 16V rail at the end of the cable run. EDIT: I'll add that on the other end it's a high-Z digital input along with that 30k pullup to 16V. Also the buffer's supply is only 3v3, driving a low Vgs nFET. simulate this circuit – Schematic created using CircuitLab AI: I suspect it's designed to be a current source, not a hard pull-down. M1 acts as a source follower with R1 as the load. The IRF530 has a VT of 2-4V, so if the buffer is producing 16V, the output current would be around 150-170mA regardless of the pull-up voltage. The other end of the line could use current sensing to determine the state of the output. That's as much as I can guess without knowing more about the rest of the system. What's connected on the other end of the line? What's the pull-up resistance? Is this a single-ended data line, or part of a differential pair? EDIT: A low-VT transistor (which is more reasonable) being driven by a 3.3V buffer suggests a current of around 37mA. The weak pull-up would source 0.53mA at most. Since the input at the other end is high-impedance, that suggests to me that the goal of this circuit is to limit the slew rate. Instead of producing a very fast (and thus noisy) pull-down, the current sink gives a gradual rise whose rate depends on the line capacitance. This reduces the bandwidth of the signal. For example, if the line capacitance is 100 pF: $$\frac {dV} {dt} = \frac {i_{out}} {C_{line}} = \frac {37\ \mathrm{mA}}{100\ \mathrm{pF}} = 370\ \mathrm{\frac {V} {\mu s}}$$ $$t_{rise} \approx \frac {16\ \mathrm V} {370\ \mathrm {V/\mu s}} = 43.2\ \mathrm{ns}$$ $$BW \approx \frac {0.34} {t_{rise}} \approx \frac {0.34} {43.2\ \mathrm{ns}} \approx 7.87\ \mathrm {MHz}$$ The bandwidth is proportional to the current, so if your transistor can sink 500 mA, that would give you a bandwidth of over 100 MHz. It's not hard to radiate at that high a frequency! The formula I used for bandwidth is a rough approximation, so don't take it too seriously. The important thing is that a 10x difference in pull-down current can give you a 10x difference in bandwidth. The weak pull-up is also interesting. It seems like the intent is for the line to stay low for a couple microseconds after it's pulled down. Based on the asymmetry, I suspect this is a reset or some kind of system-wide status signal, not a normal communications line.
H: PICkit3 MCLR to Vdd Pull-Up Resistor I'll preempt this by saying that I'm a software person whose education (unfortunately) didn't even touch on the most basics of hardware or circuits. I'm attempting to program a PIC16f1709 microchip with a PICkit3 via MPLab X v2.3. The PIC is part of a larger board, but is currently blank. The board in question is a manufactured product, not a dev board, but it does have exposed pins into which I've wired the PICKit, and I'm sure that the pins match up. I'm powering the board with its own power supply. The board doesn't have a pull-up resistor from Vpp/MCLR to Vdd. According to the PICkit User Guide, that resistor is "recommended" (see paragraph 2.4.2). When I try to program the controller, MPLab warns me first that the target ID is being read as 0x00 rather than the expected value, and then after programming it reports that the instruction at address 0x00 is 0x00 rather than the expected first instruction. My understanding is that this means that the PICkit is not able to communicate with the controller, because it is reading everything as 0x00. I've read that a freshly-erased PIC sets all bytes to one, so even if my code didn't get programmed, it shouldn't be reading zeroes. Target voltage detected Target Device ID (0x0) does not match expected Device ID (0x3054). The following memory area(s) will be programmed: program memory: start address = 0x0, end address = 0x3f Device Erased... Programming... program memory Address: 0 Expected Value: 3180 Received Value: 0 Failed to program device So my question is this: is the pull-up resistor required (as I've read from less "official" sources)? Is the fact that it is missing causing the programmer to be unable to communicate with the controller? AI: No the pullup resistor isn't required, however without it, unless there is something else to perform its function, the chip won't run under normal operation. The purpose of the resistor is to ensure that MCLR is held high so that the chip isn't in a RESET state, which allows it to run. If you don't have that resistor you would need some other system to ensure that it is held high - maybe a reset / power management chip. The fact that you can't program won't be related to that resistor, instead you should look to other connection or circuit issues, such as a lack of decoupling capacitors, not all the power pins connected, or other components on the programming pins interfering with the communication.
H: 555 as a Digital Coupler I want to read MIDI data with an arduino (specifically an ATtiny85) for use in a synthesizer. The only method I'm aware of involves using an optocoupler such as the 6N138 found on this Instructable. It seems quite practical but unfortunately I don't have one on hand right now and shipping might take too long for my project. I was wondering if it would be possible to configure a 555 timer to do the same task instead. I quickly threw together this circuit of what I'm working with. The circuit should compare the voltage difference between the relevant MIDI pins and reproduce this difference as a logic level at the output. My setup as a whole is not working but without an oscilloscope nearby I can't confirm that this is the buggy part (the other option being the code I'm using). Is my circuit a valid alternative to an optocoupler? EDIT: Removed "Isolator" from title. AI: Probably you don't know/understand what isolation means when dealing with optocouplers/optoisolators. It doesn't mean that there is some sort of circuit buffering separating the two circuits. It really means electrical isolation, i.e. there is no current flowing from one part to the other (an optocoupler does that). An optocoupler is essentially an LED and a photodiode (or a phototransistor) optically coupled in the same package. The LED and the photodiode don't share any electrical connection. They only "communicate" through the light generated by the LED. Therefore the "input section" (the LED) can be driven by a circuit which is completely isolated from the circuit connected to the "output section" (the photodiode/phototransistor). This allows two circuits that must remain electrically isolated to communicate. So the answer is no, an NE555 cannot substitute a circuit which performs electrical isolation. If you're desperate or don't need specific performance/reliability you could try building a poor man's optocoupler. And here is another attempt. Important Note: if your circuit needs an optocoupler for stringent safety reasons, using one of those home made one could pose some serious safety risks, so be warned! That is, I wouldn't trust a mains-powered heart-beat detector using a home-made optocoupler to be used on my chest!!!
H: What knowledge and equipment do I need to build this device? OK I am a Software Engineer dealing with both low level and C/C++/Java code with a Masters in Software Engineering but no knowledge of Electronic Engineering whatsoever. I wish to build a device that operates transparently at the IP level and can be inserted into a regular ethernet connection - although not a traffic logging device or a firewall this might be a good example as it is similar to what I wish to accomplish. So the device looks like this: Ethernet Connector <-----> Hardware <-------> Ethernet Connector Hardware lets say analyses at packet level but for the sake of argument does not transform/mutate the packets. I looked at Arduino type equipment but they have only one ethernet port and just as importantly are the wrong form factor (I wish the device lets say to be tubular to fit in with the cable ergonomics). Basically I need a sub sized Linux server on a chip preferably self-powered with two ethernet ports. What books can you recommend to get my up to speed in basic Digital Electronics design as to design such a board? Lets say I have a few months spare to study and learn fast (which I have right now 3 months free between contracts). Also what hardware do I need to buy to test? A Logic Analyser? Signal Generator? I have no clue really. I understand the embedded Software side quite though which I hope I an advantage here. AI: Buy an ethernet hub with 3 or more ports. Can be a bit hard to find, there are sometimes people selling Netgear hubs on Ebay really cheap. Check the photographs of the hub and make sure it says hub on it and NOT switch. A lot of people and companies use the word interchangeably and even together on the graphics branding on the product. The hub will relay packets between the input and output ethernet connectors in your 'diagram'. Take the third port on the hub and connect to the ethernet port on the arduino. Job done except for writing a bit of code. The Arduino can then see all the traffic passing between the two ethernet ports you're interested in. This will reduce the amount of software you have to write, you don't have to worry about writing software to transport the packets between the two ethernet ports, all you have to do is write the software to read packets on the Arduino and analyse them. But note the Arduino can't modify the data packets passing between the two ethernet ports, but that's what you want, the data to pass transparently and unmodified between them.
H: Replacing a piezoelectric speaker with an electromagnetic one I'm making this circuit: (source: vk2zay.net) where X1 is a piezoelectric speaker or ear piece. However, the only kinds of speakers I can get in time to finish this project are electromagnetic ones (we found one in a Father's Day card, and another from a radioshack project). The page here that explains the circuit makes it seem like the piezoelectric speaker is required, but is there a way that I can modify this circuit to use an electromagnetic speaker instead of a piezoelectric one? AI: By simply attaching a low resistance EM speaker you will not be able to keep Q2 working, see the top part of my solution posted in the image. However it is possible to accomplish with another transistor. Create a two stage amplifier where the output stage is an emitter follower. This will buffer the volume amplifier created with Q2, allowing you to drive the low resistance speaker. You will not have great battery life and your volume will tend to be less than what the piezoelectric speaker produced but it will certainly work. If you find you need more gain increase the 1K ohm resistors used to bias the base of Q3. Make sure the base equivalent resistance allows for enough dc current to keep Q3 on. PS: This solution assumes you have a greater than or equal to 32 ohm speaker. Let me know if you need more explanation or help. -Ryan
H: Why does my AY-3-8910 sound so distorted? I'm building a circuit driven by an Arduino UNO that controls a AY-3-8910 programmable sound generator. The AY has three pins labeled Channel A, B and C. Each output is an independent sound channel. I am piping the audio through an LM386 which then drives a small PC speaker. When I run a single channel, it sounds pretty good. But each time I mix another channel, things start getting really distorted. I've built this on a breadboard in the past and it sounded great so I don't know what I'm doing wrong this time. Below is a crude schematic of my circuit. I didn't draw the LM386 because it's the same with or without it and I never used it before. Any ideas how I can cleanly mix these signals? I don't remember doing anything in the past other than literally connecting all three channels together. I can provide more schematics and/or videos or pictures too. This is driving me nuts. So I'm hoping someone out there has worked with this sound generator in the past that can provide some insight. Thanks. simulate this circuit – Schematic created using CircuitLab AI: Typical circuit diagram from the manual shows a 1K resistor to ground and capacitive coupling. Both are necessary. The 10K may also be required.
H: Will an 0603 capacitor fit on an 0805 pad? I am in a pinch and just submit my final board design to a pcb house. Unfortunately, I messed up the design and had an 0805 pad in a spot where a an 0603 should have been. I will be getting the boards in this Friday with a 2-day turn. Can I get an 0603 to fit on the 0805 pad in a pinch? I know it's not ideal, I just need to know if it will work so I don't have to pay 25$ to overnight 50 cents worth of capacitors from mouser or digi-key. I did a mock up on a scale printout and it looks VERY close like it should be fine. AI: If you're using 'standard' footprints then yes, an 0603 will fit fine on an 0805 pad. The reverse works too, in a pinch.
H: Will a diode conduct if one side is floating and one side has voltage applied to it? I have a circuit that has a battery output voltage of about 7V and can also have a charger input that when plugged in will output around 9V. The battery/charger power a microcontroller (regulated down through an LDO) as well as a motor driver (simply shown here as a current limited transistor). The battery charging is taken care of separately through a multi-cell PMIC, but I need to isolate the battery positive output from the charger input when the charger gets plugged in. I figured I could use a transistor switch to control this, but I thought that an easier way to separate the supplies would be to simply put a Schottky diode from the lower voltage to the junction where they connect. I want to confirm the behavior of the diode and if this design seems reasonable (assuming the diode can handle the load current and that it is low turn on voltage). I know that when the charger is plugged in, the diode will not conduct b/c it will be reverse biased; however, I am not quite sure on the behavior when the charger is not plugged in. It would seem that the cathode side does not have a defined voltage, so I am not sure if it would be considered forward biased or not in this case. I would imagine the diode would conduct but would like a more rigorous answer. Also, do not assume the driver (represented by Q1) is turned on. simulate this circuit – Schematic created using CircuitLab AI: With the charger disabled by SW1 the diode will conduct and produce a voltage on the cathode of about 0.5Volts lower than the 7V battery. Under heavier load conditions this volt-drop might be a little higher. Under lighter load conditions this may be a little smaller. BTW diodes don't have a "V_cutoff" they have a forward volt drop.
