Split (1)

train · 1.46M rows

text stringlengths 83 79.5k
H: A passive sensor does not need any additional energy source? I am currently studying the textbook Handbook of Modern Sensors - Physics, Designs, and Applications, fifth edition, by Jacob Fraden. In this textbook, the author states the following: All sensors may be of two kinds: passive and active. A passive sensor does not need any additional energy source. It generates an electric signal in response to an external stimulus. That is, the input stimulus energy is converted by the sensor into the output signal. The examples are a thermocouple, a photodiode, and a piezoelectric sensor. Many sensors are direct sensors as we defined them earlier. In researching this further, I came across this Quora question, in which user Giulio Moro claims the following: An active sensor would be like your garage door sensors. One side has a light and the other a sensor. A beam fires across the area when the doors closing to ensure it's clear. If the light beam is broken, the door stops, and then opens wide. A passive sensor would be like a metal detector. It creates a field, and when anything crosses it, it makes a sound based on the type of obstruction. A low pitch for say a plastic bottle or piece of wood, and a high pitch for a metal object. There are two aspects of this definition of passive sensor that slightly confused with: Firstly, what is meant by an "additional" energy source? And analogously, if there is an "additional" energy source, then what would the base/default energy source? I think that this second confusion likely stems from the first, and clarification of the definitions of an "additional" and base/default energy source will probably clear this up, but I want to ask it anyway just in case. From a physics standpoint, I'm not totally convinced of this definition of passive sensor. In particular, I'm struggling to come to terms with this idea that a sensor can function without an energy source (although, again, the text does say additional energy source). The example of the metal detector also seems unconvincing, since my understanding is that metal detectors are certainly connected to some energy source (say, at airports or such)? I would greatly appreciate it if people would please take the time to clarify this. EDIT: After reviewing the answers (thank you all), one thing is obvious: There is a terribly large amount of contradiction with regards to what sensors qualify as active or passive. AI: Adding my two cents. Maybe there are no formal definitions, but the one given in your textbook seems pretty formal to me, so let's stick with that. And it is quite straightforward - no additional energy should be applied to the sensor in order for it to be considered "passive". The keyword here is "additional". The caveat is that the physics laws require some energy to be applied to the sensor in order for it to detect something. In fact, energy IS the thing that is being sensed. The thermocouple converts thermal energy into voltage, photodiode (in photovoltaic mode) converts light energy, and piezoelectric sensor converts mechanical energy. You can add a simple coil in the magnetic field to this list, which converts motion into voltage. And so on. An "active" sensor also senses energy (once again, there is no way around this). However unlike "passive" sensor it does not generate voltage on its own. Instead it changes its property (usually resistance) in response. And at this point you need additional energy to read this change and amplify it to some useful level. The same photodiode in photoconductive mode becomes active sensor, because you need to apply voltage to read its resistance. Hall effect sensor requires current across it to be able to detect magnetic field. MEMS sensor requires electrical excitation to detect motion. Your second source sounds confusing, and in fact it is. That is because it mixes complex sensor systems into the picture, and tries to categorize them by simple sensor criteria. Now, the problem is that in addition to simple sensors described above we also call "sensors" some complex devices of which the sensor itself is just a tiny component. For example, garage door sensor built on photodiode in photovoltaic mode will be an active "sensor", even though diode itself is passive sensor. Why? because you have to apply energy to the circuit that detects voltage on a diode, amplifies it and switches external relay. Not to mention energy consumed by the laser, which can be considered a part of the whole "garage door sensor" device. Same with metal detector. The coil that generates voltage in response to magnetic field is passive sensor by the definition above. However in order to induce magnetic field in a metal you need to generate alternating magnetic field of your own, and for that you need additional energy. So, if you take the entire metal detector device and call it a "sensor" then it becomes active by the same definition. I hope the above makes things clearer. Having said that [insert mischievous grin here], a bonus question for you: Is laser sensor card an active or passive device? On one hand, it detects IR radiation without any energy applied to it. On the other, it has to be recharged (i.e. energy must be pumped in) by visible light before use. UPDATE: After reading all the answers and comments here I believe the root of the confusion and the fuel of all the discussions is in the definition of a "sensor" itself. Let's get this straight: some devices can function by themselves, directly converting measured physical phenomenon (energy) into useful output signal (e.g. voltage, visible light, mechanical movement). By the textbook definition these all are passive sensors. some devices cannot produce meaningful output signal without additional energy input, whether directly used for excitation or indirectly for required amplification circuits. In this case we should call the device a "sensing element" (not a sensor!), which makes the containing device an active sensor. finally, even passive sensors can be equipped with the additional circuitry to simplify their usage (various current/voltage transducer modules, for example), which makes them active again. So, if we count only sensing elements as "sensors", then classification is straightforward. If we allow complex devices to be called "sensors" even though the actual sensing element is just a tiny part of them, then majority of the modern sensors will be active devices.
H: How can a wire have 0 conductivity and 0 resistance? First post but lurking for over a year I bought some cheap heating pads on EBay a long time ago. 4v wouldn't get all 5 in parallel hot enough and 8v was unbearably hot so they went in the pile of crap for awhile. I pulled them out to see if I could regulate the voltage with a DC voltage regulator that included a 10k potentiometer (which doesn't matter because the regulator circuit allowed it to work for my test). It did pretty well except for the need to be EXTREMELY precise while turning the pot. The wire for the heating pads seem to be sandwiched between what was sold as carbon fiber cloth. There is a 2 wire"loop" which returns to the other side of the battery. I couldn't tell if they connected inside each pad or only at the termination pad,so I tore one apart. The wire inside was so fine and the "carbon fiber" so tightly glued I never found out about the connection. I did salvage about 1 ft of the wire to test it. The wire is about 1mm thick with some type of thread in it about like a human hair. I connected the wire negative to positive on a 1.2v AA cell, a 3.7v cell and a 7.4v battery pack. All fully charged. No spark. UnNo heated wire. No heated batteries. No ohm reading on the wire itself. No battery discharge at all after 15 minutes. I took the hair sized filament out...same thing. What did I do wrong? Okay. To answer some questions. If I connect a wire from positive to negative on an 8 volt 4A battery the power should go somewhere. Heat the wire, drain the battery, short the battery...what else? Energy can't disappear. AI: I did salvage about 1 ft of the wire to test it. The wire is about 1mm thick with some type of thread in it about like a human hair. I connected the wire negative to positive on a 1.2v AA cell, a 3.7v cell and a 7.4v battery pack. All fully charged. No spark. UnNo heated wire. No heated batteries. No ohm reading on the wire itself. No battery discharge at all after 15 minutes. I took the hair sized filament out...same thing. If you can't even read the resistance on a meter, you effectively do not have a complete circuit, so no surprise you can't get it to heat up. The wire may be broken, you may not have good contact, or it may be laminated to prevent contact. Time to troubleshoot and figure out why you don't have a circuit. If I connect a wire from positive to negative on an 8 volt 4A battery the power should go somewhere. From what you've said above, you didn't actually connect the wires, so no power goes through the wire.
H: Turning on LEDs I started a small project where I want to have three LEDs which are turning on gradually (in first step just one is on, one second step two are shining and then all three). I've tried doing it via microcontrollers but isn't here any simlplier way, for instance with NE555? Thanks everyone for their answers. I'll try figure it out throught MCU and hope thant nothing will go wrong. AI: If you want oldskool analog: First, a ramp/sawtooth generator, which you can build with an opamp, or a comparator, or a 555. If you want to have it do only one run at power-up, a resistor charging a capacitor will make a voltage ramp too. Then a vu-meter circuit like LM3914 turns your voltage ramp into a bunch of LEDs lighting one after the other. If you want the LEDs to turn on gradually in turn, not just on/off, you can do that by adding another sawtooth/triangle on top of your ramp. Here's a big sawtooth plus a small sawtooth, and the output of 3 comparators with spaced thresholds driving 3 LEDs... These days, it's simpler with a micro. EDIT: More analog. Big capacitors are expensive, so I'll use only one. Power supply goes from 0V to 5V at t=0. For simpler inventory management, let's get rid of the diodes. 12V version. Less parts. BJTs going into saturation trigger the next LED. 5V supply.
H: How does interdependent state logic work? I'm deriving the logic for an up-down counter and have equations which connect bit states \$Q_{n+1}\$ to previous states \$Q_n\$. On each rising clock edge, the data bits of the counter change depending on the state of some UP/NOT_DOWN line and, as with all counters, the states of other bits of the counter. I'm a little confused on how to express this logically though (for programming a CPLD), because surely, unless every bit changes instantaneously (not possible), the logic will fall apart (as the logic for one new bit would suggest "reading" the logic of an already changed bit). Is this all part of the "timing optimization" procedure you see implementation and synthesis tools do, or is this something which needs to be considered at the design level? EDIT: I suppose another way of asking the question would be to address whether each bit needs to be stored in some temporary storage between states, in which case, why don't we see such architectures discussed often? AI: It works if the propagation delay (clock-to-Q) of the flip-flops involved is longer than the hold time of the flip-flops (plus any difference in arrival time in the clock signal between one flip-flop and another). If this is true then when each clock cycle arrives, their outputs don't change for some time (the clock-to-Q delay). By the time the output of one flip-flop changes, the next flip-flop (whose input is connected to the output of the first one) has already "decided" what it is going to change to. In programmable logic (CPLDs or FPGAs), this is often achieved using "negative hold time". That basically means that the next state of a flip-flop after a clock edge arrives is actually determined by the input a few 100 ps before the clock edge, rather than the input at the exact moment of the clock arrival. (Practically this really means the data is delayed by a few 100 ps from the notional input of the flip-flop to where it actually affects the flip-flop circuit)
H: Ultrasonic transducer circuit Can someone explain how these work ? I want to make my own ultrasonic bath, but I can't find any useful specs. Do I need some specific circuit to run them or it's just 'plug and play' ? AI: It vibrates at 40kHz so of course you need a 40kHz oscillator and a 60W amplifier to drive it. The specs say: 40kHz and 60W. But the details about the voltage and current required are missing. Maybe because it is cheap and because it is Chinese. The manufacturer might be able to provide a circuit to drive it.
H: MOSFET Common-Source Stage Confusion I'm getting very confused about drain source voltages and drain current in a simple common-source amplifier. Consider the following common-source amplifier: My understanding is that: The output voltage is Vout = Vdd - Rd*Id. If the input voltage increases by a small-signal step, then M1 will have a higher gate-source voltage, thus will increase the drain current and according the Vout equation above, the output voltage will decrease. The output voltage is the drain of M1, and so we can also say that the drain of M1 decreases. All good so far. Here comes my confusion. If I think back to the Id/Vds plot as shown below: Notice that in the saturation region, as drain-source voltage increasees, the drain current increases slightly too, due to channel-length modulation. In our common-source amplifier, the drain current increased, yet the drain-source voltage (output voltage) decreased. According to this plot, how is that possible? I would appreciate it if someone could provide a simple explanation. Where is the flaw in my understanding? AI: The MOSFET doesn't stay on the same Vgs line, or the same Vds. Those Vgs lines are just a few of the infinite number on that graph for the infinite number of Vgs possible. The MOSFET is allowed to move anywhere within that graph's space to find equilibrium. It's not trapped to one of those Vgs lines, though it may be restricted by other circuit components to traverse amongst those lines in a limited manner (like the load line drawn by the resistor).
H: Why does the output of this opamp change my analog signal? I am trying to make an inverting Op-amp circuit in LTSpice where I can invert an analog signal without any gain on my signal, I want the signal to stay the same but inverted. I have followed the basic inverting Op-amp circuits that I have seen in electronics text books. I have followed it through using the OP27 op-amp provided in LTSpice and got results looking like this: I don't know how to explain what happened here but it happens with the other models of op-amps as well on LTSpice except for the ideal op-amp. Can someone explain what happened to the signal? Is it a problem with LTSpice? or is there something missing in my circuit that could solve it? AI: OP27 is not a rail-to-rail input op-amp. Here's the relevant section of the datasheet: Notice the input range is specified with +/- 15 V supplies. This means the input voltage must be (for a typical chip) at least 2.7 V above the negative rail and 2.7 V below the positive rail for the op-amp to work as expected. This is obviously impossible to achieve with a 5 V single supply. You'll need to increase your supply voltages, and be sure to bias your input within the recommended operating range. As mentioned in comments, you will also need to have a negative supply if you want the output voltage to be able to reach below ground. Or choose a more modern op-amp designed to work with lower supply voltage.
H: RS422, RS485 Termination Resistors While going through this youtube video from TI about the RS232, RS422, RS485 differences. One thing that don't understand clearly is following slide from this video. The blue resistors that I marked on it are the termination resistors of the buses. For the case of point-to-point connection it is clear that there is only 1 receiver so we put the resistor near that point on the bus. But the other two cases, multi-drop and multi-point, I could not understand why do we put only 1 or 2 termination resistors on the bus instead of putting 1 at every branch-off point? So in the above picture there should be 3 resistors for the multi-drop case and 4 resistors for the multi-point case. AI: When you send a high speed signal down a cable, the current that initially flows is dictated by the voltage applied AND the characteristic impedance of the cable. For the types of cable recommended for RS485 and RS422, the cable characteristic impedance is circa 100 ohms. So if 1 volt is applied at the sending end an initial current of 10 mA flows and all is well until the end of the cable is reached; voltage and current expect to see a continuing impedance of 100 ohms. If it weren't 100 ohms then what does 1 volt and 10 mA make of a mismatch? The answer is that not all the power of the 1 volt and 10 mA is used at the receiving end and some power gets reflected back up the cable towards the source. For slow speed signals this isn't a problem but for higher speed signals this reflection damages the integrity of the data edges (for up to several micro seconds in some cases) and can corrupt the data. Here's a .gif file image that shows the general idea: - A pulse enters from the left and hits a transmission line impedance anomaly (as indicated by the vertical black line). A decent proportion of the pulse energy continues to flow from left to right but, there is also a reflection at the "anomaly" that flows back to the originating source. The reflection can cause data errors. Branch-off points are always regarded as having very little length and so termination resistors can be ignored but, in addition, you can't attach several termination resistors because these would alter the transmission impedance as the signal passed down the cable and this would also create a mismatch and produce reflections and corrupt data. Multi-point systems need termination resistors at both ends of the cable because data is sent bidirectionally.
H: Is this regulator a suitable replacement for original one? I am fairly new to electronics and have done my first project successfully (where it works as assembled PCB). And I am looking for the next step there. In my original project I've used SPX3819M5-L-3-3 to regulate incoming 5V (from micro-USB) into 3.3V. I am now looking to switch from micro USB to UCB-C connector, which can go up to 20V (the circuit is not the main sink, so another connected device can request up to full 20V). Given that original regulator maximum rating is 16V, I need to change it to something suitable. I've looked at TL760M33 for this purpose as it can go up to 26V and output fixed 3.3V at 500 mA max, which seems perfect for my case. Being very new to electronics I have my doubts over every single detail. So in this case my question is -- is this replacement valid and would my circuit function as I expect it to? Additionally I wonder is there any pitfalls or details I missed in choosing a replacement regulator? Original schematic: Replacement schematic: AI: The data sheet value for the thermal resistance between junction and "ambient" (when the tab is soldered to a 15 mm x 16 mm copper area of 1 oz/sq foot thickness) is 55 °C per watt of power dissipated: - If you supply 20 volts to the regulator and it drops 16.7 volts in the process of producing 3.3 volts at the output, AND the output current is 300 mA, the power dissipated is about 4.9 watts. This means that the device's "junction" (when mounted as per above) will heat up 55 x 4.9 °C above ambient i.e., in an ambient of 25 °C the junction will warm to 295 °C and this is clearly too much. You should also note that the local ambient will rise significantly above the average ambient under heavy dissipation situations. So, you need either: - A much bigger heat-sink area and/or thicker copper A lower maximum continuous load current A lower maximum ambient temperature A fan targeted at the device A switching regulator A prayer
H: Understanding purpose of a resistor in reference schematic TIDU034 Gents, I'm analyzing this schematic, but do not understand couple of things. I would be grateful if someone could explain in detail these questions: 1) R6 - shunt? What's the purpose of it? Under what conditions it is required? I believe value was chosen pretty much random but high enough? 2) R7 and R10 are input bias return paths - why do they have different values (100k vs 22k) for the same opamp part and signal level (U1A & U1B don't do amplification)? The original article where this was published is TI's Active Volume Control for Professional Audio (TIDU034). AI: R6 is needed to define a proper DC voltage level (bias voltage) at the signal node. Since the other side of R6 is connected to ground, this voltage is 0 V (zero). However if Vin has a DC voltage level of for example 1 V DC then R6 might not be able to pull the node back to 0 V DC. That's not an issue as C3 will prevent that DC voltage from reaching the rest of the circuit. But if Vin is a source with an output that has an AC coupling capacitor then the DC voltage is not defined. Then R6 will pull the DC voltage to 0 V. R6 should have a high enough value so that it does not attenuate the signal. Note how R5 and R6 form a resistive voltage divider. It is possible Vin has some output resistance and that will be in series with R5 resulting in more signal attenuation. R7's function is similar to R6's function: also to set the DC voltage. The value of R7 must be such that in combination with C3 the low frequency cut-off point is at a frequency that is low enough. R10's function is indeed the same as R7's function. My guess is that since the signal through C7 is at a lower impedance level (output of U1B + P1 = 10 kohm) which allows for R10 to have a lower value. Lower values are often preferred for less noise and less sensitivity to signals coupling into the circuit. R6 and R7 might have a much higher value (than R10) to keep the input impedance of this circuit high enough. Opamp U1A and U1C are used in "unity gain", their gain = 1 Opamp U1B and U1D are signal inverters, their gain = -1 (minus one).
H: Reduce/Remove Sparks in AC and DC Contacts of Relays I see that there are sparks being generated whenever there is a connection established between two contact points in the relays. So is there a way to reduce/remove those spark generated in the relays. AI: There are a number of ways to do this, the most generic is an R-C quench circuit. You can even get them in one package. People often use 0.1 uF and 100 ohms, but you may want to use trial and error to get the best results for your application. If the load is at line voltage you might use an X2 for the capacitor for safety. One capacitor company used to loan (sell?) R-C selection boxes to make this convenient. There are also formulas published.
H: Conservation of Power in Filters I have a simple doubt about the conservation of power in systems that include some filters. Let's consider this general situation: The filter is able to attenuate some components (of certain frequencies) of Vin. This means that the power of Vout is lower than that of Vin. I have the following questions: 1) Where does the power Pout - Pin go? If I look for instance at the following example, I'd say that it is dissipated on R, since for all frequencies such that Vout = 0, all Vin drops on R. Is it correct? 2) Is the previous reasoning true for any type of filters? If it is true, it means that it is not possible to realize filters without resistors (not even ideally): otherwise, where will the power Pout - Pin go? 3) I have noticed that filters that are used in microwave/rf circuits are considered as "reflection attenuators": they (ideally) do not have losses, since power Pout - Pin is reflected by them. But, exactly what does it mean? Which component stores or dissipate that power? AI: When considering power a filter block must have a load resistor (that's where power is delivered). Delivering power to a filter block's input side also involves a source resistor. The filter block itself may include resistors, and/or lossy reactive components. The simplest case to consider is where the filter block only includes reactive (L, C) components: simulate this circuit – Schematic created using CircuitLab The filter block of reactive components cannot dissipate real power but it can temporarily store or transfer power amongst its neighbouring reactive components. Power is either delivered to Rload or it is refused by its input. Any real power accepted at the filter block's input side must be (eventually) delivered to Rload. Referring to a filter block as a reflection attenuator reinforces this point of view. Power refused at the filter's input side can be considered a reflection of power back to the signal source where it would be dissipated by Rsource.
