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Is a bad practice make variations on power's tracks width in pcb? Its is about a circuit to source power and communicate with a car's key. I don't know exactly the current necessary, but I know that it is low, a few mA. The voltages are 5.5V and 3.3V.The frequency of the signals are not high , a few units of KHz, something near 10kHz. I would like to know if variation on tracks's width like the signalized in red on the figure below are very problematic. The signalized on the figure are GND, and VCC. <Q> I would like to know if variation on tracks's width like the signalized in red on the figure below are very problematic. <S> It's pretty common to have this kind of feature in a design. <S> In a DC or low-frequency AC power track this will cause no problem as long as the narrow section is short (you could calculate the actual resistance effect and be sure it doesn't exceed your requirements). <A> I think the significance of trace length and geometry matters for higher frequencies where intereference from reflections can be critical. <S> I think sub < 1 MHz designs shoud be still okay. <A> If u are using it for audio no problem there, right, if u wanna design for RF, I'd avoid your drawing work, I'd use soft-round corners
| Particularly when high currents are involved you might want the trace as wide as possible to minimize voltage drop, but need to narrow it down in certain areas to fit other traces or components. At 10 kHz (or probably even 10 MHz), the indicated trace width variations are not significant.
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Minimal components for timer relay A 5V LED should run for a predefined time period once a button is pressed. The LED should be illuminated for the same duration as the button is pressed, but only once the button is released and illumination should last at least 5 seconds. The time deviation should not be more than 2 seconds. Since the controller needs to be positioned on the backside of a wood panel, only limited physical space is available (the smaller the better). A 5V power source is provided for. The project is part of an art project that is to be displayed on a music festival in the summer of 2020. At least 200 of these panels+controllers are needed, so cost should be reduced to a minimum. Previously (prototype), an Arduino Nano was used. Now, cost and size need to be reduced further. What would be the minimal components required to complete this task? <Q> There are many low-cost approaches for solving this problem. <S> These chips cost several pennies from Asian suppliers. <S> A few resistors and a couple of capacitors are required. <S> Probably may also need an output buffer for the LED, depending upon LED current. <S> 2) 6-pin Microchip PIC. <S> These are also available for very little money, although my favourite part from that family (PIC10F321 or 322) does cost a bit more. <S> Again - depends on the LED. <S> You can also feed the LED with multiple pins for more current. <S> Still need a couple of resistors and a couple of capacitors but otherwise very simple. <A> ( I don't know why others are saying five minutes.) <S> I have designed analog circuits that could do this, but by far the least expensive solution is the PIC. <S> One six or eight pin PIC, a bypass cap, the switch, a resistor or two, and a FET if the LED requires more than 20 miliamps. <S> All would fit on a PCB about half an inch on a side. <S> ( Assuming the switch is located off board (and possibly the LED.)) <A> When timing doesn't need to be that exact, the schematic below is another option: it is also very simple, no programming and <S> it doesn't require more components than shown, so it satisfies the condition least components the best. <S> Maybe its dimensions are a bit larger due to the capacitor. <S> EDIT Timing requirements have been changed by OP, which makes this solution better/smaller: <S> For the capacitor now a way smaller capacitor like a SMD 0603 <S> 4.7uF 10V can be used, costing hardly nothing. <S> For a low power LED, a low power PMOS can be choosen which may be compatative in privce to the PIC solution. <S> And this solution may also have a smaller total footprint. <S> For a higher power LED, the PIC solution as well as this solution require a power mosfet, which makes this solution definitely cheaper. <S> end of edit <S> The PMOS and R2 depend on your choice for D1. <S> Also R1 and C1 values can be tweaked a bit to get closer to the 5 minutes when desired (current values should give about 335 seconds). <S> simulate this circuit – <S> Schematic created using CircuitLab
| The PIC-based solution is probably the most flexible and would also have the smallest footprint. The chip can directly supply up to 20mA or so to the LED, so a buffer may not be needed. I will list the two that I would consider: 1) CD4060 (or 74HC4060) timer / oscillator. You could literally fit the whole circuit on a PCB the size of a postage stamp. As I read your question, the LED is to stay on for the same amount of time as the switch is on, but for at least five seconds.
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Why is this clock signal connected to a capacitor to gnd? I am trying to understand the following circuit: My problem is to understand why the CLK signal is connected to the capacitor (C7). The bottom side of C7 is connected with a resistor to GND. This means that the "plate" will (after some delay) have the same potential as GND. The upper "plate" will oscillate with the CLK signal. The logic is connected to the bottom side, so I don't understand how the NAND Gate can ever get some other input than LOW on the bottom pin. This means that the NAND gate will never output LOW and thus the RAM never stores anything, which doesn't make sense. The only effect the capacitor could have in my understanding is to delay and flatten changes in the clock signal, but I don't understand the use of this. Also I don't understand the meaning of the resistor. It pushes the loading time of the capacitor, but after the bottom side is on GND potential there will be no current after that. I hope someone can help me understand this. <Q> C7 and R58 form a high pass filter, also known as a differentiator. <S> The purpose of using a differentiator in this spot is to cause a short pulse on the rising and falling edge of the clock signal. <S> This diagram shows the effect of a differentiator on a square wave (which your clock will be.) <S> As you can see, it makes short pulses on the edges of the square wave. <S> I'm not sure why that circuit needs the short pulses instead of the square wave, though. <S> Too many ICs I'd have to look up to figure out what is going on. <A> The circuit overall is a 16-byte memory bank for a homebrew computer of some sort, with manual programming capability via the switches and lights. <S> Probably the most complicated 16 bytes of memory you'll ever see! <S> The R-C combination being asked about is used only when the CPU is running — i.e., when PROG is not asserted. <S> This only works if the clock high time is significantly longer than the R-C time constant. <S> 10 nF <S> × 1 <S> kΩ = <S> 10 <S> µs <S> So presumably the clock is something less than 50 kHz. <S> Actually, the 74189 is not a slow part — the minimum write pulse width is a few tens of ns — so the R-C time constant could be much shorter, by a couple of orders of magnitude. <A> That is a high pass RC filter with a pole at 15.9kHz, which is unusual for a clock signal because it will attenuate the signal somewhat. <S> The purpose is possibly to keep the clock more than 15-20kHz, to keep the memory running above that speed as a lower clock speed would not transition. <A> As others have noted, it's a differentiator that limits the time that WE is asserted. <S> The outputs from the 74189s go into high impedance with WE is asserted, so it could be to make sure that the LEDs are not excessively dimmed, or to ensure that the output BUS_* lines are stable before the other edge of the clock. <S> The latter option might be used of the output bus is read on the other clock edge. <A> For your information, if the resistor is not there then the NAND gate will never gets a logic high in PIN 1 and always remains low and the resistor also increase the rise time and fall time so the circuit will not even act as a low pass filter, so we cannot decode the intention of the designer.
| It is probably being used to shorten the high time of the write-enable pulses going to the memory chips in order to meet hold-time requirements when the CPU is writing to memory.
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Why/How do higher amp hour batteries for tools provide more power? I have seen power tool brands have battery packs ranging from 1.5Ah to 9Ah. Their advertising indicates that the larger amp hour battery packs can run the tool longer and with more power (such as a circular saw cutting through thicker material). These are all 18v battery packs, so I can only assume that there is circuitry within the packs that are limiting the amount of current per cell. That way, the packs with more cells can provide more current. Is that correct? If it is a circuit to limit current (for short/heat protection purposes?), how are they going about doing this? Just an off the shelf solution they stick in there? <Q> Batteries for power tools are made of a set of more-or-less standard primary "Li-Ion cells", typically of 18650 size. <S> The cells came in different capacities, but overall the capacity is limited by current state of technology given the fixed cell volume. <S> There are high-capacity cells, with higher energy density, and cells with less nameplate capacity. <S> One important thing for power tools is an ability to supply high peak currents to overcome start-up inrush currents for motors, and to handle load. <S> So the battery cells must be of "high-discharge" type. <S> High discharge means smaller internal resistance (ESR) and therefore require thicker foil electrodes and current collectors. <S> As such, the overall power density (given the same volume of the cell) is smaller. <S> That's why mass-produced power tools use cells of seemingly lower capacity, say 1500 mAh. <S> Now, to increase total battery capacity they use a bunch of same cells (or series string of cells) in PARALLEL. <S> This increases not only the overall capacity (6 cells of 1500 mAh <S> give about 9000 mAh battery), but automatically it decreases the overall ESR (six ESRs in parallel result in 1/6 of the original ESR). <S> Thus the battery becomes more "powerful" automatically (can deliver more current). <S> So there is no artificial "current limiter" inside power tool battery pack (besides the fire protection), the limitation comes from property of individual primary cells used to make the battery. <A> The other thing is that for the same amount of usage, a larger pack will have drained proportionally less than a smaller pack which means it's voltage will not have decreased as much. <S> This equates to being able to run at higher power for longer periods of time, but no more power than at the peak power of the smaller pack (the above paragraph notwithstanding). <S> But if you asked me for what the answer really is, I would say the half-truths of marketing. <A> If it is a circuit to limit current (for short/heat protection purposes?), how are they going about doing this? <S> Just an off the shelf solution they stick in there? <S> It depends on the manufacture and model of battery. <S> Some have a full BMS built into the battery and possibly also a DC to DC converter (like the dewalt 18V system, the BMS is in the thing between the contacts and the battery). <S> I doubt other manufacturers are limiting the capacity with a charge controller circuit, the batteries most likely have smaller capacity cells. <S> Typical batteries have 5ish cells and sometimes have a BMS, and use more cells for more capacity. <S> Some battery packs only have cells and the BMS is in the charge controller and/or power tool (the newer 20V dewalts.) <S> That way, the packs with more cells can provide more current. <S> Is that correct? <S> The more cells you have, the more voltage you have if the cells are in series. <S> Sometimes two packs of cells are paralleled for additional capacity.
| Technically, higher amp-hours battery packs have lower internal resistance which reduces the voltage droop at higher current draws which equates to more power output for the same motor and voltage. You can easily tell by pulling one of the battery packs open and checking the datasheets from the battery manufacturer which list the capacity (and sometimes it's listed on the cells).
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How to control BTS7960 43A motor driver directions? The board has two PWM inputs, see: Do I have to use four pins to control two motors?Thanks <Q> The board controls a single motor. <S> Pins 1 & 2 - Power for the motor <S> Pins 3 & 4 <S> - Direct connections to the motor A - Ground for the 5 V controller <S> B <S> - 5 V for the controller C & D - Overcurrent indicators <S> To drive motor forward, put a pulse train up (up to 25 kHz) on pin F. <A> You can only control one motor with this board. <S> You connect your motor in the M+ and M+ connectors (The letters are written in the bottom layer of the board near the correspondent green connectors) <S> The battery is connected in the B+ and B- connectors (The letters are also present in the botton layer of the board). <S> The RPWM pin is used to rotate the motor in one direction using a PWM signal <S> The LPWM pin is used to rotate the motor in the opposite direction of the RPWM pin also with a PWM signal. <S> The R_REN <S> When is at 5V (HIGH) is used to activate the RPWM pin. <S> The L_REN When is at 5V (HIGH) is used to activate the LPWM pin. <S> The maximum voltage for the RPWM,LPWM R_EN and L_EN pins is 5V. <S> You don't need to connect the VCC pin to 5V when You use a 12V battery in B+ and B-. <S> For battery values below 12V <S> i'm not sure. <S> The GND pin is wired to B-. <S> The R_IS and L_IS are used for current sensing. <S> No need to use them in simple projects. <A> DIR signals, should go to R_PWM & L_PMW. <S> AND PWM should go to both L_EN, R_EN. <S> You can check it, it works )
| E & F - Reverse and Forward enables (can tie both to B, if desired) G & H - PWM controls To drive motor in reverse, put a pulse train (up to 25 kHz) on pin E. With a 12V motor you only need the 4 pins listed above to control a motor with this board. You can control this board with an Arduino and may good idea to isolate the control pins of the Arduino with Optocouplers.
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Ambiguity in voltage measurement So I watched this extremely fundamental video about voltage measurement and how it's measured. The one thing that confuses me is what people use a reference for zero potential. The two diagrams shown below have the exact configuration and order of batteries. The values don't change either. But I get two different values of net voltage based on where I assume the zero potential to be and this assumption is arbitrary. So if I choose the zero potential to be in the middle and if my friend chooses it to be at the end, we get two different voltages and we would both be right. So what does one do to avoid such ambiguity? Is there some pre set standard on how to measure voltage? And what does a manufacturer mean when he says a supply gives a 12V or 15V output? What would the manufacturer have used as a reference? For if I change the reference point, my voltage reading for the same supply would be different and this would have consequences, right? Edit: So what I'm doing in the first diagram is, I first connect the red lead at the top to measure 12V. I then connect the red lead to the bottom to measure -12V, getting a net voltage of 0v. <Q> So what does one do to avoid such ambiguity? <S> You can include a reference ("ground") symbol in your schematic. <S> Then when you say you measure v volts at some node "A", we know what you really mean is that the voltage between "A" and the reference node is v . <S> You can clearly say which nodes you measured the voltage between. <S> For example, "I measured v volts between nodes 'A' and 'B'" <S> (but now you'll have to be careful to communicate whether you mean <S> 'A' was v volts above 'B' or vice versa). <S> And what does a manufacturer mean when he says a supply gives a 12V or 15V output? <S> They mean there will be 12 or 15 V between the two output terminals, or between one output terminal and a designated ground/reference/return terminal. <S> For if I change the reference point, my voltage reading for the same supply would be different and this would have consequences, right? <S> Yes, but if you're buying a power supply it's usually pretty clear which terminal is the reference terminal. <A> There is no zero volts. <S> Every voltage is referenced from some point and to some point. <S> So to remove ambiguity, you say "the voltage from the battery negative to the battery positive is ...". <S> Some things are implied -- a 5V power supply is presumed to have + and - terminals, and you assume that you'll read 5V from the - terminal to the + terminal. <S> Ditto batteries. <S> If you have a schematic that's got one ground that's clearly marked, then it's assumed that "3.3V" means "3.3V from ground". <S> But if you have a schematic that has more than one ground (i.e., a "noisy" ground and a "quiet" ground), then those two grounds may have voltage between them, so you have to be careful about what ground other signals may be referenced to. <A> When surveying the land a surveyor has to pick a reference height which is his 'zero' or datum. <S> The height of every other point is measured above or below that. <S> She can, if it suits, use sea level as the standard reference. <S> Usually it's the battery negative (as on a car electrical system) but it can be the positive (famously on the old 6 V VW beetle) and <S> in some cases we need a split-rail power supply and we can call the centre one 'zero' and have a '+' and '-' with respect to that. <S> Finally, if the circuit common is connected to earth (as in The Earth) we take that as a real zero volts, in much the same way as the surveyor might use sea-level as an absolute reference. <S> Figure 1. <S> The building on the ground and shot off into space. <S> In the space situation (electrically isolated) we can call any floor the 'ground' floor. <S> [Image by @Transistor.] <S> In Figure 1 we have called the ground floor 0 (as is European practice). <S> Floors above are numbered 1, 2, 3, ... and floors below, -1, -2. <S> We can now calculate the number of flights of stairs to climb by subtracting the start from the end. <S> So to travel from -1 to +3 we need to climb 3 - -1 = 3 + 1 = 4 flights. <S> Note that we could pick any floor as "ground". <S> It doesn't matter - it's just a reference. <S> It's the same with voltages. <S> Figure 2. <S> In the upper image the ground reference is mid-way up the battery stack. <S> In the lower image the reference is at the bottom of the stack. <S> In both cases it's 24 V from the bottom to the top.
| In an electrical circuit we pick some point as being zero or ground.
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Can you replace electrolytic capacitors with tantalum ones? Due to size constraints on a custom PCB, I have opted out for using the smallest type of electrolytic capacitors (3 mm diameter.) However, capacitors in such a small diameter are almost impossible to find. If I remember my high school books correctly , tantalum caps are actually a type of electrolytic capacitors just with different purpose and their size to capacitance ratio is much better than the electrolytic ones. I know that they have other advantages, but my biggest concern is is it possible to just use tantalum capacitors instead of ordinary electrolytic capacitors? The circuit in mind is just an ordinary amplifier for an electret microphone using a simple LM386. Edit : Here is the schematic, I am asking for all the electrolytic capacitors. <Q> Yes, you can change it for Tantalum capacitors. <S> The zobel capacitor can be ceramic or nylon. <S> C7 probably you can't find any of this value 0.1uF <S> but you can change it to 1.0uF <S> this will give some aditional bass to your amplifier; however if you use any ceramic capacitor won't have any problem. <S> I've have made this preamplifiers for my projects with success handling tones for hifi amplifiers. <S> For hobby purposes you can find this kind of assortments inside an old car stereo, any old VCR, any obsolete FTA, etc. <A> The .1 uF and .047 caps can be X7R or even Z5U. <S> If space is very tight and this is not an amp for a radio you can omit R1 and the .047 cap. <S> This zobel network is not essential. <A> Yes you can, but you are far better off using the TPA321D <S> that occupies far less space and has no need for large caps as it is a bridge (differential) driver.
| Yes you can use tantalum capacitors for the 10uF and 100uF.