H: Obtain an accurate DC value in an AC+DC signal Background: My goal is to use an Arduino with two proximity probe sensors to achieve some output. It is worth mentioning that the output of the proximity probe has two components: DC output that determines the position or the GAP AC component on the top of that DC voltage to determine the vibration I want to focus on the DC part. The proximity probe sensor has a negative voltage output from 0 to -20 volt DC, as the Arduino accepts voltage only in the range of 0 to 5 volt, so I decided to use a low-pass filter to remove the AC component and get just the DC component, as this DC component is in the range of 0 to -20 volt. I used an op-amp to invert and to scale down the voltage so instead of being in the range 0 to -20 V, it will be 0 to 5 V. In this specific question my goal is to get the very accurate DC component. The circuit I implemented is very simple. Basically, it's a low-pass filter with a cut-off frequency of 1 Hz (assuming that my signal won't have a frequency component in the range of 0.00001 - 1 Hz) and then an inverting amplifier with a gain of 1/4 to invert and to scale down the DC voltage. Here is the circuit: The output is logical. I used a voltage source with AC and a DC component of -10 V DC offset to simulate the proximity probe signal, and here is the output: The green line is the DC component which is almost -10 volt, and the blue line is the output inverted and scaled down. Which is almost 2.5. The problem is with the output. It's almost 2.5 V, and it's oscillating and changing. My application needs an accuracy of millivolt, so this won't be acceptable. How do I get a more stable and accurate output? Here is the picture of a zoomed-in output: AI: Your output looks fine to me. There is always a period at start up of a filter where its output is off, and it converges to the correct output in a time period related to the cutoff frequency - a lower cutoff for a low pass filter means it will take longer to converge to the true value. This explains the "changing" output of your filter. Next, is the oscillation. That, too, looks correct. A filter doesn't remove everything out side of its passband, it merely reduces it. Since you are using a simple filter, the reduction isn't very strong. Your output has a much reduced 50Hz wave riding on the DC, which is about what you could expect from the given circuit and simulator settings (signal generator says -10V DC, 1 Volt AC at 50Hz.) You will need a better filter to reduce the AC. Some other folks here are more capable of giving you tips on that. You will probably have to live with the period at the beginning where the filter is converging to the proper DC level. I missed the bit about "millivolt accuracy." Given the Arduino has a 10bit ADC, you will only get 1024 steps for voltages between 0 and 5 volts. That's a tad less than 5 millivolt resolution (4.88mVolt.) So, you aren't gong to get millivolt accuracy just using the ADC. What you can do is to use a variant of the suggested moving average filter to improve the accuracy at the cost of time resolution. What you would do is use a low pass filter to remove (reduce, really) components above say 1Hz, then average all the ADC values for one second. You would get a value with more bits of accuracy, but only once per second. The ratio between the time period for the average and the real sampling rate tells you how many (theoretical) bits of accuracy you can gain. If you are sampling at 8000Hz, and average every two samples, you can gain one bit of accuracy but only have 4000 samples per second instead of 8000. That will help with the DC, but not the AC. I don't know what you can do there or if you also need millivolt accuracy there as well. In response to comments: I see you can do post processing in Matlab. This makes a BIG difference. Use a simple anti-aliasing filter appropriate to your sampling rate and expected signal. Store the sampled data to SD, then process in Matlab. In Matlab, you can implement the resampling to improve the bit accuracy as mentioned above, as well as using better filters to remove the AC. Since you are working offline, you can use a non-causal filter to remove the AC - this will take care of the convergence problem mentioned above. Filtering out the DC so that you can process just the AC will also be simpler - non-causal filters will also help here.
H: Opamp circuit has an unintended DC bias I was fiddling around in LTSpice for a project I'm working on. This circuit is a basic preamplifier and a low-pass filter. But this question refers solely to the amplifier stage. The opamp is the "universal opamp2" model. However, the resulting waveform has a 4.5V DC bias. Since I intend to run this from a single-supply source I was planning on biasing the input signal at 4.5V so the opamp can swing the signal to 0 and 9V. But I hadn't added the bias yet. Without C3 there is no bias and the output waveform is as expected. If I add a, say, 1M resistor from the noninverting input to ground, creating a high-pass filter, the bias is gone. The C3 is supposed to remove any DC component from the input signal. I figured this was just SPICE being stupid, but is this something that would be expected in practice? Nodes: V[n004]: Input signal (green) V[n005]: Signal at non-inverting input (cyan) V[n006]: Signal at opamp output (red) AI: Of course it is "biased": you cannot leave amp input "hanging" in the air. In real life your schematic will potentially destroy itself and everything connected to it because opamps usually have very large input impedance and will pick up static electricity from the air, resulting in complete malfunctioning of the device. You have to have the resistor between the input and ground.
H: Inductance of a relay coil I'm using a magnetic latching dpdt relay from Teledyne company. The value of inductance for the coil in the relay is needed and not given in the data sheet. The part number of relay is 422d from Teledyne. How can I know it? Is it possible to know the value without actually buying the product and testing? AI: A simple N channel MOSFET driving the relay will not care about the inductance - the final and full value of current the FET sees will be determined solely by the supply voltage and the coil resistance. All that the inductance can do is slow the current down a few milliseconds before it reaches the final value. This applies whether it is a logic gate, a BJT or a switch; it's simple ohms law: - I = V/R where V is the supply voltage and R is the coil resistance. The data sheet even tells you what this current will be so no need to do the calculation. Make sure you use a back-emf diode though or when you open circuit the coil there will likely be a big spike of voltage (due to the inductance) that can easily damage a FET or BJT.
H: Vcc pin of TVS diode array Bourns has various TVS diodes to protect Ethernet ports. Most of them (for example CD143A-SR05) have GND, VCC and I/O pins. I/O pins are for Ethernet signal line. Why do these TVS diodes have VCC pin? AI: Using a VCC rail in conjunction with unidirectional TVS diodes allows the device to ensure the data lines are clamped between the supply voltage and ground. This way, they can adapt to different supply voltages without requiring a separate device for each, and start clamping as soon as the voltage is more than a diode drop outside the rails. Another consequence of this is that if power is applied to the data line while VCC is unpowered, the TVS will conduct from the data line to VCC, just like an IC's protection diodes. You can reduce this by using series resistors to limit fault current.
H: Why does output impedance behaves this way Why does the output impedance(Ro) of a common emitter amplifier is considered in parallel with the collector resistance(Rc)...as according to the KVL--->Vcc=IcRc+Vce+IeRe they seems to be in series....?? AI: If you have two resitors which have one common node, it is IMPOSSIBLE to decide if the are connected in series or in parallel because this only depends on the node where the voltage or current source is connected. They have to be considered as parallel if the source is connected BETWEEN both resistors (common node, as in your case). If the source drives an open end of one of the resistors they are in series (one common current goes through both resistors subsequently).
H: Removing spikes from multiple onboard power supplies I have a board that is based around an STM32F303 and am using the onboard ADC/DAC, fed by a 3V power supply derived from a USB 5V input. The 5V also feeds a module, RD0515, which creates +/- 15V for opamps on the board. My problem is that the RD0515 also creates quite substantial spikes of about 2uS width and repetition rate of 20uS. These are finding their way through to the analog supply pins on the STM32 at around 100mV amplitude and screwing up measurements. The board is 4 layer (GND and 3V internal planes) and has plenty of capacitor decoupling. Is there a better way of getting the +/- 15V that will not cause such problems? AI: The RD0515 is a switching type converter. As such it is not surprising that it produces spikes in an unoptimized design. There are several things you can do to help with this situation. At the input side to the converter make sure to apply a very low impedance high value capacitor to supply the input current surges as the internal circuit switches on and off. It may also be necessary to place some inductor or power type ferrite bead in series with the 5V supply and the input of the converter. The added capacitor should be as close to the converter as possible and would be on the converter side of any series inductance added. On the outputs you may need to add a low pass filter to clean out spikes before the +/- 15V goes to other circuitry. This may need to be a multi stage filter with several series inductors/ferrite power beads with intermediate capacitors to GND. An alternate for the outputs would be to run your analogue circuits on +/- 12V and place low current linear regulators (7812/7912 types) between the converter and the analogue circuits. I can add the comment that the RD0515 is part of RECOM's "econoline". The parts are small, convenient and have attractive prices. But for this you do give up some on how much filtering that the manufacturer puts inside the converter. The econoline is most certainly designed to be adequate for many applications that do not care about some switching spikes. On the other hand if you select this type part then you take on the burden of cleaning things up to be suitable for your specific design. Also be aware that the RD0515 part does not have regulated outputs. As such the output voltages can change with load current variation and input voltage variations. If your analogue circuit designs have a dependency on the accuracy of the DC voltages and have low tolerance to power supply variations you may very well want to look at the option (3) above. Lastly be aware that this part should have a minimum load of 10% of its rated capacity in order to meet the data sheet performance specifications.
H: Purpose of input resistor in ADC circuit What is the implication of the 1 Megaohm resistor in the schematics attached? Will I be able to detect the condition of the motor (running, not running, stalled)? Part information: MCU ADC (ADC type: SAR) leakage current is 1uA. R3 (Load) is a H bridge (pair of PNPs (5CS) and NPNs(6CS)). Motor type is DC brush-less with following current rating: Normal Load Current (Running) - 80 mA Stall Current - 200 mA simulate this circuit – Schematic created using CircuitLab AI: The 1M resistor would allow high voltage spikes to exist at the motor without damaging the ADC input. By limiting the current into the ADC input protection network, even a 1kV spike would be rendered harmless (unless the resistor arced over). However most ADC inputs are not very tolerant of such a high input resistance unless they have a built in buffer amplifier- often a few K ohms is the maximum recommended. Even if there is a buffer (rare) the datasheet leakage spec (you say 1uA) is usually too high to guarantee reasonable accuracy (typical leakage might be pA at room temperature if buffered). So, the design seems somewhat dubious at first blush, though it may indeed function. It's doubtful actual spikes exceeding a few tens of volts would occur at the power to an H-bridge, so a resistor value of a few K might be fine. The diode is presumably to prevent the current sense resistor from limiting the motor stall/start current. It will kick in at around 270mA motor current. If the motor is 'plugged' it's quite possible to get around double the stall current momentarily.