H: Add LTspice IV library to LTspice XVII I have an external library which works fine for LTspice IV. To add this external library, I simply overwrote (added) the corresponding files in the lib folder in the installation directory. Hence, I have a lib folder of the following structure: lib: - cmp \|-standard.dio - sub \|-myPart.mod - sym \|-Misc \|-myPart.asy When I do the same in LTspice XVII, it doesn't work. This means, that if I open a schematics which uses myPart the error message "Couldn't find symbol(s): myPart" pops up. I also can't find this part in the "Component browser" in LTspice XVII but it is available in LTspice VI. How can I add this library in the XVII version? AI: I'm only 75% sure that this is the problem, but it's the problem I had when I moved to the new LT spice. LT spice uses a new directory, stored in documents\LTspiceXVII\lib (not in C:\program files\lt spice\lib). Make sure you modify the files there and not in the program directory which has a dual structure, but LT spice uses the files stored in documents folder.
H: Calculate maximum deviation of quartz crystal How can I calculate the maximum deviation of a quartz? This is from datasheet (12MHz): Frequency Tolerance at +25° C => +- 30ppm temperature tolerance -20°~ + 70°C => +-50ppm aging (first year) => +- 2ppm I want to calculate the maximum deviation for a specific period. For example first quartz is at -20°C then the tolerance is 50 ppm or 50 ppm + 30 ppm? So +- 50 ppm or +-80ppm? The second quartz the same but it works at 70°C. The following calculation is after one year so I add 2ppm. So when it is 82 ppm then the first quartz can operate at the worst case with 12MHz - 984 MHz and the second quartz is working with 12MHz+984MHz. Is this correct? I am not sure because the frequency tolerance is at 25°C. AI: The correct estimate is a sum tolerance stackup of all variables for the crystal. (30+50+2 = 82ppm) The 1st @ room temp is the just the initial tolerance of size of the crystal to the designed centre frequency. The 2nd is the angle of the cut of the XTAL in minutes (1/60th) of a degree that only affects the 3rd order slope at room temp, so both are additive. So a XTAL tolerance of +/82 ppm over temp in 1 year does not include shock & vibe nor C load error which adds to the room temp tolerance error and "could" be as little as 10 ppm with 10% caps but it depends on the slope of the cut and temp response. other In the early 90's we wanted to make a $1 TCXO for 928Mhz synth, so I digitized all the AT cut curves, generated 3rd order polynominals to compute any of them and created a test process to bin them using a 30 second (foam insulated copper FPC with thermistor and 1/8W heater R around XTAL) "mini oven" that tested offset @ +40'C then +70'C which is a near "linear range". Then another guy created a tool to bin varicaps in 1 % bins, then when paired with the formulae, the TCXO could be compensated with DAC to Varicap within 1 PPM over -40~ +70'C range. I generated another 3rd Ord polynomial to compute any Xtal curve based on Δf(T=40'C to 70'C). The programmer used the formulae with thermistor readings to compute Varactor DAC voltage for $1 1ppm TCXO. Now you can buy them in bulk for this price.
H: Transfer function and graphic (Scilab error) I'm trying to solve this problem with a Scilab code: I'm trying to follow this code, but in the step [b,a]=ss2tf[A,B,C,D], it shows the error ("Wrong number of input arguments.") The code of Scilab that I try to make is the following: clc; clear all; close all; A=[-1 -2;42 -0.9] B=[1.5;1.1] C=[0.7 2.1] D=[0] [b,a]=ss2tf(A,B,C,D) H=tf(b,a) figure step(H) grid How can I fix this code? AI: Sorry to point this out, but it seems that you are using the Matlab notation and have not even looked at the Scilab language reference, still, clc; clear all; xdel(winsid()) A=[-1 -2;42 -0.9] B=[1.5;1.1] C=[0.7 2.1] D=[0] H=syslin('c',A,B,C,D) H=ss2tf(H) figure t=0:0.05:5; plot(t,csim('step',t,H)) xgrid(2)
H: Input and Output capacitance I have understood that there are both input and output parasitic capacitances in real op-amps, but I am trying to understand why they are there. As for the input capacitance, there are two types: Differential: it is the parasitic capacitance between the inverting and non-inverting pins of an op amp Common-mode: it is the parasitic capacitance between each input pin and ground. Is that right? For the output capacitance, I don't know where it comes from. Also, I understand that this capacitances are almost always neglected, but they may affect our circuit if the frequency becomes extremely large, right? AI: Common-mode: it is the parasitic capacitance between each input pin and ground. Is that right? Yes. All insulators, regardless of distance and composition, will impart some capacitance to a conductor (even in a vacuum.) Any two conductors will have some capacitance between them, even in a vacuum. Insulators other than a vacuum result in more capacitance. Very, very slightly more in the case of air, significantly more in the case of plastics and such like (several times more). The ratio to that of a vacuum is often called the dielectric constant but better called the relative permittivity \$\epsilon_r\$. Also, I understand that this capacitances are almost always neglected, but they may affect our circuit if the frequency becomes extremely large, right? Not correct. Input capacitance can affect the circuit even at low frequencies, if the feedback resistance is high enough. For example, an input capacitance of 10pF with a 10M\$\Omega\$ feedback resistor has a pole at 1.6kHz, so it's pretty easy to make an op-amp amplifier oscillate when there is a high-value feedback resistor. This is a common issue in circuits designed to work from very low current, and is compounded by some op-amp designs that use scores of interdigitated input MOSFETs in parallel to minimize Vos. Output capacitances are part of the transistor structure and also caused by the geometry of the IC package and the conductors on the PCB etc. Too much output capacitance on an op-amp output can cause oscillation. Most op-amps are specified for phase margin with some reasonable amount of output load capacitance (in addition to whatever is internal to the IC), but excessive added capacitance (such as driving a few meters of cable without series resistance) can result in oscillation. The oscillation may be relatively high frequency (since the output resistance of the op-amp is generally in the 100-1000\$\Omega\$ range) but your amplifier circuit may only be designed for low frequencies or even DC.
H: Oscillators and changing waveforms I have a 40 MHz crystal oscillator that generates a sine wave. If I use a schmitt trigger on the output, would the output of the schmitt trigger be a 40 MHz square wave, suitable for digital electronics? Will a CMOS type oscillator work with digital electronics? Oscillator is Abracon ACO-40.000MHZ-EK Schmitt trigger is CD40106B AI: The datasheet (ACO-40.000MHZ-EK datasheet) indicates the oscillator is not a sine wave one and already outputs CMOS level digital signal suitable for 5v logic. You don't need a Schmitt trigger buffer.
H: Series limiting resistor? I have used a 3.3V and a 4.7V Zener diode. What value series resistor do I need for them to work properly? The test current for 3.3V is 76mA and for 4.7V is 53mA. AI: For a zener of voltage VZ fed from a Voltage source of Vin via a resistor of Rz with no other load. Iz = V/R = (Vin-Vz)/ Rz ... 1 Call this value of Iz Iz_max. Zener power dissipation = Vz x Iz ... 2 Usually zeners are rated to dissipate this power in free air - except maybe for unusual package types. In this case you do not specify Vin so Rz can only be calculated in terms of the above parameters Rearranging (1) Rz = (Vin-Vz)/Iz Vz will vary somewhat with Iz but is "close enough" to rated value for most purposes. As you draw current for a load Iz will reduce Iz_variable = Iz_max - Iload. Regulation will stop when Iload >= Iz_max Using a 9V transformer, bridge rectifier, 1000 uF capacitor. Vdc unloaded = VAC_RMS x 1.414 ~= 12.7V Actual loaded DC depends on transformer and load current, but probably 11 - 12 VDC. Assume 11 VDC For 3V3 zener with Imax = 75 mA max current. Rz must drop 11-3.3 = 7.7V Rz = V/I = (11-3.3)/.075 = 7.7/.075 = 103 Ohms. Say 100 Ohms. Power dissipation in resistor = v^2/R = (11-3.3)^2/100 = 592 mW. Use a 1 watt resistor (at least). Power in zener with no load =~ V x I = 3.3 x 0.075 =~ 250 mW. Power lost in resistor is about 2.4 x power dissipated in zener! Overall efficiency = Vz/Vin = 3.3/11 = 30%. This is the result of using a linear regulator. Using a linear regulator IC will NOT improve efficiency at maximum load but will reduce power dissipation at lighter loads. For 4.7 V zener at 53 mA. Rz = V/I = (11-4.7)/.053 = 119 Ohms. Use 120 Ohms. Max efficiency = Vz/Vin = 4.7/11 ~= 43%. Resistor power dissipation = Vr^2/R = (11-4.7)^2/120 = 0.33 watt. Use 1/2 watt or higher rated resistor. Using linear regulators will give better regulation and better efficiencies at lower loads. This is because a zener regulator always draws its design current at a given voltage - either via the zener to ground or into the load. A linear regulator uses a small amount of current to operate the regulator and then passes only the required current.
H: Alternating voltages on vintage calculator circuit board I am an electronics novice. I have a vintage calculator with six secondary wires coming out of its transformer and ending on the board, as shown below: Using the AC portion of the multimeter, I measured the voltages across the board terminals marked 1 through 6, with the results as follows: Is it normal to see AC voltages on such a board, or is this because of the age of the product (1979)? Is the AC voltage changed to DC somewhere on the board? Is this a "digital" calculator? (There is an IC chip on the board.) If so, why is there alternating current, unlike say on an Arduino? AI: Yes, it's normal on a PCB with a mains-frequency power transformer to see AC voltages from the transformer. They can be rectified (diodes or bridge rectifier), filtered (those can electrolytic capacitors) and regulated, as required. That calculator has a vacuum fluorescent display, which uses relatively high voltage for the grids and cathodes (something like 24VDC, often negative with respect to ground, is typical) and also has a filament which is typically a volt or two, and can be driven by AC or DC (possibly directly from a transformer winding in this case). Here is some more information on driving these displays. The calculator IC itself probably runs from a few volts DC. There may be an external chip to drive the display, since the voltage requirements would tend to point to an IC process that would have been costly for the calculator IC. Because of the multiple voltages, it's going to be very easy to damage the ICs by even a momentary short, so be careful probing.
H: Differential Pairs and GND I have some doubts about different configurations of differential pair I have seen in many cases: In these schemes, biasing network for the gate is not shown for simplicity. All of them are excited by an input differential signal, but In 1) two different opposite signal sources are applied to the transistors. The output is taken between the drain of one MOSFET and GND. It is clear for me. In 2) the pair receives a differential signal in a different way, by using only 1 voltage source. I have the following doubt: how can that signal be amplified by the MOSFETS, if they are not referred to GND? How can that circuit work? 3) is like 1), but the output signal is taken from both drains and not from 1 drain and GND. This allows to double the gain. But, despite that difference, is there a reason to prefer 3) on 1) or viceversa? I'd say that 3) is better because noise on GND is not transmitted to the output signal. Is it true? 4) is like 2), but the output signal is taken from both drains. I have the same doubts said about 3) and 1). Other doubts: I have seen that the current source is often replaced by a simple resistor. In this case, it is important for it to be very high in order not to have common mode gain. But, in general, which is the reason for choosing a current source between Source and GND, or an high resistor between Source and GND? Sometimes this kind of circuit works with VDD and -VDD, so with dual voltage supply. Which is the reason for choosing it or a single supply? AI: Maybe this picture will help understanding Fig. 2 and Fig. 4... ... and this - the need of a bipolar supply:
H: How to wire several 78XX series regulators? I have a 30V power source and I need to get several different voltages from it using linear regulator. 24V for op amps, 12V for relays and 5V for MCU. They all going to be installed on a heatsink. If I wire them in parallel 7805 regulator have to drop 25V and that means a lot of heat. if I wire them in series all of the current is going to pass through the 7824 regulator and that is going to generate a lot of heat too. simulate this circuit – Schematic created using CircuitLab I know linear regulators basically are not efficient but for this specific project I need a linear output. Comparing this two configuration, which one is more efficient? series or parallel? AI: They are about as inefficient either way. You have ~15mA drawn from 30V (almost half a watt) before you start drawing any current from the rails. Iq goes up slightly with voltage but it can be ignored. If you need any substantial current from this, especially from the 5V supply, you'd be far better off using a switching supply, followed by a linear regulator if that's really necessary. Either way you'll be throwing away almost 85% of the power used on the 5V bus. Also, the original uA7805s are only recommended for 25V input, and 7812s for 30V. So you would be exceeding the maximum recommended rating with the parallel connection, and are uncomfortably close to the absolute maximum 35V rating for 7805/7812. Edit: 200mA from 5V = +5W in the regulator(s) 200mA from 12V = +3.6W in the regulator(s) 200mA from 24V = +1.2W in the regulator(s) on top of the 0.5W or so quiescent So you need to get rid of more than 10W of power. Very, very wasteful and you'll need a big heatsink or a smaller one and a fan.
H: Fixing old CRT oscilloscopes... what to look for first? I've been amassing a pretty large collection of these old scopes, and it's about time I got to trying to fix them. About half of the ones I run across and buy don't have traces. I don't need to get them in perfect working order again, but I'd at least like to get a trace on them. So what do I look for first? I know resistors' values change over time, and the capacitors are big culprits, right? What about CRTs? Do they just "go bad" from sitting for 50 years? And what about the transformers? I can't really see a transformer just going bad, because you'd have to somehow manage to break one of the windings in it, correct? Anyway, thanks a heap for the help. AI: Get ahold of the service manuals first. Make sure you are experienced enough to be comfortable working safely with multi-kV voltages such as all CRT scopes use. There is quite enough voltage and current capability in a CRT scope to be lethal. If not, then stop now. Check all the power supplies- both voltage and ripple, electrolytic caps die with age, particularly the ones in the power supplies that see a lot of heat and ripple current. Measure the ESR of the electrolytic power supply caps and compare with expected values. Most older scopes will not have low-Z caps so the ESR might be a few ohms. You can usually do a quick check in-circuit if the cap is discharged. Some- the ones that use hybrids, for example, may not be realistically repairable. Most parts, other than capacitors, don't die from age, at least not very quickly, and are just about as good at 50 years old as when they were new. Beyond that, look at the waveforms in the service manuals, and see where they deviate. If the deviation isn't much out of spec, forget about it until you you find and fix the real culprit. Keep in mind that failures can snowball- a failed fuse is sometimes just a failed fuse, but often there are several other parts that failed and the fuse just protected the wiring. Inject signals at the inputs and follow them through the circuit once you get a trace etc. I once fixed a scope that had resistors that had shifted in value by a huge amount (they had high DC voltage across them and were 1% semi-precision parts). Oh, and if it becomes apparent that someone has already had a try at fixing it, you might want to leave that one for last or for the electronics recycling bin.
H: Initial pluse from schmitt trigger switch debouncer I want to use a momentary, normally open, SPST switch to act as a clock pulse on 16x2 character display. Of course, my switch has some bounce so I was looking at debouncing circuits and I always see the same one that uses an RC circuit in conjunction with a schmitt trigger inverter. For example, this one (which includes an extra diode that some folks use and some don't): The problem I see with this circuit (if I understand it correctly) is that the capacitor starts out at 0v, then charges up, which should result in the circuit putting out a high-to-low transition when it powers on the initially. Am I misunderstanding this, or do most folks not care about that initial transition? AI: The cap charges through R1 and the diode when the switch is open, and sit at a diode drop below Vcc, producing a LOW on the output of the schmitt inverter. So yes, the output is high until the cap is charged. Any system using this will need to deal with the initialization transient.
H: MCU powered from the same PSU as a LED strip I want to make a simple controller of WS2812B addressable LED strip. I will consist of a MCU (most likely Atmega328P), a couple of buttons (RESET and mode selection) and 7-segment indicator, which will show current illumination mode. Power to the strip will be delivered by a N-channel MOSFET, which gate is also driven by MCU. Max length of the strip is 4 meters, I plan to use 30 leds per meter strip, so maximum current consumption is 6A (1.5A per meter). Since this current is only required if all LEDs are white at full brightness (which I don't plan), i'll use a Mean Well 5V@5A PSU. I think even 5A will never be reached in any application. Main question is, is it OK? Or do I need to use secondary small PSU for microcontroller? This I'd like to avoid, of course. MCU together with indicator would draw no more than 150 mA. Another question is, is it OK to use only two large value capacitors (470 uF), one at the input of power, another at strip output? Do I need a separate small ceramic capacitor close to Atmega pins? It's no problem to add, but I'm not sure if it's required: the board is really small, and distance between input 470 uF capacitor and 5V input pin is just 6-8 mm. Third question is, do I need a fuse on the board? If yes, how to choose it's value properly? Below is my simplified schematic (7-segment indicator and buttons are omitted): simulate this circuit – Schematic created using CircuitLab AI: Yes same powersupply can be used. Yes you need separate 100nF ceramic at AVR supply pins, it is a local bypass for high frequency current pulses the AVR takes every clock cycle. The electrolytics are quite bad at high frequencies. Also, I am slightly worried about using FET to cut ground from the LEDs, as the data pin must then be set high so that it does not pull current when it is low. Turning the power on with FET will charge the output capacitor with a high current pulse so it might break. Think carefully if you even need to cut power from LEDs, and whether high side switching would be better.
H: Confusion about transformer spec I see transformer specs listed as 360-0-360 for a secondary of a power transformer. I understand that this means there is a center tap, and with a full wave rectifier this comes out to VRMS * 1.4142 = 360VAC * 1.4142 = ~509 VDC (assuming smoothing caps etc, beyond the scope here) But if you remove that center tap, then you could say the transformer is 360-360, or 0-700, correct? So if you then used a full wave bridge rectifier on 720VAC, you would have 720VAC * 1.4142 = ~1018 VDC. Obviously I'm missing something crucial here about how transformer specs are defined. I certainly don't want to be off by a factor of 2 when I'm getting my transformer manufactured. AI: It's a 720 VAC secondary, with a center tap it makes 360-0-360. You can get a 1018 VDC out, or by using the center tap as 0V, you get +509 and -509 VDC. Same thing.
H: Heatsink design, factors to consider? I'm trying to design an small but effective heatsink (DIY type), the design that I came up with so far is a 4x4x9cm aluminum square tube with a 12V fan under it blowing up: poor designing skill... The size of the heatsink is choosen based on the box space limit and I only have access to one side of the heatsink to install the regulators. I want to install 4 inefficient linear regulators on it, which in worst case will waste 20 watts of power in total. Does vertical heatsinks have any advantages over horizontal heatsinks? How can I calculate the cooling capability of this heatsink? That regulator which is going to dissipate more heat, should be at the bottom side of heatsink or top? Where should I install the temperature sensor to monitor the temperature and change the fan speed? AI: Does vertical heatsinks have any advantages over horizontal heatsinks? For this small scale and low temperature delta, and with forced airflow, there will be no appreciable cooling advantage. Convection will be dwarfed by the airflow from even the weakest of fans. How can I calculate the cooling capability of this heatsink? "There's an app for that." Seriously, there are many good calculators online. Find one that you like. Many of them link to articles explaining how the calculations are done, in case you want to go down the rabbit hole. Here's a search link. https://www.google.com/search?q=heatsink+calculator That regulator which is going to dissipate more heat, should be at the bottom side of heatsink or top? I would place the hottest component near the middle where there's more aluminium to spread the heat in all directions. And perhaps place the less hot components some distance away from it. It will of course have to be a compromise. Where should I install the temperature sensor to monitor the temperature and change the fan speed? I would keep the sensor out of the airflow, but close to the hottest component, with good thermal contact to the heatsink.