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DC-DC converter from low voltage at high current, to high voltage at low current I want to build a DC to DC converter with: 25 V input side 400 V output side 4000 watts of continuous rating That means (obviously) 160 amps on the input side and 10 amps on the output. I can’t find anything remotely close, that has such a high voltage gap with such a high wattage rating. Is this feasible? I can’t see why not. What would be the best way to go about this, or a reference design somewhere that I can scale up? I don’t mind spending the money on quality products, but I prefer to DIY it so I can make changes etc. I have an adequate voltage source for the input side that can handle the current so that’s not the issue, but I’m failing to find more info and even big companies like Vicor, Lambda, etc. all have 400-600 max rated units. Therefore I have to build it. <Q> 160A at 25V will not give you 4kW out. <S> If it is very well designed, you'll get around 3.2kW. <S> The rest is wasted as heat. <S> As you're just setting out to do this, and you're trying to design it yourself, you need to model it well and simulate to work out where your losses are going to be, and how you're going to cool it. <S> This is a perfectly do-able boost converter project. <S> I have done a 5kW output DC-DC (admittedly that was 48V), and that required a full automotive style liquid cooling system. <S> That was using a standard synchronous DC-boost converter, 48V came in, and we got up to 200V out. <S> First things are cooling and component ratings, those are the hard bit to do. <S> 25V is low for 4kW, so you'll quickly see (once running the numbers) why it is that higher voltages are chosen for these kinds of power. <S> At 25V in, 4kW out, 80% efficient means around 200 Amps in, plus a 50% overhead safety factor for your FETs, so you need to find FETs rated to 300A, 800V (high voltage due to high output voltage requirement). <S> Don't forget to de-rate for temperature, and check your simulations for junction temperature rise. <S> I say your FETs, but it doesn't have to be FETs, I've used FETs and I've used IGBTs for this kind of work, some people say GAN transistors would also work (but I've never used them). <S> Depends on the voltages, currents and switching frequencies you're looking at. <S> But that in turns depends on your application, budget, size constraints, development time allowed etc. <S> You'll also need an inductor, rated at similarly. <S> But these are probably easier to find. <S> You could split the power down, using multiple channels in parallel, each channel doing a part of the current (I used 3 channels on my 5kW system). <S> But still, cooling will be your biggest challenge. <S> You can then build it, find out what fails <S> , what mistakes you made in your simulations, and then repeat the process until it works how you want it to work. <S> In summary: It will get hot <S> You need to simulate it <S> Don't underestimate how hot it will get <S> You can just scale up a standard DC boost converter <S> Watch <S> your cooling Simulation is vital <A> The inductive coupling must be modeled. <S> As well as eddy currents. <S> 200 amps switched in 200 nanoseconds (for high efficiency, fast switching must occur) and wired to be 1cm away from a 1cm-by-4cm servo-regulator loop, will induce this error voltage: <S> Vinduce = <S> 2e-7 <S> * Area/Distance <S> * dI/dT <S> Vinduce = <S> 2e-7 <S> * 1cm*4cm / 1cm <S> * 1Billion amps/second Vinduce = <S> 2e-7 <S> * 0.04 <S> * 1.0e+9 <S> Vinduce = <S> 2e-7 <S> * 4e-2 <S> * 1.0e+9 = 8 volts. <S> To be completely accurate, you need to write the integrals and extract the equation that uses NATURAL_LOG. <S> And you need to model the eddy currents. <S> At these levels of dI/dT , ground planes will NOT be ground planes. <S> There will be large differences in voltages across the plane, because of eddy currents. <S> The math suggests shields and planes (VDD or GROUND) will have EIGHT volts of gradients. <S> I was brought in to diagnose the failures on a 15,000 horsepower speed controller. <S> A loop of wire, to sense the magnetic field , held near the ground-plane, indicated 2 volts per square-inch. <S> Ground Was Not Ground. <A> Some crazy variable speed drive system with a 24V supply and programmable voltage might even work out of the box. <S> Starting to design one from scratch for a once-off will cost much more than you think, hoping to save money this way is futile. <S> I would spend a week making calls to every VFD and inverter manufacturer with your loosest tolerable specifications and see if anyone can configure OTS parts.
| A programmable voltage three phase inverter with a full bridge (or two that are wired with a 30degree phase off-set) (to gain minimum ripple) with some inductance and capacitance to follow that are designed for 24V input might just be able to be found semi custom.
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Connecting the Block AIM 5.0/2.0 auto-transformer to UK three-wire plug I am not electrically very knowledgeable person. At the moment I am trying to connect a Block AIM 5.0/2.5 Auto-transformer to a piece of very sensitive medical equipment. The auto-transformer was purchased to regulate and supply a constant 220 V (to avoid any fluctuations in the voltage as the medical equipment is very sensitive and recommended to run only at 220 V) to the equipment here in the UK. Now, very recently the company that checks our equipment for electrical safety has labelled this auto-transformer as FAILED due to "No Earth". Please see the picture where there are only two input (L and N) wires and two output (L and N) wires connected. The ground (earth) wire of both input and output are left open. Now, what and how can I make this device electrically earthed (or have it pass the electrical safety test)? Please let me know. I would really appreciate your help. Thanks a lot. <Q> Product specification: Block AIM 5.0/2.0 Autotransformer . <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> An isolating transformer and a non-isolating auto-transformer. <S> At the moment I am trying to connect a Block AIM 5.0/2.5 Auto-transformer to a piece of very sensitive medical equipment. <S> That's going to have high safety specifications. <S> The auto-transformer was purchased to regulate and supply a constant 220 V (to avoid any fluctuations in the voltage as the medical equipment is very sensitive and recommended to run only at 220 V) to the equipment here in the UK. <S> The transformer will not "avoid fluctuations". <S> Its output voltage will fluctuate exactly in proportion to the input voltage. <S> It is much more likely that the transformer has been recommended to reduce the UK 240 V supply to 220 V to suit the equipment. <S> This is done by setting the 'tap' or tap-off point on the auto-transformer 220/240 = 11/12 of the way up from the bottom of the coil as oriented in Figure 1. <S> Now, very recently the company that checks our equipment for electrical safety has labelled this auto-transformer as FAILED due to "No Earth". <S> This is, presumably, because the equipment to be powered <S> requires an earth. <S> Please see the picture where there are only two input (L and N) wires and two output (L and N) wires connected. <S> The ground (earth) wire of both input and output are left open. <S> Now, what and how can I make this device electrically earthed (or have it pass the electrical safety test)? <S> The transformer doesn't require an earth because it is fully isolated. <S> It's the load that requires an earth. <S> simulate this circuit Figure 2. <S> Standard industrial practice. <S> If this were an industrial application the solution of Figure 2 would be acceptable. <S> Your equipment test company should be able to advise or the medical equipment supplier. <A> If it were just domestic equipment, I might suggest ways to do it. <S> But it's medical equipment, and doing it safely and correctly is important. <S> Hand it to a qualified electrician and get them to do it properly. <A> Autotransformers do not need an Earth Bonded ground connection, but your appliance may need it. <S> simulate this circuit – <S> Schematic created using CircuitLab Orange or Yellow should be the proper gauge. <S> Strip 1cm of insulation and twist until firm. <S> and wrap electrical tape around to prevent turning. <S> The other leads need to be shortened to allow the gnd leads to get close to each other.
| Essentially, you just connect the incoming earth wire to the outgoing one, rather than just chopping them off and leaving them floating around as at present. Because it's a medical application you shouldn't be prepared to take that risk and run on uncertified equipment. That means that it is effectively on a European supply voltage and will tolerate normal fluctuations in supply.
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Breakdown Voltage of relay contacts in series I am working on a module that allows a small internal-combustion engine to be shut down remotely. These small gasoline-powered motors are of various makes and models and all have magneto ignition systems. The ignition systems can have either electronic or Kettering (points and condensor) ignition. The most-common method of stopping these motors is to ground a node within the ignition system. For Kettering systems, one simply grounds the points. For electronic ignition systems, a node is provided that has the same functionality. My customer wants me to provide SPDT relay contacts. I prefer to use Telecom-grade relays for this application because there is the possibility that the relays may need to switch low-current / low voltage-level signals. Problem is that the maximum voltage present at the "kill engine" node is a spike that approaches 400V peak. All of the Telecom relays that are readily available to me have maximum voltage rating of 250V. However, these relays are all DPDT. Theoretically, I should be able to connect the relay poles in series and get double the maximum voltage rating. Theoretically. My question is: is connecting relay poles in series to get higher maximum voltage a reliable solution? [Edit] It's been pointed out that "All the pain is taken by the last to close and first to open." I agree completely. But please follow my logic here and see if the following makes any sense at all. First: the relay operates only to either allow the engine to run or force it to stop. When the relay operates, the poles do NOT operate simultaneously. There is a lag of some hundreds of microseconds between when each contact changes state. But: does that matter? The only time this is a consideration is when the relay is changing state. When the engine is supposed to be running, you have two sets of contacts in series with, theoretically, double the rating of each individual contact. When the relay goes to the state that will allow the engine to start, the engine is currently stopped. There is currently NO ignition present, and therefore no voltage on the contacts. When the relay goes to the state that shuts the engine down, one contact closes first, then the other. It is possible that the contact that closes last has a single spike that could breakdown the gap. Before the next spike can occur, the other contact has closed and there are no more spikes. The total energy contained in a single spike is really quite tiny. I don't have hard numbers right now but I will get them when I start testing with real engines. But I don't believe there is enough energy in the spike to damage the contact. And, I do suppose that I can monitor the voltage across the open contacts (both open) and allow the relay to change state only immediately after the spike has decayed to a low value. That completely gets rid of the voltage across a single contact. But my original question stands: can I count on the breakdown voltage to be double that of a single contact if two contacts are wired in series and not changing state? <Q> Quoted relay voltages are usually set by the switch-opening process. <S> Not by open-switch standoff voltage. <S> In other words, your 250V relay can tolerate the severe arcing which occurs whenever a 250V circuit is suddenly opened. <S> The same relay probably can handle far higher voltages before it would arc across, as long as the contacts weren't opening during the moment the 400V spike occurs. <S> (And even then, it might work OK) <A> simulate this circuit – Schematic created using CircuitLab <S> In theory with 1k to 3kV/mm for air breakdown smooth contacts, and there is some tolerance for gaps and environmental contamination that may reduce this to 20% of these values. <S> Conclusion Use a better, sealed relay rated for 1kV for reliable opeartion. <S> mercury wetted reed relays are not expensive. <S> 2A 1kV <S> $5typ <S> https://m.ebay.ca/itm/GI-Clare-Mercury-Wetted-Relay-type-CUP-P001A505-1-form-A-one-NO-2A-1000V-coil-5V/253552671880?hash=item3b08eabc88%3Ag%3APfcAAOSwY45UOw2G&_sop=15&_sacat=0&_nkw=mercury+wetted+relay&_from=R40&rt=nc <A> Magneto service is fairly easy duty for relays. <S> Think about it <S> ; with a magneto ignition you leave it open (disconnected) while the engine runs, and close it (short it to ground) to ground out the coil and stop the engine. <S> Think about what that does to relay states. <S> In closing , the contacts may arc as they come together. <S> However the arc is instantly extinguished by the contacts closing in full. <S> Opening will never face a load. <S> Because the engine isn't running if the contact is closed. <S> Ergo very little magneto speed, ergo weak spark. <S> (Normally opening is the worst case for a relay, because an arc will strike as the contacts start to open, and could then sustain once they're fully open. <S> Not an issue here .) <S> Holding closed , the only risk is arcing between contacts and coil or chassis, and that's pretty easy to insulate. <S> Holding open , the contacts need to simply be far enough apart to prevent a cold leap. <S> Better, magneto voltage acts like AC, because it self-interrupts periodically which means arcs will self-extinguish like AC (better, actually), but again arcing-on-open isn't a big problem here. <S> Now about double-pole, you only want to do that <S> so you can put the contacts in series for double the voltage resistance. <S> You're giving us an XY problem there; <S> your real problem is you want to interrupt 400V intermittent. <S> So take that on de novo . <S> I just asked mouser.com for electromechanical -> <S> relays -> contact voltage 400VAC and up. <S> There's a sea of them. <S> It's not a problem . <S> I know you say you want telecom relays, but <S> that's just you being comfortable with a particular type. <S> That's nice <S> but it is not appropriate here . <S> Honestly, itll probably work - <S> like I say, electrically <S> this is easy service for a relay. <S> But mechanically, the vibration is very worrisome, and telecom relays are totally not made for high-vibration situations. . <S> I would be looking for vibration resistant relays. <S> Particularly you can't have relays where the vibration closes the relay, because that'll make the engine miss.
| Or use a Reed Relay.
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What's the output of a record cartridge playing an out-of-speed record I'm very interested in vinyl records and analog music, and the belt of my turntable got loose. Upon such situation it piqued my curiosity, what is the output signal at the end of the arm cartridge wires for a known waveshape if the speed is not the correct one. Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave? I'm not considering the filters that the cartridge might apply on the signal, whether it is a low pass, band pass, or high pass nor any other impedances that might alter the signal in any circumstance, just a supposedly ideal stylus and cartridge. <Q> Grooves are cut with frequency correction according to RIAA equalization . <S> Playing the record off with wrong speed increases all frequencies by the same factor (corresponding to a shift left <S> /right on the frequency axis of the doubly logarithmic transfer function diagram). <S> Since the frequency correction is not a straight line, this does not just result in a frequency shift but also in an uneven frequency response due to recording and replaying correction no longer being proper inverses. <S> In addition, the equalization is done in order to reduce excessive signal amplitudes on stylus and pickup. <S> Counteracting this by wrong speed may lead to either excessive amplitudes (electrical or mechanical) or too low signals overlaid with a relatively higher noise floor. <A> Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave? <S> The pitch and tempo will change in proportion to the speed change. <S> At 33 RPM it would already be musically flat as the correct speed is 33 1 /3 <S> RPM. <S> A 1 kHz test tone - common on test records - would, at 33 RPM, give off \$ \frac { <S> 33}{33.33} \ \text {kHz} \$ . <S> The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch. <A> Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else. <S> A sine wave with the time axis compressed or expanded is still a sine wave. <S> In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time. <A> To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. <S> (This ignores stereo, and any companding, but it answers your question). <S> Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove. <S> If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. <S> Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk. <A> I have practical experience with this--record players with variable speed drives used to exist. <S> These were specialty systems intended for blind people--they allowed the listener to speed up the records. <S> They were made variable because not everyone wanted the same speed. <S> Obviously, for music this would be insane but these units were intended for playing voice--magazines read aloud onto special 8 1/3 rpm 9" flexible plastic records. <S> They were not durable at all (but neither are magazines) but did their job at a much lower cost than other technologies of the day. <S> Other than the variable speed drive, the low speed settings (their highest was 33 1/3), and the ability to survive being mailed as is they were ordinary players. <A> Often 5 millivolts output at 5 centimeters per second needle movement. <S> This would be for MOVNG-MAGNET, with thousands of ohms because of the very tiny wires in the FIXED coil. <S> The high resistance causes a high random thermal electron noise floor. <S> The lowest noise cartridges are MOVING_COIL, often with resistance under 10 ohms. <S> But serious amplification, at low noise, and low VDD trash injection, and thorough shielding, is needed. <S> The output voltage is often 0.2 millivolts or even less,at that stated needle velocity, at 1KHz. <S> [edit] <S> Some RIAA preamplifiers use common-source JFET amplifiers, and those type of gain stages have ZERO power supply rejection. <S> Thus large RC filters are needed for VDD inputs, and regulators are SHUNT designs, with approximately 1nanoVolt/rootHertz noise density.
| The output voltage of record cartridges is given at 1KHz at some standard velocity of the needle.
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Overvoltage protection with zener - How do you get a sharp knee clamp for analog input? I'm working with an MCU that sees a 0-5V input range. I have a possibility where they may accidentally see 12V (user error). My problem is this analog input using a 5V Zener has a knee that isn't as sharp as I'd want and can load the sensor reading giving a bad measurement. How can I get a sharper knee for around 5-5.5V or so? <Q> Here is one method: simulate this circuit – Schematic created using CircuitLab <S> If the errant user leaves the +12 on there the resistor will see almost half a watt and the transistor <S> a few hundred mW. <S> You could increase R1 to a few hundred ohms to reduce both those numbers. <S> Note that the input current is diverted to ground through Q1 rather than into the +5V supply. <S> The latter can cause unwanted side effects. <S> This is still dependent on the characteristics of a PN junction so the clamp voltage will change with temperature and there will be some leakage as you approach the clamping voltage. <S> but it's far from perfect. <S> If you want something more precise, especially over temperature, you'd probably have to combine an active op-amp clamp with something like the above to handle fast transients. <S> Edit: As Jack Creasey mentioned in a comment, in this case R2 can be replaced with a diode or a transistor with collect tied to base (ideally in a dual NPN-PNP transistor array) to give better temperature compensation. <S> You can add a small resistor in series to give a bit more headroom if you want to be able to go closer to the supply rail. <A> For input protection of MCUs, they can usually tolerate a small over-voltage. <S> A diode to the positive rail is usually good enough. <S> You can choose between a low leakage silicon device that clamps around 0.7v, or a higher leakage schottky that clamps at around 0.3v. <A> How can I get a sharper knee for around 5-5.5v or so for MCU port signal clamp? <S> Your definition of the clamp voltage as 5-5.5V may be too broad as well. <S> Most MCU's have internal conduction increasing rapidly at VDD + 0.3V and VSS - 0.3V <S> (intrinsic diodes), so you really need to clamp at LESS then 0.3V above VDD or below VSS. <S> I assume that you have multiple inputs (both digital and A/D inputs) on your MCU to protect, so the following may be effective for you: <S> simulate this circuit – <S> Schematic created using CircuitLab Note: <S> The negative clamp is simply the diode <S> Vf <S> so varies considerably with temp. <S> The positive clamp can be adjusted to clamp just above VDD but before the internal intrinsic diode conducts. <S> If there is the possibility of multiple inputs being wired to 12V (if this is a student environment this may well happen), then you have to account for 70mA for EACH clamp. <S> This adds up, so you may need a small heatsink for the TIP42 which may need to sink several amps in total. <S> Note: This clamp circuit could drive about 30 inputs or more depending on the number of simultaneous errors to 12V <S> you want to allow.
| You can play with R2/R3 ratio to change the clamping voltage and even try to make it temperature dependent with a thermistor or diode to a higher voltage source No Zener diode will be effective in this type of application, the knee voltage curve is just too broad. If your 'user error' includes +12V as a connection possibility, then perhaps you should also include -12V as a possibility.