H: Calculation of a current in the Laplace domain In the circuit below (working in the Laplace domain), I have to calculate \$I\$. $$\begin{align} &I=\frac{-V_L-V_C}{sL+\frac{1}{SC}}=\frac{-V_L - \frac{V_0}{s} - I \frac{1}{sC} }{sL+\frac{1}{SC}}\implies\\ &\implies\ I\Big(1+\frac{\frac{1}{sC}}{sL+\frac{1}{sC}}\Big)= \frac{-V_L-\frac{V_0}{s}}{sL+\frac{1}{sC}}\implies\\ &\implies I=\frac{sL+\frac{1}{sC}}{sL+\frac{2}{sC}} \frac{-V_L-\frac{V_0}{s}}{sL+\frac{1}{sC}}=\frac{-V_L-\frac{V_0}{s}}{sL+\frac{2}{sC}} \end{align}$$ By seeing the solution on my book, it gives $$I=\frac{\frac{V_0}{s}}{sL+\frac{1}{sC}}$$ Why is that? Where am I wrong? simulate this circuit – Schematic created using CircuitLab AI: Your first equation is wrong, or I don't grasp your notation. I equals the applied voltage divided by the impedance, so the first formula you use is not quite right. The third term if different from the second, and should make zero since you are summing all the voltages along a closed loop. The proper way to tackle this problem is apply ohm's law: $$ V = I\cdot R \rightarrow I = \frac{V}{R} $$ The law is still valid in the s domain if you use the impedances: $$ I(s) = \frac{V(s)}{Z(s)} $$ In your circuit you have an impedance that is the series of a capacitor and an inductor, thus: $$ Z(s) = sL + \frac{1}{Cs} $$ While \$V(s)\$ is given. Finally: $$ I(s) = \frac{\frac{V_0}{s}}{sL + \frac{1}{Cs}} $$ You can have a little more fun and come to: $$ I(s) = V_0\frac{C}{s^2LC+1} $$
H: PCB, Plated hole with annular ring on only one side I need my pads for connectors to be plated so that the pins from the connectors can be soldered. But the bottom side cannot have copper pad/annular ring. I have made the pads as shown in the figure below. Bottom pad size is same as the hole size and the hole is Plated. Is this the proper way? Will there be risk that the holes will be unplated my the manufacturer? AI: Well as comments have said only your fab house can tell you for sure. You should try to keep a close relationship with them as they're basically your partner in all your designs :) Now this question just happened to come up for me as well a few months ago on a really dense board. I use Via Systems when in the US, a very large and capable manufacturer. I asked them a similar question, could I make vias that only had an annular ring on one layer. Their reply was no don't do that, we can't plate the via properly that way. Your shop may tell you it's ok but I would expect they won't like it. There's always surface mount connectors. Or and you'd have to ask about this to but I imagine you could just not plate those vias and only route to the top layer where the pad is. There really wouldn't be much holding the connectors on the board at that point and I'd look at mechanically securing them to the board. Maybe epoxy or screws or something like that.
H: How can I prevent reverse voltage in my stop/tail light design? I'm looking to modify the rear lights on my 72 VW beetle to enhance ascetics and improve safety. I intend to use LEDs to achieve this and feel that both the indicators and reverse lights should be pretty simple circuits. Once I've figured out how many LEDs I need I can figure out the correct resistors required and wire them up in parallel. I'm new to electronics so decided to find a testing platform and discovered Yenka which I've been using to try and figure out the more complex system I want for the stop/tail lights segment. I decided I wanted an L shape illuminated when the car lights are on, then when the brakes are applied I want it to be one giant block. Due to the nature of the tail lights I have, to make things look neat I decided that when the brake lights are applied 2 LEDs from the tail lights would actually go off. (See picture) Click here to see large view schematic I've created the circuit in Yenka, I've figured out that I can use a relay to give the behaviour that I am after. I can turn the side lights on and off with no issue but when I come to the brake lights it's a different matter. The lights turn on fine and all light up as I want, but when you turn the brake lights off it blows all the LEDs with an error message saying "Exploded Object - The peak reverse voltage was 178.91V. The maximum rating is 9V." I've had a quick look into reverse voltage and know that a diode can be used to prevent it, which is why I have a diode separating the top 2 LEDs in the sidelight circuit to prevent them from getting power when the rest are illuminated. The resisters are currently set to 350 ohms but haven't been calculated yet as I haven't worked out how many LEDs I'll actually use in my circuit. Whats causing this reverse voltage? Why does it not affect the side lights in the same way? What causes the reading to be so high? Please be patient with me if I ask stupid questions, and thanks for any help offered. I'm sure this is probably not the best solution or the most efficient. AI: Here's one with no relays and no spikes, and here's the LTspice circuit list if you want to play with / simulate the circuit: TAIL LIGHT: In operation, S1 is the tail light switch, shown on the simulator schematic as electrically operated, but in real life it's just the switch that turns on the tail lights. When S1 and S2 are open, Q1 and Q2's bases are connected to GND through R1 ,R2, and R8, turning OFF Q1 and Q2, which will prevent current from flowing through the LED loads represented by R6 and R9 and illuminating them. When S1 is made, however, Q1 and Q2's bases will be forward biased, which will turn them ON and illuminate the 13 LEDs comprising the 'L' shaped tail light. Note that when S1 is made, D1 will be reverse-biased and will prevent the 12V on the output of S1 from affecting the brake light circuitry. BRAKE LIGHT When S1 is open and S2 is made, D1 will be forward biased, which will turn on Q1,illuminating the 11 LEDs common to the tail light and the brake light. Q2 would also be turned on by the current through R2 into its base, but the current through R3 turns ON Q3, which sinks the current that would otherwise flow into Q3's base to ground, preventing Q2 from turning on, which will keep the two tail light LEDs dark. At the same time, current will flow through R4 into the base of Q4, illuminating the 15 LEDs which only turn on when the brake light switch is made. If the tail lights are on when the brake pedal is pressed, all of the LEDs except for D4 and D5 will illuminate, and when the brake pedal is released D4 and D5 will come back on again, resulting in the 13 tail light LEDs being illuminated and the 15 brake light LEDs going dark.
H: Statistically, which embedded system is the most used in Cryptography/Security? I am considering to buy an embedded system development platform (electronic board + software) for Cryptography and Computer Security. Statistically, which embedded system is the most used in Cryptography/Security in Industry and personal learning purposes? I would use them for hardware implementation of Cryptography. AI: Many manufacturers of microcontrollers -- Atmel, Freescale, Microchip, NXP, Silicon Labs, STMicroelectronics, Texas Instruments and others now include hardware encryption/decryption engines in some of their microcontrollers. They are packaged a as separate module showing up as a set of registers, like I/O but without any pins, similar to a CRC-32 module might be included. All of the ones mentioned below do AES, many do DES and 3DES as well, along with RSA, SHA etc. Atmel: AT91SAM family Freescale: Kinetis family, i.MX Microchip: PIC24F "GB2" family, e.g. PIC24FJ128GB204, also PIC12F635/PIC16F636/639 NXP: A710x family (8051) Silicon Labs: EFM32 family STMicroelectronics: ST33 family, STM8L162M8, STM8L162M Texas Instruments: CC430 family, AM3894/3892 This by no means an exhaustive list! But it shows that there is support for hardware encryption/decryption not only for 32-bit processors, but 8-bit and 16-bit as well -- yes, hardware AES support for 8051s and PIC12s. Who'd have thought. This article, "Secure Microcontrollers Keep Data Safe" covers many of the processors I've mentioned here, plus many some others. I have no idea which is the most popular though. As much as I like Microchip, your best bet would probably be to pick one of the 32-bit ARM implementations.
H: SPI communication with multiple slaves I am working on a project that is using an ATtiny861A and this only has one SPI. If I want to hook up a radio and an external memory to the same SPI, will there be an issue sending data from the external memory to the radio? I know I can hook up the external memory and the radio to the same SPI but I am not sure if the radio will be able to send the data saved on the external memory. AI: Your micro is the master and as such you can't get this type of bus to interact slave-to-slave. It's a case of reading the memory into the micro and then forwarding this to the radio i.e. the micro must buffer the data when sending it between slaves.
H: Altium: Surface Mount Test Pad Passing Through PCB I'm working on a PCB in Altium that will sit inside an assembly above another board. The board on the bottom has a test pad, but once everything is assembled it will be inaccessible. Therefore, I am attempting to add two pads to the top PCB (one on the bottom layer, one on the top) that are connected through vias around the edge (No, I cannot simply use a through-hole pad). The issue is this: In order to connect the vias to the top and bottom pads, they must all be connected to the same net. However, there is not a designated net for these pads, as they are simply fed through the PCB and are not used by it. It is a PCB-only feature and is not part of the circuit, so I don't want to add it to the schematic. How might I go about connecting these top and bottom layer pads through vias? Here's an image showing the top side of the PCB: The bottom layer pad is directly beneath the one shown in the image. AI: You can manually add a net using Design->Netlist->Edit nets, then select each pad and assign that net to it. But I don't see why you don't want to do this in the schematic.
H: How to find voltage across X and Y? This is the schematic diagram of the circuit This is a homework question , which belongs to mcq type It is about finding voltage across X and Y? Is there voltage between x and y ? I wonder if there is a voltage across x and y, because they aren't connected , so how can a current flow ? But non of the answers given are zero,so what is the way to do this? AI: Think of it this way: Since the middle branch (containing the "top" 50 ohm resistor) is open, no current will flow through it. Therefore, you're dealing with a single series loop, and you're actually measuring the voltage across the "bottom" 50 ohm resistor. All you need to do is find the total current through the main loop, and use Ohm's Law to find how much voltage that "bottom" resistor drops
H: Voltage transfomer without ground connection today i bought a transformer 200v to 110v for my soldering station hakko fx-888D however it doesn't have a ground connection. Is there any problem if i connect the soldering station to it? AI: Usually the ground connection on soldering irons are used for grounding the tip, preventing electrostatic discharge to potentially sensitive components. If the Hakko iron has this feature, and you use a transformer without a ground connection, then the feature won't work and you will endanger your static-sensitive components. You could connect the ground from the iron straight through to mains ground without connecting to the transformer, but it would be even better to have it connected to the transformer core. Then make sure you use a 3-prong plug (with a ground prong) and connect that pin to the transformer core as well. Check if your transformer is already mid-point grounded--that's where the transformer part number would come in handy.
H: Energy absorbed by the condenser If the current \$i=0\$ the total energy absorbed is: \$E_{tot}=E_C+E_L\$, where \$E_C\$ is the energy absorbed by the condenser and \$E_L\$ that absorbed by the inductor. Now: \$E_C=\frac{1}{2}C v^2_C(t)=\frac{1}{2} C \Big(\frac{1}{C} \int i\ dt\Big)^2=\frac{1}{2C}(\int i\ dt)^2\$. Since \$i=0\$, it should be \$0\$ too, but my book says it isn't zero. Why? \$E_L=\frac{1}{2}L i^2_L(t)=\frac{1}{2} L \cdot 0 =0\$ simulate this circuit – Schematic created using CircuitLab AI: First of all you have your energy formulas wrong! \$E_C(t) = \dfrac 1 2 C \, v_C^2(t) \$ and \$E_L(t) = \dfrac 1 2 L \, i_L^2(t) \$ But this inconsequential. Your error lies in the integration, because you are neglecting initial conditions on the capacitor. The problem is that the correct formula to express the voltage on a capacitor given the current is: \$ v_C(t) = \dfrac 1 C \int_{-\infty}^t i(\tau) \, d\tau = \dfrac 1 C \int_{t_0}^t i(\tau)\,d\tau + v_C(t_0) \$ where \$t_0\$ is some arbitrary time instant before the time you are evaluating the voltage for, i.e. \$ t_0<t\$. So, even if the current is 0 in the interval \$[t_0,t]\$, you still have a voltage \$v_C(t_0)\$ across the cap which accounts for its stored energy. The current in the capacitor accounts only for the voltage change between the two instants \$t_0\$ and \$t\$. If the current is 0, the voltage won't change, but if it was initially non-zero it will remain the same, as the energy stored in it (bar leakage, in practice). EDIT (To be more precise, in response to a comment) \$ v_C(t_0) = \dfrac 1 C \int_{-\infty}^{t_0} i(\tau)\,d\tau \$ The integral represents the net charge accumulated on the reference plate of the capacitor in the interval \$(-\infty,t_0]\$. This model implicitly assumes that the capacitor was not charged initially, i.e. looooong time before \$t_0\$. To be more specific, it is assumed that \$v_C(-\infty)=0\$. If you wonder where this relation comes from, it is from the "inversion" of the capacitor V-I relationship: \$ i(t) = C \dfrac{dv_C(t)}{dt} \$ which, assuming i(t) is a known function, is an ordinary linear non-homogenous 1st order differential equation of simple type. You can solve it using the method of separation of variables: \$ dv_C(t) = \dfrac 1 C i(t)\, dt \$ Which can be integrated over the interval \$(-\infty,t]\$, yielding: \$ v_C(t) = \dfrac 1 C \int_{-\infty}^{t} i(\tau)\,d\tau \$
H: Serial Communication with rs232/485 and rj45 I am working a project where I want to connect and control, using Labview, TDK-Lambda Power Supply Process Vacuum Controller Ion Gauge (MODBUS interface) They both have rs232/rs485 capabilities, just depends on a switch on the back of the terminal and both connections are with a rj45 jack (I will attach relevant manual). And connect via a RJ45 cable, and I want to connect it to a computer. Now, I've set up a .vi that should be able to connect to, or at least discover, the devices but to no avail. It might be because I am using a regular Cat5E Patch/Ethernet Cable and it therefore not the right pinout. I've tried to read some post and/or white papers about the subject, but these devices are specific and have different pinouts. So my question is, how to I go about using an rj45 cable for serial communications? Can I simply connecting a cat5 cable to my laptops ethernet port and to one of the devices serial rj45 jack? Or am I missing something major? This is my first attempt at anything like this, but I have used Labview to connect to serial DB9 to usb adapters and able to communicate. Here is a link to the PVC Ion Gauge Manual And the TDK-Lambda Power Supply I know I am asking for a lot of help here, but I am super lost and any help is much appreciated. AI: You are missing something major. You CANNOT connect a serial connection to an ethernet port and get anything useful out of it. You will need an adapter from RJ45 to an RS232 conector (usually Sub D9) or to an RS485 connector (Not sure what this looks like.) You will then need the appropriate interface on your PC - that would be a serial port for the RS232, or a special interface for the RS485. Both of these can be had as external units that connect to your PC via USB. You will then need a VI for Labview that can use the adapter. RS232 should be standard, I don't know about RS485. You can probably purchase an adapter from National that will include the correct VI and or drivers. Both manuals include information on how to wire the RJ45 to a serial cable. You will have to either build them yourself or try to order them from the equipment manufacturer.