H: Negative loop gain? I was dimensioning the CE capacitor. To do that I was looking at some formulas seen in class. The formula (freq_C=1/(2piReqC)) involves the knowledge of the loop gain (open loop gainbeta). It appears to be negative in my schematic. What does it mean? Is something wrong? In other exercises I've got a positive number: I though positive numbers were to be expected. This is the audio amplifier circuit I'm working on: -- update -- @Tony Stewart Sunnyskyguy EE75 I have upgraded the circuit but I haven't applied the 3:1 resistors value ratio in the Vbe multiplier since the quiescent current on the output resistors was several hundreds mA... The circuit is behaving better now though. Do you think that something is wrong? The fact that I have good phase margin (+90.8deg) and good gain margin (-37.8dB) (do you agree?) isn't enough to assure me stability? It looks like I still need miller compensation capacitor since I've got peaking. Why do i have peaking though? AI: Your output is saturated open loop due to offset between sensitivity to R5, R9 values and hFE assumptions. I can tell by your choice of R9 = 595 Ohms and Vout = 14.3V Increase R5 to say 4k7 until you get in the linear output range and test Aol with 1uVpp input. You don't have to null it. Also your 4 Vbe output bias circuit needs to be more than 3:1 R ratio to make 3+1=4. Vbe=0.6V @ 1mA and 0.65V @ 10mA so the crossover output bias should be not 1.4 AMPS. let R3/R4 >=3:1 consider R3/R4 = 3.0. With 0.6V across R4 you expect 3x this amount across R3. you might expect Aol > 90dB
H: How high do you overspec your resistors? I have a resistor in a circuit that I've calculated will dissipate 225 mW, worst case, calculated for tolerance variation. Most resistors are rated to dissipate 250 mW. This seems a bit close to the spec, but is the spec already given conservatively? How close to the spec do you allow before you jump up to 500 mW resistors? AI: the power spec is not really a power alone specification per se, you gotta think of it as heat as well, if the resistor has proper cooling it should be fine. However, depending on the application you spec everything with a certain safety margin, I think for high grade circuits that would not be acceptable since they are in the range of 50% or so Now, you gotta consider this can you afford the circuit going down and needing that resistor replaced? can your other components handle a short at that point given that resistor burns out? do you have enough space for 500mW instead? what is the role of that resistor to begin with... it takes a lot of considerations. you are right at 10% safety margin, which for most common uses would be okay in my opinion, for something like military or circuits for automobiles it would not be acceptable design though. you gotta look at each situation and assess the cost, risk, reliability and decide the margin with all of that in mind. Blindly using a margin is dangerous as well.
H: How do I best visualize this voltage data for a science project? I'm helping my son with his 7th grade science project. We've had a good deal of fun with our experiments with Solar Arrays and charging 12 volt UPS batteries! But, I am not sure how to interpret the data! Our original hypothesis was that the length of the wire between the solar panel and the battery would affect the voltage charge the most. We do NOT think that was true! We just need to get some good charts out of the data to show something that we learned! AI: The 7 steps in conducting a scientific experiment are:- 1) Pose a Testable Question: 2) Conduct Background Research: 3) State your Hypothesis: 4) Design Experiment. 5) Perform the Experiment. 6) Collect Data. 7) Draw Conclusions. Now you are at step 7, and you have a quandary. The data doesn't appear to support your hypothesis! But hold on a minute - exactly what were you expecting, and why? Your research should have told you how and why wire length might affect the charge voltage, and how to design an experiment which would give meaningful results. So let's examine the experiment to determine what results you could expect from it. Then you can organize the data to see if it supports the theory. The factors that might have affected the end of charge voltage are: Light intensity. Charging time. Wire length. battery voltage before charging. Battery voltage increases roughly proportionally to the amount of charge put into it, and the solar panel is capable of putting out a current proportional to light intensity. But the resistance of the wire might restrict charging current to a lower value, which should result in lower battery voltage. Wire resistance is proportional to length. You are hoping the increased resistance of the longer wire will have a measurable effect. So you could plot a graph with wire length on the X axis and battery voltage on the Y axis. But to get meaningful results you need to separate out the confounding factors of light intensity, charging time, and battery starting voltage. For voltage you can subtract start voltage from end voltage and show the difference. To remove charging time from the result, divide the voltage increase by time to get eg. volts per minute. To account for light intensity you could plot 3 curves, one for each intensity (mildly cloudy, clear sky, sunny). If the graph shows a clear relationship between wire length and voltage increase for all three light intensities then the results support your theory. A stronger effect at higher light intensity might indicate that the panel is restricting current in lower light. If results appear to be 'random' then perhaps the wire has too little effect, or some measurements are not sufficiently precise (light intensity) or not a good proxy for charging current (battery voltage). Ohm's Law says that Resistance = Voltage / Current. The higher the resistance the less current will flow in a circuit, so a longer wire should reduce current and increase the length of time needed to charge the battery. But the solar panel and battery also have resistance. If that is much higher than the wire resistance then wire length will have little effect. If the experiment does not show a 'positive' result it doesn't prove that the wire length has no effect, but just that it is not significant in this case. This is good news, because it shows that the panel can be positioned a considerable distance away from the battery without adversely affecting charging time. Therefore if this is the result then the experiment was a success!
H: Is there a relay package which can be triggered by a numeric coded pulse? I would like to design a simple relay which is triggered by a series of pulses rather than just a high signal. e.g. the relay would have a serial data input. The relay would need to be "preset" with the code which will trigger it - e.g. 119. Then if you send it 01110111, the relay latches. Then sending a reset code (or the same code) could unlatch it. I know this would be a fairly simple logic circuit to make, to drive a standard relay, but I'm wondering if such a product already exists, what it would be called so I can find it in component catalogues. Update: the signal would be sent from an existing system which can be programmed to send pretty much any digital signal sequence (e.g. faking RS232 or other encodings) but only has one output line, and I can add line level conversion if needed. Update 2: as I only have one driving line, I cannot provide a clock signal, so it must be asynchronous serial. AI: I know this would be a fairly simple logic circuit to make, ... but I'm wondering if such a product already exists, ... It probably doesn't exist. When you have the luxury of one or more wires you'd typically use a versatile or robust solution. One which offers more than just unsolicited command shouting, such as I2C. One that isn't as susceptible to noise as a single ended wire. Such as RS-485. One which layer 2 protocol contains data integrity checks, such as CAN-bus or MODBUS-RTU. Shouting 01110111 over a wire won't be a suitable solution in the industry for obvious reasons. It is however widely used in cheap 433 MHz remote control sockets. A set of dip switches then configures the numbers it will listen to.
H: What is this inductor symbol from a video circuit? This is from the datasheet for the TDA8217 PAL decoder and video processor. This is the "Applicaton Diagram". The circled item is unfamiliar to me, I asked at my local electronic components distributor and no-one there recognized it either. It looks like an inductor, but seems to be connected to ground maybe? AI: Note how it says "330ns", a proper inductor would have a value in Henries. My guess is that this is an Analog delay line like this: Note how it has 3 connections of which one is ground which is connected to the tube and the shield. I actually came across one of these many many years ago, probably in some video equipment I scrapped for parts. At the time didn't realize it was a delay line. The "330 ns" and a search for "delay line" led me to the Wikipedia page. The delay line in the picture has 450 ns delay so 330 ns looks like that could be achieved with a similar component. Sidenote: there are also Piezo electrical delay lines (see an example here) used in analog video equipment and I have seen lots of these and most have a 64us delay time which relates to the old PAL television standard (64 us is one horizontal line of the image). But 64 us is much larger than 330 ns so I knew that these were different types.
H: ICL3232 and SN75C3232 difference I'm using ICL3232EIV-16Z-T in one of my projects for UART communication. As per the datasheet, the max speed is 500Kbps. But I've tested this IC for baud rate up to 921600 and there is no data loss. Also, I've tested with SN75C3232PWR which offers the max speed 1Mbps. Also, it is tested at 921600 baud rate and there is no data loss observed. In both cases, the time taken remains the same. Kindly suggest which is the better IC among these two and What is the advantage over the other? AI: Looking at the ICL3232 datasheet we see:- Maximum Data Rate RL = 3kΩ, CL = 1000pF, min 250 typ 500 kbps and in the SN75C3232 datasheet:- Maximum data rate RL = 3 kΩ CL = 1000 pF min 250 kbit/s So it seems there isn't much difference between them. The SN75C3232 also specifies a minimum of 1000 Kbps with a supply voltage of 4.5 to 5.5 V, so if you are running it from 5V the SN75C3232 is guaranteed to meet your requirements (the ICL3232 may do as well, but you can't complain if doesn't).
H: On the possibility of steel nails as EMI filters I live in a place where ferrite beads for EMI suppression are hard to get. I thought steel had a very high iron loss, especially at high frequencies. So is it possible at least theoretically to use hard steel nails to attenuate high frequency noise in, say, power supplies? AI: I thought steel has a very high iron loss especially at high frequencies. So is it possible at least theoretically to use hard steel nails to attenuate high frequency noise in say power supplies? No, not really. A ferrite bead (for instance) relies on the outer ferrite material (not the through-going wire) being both a poor electrical conductor at low frequencies but, at high frequencies, becoming a lossy capacitor and capable of turning EMI into heat. Here are a few examples from Murata: - As you should be able to see, FBs are designed to target a specific range of frequencies and, different values in the same model range can be chosen to give better attenuation at certain parts of the spectrum whilst maintaining reasonably low losses for signals that should not be significantly attenuated. I live in a place where ferrite beads for emi suppression is hard to get A nail doesn't have one of the vital characteristics of ferrite that make it very useful as an attenuator namely; that it acts as a lossy capacitor as frequency gets higher and therefore resonates with the parallel inductance of the through-going wire. There is a good document from Analog Devices that explains things in more detail and that document shows the developed model for a Tyco Electronics BMB2A1000LN2: - R1 and C1 represent the lossy dielectric of the ferrite material and you just won't get that with a regular piece of iron or a nail. In case anyone notices the typo in the ADI picture above (L1 = 1.208 uF) should read 1.208 uH thus producing a peak resonance at around 112 MHz.
H: 2 band inductor value? An old circuit board I have (that is very old) has these inductors on it. I can't find any mention online of a 2-band color code. What value would these inductors be? I thought maybe one of the colours had faded but even on close inspection I can't see any indication there used to be another band anywhere. ]1 AI: They are 0 Ohm resistors, basically a trace/connection between 2 parts on the PCB. This can be used if you want to seperate 2 parts of the PCB with only a small pathway to control the flow of the current. Or to disable/enable certain parts of the PCB if it's multifunctional and not everything is needed.
H: PMOS Current Mirror I'm being having a lot of trouble fully grasping the interaction between NMOS and PMOS current sources. Is my understanding below correct? Consider the following circuit in two phases, Assume that M3 and M2 are at a 1:1 ratio. The bias voltage for M1 Vb is chosen to allow for 100uA. Phase 1 Iref = 100uA. Due to the 1:1 ratio between M3 and M2, 100uA flows through M2 and M1. Phase 2 Iref increases to 200uA. Due to the 1:1 ratio between M3 and M2, 200uA flows through M2 and M1 M1 has a bias voltage on it's gate to support only 100uA, so the drain-source voltage of M1 (=Vout) must increase to support this new 200uA current. This drives M1 further into saturation and M2 towards the linear/triode region Phase 2 Alternative Understanding Iref increases to 200uA. Due to the 1:1 ratio between M3 and M2, 200uA flows through M2 and M1 As M1 has a fixed gate-source voltage, it can be seen as a fixed ressitance with resistance of ro1. A higher current in the right-branch means, more current through ro1 and thus by Ohm's Law, a higher voltage drop and hence an increased Vout. This drives M1 further into saturation and M2 towards the linear/triode region Is this understanding correct? Am I thinking about these circuits in the correct way? I've been following Razavi's Design of Analog Integrated Circuits Book. AI: You need to realize that M1 and M2 both want to behave as a current source. Also, M1 and M2 are in series so their current must be the same. Which device (M1 or M2) determines the current depends on which device wants to make the smallest current. If M1 wants 100 uA to flow but M2 wants 110 uA to flow then M1 will "win" and 100 uA will flow. Then M2 will be forced into linear mode. Theoretically there can be a situation where both M1 and M2 are in saturation and both want the exact same current to flow. But this is purely theoretical, in the real world the currents are never identical so in practice one transistor is in saturation and the other is in linear mode (it is possible to use a DC feedback loop to force both transistors in saturation but that requires a more complex circuit). Phase 1 Iref = 100uA. Due to the 1:1 ratio between M3 and M2, 100uA flows through M2 and M1. That's not entirely correct, M2 wants to make 100 uA flow, it depends on M1 if that's going to happen. If M1 is set to slightly more than 100 uA, for example 101 uA, then M2 will "win" and 100 uA will flow. M1: linear mode, M2: saturation mode If M1 is set to slightly less than 100 uA, for example 99 uA then M1 will "win" and 99 uA will flow. M1: saturation mode, M2: linear mode Phase 2 Iref increases to 200uA. Due to the 1:1 ratio between M3 and M2, 200uA flows through M2 and M1 M1 has a bias voltage on it's gate to support only 100uA, so the drain-source voltage of M1 (=Vout) must increase to support this new 200uA current. If a transistor saturates at 100 uA then you cannot and should not (try to) increase its \$V_{DS}\$ to a higher value so that more current will flow. You would need more than the breakdown voltage of the transistor and that could potentially damage it! Usually the maximum \$V_{DS}\$ you can apply is the supply voltage and usually this supply voltage has a value of less than the breakdown voltage. So: you cannot make 200 uA flow, 100 uA will flow because that's what M1 allows. This drives M1 further into saturation and M2 towards the linear/triode region M2 will try to make 200 uA flow but M1 limits the current to 100 uA so M2 has no choice other than to go into linear mode. Phase 2 Alternative Understanding Iref increases to 200uA. Due to the 1:1 ratio between M3 and M2, 200uA flows through M2 and M1 As M1 has a fixed gate-source voltage, it can be seen as a fixed ressitance with resistance of ro1. No this is not the case, A fixed \$V_{GS}\$ does not mean that the MOSFET has a fixed resistance! Also, ro1 is a small signal parameter which is irrelevant now since we're dealing with large signal or DC behavior. Your ## Phase 2 Alternative Understanding ## is wrong and not a proper explanation of the circuit's behavior.
H: What is this symbol used in this D flip flop circuit? I'm studying some notes on latches and flip flops. After the introduction of the D flip flop, the notes go on to explain its function - how it works. However, the first circuit diagram contains this weird circuit symbol : Does anyone recognize it? In case it helps, here's the full circuit diagram : Could it be just a symbol replacing the whole D flip flop? In that case the inverters seem to be there so that we can get the 2 complementary Q outputs. I don't get the whole feedback thing though. AI: Thats the symbol for a transmission gate. It's basically a switch controlled by a digital input. It consists of a CMOS-pair for the actual switch and another pair for an inverter for the control signal. The inverter is needed because the P-MOS and N-MOS transistor need inverted levels to turn them on. Note: the whole latch consist of two transmission gates and two inverters. A transmission gate can be used as an analog switch but of course it also works with digital signals like in this case. Note also, to be exact: the circuit shown is not a D-flip-flop as stated, but a (transparent) latch.
H: Measurement current consumption of low power circuit I have a circuit which has uC, Bluetooth module and some other ICs. Current consumption of it varies between 100uA and 20mA. I want to see current waveform using oscilloscope. I worked before high current measurement(above 1Ampere) using low-side sensing technique but not so small this. How to setup a small circuit to see current values of each phase(sleep, active, transmission) separately? Note: The circuit is powered by battery AI: Use "shunt" resitor (low ohm resistor) in power supply line and measure voltage drop on it. Current is low, so resistor dont have to be real shunt resistor.
H: Can I decrease voltage in a battery to get higher amp hour rating? I am willing to use my 7.4 volts 1000mAh lithium polymer battery to power a motor that works with 3.7 volts, so I will input the 7.4 volt to a high efficiency buck converter to get an output of 3.7 volt, which is connected to the motor, Assuming that the motor draws 1 Amp, my question is will I will still get 1 Amp for only 1 hour from the output of the converter or I will get 1 Amp for two hours ?? The total watt hour for my battery is as I know 7.4 watt hours (VAh) but my motor draws 1 Amp at 3.7v so watt hours of 3.7 watt, so If I divided the watt hour of my battery by the watts my motors draw I get 2 hours, Is these calculations correct or there is another rules or factors that affects it ? I am a school student by the way so, please if you answer me, answer me in full details as if I don't know anything. Thanks in advance. AI: Assuming that the motor draws 1 Amp, my question is will I will still get 1 Amp for only 1 hour from the output of the converter or I will get 1 Amp for two hours ?? You'll get 1A for 2 hours (approximately). The total watt hour for my battery is as I know 7.4 watt hours (VAh) but my motor draws 1 Amp at 3.7v so watt hours of 3.7 watt, so If I divided the watt hour of my battery by the watts my motors draw I get 2 hours Correct, but you have to substract losses in the DC-DC converter. If your converter is 90% efficient and uses a 7.4Wh battery as power source, 7.490% will be available on the output, or 6.66 Wh. Basically, conservation of energy (input energy = output energy + losses). Note that you don't need a DC-DC converter. You can use a MOSFET instead: A permanent DC motor does not require accurate voltage regulation, and it is quite inductive so PWM works very well. If the question was about using a 110V rated motor on 230V we should definitely check if the wire insulation would withstand the extra voltage, but for 7.4V I don't think this would be a problem. This simple schematic uses PWM to apply 7.4V to your 3.6V motor with 50% duty cycle, so the average voltage on the motor is 50% of supply voltage, or 3.6V. This saves a buck converter. Efficiency calculation: suppose the PWM is fast enough so we can neglect the current variations in the motor, it will be an average of 1A. Since DC motors have quite large inductance it will smooth the current. Duty cycle 50%, so D=0.5 When mosfet conducts: it has resistance RdsON, and dissipates \$ R_{dsON} I^2 \$ When diode conducts, it has forward voltage Vf, and dissipates \$ V_f I \$ Note that mosfet conducts with duty cycle D and diode conducts with duty (1-D), so: Total losses: \$ D R_{dsON} I^2 + (1-D) V_f I \$ You can add switching losses which are estimated by (Switching time) Frequency * I*V but at motor PWM frequency they will be pretty small. With realistic values RdsON = 0.02 ohm, Vf=0.5V using a schottky, D=0.5, Losses=0.26W, Motor power=VI=3.6W, efficiency=93%. Another option is to wire the battery in accordance with the motor voltage. Since your battery is made of two 3.6V elements in series, if you wire them in parallel instead, you got 3.6V 2000mAh instead of 7.4V 1000 mAh (the energy capacity in Wh stays constant of course). But you may need a new charger, protection pcb, etc, so that may not be an option.
H: Mux UART with 74xx Multiplexer I have a legacy LPC MCU with 2 UARTs only. One of these UARTs is reserved for different functions such as IAP and RS232. So basically, I have one UART available only. For an experimental project I need 5 to 7 UARTs to add 5 to 7 SOC modules. So, I plan a new hardware with any controller, but due to PCB space and controller limitations I do not expect to have many possibilities to add unlimited UARTs. Planning some type of extender, I had the idea of using two 1:8 74xx multiplexers in order to switch between UARTs and a common VCC/GND shared by all of the participants in my SOC/MCU chain. Can you switch safely between UARTs using analog muxes? Or are there ramps/peaks that might lead to UART periphery problems? AI: If all UART comms protocol originates at the Port of the MCU then this may be a workable solution to MUX to the various target devices. On the other hand if UART comms can asynchronously originate from any of the target devices at any time then this multiplexing scheme is a non-starter.