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Why doesn't this battery voltage measurement circuit work? So, I built this simple circuit for measuring voltage from a battery pack to an Arduino. The thing is, I want to measure the first cell and then the voltage from the whole battery pack: using two transistors for changing the voltage divider input (check the image so that you know what I mean.) I do not want to apply a constant load to the battery, that's why I'm using transistors to turn on and off the voltage divider circuit. Here's the circuit: I'm using a simple calculation when reading from pin A0: 0-1023 (analog range) reading to 0-25 volts (as the voltage divider is at a 5:1 ratio.) My question isn't about the code. Here's the situation: When both transistors are off, I get a 0 Volts reading (as it should). When the first transistor is on, and the second is off, I get a ~4.2v reading (as it should.) The problem is when the second transistor is on (and the first is off). I get the SAME reading (~4.2v) instead of 8.4v (the whole battery pack voltage.) Adding diodes between each collector and the voltage divider for protection didn't alter my results. I'm really confused about this. UPDATENote: Each battery cell is rated at 4.2v. Real Schematic: <Q> You can keep most of what you have if you just change to a high-side switch using PNP instead of NPN transistors. <S> Something like this: simulate this circuit – Schematic created using CircuitLab <A> Your problem is that the emitter voltage on the transistor can't go higher than the base voltage. <S> You're measuring the base voltage, minus two diode drops, not the collector voltage. <S> You could try using an "analog switch" or "analog mux" IC instead of a single transistor. <S> Pay attention to the allowed voltage range for the signal pins. <A> Figure 1. <S> Circuit from OP. <S> This won't work. <S> Your transistors are wired as voltage followers. <S> The emitter voltages will be about 0.7 V below the base voltage. <S> This is confirmed by <S> When the first transistor is on, and the second is off <S> , i get a ~4.2v reading (as it should.) <S> The problem is when the second transistor is on (and the first is off). <S> I get the SAME reading (~4.2v) instead of 8.4v (the whole bpack voltage). <S> The emitter of either transistor can't go above about 4.3 V. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> A simple solution using two analog inputs. <S> This will reduce the voltage by a factor of two on each measurement input.
| The issue is that you're using a low-side switch configuration in a high-side switch application.
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Why do we use polarized capacitors? I want to know if the polarized capacitor has some advantage that they are used in some circuits? For example, in a schematic of the BISS001 PIR controller IC, in some places, a polarized capacitor is used and in some places a non-polarized capacitor one. Can I use a non-polarized capacitor with the same voltage and capacitance instead of these polarizing capacitors? Reference Docs: BISS001 datasheet HC-SR501 PIR MOTION DETECTOR datasheet Grove - PIR Motion Sensor or EasyEDA link What I've understand from your answers is why the electrolytic capacitors are used, and why these are polarized. But the designers of this circuit could have used a non-polarized capacitor or even polarized tantalum capacitors. Is it true? As the ( Grove - PIR Motion Sensor ) module uses polarized tantalum capacitors. I want to know if the polarized capacitors are being used for circuit protection or is there any other reason (regardless of the type of capacitor.) Is there a problem if these capacitors are replaced with non-polarized capacitors in these circuits? <Q> The physical size of a capacitor is a function of the thickness of the dielectric (among other things). <S> Early on, it was discovered that the oxides of certain metals (aluminum and tantalum in particular) made good dielectrics, and could be made very thin through a chemical process — orders of magnitude thinner than other dielectrics such as waxed/oiled paper and plastic film. <S> Therefore, the electrolytic capacitor was invented to provide high capacitance in a reasonable volume. <S> Unfortunately, the chemical process requires that the voltage across the capacitor must have only a single polarity, so these capacitors are "polarized". <S> Reversing the polarity degrades and eventually destroys the oxide layer. <S> It's something we just have to live with in order to take advantage of this technology. <S> There is generally no problem with making this substitution, although you may need to consider some of the quirks of the technology you're switching to. <S> For example, some high-K (high dielectric constant) ceramics exhibit significant capacitance changes with voltage. <S> This might be acceptable in a coupling or bypass application, but completely unacceptable in a filter design. <A> Can I use a non-polarized capacitor with the same voltage and capacitance instead of these polarizing capacitors? <S> Electrically speaking, non-polarized capacitor is always better than a polarized one. <S> Yes, you can always replace with a non-polarized capacitor with exactly same rating. <S> But there is an assumption hidden here: Provided you can find one that's physically small enough to fit on your board and cheap enough to fit in your budget. <S> And the fact that you can't is the only reason we use polarized caps. <S> I assume that, if we ever learn to make non-polarized caps that are as cheap and dense (capacity-per-volume) as electrolytic ones, the polarized capacitors will vanish. <S> Side note - voltage and capacitance are not the only electrical parameters of a capacitor. <S> They would suffice for an ideal capacitor, but real world brings other, ugly metrics. <S> Like ESR, capacity coefficient with temperature or voltage, frequency response, etc. <S> Circuits designed around quirks of particular tech can fail if the substitute differs there. <S> Even being too good can cause trouble, eg. <S> high-ESR caps naturally keep peak current in check so substituting with a theoretically superior low-ESR part can cause the whole thing to blow up. <S> Adding ESR is trivial - but that's no longer a drop-in replacement, but rather a circuit redesign. <S> So we don't replace electrolytics with something else not because polarization is important, it's just a nuisance. <S> We keep them because of many other parameters, less obvious than C, V and polarization. <A> Since you mention protection I'll add that polarized caps should not be used for reverse polarity protection . <S> They will react on a reverse voltage very slowly (seconds or minutes), while typical sensitive components which are worth protecting will be dead within milliseconds. <S> And once a polarized cap starts to absorb the reverse voltage, it may vent, explode or catch fire which (apart from the obvious problem with smoke and fire) can make it non-conductive again, exposing your circuit to the reverse voltage once more.
| The ability to produce high-value capacitors in nonpolarized technologies such as multilayer ceramic means that it is now possible to use them where only a polarized capacitor would have been previously available.
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Peak TX current of LoRa RFM95W I'm running into some issues powering a test board using a LoRa chip (RFM95W) with a CR2032 battery. I'm well aware of the peak current limitation of such battery and I can see a major voltage drop across the battery when too much current is asked during LoRa TX (around 80mA), even using a 100uF capacitor in parallel; the voltage drops down below 1.5v which make my MCU brown-out and reset. Thus entering an infinite loop as my MCU is setup to send a measurement on start-up. I'm using a power gating timer (TPL5110) in order to minimise the current consumption between measurements, 1 mesure every two hours. The question is: how do people manage to use CR2032 coin cell battery alongside LoRa RFM95W as I assume the peak TX current would be fairly similar in all these other IOT boards I can see around internet. I'm using default firmware mode of communication under LoRa-WAN standard of The Things Network and my support hardware around the RFM95W does not account for more than 5mA of current. I'm not sure my question if well formulated, but mainly I'm looking for any advices on how to use a CR battery and still manage to power my device. In the meanwhile I'm also looking at replacing by 2x AAA battery, but space is precious. Thank you for your time you might take to answer. <Q> To be blunt, that's really not a great idea, as the current demand is far in excess of what a coin cell can supply. <S> Typically sensors are designed to run for a long time, and so will use higher capacity cells, which are typically able to support proportionately higher peak loads as well. <S> But realize that packets of hundred of milliseconds are common (or well over a second in Europe), so that's going to be a fairly large capacitor. <S> You may even want to go so far as to to have MCU logic which pre-charges <S> the capacitor only in preparation for a transmission - typically the packet duration ends up being calculated anyway, so you could anticipate the overall energy requirement and factor that into the preparation. <A> If you look at my CR2032 equivalent circuit below can you spot your issue? <S> What is your equivalent load current? <S> simulate this circuit – <S> Schematic created using CircuitLab Battery ESR is generally inverse with mAh rating in all batteries but differs for Lithium, Li-Ion, Alkaline, AGM etc. <S> Battery ESR also rises rapidly after 80% DoD until drained. <S> then it is > 100x fresh ESR. <S> I did not bother to model the effective capacitance for the mAh and V rating but this can also be done or define its equivalent storage charge Q or stored energy E to cutoff. <S> Z load needs to be > 300 Ohms to minimize drop to 10% unless there is a large storage cap in theory. <S> In practice, choose a bigger battery with lower ESR. <A> suppose you need to store a watt-second of energy. <S> 2 farad at 1 volt will work or 0.02farad at 10 volts or 0.002 <S> farad at 31 volts its your choice
| If you really want to try to design something that can achieve a very short number of transmissions off an undersized battery, you probably need to use a large capacitor to provide the transmit current.
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Supply current of negative rail I'm using an LTC1983ES6-5 to provide a negative rail for both a MAX4311 and a MAX4395 . LTC1983 provides up to 100 mA, but I cannot find in the other datasheets how much current is required for the negative supply rail of the components. What parameter(s) I should look to? <Q> The datasheet does tell how much the chips itself use. <S> It's the quiescent current section. <S> But what answer you really want depends on how much current is used by the loads connected to these chips. <S> Only you can answer this. <A> The datasheets for both ICs have a section called "output current." <S> Both datasheets give the conditions for that output <S> current: <S> MAX4311: <S> That's 95mA into a 30ohm load when operated at 5V, with 2.5 V output. <S> MAX4395: <S> Again, 95mA. <S> But, into a 75ohm load operated on +- 5V. <S> You have to add in the quiescent current, of course. <S> Also given in the datasheets. <S> It looks to me like <S> your LTC1983 is a little on the weak side (100mA opposed to the 200mA both amps can draw,) at least if you are driving anywhere near the rated loads - <S> and I expect you are, since both ICs are intended to drive video signals into typical video cable impedances. <S> You know your circuit better than we do. <S> If you are driving lower loads, you might get away with just one LTC1983. <S> If you are driving typical video signals, you'd do better with two LTC1983 (or a single converter that can supply more current.) <A> You can start by assuming the supply current for these ICs flows from the positive to the negative supply. <S> Thus all positive supply current must be sunk by the negative supply. <S> This assumption is not correct if the ICs are sinking or sourcing current on other pins. <S> Sometimes current is directed from one pin to another (non-supply) pin. <S> I haven't looked at the data sheets but you should be able to find the supply currents in there.
| Current from other pins is generally directed to either the positive or negative supply.
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In microwave frequencies, do you use a circulator when you need a (near) perfect diode? I am confused with the "diode concept" in microwave frequencies. As far as I could read, I found that even small signal Schottky diodes have a non negligible capacitance, that may perturb the signal from 1GHz and more (well I've seen a few that work up to 20GHz but they require special mounting etc.). On the other hand, there are circulators with operating frequencies from 1GHz to 40GHz, and I think they can be used as a diode since a signal entering at port 1 output at port 2 but the inverse path is blocked. Hence my question: is it what microwave engineers use? <Q> Diodes and circulators are totally different components. <S> Microwave engineers use one when they want one, the other when they want the other. <S> They use diodes when they want to rectify signals, and circulators when they want to isolate ports. <A> A circulator can't perform the same functions diodes can. <S> For example, diodes can be used as an RF switch (pass RF when supplied with a DC bias, block RF and leave a high-impedance output when no bias is applied), which a circulator cannot do. <A> At microwave frequencies one might use a biased pin diode in a duplex port to protect (Block) <S> the Rx LNA input while Tx out on the same coax to the antenna. <S> However, using a Circulator is somewhat like a Directional coupler except it <S> is undirectional <S> ( in only one direction of the circle ) or "non-reciprocal" feeding only to the next port. <S> They are made in 3 port and 4 port version. <S> To achieve isolation and amplification one might choose tunnel diodes or Gunn diodes which due to their avalanche effects have a Negative Differential Resistance <S> Then the diode can block the Rx during Tx mode and amplify during Rx mode with proper bias on the diode in a circulator. <S> This is a half-duplex <S> Rx/Tx microwave method used for higher bands of microwave. <S> [Ref] 2 Ref Ref
| In short, no, one would not use a circulator when one needs a diode.
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Why Does this Short? simulate this circuit – Schematic created using CircuitLab I'm new to electronics and wired a dc motor to a battery and switch. After the battery got really hot, I realized that it was probably a short circuit, but I don't understand why. the positive wire from the battery and to the motor where connected to one side of the switch, while the negative of both where connected to the other side of the switch. The motor turned very slowly when the switch was off and quickly when it was on. Can someone help me understand what's going on or point me to good resources? Thanks! <Q> That is not how you connect a motor to a switch. <S> You connect everything in series like so: simulate this circuit – <S> Schematic created using CircuitLab <S> Such that opening the switch stops current from being able to flow into through the motor. <S> In your circuit, when the switch is closed the switch is much much lower resistance than the motor so most of the current goes through the switch instead of the motor. <S> The current tries to take the easiest path through the circuit. <S> Therefore, the motor gets less power. <S> It short circuits because the battery tries to drive enough current through the switch and motor parallel combination so the voltage across it equals the battery's own voltage. <S> That is how equilibrium is reached. <S> With the small resistance of the switch, this takes a lot of current. <S> More current than the battery can safely provide but it still tries to do it anyways (and heats up in the process). <S> The battery tries so hard that it starts to buckle and the voltage sags. <S> You can now look at why the motor is running with less power from two different perspectives: (a) <S> The battery voltage sagging means less voltage across the motor so less current can be pushed through the motor. <S> (b) Most of the current produced is diverted through the switch around than the battery. <S> So it's really <S> the same effect looked at from two different perspectives since you need sufficient voltage to push a certain current through the motor (or any load). <S> They're not independent and you cannot have one without the other. <S> When the switch is open, all the current flows through the motor <S> so it operates normally. <A> simulate this circuit – Schematic created using CircuitLab Note <S> this is assuming ideal parts. <S> Reality check: every part has resistance. <S> Battery (ESR) , wire {mOhms/meter} Motor DCR coil resistance and switch. <S> Each have current limits <S> Current in loop <S> I = <S> V/R on startup then reduces and depends on mechanical load and RPM. <A> simulate this circuit – Schematic created using CircuitLab <S> In other words, the switch bypasses the motor, thus causing a short since there is 0 resistance between the + and - terminals on the battery, but there will always be greater than 0 resistance on the motor. <S> Remember, electricity flows through the path of least resistance. <S> Good question, Benyomin!
| If the switch were zero resistance, all of the current would go through the switch and none would go through the motor. The easier the path is relative to the alternate paths, the higher the proportion of the current will take that easy path.
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Can I connect a DC high voltage booster directly to my Raspberry Pi? I have one of these modules: Is it safe to connect it to two pins to my Raspberry Pi, directly, and turn it on by emitting HIGH on one of the pins (the other one being the ground), which is around 5V, if I am not wrong? Will that use too much power from the Raspberry Pi or maybe burn it? <Q> The advert says it needs 5 Amperes. <S> That’s hundreds of times more than a Raspberry Pi GPIO can supply. <S> You would need a driver, preferably with isolation, and a separate power supply capable of at least 5A. One solution would be a CPC709J , with a suitable series resistor for the LED. <S> You may well cause disruption or damage even with that, but the chances are better. <S> You’re essentially putting a multi-watt spark-gap transmitter in close proximity to an unshielded microcomputer. <A> Apart from not having enough energy to drive that circuit: Generating sparks near a piece of electronics is never a good idea. <S> Although the circuit has protection on some of the I/O ports, especially the HDMI and USB interfaces, there is non on the GPIO pins. <S> Those have the standard ESD protection which is designed for, well... ESD. <S> It is NOT designed to handle the energy which comes from huge voltage sparks. <A> The micro-lightning arc HV noise generator needs 1 or Li-Ion cells to power it. <S> If should never be operated near any computer.
| If it operates near an R-Pi, it will cause functional failure and possible damage to signal ports on cables acting as an antenna.