H: why my LEDs burned with the right driver? I am trying to connect, in parallel, 18 LEDs that have these specs: LED size: 10mm; Voltage: 3.2V - 3.4V; Current: 20mA. Driver Type/Model: MeanWell LPC-20-350 (datasheet: http://www.meanwell.com/search/lpc-20/LPC-20-spec.pdf) From my calculation, by putting 18 LEDs in parallel I need a total current of 360mA, so by driving a (max) current of 350mA i do not need a resistor (to limit the current) and it should not burn the LEDs, however after a blink they all burned. What am i missing? AI: You have two problems. Voltage - the power supply you are using provides a constant 350mA by varying the voltage. Unfortunately, it can only go down to 9Volts. That is too much for one LED. You would need (at least) 3 LEDs in series for that to work. Parallel LEDs - this is a bad idea. LEDs don't all react the same. One will switch on at a slightly lower voltage than the others, and will get most of the current which will cause it to go BOOM in short order. Then the next LED wil get overloaded and the next and the next, and so on until you are out of LEDs.
H: Eagle PCB design issue I have finished the schematic layout for my design which contains transformer. I also found the optimized design 100%, but the transformer background should be black while I'm having only portion of it black, any idea on how to do it? AI: Your polygon pour is filling in (showing as blue). To make your "background black" under the transformer, you need to change the shape of your polygon. When you created the polygon, you had to first draw the outline. All you have to do is click the outside edge of the pour and it should highlight the edges of the whole polygon. You can then drag the sides to change its shape. Not sure why you don't want the pour under the transformer though. Assuming it's insulated with solder mask and your transformer is in a protective case, having the polygon pour completely shouldn't be a problem.
H: Charging a 36V battery set with a 255W solar panel I have directly connected a 255W solar panel (with Vmpp of 49.5V but can peak at 56V in full sun at mid-day) to the 36V battery set of my golf cart. This seems to have been working fine for months, as I am a light user of the golf cart (a few miles per day) in a sunny state. In other words, I have been able go for months without connecting to the grid. However, my research certainly leads me to believe that there is a risk of overcharging (if there isn't a good balance between my driving and the sun-hours, I pressume). My question is whether the overcharging would be due to excessive Amps or excessive Volts? (depending on the answer, I may fashion the correct protection). AI: More info, e.g. battery specs. I would almost certainly guess the batteries can handle much more input current than the panel will ever be able to put out, though. You'd be concerned about overvoltage. From this SO Answer (referring to a single 12V, but you can extend it here): "The maximum is typically around 14.4V to 14.8V at 21C temperature. If the maximum is exceeded for minutes to hours, the battery will be permanently damaged. The damage occurs around the same time that flammable gas is generated and vented, so there is a risk of gas explosion near the battery. Your setup would require constant monitoring, especially if the load is disconnected or turned off. Check the voltage every hour, and disconnect the solar panel as the voltage approaches the maximum. To find the maximum voltage, check the manufacturer's label or datasheet for the battery." (Emphasis added) Based on the three stages of 36V battery charging described here, it seems a reasonable maximum voltage for your setup might be 40.5V. I don't know your batteries though, and I am also assuming that you are using a lead-acid chemistry.
H: How to generate a 1KHz singal on a PIC18F4450? I'm looking for some tips and hints to move on. The 1KHz signal should exit the microcontroller at port RD0. Thanks in advance. AI: As requested, here are tips and hints: Go to http://www.microchip.com and download the datasheet. Read it. No, really, actually read it, particularly the sections on timers. Read section 13 on timer 2 again, but this time notice it has its own period register. Stop and think how this hardware might be used to create a 2 kHz periodic event, at which you can toggle a pin. Since the pin is toggled at 2 kHz rate, its frequency will be 1 kHz. Now read section 15 on the PWM module again. This time think about how it can be used together with timer 1 or 3 to produce a square wave at a fixed frequency divided down from the clock. Muse over which of these methods would fit better within your overall firmware, given what else it has to do and what hardware is available to help do it. Write the code.
H: How to wire screw type banana jacks I recently purchased some of these: Digi-Key Banana Jack. Unfortunately, I'm new and there isn't the supporting documentation I need to be able to determine how they're to be used. Unlike other banana jacks, there is not a hole to feed the wire through and screw down on. I understand the large nut is for holding it against the enclosure, but I don't know what the intention is with the threaded bottom section and the two nuts. I'm guessing that the wire is simply looped around and a nut is tightened down against it, but this doesn't seem very durable. Why the double nuts, what is supposed to be done with them? Edit To clarify, I understand that they are the female piece of a banana connector. I have male banana connectors that I am using these with. My lack of understanding comes from how to wire this female piece to the... for lack of a better word "connection wire". I.e. how to wire this female portion into a circuit so that the circuit is accessible through it. AI: Run down to your local hardware store and get matching ring crimp terminals and non-slip washers. Crimp the terminals onto wires, then slip them in between the nuts on the bottom with a non-slip. This will provide a fairly secure connection, dependent on how well you crimped the terminal of course.
H: Voltage Drift on an Open Circuit (and how to get rid of it) I'm trying to create a custom lick-port to detect mouse licking behavior. I am following a design published here that uses the junction potential between the mouse's tongue and the metal water spout to detect when a lick occurred, see image below: To do this, I'm acquiring the voltage signal as an analog input to a National Instruments DAQ card (PCIe-6363) via an NI breakout box (model SCB-68). I'm reading out the voltage using Matlab R2014a, using the DAQ toolbox to read out voltage from the analog input (ai) channel on the NI DAQ card. My problem is that when the circuit is open, the voltage is slowly drifting to a particular value over about 8 minutes (it looks to be around 2.8V): What I really want this system to do is have a voltage of 0 until the circuit is closed, at which point I can detect a junction-potential that signifies a lick. So, why is this drift occurring? What can I do to prevent it during the open-circuit condition? Here is a sample of what happens when I hold the ground in one hand, and touch a wetted finger to the + wire. Each sharp drop coincides with the circuit closing: Now, I could still potentially use this, just by calculating the derivative and seeing when that exceeds some threshold, but I want to know why this isn't working as expected. Thanks! AI: The slow drift may be due to leakage currents in your sensor. Since the input is open-circuit (theoretically infinite impedance), even a tiny leakage current can cause a significant offset voltage. The simplest solution is to put some resistance across the sensor. 1 MΩ will likely swamp the leakage signal. 10 MΩ may work too. It depends on the magnitude of the leakage and the input impedance of whatever is receiving the raw signal. You want to make the resistance low enough to bring the offset you are seeing down to don't-care levels, but not so low as to significantly load the mouse-tongue-on-metal battery. I suspect the latter has significantly lower impedance than 1 MΩ, in which case that's a good value to use.
H: DIY rubidium clock construction I recently found my way onto a Datum x72 rubidium standard and had half a mind to turn it into a homebrew atomic clock, since I'm not exactly using it for a garage-based satellite business. It seems there are plentiful examples of building atomic clocks using AVRs and PICs online, but most of these appear to feed one of the source-derived inputs (for example, the 1pps signal) into a timer that counts. On the interrupt, there is software action that drives the clock. I suppose the main question is - is there a better, more direct way to do this? I haven't found any RTC modules that would accept a 10MHz signal (seeing as that would drive power draw up, voiding the concept of an RTC), but is there any sort of clocking device or IC out there that could reference the rubidium signal? The software clock-timer system tends to feel like a bit of a hack to me, so if someone could break the components of this project down for me a little better it'd be much appreciated. AI: If the input is down to a 1PPS, then one method of keeping the time calculation at ISR deterministic (neither fast, nor slow, but consistent) would be to actually calculate time one second into the future elsewhere, then use the ISR to update time to the "one second ahead" time. In this case, even relatively slow periodic code could: Keep a previous time and see the ISR advance the time Use the change of time to calculate the "next second's time" to prepare it for the ISR. If the timing of the interrupt is faster than 1PPS, that simply means less time to detect and calculate the 1/x time into the future somewhere else.
H: PIC Microchip Keeps Resetting I have a PIC16LF1709 20-pin 8-bit microchip MCU. I am programming it using a PICkit 3. I am able to successfully burn a program to the chip and run it, but it always seems to reset part-way through execution. While programming the PIC, I'm letting the programmer supply the power, but I can also unplug the programmer and run the chip on its own 5V supply. As recommended, I've run a 50kOhm pull-up resistor from MCLR to Vdd to prevent that from resetting the device. I have five LEDs connected to RC0-RC4, and I've written the test program below to demonstrate the problems I'm experiencing: #include <xc.h> /* XC8 General Include File / #include <pic16lf1709.h> / Definitions of I/O pins */ void main(void) { TRISC = 0x0; // port C (all are LEDs or NC) is output PORTC = 0x0; // turn on all LEDs for(int i=0; i<500; i++) _delay(250); // delay for about a second PORTC = 0xFF; // turn off all LEDs while(1) { // do nothing, forever // lights should remain OFF. } } I expect the lights to turn on briefly, then turn off and stay off until I power-cycle the device. Instead, the lights are blinking. This makes me think that the PIC is being reset about every 1 - 2 seconds. I've tried including the following configuration line (other than this, I didn't specify any configuration) in an attempt to disable some of the features that might automatically reset the device. The behaviour was unchanged. #pragma config MCLRE = OFF, STVREN = OFF, LPBOR = OFF I've also tried letting the execution run while powered by the 3.25v supplied by the PICkit (as opposed to the 5v supplied by my main power supply), but that made no apparent difference either. What could be causing this behaviour? AI: As Eugene Sh. correctly pointed out. The most likely culprit is the watchdog timer. If you look at the specification sheet (http://ww1.microchip.com/downloads/en/DeviceDoc/40001729A.pdf) page 99, you can see that the default timeout for the watchdog timer reset is 2s. Also, on page 48, you can see that the default configuration bits enable the watchdog timer regardless of runtime software setting. Your issue may be solved with the addition of the line: #pragma config WDTE = OFF If that does not solve your issue, you can get the source of the last reset by checking the value of the PCON and STATUS registers ( see section 5.12 Determining the Cause of a Reset in the specification sheet at page 58 ).