H: Why won't zones flood into vias or other plated through holes that are part of a footprint in Ki-CAD I am using KI-CAD. Visible in the picture is a GND_USB plane (yellow) on layer 2, and also the top layer (red). I have a QFN package with an exposed pad in the center. For some reason the ground plane won't flood into the thermal vias. In the picture you can see that the ground plane floods over some other vias. It just won't flood over the ones that are part of the SMD footprint. In general I can't get zones to flood over any plated through holes in any footprints. Each via in the footprint shows what net name is assigned to it from the schematic. They all say they are assigned to GND_USB (the same as the plane). The vias are defined in the footprint as plated through holes, and exist on all copper layers. What might be possible causes for why the ground plane won't flood over vias that are part of a footprint? Note that the plane is staying exactly 20 mils from the vias, which is the clearance setting for the plane (zone). I confirmed this by reducing the clearance and watching the flooding get closer. So apparently the zone thinks it needs to maintain clearance from those vias even though they are the same net. EDIT: Well Seth found the setting. The problem was in the pad properties in the footprint. For some reason the "Pad Connection" property was set to "none" by default. I changed it to "Solid" and it works now. That was a little confusing because all the SMD pad types do connect by default so I never bothered to notice that setting. AI: The zone settings for your connection have a 10 mil clearance and 10 mil spoke width. This prevents the fill from being able to form a thermal spoke to the via. One way to adjust this is to change the settings of the thermal via pads to be solid connection like this: You will need to do this for all thermal vias in the footprint. Alternatively, you can reduce the clearance size to smaller than the thermal via spacing in your zone settings. -- Edit -- As Rene notes in the comments, you could also set this to (or leave it at) "From Parent" and ensure that the footprint Local Clearance and Settings is taking the connection type from the zone rather than forcing thermal vias.
H: LM7805 heating up I maked a schematic for trigger a siren 12v by a signal from Raspberry Pi (GPIO18 PWM 3.3V). All the circuit is powered by a power source of 12v 15A (see the image below). SCHEMATIC UPDATED: It's works fine (although i prefer not power on by much time), but the LM7805 voltage regulator is heating up a lot! This is why I don't let the circuits run for long time. Is it normal the LM7805 is heating up? If not, what i making wrong? Thank you! AI: The 7805 is a linear regulator. When regulating 12V to 5V, it will be passing all the required current with a 7V drop across itself. It will thus dissipate about 140% of the power that everything on the 5V supply dissipates. It's not unexpected, but it may require heatsinking.
H: Op-amp envelope detector - capacitance load issue I have setup the below circuit; a basic envelope detector, however I believe that the op-amp is struggling with the load capacitance. With the 100nF cap removed, then the output of the op-amp behaves as expected following the input voltage (AC around 1.6v DC). The output of the op-amp is driven higher to make up for the diode voltage drop. (Below scope trace, blue input, red output). However when the 100nF is added, then the op-amp is not drving the output high enough. It just stays around 1.6V. In theory the output voltage should rise to around 1.9V. I added the 2.2K output series resistor to try and get round the issue, but it didn't help. Are there any obvious errors or ways around the issue? It may be that I need to use a different op-amp. Thanks AI: The opamp will saturate at the negative rail unless you add some more components. The opamp can take a while to come out of saturation affecting the circuit operation. You may need to lower R1 if you are running at a high frequency. Note that R2 and R3 will form a voltage divider, so if this is undesirable, make R2 very small. I try to avoid opamps hitting their current limit, which it will if R2 is small. simulate this circuit – Schematic created using CircuitLab
H: Signal 'bounces' back to high I transmit a signal using LVDS over a CAT6 ethernet cable. However, instead of coming through clean, it always 'bounces' back to the high state: (Pink is the input and yellow is the output) (Sorry for the glare and angle, there is a window behind me so it's impossible to take good photos) When the input is turned high, pretty much the same thing happens to the output. The setup is as follows: The SN65 chip takes the sync signal (left, 6) and puts a differential output on any channel that is enabled (left, 9..16). Then the differential signal is routed to the RJ45 port (right). Then there is 5 meter of CAT6 cable. On the other side there's another RJ45 port that is hooked up quite the same. The differential signal leads to this DS90 chip which outputs the sync signal again. Note that there isn't termination. The designer of this circuit assured me that it's not (or at least shouldn't be) necessary. However, when I manually press a 100 Ohms resistor to the contacts, then the signal is better, but not fixed. Can you spot a mistake that would cause this signal going back to high to happen? Do you have tips on how to improve this situation? The sync signal is not a clock or anything. It will just be occasionally toggled (once every minute maybe or only at bootup. Depends on what the system will need in the end). AI: The sync signal is not a clock or anything. It will just be occasionally toggled (once every minute maybe or only at bootup. Depends on what the system will need in the end). Your differential signal is passing through isolation transformers in your RJ45 magnetics. You cannot pass low frequency signals (or DC) through isolation transformer because transformers like these don't work well at low frequency. Note that there isn't termination. The designer of this circuit assured me that it's not (or at least shouldn't be) necessary. It may not be necessary for slow speed signals but it's a moot point because the transformers cannot handle slow signal rates.
H: Getting started with AVR programming concepts I am trying to get started with programming the AVR. I have chosen to use Atmega328P (which is in my Arduino Uno). When I did web search regarding the initial steps for learning the AVR programming, I came to know that the microcontroller requires a special program called "bootloader" to initailize the sketch we have created using the Arduino IDE. The bootloader checks if data (program) is present in RX pins. If the data is present, the bootloader grabs the program and places it in the flash memory. This process of loading the program into microcontroller takes place through UART protocol. In other words, the bootloader lets the program to be loaded into the microcontroller. I think I have understood the concept upto now. Any corrections are heartly welcome. My actual confusion starts when I don't have any bootloaders in my microcontroller. I have my program ready to be loaded into the flash memory.But there is nothing to let the program to be loaded into the microcontroller. There is no bootloader, so there is nothing to let the program into the memory. This is where the programmer comes in action. The programmer takes .hex file from the PC via USB protocol. Then it sends this file to the microcontroller via SPI protocol(Since MISO-MOSI-SS pins are used). My question is: Q1- Why don't I need bootloader to upload the program via SPI protocol. Why can't I upload the program to the microcontroller via UART protocol when there is no any bootloader present? What lets the microcontroller load program into its memory without bootloader when SPI protocol is used? I went through various resources available to grab a concept of these things but I often get confused one way or other. I was studying "The Art of Electronics" and I am getting even confused regarding my first question. The book says: "Contemporary microcontrollers use internal nonvolatile (flash-memory) storage for program code, which you load (while the μC is in-circuit) by one of several methods. You usually do the loading55 with a commercial “pod” (officially called a “device programmer”), which you buy from the chip manufacturer or third party. If you buy a development kit (Figure 15.24), it will often include a programming pod, along with software (for compiling, simulating, assembling, and loading), and with a circuit board on which there’s a microcontroller and other hardware (digital and analog ports, LEDs, a serial port, a programming header, and perhaps some display device). Here are the several loading protocols: UART serial-port bootloader, SPI serial-port bootloader, JTAG serial-port bootloader, Proprietary serial-port bootloader, USB serial-port bootloader, Parallel loading." Is the writer trying to tell me that the program I am uploading via SPI protocol is the bootloader? I can't rule out this possibility because the bootloader is uploaded into the microcontroller by the use of SPI protocol as well. Here's a tutorial from Arduino website. Next part I have been studying is regarding the process of uploading the program into microcontroller. I have made a simple LED blinking program in Atmel Studio 7 and I have it's hex file ready to be transported to the microcontroller. I am taking this analogy to understand how hex file is uploaded into the microcontroller. I have considered my hex file to be a product to be delivered to customer (microcontroller) by the seller (PC). The vehicle to be used is a delivery van (AVRDUDE). But the problem is the river between the seller and customer. So I need a bridge to connect them. This bridge is the "programmer". Is my analogy correct? If not, please help me understand what is the propose of AVRDUDE and the 'programmer'. AI: AVR microcontrollers (and most modern µC) contain special hardware specifically for programming directly using a combination of the SPI port (called ISP for this special case) and the reset pin. This is a very convenient feature because you don't need to worry about bootloaders or memory offsets or anything else. You can program the chips separately or after they are installed. To use a different method for programming the flash, you need to use a bootloader that instructs the chip on how to accept this. UART, for example. Some AVRs can use other methods as well, high voltage programming for instance, that do not use a bootloader. The choice comes down to what is most convenient/useful for you. One nice thing about a bootloader, is it potentially allows an easy way to apply field upgrades to your firmware. As for your analogy: The hex file is the program you are loading into the flash on the microcontroller. The programmer is the physical hardware that will stream this data. AVRDUDE is the software that runs this process.
H: What is the gain of this capacitance measurement circuit? To sense the capacitance value of a sensor, I found the following circuit. My knowledge on analogue electronics isn't that great, so I'm not sure how I'd determine the gain of this circuit. Vo is the output voltage I need, the peak value detector and Vdc can be ignored. So what is the gain, Vo/Vs, in this circuit? AI: If you pick Rf to have a high resistance relative to the impedance of Cf at the input frequency, the gain before the peak detector will be approximately -Cs/Cf. That's a consequence of the feedback impedance being XCf \|\| Rf ~= XCf and the normal equation for the closed-loop gain of an inverting op-amp amplifier. One would normally pick the resistor in such a way- typically the purpose of Rf is to provide a path for the bias current of the op-amp. At very low frequencies the gain will be less as the resistor influence manifests, and you can easily write an equation for the gain including the resistor and including the op-amp open-loop gain too, if you want.
H: Accessing structures concurrently modified by ISR I have the the following situation: An STM32L476RG (ARM Cortex M4F @ 80MHz, 1MB Flash, 128kB RAM) parses some string data coming in over UART. The project uses mbed-os. The parsing of the data is done in an interrupt-based manner, meaning an ISR is in place which reads the just received character and has some stateful parsing logic to extract data from it. While doing so, it modifies an working-object which gradually gets filled up with the parsed values. When the parsing process is done (all values filled), the working copy is saved into another global object of the same type. This is so that the application has an object with valid values and doesn't have to access the working copy of the ISR-based parser, which might change anytime. The structure in question is about 204 bytes big. Code looks like (high level) struct MeterState { double valueA; double valueB; double valueC; char id[64]; //more.. } volatile MeterState isrWorkingCopy; volatile MeterState lastValue; /* called in an ISR context by mbed-os whenever something is received on the serial / static void SerialParserHandler() { uint8_t data = (uint8_t) meterSerial.getc(); parse_data(data, &isrWorkingCopy); //check parsing finished.. if(parsingFinished) { //save finished object lastValue = isrWorkingCopy; } } / called by application to get last valid data / void MeterState GetLastValue() { return lastValue; //copy structure to as return value to stack } However, here comes the crux of the problem. Due to length of the object and given some certain timing, it may be possible that: Appcode calls GetLastValue() Function starts copying the huge structure to the stack of the caller. Mid-term gets interrupted by the ISR ISR overwrites lastValue with new data when parsing is finished GetLastValue() is resumed after ISR Resulting returned value is half the old half the new value See e.g. compiler output for GetLastValue(), it is implemented by a memcpy(): -- 08015f3c <_ZN13ElectricMeter16GetLastValueEv>: 8015f3c: b508 push {r3, lr} 8015f3e: 22d0 movs r2, #208 ; 0xd0 8015f40: 4901 ldr r1, [pc, #4] ; (8015f48 <_ZN13ElectricMeter16GetLastValueEv+0xc>) 8015f42: f004 fa5f bl 801a404 <memcpy> 8015f46: bd08 pop {r3, pc} 8015f48: 20001d38 .word 0x20001d38 Basically overwriting and getting the value is non-atomic. I'm not quiet seeing how it's possible to make it atomic so that the above scenario doesn't occur anymore. I cannot just use a Mutex-based locking mechanism to atomatically exchange the lastValue because I cannot lock a mutex inside an ISR. The ISR must be non-blocking. I also thought about temporarily disabling interrupts around the getter method as such MeterState GetLastValueSafe() { __disable_irq(); MeterState s = GetLastValue(); __enable_irq(); return s; } However, doesn't this have the possibility (as small as it may be) that the parser misses a character (which might screw up parsing and constitute a lost data value)? If the IRQs are disabled and the handler doesn't get invoked, the character is basically lost, as I understand. Is there any sane way to solve this general problem of ISR concurrency with large structures? AI: Assuming that your parsing routine is fast (you really should minimize time spend inside an interrupt), I'd consider modifying your code, as follows. struct MeterState { double valueA; double valueB; double valueC; char id[64]; //more.. } MeterState MeterBuffer[2]; MeterState filled = &MeterBuffer[0]; volatile MeterState unfilled = &MeterBuffer[0]; #define Increment(a) do { MeterState z= (a); if ( (z += 1) == &MeterBuffer[2] ) z= &MeterBuffer[0]; (a)= z; } while (0) /* called in an ISR context by mbed-os whenever something is received on the serial / static void SerialParserHandler() { uint8_t data = (uint8_t) meterSerial.getc(); parse_data(data, unfilled); //check parsing finished.. if(parsingFinished) { Increment(unfilled); / Problem here if both buffers become filled before GetNextValue is called / } } / called by application to get last valid data / void MeterState GetLastValue() { while ( unfilled == filled ) ; MeterState result= filled; Increment(filled); return result; } The above technique only declares one pointer as volatile; the one that can be modified by the interrupt routine. There's no need to declare anything else in that way. The above code also makes the interrupt routine solely responsible for updating the unfilled variable. No one else has any right to modify it. They are only allowed to observe it. That's all. (It's volatile, of course, since the interrupt might occur at any time and update it.) It also makes the GetLastValue function solely responsible for updating the filled variable. No one else (not even the interrupt code) has any right to modify it. Others may only observe it. There's no need for it to be volatile as any modification by GetLastValue is done outside of any interrupt events (unless you do something to hook it into one, I suppose.) By separating ownership of these pointers, there's never a problem with processing and updating their values. I added an Increment() macro for clarity. You can expand it inline if you want. There is a potential problem in the above code, highlighted by my comment. If the interrupt routine executes sends out two filled buffers before GetNextValue() has a chance to fetch one, then the double-buffer is insufficient and there really isn't a good way to recover from that. Which is why I suggested a circular buffer (of larger size.) You can modify the above code easily by increasing the number of MeterBuffer's and reduce (but never eliminate) the risks here. Only you can work out what you want to do if this problem happens. So I leave that to you to worry about.
H: Trigger PLC multiple times with delays I am brand new to PLC's (don't even have a single device in use yet), but I have been programming for more than 20 years and haven taken classes in electrical controls, so I understand machine logic very well. I am planning an application where a machine will make a simple movement with a preset delay after being triggered. Specifically, there are many items on a conveyor belt and some of them will be diverted off the belt. The sensor that triggers this is near the beginning of the belt, and the device to remove them is at the other end. There will likely be multiple triggers/inputs occurring prior to the end of the previous delayed output. Is this something most PLC's will readily handle and is there anything special I would need to do to accomplish this? AI: Traditionally we use a shift register to track parts through a machine. This works well for discrete moves but is more problematic for continuous motion such as on a conveyor. If, for example, your product is 200 mm long then I would index the shift register every 50 mm or so and that way you're guaranteed that each product will be seen at least once and tracked in the shift register. The number of stages required will be the distance between sensors divided by the step. So, for a 1500 mm conveyor you would need 1500 / 50 = 30 stages. simulate this circuit – Schematic created using CircuitLab Figure 1. Pseudo ladder logic code for a 30-bit shift register. In the pseudo code above a 30-bit shift register has been set up starting at F200 (F for 'flag'). If the photoeye is on during a rising edge of the CLK then a '1' is loaded into the shift register at F200 after the existing data has been moved up one position. Any time a '1' reaches F229 the ejector output will turn on. The 50 mm pulse would ideally be triggered by sensing the rotation of a shaft on the conveyor but if the speed does not change it could be done on a timer. From the comments: The important part of the question is the concern that there will be several items on the conveyor at a time in between the sensor and the ejector, will the logic still work as is? Yes. \| sensor \| ejector _______ ________ ________ ________ \|_______\| \|________\| \|________\| \|________\| >>> ___________________________________________________________ (___________________________________________________________) Conveyor 1 1 0 0 0 0 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0 0 1 1 1 1 0 mm 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0 0 1 1 1 50 mm 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0 0 1 1 100 mm Figure 2. Shift register contents.
H: What is the reason for multiple ground connection and multiple testpoints on this PCB? Please, have a look at the image: The biggest pad at the top is the negative terminal of a battery holder. Through 0 Ohm resitor above it goes to a capacitor across a DC motor (not shown). Below the diode and off the image it goes to a ground of a charger connector. Just below the pad there is a thin trace, leading to a small capacitor and a 5 pin SOT-packaged something. And a bit lower there is yet another ground connection, that is leading down to transistors and below. I do agree on keeping DC motor noises as far as possible and it it well demonstrated here via 0 Ohm resistor up. But it is a mystery to me, why would they split ground connection more and more? They could've taken the first split and led it immediately to transistors. Or they could take the second split lower -- near a diode bottom left. Or they could lead a thick trace instead of split. Is there any real need to split it so many times? I also wonder, what is the purpose of so many test-points? This ground has 3 testpoints, traces near also have 2 test points. If they are at the same trace, at the same potential, what's their purpose? AI: You are on the right track with your comment about the "same potential." PC board traces have non-zero impedance, but have resistance and (more importantly in this case) inductance. This means that when you are running a load with potentially high, switched currents, the instantaneous potential is not the same at all points along the current path. Your board has established multiple current paths; some are low-current and these will be pretty close to the same potential because there is little voltage drop associated with the current. The ground for the circuitry on this branch will be the more or less the same between devices, making this ground branch useful for a reference for analog signals and for digital thresholds. The path containing high, switched currents can have relatively large instantaneous voltage drops along its length; one volt or more is not uncommon at the switching edges. Imagine the ground reference on a processor moving one volt with respect to the digital input signal levels and you can see how this would be a problem. The strategy is therefore to design so that the high-frequency, high amplitude currents are not flowing in the sensitive, low power circuitry. Similarly, to measure voltage on a device, the test point used for ground reference should be local to the device. You can experiment with an oscilloscope, measuring a voltage at one point in the circuit while moving the probe ground to different points and you will see the effect.
H: 5 m, 5 V LED strip 20 A PS = Fire? I've got a 5m LED RGB strip to go up the stairs and will be programming it with arduino. eBay link. They state 60 LEDs/m @ 18 watt/meter, thats 3.3 W per LED? It says current rating of 3 A, so that means I can only power 9 LED's at a time wiring it from both ends? That doesn't sound right. By the looks of it if I wired in my 5 V, 20 A supply to both ends it would still set fire. I've seen these questions asked with some good answers, but I don't understand how the small cables I see people use don't simply go up in flames. So I would have to run it in batches of 9 LEDs using my calcs, which would be an absolute nightmare, surely I'm wrong. AI: They state 60 LEDs/m @ 18 watt/meter, thats 3.3 W per LED? 0.3 W per LED, they're assuming the standard 60 mA per diode for RGB 5050 SMDs. By the looks of it if I wired in my 5 V, 20 A supply to both ends it would still set fire. 20 A is how much current the supply can provide, not how much it will actually give when hooked to the load. Those tiny traces have a lot of resistance, so if you try to hook up 5 m, you will get a lot of voltage drop, many or even most of the LEDs won't turn on and the actual current load will be much lower. Hence, no fire (and not much light). To use that many LEDs you will probably have to wire a bunch of shorter strips in parallel, each using a current of a few amps.