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Tradeoffs for creating higher order butterworth filters I am working on a bandpass filter to filter the output from a hydrophone. This is my first time designing a filter and using a cookbook from TI , I built a 4th order Butterworth filter with two partial 2nd order MFB filters. The filter is working as expected when tested in ltspice but the rolloff is not as steep as I want it to be. To get a steeper rolloff, I am thinking about cascading 2 or more of these 4th order filters and create higher order ones. Is that ok or are there any tradeoffs for doing so, other than the increased complexity of the circuit? Will the PCB layout difficulty increase? Are there any better approaches to get both a flat response at mid frequency (the bandwidth is only 10% of the mid frequency) and a good rolloff? Is using a Chebyschev optimization instead of Butterworth a better idea for my use case? <Q> Cascading two 4th order <S> Butterworth filters does not give you an 8th-order Butterworth filter. <S> If you want the benefits of a Butterworth filter (maximum passband flatness), then you should go back and design an 8th-order Butterworth filter. <S> Is that ok or are there any tradeoffs for doing so, other than the increased complexity of the circuit? <S> Will the PCB layout difficulty increase? <S> More components means more PCB area. <S> As another answer points out, more stages means more noise. <S> Is using a Chebyschev optimization instead of Butterworth a better idea for my use case? <S> A Chebychev filter will let you trade off passband ripple for steeper roll-off. <S> Presumably in the limit that you specify minimum ripple <S> you will just get the Butterworth filter ( <S> but I'm not 100% sure on this point). <S> Only you can decide what is the correct balance between roll-off and ripple for your application. <S> Are there any better approaches <S> If you are targeting some well known application where there are many customers who need the same type of filter, there may be a drop-in filter component available to suit your needs. <S> This could be a passive resonant structure that can act as a high-order filter in very little board area, and with no noise sources. <S> The suitable technology and availability will depend on your exact center frequency and pass-band requirements. <A> The main thing is the number of amplifiers that must be used to create the filter. <S> In general adding more amplifiers will add more noise. <S> Do a noise analysis in spice to get a basic idea for what the noise is (and be careful because some manufacturers models do not accurately reflect real world noise values or noise values found in datasheets). <S> If SNR is a concern then the noise added by filtering must be considered. <S> Other than that, adding stages to a filter and adding amplifiers will also increase the obvious design parameters as you've mentioned like PCB area, power consumption, ect. <A> Because of tolerance issues in capacitors, high-order multipole filters are notoriously difficult to actually build, despite working well on paper. <S> For this reason, the industry provides us with filter building blocks that can be tuned entirely with resistors. <S> The UAF42 is one such device. <S> This makes filter design much easier. <A> At high frequencies, the opamps can no longer control their output signals, and the attenuation likely becomes ----- not what you are expecting. <S> Make sure the tools are modeling the gain-rolloff of the OpAmps.
| To get a steeper rolloff, I am thinking about cascading 2 or more of these 4th order filters and create higher order ones.
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What is the safest way to trigger an output? I need to trip a 12V output with a signal. In other words, I will send a signal through the output of my microcontroller and this signal will activate its output. I've figured out several ways to do this, the most common ones being activation with relay or activation with Mosfet. As in the examples below. My concern is reliability, I need to make sure that the output is triggered. How many drives per day will be needed to ensure that the system supports a large number of activations? How can I know which of the forms of activation is the most reliable? Which one will guarantee the activation of the output without fail and which supports the largest number of activations? EDIT- The load connected to "output12v" is an electronic lock that consumes around 1A. simulate this circuit – Schematic created using CircuitLab simulate this circuit <Q> Either circuit has failure modes and you'd have to analyze possible abuses such as transients on the 12V line and possible shorts on the output to have some hope of deciding between them. <S> MOSFETs can fail at any time, but in a benign environment, and if minimally stressed (lots of margin on voltage ratings, minimal temperature cycling from self-heating etc.), have few wear-out modes. <S> Relays tend to fail 'off' when they wear out (but they can stick) and MOSFETs tend to fail 'on' (but not always). <S> If the consequence of failure is an unsafe condition for people or significant property damage, you should not use either circuit alone. <S> Either add redundancy or acceptably mitigate the damage with other means such as fully independent limit controllers etc., preferably using a different physical mechanism to operate. <S> For example, I have a molding machine. <S> The consequence of the platen closing unexpectedly could be a crushed operator. <S> There are three interlocks- one electrical, one hydraulic and one a mechanical rod that physically blocks the clamping action. <A> Relays are good for galvanic isolation. <S> But if you don't need to isolate switched voltage from voltage that powers your microcontroller, then I thing the more suitable <S> is solution with MOSFET. <S> Also, you can easily find MOSFET that can hold 1 A. <A> I am in the MOSFET camp. <S> If you want high reliability you can parallel two MOSFETs and drive them with separate signals -- one as the primary switch, and one as the backup. <S> I would also suggest a PGOOD (power good) type signal as feedback to the MCU that the +12 V has indeed been supplied. <S> simulate this circuit – <S> Schematic created using CircuitLab
| Relays are electro-mechanical components, so thay could break after less switchings than MOSFETs. Relays are more rugged but have a limited and fairly predictable life.
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Why is this camping light deadly? I am learning, so go easy on me, if you can. In a video with BigClive, he takes apart a camping light and says it's deadly when you have mains plugged in and usb plugged in at the same time. In that situation the USB can be live. Here's a picture of the circuit he draws in the video. I am assuming it is deadly because it uses a capacitive dropper and if the plug is inserted backwards it doesn't drop the voltage and the USB is sitting at 240v? Or is there another reason? <Q> Most outlets aren't polarized. <S> Even in places where the outlets are polarized, mistakes happen. <S> There's a straight connection between the outlet and the ground of the USB port. <S> All there is in between the two is a diode. <S> If you plug it in the other way around (or the polarized outlet is wired backwards,) <S> then you have the full line voltage on the USB ground. <S> That can be deadly. <S> Additionally, even if it is connected the right way around, all that's between you and line voltage <S> is a capacitor. <S> If it is a Y rated capacitor and it fails, then you are safe <S> - Y rated capacitors are supposed to fail open. <S> They almost always fail open, but do you want to bet your life that a company that builds a capacitive dropper powersupply with user accesible connections will also think to use a Y rated capacitor? <S> Even if it doesn't kill you, it can destroy any laptop or other USB source you connect to it. <A> The impedance of that capacitor at 50Hz is about 3200 Ohms. <S> Which means a current of 75mA if I have done the maths correctly. <S> The problem is that the voltage at the user end will rise until it either gets to 240VAC or it drives that 75mA through whatever - including you. <S> Which is quite possibly fatal. <S> There is nothing to "tap off" that current to keep the voltage within sane limits as far as I can see. <S> Normally there would be something like a zener to do that. <S> They seem to be relying on the LEDs <A> Is it only deadly when plugged in incorrectly? <S> Neither the mains plug nor the inlet connector appear to be polarised, there appears to be no indication of which way around is correct <S> is no indication given of which way round is "correct". <S> I am assuming it is deadly because it uses a capacitive dropper and if the plug is inserted backwards it doesn't drop the voltage and the USB is sitting at 240v? <S> If the plugs are inserted such that the live connection goes to the rectifier and the neutral to the capacitor then everything connected to the output of the rectifier will be live. <S> If the plugs are inserted such that the neutral connection goes to the rectifier and the live to the capacitor then you don't have the immediate danger, but you still have a product that would not pass safety standards. <S> If the neutral were to break, the components that absorb the current were to fail open-circuit or the dropper capacitor <S> were to fail short <S> then you are back to having a live output.
| If that wire is connected to neutral in the outlet, then nothing bad will happen if you touch the USB ground.
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Series resistor on digital signal lines I have seen many schematics using 22 ohm resistor in series with digital signal lines. I was wondering is this always a MUST or there are there some special scenarios where we don't need them since they lower the signal's bandwidth. For instance, are they needed on 48MHz signal line which is cca. 2cm long? Is this somehow connected to STM32 output speed capabilities (low, high, ultra high speed, etc.)? <Q> Here is a simulation of 50MHz Clock on 2cm track of 8:1 L/W ratio for the trace. <S> L,C is distributed track inductance and the capacitance ratio for a known 3.3V logic <S> Vol/Iol=Zo driver impedance. <S> are they needed on 48MHz signal line which is cca. <S> 2cm long? <S> Maybe. <S> Do you feel Lucky? <A> These are series termination resistors. <S> Yes, the STM32 IO output speed can alleviate the need for these on corresponding I/O. <S> In Henry Ott's book "EMC Compatibility" he recommends that any clock traces faster than 20MHz should have a series termination. <S> He also recommends that even on very short clock traces, you should still add a series termination resistor (a ferrite bead works too) so long as it does not make the trace longer. <S> My board has a 24MHz oscillator literally cannot get closer to the MCU pins without being on top of them. <S> It has no such termination and it works just fine. <S> At 48MHz it probably won't stop your system from working, but your system will make more noise than it needs to without it. <S> He says the typical value is 33 ohms. <S> The ideal value depends on your the characteristics of the output pin, the input pin, and your PCB layout. <S> The easiest way is to try various values (including zero ohms just so you have a reference) then scope the trace at the resistor nearest to the input pin and <S> the input pin itself and observe the waveform integrity. <S> Your scope needs to be of sufficient high bandwidth relative to your clock frequency to see what you're looking for (ringing, overshoot, etc). <S> BTW the series termination resistor needs to go as close as possible to the DRIVING end of the line. <S> This is important. <A> Although they are somewhat related, the edge rate is more important than the frequency. <S> A low frequency line may still need a terminator if the edges are fast. <S> There are some rules of thumb that I don't remember off the top of my head, for 20+ years, other designers have been handling these details for me (edit, I looked it up in the book, see below). <S> The other important factors are the trace length, and how tolerant the receiving end is to overshoot/ringing. <S> Electrical noise generated from ringing is usually a secondary consideration. <S> A clock input is usually not very tolerant to ringing (can double-clock). <S> On any input, the overshoot may exceed the recommended input voltage rating of the part. <S> In some cases, the part can be damaged or malfunction. <S> Other parts seem to work just fine when the voltage rating is exceeded. <S> If you are designing something critical, you should follow all the rules, if you are hacking on a home project, you can bend the rules more. <S> For a series terminator, the source impedance of the output plus the terminator resistance should equal the characteristic impedance of the trace. <S> Typical values are 20, 30, and 50 ohms respectively. <S> An excellent book is: High-Speed Digital Design, A Handbook of Black Magic, by Howard Johnson and Martin Graham. <S> According to the book: If the line length exceeds one-sixth of the electrical length of a rising edge, terminators are needed. <S> Propagation speed in a typical PWB is 2 ns per foot, or 1 ns per 6 inches. <S> So, a signal with 1 ns rise time should have a terminator if it is longer than 1 inch. <S> A 48 MHz square wave probably has a rise time of about 2-4 ns. <S> Much longer and it will be a really crappy square wave. <S> It could be shorter, look at the specifications of the driving component. <S> Assuming 2 ns, 2 cm is quite a bit shorter than 2 inches, I don't think that a terminator is necessary.
| They are there to manage transmission effects line such as ringing and oscillations from signals that have fast rise/fall times relative to the length if the trace they are running down. In your case, at 48MHz and 2cm, you probably have enough trace length to add a series termination without making it longer so you should have it there.
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Current Direction in Mosfet As I know, the direction of current in N channel mosfet is from drain to source. Now, how this current can be pass in this circuit????SO, why most of the reference books say, in n-channel current flows from drain to source, and for p channel from source to drain?? <Q> Similar effect takes place in a P-Channel transistor. <S> In the shown circuit the N-Channel FET diode prevents current to flow from the output back to the input in a paralleled supplies configuration. <S> The FET get's turned OFF when the input voltage drops below the output voltage minues some small voltage. <S> In normal operation the FET is ON, bypassing the internal diode and ensuring minimal losses. <A> No, current can flow either way between drain and source, as long as V GS is above the threshold voltage. <A> Next to the answer of Dave Tweed, the body diode also conducts from source to drain, independent of the gate voltage. <S> The body diode cannot handle huge currents, therefore in shown application it is 'shorted' by a better diode which will take most current into account. <S> As soon as the LM5050 senses current is flowing through the diode(s), the LM5050 opens the mosfet, allowing for smaller voltage drop of <S> \$I <S> \cdot R_{DSon}\$ instead of the diode's voltage drop. <S> The LM5050 has fast response comparator to turn off the FET when current flows in the reverse direction. <S> For applying ideal diodes, it is important to sense if the input voltage drops below the output voltage and to close the mosfet. <S> Otherwise, the ideal diode will act less ideal than you thought, i.e. conducting the current both ways instead of only one way like an ideal diode would. <S> Regarding the textbook stating current will flow from drain to source: I don't know if that statement is true. <S> Mosfets are normally applied such that the body diode will not conduct. <S> This is the case when the drain voltage is higher than the source voltage. <S> The latter implies current will flow from drain to source when enough gate-source voltage is applied.
| The reason reference books say that N-channel current flows from drain to source is that when the transistor is OFF the intrinsic diode of the transistor pevent the current from flowing from source to drain. When the transistor is ON, the current can flow in either direction as the diode is effectively shorted by the Rds(ON) of the drain to source channel.
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Confusion about Ohm's law and electrical power (basics of electrical engineering)..? I am studying electrical engineering, but I still have some doubts about Ohm's law and electrical power. According to Ohm's law: \$I=\dfrac{V}{R}\$ So that means voltage is proportional to current. Higher voltage = higher current. Now I was taught that in order to transmit on power lines, the voltage needs to be transformed higher so that according to \$P = U * I\$ , current decreases and there is less heat loss. But wasn't current proportional to voltage? How can a higher voltage not induce a higher current? For another example let's take a 100 W light bulb. I was told that if I have a 10 V Voltage, the light bulb "will drain" 10 Amps in order to function. Is this wrong? Is 100 W the amount of maximum power the light bulb can handle or the power it needs, and how can it "take" 10 Amps if current depends on the applied voltage and bulb resistance? I don't understand how current doesn't depend on the applied voltage according to Ohm's law. <Q> You are confusing some things (which i understandable). <S> In a simple resistive circuit, then with some voltage source V and a resistor R, the current is indeed \$I=\frac { <S> V}{R}\$ <S> Increasing the voltage will indeed increase the current. <S> If we need to do a transform to transmit power, we use a transformer: simulate this circuit – <S> Schematic created using CircuitLab <S> A transformer is passive; apart from inefficiencies, the output power and input power are the same, so if we had 1kV input voltage and a 10:1 step up transformer <S> we would have a 10kV output. <S> (We will get to the current shortly). <S> At the other end, if we had a 100:1 step down transformer, we would have 100V out. <S> Let's say we have a 100W light bulb here, which would draw 1A. <S> Now work backwards to see what the input current is: Input to the step down transformer is 100W at 10kV = 10mA and the input to the step up transformer is 100W at 1kV <S> = 100mA. <S> Ohm's law still holds. <A> Because transformers aren't resistors. <S> V=IR only works for resistors. <S> It is meaningless for any other sort of component, including semiconductors, transformers and the like. <S> It's even misleading for resistors that get hot, such as tungsten filament lamps, as the resistance changes with temperature. <S> In the case of power lines, then the lines themselves do have a resistance. <S> But if there is a transformer at the other end of the line, then just looking at the resistance of the line itself <S> gives the wrong answer. <A> All of the answers above are good ones. <S> My only contribution will be to point out that the raison d'être for a power line is to maximize power transmission and minimize losses. <S> Ohm's law not only gives us the simple E=IR relationship, but can be extended to power. <S> Of particular interest for us is the variant, P= <S> I^2*R. <S> In other words, the power lost through the transmission line is equal to the square of the current times the resistance. <S> In general, there's only so much that can be done to minimize the resistance of the wires. <S> But setting that aside, the impact of reducing the current through the lines is tremendous as the power loss is reduced by the square of the current reduction. <S> Since P= <S> IE, you can transmit the same power either at high current and low voltage or at low current and high voltage. <S> So by multiplying the voltage by 100, you can reduce the current by a factor of 100, and then the lost power is reduced by a factor of 10,000 (assuming all else is kept equal. <S> It won't be, though, because less current means you can use thinner conductors and higher voltage <S> means you have to use thicker insulation). <A> If you double the transmitting voltage you only need half the current pulled through the wire in order to transmit the same power (P=VI). <S> Half the current through the wire means the power dissipated in the wire has gone down by a factor of 4. <S> (P=I^2.R). <S> Or alternatively, if the current through the wire has halved then the voltage dropped across the wire has also halved which also gives a factor of 4 reduction in power dissipation in the wire (P=V^2/R). <S> And the current it supplies is that pulled by the consumer. <S> With regard to the light bulb, The light bulb has a certain resistance and so at a certain supply voltage it will pull or draw the current it needs in order to satisfy Ohms Law. <S> The power dissipated in light and heat will be V.I, I^2.R or V^2/R.
| Your confusion is that you are thinking about the transmitting voltage as though it is all dropped across a resistor which it is not, it is just a generated voltage.