H: How to make USB smartphone charger? I want to make a USB cell phone charger. I know USB operates in 5 volts which means i need to drop down voltage if i have battery potential greater than 5 Volts. I have 12 volts sealed lead acid battery (SLA) of capacity 9AH. I need to drop down the voltage for which i am using LM2596 buck regulator ( Switching power supply) and it can deliver 3 amperes of current. But my cell phone charger(came with the product) is rated for 2A. So i was wondering how can i limit the current so that it doesnot blow my Cell phone. Thanks AI: The cell phone charger spec is a maximum. I.e. the maximum current that can safely be drawn from the charger is 2A. If the buck converter is rated for 3 amps, then you have more than enough current available to charge your phone. Your phone will draw the current it needs. Some phones (especially Apple devices) are finicky about the data lines. You can't just leave them floating (with nothing connected). Adafruit has a good discussion about reverse engineering the voltages that are required on the USB data lines for charging Apple devices.
H: Smoke detectors replacing - why? Lifetime of components? I often see that smoke detectors "should be replaced" every 5, 10 or 20 years. Like on this website for example: http://modernsurvivalblog.com/health/did-you-know-smoke-detectors-expire/ The U.S. Fire Administration says most smoke detectors installed today have a life span of about 8-10 years. After this time, the entire unit should be replaced. I wonder why? It probably has nothing to do with half-life of Americium 241(because it is 432 years), so what is the reason? I though about electrolytic caps, but mine for example has only one and its lifespan is often much higher than 10 years(if it is not piece of crap). I don't see any other electronic-related problems with the circuit in this things, it is pretty easy and hardly anything can break inside if it just sits on the ceiling doing nothing 99% of the time. Is there logical reason or is it "just in case"? AI: As you suspected, apparently it has nothing to do with the radioactive material in those alarms that use it. According to the Reliability section for smoke detectors: The NFPA strongly recommends the replacement of home smoke alarms every 10 years. Smoke alarms become less reliable with time, primarily due to aging of their electronic components, making them susceptible to nuisance false alarms. In ionization type alarms, decay of the 241Am radioactive source is a negligible factor, as its half-life is far greater than the expected useful life of the alarm unit. So it is just a matter of component reliability. Although we are used to some electronic devices working for decades, we also accept that sometimes (but very infrequently) they just up and die, and perhaps on a statistical basis 10 years has been found to be a good marker for these alarms where one can be assured some large number of units (99% or higher?) will still be working. Anecdotally, the house we are currently living in was built 20 years ago. We bought it 10 years ago. It has 10 smoke alarms in it. I never knew about the "replace every 10 years" rule. About two months ago, some of the alarms started to go off on their own (false alarms). After this happened a couple of times, I replaced them all.
H: Doubt about potential difference through a resistor? First of all let me give you the schematic diagram of the circuit, Here, how to find the current passing through 150 ohm resistor? These are the things I have tried, Since I want to find the current through 150 ohm resistor I need the potential drop across that resistor , therefore in order to do find the potential drop across 150 ohm resistor what I did was I marked the nodes as shown below After that I took the voltage through-out (green)node as 10V(10V battery is on green node) , likewise I took the voltage throughout red colour node as 5V. Then I could find the voltage drop through 150 ohms, so let's assume it as K Therefore 10V - k = 5V So K=5V Then the current through that resistor is , V=IR 5V/150 I = 0.03A But the answer is wrong, The correct answer given is 0.1A. What is the mistake I have done? Please help me AI: An alternative method to solve this would be using the superposition theorem. It isn't really required in this case, but in more complicated setups with multiple sources it is certainly useful. Here is how you do it. 10V source 5V source 1. simulate this circuit – Schematic created using CircuitLab Looking at the diagram above it is clear that the current through R3 is $$ I_{R3}=\frac{V}{R_3} = \frac{10}{150}=\frac{1}{15}\text{Amps} $$ 2. simulate this circuit Because of the short R2 and R1 are in parallel and become zero and therefore the current through R3 is: $$I_{R3}=\frac{V}{R_3}=\frac{5}{150}=\frac{1}{30}\text{Amps}$$ Now because the direction of current assumed in part 1 and 2 is the same, both the currents are additive. Hence: $$I = \frac{1}{15}+\frac{1}{30} = 0.1\text{Amps}$$
H: How electric devices can sustain at such a huge voltages? This is a common experience.When we buy any electricity run device ,it is given with it's specifications like it's operating voltage,current ranges. Also,normally power supply is 240 V , 50 Hz. But when we turn ON the switch, the electric devices have to face very high voltage level for initial microsecond or millisecond time interval. Please consider the below equations. current=dq/dt ( where q= electric charge) and voltage =current resistance. So for extremely small time interval,the voltage is very high. So, I think that it is higher than operating voltage range of the device. So my question is how any device (for example light bulb) can sustain at such a high voltage ? can anybody explain with the help of mathematics and voltage equation,what happens for that extremely small interval of time? AI: Your assumption is wrong. A high inrush current does not mean there's a high voltage. In electronic circuits, inrush current is caused by capacitors charging up. In a light bulb, the filament's resistance is lower when the bulb is cool, which allows more current at startup. The input voltage from the AC power line stays at 240 VAC. EDIT: Here's an example circuit. I'm using DC to make it simple, but the same concepts apply to AC. simulate this circuit – Schematic created using CircuitLab When the voltage is first connected, the capacitor acts like a short circuit. This means the current is: $$I_{in} = \frac {240\ \mathrm V} {0.1\ \Omega} = 2400\ \mathrm A$$ That's very large! But after the capacitor charges up to 240 V, it acts like an open circuit. Then the current is: $$I_{in} = \frac {240\ \mathrm V} {0.1\ \Omega + 1000\ \Omega} \approx 0.240\ \mathrm A$$ That's much smaller! But the input voltage never changed. Ohm's Law (voltage = current resistance) only works for resistors. There are other kinds of devices that don't obey Ohm's Law. Here's a light bulb example. According to Wikipedia, the resistance of a 100 W, 120 V bulb when the bulb is cold is about 9.5 ohms. When the bulb is hot, the resistance is about 144 ohms. So when you first flip the switch, the current is: $$I = \frac V {R_{cold}} = \frac {120\ \mathrm V}{9.5\ \Omega} \approx 12.6\ \mathrm A$$ After a tenth of a second, the current is: $$I = \frac V {R_{hot}} = \frac {120\ \mathrm V}{144\ \Omega} \approx 0.83\ \mathrm A$$ (I'm leaving out some differences between AC and DC circuits, but this should give you the right idea.)
H: Considerations when repairing dimmer switch I originally asked this question on DIY StackExchange. So far no answers and a suggestion to try on Electronnics, so here goes: I have an IKEA lamp with dimmer switch which took a knock and immediately started outputting full brightness. I've read a little about how dimmer switches work and that this is likely caused by the triac breaking. I would like to repair this by replacing the triac (/diac combination?) but want to make sure that I have the right idea and check if there are any risks in doing so. Photos: As I understand it the block in the middle with the metal cover is the triac, possible also a unit containing a diac. Then the obvious variable resistor connected to the outside, a fuse (looks like a second fuse attached under the metal hood -- and labelled on the board -- but can't see it properly until disassembly), some resistors (baseline resistance? "taper"?) and finally the inductor and capacitor smooth out the voltage cut, reducing stress/noise on/from the load. I plan to desolder the triac unit and replace it with an equivalent part (if I can find one). Is there a good way to confirm the triac is broken in advance? Are there any safety concerns? UPDATE: I have since confirmed that the metal cover in the middle contains a triac (including make/model) and a thermal cutoff. AI: To test the triac out-of-circuit you can use an ohmmeter on the lowest range. You need to find the datasheet. If it's the type with a built-in diac you won't be able to easily test the gate with low voltage. Test for continuity between MT1 and MT2, in both directions. If you find continuity in either (or both) direction(s) then the triac is knackered, end of story. This is the most likely failure mode. It is typically caused by an overcurrent surge even such as arcing during the failure of an incandescent bulb. Another test that can be done is to hold the meter terminals (using the lowest ohms range) on MT1 and MT2 and momentarily touch MT2 to gate without breaking either of the connections. The meter should show continuity as long as the gate is touching, and may continue to show continuity after the the gate connection is broken. You can test it both ways. A more functional test that is a bit more trouble to set up is to get a 12VAC transformer and a 12V few-watt light bulb (maybe an automotive bulb) and connect the triac in series with the bulb to the transformer secondary. Use a 100 ohm resistor to momentarily connect gate to MT2. The light should go on as long as the gate is connected. If the light does not turn on or the resistor gets really hot the triac is bad. If the light goes on without the resistor (full or partial brightness) then the triac is dead. Personally, I would order the triac immediately without testing. There is no other plausible failure that will cause the light to remain on at full brightness, and the triac failure is likely the only failure.
H: MOSFET driver IR2301S I am trying to implement a brushless motor driver using an IR2301S driver. I started with testing the driver without MOSFETs. Here is schematic: In the datasheet of the IR2301S is this picture: If I understood correctly, I should see oscillating on LO if there is an oscillating signal on LIN and unconnected HIN, VS. I got it. But when I send an oscillating signal on HIN while LIN and VS are unconnected I don't see oscillating on HO; there is stable 6 V on HO. Input signal amplitude is 3.3 V. I've tried to change VCC up to 12 V, change the capacity of C1 and C4 from 1 μF to 100 μF without luck. What am I doing wrong? AI: Here is the block diagram: - Without connecting Vs to 0V and Vb to +V you will see nothing on HO
H: Making function using 2x4 decoder and one or gate I frankly don't understand how to make it. I try to make but, I think it's wrong. How can I make it ? Function is: $$f(a,b,c) = ∑(0,1,2,3,7)$$ My trying: In addition using 3x8 I make: AI: The problem stated in the question looks to be impossible to solve. I will try to prove it here. The function required is $$F=\bar{A}+ABC=\bar{A}+BC$$ First let's look at two different(and the only possible) configurations: 1) The OR gate's output is connected to one of the sel inputs of the decoder. 2) The OR gate's output is the function output. The first configuration assuming two of the function inputs to be connected to the OR inputs, and the third connected to the decoder input (and might be connected to OR as well): simulate this circuit – Schematic created using CircuitLab For the 2-input OR variant the outputs are given as follows: x y z Y0 Y1 Y2 Y3 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 0 1 0 0 0 1 0 0 1 0 1 0 0 0 1 1 1 0 0 0 0 1 1 1 1 0 0 0 1 As we can see neither of Y can be represented as SOP of 4. So this configuration won't work. For the 3-input or variant: x y z Y0 Y1 Y2 Y3 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 0 0 0 0 1 1 1 1 0 0 0 1 Only \$Y_3\$ is SOP of four, but it's obvious that the function is \$Y_3=x\$, and no assignment of the original variables to \$X,Y,Z\$ will make it the required function. As for the second approach, while the OR is connected to the Decoder outputs, there is only two inputs on the decoder, then the third one has to be connected to the OR itself. But then the output will be always 1 when that input is 1. But, as can be seen from the truth table, there is at least one case for each input \$A,B,C\$ where it is 1, but the output is 0. Then this configuration doesn't work as well.