H: Why did Axial Capacitors Fall Out of Use in the Industry? When inspecting the boards from electronics made in 1980s and earlier, one distinct feature is the extensive use of axial, electrolytic capacitors as power supply filter. Axial ceramic decoupling capacitors were used as well, to a lesser extent. For example, this is a C64 motherboard. Source: Wikimedia Common, by Gona.eu, license: CC BY-SA 3.0 This is a Tektronix 1720 vectorscope board. Source: Flickr, by Toby Thain, license: CC BY-NC 2.0 However, although they are still being manufactured, it seems that axial capacitors largely disappeared in most devices since the 90s. Almost none of electronic devices we commonly see have a single axial capacitor. And it's certain that one is going to find something similar to this in a modern device... Source: Wikimedia Common, by Dave Jones from Australia, license: CC BY 2.0 Question Why did Axial Capacitors Fall Out of Use in the Industry? I can imagine that axial capacitors were optimized for point-to-point wiring back in the pre-PCB era, not PCB assembly, and that the introduction of SMT was another shot. But it was just my imagination, backed by nothing. What were the exact sequence of events and/or rationale that led to the disuse of axial capacitors? AI: PCB area. Axials pre-dated PCBs, their construction was ideal for wiring to tag strips and valve bases, and they were adopted for PCBs because that's what was available. Example of tag strip construction below. (There WERE radial caps in the valve days : they were generally designed for chassis mounting via a ring clamp, and had tags rather than wire leads. The round object bottom centre is the base of one such capacitor) It's actually quite surprising they lasted as long as they did alongside radials, into the 1980s. Radials use much less PCB space, and standing axials on end is a poor compromise, with a long exposed lead (or the added assembly step of sleeving it) as well as being much less robust.
H: Why is it desired to divert the surplus PV power into a resistive load? From wikipedia/Maximum_power_point_tracking: When the batteries in an off-grid system are fully charged and PV production exceeds local loads, an MPPT can no longer operate the panel at its maximum power point as the excess power has no load to absorb it. The MPPT must then shift the PV panel operating point away from the peak power point until production exactly matches demand. (An alternative approach commonly used in spacecraft is to divert surplus PV power into a resistive load, allowing the panel to operate continuously at its peak power point.) I can't imagine why is it desired to divert the surplus power into a resistive load instead of simply ignoring it. What is the advantage of keeping PV module operate at its maximum power point at all times? AI: Spacecraft aren't air-cooled. Simple as that. Solar cells operate in full sunlight, and get hot. Dissipating 30% of the incident power somewhere else helps keep the temperature manageable. The unlit side tends to get very cold, for much the same reason - so there may be benefit to using the heat in another part of the satellite.
H: How do I wire up this item (BM70 - Microchip BTLE) for prototyping I've just received my BM70 (Microchip BLUETOOTH 5.0 BLE MODULE) via DigiKey and I'm wondering how I can wire it up. According the BM70 Datasheet the pad holes are (are .7mm (outer) and .5mm (inner)). Those are smaller than items I've worked with on common breadboards so I am just wondering best ways to deal with them. Is there a common socket for chips of this (rectangular) dimension, which would allow me to pop the entire BM70 into it and then use with a breadboard? Is it just a matter of soldering (very precisely) to each small pad? Is this item generally only used by soldering on a PCB? AI: This board is designed as a daughterboard, to be soldered from both sides to a larger PCB. But it's possible for a hobbyist to prototype with it. Every hobbyist eventually learns the limitations of breadboards: as simple as they are, there are times when they simply won't do the job. There are mainly two possibilities. First of all, don't care how ugly the construction is, the first goal is to make it work. A prototype is subjected to numerous changes on-the-fly anyway. Use a Perfboard A perfboard is a single or double-sided empty circuit board with a matrix of holes. There are no connections on the board, and one is free to solder the board in all possible ways. Although it's not designed for SMT, it's certainly can be done. An extreme example is shown here. Source: @Philips_NE555 on Twitter, fair use You can either choose to assemble the entire circuit on a perfboard, or you can cut a small piece of perfboard, buy a row of standard 2.54mm connectors, and solder your module on the perfboard. Once assembled, you can hook up wires, just like a standalone module. I recommend: If you don't have any experience with perfboards, purchase some cheap electronic devices (like a 555 timer in surface mount package) and build some circuits for practice. Also, once the circuit works, try desoldering every single component from your board without damage (if inexperienced, a lot of damage can be created in this process). Particularly, a double-sided board is sometimes challenging to desolder, those plated through-holes can be nasty, and need practice. I recommend a single-sided perfboard, although you cannot solder from both sides, it's easier to (re)work with. Use AWG-30 wirewrap wire or magnetic wire for interconnection. Important: Solder some 0.1 uF ceramic decoupling capacitors across the power and ground on the perfboard. Solder some electrolytic (100 uF or so) filter capacitors. It ensures the power integrity of the daughterboard. Also, keep all signal wires as short as possible. More information: Wikipedia: Perfboard Dead-bug Construction In a dead-bug construction, a bare copper board is used. The chip is flipped to its back, with leads and wirings hanging in the free-air. Sometimes, tapes are used to stop the chip from touching the copper board. A dead-bug construction is essential for high-frequency digital electronics and radio electronics, as it provides a continuous ground plane as a low-impedance return path. However, it seems that your BM70 module is already a self-contained module, and it's not mandatory to use a copper board, but you may find it useful. Source: IEEE Spetrcum, fair use More information: Hackaday: Getting Ugly, Dead Bugs, And Going To Manhattan IEEE Spectrum: With the “Dead Bug” Method, Hobbyists Can Break Through the High-Frequency Barrier Homebrew PCB, or Design and Buy a PCB It depends on the circumstance, but sometimes a homebrew PCB is an easier plan to execute than a prototype board, a lot of wiring time can be saved. If you are not sure about the circuit, you don't have to design a PCB that contains the full circuit, it can just be a breakout board: a row of connectors, a bunch of traces, and a row of pad for soldering the module. This is a computer system I was building at home. There are various resources online for homebrewing PCBs, just search it. The photoresist process is the most common. But you need some tools. Buy UV-sensitive prototype circuit boards, these boards are just regular copper boards with a layer of UV-sensitive chemical (cheap). Buy chemicals (cheap). Buy some plastic containers (cheap). Buy etchant (an acidic solution, cheap). Buy safety goggles (cheap). Buy a set of PCB drill bits (cheap) Buy a blacklight UV-exposure box (30-50 USD). Buy a drill press (the most expensive part, a cheap model can do the job, if you are just making a single, drilling the holes manually might work as well). Design your PCB in a CAD package on a computer, as usual, but single-layer. Print out the layout on a film, using a printer. Put the film on the board, expose board under UV light. Put the board in a exposure solution, the covered portion remains, but the uncovered portion is washed away, leaving a layer of exposed copper. Put the board in the acidic solution, the copper dissolves in the solution, forming the circuit pattern. Drill. Solder. More Information: How to Make a Printed Circuit Board (PCB) Using the UV Light LED Method. Design your PCB, and Send it to a Factory Alternatively, if you don't want to make PCBs at home, you can design your own PCB and outsource it to a manufacturer. There are plenty of manufacturers from China that take orders online, you can tell a factory to make a board for you in a few mouse clicks. And it can be as cheap as 10 dollars. The disadvantage is time. You'll need to wait for 3-5 days, but you can have professional-quality boards. A faster order from a local business (e.g. OSH Park in the USA) is possible, but makes it more expensive. More Information How to Order Cheap Custom PCB Online
H: Is noise density over a bandwidth a measure of the noise's standard deviation? Suppose I have a sensor which has a dominant white gaussian noise source, a \$1\textit{unit}/\sqrt{Hz}\$ noise density and a bandwidth of 100Hz. Then the noise becomes 10\$units\$. Is this a measure of the standard deviation of the noise? What exactly is \$10units\$ a measurement of? In essence my question is, assuming a AWGN source with noise \$10units, \$ is it proper to say that this implies that the noise \$w \sim N(0,10^2)\$? AI: The RMS signal level for Gaussian white noise is measured in units per square root of bandwidth. And RMS has the same formula as when calculating the statistical standard deviation of a series of points. So, for instance; For each point, calculate the distance from the mean value (call it 0 volts), square the distance. Repeat for all other points and sum all the squared distances. Then divide by the number of samples (take the mean of the squares) then, take the square root. Sound familiar?
H: LED Lamps in series I have some G9 220-240V 50Hz 24mA 2.8W LAP LED lamps. I assume that in order to power one I need a supply voltage of 220 - 240V AC able to provide 24mA, which is 5.28W. Why is it rated at 2.8W, does it actually only need a supply current of 17mA? If 5 of these were in series would the power supply still only need to supply a current of 17mA, but at 1100V to work? Would the total power provided by the supply need to provide - 2.8W x 5 = 14W or 1100V x 17mA = 18.7W or 1100V x 24mA = 26.4W What is the minimum power these specific LEDS can run at?, and how would I be able to find this value experimentally? AI: I assume that in order to power one I need a supply voltage of 220 - 240V AC able to provide 24mA, which is 5.28W. Why is it rated at 2.8W, does it actually only need a supply current of 17mA? This lamp is not a resistor, its current does not follow the I=V/R law. In AC, P=VI only works if both V and I are sine waves and in phase. Since this lamp will use a bridge rectifier and won't have a power factor correction circuit because it is too tiny to fit one, current won't be a sine wave. So "24mA" rating is not the RMS current, it will probably be the peak current drawn by the lamp, on the peak of the voltage sine wave. If 5 of these were in series would the power supply still only need to supply a current of 17mA, but at 1100V to work? Most likely not. Every LED is a little bit different due to fabrication tolerances, and every lamp will use a bit more or a bit less current than the next. The current regulator chip, if there is one, also has tolerances. When wiring loads in series, the one which has higher impedance (ie, draws less current) will end up with more voltage than the others... it acts like a voltage divider. And the one getting the highest voltage could blow. There is no way to say what will happen though, and ironically cheaper lamps may work better in series. If the lamps are really cheap they won't have proper current regulation, just LEDs so they would act a bit like zeners and work well in series. If the lamps have proper current regulation, then the lamp which has the lowest current will end up with more voltage and blow. In other words, meh. Would the total power provided by the supply need to provide - 2.8W x 5 = 14W or 1100V x 17mA = 18.7W If it worked, then yes. What is the minimum power these specific LEDS can run at?, and how would I be able to find this value experimentally? You can actually do this experiment if you have a variac. Simply step up the voltage until they light. But the result you get will depend on whatever bulb you test, so don't design a fixture based on that, or it will only work with the bulbs you tested... Note G9 bulbs are trash. They are usually too small to fit a capacitor, so they will flicker at twice mains frequency. They are also too small to fit a properly efficient DC-DC switching converter, so they will be very inefficient.
H: Capacitor on Op-Amp power supply pins What is the purpose of the capacitors between 5V and ground on the power pins of the op-amps? AI: It looks like these are just some decoupling capacitors. Their purpose is to clean out any noise from the power line (on the pcb caused by other parts) and from the device itself. Technically the op-amps would work without them (only do this for testing purpose) but it is always good practice to put them on any integrated circuit you have in your design. Some times the ic manufacturer will give you instructions about the capacity, but 100nF (0.1uF) is a very common value and my goto value if nothing is specified. These capacitor will not dramatically increase your BOM you can get them for around 0.1USD (single pice) or even cheaper when buying a higher amount. But they will increase the reliability of your design. You can find some more information here: https://en.m.wikipedia.org/wiki/Decoupling_capacitor
H: Write endurance: Is there a difference between eMMC and Micro SD Card? I have seen several forum posts (link 1, link 2) debating whether eMMC offers an endurance advantage over SD cards for embedded applications. My understanding was that the write endurance problem is an inherent feature of all Flash storage technology, and that there is only a nominal control interface difference between Micro SD cards and eMMC chips. I am aware that there are differences in write endurance due to the structure of the Flash storage itself. As I understand it, this is essentially a trade-off between storage density and write endurance, where Triple Level Cells (TLC) have higher capacity than Single Level Cells (SLC), and Multi Level Cells (MLC) somewhere in the middle. I can understand how write endurance is tied to this internal flash structure, but I don't think the write endurance has anything to do with whether there is an SD or eMMC interface on the front end. Is this correct? Or am I missing something? AI: I can understand how write endurance is tied to this internal flash structure, but I don't think the write endurance has anything to do with whether there is an SD or eMMC interface on the front end. The interface is determined by the controller the NAND cells are hooked up to, so putting an eMMC controller on a flash cell doesn't make it more reliable than if you'd hooked up an SD controller. Each cell still has the same endurance, although different controllers can of course implement different wear leveling. The usual concern with SD cards however is the large number of fakes out there, which sometimes find their way into even legitimate products. In that sense, if you're buying higher endurance parts, it may be easier to find what you're looking for in an eMMC package or at least to be sure that your product has the part you think it does once it goes into production.
H: How do I make water tank control circuit? I would like to make electronic circuit to control water tank. Its works expected as below: Once water inside the tank is lower than its minimum level, then the mechanical sensor A will trigger to activate the electronic circuit C to power electric water pump connected to it. Electric water pump would be connected by relay, to make it separate from the electric circuit. Once the circuit is triggered by sensor A, it have to be REMAIN ON even the mechanical floating sensor no more pulling the trigger. That is required to make sure the pump motor keep working to pump water till the tank is full. Once the water reaches the maximum or full level, the mechanical sensor B will trigger the circuit to power off the water pump motor (to power of the relay). If during the filling water to the tank power grid is OFF, then it will be assume as the sensor B as active, the pump is powered off. My question: How to make the circuit active (water pump is working) after the mechanical sensor A is no more working as it start float. How do I make the circuit? Motor is put separate and controlled through relay. For this purpose, circuit is avoided to use logic gate. Electric power to the circuit could be DC or AC, but preferred is AC. AI: The simplest way of controlling the pump I can think of is as in this schematic. Once the lower switch is activated, the relay will latch on. Only when the upper switch is broken will the relay switch off. simulate this circuit – Schematic created using CircuitLab
H: Instantaneous and active power in an ideal switch powered by real source and squared wave This is a somewhat theoretical question, but one that has some impact on power theory. The circuit in the figure is composed of a real DC source (with non-negligible internal resistance) and feeds a non-linear load, composed of an ideal switch that switches every T/2. This problem is proposed in https://doi.org/10.1109/EPE.2019.8777983 by L. Czarnecki (a renowned power system engineer). It seems clear that the instantaneous power is null, since the voltage is 0 when the current flows and the current is also 0 when the switch is open. It seems quite intuitive. Mathematically, the instantaneous power is defined by \$p(t)=u(t)i(t)=0\$ and the active power \$P\$ is the average value, so it should be also 0. The problem arises if you use the Fourier series for a square wave. In this case (\$\omega=\frac{2\pi}{T}\$), $$u(t)=\frac{100}{2}\left(1+\frac{4}{\pi}\sum_{n=1,3,5...}^{\infty}\frac{1}{n}\sin{n\omega t}\right)$$ $$i(t)=\frac{100}{2}\left(1-\frac{4}{\pi}\sum_{n=1,3,5...}^{\infty}\frac{1}{n}\sin{n\omega t}\right)$$ In this case, it can be seen that the power is no longer equal to 0, as there is at least one DC term $$p(t)=\frac{100}{2}\frac{100}{2}+...=2500+... \quad \text{[W]}$$ What do you think about this? For me, it makes more sense physically the first proposal, but I think there's something that I miss... Update: I update the summation index to reflect odd sequence. SOLVED: Indeed, all the terms sum up to zero. The key is that the cross products between 1 and the summations cancel out and the product of summations is $$\left(\frac{100}{2}\frac{4}{\pi}\sum_{n=1,3,5...}^{\infty}\frac{1}{n}\sin{n\omega t}\right)\left(-\frac{100}{2}\frac{4}{\pi}\sum_{n=1,3,5...}^{\infty}\frac{1}{n}\sin{n\omega t}\right)=-2500 $$ so \$p(t)=0\$ in both time and frequency domain AI: It seems clear that the instantaneous power is null, since the voltage is 0 when the current flows and the current is also 0 when the switch is open. It seems quite intuitive. No, when 100 amps taken by the resistor (switch closed), one side is shorted to 0 volts and the other is at 100 volts hence, the power dissipated during that closure is 10 kW. What you seem to be considering is the power dissipated by the switch and, of course, for an ideal switch, that will be zero. In this case, it can be seen that the power is no longer equal to 0, as there is at least one DC term Just because there is at least one term that is DC positive does not mean that there are not other terms; such as sin(nwt)*-sin(nwt), that produce DC negative terms that cancel the positive term. Do the math.
H: Circuit for stabilizing the mini - 3 phase wind turbine output to 5 Volt using 7805 IC I am trying to get some stable output from a Brushless power generator/motor/turbine which i purchased from Aliexpress Mini Micro Small 3 Phase Turbine To check if it works i tried to rotate it and LED blinks. Added some toys gears in front of it and running it fast i am able to run over 7 3/5 mm leds so looks like it is generating around 10 Volts with a current of around 150+ milli amperes. Now i want to get a stable voltage something like 5 Volt so i added a Voltage IC 7805 and then added couple of LEDS. I still didn't added a capacitor as was not sure how much value of capacitor i really needed. So it is like this Turbine -> 7805 -> 5 mm Led - no other electronic component In this case when i run the motor - the LEDs just blink for once (0.5 seconds) and then unless the motor fully stops then didn't blink at all. And if it run it again it again just blink for a second. Is i am missing only the capacitor and what about what value? or something else is needed more to prevent damage to IC/Motor like some diodes etc? I am also curious that my cheap Multimeter (under $10) does not show any voltage out of the IC and of motor but works for AAA/AA etc. Any ideas are appreciated AI: A 3-phase machine produces AC: I am also curious that my cheap Multimeter (under $10) does not show any voltage out of the IC and of motor but works for AAA/AA etc. Your multimeter is set to DC when it can measure the voltage of batteries. DC is not AC. LEDs and the 7805 need DC. You don't have a rectifier. Hence, this doesn't work. Build a rectifier. If you build a 3-phase rectifier, you will leed relatively little capacitance to stabilized the output voltage. How much you'll really need depends on your requirements for voltage accuracy, and the current you draw and the current the generator can supply. PS: You're operating the 7805 outside its specification by providing it an inverted supply voltage half of the time; that's why it shuts down. Chances are that doing that damages the IC.