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How do PCB vias affect signal quality? Is it bad practice to route high speed signals (like a SPI bus clocked at 4MHz) through PCB vias? I've noticed a good bit of noise (+-300mV) on my SPI bus signals with 3.3V levels. The signals traces are only about 5cm long but they go through about 5 vias each on the way to their destination. The board has only 2 layers which is why there are so many vias on these lines. What kind of noise can I expect (if any) to be introduced by a PCB layer change via? Lots of good information in the answers. It's going to be hard to pick only one. Given that a PCB via introduces about 1.2nH of inductance and 0.4pF of capacitance the consensus seems to be that the 5 via's wont affect a 4MHz signal in any significant way. <Q> I'm a novice when it comes to higher speed signals, but it just so happens I was researching signal integrity when you asked the question. <S> One source I am referencing is Right the First Time by Lee Ritchey . <S> You will want to check out chapter 25, Right Angle Bends and Vias: Potential Sources of Reflections and Other Problems . <S> I don't believe the vias will cause any problems in your design. <S> Here is an excerpt from the source: <S> Vias, when used in traces, are capacitive, not inductive. <S> The capacitance value of a via is small compared to the capacitance of a trace (3.5pF/inch for 50Ω). <S> In general, vias are not visible to signals with edge rates slower than 0.3 ns. <S> The chapter goes on to discuss reflections due to PCB layer impedance mismatches, however this appears to be a case when manufacturing tolerances are not met. <A> 300mV is a lot for a 3.3V bus. <S> Transmission line effects don't become noticeable until 50MHz, so 4Mhz should be fine. <S> The most common problem on two layer boards is common mode noise from improper grounding (daisy chaining grounds) and common mode noise. <S> So I would first look at the grounding system in the design, make sure that currents don't create common mode noise through small traces that are daisy chained. <S> The other problem might be with grounding and where the scope ground is placed. <A> The issue is not the SPI clock being too high frequency (4 MHz). <S> It could be 0.1 Hz and the signal edges would still ring, as it's the edge rate that defines the bandwidth. <S> Typically microcontroller IO pins are moderately strong, and can drive for example a 30pF capacitive load with 4ns rise time or 10pF capacitive load with 2.5nS rise time. <S> That's strong enough to drive 100-120MHz signals out from a MCU, according to STM32F207 datasheet. <S> What you may be missing is that if your MCU does not have settable pin drive strength, you can slow down the rise/fall times to sane levels by putting for example 33 ohms series terminating resistors at the device that is driving the pins. <S> This way the edges need less bandwidth and there is less ringing. <S> 4MHz SPI running for 5cm of length should not be an issue, but do check what rise/fall times your chips need to work. <S> Another issue is that your oscilloscope might show ringing for signals <S> just because the scope or probes have 100MHz BW limit and signal edges are fast enough to go over 100MHz BW limit. <A> 5MHz is slow. <S> But the bandwidth of the signal depends on risetime. <S> BW=0.35 <S> / <S> Tr <S> so it is 10ns=0.01us <S> the BW= 0.35/0.01us = <S> 35MHz <S> But if the signal was HDMI or CML logic or even just 1ns risetime, then ; BW= 350MHz <S> Then we have two Rules of Thumb more maximum path length to ignore reflections from vias or long traces; 1 <S> : 1/10 Lambda <S> the 1ns rise time is using v=c <S> /sqrt(Er) <S> - max path length is 8.5 cm Slewrate /4 <S> max path length is 4.5 cm <S> For better analysis use some calc tools like Saturn PCB.exe or analysis tools using the ESL,ESR,C(pf) of your via inductance and capacitance into a model to see the result using the VOl/Iol=Ron driver impedance. <S> Then model into your favorite simulator. <S> Mine is Falstad's Your results are ONLY as good as your model values as FALSTAD uses ideal voltage sources and wires are ideal. <S> So you add R,L,C values to suit your model.
| Vias will not cause a problem as a via only adds a few nH of inductance and if the capacitance on either end is lower than 100pF and a trace that short would be under 0.1Ω which would make an RLC resonator at around 1GHz, and you won't see it.
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Modelling the equivalent series resistance's voltage in a circuit Say a 2H inductor has an equivalent series resistance of 2000 ohm at a particular frequency. Logically to me, that inductor would be modeled as an inductor in series with a resistor. However, that doesn't quite work in my mind. For example, that would imply that the resistor has its own voltage across it, when (I believe) it is the same as the voltage across the inductor itself. If that's true, then it would be a parallel resistance, not a series resistance. I obviously misunderstand something. How do you model it? <Q> The "equivalent series resistance" in the model is not physically localized like a real resistor, so it doesn't have its own voltage drop. <S> The point of including it in the model is that, if you treat the inductor as a black box (i.e., you can only access the terminals, not the internals) it behaves the same way as an ideal inductor in series with a resistor. <S> It makes the circuit analysis easier, but that doesn't mean it's literally true... <S> it just seems that way until you look inside. <A> The resistor does have the same voltage as the inductor. <S> The only way it would have a different voltage accross it would be for the inductor to have a resistive component. <S> In the model of a series inductor and resistor the inductor is ideal (ie only reactive impedance <S> no resistive) <S> therefore the voltage at the resistor is the voltage at the inductor. <A> Consider an inductor in the real world it is a coil of wire <S> and this wire has resistance distributed along its entire length. <S> Similarly if you add up all the resistances you get the ESR (Equivalent Series Resistance). <S> You therefore can never measure the inductor voltage or the resistor voltage because there is no point in the circuit to measure it. <S> We can only measure \$ <S> V_R + V_L \$ . <S> Mathematically we can model this as a single inductor in series with a single resistor. <S> For almost all cases this is good enough; only if we are trying to model the H and E fields around a component to we need any more detail. <S> The voltage across the ideal single inductor \$ V_L \$ and the ideal single resistor <S> \$ <S> V_R \$ will be different and frequency dependant. <S> We must not confuse this with the AC impedance however \$ <S> X_L = 2 <S> \cdot <S> \pi <S> \cdot f <S> \cdot <S> L = <S> \omega <S> \cdot <S> L \$ . <S> This is sometimes modelled as a resistor but does not account for the \$ <S> 90^o \$ phase shift.
| The best model is therefore to consider a tiny inductor in series followed by a tiny resistor and this is repeated infinitely many times such that if you were to add up all the inductances you get the total inductance.
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Power Ratings of RF Attenuators In Series If I connect RF attenuators in series, does the power rating of the attenuator closest to the antenna have to be high enough to handle the expected RF power on its own? Or does the placing attenuators in series limit the amount of power each attenuator must dissipate by spreading it out amongst attenuators? For example, if my expected RF power is 2W, do I need to use a 2W attenuator nearest to the antenna even if there is going to be a 1W attenuator between it and the receiver? Or can I get away with a 1W for the attenuator nearest to the antenna? <Q> A 2W attenuator is not designed to dissipate 2W, it's designed to receive 2W. <S> The amount of power at its output will be determined by its attenuation. <S> A 3dB attenuator would output 1W. A 10dB attenuator would output 200mW. <A> Yes, the initial attenuator still must handle the power level present. <S> Attenuators are not just series resistors, they are dividers with shunt resistors as well, so they do not combine like resistors in series. <S> It is convenient that you can add the logs of the power ratios when combining attenuators - but the actual power converted to heat depends on the absolute power level input, not just the ratio of the input to output. <S> To the degree that each component is ideal, the power the first attenuator converts to heat is the same regardless of what follows it, as long as what follows matches the impedance in a purely real way. <S> In a mismatch case it could actually be slightly higher. <A> Just to concrete a little bit more answers above, if you want to handle 2W at the input of your first attenuator, you should look at the máximum input power level figure in the Maximum Absolute Rating section of your attenuator, and not to choose one with a 2W value but higher.
| If you cascade attenuators, each attenuator only needs to be rated for the expected output power of the previous one.
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High Q peak in frequency response means what in time domain? Reading Linear Circuit Transfer Functions and one of the graphs got me curious. I've recreated the circuit (series RLC) and plotted the frequency response for a Q of 7. We have a peak of ~16.3 dB when Q is 7 @ 10Khz. Can this value be used (16.3 dB) to accurately predict something in the time domain - such as the value of Q or how long the oscillatory decay would take, the amplitude of the oscillations etc.. ? Added in case its relevent <Q> Q is (among other definitions) <S> the voltage gain at resonance, and a voltage gain of 7 times is $$20 * \log(7) = <S> 16.9dB$$ <S> which seems close enough as your cursor is clearly not actually on resonance (phase would be -90 not -93). <S> So dB of resonant gain is trivially converted to or from Q. Q gives you risetime and whether the circuit is over/under or critically damped in the time domain, as well as how well damped the ringing in an under damped circuit is. <A> The above answer ("Q is..... <S> the voltage gain at resonance") is definitely wrong. <S> The relation between the quality factor Q and the magnitude peak in the frequency domain for a 2nd-order lowpass/highpass is as follows: Amax=(Ao <S> * Q)/sqrt[1-(1/4Q²)] with Ao=DC gain. <S> For a bandpass filter the Q value defines the 3-dB-bandwidth of the circuit. <S> TIME DOMAIN <S> In the time domain, the Q value determines the step response as follows: (1) <S> For Q>0.5 the step response shows an overshoot "gamma" above the final value (when the transient has settled). <S> This "gamma" value is given in % about the final value. <S> "gamma"=100 * exp[-3.14/sqrt(4Q²-1)] Examples (gamma values in brackets): Q=0.5(0%); Q=0.7071(4.3%); Q=1(16.3%); Q=10 (85.4% (2) <S> The oscillatory decay is determined by the real part ("sigma") of the pole position only: exp(-|sigma|t) . <S> The relation between "sigma" and the Q value is |sigma|=wp/2Q with wp=pole frequency. <A> I would give a slightly different definition for the quality factor \$Q\$ : \$Q=2\pi\frac{(stored\;energy)}{(energy\;dissipated\;per\;cycle)}\$ <S> The series resistance represents the circuit losses. <S> When the circuit is energized with the input stimulus, oscillations take place as an energy transfer swinging back and forth between \$L\$ and \$C\$ . <S> If the resistive term is 0, there are no losses and oscillations keep for ever: <S> \$Q\$ is infinite and poles are imaginary with no real parts. <S> As \$R\$ increases, you start dissipating energy in heat and you damp the circuit, bringing oscillations to 0 after a few cycles. <S> The measurement of the peaks and valleys via the logarithmic decrement \$\delta\$ <S> lets you also compute the quality factor: \$Q=\sqrt{(\frac{\pi}{\delta})^2+\frac{1}{4}}\$ <S> These peaks and valleys are directly linked to the energy dissipated in \$R\$ .
| There is only one single definition: The quality factor Q is the so called "pole-Q" - defined by the pole position in the complex frequency domain (s-plane).
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Active filter with series inductor and resistor - do these exist? I have seen (online) and in books basic active filters (involving an op-amp) with a series R C branch at the input side. This series R C branch at the input side would make the circuit a high-pass filter. Series RC configuration at the input is quite commonly seen online. However, there appear to be no circuit diagrams seen online involving series R and L (resistor and inductor in series) for the input branch, as shown below in the drawing. There is likely some reasons for no diagrams like this seen online - anywhere. I don't know those reasons, and definitely interested to know why. Anyone know why op-amp configurations with series RL inputs are not seen? Thanks in advance for any comments! <Q> Inductors are generally more expensive, more bulky and less ideal than capacitors <S> so you'll usually see a strong preference for capacitors over inductors. <S> Only very tiny values can be put into an IC, whereas useful values of capacitors can be integrated. <S> Sometimes as inductors, sometimes as ferrite beads (which are inductive over a range of frequencies) and sometimes as common-mode chokes. <A> To add to Sphero Pefhany's reasons, the circuit as shown has a gain at DC, which may cause issues with the following circuitry because of any DC offset voltage in the amplifier. <S> Also, if the power supply is disconnected at the instant while current is flowing through the inductor, the induced (possibly large) voltage spike may damage other components. <S> Adding protection diodes etc in a simple way would most likely make the response of the circuit nonlinear. <S> Circuits using capacitors are immune from issues like the above. <S> Note that using active circuit elements like op amps, it is possible to create a "virtual inductor," as in this example circuit. <S> Such virtual inductors can have a wider range of values and be closer to ideal components than real inductors, are unaffected by stray magnetic fields, and do not have the same issues with high back EMFs. <S> Unless the "non-ideal" and/or non-linear behaviour of a real inductor is essential for a circuit to function correctly, it is not necessary to use an inductor at all. <A> The primary reasons for not using inductors in filters is that it is bulky. <S> It takes up considerable amount of space as compared to resistors and capacitor if fabrication is done. <S> Cost is also a factor. <S> There is also a fundamental reason for not using inductors. <S> For an inductor, impedance goes up with frequency. <S> It behaves as a short circuit at low frequencies, and an open circuit at high frequencies which may be pose a problem for the user. <S> So, operating frequency has to be selected properly for proper operation. <S> Other reasons includes inductors are known to show parasitic effect i.e if current is passed through inductors, magnetic field is generated which may change the impedance of other components and as a result the net impedance of circuit is changed. <S> So, cutoff frequency also changes. <S> Also, inductors used in circuits do not possess much of an inductance. <S> So, if they are to be used in filters, cutoff frequency do not change a lot. <S> So, presence of inductors in filter do not make much of a difference. <S> Inductors are severely nonideal as compared to resistors and capacitors and this nonlinearity may pose problem when deriving the transfer function of the filter. <S> Lastly, inductors generate electromagnetic interference(EMI) than resitor and capacitor and also they are more susceptible to EMI. <S> I could find a parallel R-L connection in a filter which is a 1st order active HPF and also an inductor in feedback present in a 2nd order active cascade HPF. <S> References https://www.cs.cmu.edu/~tdear/ee/filters.pdf <S> (pictures) <S> http://www.swarthmore.edu/NatSci/echeeve1/Ref/FilterBkgrnd/Filters.html http://alignment.hep.brandeis.edu/Lab/Filter/Filter.html <S> https://www.allaboutcircuits.com/technical-articles/inductor-out-op-amp-in-an-introduction-to-second-order-active-filters/
| However, series inductance is frequently used where conductors enter a shielded enclosure or signals enter a PCB.
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What kind of sensor that can detect prolonged contact? i'm currently working on project that needs an implementation of certain touch/force-presence sensor on human body surface. I want to place the sensor on my backpack shoulder straps, so that the electrical system of my smart backpack will know, whether the backpack is currently being worn or not. I have already tried to use piezo electric sensor, but it seems like it can only detect an impulse signal; output signal HIGH when a force is momentarily detected, then go LOW even if the force is still detected (only detect the 1st contact). I'm curious, is there any type of sensor that can detect prolonged contact, so that the sensor will always give HIGH signal as long as the force is detected? Thanks!! Edit: There's also space constraint, so i need the sensor to be as thin and small as possible <Q> You could use a force-sensitive resistor . <S> They come in a variety of shapes and sizes . <S> Their resistance will vary based on the actual force, so with appropriate thresholds you should be able to detect whether the backpack is being worn. <S> There are also sensors that detect bends, known as flex sensors , but I'm not sure they would be appropriate here. <A> I would suggest you fit a thin tube in the contact region that is closed one end and terminated at the other end in a pressure transducer. <S> The transducer can be remote and protected and it will measure the compression on the tubing. <A> A different approach would be to use an accelerometer to detect when the backpack is moving around, and assume that means that someone is wearing it. <S> More of less clever algorithms could be implemented to detect if the person stands stills or is walking with the backpack. <S> Bonus: <S> This will inevitably be much more robust and less error prone over time as there is no mechanical slidtage compared to the more mechanical/physical solutions suggested. <A> You could measure the presence of the human body using a capacitive touch sensor. <S> A backpack which is worn by a human is going to have characteristic orientation with respect to vertical. <S> It may also "see" characteristic motion. <A> Disclaimer: this response is based on a 30 year old memory. <S> 3M used to make a pressure sensitive conductive film (actually a sandwich of two films). <S> The harder you pressed, the lower the resistance. <S> Other than samples, I don't think there were any standard products; they produced it only in volume for custom designs. <S> I suggest looking into this technology.
| You could also use an accelerometer as an additional sensor. A capacitive sensor can see the difference between the presence and absence of the human body even when it separated from it by several layers of fabric.
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Truth table with logical gates for a traffic light I want to create this truth table: So I tried this but I was not able to get the last row, which is: While a =1 and b =1 ,yellow = 1 red = 0 and green = 0. This is my design for it, but I could not make the last row. Does anyone have a sample or a tip about what I can do to fix this? Thanks! <Q> YELLOW = \$B\$ . <S> RED = \$\bar A\$ . <S> GREEN = <S> \$\overline <S> {RED + YELLOW}\$ Purple is the implied logic for the red light, orange is the implied logic for the yellow light and green is for the green light. <A> You can write the equations easily from the truth table <S> $$ <S> Red = <S> \bar <S> A $$ <S> $$Yellow <S> = <S> B$$ <S> $$Green = <S> A \land <S> \bar <S> B$$ <S> They can be built the following circuit : <A> I do have a tip for you! <S> Work column by column, not row by row. <S> First, create a circuit for the red light, ignoring the other two lights. <S> Next, do the same for the yellow light. <S> Finally, do the same for the green light. <S> After you've created these three circuits, you can simply combine them into one. <A> I will be honest and say, i am not all that good at designing logic diagrams, way too long since i had it in school. <S> But there is one thing i remember: Logic Friday. <S> Its a freeware program that is an immense help when designing such diagrams and solving the equations. <S> Using Logic Friday, i got these results: A - input 1 <S> B - input 2 F3 - red F4 - yellow F5 - green Logic input gave me this truthtable after minimizing: F3 = <S> A'; <S> F4 = B; F5 = <S> A B'; <S> And this diagram followed: <S> I know a program like this is no substitute for real knowledge, but it works for a home fiddler like me. <S> Good luck :)
| One OR gate and two inverters would do the trick.