H: What proximity detection technologies can work through glass? I'm looking for sensor options for proximity detection through a pane of glass. Something that will sit 6-24" on one side of a pane of glass, and be able to detect presence the same (roughly) distance on the other? I assume ultrasound won't work, but am not sure about infrared. If not, are there other options? Edit per this link from the comments, most glass is opaque to IR, so no, it won't work. That leaves the second part of the question - are there ways to detect presence / motion on the other side of a pane of glass. Video-based motion detection won't work because I need to be able to restrict distance (i.e. something present/moving within a fairly tight distance of the glass). Use case is something along the lines of a home security motion sensor looking out the window. AI: You could use a microwave presence sensor e.g. this one which mentions glass specifically In general, they will work through glass, without a big reduction in range. They are often combined with PIR detectors to reduce false triggering (from sources of heat), but usually configurable so you can only use the microwave part.
H: CC2520/ATtiny861A pin configuration I am looking to hook up a radio with an ATtiny861A microcontroller. The radio that I will be using is the CC2520 from Texas Instruments. The RESETn pin(on the CC2520) will allow me to put my circuit into a sleep mode and this requires me to have the reset pin attached to the MCU(The datasheet says that I could reset it through SPI but I want to know if I can use one of the pins that I have left over). The RESETn Pin says it is a TBOUTH/SVSOUT. Leftover on my 861A MCU I have pin 4 which is OC1B/PCint11, pin 17 which is AREF/PCINT3, and pin 9 which is ADC9/INT0/T0/PCINT14. Am I able to connect any of these pins to the RESETn from the CC2520 radio? Thanks AI: From the CC2520 datasheet it looks like all you need to do is control the resetn of the CC2520 with any GPIO pin. As long as you can drive the pin high and low you should be fine. Any of those pins should work provided they are configured as outputs. I might consider adding an external pullup resistor on that resetn line if you don't want the CC2520 to be reset when the MCU gets reset, otherwise it shouldn't really matter.
H: CS/SS pin configuration I am using an ATtiny861a and I want to enable it to use SPI. In doing this I need to assign not one, but two CS/SS pins in order to communicate with a radio and an external memory. Does the CS/SS pin need to be connected to an ADC pin or could it be connected to a AREF, PCINT, or OC1A/OC1B pin? AI: I believe you can use any GPIO pin as CS/SS. Keep in mind you will need a different pin for each slave device. For example say device0 SPI CS is connected to PORTD pin 5, and device1 SPI CS is connected to PORTD pin 6: //init ports DDRD \|= (1 << 5) \| (1 << 6); //sets PORTD bits 5 and 6 to output PORTD \|= (1 << 5) \| (1 << 6); //sets PORTD bits 5 and 6 to high (spi CS inactive) //spi device0 PORTD &= ~(1 << 5); //resets device0 CS (makes it active) //SPI send/receive stuff here PORTD \|= (1 << 5); //sets device0 CS (makes it inactive) //spi device1 PORTD &= ~(1 << 6); //resets device0 CS (makes it active) //SPI send/receive stuff here PORTD \|= (1 << 6); //sets device0 CS (makes it inactive)
H: I need help finding the equivalent resistance Hello guys , i'm new in the amazing world of electronics , and im very confused with this circuit . Im supposed to find the Req (equivalent resistance , but dont know how in this specific circuit), plz help me out . AI: The reason that this is difficult upon initial inspection is because of the way that the circuit is drawn. First, look for common nodes, and re-draw the circuit to clearly show the parallel and series sets of resistors. This will make it much easier to simplify further: simulate this circuit – Schematic created using CircuitLab I will not solve it for you, because if this is homework, that defeats the purpose. I hope this helps!
H: Current source with op amp and transistor I'm building current source with OP AMP LM358 and MOSFET transistor IRF5305. Here's schematic diagram of the circuit: I have two questions about it: 1) Maximal gate-source voltage of IRF5305 is +/-20V. Since V+ is 24Vdc, is it a good idea to add 10V zener diode between source and gate to protect MOSFET? 2) Why they use voltage divider connected to non-inverting input and apply reference voltage to inverting input ? Here's an example of current sink schematic diagram with the same OP AMP. I'm assuming current sourcing can be done by "inverting" this circuit. Is it so ? Thank you ! AI: is it a good idea to add 10V zener diode between source and gate to protect MOSFET? It wouldn't hurt. You'll want to make sure that the maximum current through the Zener isn't enough to burn out the op-amp if this protection circuit ever operates (for example if RL is broken giving an open-circuit output). You could also just add a resistor between gate and source, so that the divider between the existing 100 ohm resistor (whose value you'll want to increase and your new resistor never allows Vgs < -20 V. You'll slightly degrade the accuracy of the output current since the current through the resistor divider will be counted by the sense circuit but not go through your load. You might also need to recheck the circuit's phase margin with this arrangement. I'm assuming current sourcing can be done by "inverting" this circuit. Is it so ? You could invert the current sink circuit, but it would make the control signal more complicated. To get 100 mA output, for example, you'd need to provide an input 100 mV below V+, instead of 100 mV above ground. If you're sourcing the input signal from a ground-referenced DAC, for example, you'd need to jump through some hoops to get an accurate output current.
H: How long can a brand new PCB last before additional protective coating is required if any? I have just made several PCBs for a large project using the photoresist method. (I ran the boards through sodium silicate developer then through ferric choride) I then rinsed off all chemicals and dried the boards. I believe there is some chemical electronic stores sell (conformal coating? I don't know) that would protect the circuit board from any sort of corrosion (which can include introduction of open circuits). I know the next steps are to drill holes and solder components in, but after that process is complete, how long can I leave the board like that before it is absolutely necessary to add some sort of protective coating to it? I want my board to last forever but at the same time, I don't want it to corrode (or have traces break up) as the years go by. AI: It will last as long as you keep condensation away from it or other solvents. Ionic solutions are what degrade metals, so the easiest way would be to keep it from those things. That being said, feel free to apply your own soldermask Polyurethane (humiiseal) also may go a long way toward keeping moisture out. Source: https://learn.adafruit.com/how-to-solder-mask-pcbs/overview Make sure you either, mask the metal you still want to solder to before applying the soldermask. And conformal coatings will coat everything and make for hard reworking and should be applied after assembly.
H: LTC1775 Buck converter component question I am trying to design a 12v,10A buck converter using a with LTC1775 and have gone through the calculations assuming a 40% inductor ripple current and a switching frequency of 150kHz. This gave an inductor value of 10uH and a mosfet with Rds(on) of 0.014ohm. So based on these i have selected IRLR3915 NMOS and 7443558100 inductor. My main concerns are the following: Are these two good choices with regards to the output current or they are overkill? Should i attach a flyback diode across the output incase I have an inductive load like a motor? Are the Cin and Cout capacitors adequate? Based on datasheet formulas our output voltage ripple is supposedly ~60mv but the simulation only shows ~10mv here is the circuit schematic: and here is a inductor current on LTspice, as you can see the current does reach ~15A: I can attach the calculations if its needed. Thank you, I really appreciate your guys help. AI: Are these two good choices with regards to the output current or they are overkill? Use LT powercad (almost all linears DC DC conveters are there) to find optimal components, it is much easier to use and quicker (I'd run this design through myself but haven't the time). Should i attach a flyback diode across the output incase I have an inductive load like a motor? Yes, you may also want to consider a filter, the DC DC compensation loop may also present problems from the inductance of the motor. Extra capacitance\filtering may be advantageous with a motor. Because the motor impdeance is hard to determine (and for most motors non-linear) it may require experimentation to get the right compensation\filtering and to avoid voltage dropouts. Are the Cin and Cout capacitors adequate? Based on datasheet formulas our output voltage ripple is supposedly ~60mv but the simulation only shows ~10mv You have a fixed load, switch the load in the simulation to determine if it really is adaquate (unless your load really is fixed).
H: Calculate battery duration Maybe this is a trivial question, but I am struggling with it for a while. I have an alkaline AA battery which is consuming 80mA for 8 seconds each 30 minutes. I need to calculate how many days it will last before reaching 1.1v. I know it depends a lot on the battery type, but let's say it is a Duracell battery. I have problems in calculating the "average" consumption per hour, as the battery is drained for just 8 seconds (80mA) each 30 minutes. Thanks for any help. AI: Average current is given by $$ I_{AVG} = \frac {80m \times 8}{30 \times 60} = 0.35 \ \text {mA} $$ Figure 1. The Duracell AA Plus Power quotes for currents down to 5 mA continuous. This shows that at 5 mA you can expect > 500 hours. You'll have to extrapolate for your load conditions.
H: Mesh Current Polarity, Equation Setup Practicing Mesh Analysis I came across this problem and te answer is really alluding me and I believe is just one polarity that I'm not understanding, the problem is the following... Vs=10V, R1=100Ω, R2=50Ω, R3=25Ω, Is=2A. Calculate the current I1 in amps that goes through R3 from left to right. Enter >only the numerical answer for I1 in the text box, omit the units. Note: this >problem can be done different ways, but try using Mesh Analysis. Now, I'm setting up my equation using the passive sign convention as following. Vs voltage rise R2 voltage drop R3 voltage rise (since current is flowing clockwise entering the "negative" terminal of R3) so it would look something like this: -Vs+R2(I1-I2)-R3(I1)=0 -10+50(I1-(-2))-25(I1)=0 Now the answer to the question is 1.2 which I could achieve if the voltage across R3 was in fact " 25(I1) " instead of negative as in my equation, the thing is that I don't understand what I'm doing wrong and I don't Understand why it is positive if it is a voltage rise? AI: When you assume the current direction in the loop you automatically set the voltage polarity across the resistors (current flow from + to - in the resistor). Hence, you assumed the clockwise flow. Therefore this forces you to stick to this assumed direction and the voltage across the resistors. And you should forget about the VR3 polarities shown on the diagram. Case one: And the equation (notice that in this case only one equation is needed) $$(I_1+ I_S) R_2 + I_1R_3 - V_S = 0$$ And the solution is \$I_1 = -1.2A\$ which means the \$I_1\$ current is flowing in opposite direction than we have assumed. Case two $$V_S + I_1 R_3 + (I_1 + I_S)R_2 = 0$$ Additional we see that \$I_S = -2A\$ So, the solution is \$I_1 = 1.2A\$ EDIT For each individual mesh, you can pick the loop current direction arbitrarily. Look at this example Loop one and two have the same loop current direction (clockwise). So for loop one we have I start at point B $$I_13\Omega + 2V + (I_1 - I_2)10\Omega + I_14\Omega - 10V = 0 $$ (notice that I1 is first here (I1 - I2)10 ) And the second loop (start at point A) $$ I_28\Omega - 15V + (I2 - I1)10\Omega - 2V = 0$$ In this case loop I2 "is first" (I2 - I1)10 And the solution is: \$I_1 = 1.52427A\$ , \$I_2 = 1.79126A\$ And now in this example, I pick the loop current direction this way: As you can see I1 is clockwise but I2 is counterclockwise. And the equations look like this: Loop one $$I_13\Omega + 2V + (I_1 + I_2)10\Omega + I_14\Omega - 10V = 0 $$ Do you see the defense? Loop two: $$2V + (I_2 + I_1)10\Omega + 15V + I_28\Omega = 0$$ And the result is: \$I_1 = 1.52427A \$ \$I_2 = -1.79126A\$ And this minus sign in the final result tell us the I2 current is, in fact, flowing in the opposite direction then I assume.
H: Connector pins protection from overvoltage I have a connector which has the following lines: 12V 3.3V GND SDA (I2C) SCL (I2C) The connector might get wet with water and I would like to protect the circuit from the resulting over-voltage in case the 12V line gets connected to the other lines. Based on this schematic from this question Why is a resistor needed in zener protection circuit?: I've designed this connector using a 3.6V zener diode and 330Ohm resistors. I don't understand why R1 and R2 are required and I've removed them in mine: The connection would look like this: Would this solution protect the connections from the 12V line? Is there a better approach? AI: Pull-ups are required on SDA and SCL to ensure that they are high when no signal is present. See the I2C details (try Wikipedia).