H: How to make sense of the signal trace of differential signal I am trying to understand following signal trace picture that is shown on the wiki page of 1553 Bus. The Bus has balanced differential pair wires. As the 1553 bus has differential signals on the twisted pair of wires it means that the physical arrangement of the oscilloscope probe was that the probe SIGNAL and GND pins would be connected on the differential wire-pair. As there is no 'ground' wire in the 1553 bus so the oscilloscope probe GND will not be connected to any 'ground' potential but one of the two differential wires will act as the reference potential point (ground) for the oscilloscope input signal. In the picture I can see 3 voltage levels, Low, Mid and High. My understanding is that when the Bus is quiet then both lines are at 0 volts w.r.t their signal ground so their difference will also be 0 volts. It looks like this is the 'Mid' signal level case. I could not understand how the Logic '0' and '1' is translated as 'Low and 'High' in the above signal trace on the oscilloscope? If I want to draw the individual signals on the differential lines w.r.t. to the signal ground then what would be those 2 signal traces that produces the trace as shown in the above picture? AI: This bus is Manchester encoded. This means that one symbol (bit) is encoded as a low transitioning to a high, or a high transitioning to a low. These correspond to 0 and 1 respectively. Therefore, if you want to send: 1 0 0 1 0 1 1 1 the sequence is: HLLHLHHLLHHLHLHL This encoding moves the center frequency up to the bit rate, eliminating the low frequency content, and makes clock recovery easier.
H: Can I replace this oscillator circuit with oscillator IC? Disclaimer: I am programmer and when it comes to analog circuits I have no idea what I am doing. I have a board with STM32F103 that uses external oscillator made by crystal and two capacitors: The crystal is CD05M008000RD1 / 49SMD-8-20-20 described as 8Mhz, 20pF, 20ppm, ESR 60Ohm. Everything seems to be working fine, but I am trying to reduce the board size and the crystal is rather big. I am looking for either smaller crystal or a complete oscillator in small package. I found for example this: O 8,0-JO32-B-1V3-1-T1-LF described as XTAL OSC XO 8MHZ HCMOS SMD, 50ppm. Can I use it so that it will look like this? I see that it is less precise (20ppm vs 50ppm), but for my use-case it should be well within the limit. It is still way better than the internal oscillator. Are there any other differences that I am not seeing? AI: From the datasheet both types can be used. Just select the type in FW.
H: Automatic watt/volt/amp switch Google has failed me so I'm asking here. I am looking for an automatic switch, which will change state depending on the number of Watts/Volts/Amps flowing through the circuit. As an example: When the amount of Watts flowing through the switch is less than 40, the switch remains in the default position. When the amount of Watts is greater than 40, it changes position and opens a different circuit. Does this type of device even exist? (Forgive my ignorance, I am a beginner) EDIT The problem I have: I have the following circuit, where the sensor is generating enough Watts in its off state, that the LEDs turn on even then. I am thinking of a workaround like this: Where an automatic switch would workaround the issue and only connect the LEDs when the sensor has 'sensed' something. AI: The crux of the problem This is a bog-standard problem seen over and over in the DIY stack, when obsolete/cheap dimmers, motion sensors or other smart devices are used with efficient, driver-based LED lighting. Older switch loops do not have a neutral wire. To solve this problem, older/cheaper, obsolete dimmers, motion sensors and other smart devices use a characteristic of incandescent lights - their extremely low impedance when off. They leak their operating current through the "off" incandescent filament. This doesn't work on LEDs; they are so efficient they turn all current into light. The proper way to solve this problem is bring neutral to the switch, and have the smart switch power itself between supply hot and neutral. Like a normal load. The direct solution The problem with your idea is you're not allowed to horkle-dorkle "hacks" like this where mains wiring is concerned. You must follow your local Electrical Code, which says every device must be approved for use in mains wiring, and be installed according to its instructions. Here's what you can do: a relay that is approved, such as an Aube or RiB. You wire the hunk-a-junk dimmer in series with a relay coil, with the coil returning to neutral. Separately (it can even be off a different circuit), you take supply hot to one relay contact and connect the other relay contact to the lamp(s). Now, the relay coil may be too high impedance for your obsolete sensor. If so, you can add an approved "dummy load" in parallel with the coil, such as the Lutron LUT-MLC, which is approved for this purpose. It acts like a resistor. Actually, it's a capacitor tuned for your mains frequency. Or, just fit the dummy load directly across your lights Like I say, dummy loads like the Lutron LUT-MLC are designed and approved for the purpose of shunting across LED lights to provide an alternate current path to get obsolete dimmers to work. So you could just fit this directly to your lights. Problem solved. This method will work no matter what, even if you have an old-school "switch loop" which has no neutral at the switch. Never misuse safety ground for neutral, it defeats the purpose of having a ground wire. Indeed, one of the diagnostics we give people who report this problem is "Replace one of the bulbs with incandescent, does the problem go away?" The incandescent acts like the dummy load, or to be more precise, the reverse. Or get a modern motion sensor Get a motion sensor that has a neutral wire. It powers itself via the supply hot and the neutral. It places no demand on the bulbs at all, and will in fact work properly with no bulbs connected.
H: Why do my all Atmega328P chips stop responding suddenly in a 12v circuit with Zener diode? I have noticed this strange thing where my atmega328p stops working and when checked through Arduino IDE or AvrdudeSS or ProgISP, they show programmer not in sync, rc=-1 initialization error (while the same circuit recognizes a new chip). EDIT3: Zener removed form schematic as the problem exists even without the zener. My circuit has a 12volt input coming from an adapter, then to 7805 and 7805's 5v output to my atmega chip. The circuit worked all fine until 7805 was moved a bit by jerk and all becomes hell, atmega stops. Is it permanent damage? I am unable to understand what is it and my money is wasted on buying new chips and demoralizing me to continue. EDIT: I must also mention that one of my chips died in the same way while I just had an led tested with the chip's 5volts. All I noticed is a "jerk" to 7805 did it (switched-off the led and then the chip stopped responding to anything) EDIT2 - Is it possible that fuses go wrong if chip resets due to jerk on voltage regulator or anything similar? I know it sounds silly. I always burn bootloader on my new chips using usbasp and Arduino IDE 1.0.1 and then set fuses using AVRDUDESS default fuses for Arduino Uno (L 0xFF H 0xDE E 0xFD, after which blink program blinks at normal rate). AI: I noticed is a "jerk" to 7805 did it Any interruption in the 7805's ground connection would result in the 12v input being applied to the downstream circuitry, resulting in immediate damage. You really need to arrange things such that a "jerk to the 7805" is not possible, ie, solder the connection or use a good connector in a way that it is not under mechanical stress. Generally speaking, you're better off with a regulated power supply built in some lasting way, not on a breadboard or temporary improvisation.
H: Miller compensation capacitor sizing How can I size the circled capacitor in order to get rid of the peaking? Is there some formula? Cheers AI: Given; uncompensated discrete amplifier with closed loop gain of 20 dB , (R1+R2)/R1= 10x high Q resonance at 5MHz, ~20dB/decade rolloff >10MHz 0dB gain at 50 MHz which becomes the GBW product. We could compute the open loop gain and determine the ω/RC breakpoint for some circuit Req and added Miller C to achieve 0 dB gain and then expect 45 deg phase shift at unity gain at GBW/G = 50MHz for phase margin. example if \$a_v=60dB\$ and 1st order = -20dB/decade above 50MHz the breakpoint "at least" must be less than 3 decades lower or 50kHz. ( if 80 dB then 4 decades lower) the Req=R29//hFER24 + R25 your C1 is then C1=1/(2π fReq) There will be tolerances for each transistor GBW in this design which determines the extra margin to reduce breakpoint. In typical BJT Op Amps , the breakpoint is 10Hz. For video Amps not unity stable it may be > 10kHz + e.g. The goal is to make the slope 1st order or phase shift <=90 deg at unity gain and eliminate the cascade stage higher order effects at high gain where negative feedback almost shifts into positive feedback "low phase margin" ref
H: Why is the following assumption valid? For the following AB power amplifier VBB/2 is large enough such that both Qn and Qp are on for no input signal Vi, if output voltage is large and positive, it may be assumed that most $$i_{L} \approx i_{N}$$ What justifies this assumption? AI: For the positive side of v1 all load current is flowing through Qn and Qp is open. In= IL and Ip=0. And reversely for negative side of vi: Ip = Il and In=0. The purpose of Vbb/2 is avoiding discontinuation of the output signal when v1<0.6 V. It's the enhanced version of the pur AB class amplifier. Don't know if this help you.
H: Calculation of a feedback resistor in an inverting shmitt trigger This is taken from a frequency counter circuit based on a schmitt trigger comparator . The frequency counter has a noisy immunity threshold of 100mV. How would you calculate the Rfeedback resistor value and what is the purpose of an immunity threshold? AI: To answer your second question first, if you have a signal with noise on top, and that signal has a relatively slow slew rate through the point where the comparator switches, you can get many transitions where you would prefer only one. Image below from here By adding some hysteresis, more than the noise peak-to-peak voltage, you can get just one transition per cycle of the underlying signal and the noise will be ignored. With the circuit you show, if you consider the Thevenin equivalent of the part of the circuit to the left of the comparator, you have a 2.5V source and 5K resistor, and there's a voltage divider with the series resistance Rfeedback to the output. If the output changes by 5V (it may not!) then the hysteresis is just: Vhyst = \$ 5\text{V} (\frac{5\text{k}\Omega}{5\text{k}\Omega + R_{feedback}})\$ The hysteresis will be centered above the 2.5V level by half the hysteresis since the output is assumed here to go from 0V to 5V. That means that if you change Rfeedback, the center of the hysteresis band shifts so you can't independently adjust the hysteresis by changing one part value without affecting the center of the band.
H: 555 timer frequency is way off I'm currently designing a 555 timer to provide a 38kHz square wave to drive an IR LED as part of a beam interrupt detection system. This beam interrupt is a part of an embedded product where we cannot easily modify the existing processor software, and do not want to add a second MCU with a separate software installation step just to spit out a 38kHz signal. The goal is to use the VSOP38338 as the receiver, so 38kHz +/- 2kHz should be sufficient for this application. I am having trouble getting the output frequency of the 555 to match the calculated value I expect. Here are some value combinations I've tried so far: R48 R49 C11 Fcalc Fobs 1k 191k 100pF 37.67kHz 31.4kHz 150ohm 1.82k 10nF 38.07kHz 31.9kHz 10k 10k 10nF 4.81kHz 4.68kHz 1.82k 1.82k 10nF 26.43kHz 23.5kHz Regardless of the R/C combinations I've tried, the output appears to be significantly off from what I would expect. For reference, the resistors are 1% tolerance and the capacitors are 5% tolerance. The specific 555 chips I ordered are from Texas Instruments and all of these parts were ordered new within the last 3 weeks. I have verified the resistor and capacitor values with a multimeter out of circuit before installation. Also note that there is not currently an IR LED plugged in to connector J4. Am I missing something obvious? I would expect some amount of variation due to the part tolerances, but the deviation seems unusually large. I have tried replacing individual Rs and Cs as well as replacing the 555 chip itself. This has not led to any noticeable performance changes. I have noticed that most examples decouple Pin 5, CONT, to GND with a 0.01uF cap. I am using an 0.1uF cap instead, because it was already on my BOM. As this is a decoupling cap, I didn't think going a bit bigger would make a difference. Any suggestions on what might be causing the high level of frequency error? Here is my schematic and layout for the 555 timer: Here is an oscilloscope capture for the first arrangement, with the 100pF capacitor: And here is an oscilloscope capture for the second arrangement, with the 10nF capacitor: AI: A 555 is not really a precision device, and as you've found, if the resistor values or the capacitance value is too low you'll see relatively large deviations from the theoretical values, especially with the old bipolar type operated from 5V. Stray capacitance affects a 100pF capacitance value, and 150 ohms is low. Bypassing pin 5 has an effect on the frequency that is not covered by the equations. Typically the frequency is lowered by as much as 10% depending on how good your bypassing is. You'll probably do better with a 1nF NP0 capacitor and the CMOS version of the 555. Consider feeding back the voltage from pin 3 with a single resistor rather than using the discharge pin which forces a rather low value for R48 to get close to 50% duty cycle. If you stay with the bipolar version, in particular, increase the supply bypass capacitor C10 to at least 1uF and reduce the pin 5 bypass to 10nF. The bipolar version draws rather nasty current spikes when switching.
H: Do small 1:1 isolation transformers exist? I am building a Nixie clock that will be powered from 110V wall power. The design is based around having this 110Vac as the input voltage, but I wanted to put an isolation transformer inline for safety. This thing will maybe go up to 10W, so I don't need a huge high power transformer. But I cannot find a 1:1 small transformer. Are these not a thing? Am I looking in the wrong place? Or do I need to redesign my circuit to use a different input voltage? AI: If you're scavenging parts, you could consider using two similar transformers such as 120:12 and then 12:120. Or follow the suggestion of The Photon and search for a new in-stock part- there's a 12VA "Signal" (the manufacturer) part for about USD 11. in one-off. It has a nice split-bobbin construction.
H: Is plugging a wrist strap in enough to prevent ESD? I've been told by a few people that if you plug in a PC and switch off its power supply that it is enough to prevent ESD. But I was taught differently. I was taught that it's more about difference in potential rather than grounding. The way I was taught was that everything must be unplugged, placed on a matt with a grounding strap for the matt and the installer. If a person follows method 1 will it still cause ESD damage? AI: I've been told by a few people that if you plug in a PC and switch off its power supply that it is enough to prevent ESD. No, it's not enough to switch off the supply. For most PC's the working voltage has a maximum of 12V, using the Human Body Model (HBM) for Electro Static Discharges (ESD) ranges from 2kV to 8kV, which is hundreds of times more voltage. Whether a circuit is powered on or off will in most cases have very little effect on where the current from an ESD event flows. The point of preventing ESD is not to 'float' a circuit and remove it from ground (which doesn't happen when a PC is turned off, it should always be grounded when proper electrical codes are followed), it is to stop the charge from accumulating in the first place. There are transistors (mosfets) that you can destroy simply by waving your hand over them. most of them are older, and many newer electrical components have ESD protection in their inputs, but that gives you an idea of how easy it is to kill electronic parts with static electric fields. The way to prevent ESD is to strap in, with a 1MΩ resistor on the strap to prevent electrocution. Other ways are, wear an ESD smock, use an ESD mat and to keep humidity high-ish (like 60%) as water vapor increases charge transport through air. Another thing to keep in mind is that some materials like plastic and paper and clothing readily create\buildup charge so only use ESD compatible materials near sensitive parts.
H: LM324 output voltage gets low in difference amplifier configuration when UPS supply is turned on I am using lm324 as difference amplifier to measure the string voltages of lithium ion battery pack ( four cells in series). I am using all single package of LM324 that have 4 Op-amp. The Lithium ion battery pack is being charged by the UPS ( uninterruptible Power Supply). The UPS voltage have some noise and you can see in the attached picture. The supply voltage of LM324 is also connected in parallel with UPS and Battery terminal. The problem is that , whenever the ups start charging battery, the output voltage of LM324 is disturbed. Without charging of UPS, the voltage of each cell in the string is 3.3V as nominal. But when the UPS start charging, the same voltage becomes low as 1.6V. We have connected decoupling capacitor with LM324 but it didn't work. Can you please recommend any suggestion. UPS voltage difference amplifier circuit configuration. Complete circuit of LM324 as difference amplifier. Output Voltage of LM324 with Oscilloscope. AI: The tolerance of resistor is 10% If that is the case then THAT is a fundamental problem in your design. On the left (below) is a model of your op-amp circuit using perfectly exact values. On the right I've increased some resistors by 10% and decreased others by 10% to make a worst case tolerance scenario. The good circuit produces 3.3 volts at the output and the bad circuit produces 1.1 volts at the output. Just in case you are wondering about "my op-amp", I've used a VCVS with a gain of 100k to mimic an op-amp.
H: Capacitor on sensor output purpose I'm currently using a demo-board from sparkfun for quick prototyping of a hall-effect sensor application. The specific chip I'm using is the ACS723-5AB, a Hall-Effect sensor that can measure between -5A and +5A. Now that I'm done using the demo-board from sparkfun I'm looking to just directly use the sensor chip in my final circuit design. One of the improvements I wanted to make was to add a low-pass filter on the sensor output for smoothing the signal a bit by adding a simple RC filter, but when I look at the circuit-design of the SparkFun board, they don't use an RC filter but just a capacitor to ground on the output line. I can't imagine anything drawing current from the sensor output, so decoupling doesn't seem likely. It doesn't really create a filter without a resistor (as far as I know?) My question is: What is the purpose of capacitor C2 on Viout? link to circuit diagram of daughterboard Link to sensor datasheet AI: It doesn't really create a filter without a resistor (as far as I know?) It probably does because it's likely that the output amplifier will have some output resistance that isn't trivially small. The data sheet tell us two things here: - Output Resistive Load from VIOUT to GND is 4.7 kΩ minimum Output Capacitance Load from VIOUT to GND is 10 nF maximum This fairly convinces me that the output impedance is not insignificant else why have such a high minimum load AND why have such a high limit for output capacitance; most op-amps would probably go unstable with a load capacitance as high as 10 nF so, I expect that there is an internal output resistor. What is the purpose of capacitor C2 on Viout? An extra means of reducing high frequency output noise.
H: What are these wiry protrusions on a TV antenna? This photo shows part of a large antenna assembly which has a series of thin wire protrusions: (Source) This antenna is apparently used for TV transmission. What are those protrusions for? I'm guessing something to do with lightning or static? The photo is from near the antenna's base, it was hard to tell how far up they go. (Interesting video, BTW as the giant antenna has to be replaced by helicopter at the top of a 1500 ft tower) AI: Looks like they're what's commonly referred to as lightning rods, though their purpose is to provide a means to dissipate and equalize the charges that may build up between clouds and the earth and thus prevent a lightning strike from occuring. Those little bristles provide for the charges to leak off. Tiny points at the end of a piece of metal is an effective way to to this. Notice how most "lightning rods" are pointed at the tip? Another key is that the cable those bristles come off of are clamped to what looks like a big grounding cable, that runs down the mast to earth ground.
H: Finding 70% Voltage Point Let \$v_1\$ be the input AC voltage and \$v_2\$ be the output AC voltage such that gain \$\dfrac{v_2}{v_1} = \dfrac{A+jB}{D+jE}\$ where \$j = \sqrt{-1}\$. How should I find the 70% point (the point where the gain is reduced by a factor of \$\sqrt{2}\$)? I was told that I must set \$D = E\$. In that case, \$\dfrac{v_2}{v_1} = \dfrac{A+jB}{D(1+j)}\$. The explaination I was given was: since \$\|1+j\| = \sqrt{2}\$, the ratio is decreased by a factor of \$\sqrt{2}\$. But what I don't understand about this is that, initially when \$\dfrac{v_2}{v_1} = \dfrac{A+jB}{D+jE}\$, the magnitude is \$\dfrac{v_2}{v_1} = \dfrac{\sqrt{A^2+B^2}}{\sqrt{D^2+E^2}}\$. When \$\dfrac{v_2}{v_1} = \dfrac{A+jB}{D(1+j)}\$, the magnitude is \$\dfrac{v_2}{v_1} = \dfrac{\sqrt{A^2+B^2}}{D\sqrt{2}}\$. But $$\dfrac{\sqrt{A^2+B^2}}{\sqrt{D^2+E^2}} / \dfrac{\sqrt{A^2+B^2}}{D\sqrt{2}} = \dfrac{D\sqrt{2}}{\sqrt{D^2+E^2}} \neq \sqrt{2}$$ Clearly the gain has not decreased by a factor of \$\sqrt{2}\$. Is there something that I'm missing? AI: IN GENERAL the whole approach of setting \$D=E\$ is incorrect. It fails e.g. in the quite obvious example of \$A=D\$ and \$B=E\$. $$$$ The approach is only valid when D is constant and E is variable (or vice versa) the variable can become very small or very big with respect to constant $$$$ Let D be the constant and E the variable (think of E being frequency dependend). Let E be originally very small The magnitude of \$\dfrac{v_2}{v_1}\$ is $$\dfrac{\sqrt{A^2+B^2}}{\sqrt{D^2+E^2}}$$ When \$E \ll D\$ the magnitude becomes $$\dfrac{\sqrt{A^2+B^2}}{D}$$ Then, E increases such that \$E = D\$ the magnitude becomes $$\dfrac{\sqrt{A^2+B^2}}{D\sqrt{2}}$$ Next, you should divide the new magnitude by the original magnitude (not the other way around as in the OP!!), so $$\dfrac{NEW} {ORIGINAL} = \dfrac{ \dfrac{ \sqrt{A^2+B^2} }{ \sqrt{ D^2+{(E\|_{E = D})}^2} } }{ \dfrac{ \sqrt{A^2+B^2}}{\sqrt{D^2+{(E\|_{E \ll D})}^2}} } = \dfrac{ \dfrac{ \sqrt{A^2+B^2} }{D\sqrt{2}} }{ \dfrac{ \sqrt{A^2+B^2}}{D} } = \dfrac{1}{\sqrt{2}}$$ conclusion write out the transfer function for finding useful point (for e.g. bode diagram) find for both the numerator as well as the denominator: the frequency for which the real part is significant larger than the imaginary part the frequency for which the real part is equal to the imaginary part the frequency for which the real part is significant smaller than the imaginary part e imaginary part
H: Why are capacitors used parallel to transistors In the circuit attached. There are two capacitors. One is parallel to "PWR LED". Although I understand it delays the turning on and off of the LED, why would you need that? Also, the other capacitor is connected in parallel to the phototransistor. This connection I completely fail to understand the purpose of. This circuit diagram is of a flame sensor, which generates an output when the phototransistor detects IR. AI: The capacitor that is parallel to the photo-transistor is used to extend the time the DO_LED is on after the flame has disappeared or momentarily ceased. The recharging of that capacitor (100 nF) is via the 10 kohm resistor hence the CR time is 1 millisecond. The only other capacitor is across the power rails and this is a requirement for most op-amps to ensure stability and/or correct operation.