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PIC IOC vs INT and the interrupt function I am using PIC18F26K83 and I want to use interrupts for a switch that will be connected to one of the pins of my PIC. So I have some related questions: There are 2 interrupt options in PIC18F26K83: IOC vs INT. What is the difference? I heard that is quite easy to miss edges with IOC though. Can I use both of them at the same time? Other question is related to interrupt function. If I have 2 interrupt functions : interrupt i1(void) and interrupt i2(void), in case of any interrupt, does the software enters all the interrupts one by one? If not how does it get which function is related to coming interrupt? For example I am using CANBus interrupt, I2C interrupt, timer interrupt and one external interrupt coming from a switch. If I press the switch, how will software know which interrupt function is related to switch? Thanks beforehand. <Q> There is an interrupt flag associated with the interrupt you choose to use. <S> That will tell you which pin was responsible when you service the interrupt (it is set automatically by the pin change). <S> You will reset it in your ISR. <S> That ensures you won't miss an pin change interrupt if your ISR is properly written. <S> But if the switch is bouncing you'll get additional interrupts when the interrupts have been re-enabled. <S> Maybe sometimes, maybe not other times depending on what the processor is doing. <S> Maybe you'll reset the flag and it will immediately be set again by the hardware before you even exit the ISR, then the interrupt will occur immediately when you exit. <S> You could put a (blocking) delay routine in your ISR <S> but I think most real-time programming folks would be nauseous or possibly physically ill by that point. <S> Generally manual switches should not be serviced by interrupt routines. <S> Institute a periodic interrupt (typically no faster than 1kHz, and maybe a bit slower depending on the switch- <S> you can also check every n'th time through with a 1kHz interrupt) and you can check and debounce the switch input by polling. <S> You would have the current state of the switch, the previous state and the previous stable state. <S> If you have two readings the same (in a row) that differ from the previous stable state <S> you know you have an edge, and you can discard the polarity of edge <S> you're not interested in (say the release of the switch). <A> When you will compile your code in it's output data will be a portion of data called an Interrupt Vector Table where are store what function to call for which interrupt. <S> Processor will carry out the priority and sequence of execution of Interrupt Service Routines. <S> Update <S> I have just installed MPLAB X IDE and there are embedded examples for PIC18 processors with interrupt.c file: <S> / <S> * High-priority service <S> */#if defined(__XC) <S> || <S> defined(HI_TECH_C)void interrupt high_isr(void)#elif defined (__18CXX)#pragma code <S> high_isr=0x08#pragma interrupt high_isrvoid high_isr(void)#else#error "Invalid compiler selection for implemented ISR routines"#endif{ / <S> * <S> This code stub shows general interrupt handling. <S> Note that these conditional statements are not handled within 3 seperate if blocks. <S> Do not use a seperate if block for each interrupt flag to avoid run time errors. <S> */#if 0 <S> / <S> * <S> TODO Add High Priority interrupt routine code here. <S> */ <S> / <S> * Determine which flag generated the interrupt * <S> / <S> if(<Interrupt Flag 1>) { <Interrupt Flag 1=0>; /* Clear Interrupt Flag 1 <S> */ } else if (<Interrupt Flag 2>) { <Interrupt Flag 2=0 <S> >; /* Clear Interrupt Flag 2 <S> * <S> / <S> } <S> else { /* Unhandled interrupts */ }#endif} <A> The PIC 18 has just two interrupt vectors, one for high and one for low priority interrupts. <S> You had to assign the correct interrupt by your own in software. <S> e.g. void interrupt tc_int(void) <S> // <S> High priority interrupt{ if (TMR1IE && TMR1IF) <S> { TMR1IF=0; ++tick_count; TRISC=1; LATCbits. <S> LATC4 = <S> 0x01; }}void interrupt <S> low_priority LowIsr(void) //Low priority interrupt{ if(INTCONbits. <S> T0IF && INTCONbits. <S> T0IE) <S> // <S> If Timer flag is set & Interrupt is enabled { TMR0 -= 250 <S> ; // <S> Reload the timer - 250uS per interrupt INTCONbits. <S> T0IF = 0; // <S> Clear the interrupt flag ADCON1=0x0F; TRISB=0x0CF; LATBbits. <S> LATB5 = 0x01; <S> // <S> Toggle a bit } if (TMR1IE && TMR1IF) <S> { TMR1IF=0; ++tick_count; TRISC=0; LATCbits. <S> LATC3 <S> = 0x01; }} <S> That's different compared to the greater PIC24 which has one vector for each interrupt.
| According to Datasheet some of Pins can be configured for both types of interrupts - it is your choice.
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Can a USB port passively 'listen only'? I have a small dedicated controller built around an Arduino that accepts commands from a PC over USB (serial). The communication is one-way -- the controller never sends data back to the PC. Now I need to extend the controller to handle more duties, but there's no room physically or program-wise, so I'm adding a second box with another Arduino that will accept the extended commands. I can't alter the controlling PC application or add another USB port to the PC. The application controls the hardware and is in turn driven by a script. It can only talk to one serial port (physical or virtual) at a time. We can alter the script but not the hardware/application. So my hope is that I can just bridge across the D+/- lines to feed the second controller. If this were straight serial I'd have no problem, but with USB in the middle I see issues ahead, because the PC will want to enumerate both Arduino USBs, right? Is there a solution here I'm overlooking? Can I use USB passively, as a listener only, with the second controller? <Q> USB CDC serial is quite a complicated protocol. <S> Even when the virtual serial bus is quiet, the usb host is continuously asking the usb device "do you have anything to send" and the device says "no". <S> Consider sending the phrase "hi" to an arduino, and getting "hello" back. <S> On the USB bus, it will look something like this: <S> Host: <S> Do you have a message? <S> Device : <S> No. <S> Host: <S> I have a message for you Device : <S> OK <S> Host: <S> hi\n <S> Device : <S> OK <S> Host: <S> Do you have a message? <S> Device : <S> No. <S> Host: <S> Do you have a message? <S> Device : hel <S> Host: <S> Do you have a message? <S> Device : <S> llo\n <S> Host: <S> Do you have a message? <S> Device : <S> No. etc. <S> etc. <S> These messages will also be mingled in with messages sent to other devices on the same hub (even if you don't have a USB hub in the system, there may well be one inside the computer. <S> So you could easily see messages to your mouse and keyboard mixed in). <S> The USB protocol is extremely complicated, so splitting it in the way you mention is not going to be practical. <S> You can "sniff" it though. <S> If you connect the D+ and D- lines, it is possible to see the traffic on the bus. <S> As long as the sniffer doesn't try to manipulate the D+/D- lines, the USB bus will keep working. <S> The sniffer would then need to identify which messages were relevant and decode them. <S> Building a sniffer out of an arduino would be difficult but probably not impossible. <S> It almost certainly isn't the best solution for your problem. <A> No you can't split USB, but what prevents you from connecting the UART receive pins of two Arduinos together <S> so both receive same UART data, or making the first Arduino to send commands it does not understand to the other Arduino? <A> Can I use USB passively, as a listener only, with the second controller? <S> Technically <S> yes, you can build a device that is passively looking at USB traffic and be able to get information from it and take any actions. <S> In fact, devices like that do exist. <S> They are called "USB Protocol Analyzers". <S> Typical architecture of such devices consists of an non-invasive sniffer on D+/D- lines (high-impedance voltage divider plus gain compensating wideband amplifier), which feed a standard (stand-alone) USB PHY. <S> The serial-parallel output interface from the PHY (ULPI or UTMI or PIPE3) is then watched/logged by a sizeable FPGA that has decoding tools and can trigger off USB protocol events. <S> Examples: Teledyne-LeCroy , Ellisys . <S> So you can watch USB traffic and get data. <S> To get them in real time you will need to program FPGA and run it at sufficient clock rate to implement most elements of USB Serial Interface Engine (except forming ACK/NAK replies). <A> One solution could be to essentially implement a daisy-chained local serial topology of Arduinos. <S> One of these at the 'front' of the chain would expose its USB CDC interface. <S> Then you'd wire a pair of pins configured for SoftwareSerial to the next Arduino's Serial pins (D0/D1) and so forth. <S> Then you could echo the serial traffic received up and down the chain as necessary, with every member of the chain deciding whether it should respond as appropriate.
| Even if you are only sending serial commands one way, there is USB traffic going backwards and forwards all the time.
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Battery systems for 440v industrial motors Are there systems utilizing batteries to start large (25hp - 50hp) 440v 3-phase electric motors, thereby eliminating spikes to the supply grid? <Q> The common solution is not to use battery power inverters but AC to DC to AC inverter "variable frequency drives" or VFD's. <S> These start from DC and accelerate line frequency ( or less to the desired RPM ) by raising sinusoidal 3 phase V with f to keep V/f constant and thus increase current with RPM to prevent high acceleration currents that occur from full voltage to start an induction motor. <S> These can give constant torque with acceleration constant rise to target speed. <S> (RPM) or slightly higher than constant torque with faster acceleration such that peak/load power can be kept as close to unity as required. <S> This also reduces mechanical step power resonances to pumps and other mechanical systems by reducing the slew rate of speed and thereby reducing the bandwidth of mechanical spectrum of shock defined by the rate of acceleration to a mechanical load which may have some natural frequency of vibration. <A> When a "soft start" is required the most common method is to use one of several soft starting techniques that reduce the voltage applied during starting. <S> That is quite effective and not as expensive as a variable frequency drive (VFD). <S> Using a VFD allows starting at 100 to 150 percent of the motor's rated torque while drawing only about 100 to 150 percent of the rated operating current. <S> VFDs are usually used only when variable speed is required, but they are sometimes used just for starting when the load is particularly difficult to start. <A> If you needed a 3-phase motor to rapidly start without impacting the grid you could have a larger motor idling nearby, with a flywheel attached to the shaft.
| In the world of industrial motors, 25hp - 50hp is a medium sized motor and does not require limitation of the starting current surge at least not in the USA where motors are rated in horsepower and 440 (or 460) volts is a standard 3-phase supply voltage.
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Voltage error when connecting a zener diode I am working on a project where i need to measure the battery voltage and send the measurement to a uC, which is powered by 3.3V. The voltage divider is projected as shown before: As input we have the battery voltage (4.2V maximum) and as output, our goal is to have 3.3V (maximum) that is read by the uC. The problem is that when I connect the zener diode, as output voltage I have 1.55V instead of 3.3V, while if I connect only the resistance and capacitance I have 3.3V. Does anyone know the cause of this? The electrical characteristics of zener dioce are in the following picture: <Q> The values of resistor R17 is too high, this means only small currents can flow. <S> The zener diode isn't ideal, it also leaks current. <S> Perhaps the current through R17 and the zener's leakage current are in the same order of magnitude preventing "good zener" behavior. <S> Let's calculate: When Vbat = 4.2 V and the zener would work there would be 3.3 V at the output. <S> So across R17 we get 4.3 - 3.3 V = 0.9 V. <S> That means a current of 0.9 V / 127 k = 7.1 uA. <S> Hmm, only 7.1 uA, that's quite close to the leakage current of the zener diode. <S> See \$I_R\$ in the table in your question, it is 5 <S> uA <S> so quite close to the 7.1 uA that would flow through R17 if all worked well. <S> Also note the value of \$I_{ZT}\$ in the table, it is the current which the zener diode needs to have a proper zener behavior. <S> Using a higher current is also OK. <S> To get the 5 mA from the table, the value of R17 needs to be substantially lower. <S> In the order of 180 ohms ! <A> The 3.3 V Zener diode does not have always 3.3 V on it. <S> The nominal voltage on the zener diode is only achieved at a specific current – see \$I_{ZT}\$ being 5 mA in your image. <S> If the current is smaller, the reverse voltage of the diode will drop too. <S> Looking at R17, there can be no more than 25 uA going through your diode. <S> Most datasheets show a graph between current and reverse voltage. <A> At 4.2V you have ~7uA of current flowing into the resistor divider. <S> Look at the reverse leakage current Ir@Vz for that diode. <S> The datasheet says it can leak 5uA at 1V. <S> This is reducing your voltage because it is leaking current to ground. <S> Zeners are a bad choice for this. <S> I'm assuming you are trying to protect the micro controller. <S> You are better off adding a low leakage diode BAV199 with the anode tied to the voltage divider node and the cathode tied to the supply voltage of your micro controller. <S> As BAS40 might work as well if the board doesn't get too hot, leakage will be <S> higher but forward voltage is lower and it will clamp sooner. <S> Anyway, this will clamp the voltage to 3.3V plus the diode drop and keep the micro controller ESD diodes from forward biasing and damaging your part. <S> Most micro controller datasheets tell you the absolute maximum a pin can take (See Absolute Maximum Ratings in the datasheet). <S> It is usually Vcc plus 0.3-0.5V. <S> This is because they don't want the internal ESD diodes forward biased all the time. <S> They are designed to take short ESD hits, not be forward biased all the time. <A> This is how Zeners ( and LEDs) work. <S> The Vf depends on the applied current and the tolerance on bulk resistance, Rs above some knee current =Izk. <S> If you know the characteristics, one could do this with a Zener or LED or a couple signal diodes using the Zener specs which I have converted to a linear formula that works < 10:1 current range. <S> This just shows a 2V Zener or LED. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This is just another way to measure voltages above the max input range, over a limited range. <S> I assume you may want to minimize this current from battery drain, so factor the reduced voltage on the diode using the diode linear resistance down from rated current. <S> Vzt ( Zener Test voltage) @ <S> 5mA Vzk <S> ( Zener knee voltage) @ <S> 0.5mA <S> Reducing the Zener current to 0.5mA also reduces the Zener knee voltage by 0.5V from the rated nominal value and is close to the 2.8V.
| One can scale the battery "voltage guage" with an offset to improve resolution or use an R divider with a clamp.
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Is there a kind of relay that consumes power only when switching? The relay must stay in its state. When an electrical impulse is sent to it, it should switch its state and should maintain its status. So I am really curious about it. Is there such a thing for saving lots of power? <Q> Common in small ( telecom style 2A or less ) and also some power relays have this function. <S> There are various methods of actuating them, the small ones typically have either two coils (pulse one coil for 'on', pulse the other for 'off') or single coil (reverse polarity pulses for on vs. off). <S> Some of the power ones are alternating (pulse on, pulse off). <S> One application of the high-power relays is for electrical metering where the power can be remotely switched off in case of bill non-payment by the customer. <A> But latching relays continues to draw energy for coil inside of it after one o hit the button ... <S> You are getting confused with a relay circuit that electrically latches the relay on. <S> A latching relay is bi-stable. <S> It has two stable positions. <S> It uses two coils to switch it - or one <S> but you have to reverse the polarity. <S> Figure 1. <S> A latching relay will stay in the last energised position when power is removed. <S> Source: Homofaciens . <A> Example of a mechanical latching relays can be found in early telephone exchanges. <S> The stepping relay, or Strowger switch , was used to establish a switched circuit through a telephony exchange. <S> Power was used to advance the relay for each click of the dial pulse, but once the digit was complete, the switch held its position. <S> More advanced exchanges used crossbar switches which connected more points in less space. <S> These were also latching and did not require power to maintain each specific connection. <A> In a way, each cell in non-volatile memory (NVRAM) is a latching relay. <S> Each cell stores its state when power is removed and uses no power to remain as it is. <S> Only when the state is changed does the "bit" use power. <S> In many implementations of NVRAM, the state is stored as an island of electrical charge stored within an insulating framework. <S> The stored charge influences the ability of current to flow through an adjacent semiconductor channel. <A> If you want a readily available example, GE makes a line of latching relays called RR7. <S> They are used for commercial lighting. <S> They work exactly as you wish, and run on 24 VAC or DC or voltages in that ballpark. <A> Also the Half-Bridge FET Relay only consumes power during the switching transition which is proportional to the Q charge on the Gate input and Drain output during switching. <S> However, in normal operation, they are used to also vary voltage with PWM which draws switching power at a higher rate. <S> But the high side of the dual half-bridge or "full-H bridge" bridge is used to change directions of current after flow has stopped with very little power. <S> However, IGBT's with FET input are more suited to AC line voltage and require costly protection for line faults or surges. <S> Similarily , the Thyristor family of parts only need a pulse to latch-on for the next cycle.
| Yes, they go by various names such as a bistable relay, latching relay, or impulse relay.