H: KiCAD 5 custom footprints I want to use the TPA3251. Package is TSSOP 44 pins, 0.635mm pitch. KiCAD's built-in library does not have it. (as the subject indicates, I'm using KiCAD 5). I created the library symbol without any trouble. Well, sort of no trouble — I had to change ownership of everything under /usr/share/kicad so that I could modify the file /usr/share/kicad/library/Audio.lib (I suspect that this part is already not quite right). I added it as an entry inside the Audio library. This is what it looks like: # # TPA3255 # DEF TPA3255 U 0 40 Y Y 1 F N F0 "U" 0 500 50 H V C CNN F1 "TPA3255" 0 50 50 H V C CNN F2 "Package SSOP:SSOP-44_7.5x13.3mm_P0.635mm" 0 0 50 H I C CNN F3 "" 0 0 50 H I C CNN DRAW S -450 450 450 -1850 0 1 0 f X GVDD_AB 1 -550 350 100 R 50 50 1 1 W X OSC_IOP 10 -550 -550 100 R 50 50 1 1 O X DVDD 11 -550 -650 100 R 50 50 1 1 W X GND 12 -550 -750 100 R 50 50 1 1 W X GND 13 -550 -850 100 R 50 50 1 1 W X AVDD 14 -550 -950 100 R 50 50 1 1 W X C_START 15 -550 -1050 100 R 50 50 1 1 O X INPUT_C 16 -550 -1150 100 R 50 50 1 1 I X INPUT_D 17 -550 -1250 100 R 50 50 1 1 I X ~RESET 18 -550 -1350 100 R 50 50 1 1 I X ~FAULT 19 -550 -1450 100 R 50 50 1 1 O X VDD 2 -550 250 100 R 50 50 1 1 W X VBG 20 -550 -1550 100 R 50 50 1 1 w X ~CLIP_OTW 21 -550 -1650 100 R 50 50 1 1 O X GVDD_CD 22 -550 -1750 100 R 50 50 1 1 W X BST_D 23 550 -1750 100 L 50 50 1 1 W X BST_C 24 550 -1650 100 L 50 50 1 1 W X GND 25 550 -1550 100 L 50 50 1 1 W X GND 26 550 -1450 100 L 50 50 1 1 W X OUT_D 27 550 -1350 100 L 50 50 1 1 O X OUT_D 28 550 -1250 100 L 50 50 1 1 O X PVDD_CD 29 550 -1150 100 L 50 50 1 1 W X M1 3 -550 150 100 R 50 50 1 1 I X PVDD_CD 30 550 -1050 100 L 50 50 1 1 W X PVDD_CD 31 550 -950 100 L 50 50 1 1 W X OUT_C 32 550 -850 100 L 50 50 1 1 O X GND 33 550 -750 100 L 50 50 1 1 W X GND 34 550 -650 100 L 50 50 1 1 W X OUT_B 35 550 -550 100 L 50 50 1 1 O X PVDD_AB 36 550 -450 100 L 50 50 1 1 W X PVDD_AB 37 550 -350 100 L 50 50 1 1 W X PVDD_AB 38 550 -250 100 L 50 50 1 1 W X OUT_A 39 550 -150 100 L 50 50 1 1 O X M2 4 -550 50 100 R 50 50 1 1 I X OUT_A 40 550 -50 100 L 50 50 1 1 O X GND 41 550 50 100 L 50 50 1 1 W X GND 42 550 150 100 L 50 50 1 1 W X BST_B 43 550 250 100 L 50 50 1 1 W X BST_A 44 550 350 100 L 50 50 1 1 W X INPUT_A 5 -550 -50 100 R 50 50 1 1 I X INPUT_B 6 -550 -150 100 R 50 50 1 1 I X OC_ADJ 7 -550 -250 100 R 50 50 1 1 B X FREQ_ADJ 8 -550 -350 100 R 50 50 1 1 O X OSC_IOM 9 -550 -450 100 R 50 50 1 1 B ENDDRAW ENDDEF I can use the symbol. However, the footprint won't work. I tried several ways; in the end, this is what I got with the footprint editor, using the footprint wizard. The following is the contents of the file /usr/share/kicad/modules/Package_SO.pretty/SSOP-44_7.5x13.3mm_P0.635mm.kicad_mod: (module SSOP-44 (layer F.Cu) (tedit 5B815739) (attr smd) (fp_text reference REF** (at 8.89 -12.1775) (layer F.SilkS) (effects (font (size 1 1) (thickness 0.15))) ) (fp_text value SSOP-44 (at 8.89 -3.81 -90) (layer F.Fab) (effects (font (size 1 1) (thickness 0.15))) ) (fp_line (start 6.39 -11.31) (end 11.39 -11.31) (layer F.CrtYd) (width 0.05)) (fp_line (start 6.39 3.69) (end 6.39 -11.31) (layer F.CrtYd) (width 0.05)) (fp_line (start 11.39 3.69) (end 6.39 3.69) (layer F.CrtYd) (width 0.05)) (fp_line (start 11.39 -11.31) (end 11.39 3.69) (layer F.CrtYd) (width 0.05)) (fp_line (start 6.54 3.5575) (end 6.54 -10.3775) (layer F.SilkS) (width 0.15)) (fp_line (start 11.24 3.5575) (end 6.54 3.5575) (layer F.SilkS) (width 0.15)) (fp_line (start 11.24 -11.1775) (end 11.24 3.5575) (layer F.SilkS) (width 0.15)) (fp_line (start 7.34 -11.1775) (end 11.24 -11.1775) (layer F.SilkS) (width 0.15)) (fp_line (start 6.54 -10.3775) (end 7.34 -11.1775) (layer F.SilkS) (width 0.15)) (pad 22 smd roundrect (at 5.14 2.8575 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 23 smd roundrect (at 12.64 2.8575 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 21 smd roundrect (at 5.14 2.2225 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 24 smd roundrect (at 12.64 2.2225 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 20 smd roundrect (at 5.14 1.5875 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 25 smd roundrect (at 12.64 1.5875 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 19 smd roundrect (at 5.14 0.9525 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 26 smd roundrect (at 12.64 0.9525 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 18 smd roundrect (at 5.14 0.3175 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 27 smd roundrect (at 12.64 0.3175 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 17 smd roundrect (at 5.14 -0.3175 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 28 smd roundrect (at 12.64 -0.3175 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 16 smd roundrect (at 5.14 -0.9525 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 29 smd roundrect (at 12.64 -0.9525 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 15 smd roundrect (at 5.14 -1.5875 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 30 smd roundrect (at 12.64 -1.5875 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 14 smd roundrect (at 5.14 -2.2225 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 31 smd roundrect (at 12.64 -2.2225 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 13 smd roundrect (at 5.14 -2.8575 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 32 smd roundrect (at 12.64 -2.8575 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 12 smd roundrect (at 5.14 -3.4925 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 33 smd roundrect (at 12.64 -3.4925 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 11 smd roundrect (at 5.14 -4.1275 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 34 smd roundrect (at 12.64 -4.1275 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 10 smd roundrect (at 5.14 -4.7625 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 35 smd roundrect (at 12.64 -4.7625 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 9 smd roundrect (at 5.14 -5.3975 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 36 smd roundrect (at 12.64 -5.3975 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 8 smd roundrect (at 5.14 -6.0325 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 37 smd roundrect (at 12.64 -6.0325 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 7 smd roundrect (at 5.14 -6.6675 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 38 smd roundrect (at 12.64 -6.6675 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 6 smd roundrect (at 5.14 -7.3025 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 39 smd roundrect (at 12.64 -7.3025 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 5 smd roundrect (at 5.14 -7.9375 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 40 smd roundrect (at 12.64 -7.9375 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 4 smd roundrect (at 5.14 -8.5725 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 41 smd roundrect (at 12.64 -8.5725 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 3 smd roundrect (at 5.14 -9.2075 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 42 smd roundrect (at 12.64 -9.2075 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 2 smd roundrect (at 5.14 -9.8425 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 43 smd roundrect (at 12.64 -9.8425 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 1 smd roundrect (at 5.14 -10.4775 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) (pad 44 smd roundrect (at 12.64 -10.4775 270) (size 0.4 1.8) (layers F.Cu F.Paste F.Mask) (roundrect_rratio 0.25)) ) Problem is, the footprint won't show up when assigning footprints — it initially does show the package indicated in the library file, but nothing shows on the PCB editor. I try removing the footprint associations to assign manually; the SSOP-44 0.635mm pitch won't show on the list. I try filtering typing SSOP, typing 44, typing 0.635. Nothing. What am I doing wrong? AI: Let's start with library setup (probably not the base issue but can't hurt). Make your own libraries Do not modify the base libraries included in your KiCad package. The base libraries will be overwritten by upgrading your kicad package or downloading new copies of the libraries. Make a directory in a location that has r/w permissions for you called kicad_libraries under which you can create symbols and footprints directories. Make a Symbol Fibrary Open Symbol Editor Right-click on the library list on the left and choose "New Library" Choose the kicad_libraries/symbols location you made earlier If you are creating symbols that you'll want to re-use then choose the Global library table when prompted. Otherwise choose Project. Now right-click on the library you created in the list and choose New Symbol to create your new symbol. or pick a symbol that is close, right-click on its name in the full list, copy and then paste the copied symbol into your new library before editing. Make a Footprint Library Open Footprint Editor Create your new footprint Or, in your case, you can import the footprint you already made Click on the button that looks like a library Choose the kicad_libraries/footprints location you created earlier and specify a new library name. Choose Manage Footprint Libraries from the top menu bar. Select either the Global Libraries or Project Specific Libraries in the top tab as before depending on whether you want to re-use this footprint or not. Click the Browse Libraries button and navigate to your new footprint library and select OK Using cvpcb to Assign Footprints The reason your footprint isn't showing up is almost certainly due to filtering. Look at the top button bar and see which buttons are selected. The first button on the left filters by keywords in the schematic symbol. The next button filters by pin count (do the number of pins in the symbol match the pins in the footprint?) The next button filters by selected library in the left pane of the window. The final button allow filtering by type-in strings If you unselect all buttons, you will see all footprints.
H: Trying to solve a RL circuit by hand but simulator calculates a value I do not understand I have this circuit. V3 is just a trigger to close the switch, just ignore it. The switch is open for a long time. So, the left part of the circuit is working for a long time and the inductor is fully energized. Current is maximum on that part. At t=0 the switch is closed. I want to know the voltage across the inductor 3ms after the switch is closed. I put this circuit on a simulator and it gives me the voltage across the inductor at 3ms as -11.3022 V. Strangely, the simulator shows a peak of -12V when the switch is closed and 3ms later the voltage is -11.3022 V. -12V? Where is this number coming from??? I am trying to calculate that voltage by myself, but the result I get is completely different. This is my solution. Before the switch is closed the voltage across the inductor is zero. The switch closes. Now, a -15V (V2 is reversed) is connected with R2 in parallel with the inductor. What I thought was this. Voltage across the inductor was zero before, now it is -15V. So it will discharge from -15 to zero. I was expecting to see a -15V peak when the switch is closed but simulator says the peak is at -12V. I would use the inductor formula having -15V as a start but simulator apparently uses -12V. How do I solve this? AI: I give you a tip. Before the switch is close the is a current in the inductor. And this current is equal to \$I_L =\frac{20V}{200\Omega} = 100\textrm{mA} \$ So at the very first moment of time (t=0) when we closed the switch. The equivalent circuit looks like this: simulate this circuit – Schematic created using CircuitLab As you can see I replace an inductor with a current source. Because this is how the inductor will behave at time t = 0 when you closed the switch. because the inductor current cannot change instantaneously. And if you do the calculation you will get that the voltage at the \$V_X\$ node is indeed equal to \$-12V\$ And the circuit time constants is \$ \tau = \frac{L}{R_1\|\|R_2}\$.