H: I2C bus switches/multiplexer vs analog switch/multiplexer I'm refactoring a board that needs to switch the communication between an I2C master (M1) and two identical I2C slaves (S1 & S2 with same address). The first design of this board was using a TCA9545A. This is a nice IC but, like many other I2C switches, it is controlled through the I2C bus. In my application this is an unwanted feature because the only I2C master I can control is not M1 and if I connect it to the same I2C bus, I'm going to create a multi-master topology that's not accepted from S1 and S2 devices. My thought was to replace the TCA9545A with another I2C switch, which is NOT controlled through I2C (i.e. through a digital controller). But I have to admit that it's not an easy task, because after having searched in all catalogs of main IC companies, I haven't found a single component matching such features. The other option was to use a standard bidirectional switch/mux like ADG1636 or similar. Is there another option? Is this a good replacement to fulfill my requirements? --- Schematics update ---- AI: I²C switches are called "I²C switches" because they can be controlled through I²C. The I²C signals themselves are just plain digital signals, which are analog signals. So for switching I²C signals controlled with GPIOs, two-channel SPDT analog switches are the correct choice, and commonly used (because most of them are quite cheap (unlike the ADG1636)).
H: Is there any device for measuring phase and gain of a device in frequency range as low as 0.1 Hz? I need to measure phase and gain of a circuit starting from 0.1 Hz to 10 MHz. Most of VNAs are for frequencies of several MHz to GHz. I've just found one VNA for 1 Hz. Is there a device or method to measure even lower frequencies? AI: If you can get a hold of (rent or buy used) an HP/Agilent 3562A dynamic signal analyzer, it can handle the job from 64uHz to 100kHz. You can use a conventional frequency response analyzer for the higher frequency range. I got one years ago for $400 on eBay, but there are ones in better shape easily available (for more $$): 3562A Of course your measurement time for sub 1Hz testing is going to be long if you need a lot of points.
H: Can the 1D wave equation be recreated in a conductive wire? I am ignorant of EE, but not the wave equation, so bear with me. The (scalar) wave equation \$d_{tt} - c^2d_{xx} = 0\$ describes the displacement (and velocity) of some medium through space and time. It works for air, EM fields, and many physical quantities. I am interested in recreating transport phenomena modeled by the wave equation within a conducting wire, so the voltage (or perhaps something else) at any point of the wire would be described by the wave equation. My main interest in this question derives from the difficulty of experimentally (rather than theoretically or computationally) modeling more complex wave equations \$d_{tt} - c(x,t)^2d_{xx} = 0\$ with variable wave speed. If there were a straightforward way of recreating variable coefficient wave equations using EE, it would be helpful to me! AI: Certainly To quote wikipedia; The telegrapher's equations (or just telegraph equations) are a pair of coupled, linear partial differential equations that describe the voltage and current on an electrical transmission line with distance and time. The equations come from Oliver Heaviside who developed the transmission line model in the 1880s. The model demonstrates that the electromagnetic waves can be reflected on the wire, and that wave patterns can form along the line. The theory applies to transmission lines of all frequencies including direct current and high-frequency. Originally developed to describe telegraph wires, the theory can also be applied to radio frequency conductors, audio frequency (such as telephone lines), low frequency (such as power lines), and pulses of direct current. It can also be used to electrically model wire radio antennas as truncated single-conductor transmission lines. To clarify, the "telegrapher's equations" are a set of equations that give you the voltage and/or current, present at any length along a transmission line and at any point in time. A transmission line can be anything which propagates voltages/ conducts current along its length, a close to ideal transmission line would be a good quality coax cable, but even a loose bundle of wires can in theory be modeled as a transmission line. The telegrapher's equations used to model transmission lines are not wave equations, but they can be rewritten to be of the form of wave equations, here is a link I found which explains how; https://www.google.com/url?sa=t&source=web&rct=j&url=http://www.ittc.ku.edu/~jstiles/723/handouts/The_Transmission_Line_Wave_Equation.pdf&ved=2ahUKEwj4kc_W9_bmAhXlzcQBHdSCC5cQFjAHegQIBBAC&usg=AOvVaw1VnVhK-_iH9CWsDbR7WLeA&cshid=1578588439100 I Hope this helps. As a side note; thinking of electricity as 1d (scalar) waves on a wire is actually the correct physical way of thinking of electricity, but rather than just having one wave with one amplitude we have two perpendicular waves with two different amplitude, the voltage (or electric field wave) and the current (or magnetic field wave), the proportion of the electric wave to the magnetic wave is what we call the characteristic impedance, in units of ohms (voltage over current). Another side note; the thermal noise generated by a "Hot" resistor is actually the 1d equivalent of the Black body radiation generated by a three dimensional "Hot" object
H: What is a proper way to send information from an HC-05 (Bluetooth module) to a phone? I have 12 simple touch sensors and each of them produces an analog output. I want to send that information to a phone (or some other device). To do that I would need to use the HC-05 Bluetooth module. I don't want to use an Arduino. I looked up some alternatives and I can program an ATtiny85 chip (using the Arduino) to somehow function with the HC-05 and from there send the information to the phone. I don't know if that's the right way to do it. Also, I have 12 analog states that I want to send and only 6 ATtiny85 pins. The project is small so I don't want to use any big components or chips. AI: The usual way of doing this is to use a microcontroller to interpret the inputs and then send your information to the HC-05 module via UART. You need to make the communication protocol yourself, which is probably a good thing. Since you want 12 analog inputs you might be facing some difficulty finding a small microcontroller with that amount. My suggestion for this task would be to use an analog to digital converter chip (ADC) for the inputs and send this to the microcontroller, which interprets the information and codes it according to your communication protocol and then sends it to the HC-05. An 8-channel ADC can be seen below. Such an ADC connected to the ATtiny85 communicates with SPI or I2C, which are bus protocols needing few wires. Alternatively, you can use something like a Teensy 3.6 as it already has enough analog inputs. Of course it's bigger and way more powerful than what you need for your purpose. This will be up to you and how much convenience you want in your project. Teensy 3.6 pictured below: If you want real good convenience Adafruit designs a series called Feathers. These are Arduino compatible microcontrollers with built in Bluetooth functionality that are easy to use. The Feather M0 features 10 ADC channels, which is almost what you want. By searching their range of products, you might be lucky to find a 12-channel one or you could simply use the Feather M0 and a small 2-channel ADC so you have 12 channels. https://www.adafruit.com/product/2995
H: What type of capacitor is most stable for use in 555 timer circuit? I am building a 555 timer circuit with a frequency of 15, 30 and 60 Hertz. I will most likely use a 1µF capacitor but I was wondering what kind of capacitor would be best for keeping a stable value even when the temperature changes (goes down most likely)? I will be putting potentiometers so that if the values do change they can be calibrated again, but that will only be done before we deploy the systems. Once deployed, we cannot calibrate again until retrieved days later). Basically, I am asking what kind (ceramic, electrolytic, etc.) of capacitor will keep its capacitance constant at a value of 1µF and with varying temperatures. AI: As DKNGuyen said, if you want a stable capacitor, use C0G/NP0 ceramic. However, if your actual goal is to have a stable frequency rather than discuss capacitors, then your original idea of using a 555 timer is not the way to go, as this chip will have worse drift than the C0G capacitor. A much better option would be to use a quartz oscillator, for example a 2.4576 MHz oscillator which will cost about €1 in 50ppm/°C stability, then divide by 4096 using a 74HC4040 ripple counter. You could use a 74HC4060 too, with a crystal instead of an oscillator. Also the quartz oscillator is pre-calibrated and you don't need to adjust it. EDIT I misplaced a decimal point... I mean, a 1.2288 MHz oscillator followed by 74HC4040 to divide by 4096, resulting in 300Hz. Then a 74HC390 or similar which can divide by 5 then 2 then 2 which gives 60Hz, 30Hz, 15Hz. Or a 192kHz oscillator then the same two 74HC chips, but this oscillator is only available on digikey in tiny MEMS flip chip so maybe not the best option. Anyway you get the idea, pick a frequency and a convenient division ratio with 74HC chips... The BOM cost for both solutions will be less than a precision capacitor and a potentiometer, and that doesn't count the salary of the intern who gets to tweak the frequency just right... If you need better stability than what a cheap XO will provide, you can use a TCXO. That will cost a bit more (a few €) and it will be available in less convenient frequencies. So you can use, for example, a 12.288MHz TCXO and divide by 60000, you'll get a few ppm stability over temperature. Note a cheap microcontroller makes a nice programmable divider if you need one. If you already have a microcontroller in your project, why not use that?
H: Controlling a 0-20mA transmitter from an isolated PWM signal I'm trying to generate a 0-20mA signal for controlling an industrial actuator (the position of a 3-way valve) from a Teensy 3.2 system. Preferably, the mA-output should be galvanically isolated from the microcontroller to protect it properly. In order accomplish that, I currently have a setup, where I am sending a PWM signal (at ~488Hz) from the teensy through a 4N26M optotransistor (4N26M datasheet). The output of the 4N26M is sent through a RC filter into a TI XTR117 Current Loop Transmitter (XTR117 datasheet). This chip controls the current in a separate current loop sourced from a 24V PSU coupled in series with the actuator, which is controlled with the current signal. I measured an internal resistance of 100 ohms in the actuator. According to the datasheet, the XTR117 generates an output current which is a factor 100 higher, than the input current going into (pin 2) I_in. Therefore, the input on pin 2 needs to be between 0 mA and 20mA/100 = 0.2 mA to provide an output of 0-20mA. The XTR117 features a regulated 5V line (V_reg) on pin 8 relative to pin 3 (I_ret). A pull-up to this line ensures a 5V output of the optocoupler, when it is in the ON mode. The collector to emitter saturation voltage of the optotransistor is 0.5V according to the datasheet. When the optotransistor is "open", a voltage of 4.5V at pin5 of the 4N26M is expected. This means, that a current through R3 and R4 in the RC filter at 4.5V/22.5kΩ = 0.2mA is expected, which means an output of 1000.2mA = 20mA in the main current loop. See the schematic below: However, there are two problems in the circuit: When the PWM duty cycle is 100% (LED on in optocoupler) an output of approx. 2mA is generated, and not 0. I suspect that this is due to the saturation voltage of 0.5V in the optocoupler. At 0.5V, the current into XTR117 pin3 should be 0.5V/22.5kΩ = 0.02mA which should correspond to 1000.02 ≈ 2mA on the output side. However, I can't think of a good design to avoid this scenario. What can be done? When the duty cycle is 0% (LED off), a measured output of 19.9mA is generated, which is acceptable in this scenario. However, the signal seems to be unstable, as the controlled valve is turning a bit back and forth all the time. But when I touch with a finger at the joint between R3 and R4, the problem is not visible (valve position is stable). This effect can be replicated every time. So what is the effect of touching it, and how can I implement that in the circuit? I hope I'm making sense! Also, if you suggestions for completely different components, which can generate the desired signal, they are more than welcome! AI: Add a CMOS logic gate after the optocoupler/R2. That will buffer the output and the voltage will be more accurate and the time constant will be the same for on vs. off. split R4+R3 approximately evenly (11K-ish) and put the capacitor to IRET in the middle. You're not really filtering the current input well by putting the capacitor where it is. The time constant will be the parallel combination of the two times the capacitance, so about 5500 ohms * 1uF or about 5 or 6ms. Adjust capacitance as required. I'm guessing 47uF or 100uF might be a better choice for a mechanical valve actuator. You probably want the ripple to be less than the hysteresis in the actuator control loop. Consider replacing the optocoupler with a logic-output type (eg. 6N137) which will be much faster (so a more true representation of the PWM on the isolated side), but retain the above two suggestions (unless the optocoupler has a CMOS output- the cheaper older ones tend to be bipolar). If you use the logic output optoisolator, you might want to consider upping the PWM frequency to a few kHz to reduce the size of the filter capacitor so a ceramic part can be used (and it will respond a bit faster).
H: What kind of plug does this machine take? I was in my attic and I found what seems to be a light projector, it appears to have no power as the on/off switch won't work, so I think it may be out of battery, the only thing I found related to battery was that plug on that back, I searched all around my house but didn't find any plug that would work with it. Could you guys help me find out what plug it is so I can purchase one at my local store? (European by the way, I don't know if that helps anything) Images of the projector: After brute forcing the fan open, this is how it looks like: AI: The barrel jack mates with a wire from a "wall wart" wall plug adapter most likely, and most likely it supplies DC voltage. If you can read the voltage on the cooling fan label it's a fair guess that the input voltage matches that. Probably something like 12VDC. Center pin is usually positive, but again no guarantees (wrong polarity or voltage will or can damage it). You can open it up and trace the connections- if the fan goes directly through a switch to the input power that will give you the voltage and polarity. Current requirement is a complete guess but it could be measured. To spec a replacement wall wart you need the connector, polarity, voltage, and current requirement (at a minimum).
H: Current divider usage I was solving this exercise where I have to find the current through each of the circuit branches. Sorry if the image is a bit confusing but after determining the current I I used the current divider formula to find the current in each of the branches. Apparently I got the wrong answer but I can't understand what I have done wrong. (Original) AI: Here's your schematic: simulate this circuit – Schematic created using CircuitLab (By the way, if you plan to use this site again you should learn to use the included schematic editor. It automatically numbers the parts for you and that provides us with an easier, more specific way to address ourselves to your schematic and questions. It's not perfect. As you can see above, it insists on adding "F" and "H" where it's not wanted. But we can all live with these warts.) For more explanation about why I laid this out as I did, see Note below. I completely agree with your approach in finding the total \$Z\$ and the total \$I\$ for the circuit. You have \$R_1\$ in series with \$L_1\$ in series with the parallel impedance of \$R_2\$ and \$C_1+L_2\$. To compute \$I_{_\text{TOTAL}}=\frac{V_{_\text{S}}}{Z_{_\text{TOTAL}}}\$, you need to find \$Z_{_\text{TOTAL}}\$: $$\begin{align} Z_{_\text{TOTAL}}&=Z_1+\left(R_2 \mid\mid Z_2\right)\\&=\left(4\:\Omega+j\,20\:\Omega\right) + \left(16\:\Omega \mid\mid \left[-j\,14\:\Omega+25\,j\:\Omega\right]\right)\\ &=9.13527851 + j\,27.469496\Leftrightarrow 28.9486878 \:\angle\: 71.6048952^\circ\\&\therefore\\I_{_\text{TOTAL}} &=0.130811304 - j\,0.393345488\Leftrightarrow 0.414526561 \:\angle\: -71.6048952^\circ \end{align}$$ You found similar results. So that's good. I think where you went wrong (and this is offered with respect, because I think you were actually pretty smart here -- just that you got it backwards) is that you constructed two ratios, \$\frac{Z_2}{R_2\mid\mid Z_2}\$ and \$\frac{R_2}{R_2\mid\mid Z_2}\$, as a means of dividing up the total current in the circuit. But what I think you forgot here (the thing you got backwards, so to speak) is that you really wanted the ratio of each branch admittance to the total admittance. This is why I think your brain is working really well. It's just that you forgot to realize to first convert the impedences to admittances. So you were very close. Let's see what happens when we apply your concept but instead use the correct objects, namely the admittances: $$\begin{align} i_1&=I_{_\text{TOTAL}}\cdot \frac{\frac1{R_2}}{\frac1{R_2\mid\mid Z_2}}=I_{_\text{TOTAL}}\cdot \frac{R_2\mid\mid Z_2}{R_2}\\\\&=0.225615314 - j\,0.0651777575\Leftrightarrow 0.234841245 \:\angle\: -16.1134182^\circ\\\\ i_2&=I_{_\text{TOTAL}}\cdot \frac{\frac1{Z_2}}{\frac1{R_2\mid\mid Z_2}}=I_{_\text{TOTAL}}\cdot \frac{R_2\mid\mid Z_2}{Z_2}\\\\&=-0.0948040109 - j\,0.32816773\Leftrightarrow 0.341587265 \:\angle\: -106.113418^\circ \end{align}$$ And those are the right answers. Your basic idea was on the right track. You just got mixed up about the objects upon which your idea applies. We've all made such mistakes and I think you are going to do just fine on this stuff. And this kind of question you wrote is perhaps the classic kind of student question we should see presented here. You had a clear description of the problem you were working on and you showed all of your work. In addition, you showed enough about how you applied your thinking so that we could work out where you went wrong. If every student question were like this, I believe this site would find far more joy and readiness in helping answer them. So +1 for that! Note I've arranged things so that the divider in a way that eliminates having to add the source voltage. It's not necessary to see it or to see the wires busing power between it and the parts; and those wires just get in the way of seeing the remaining important bits. Also, I've arranged things so that we can use a conventional schematic drafting sheet of paper where currents flow from top to bottom (yes, I know this is AC) and where signal flows from left to right. (In this case, the schematic doesn't have a signal to worry about.) If you consistently draft your schematics to follow rules like this, you'll find them to be more understandable. I like to think of it like a curtain or sheet where currents flowing from top to bottom are like the flow of a river downstream. Signals then propagate across that river from one side to the other side. These signals require that flow in order to operate, so to speak. It's a picturesque way of seeing things and it pretty much always works better than other alternatives. But there are times, such as with some power supply schematics, where the busing around of power is the important thing to show and highlight to make the schematic better understood. So it's not a hard and fast law. The main point is to have the schematic communicate with others and to do that you need to emphasize the parts and sections to which attention must be drawn and to avoid wasting "ink" on stuff that doesn't matter.