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Circuit to "zoom in" on mV fluctuations of a DC signal? I have a signal that is roughly 0.2V + noise fluctuations of order 0.1-2 mV. Ideally I want to amplify this signal such that the mV fluctuations become about 1V. In other words I want to amplify the signal by about 1000x. However, if I flat out amplify the signal, the total signal becomes 200V + 1V fluctuations, which I can't reasonably read on some bench top DAQ (0-10V range). Is there some combination of circuit elements that can take my input 0.2V + 1mV signal and spit out only the amplified fluctuations (i.e. 0V + 1V fluctuations)? I should say that these fluctuations are controlled by me physically squeezing a pressure gauge, so they aren't necessarily high frequency. Basically the signal rises to 0.202V when I squeeze, and 0.200V when I let go. I want to see that excess 0.002V blown up to 1V, but I may be squeezing and letting go slowly in general. <Q> Capacitors block DC and pass AC. <S> You can use a series capacitor into an opamp with whatever gain you need. <S> Even better might be a simple RC high-pass filter... <S> One capacitor (series) and one resistor (to ground) in front of your amplifier. <S> Like this: simulate this circuit – Schematic created using CircuitLab R2 and R3 set your gain. <S> C1 and R1 set your low frequency cut-off. <S> The formula you use to find the cutoff is: $$F\text{(Hz)} = <S> \frac{1}{2 <S> \pi R C}$$ <A> Here's something inspired by the first 2 answers. <S> Make a 10-second low pass filter of the input signal and feed that into an op-amp's non-inverting input (+). <S> Then take a 1-second high pass filter of the same input signal, and feed that into the inverting (-) input of the same op-amp. <S> Fluctuations get subtracted from the average and amplified a lot . <S> If it's too much amplification, a resistor in series with C2 will lower the gain. <S> This also inverts the fluctuation signals. <S> If you want them non-inverted, follow this with a gain of -1 inverting stage. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Digital designer here <S> so I'm not certain, but... <S> The other answers assume high-frequency fluctuations. <S> Instead you want to subtract the 0.2 V and amplify that. <S> You can use a summing amplifier to subtract the offset, if you've got positive and negative supply voltages. <S> I think you can also use an inverting configuration where the non-inverting input is at 0.2V instead of ground. <A> Sure, just an ordinary inverting op-amp can do that: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Remember that an op-amp wants to make its inputs the same. <S> So if you put 2V on the non-inverting input, and the signal input is also 2V, the output will be 2V. But say the signal input is 2.1 V. <S> The op-amp wants to make the non-inverting input also 2V, and will have to drive the output higher than 2V to make that happen due to the voltage divider action of R1 and R2. <S> The selection of these resistors thus sets the gain. <S> Keep in mind any source impedance will effectively add to R2, so if your sensor doesn't already have a low-impedance output, you may want to buffer it. <S> You have a couple options for realizing V2, since you probably won't want to find a 2V battery. <S> Of course this will make the circuit somewhat dependent on the supply voltage, and the small but non-zero input current to the op-amp will introduce some error, so if you require high precision you might find an adjustable voltage regulator more suitable. <A> Use a coupling capacitor prior to the amplifier. <S> The DC signal will be blocked but the fluctuations will pass through.
| Since the op-amp's input impedance is quite high, this doesn't need to be a low impedance source, so it could be as simple as a potentiometer across the power supply.
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How to alleviate the negative effects of using parallel LEDs with a common resistor? We all know the rookie mistake of using LEDs in parallel and not giving them separate series resistors. They are often doing it for ill-founded incentives like "I want to save money on resistors" , "I don't want to solder that much" , and "They all have the same color anyway so why bother?" . The correct answer in such a situation is to persuade the OP to use separate resistors. Forums and QA sites (including this one) are full with such examples. However, consider a case when it's completely impossible to avoid using a single resistor for separate parallel LEDs. I know it's bad. I know it's evil. But in a case it cannot be avoided, what methods are there to at least try to slightly alleviate the problems such a circuit can will cause? Consider a case with 2 to at most 5 LEDs of same color from the same production batch. The LEDs don't have to shine with their maximum brightness, so the current can be smaller than the maximum allowed current (although 1/x for the case of the device with x LEDs would probably be too low). The LEDs are inside a proverbial box, in parallel, and the box can not be changed. With the two wires hanging out from it, I can do whatever I want. Still, I would prefer simple solutions with few and only passive elements. (of course, complex microcontroller-driven solutions will also be appreciated, especially if no simpler solutions can be found) Clarification : By "alleviate" I mean to reduce the brightness differences as much as possible, and, of course, to avoid them being damaged. <Q> Have R so big that the current wouldn't be too high even in case where only one led actually conducts. <S> This prevents smoke-effects, but doesn't fix the possible non-uniform current distribution. <S> The leds can all seem to illuminate acceptably, but do not expect it stay true when the parts get older or the temperature changes. <S> My suggestion: Get a can opener and fix the circuit. <A> You accept the fact that not all of your LED's will be the same brightness, and some of them might not light at all. <A> "I want to save money on resistors" Do you really think you can find an alternative that allows you to replace resistor that cost 0.05 <S> $ or € each? <S> "I don't want to solder that much <S> " Do you really think you can find an alternative that does not involve soldering? <S> "They all have the same color anyway <S> so why bother?" <S> We bother because we have to, improper current limiting WILL lead to the destruction of your LEDs, color does not change anything. <S> But to try to answer the question nonetheless, you could Select your LED and group those that have the closest Vf; <S> Make sur <S> they are evenly cooled, to avoid thermal runaway of an already unstable system; Route the PCB in a way that creates small resistor in the tracks (using similarly long and thin tracks for each LED And last option (and this is really pushing it) <S> would be to magnetically couple your LEDs, and drive them with an AC signal. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Better solution would be to have the LEDs in series, then they would all get the same current. <S> Of course, the source voltage needs to be higher for them to all turn on. <S> If one fails open, they all turn off. <S> If one fails shorted, the current for the rest goes up until they fail too. <S> Example, 12V source, Vf = 2.7V, <S> 15mA: <S> (12V - 2.7V - 2.7V - 2.7V)/.015A = 120 resistor for current control. <S> One LED fails shorted, then (12V - 2.7V - 2.7V <S> )/120 ohm = 32.5mA, above the normal 20mA continuous rating of a typical LED. <S> In parallel, if one fails shorted, the current jumps up and possibly takes out the resistor while the other LEDs simply don't turn on. <S> If one fails open, the others see what? <S> Their Vf is still the same, so one could argue the current draw from the source actually drops. <S> I made a timer box with MAX7219 driving 7 segment displays, each segment of a digit was made of 3 individual LEDs in parallel. <S> They were very bright LEDs, rated for thousands of mcd at 20mA, something like this <S> http://www.dipmicro.com/store/LED5W <S> , so I ran the MAX7219 at a reasonable current level (I think the current limit resistor selected only allowed 15mA) and all three LEDs drew similar current so they shared that current, and were still very bright (not like the old indicator LEDs that needed 20mA and output just tens of mcd). <S> After 10 years of use several days a week for several hours at a time, all the LEDs were still going with no failures. <S> So the answer to <S> "Can I run xx LEDs in parallel? <S> " could be "It depends", and not a hard & fast <S> "No, Never". <S> If the LEDs were run in pulsed mode (like from a MAX7219 which multiplexes across digits at 800 HZ rate), the LEDs are not on continuously and thus don't develop the heat that high current continuous on creates that typically kills an LED. <S> If to be used as on continuously than the smart design would have a resistor per LED, or ensure that the max current allowed by the current limit resistor did not allow a single LED to go into over current situation. <A> There is no "magic" solution. <S> But you may not need one. <S> Most of the posters here will tell you that LEDs in parallel will never work. <S> But every Halloween and Christmas, where I live, the shops are full of battery-operated strings of LED "fairy lights". <S> Each consists of a small battery box, one resistor, and a whole string of LEDs wired in parallel. <S> Sometimes, for blue or white LEDs, the manufacturer won't even bother with the resistor, and will rely on the internal resistance of two AA cells. <S> Despite the crude design, they do actually work. <S> The resistors are all the same, and are likely to all be at much the same temperature. <S> The curves of voltage against current will be very similar for each LED. <S> Because of this, the current naturally tends to balance out between them. <S> The only thing you may see is that if you run the LEDs at a very low current, some may be obviously dimmer than others. <S> Increase the current, and they will all be reasonably bright. <A> Mount the LEDs on an array of individually-controllable thermoelectric Peltier devices such that you can independently bias each LED die warm or cold. <S> Position a calibrated camera in view of the LEDs such that you can estimate how much light each is putting out. <S> Wrap a feedback control loop around it, adjusting the heating/cooling to achieve even light distribution. <S> (ask a silly question, get a silly answer)
| If there's already a substantial brightness difference, you can try paint the brightest one slightly to make it dimmer. There is no other way, other than using wire cutters along the green line, and replacing the part that can't be changed with a properly built circuit.
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LM35 max input current - I have a wrong output from the sensor I’m currently doing a small project. I’m trying to make my first pcb. That’s for a light I made. But the LM35DZ outputs a wrong voltage. My LM35 puts out 4.6V instead of 380mV. With a 12V - 2A power supply. With the datasheet I can’t find the maximum input current. Do you have an idea? This is my circuit: Datasheet : https://www.mouser.fr/ProductDetail/Texas-Instruments/LM35DZ-NOPB?qs=QbsRYf82W3F5RpWTxhXHxA%3D%3D The component is wired to the CN1 jst R6 : 47 ohmR5 : 330 ohm With a 3,3V linear regulator. <Q> The LM35 was damaged. <S> miss wiring. <S> Because of that the LM35 ouput a wrong voltage when it was correctly wired. <S> Thank you for yours helps. <A> The Mouser link clearly indicates Output Current: 10 mA max <S> However, it will not pull down negative bias current since the sensor uses an NPN emitter follower output inside a feedback loop with low quiescent current. <S> Therefore put a resistor on the signal input of the floating LM324 (PNP inputs) to ground e.g. 1k to 10k. <A> I don't see anything wrong with your circuit up to the LM324 input. <S> Check it with a meter. <S> If the LM324 GND is open, the several mA current will exceed the ~1uA sink capability of the LM35. <S> Failing that, perhaps you have a mis-wired or damaged LM35, but you have not provided a schematic for that part of the circuit.
| Check that the LM324 GND pin is actually connected to ground, and check the wiring to the LM35.
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How to properly use STM32 flash memory as an EEPROM? I have to save certain parameters to the flash memory and these values should be non-volatile. But my program will change these parameters and I want the changes updated to the flash. I understand to overwrite a flash page, you have to erase it first. And the granularity of erase operation is a page. Considering all of this, how can I properly implement it in software? Whenever a parameter changes, should I erase the entire page? The page size is 1 kB, which is much space than all of my parameters put together. Is this the only way to go? <Q> Yes, as you've already discovered, you can't erase less than one page at a time. <S> However, if your parameters take up much less than the size of a page, you could consider creating a scheme where instead of writing to the same address every time and erasing in between each write, you write to a different address within the page at each write. <S> So when you want to read your parameters, start reading at the beginning of the page, and continue reading until you get to a parameter bock full of 0xff entries. <S> You'll then know that the previous block was the last one you wrote. <S> There are certainly many other ways to do this kind of thing, for example you could use a bit-field at the start of the page to indicate which blocks within the page have been written instead of having to potentially scan through the whole page. <S> But this may depend on hardware support. <S> Some micros - like the STM32L0, won't let you write anything other than 0x0000 to a flash location if it's not currently fully erased to 0xffff, preventing you from using it as a bit- field and clearing 1 bit at a time. <A> Yes, an entire page must be erased (set to 0xFF) before you can start writing to it. <S> With most external flash memories, you can actually write to a page multiple times without erasing as long as you are writing to previously-unused byte locations. <S> Please see this answer . <S> However, the internal flash memory controller in the STM32's won't allow any writes unless the entire page is cleared. <S> If you want to go the easy route, ST has a freely-available software solution which provides an EEPROM emulation layer using an area of internal flash memory. <S> It provides a simple set of functions, and handles all the complexities "under the hood". <S> It allows for single-byte read and write granularity, and handles the erasing for you. <S> I don't know which microcontroller you are using. <S> Here are the EEPROM-emulation docs for the STM32F0xx <S> and STM32F10x microcontrollers. <S> For example, you write a byte using EE_WriteVariable() . <S> The software maps this location to a flash page, reads that page, inserts your byte where appropriate, then programs an entire new page onto another flash page . <S> It bounces back and forth between pages, and keeps all this hidden from you. <S> However, this is pretty time-intensive. <S> Not only does it take a while, but your memory bus can stall while waiting for a flash write to complete, <S> so you can't do this on timing-critical applications. <S> If this software doesn't work for your application, you can build up as complex a solution as you need. <S> Once I wrote a large system for handling mission-critical configuration data, which data could change on the fly. <S> It used multiple sectors, redundant locations, crc-verification, wear-leveling, etc. <S> I couldn't rely on a table-of-contents, because what if the system powered down in the middle of the TOC update? <S> So it had a routine to discover <S> the "active" (read: "most-recently-written") flash configuration bank upon initialization... etc etc. <A> Yes, that is the way to do it as long as you do not do it often because they have only a few thousand write cycles before the flash wears out. <S> Everything is done by the page. <S> Do NOT try and use the flash as RAM
| Only erase the page when its full.
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What does the power rating of 10W mean? When hair clippers or similar appliances are rated at, for example, 10W, does that mean 10W per hour or second? I'm trying to determine power of various hair clippers. <Q> Neither. <S> You're confusing energy and power. <S> Power is energy per unit of time, and one watt is one joule per second. <S> Now, the issue with electric energy is, that joule, as a unit, is quite small. <S> Therefore, we use larger units, such as wat·hour or kilowatt·hour. <S> Note the multiplication there! <S> Hour is 3600 seconds. <S> So one <S> watt·hour is 3600 J, and one kilowatt·hour is 3600000 J. <S> So your 10 W device will spend 10 watt·hours, if it works for one hour, assuming that it always uses same amount of power. <S> Now, the confusion comes, when people start dropping the time, and start referring to kilowatt-hour as kilowatt. <S> Don't do that, and be careful when others do, since they might now understand what they're talking about. <A> "W" stands for Watt , which is the measure of the rate of consumption of energy, in Joules <S> (energy) per second. <S> Adding in an additional time component is commonly done to identify the total energy used. <S> If you used your 10W appliance for 10 seconds, then you would use \$10W\times10s=100J\$ energy. <S> A common use for this form is used by the electric company to bill you for power. <A> 10 watt means it needs 10 watt (volt x current) to operate itself. <S> If your clipper is designed to run on a 5 V supply and is attached with a source of 5 volt it needs 2 ampere of current to operate with its rated power.
| So if you have a 10 W device, that means, that it's consuming 10 J each second.
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Why system energy is equal to 1/f? I am studying Intel speed shift. I see a slide: Reference I don't understand, why the green line 1/f means system energy, I think more performance should need more energy as the red line (compute energy). Original site: The Intel Skylake Mobile and Desktop Launch, with Architecture Analysis <Q> Power Management <S> Power spectral density is inversely proportional to the frequency of the signal. <S> This diagram might help illustrate this effect a bit more: <S> Imagine that as the CPU's over frequency increases, or as frequencies of the RAM or various busses increase, the amount of power available to leverage will be in a relationship that will be most like the graphic showing the frequency domain (lower right). <S> NOTE1: <S> the relationship of 1/f is very common in nature and is typically called fractal noise" or "pink noise". <S> NOTE2: <S> This is also why Fourier transforms are important in our field. <S> Additional details <S> The main author of the slidedeck you reference, Efraim Rotem, also wrote this paper titled: H-EARtH: <S> Heterogeneous Multicore Platform Energy Management . <S> In this paper he mentions this: Platform components consume fixed runtime power with energy proportional to runtime (1/f). <S> All other components are ignored. <S> The two opposite energy trends might have a global minimum (see Figure 1). <S> References <S> These are helpful in understanding many of the concepts that are key to grasping all this. <S> 1 <S> /f like noise and self organized criticality <S> Pink noise Colors of noise <S> Chapter 6 - ENERGY MANAGEMENT TECHNIQUES FOR SOC DESIGN Learn how to create "pink" noise <S> But what is the Fourier Transform? <S> A visual introduction <A> It looks like this graph attempts to explain how the energy per clock cycle typically behaves for a digital system. <S> It is a bit stylized. <S> The green curve relates to power dissipation that is constant and doesn't change with clock frequency. <S> ("system power" would then need to be understood as more or less constant leakages, quiescent currents, etc throughout the system). <S> To go from power [J/s] to energy [J] per clock cycle you divide power by the clock frequency f [1/s]. <S> For a constant power this gives a 1/f slope for the energy per cycle spent by static power dissipation. <S> For the red curve the exponents given for the "Compute power" seem like very rough estimates. <S> Switching power is \$P_d \propto fC_LV^2_{DD}\$ for a constant load (switching every cycle). <S> For higher clock frequencies the supply voltage is scaled up as a higher voltage is necessary to speed up the digital switching - enough to support the higher frequencies. <S> However the speed-up you get is typically better than linear with \$V_{DD}\$ , so <S> \$P_d \propto <S> f^a\$ with <S> \$a\$ <S> somewhere a bit above 2 is more likely in my opinion, although \$P_d\$ can be dragged up by other leakage contributions which show a supply voltage dependence (leakage, short-circuit leakage, etc.). <A> The platform energy is proportional to the run time of the workload and therefore inversely proportional to CPU frequency.
| That diagram is in the section of the slidedeck where they're talking about power management is stating that the energy efficiency for System's on a Chip (SoC) will behave inversely proportional to the frequency of the CPU as it increases.