H: Laser pointer circuit I've got a small red laser pointer from a chinese ultrasonic distance meter, which has a small regulator circuit (6mm x 7mm) with SMD components. I guess that is a constant current to drive the laser, but I was unable to understand the circuit. I've used a microscope and a continuity tester to check all connections several times, as I was thinking I got some wire incorrectly attached. But I was not able to find any error. The transistors are labeled L6 in a SOT-23 package, and I assume they are KST1623-L6 or 2SC1623-L6, both similar NPN transistors. I was not able to measure the capacitor without desoldering it, and I'm not confident in being able to take it apart and put it back being so small (~ .5mm x 1.4mm). I powered the circuit using my Arduino regulated output at 3.3V during about 10 seconds, and the laser lights up, without any overheating observed. I don't understand how Q2 works, as the base is attached to GND with 39 Ohms resistor, so as far as I know, ideally, it should be in cutoff mode. I'm curious how this circuit can work and why they decided to use it. AI: simulate this circuit – Schematic created using CircuitLab Figure 1. A possible correct version of the circuit. Have another look and see if it's actually the circuit of Figure 1. How it works: R1 provides bias to Q1 which starts to turn on. D1 starts to light up. When the D1 current through R2 reaches about \$ I = \frac {V}{R} = \frac {0.7}{39} = 18 \ \text {mA} \$ Q2's base will be forward biased, it will turn on and steal the bias away from Q1. (The turn-on point is 0.7 V.) The circuit will stabilise at 18 mA. This seems reasonable and will result in consistent operation across a range of battery voltage.
H: Power supplies with or without 'COM' and without ground port at all This is probably something really simple, but I wasn't able to find the answer yet. I've been using a power supply with both COM and earth ground, like the first image below. To power an op-amp, I was using COM for grounding. Then, I came across a power supply like the second image below, which doesn't have COM. Also, I've seen a power supply with no ground port whatsoever like the third image below. I heard the earth ground is noisier. Why do some power supplies have both COM and earth ground, but others don't? And if I have to use one without COM or no ground port at all, what are my alternatives? AI: simulate this circuit – Schematic created using CircuitLab Figure 1. Various options. Photo 1 The first photo shows a PSU with configuration of Figure 1a. There are two isolated supplies - isolated from each other and from mains earth. Normal use would be to connect 5V- to COM and now you would have a dual variable supply for the analog electronics - typically +/-12 V - and a 5 V supply for the digital logic. If the circuit requires mains earth for any reason then connect the green post to the relevant point. Typically this is the COM. Photo 2 This power supply has remote sense inputs. These allow the power supply to compensate for voltage drop in the wires to a remote load. If not required then wire as shown in Figure 1b. Note the shorting links in your photo. If remote sensing is required then open the links and wire as shown in Figure 1c. The voltage between the + and - terminals will vary with load but the voltage across LOAD2 should remain at the setpoint. Again, if an earth reference is required then this can be achieved using the green post. simulate this circuit Photo 3 Figure 2. A dual supply can be used in multiple configurations. This has two independent supplies but without the earth option. These can be used independently, as a symmetrical supply or, for example, as a +12 V and +5 V supply. Note the connection (or lack of) between them in each case. From the comments: So if i were to use a power supply with remote sense inputs, and if i went to use it like the first power supply with COM, i would connect what would've been connected to COM to the green post. I think you are confused. The Photo 2 PSU has only one output with + and - terminals. It is not a dual supply as shown in Photo 1. You can think of it as a variable voltage battery with an optional earth connection. You always connect the load to red and black and add an optional link from either to the earth terminal. Have a look at my answer to Actual electric potential at terminals of battery and it may help your understanding.
H: Current source with unlimited cap drive op-amp oscillating I've been trying to make a current source, and have finally gotten my hands on some LM8261's which should be an 'unlimited cap drive' op-amp. The problem is, it is still oscillating. This is the schematic and PCB layout: I have tried to route the ground inside the loop area, close to the 5V connection for the opamp, so I could place the bypass capacitors as close as possible. Although the bigger 10uF cap has one of its legs at the groundplane (forgot to include this cap in the schematic picture.) This is what it looks like when I probe the gate of the MOSFET: It's all over the place. I tried adding a 10uF capacitor across the DUT, so from VDD to ground. This removed a lot of the wierd noise and made it look almost like a sinewave: I've tried with and without the gate resistor R1, it didn't seem to make any difference. What is going on? Any clues on why this just doesn't work? EDIT: Solved it! With the help from @Andyaka below, I used his solution of inserting a BJT emitter follower in between the op-amp and the MOSFET, together with an integrator and got some great results! Final circuit: The simulations can be seen here: https://i.stack.imgur.com/1jULS.jpg The gain/phase bode plots I made on the PCB: Bandwidth of approximately 350kHz! Very pleased. AI: I've been trying to create a current source, and have finally gotten my hands on some LM8261's which should be an 'unlimited cap drive' op-amp. Finding a chip that can drive a capacitive load is one thing but then using a 1 kohm in series with the output to drive that capacitive load is asking for trouble. Reason: the 1 kohm and the MOSFET gate-source capacitance form a low pass filter within the feedback loop and push the phase margin of the op-amp to 0 degrees at a moderate to high frequency turning the circuit into an oscillator. Look at the phase margin in the data sheet and note that if you factor in the 1 kohm resistor and about two-thirds of the gate source capacitance, the phase margin graph turns into the blue line I drew below: - And, the phase margin crosses zero degrees (i.e. it becomes an oscillator) at about 1 or 2 MHz (magenta circle). How did I do this you may ask? Well even though there is a resistor in the MOSFET's source it doesn't do very much for reducing the gate capacitance - it might reduce it to about two-thirds so, 900 pF GS capacitance and a 1 kohm resistor form a low pass filter with a 3 dB point at 265 kHz. At that frequency the extra phase introduced is 45 degrees hence I drew a red dot 45 degrees lower. Then I considered a frequency that is five or ten times higher just so I could roughly pin-point where the added phase shift limits at about 90 degrees and drew the 2nd red dot. Then I joined up the two dots in blue and drew a magenta circle where phase margin is modified to 0 degrees (the point of closed loop oscillation where negative feedback becomes exactly positive feedback. It's not a massively accurate technique but can tell you whether you are going to hit trouble.
H: Is it possible to measure frequency upto 500Mhz with 50Mhz oscilloscope? Is there a way to measure higher frequency measurements with a 50 MHz oscilloscope (eg rigol ds1054z)? Perhaps some post processing of captured waveforms? Or with the help of some adapters? I have a Rigol DSO1504z digital oscilloscope. Link. http://www.scientechworld.com/test-and-measurement-solutions/digital-oscilloscopes/50mhz-digital-oscilloscope-ds1054z AI: Included mostly for interest, as this is not a beginner project. There's a way, but it's not all that simple. Build a sampling front end for an oscilloscope. The above circuit will allow periodic signals with 1GHz bandwidth to be measured on a 10MHz or better analog oscilloscope. It's no good for one-shot signals.
H: What is this connector called that looks like a small mono audio jack? The thermocouple from my oven thermometer has a connector that looks somewhat like a mono audio jack, but smaller: The connector is around 2.4 mm thick in the long, extended section and around 11.2 mm long. Does someone know what this is called? Or is it proprietary? AI: It Looks like a standard 2.5 mm audio jack plug.
H: What type of power connector is this? I have one of those portable 16 Ah car jump starters that I want to use as a power pack for some 12v stepper motors. Rather than canabilize this plug, I am wondering if someone can tell me what it is called so I can see if I can find an unwired one on ebay or something to use. My current needs are very low at around 1 amp. Many Thanks AI: I'ts an EC3 connector, available at dhgate.
H: Mains earth and op amp ground I was trying to power a transimpedance amplifier built with an op amp on a protoboard. I only had a power supply that looked like the first image below (no COM, which I would have connected to GND of the op amp). The answers I got for my initial question posted on this website were really good and perfectly explained what I was supposed to do. (I initially asked questions about different kinds of power supplies) I made a mistake of connecting the mains earth of the power supply (circled in the first image below) to the + pin of the op-amp, where GND (COM of a power supply) was supposed to go. I understand why I was not supposed to do it. The mains earth is ill-defined since it is disconnected from both positive and negative voltage outputs of the power supply. It seems I broke my op amp when I connected the mains earth to + pin of the op amp where GND (so COM) was supposed to go. I see that I was not supposed to do it. My question is how did I break the op amp by connecting the mains earth to the + pin of the op amp? AI: simulate this circuit – Schematic created using CircuitLab Figure 1. What you've done. Figure 2. What you're not accounting for. The power supply has capacitive coupling between the high-voltage side of the transformer and the low-voltage side. This occurs through the transformer capacitance and the deliberate capacitance circled in Figure 2. The effect of this is that the PSU output has a weak alternating voltage on it. You might be able to read this with a digital multimeter between one of the terminals and the mains earth. simulate this circuit Your op-amp has a very high input impedance and is very sensitive to over-voltage. Most likely is that the capacitively coupled voltage exceeded the maximum allowable input to the op-amp.
H: wind turbine voltage multiplier I am building a wind generator with a car alternator without a gearbox. I know that the alternator is designed to spin at 5-10k RPM and the blades runs at 60-120 RPM so the alternator generates less than 1 volt. My question is: Can I use a voltage multiplier such as a Cockcroft-Walton generator to get useable voltages at reasonable efficiencies? AI: You probably need to connect your field winding up to get more output but running at this dismal speed is going to be a poor generating solution. No, a cockcroft walton multiplier won't help much here because they need several volts p-p to make any headway due to the internal diode volt drops. Drop the idea of the CWM and stick with understanding how to set the field winding current. I believe a lot of automobiles use the field winding current to control output voltage hence the final alternator output is regulated. If you are not using the field winding then you are relying on residual magnetism in the rotor for generating your output voltage so it will be low.
H: What is a (side) current alarm? On the technical sheet of a motor controller integrating two BTS7960B (picture below), the description of the 8 PINs on the board is as follows: RPWM: Forward level or PWM signal input, active HIGH LPWM: Reverse level or PWM signal input, active HIGH R_EN: Forward drive enable input, HIGH enable, LOW Close L_EN: Reverse drive enable input, HIGH enable, LOW Close R_IS: Forward drive - side current alarm output L_IS: Reverse drive - side current alarm output VCC: 5V power input GND: Ground I tried to search for "side current alarm" and "side current" but all I found was about "high/low-side current sensing". Being a newbie in electronics, I'm not sure what is the connection between all those terms. What is a "side current alarm"? How is it used? AI: I think that the text is badly typed. If we change R_IS: Forward drive - side current alarm output L_IS: Reverse drive - side current alarm output to R_IS: Forward drive-side current alarm output L_IS: Reverse drive-side current alarm output it implies that the alarm is on the "drive-side" of the board as opposed to the control side. The BTS7960 datasheet says: The status pin IS is used as a combined current sense and error flag output. In normal operation (current sense mode), a current source is connected to the status pin, which delivers a current proportional to the forward load current flowing through the active high side switch. If the high side switch is inactive or the current is flowing in the reverse direction no current will be driven except for a marginal leakage current IIS(LK). The external resistor RIS determines the voltage per output current. E.g. with the nominal value of 8500 for the current sense ratio kILIS = IL / IIS, a resistor value of RIS = 1kΩ leads to VIS = (IL / 8.5 A)V. In case of a fault condition the status output is connected to a current source which is independent of the load current and provides IIS(lim). The maximum voltage at the IS pin is determined by the choice of the external resistor and the supply voltage. In case of current limitation the IIS(lim) is activated for 2 * tCLS.

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