H: Comparator problems with output I am kinda new to electronics design and I am having some trouble with the Analog Devices LT319AN chip. I am trying to do simple 250KHz PWM which worked on a lesser LM393 chip in this exact setup. I know I am probably missing something simple due to the lack of some basic knowledge, any help would be much appreciated! Anyway I have added a schematic diagram and oscilloscope pictures so you can see what I am dealing with. Edit: Forgot the Schematic! Edit #2: Thank you @DKNguyen! Per your input I added 10k OHM resistors between +5vDC and ground to bias the input's and this is the output am now getting: https://youtu.be/hdRFULSaPc0 Thank you for your help! AI: Add ceramic decoupling capacitors across the power pins of the comparator. Input bias currents need a DC path to flow. You can't just leave them DC floating like that. You can normally handle this by sticking resistors from each input to gnd. But in your case, you should stick each input in the middle of a resistor divider instead to DC bias in the middle of the input common mode range the because your input is bipolar but the comparator is not (it only has a unipolar supply). But this should have been an issue with the LM393 too. simulate this circuit – Schematic created using CircuitLab
H: Can I send an SSH command via Nucleo Board? I'm in a very embryonal phase of my project, so I'm just putting the pillar on possible solutions. I'd like to have an opinion or suggestion by experts about this question: For the above-mentioned project, I should send an SSH-based command to a final device. Note: the final device can communicate only via ssh. Although send an SSH command can be simple by using a computer (e.g., by using pink or PuTTy), is there a way to do the same via the Nucleo L476RG board? Ultimately, can I open an SSH connection by using a Nucleo board and send commands? Note: The project could admit the purchase of other hardware components if necessary (e.g. wifi modules). AI: Yes. You theoretically can. All you need is for that Nucleo board to have some sort of network connectivity an IP stack, a TCP layer that uses that network hardware an SSH program that uses the above TCP/IP stack to connect to your server. That's per se possible, but the way you asked indicates you might not be experienced enough to pull off the integration/software porting in reasonable time. Also, there's other crypto protocols than SSH that are already available for such microcontrollers as yours – there's compact TLS implementations, for example. Building up an SSH connection for well-defined machine-to-machine comms also doesn't sound so effective. So, honestly, hardware is hard and software usually is easier, so change the whatever runs your SSH daemon to accept commands e.g. via a TLS endpoint, and you're removing a lot of the complexity. You'd still need to add networking hardware to your microcontroller, and make your TCP/IP stack talk to that, and you'd still have to learn how to use a TLS library in your microcontroller board, so that's really a lot of effort... The project could admit the purchase of other hardware components if necessary (e.g. wifi modules). Generally, this sounds like a job far better suited for a thing that comes with support for "fully fledged" operating systems, like a Raspberry Pi Zero, for example. There, you'll have no problem to just get an SSH client to do what you want. So, consider that the hardware component you'd want to buy. It can possibly either replace the Nucleo board, or run alongside it, maybe even powered off, until communication becomes necessary.
H: High output of NCP5104 (half bridge driver) is active in dead time zone. How can I solve it? I try to drive BLDC motor. I have a problem in gate driver circuit. I use three NCP5104 gate driver. NCP5104 has 520ns internally fixed dead time. "The dead time is the time between the high side and low side output signal to avoid any cross conduction."@Delphesk But in my circuit, high output of NCP5104 (half bridge driver) is active in dead time zone. How can I solve it ? AI: But in my circuit, high output of NCP5104 (half bridge driver) is active in dead time zone. How can I solve it ? Look at "PHASE" voltage - it also goes high by the amount that "Driver high out" goes high. This means that MOSFET Q18 has the same voltage on the gate as the source hence, MOSFET Q18 is actually off (despite your protestations). What you are probably seeing is an effect of the back emf from your BLDC motor. It certainly doesn't mean that Q18 is "on" during the dead-band period. The bootstrapping in the driver chip is ensuring that \$V_{GS}\$ is zero in that dead-band period.
H: Input bias current existing even when no input is given While I was studying the DC characteristics of op-amps I came across input bias current. I understood the concept that it draws some current when input is applied but in proof they put both the voltages at ground and they were still considering input bias current. I'm a little confused that even when there is no voltage at input from where the op-amp is drawing current? AI: but in proof they put both the voltages at ground and they were still considering input bias current.I'm a little confused that even when there is no voltage at input from where the op-amp is drawing current? Consider the LM324's equivalent circuit: - If either input is grounded or taken to the negative rail then current will flow from the base of Q1 and Q4 to the tune of about 45 nA.
H: Can we ground all the nets in a printed circuit board? I am a newbie in electronics and I am designing a printed circuit board. The IC I am using has some pins that I don't use in my circuit. Is it okay to ground those nets or leave them as they are? I am asking this because I have heard that sometimes these nets can act as floating pins and could cause noise and errors. I am not sure whether this is true or not so please clarify my doubt and suggest me the right way. AI: If you check the HW Design Manual then you'll see that it suggests the unused pins should be left floating. Please refer to Table 4 on p.15 and p.16.
H: Is there a reason to use 4 diodes instead of 2 when controlling a motor with L293B? I am looking at the datasheet of L293B and there are some usage examples provided in it: Is there a reason to use 4 diodes instead of just 2? Why are these connections needed here in the first place? What would be wrong with just connecting Output 1 and Output 2 alone to the motor? AI: What would be wrong with just connecting Output 1 and Output 2 alone to the motor? In your 2 diode diagram you have the motor completely shorted out: - 4 diodes are needed when you have a full bridge driver to ensure motor back emfs adequately push excess motor energy (due to inductance) back to the supply rails. You also need a reasonably good reservoir capacitor on those supply rails too.
H: How to bias an AC signal correctly? I have an AC signal and want to give it a new DC offset. There are some questions about my experiment results. (The original signal bias and terminal output bias are the same because I want to know if there is the attenuation or phase shift or not.) (yellow-TP,blue-V2) Can I apply these two circuits to provide DC bias? Why are the waveforms changed in each circuit with different capacitors? Similar question, but this time is with different resistor. AI: You're creating a high pass filter between your capacitor and whatever resistance you're connecting to Vref with. In the first circuit, I can't see the impedance to Vref, but it's apparently not zero. In the second circuit, it's the output impedance of U_2; again, it's unknown to me. I might be able to estimate them if I knew the frequency of your test signal. In the 3rd circuit, the 10K with the 0.1uF gives you a corner of about 159Hz; using the 330K brings the corner down to about 5Hz. This leads me to suspect your signal is in the 50-80Hz range. The 4th circuit pits your 0.1uF against a 50K load, giving you a corner of about 32Hz. Changing to 680K resistors brings the load up to 340K (they're effectively in parallel from a loading standpoint) and again close to 5Hz corner. You need to pick your R and C so that the lowest frequency of interest is higher than the 3dB point of the filter; you might want to make it about 3-5 times higher if you're trying to minimize distortion.
H: Identifying strange amplifier types of amplifiers in a 1979 CB radio I'm trying to fix an old Midland 3001 CB radio and I came across a few strange types of amplifiers in the receive stage of the radio. I was wondering if anyone could identify the type of amplifier used. In the screenshot the direct input from the antenna is coupled through C101 on the left I believe. Then there are some diodes which I assume are used to protect the receive stage. The L101 is a variable transformer that couples the signal to the transistor Q101. What kind of amplifier circuit is the Q101 transistor involved in (is it common emitter, common collector, or something else entirely)? What is L102 doing at the output (I believe it is a tunable LC filter)? I have also attached the full schematic diagram for the radio. AI: D101 along with D105 do protect the receiver input from huge transient events, as OP surmises. They also limit amplitude of the transmitted signal going into the receiver. But D103,D104 are DC-biased with current from the RF GAIN control, to attenuate merely large input signals while receiving. Common-emitter RF amplifier transistor Q101 has resonators (27 MHz) at both input (L101) and output (L102). C101 (33pf) resonates with L101 @ 27 MHz with fairly low-Q. A small-value neutralizing capacitor (C104) keeps this amplifier from oscillating. It compensates for collector-to-base capacitance inside the transistor. L102 is a tuned circuit (27 MHz), part of a double-tuned resonator along with L108, top-coupled with a small coupling capacitor (C107, 3pf). This double-tuned resonator has a bandwidth wide enough to pass the whole CB band on to the down-converting mixer. A noise blanker (NB) module (possibly an optional addition?) also takes its input from this double-tuned resonator.
H: Switch to turn off motor and another to turn the motor on I have a DC motor that turns a gear which then has an arm that will hit a switch which turns the motor off. Is there a way for me to build this without using a microcontroller? Like I start the motor with switch1 then the motor hits switch2 which then turns the motor off, so to activate it again I need to press switch1. AI: A possibly simpler method is as follows: Set the motor up with a wheel which engages a microswitch (or similar) Wire the microswitch as SW1, the starting switch as SW2: With SW2 unpressed, the motor will go until the tab engages the microswitch, breaking the normally-closed side. It stays there until you press the start switch SW2 for long enough to move the wheel off the microswitch. I once made a doorbell like this, out of a CD-ROM motor and a music-box mechanism. It was arranged to play the whole of the Pink Panther tune. All was fine until the tab broke and it played for about an hour: everybody was hoping their patience lasted longer than the battery. The tab was glued onto the wheel, marked 2 in Wikipedia image:
H: Amplify AC signal only I amplify a signal which including AC and DC.Why is the output waveform changed when C1 is 0.1u?And how to decide the value of C1? AI: When Capacitor Reactive Impedance, Xc(f) rises to affect impedance ratio with R such that they are equal, you can measure many responses; the voltage amplitude divider ratio (gain) reduces times \$\dfrac{1}{\sqrt(1+1)} = 0.707 = -3~dB\$ the current phase shift changes by 45 deg, (=trig. angle with equal sides of impedance) which as already started to shift with 0.1uF that raises the impedance of Xc(f) to 26.5 kohm the -3dB breakpoint is \$ω_1=1/(R_1C_1)=2\pi f\$ you can visualize all of this with a log impedance RLC nomograph and measure the impedance ratio of any XL(f) or C(f) vs V as the division ratio becomes a difference on a log scale. like 1000 is a difference of 3 decades R+ C Series Impedance = \$Zc(f)=R+jXc(f)\$ Also with this you can "ballpark" estimate corner frequencies and LC resonant circuits and Q gain factors for LC intersections with R differences (Ratios). e.g. L//C//R is high impedance so $$Q_p=R/X_L(f)=R/X_C(f)$$ at LC value intersection (=resonant f) then series resonant f (L+C+R) $$Q_s=\frac{X_L(f)}{R}=\frac{X_C(f)}{R}$$ (using absolute values or ignoring phase due to j Here is one example of the RLC Impedance nomograph.
H: Which parameter to use to specify the relay in a CAN line communication? I need to use SSRs (solid state relays) to comute lines of comunication to a vehicular OBD-II interface. I will use SSRs to choose wich lines will be used at the pins of the connector. The lines wil not be commuted during comunication. The signals I need to comute are CAN. The CAN transceiver is the SNx5HVD251, so, In order to specify the SSR electrically, I tried to very the maximum current the SNx5HVD251 supports. In the datasheet says, in the operational conditions, that the driver output current, IOH is at most 50mA or -50mA. But There is another parameter, Short-circuit steady-state output current that, accordingly with datasheet, is 200mA. So, this is my doubt, wich parameter to use ? AI: There are several parameters of interest; Impedance of transmission lines and terminators Signal currents Short cct. currents to +12, -7V Internal Switch Ron or RdsOn Off switch capacitance. The currents usually follow Ohm's Law if you have an accurate model for Ron and drive levels for PFET and NFET to short voltage for Hi and Lo. The CANBUS is already a 30 port MUX with stubs on a bus terminated by 120 Ohms. But you may want to consider all options such as FETs and miniature Reed Relays to MUX to many busses. Start with good Specs.
H: Positive and negative rail for op amp circuit I need a positive and negative rail to power an op amp Since it's an op amp for a voltage reference, would using a circuit like this one be enough? Input voltage is unregulated 37 volts, with 4.7k resistors; current is about 7.8 milliAmps Since the op-amp is needed as the error amp for variable output voltage, the op-amps supply voltage will be unregulated. Are there any pitfalls that I should be aware about? Any other topology? AI: The biggest potential issue is that your negative supply is only negative with respect to that wonky quasi-split-supply point and usually that's not what you want in a DC-coupled circuit like a regulator that needs to provide an output and accept an input relative to ground. It might be okay enough for an AC-coupled (both input and output) amplifier of some kind.
H: What stepper motor is this? I am so sorry, I have to ask such a stupid question. I got these stepper motors from a thrift shop, the owner had no idea what these are. And these have all labels damaged. Can anyone help me figure out the name, model, or wire codes of this. Thanks in advance It has pink, yellow, green, and blue wires. Here is its image: https://ibb.co/HDm7BKj AI: (1) OP Label seems to say ... C720-02 (2) Hmmm.... Google... C720-02 stepper motor ... (3) Profit! (as they say on Slashdot). I have seen 95 Euros and 250 dollars US on different websites. After step (2), I took another look at the OP's label, and I'm pretty sure I can see the 23LM at the beginning now. Anyone else agree? Informative brochure: Also... They all seem to have a standard wiring scheme. Read the brochure... lots of info... stepping sequence on page 31. Use this as a start for your own research!
H: Given desired overshoot graphically find k from the root locus (Scilab) I want to find \$k_1\$ that leads to the feedbakc system having poles in the 53º line (to have overshoot \$\leq 10\%\$). From the root locus, the poles move along the vertical asymptote centered on -1. How can I find \$k_1\$ using Scilab and how can I draw diagonal lines representing the 53º line in which I want my poles to be at? So far, I made this graph in Scilab, Code: s=%s; num=2; den=s(s+2) t=syslin('c',num/den); clf; evans(t); mtlb_axis([-10 10 -2 2]); Obtained graph: Desired graph: AI: First, drawing two more lines with angles 53⁰ and -53⁰ should be easy, just do another plot while keeping the root locus. The lines should be, $$ y = \tan(53^\circ)x$$ and, $$ y = -\tan(53^\circ)x.$$ Once you have those lines, set some \$k\$ to be the maximum value of the evans function and change that value until you find the one that lies on the line you plotted. I would use a binary search to pick the values of \$k\$. s=%s; num=2; den=s(s+2) t=syslin('c',num/den); clf; //test your k by changing this k = 100; evans(t,k); mtlb_axis([-10 10 -2 2]); //draw the 53 deg line r = linspace(0,-3,10); up_line = tan(53%pi/180)r; dn_line = -tan(53%pi/180)r; plot(r,[up_line; dn_line],'r-.') //pzmap of the system with feedback figure plzr(1/.(tk)) //draw the 53 deg line r = linspace(0,-3,10); up_line = tan(53%pi/180)r; dn_line = -tan(53%pi/180)*r; plot(r,[up_line; dn_line],'r-.')
H: Wiring a 6 pin On/Off/On toggle switch (DC) 2 Pole I bought a 2 pole 6 pin On/Off/On toggle switch that's for DC power, (25A 12V) see here: Toggle Switch The only issue is that I have no idea how to wire it, as mentioned, it has labels on the back, 1, 2, 3, 4, 5, 6 with no indication of what's positive and negative. What I would like to do with it, is connect two power sources, and a main device, is that even possible with this switch? Basically, I have a Monitor that takes 12V DC, and I want to use, USB-C (Which is converted to 12V through a buck converter) and a DC power adapter, but the switch is used to switch between them, so flicking to the first On will use USB-C, and then the second On will use the adapter, again, is it even possible to do that with this switch? If it is, could anyone help me with wiring? Because there isn't any schematics or diagrams that explain how this switch is wired. Thanks in advance. AI: Figure 1. The common on one pole of the switch is connected to the upper or lower tag depending on switch position. The other pole of the switch is the same and is electrically isolated from the first. Connect the monitor + and - to the common tags. Connect one power source to the top tags, observing polarity. Connect the other power source to the bottom tags, observing polarity. Figure 2. Cross-section of a switch that was a serious problem on Apollo 15 due to a piece of wire floating around inside the switch. Source: Apollo 15 Day 2: Checking the SPS. Don't buy stuff that doesn't have datasheets.
H: What is this PCB component? I took apart a bank token (Digipass 550 manufactured by Vasco) and this is part of the board, you can see the screw hole with the gold plating together with what looks like a PCB inductor and something that looks similar to a spark gap? Edit: Device stopped working after removing these screws, called the bank and they mentioned that once taken apart the device auto-destroys AI: The circular trace is an inductor, what it's doing in the circuit is hard to tell, without looking at the other components in the circuit. It looks like it's connected to a ground plane along with the screw. They removed the ground plane around the inductor to avoid capacitance with the ground plane. The thing could be an antenna. The "spark gap" really depends on if it is there is a gap (which isn't apparent from the picture. It's either for ESD suppression (which might make sense if its attached to a screw. If there is solder mask (darker green hourglass shape) with copper underneath, then it's simply to ground out the fastener .
H: Antenna connector placement I am currently redesigning a PCB utilizing the Xbee 868 Mhz module with external antenna connection on the pad ( https://www.digi.com/products/embedded-systems/digi-xbee/rf-modules/sub-1-ghz-modules/digi-xbee-sx-868 ). I have followed the rules from the datasheet and calculated the proper antenna trace width and distances for 50 ohm impedance. From Xbee SX 868 User guide: In my application the question is, either it is OK to place the SMA antenna connector not in the exact corner of the PCB: As you can see from the screenshots, the connector is on the edge of the PCB, but only horizontal. Vertical wise it is closer to the middle. The bigger issue is that in close proximity to the connector (counter side of the PCB to the antenna trace) there are traces and components placed. Even though there is going to be solid ground and shielding between the connector (GND polygon not poured on the screenshot) and the rest of the traces, I fear that the signal quality might get altered by the digital circuitry in the close proximity. Are there any rules I can follow and a way I could check what is the desired minimum distance of other circuitry to the antenna trace? EDIT: A screenshot of the top layer with GND polygon poured: AI: Let's suppose that the effective relative electric permittivity is above 3 (I don't have the dimensions of your stripline so I'll do some estimation). Anyway with an Er of 3, you'll have a wavelength of roughly 8 in, this is much much longer than the stripline length, so transmission line effects will not be noticeable from the short stripline. In general making transmission lines short will reduce the effects from them, and is actually a way I use to simplify designs (in some cases). You don't need to worry about noise from the Xbee SX 868 affecting the signal around the SMA as the SMA connector has ground bulit into the case and serves as a shield, the limited area that is exposed will provide minimal capacitance for coupling RF/noise. In many respects, moving the SMA closer to the source is better, make sure you leave the ground plane below the trace even if it is shorter. And I also like to run the trace around all four posts of the SMA on the top layer for the sake of betterness (even if it is minimal, might make it slightly harder to solder though).
H: Is it possible to have a non-integer multiple of minimum length in MOSFET technology? I need to design an OTA with 0.25 \$\mu m\$ CMOS technology. Can I choose to have a transistor with a channel length of 0.60 \$\mu m\$? I don't know if 0.25 is the resolution of our technology process and so it means that each transistor must have a channel length (or the width) that is an integer multiple of 0.25. AI: It's always been fine for technologies I've used to have a length just slightly greater than the minimum length; it does not need to be an even multiple. I wouldn't count on being able to have the width be 0.25 μm, it is likely some greater value. You will need to get your hands on the design rules for your process. They should be automatically checked by DRC. There are reasons to use multiple copies of a transistor in parallel instead of just increasing the W/L ratio. For example, if you need one transistor with 5 times the W/L of another transistor, it's better to use 5 copies in parallel for the second one. But it's not required, just a strategy for better matching.

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.