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Replace some LEDs with diodes in relay circuit I need to know if I can replace the input LEDs (IN1, IN2, IN3, IN4) by a diode or resistor and if this can help me to reduce the voltage consumption generated by these LEDs. The circuit is a 4 relay module for use in Arduino and RPi projects. I have this module connected to my RPi, but the RPi show me a "LOW POWER" message when I connect the module to GPIO pins. The RPi is powered with their appropriate power supply. (*) All pics taken from SUNFOUNDER . For more info about the relays module visit SUNFOUNDER . <Q> If you are unable to power the relay module with Raspberry Pi (i.e. the relays do not switch properly), it is because the module requires 5V input but <S> RPi GPIO can only output 3.3V. <S> This problem can be solved by powering the relays with external 5V voltage source. <S> Follow this diagram: <S> The Low Power message is due to the relays drawing quite a lot of current from RPi power supply. <A> Removing the LEDs won't really help. <S> The optocouplers require a certain current through their LEDs to turn on correctly. <S> The manufacturer of the board will have designed it so that the combination of a 1K resistor and an external LED results in the correct current. <S> You could replace the LEDs with resistors. <S> But you would have to choose resistors that give the same current as before. <S> The end result is just as much current draw, and no indicators to tell you which outputs are on. <A> assuming that by their appropriate power supply <S> you mean this one: https://www.raspberrypi.org/products/raspberry-pi-universal-power-supply/ <S> that should have plenty of power to spare for 4 relays. <S> so I'm guessing you're trying to operate the relays of 3.3V. <S> this may kind-of work,but <S> the raspberry pi probably can't spare enough current from it 3.3V source Connecting the VCC to +5V <S> seems wrong as the GPIO is not 5V tolerant. <S> but the optocoupler will drop about 0.9V and the LEDs will drop about 1.6V so the voltage seen at the GPIO will be well within the range allowed.
| Removing the LEDs won't solve the problem as those LEDs don't consume much current.
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Why are current probes so expensive? I notice that clamp-type current meters range in price from a few dozen dollars to a few hundred, but current probes for oscilloscopes cost significantly more, with many close to $1000 and some well over $4000. Why are the oscilloscope current probes so expensive? Are they built by princes? Do they contain coils of solid gold wire? I get that these are fairly low-volume items, and that they need to be calibrated and probably have circuitry that compensates for various errors, but isn't that also true of current meters? Is there something special about the probes, or just market forces at work? <Q> I believe there are two components at play here: 1 - Clamp-type current meters are much simpler devices because they only need to measure the amplitude of an AC current at a particular low frequencies (some measure DC, but are more expensive). <S> Current probes (especially those with ability to measure DC) are much more sophisticated, having to provide flat response over a decent frequency range and being much more susceptible to all sorts of calibration problems and drifts. <S> 2 - The market for clamp-type current meters is way bigger than the market for current probes, since it's used by electricians, which is a much bigger universe than the one represented by electronic engineers and technicians. <S> Hey... maybe there is a market opportunity here. <S> Whoever comes up with a more affordable current probe may find a good market among hobbyists and small businesses. <A> But apart from that, there are three reasons: Bandwidth Bandwidth Bandwidth <S> High bandwidth AC-DC probes work by sandwiching a Hall effect sensor in a magnetic core. <S> If you want 20 MHz of bandwidth, you need to find a Hall effect sensor with 20 MHz bandwidth, or, you need to do some fancy blend of inductive coupling at high frequencies plus Hall effect at low frequencies, and maintain accurate response across the whole range. <S> A low-cost AC-only probe with limited bandwidth may just be a current transformer. <A> Terms like upper/lower frequency, peak A range, sensitivity and linearity can be very cheap in Hall Effect Sensors , and Op Amps with high gain <S> -BW are cheap . <S> We know the cost increases greatly to get a DC response in a clamp with very high sensitivity. <S> We know there are some tradeoffs with dynamic range, calibration, saturation, linearity and Remenance. <S> Why/how are ferrite probes better? <S> and why more expensive? <S> How difficult is the ferrite air-gap for high permeability ferrite cores for measurements of current <S> how does this relate to <S> $ <S> vs <S> \$(gain\cdot <S> BW\cdot Amps)/(sensitivity\cdot accuracy)\$ <S> Let's consider a few of leading Keysight current-probes in this $ range for a measure of sensitivity to these parameters for market price. <S> Firstly, organize a list of parameters. <S> KEYSIGHT CURRENT PROBES, US$ for output to 1 MΩ BNC N7042A Rogowski AC <S> $ 1,881 <S> 9.2 Hz ~ 30 MHz 20 mV <S> /A <S> 300 <S> Apk N7041A <S> Rogowski AC <S> $ 1,881 12 <S> Hz ~ 30 MHz 10 mV <S> /A <S> 600 <S> Apk <S> N7040A Rogowski AC <S> $ 1,881 3 <S> Hz ~ 23 MHz 2 mV <S> /A 3000 <S> Apk N7026A <S> AC/DC clamp <S> $ 5,016 150 MHz 1000 mV <S> /A <S> 40 <S> Apk 30 Arms1146B <S> $ 685 0.1 MHz <S> 100 mV/ <S> A <S> 100 mA <S> ~ <S> 10 Apk 10 mV/A 1A <S> ~100 <S> Apk N2893A <S> AC/DC $ 3,999 100 MHz <S> 100 mV/ <S> A 30 <S> Apk <S> 15 <S> A 1147B <S> AC/DC $ 2,526 50 MHz 100 mV <S> /A <S> 30 <S> Apk 15 <S> A <S> N2821A <S> AC/DC $ 3,226 3 MHz 1 V/A 50 uA - 5 <S> A N2820A AC/DC 2-ch <S> $ 4,302 3 MHz 1 V/A 50 uA - 5 <S> A N2783B <S> AC/DC $ 3,221 100 MHz <S> 0.1 <S> V/ <S> A <S> 50 <S> Apk <S> 30 ArmsN2782B AC/DC $ 2,840 50 MHz 0.1 <S> V/ <S> A <S> 50 <S> Apk <S> 30 <S> ArmsN2781B <S> AC/DC $ 4,333 10 MHz <S> 10 mV/ <S> A <S> 300 <S> Apk 150 ArmsN2780B AC/DC $ 5,358 2 MHz <S> 10 mV/ <S> A <S> 700 <S> Apk <S> 500 <S> Arms <S> (this answer will be a work in progress, and this text deleted when completed) in the meantime , some may reconsider that BW is the only driver, <S> why is the most expensive only limited to 2MHz? <S> yet a less costly Rogowski coil good up <S> 30MHz <S> ref: https://www.keysight.com/en/pc-1659326/oscilloscope-probes?pm=SC&nid=-32553.0&cc=US&lc=eng <A> Cheap clamp current meters can only measure quasi-constant current with variations noticeable on a human timescale (seconds or single-digit Hertz). <S> Current probes have frequency responses starting at hundreds of kHz and going up to hundreds of MHz for the expensive ones. <S> That's a difference of 3 to 6 orders of magnitude.
| Well, first of all, DC-capable probes are a bit more expensive because they have to use Hall Effect sensors and deal with small offset voltages.
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SPI and I2C connections by using mosfet (high side, low side) I need to design sensors card in altium for sensors connected I2C and SPI. When I searched modules on internet, while some used mosfet to handle high and low side others did not use mosfets. I could not understand clearly for which purpose they used mosfet. Could you help me, should I use mosfet connections for my card? To explain my question well, I added two diffirent photos for the same sensor type. <Q> They are using the MOSFETs as level shifters. <S> If your devices do not all use the same voltage signalling levels, you may need to use them if your signals are bi-directional. <S> https://www.nxp.com/docs/en/application-note/AN10441.pdf <A> The circuit shown is a common level shifting circuit for I2C. <S> It is used because it can translate bi-directional open-drain signals as used in I2C. I wouldn't recommend this circuit for use with SPI, because SPI doesn't require bi-directional translation. <S> This circuit is slow (but it's difficult or at least tedious to predict how slow). <S> That's okay for early generation I2C that runs at 100 or 400 kHz. <S> But it's not adequate for SPI busses that might be expected to run at 1 - 10 MHz. <S> For SPI I'd recommend an off-the-shelf level translator IC to get predictable output levels and propagation delays. <S> Even for I2C if your application can accept a few $0.10's of extra cost <S> I'd recommend paying for an off-the-shelf <S> I2C translator IC to get predictable performance. <A> STM32 series microcontrollers can work directly with 5V peripherals just fine , without any MOSFET or other kind of level shifter. <S> Check the pinout table in the datasheet for your controller. <S> Most pins are marked as FT, meaning F ive-Volt T olerant. <S> These pins can safely take 5V input voltage when used as inputs (e.g. as the MISO line) or as open-drain outputs (SDA and SCL lines). <S> Although they will output only 3.3 Volts when used as push-pull outputs, most 5V parts (check the datasheet of yours) would interpret a voltage above 2 Volts as a logical high level. <S> See also this answer
| Some of the available chips will also give a speed boost and/or allow you to power down one side of the bus without blocking communication on the other side.
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What does voltage really do? I know there are many videos (many of which I have watched) and many posts (many of which I have read) about voltage, but I feel like a few points are left out that should be explained. What is voltage really doing ? Is it making the current move faster by using more energy (measured in joules) to move one coulomb, so is it just making current faster, which is where the pressure analogy comes from? What is electrical potential, and how does it tie in with voltage? Do voltage dividers limit current? And if so, how do you find out how much current is per volt? What happens when voltage is increased, does more current flow? I bet that I am misunderstanding something so if you could also explain that, it would be greatly appreciated. This is the last question I’m going to ask for a really long time, so I hop it sends voltage home for me... Thanks in advance! <Q> This image explain pretty neatly whats is happening without much technical terms <A> As has been explained multiple times in the answers and comments to your previous questions you are confusing a physical property (energy) with the units of measure (joules). " ... making the current move faster by using more joules ... " should read "... using more energy". <S> What is voltage really doing? <S> Is it making the current move faster ..., <S> so is it just making current faster ... <S> Raising the voltage will, in a resistor circuit, increase the current. <S> Current is the flow of charge so, for a given conductor cross-section, increasing the current must increase the charge velocity. <S> The current, however, moves simultaneously all around the circuit the same way a bicycle chain moves all links in the circuit simultaneously. <S> What is electrical potential, and how does it tie in with voltage? <S> As explained in my previous answer , "Voltage electric potential difference, electric pressure or electric tension is the difference in electric potential between two points." <S> Do voltage dividers limit current? <S> Voltage dividers are usually made of resistors. <S> Resistors limit current. <S> And if so, how do you find out how much current is per volt? <S> Use Ohm's Law. <S> \$ <S> I = \frac { <S> V}{R} \$ . <S> What happens when voltage is increased, does more current flow? <S> Use Ohm's Law. <S> \$ <S> I = \frac { <S> V}{R} \$ . <S> From the comments: <S> 1) Why does the current only increase in a circuit with resistance? <S> I mentioned resistance as an example because they are linear. <S> Other devices don't behave in a linear fashion; LEDs, transistorised circuits, constant current circuits, etc. <S> 2) <S> How does having more charges (electrons in this case -> amps) increase the velocity of the charges, and why would velocity increase the amount of charges electrons in a certain space rather than just speeding up the electrons? <S> To increase the amount of water going down a channel (the current) you have to speed it up or make the channel wider so that it goes at the same speed. <S> To get more current through a point in a circuit you increase the current density (and therefore the speed of the charges) or you increase the conductor width and maintain the same density (current per unit area - measured in A/m 2 ). <A> Voltage is analogous to water pressure, current is analogous to water flow. <S> A high voltage (pressure) wants to find a path to lower voltage (pressure). <S> If it finds one, current will flow. <S> How much current is dependent on the resistance, per ohms law. <S> Electric potential is the same as voltage. <S> Resistance limits current. <S> Using two resistors will create a ratio of voltages. <S> The current is dependent on the sum of the resistors.
| Voltage is electrical potential.
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Controlling 1 fan with 2 different computer power supplies Ok here is my scenario: Because our 2 gaming computers are pretty noisy, we moved them to another room in a closet. There are vent holes to the outside of the building, and now that the summer comes I want to add a small computer fan (12V 0.3A) to push the warm air out. The plan is that if one of the 2 computers is turned on the fan should spin. What I have done so far... and it works: I extended cables from both computers from the PSU (GND and +12V) to a little box. I added 2x 1N5059 diodes in parallel on each +12V from the PSU going to the fan. I know now the diodes can handle up to 2A, so one of them on each +12V line should have done the job. As I said, it works as expected and does what it should, but, and this is my actual question: I know diodes allow a very small amount of current in the other direction. Do I need to be concerned that one PSU could break the other one in long term use? <Q> If you remain concerned you could do these things: Remove one of the paralleled two diodes since it is not needed. <S> This will cut the leakage in half. <S> You could add a load resistor on the supply side of your diodes to ground to assure that there is a path to ground for any reverse leakage when that particular supply is off. <S> Just equip your vent with two fans instead of the one. <S> These would be powered separately by each computer. <S> This could be a wise move because when both computers are on you actually have more heat generation in the closet. <A> There is no chance that 1 PSU will feed the other PSU via the diodes. <S> So no chance of break it!Did you put an fuse in place at the (+12V) FAN side just in case?You <S> wil loose about 0.6V via the diode (see datasheet 1N5059) to your FAN <S> but that will not be an issue I think. <S> You can add an picture just in case so we can check it. <S> best regards,Erwin. <A> So, no diodes or losses from diodes, but do need a power supply to the fan, but that is easy...
| I would have used the supply you took from each computer to control a relay, one relay for each computer will give total isolation (excepfor the shared power line) and thise relays can drive the same fan without problem. The small amount of reverse leakage current in the diodes is not going to be a problem for the power supplies. The loads on the power supplies will sink the leakage current to ground.
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What kind of capacitor is this in the image? The power supply of my model railway got broken. I think the problem is the capacitor shown in the image. So does somebody know which kind of capacitor this is? <Q> Aluminum electrolytic capacitors do not, in my experience, use the 3-digit system. <S> From this datasheet: <S> I see what appears to be flux on the PCB, but often caps will bulge up if they are dying. <S> If it has actually leaked from the cap, then it should be replaced. <A> That is an aluminum electrolytic capacitor. <S> 100 microfarad, rated for 35V. <S> I'm not sure there's anything wrong with it, though. <S> It has score marks across the top. <S> If it had gone bad, then it would have burst along those lines. <S> It looks like there's some glue on one side, but that's about it. <S> The trace to the left of it looks burned. <A> The capacitance value I state is based on the assumption that the "100" marked on the capacitor is a three digit code as apposed to the literal capacitance of the capacitor. <S> This may not be the case, <S> I am leaving my "answer" in the hope that someone can clear this up. <S> -Edit Second Edit. <S> I did more digging <S> and it looks like indeed this cap is 100uF, and <S> what really sealed it for me is as Sphero Pefhany pointed out, <S> a 10uF rating would only need a two digit marking as would any value up <S> too 99uF. <S> So it must be 100uF <S> -Edit <S> It is rated for 35V with a capacity of 10.0 uF. <S> The "100" is a code that translates as first digit followed by a second digit finished by a power of ten digit. <S> The scale starts at uF, for example to get 47uF <S> the code is: 470, or if you have a code of 331 <S> that's <S> 33 <S> * 10^ 1 uF or 330uF capacitance. <S> A chart with capacitor codes can be found in this document. <S> link
| It's a 100uF/35V aluminum electrolytic capacitor. That is a surface mount aluminum electrolytic capacitor
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Please help me understand how to add AC power to my DC device I have a device that runs off of three D batteries. I would like to be able to add a power cord so that I can plug it into the wall and run it off of that instead. I would also like to still be able to automatically run it off the batteries on the off chance that the power goes out or I need to move it, but that is secondary and not as important. What would I need to get to do this? In my case, the device has an input of "DC4.5V 500mA" and the cord that I am planning on using is from an old phone charger and it has an output of "5.1V ⎓ 0.7A". I would imagine that if I just wanted to power it from the cord, I could just cut the wires from the battery case and solder the red to red and the black to the black. But, because the cord outputs 5.1V, would that short out the 4.5V input? If I put in a resistor, could that prevent it from happening? How many ohms would it need to be? That would still not help me with the lesser issue of the power going out. How could I wire it so that I would still have the batteries but not run off the batteries until needed. I think I would need a logic gate, but I don't know enough to know what to use and then how I would need to wire it up. Any advice for a novice would be greatly appreciated. <Q> Broadly speaking there are two options. <S> The simplest is to use a jack with a switch in it. <S> The downside of that is that it doesn't provide automatic changeover. <S> If the mains power fails your device remains off until the plug is removed. <S> Since your AC adapter is higher voltage than the batteries (about 1.65V per cell for fresh alkalines or 4.95V total) , you could use two diodes. <S> Schottky diodes (eg. <S> 1N5817) would be preferable to minimize the drop. <S> That will give you automatic changeover. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You can also do something like this with a 5V relay and a flyback diode. <S> (add a series switch on the battery or on the output depending on what you want, the relay will consume power when the mains is present regardless): (the voltage drop is very low compared to the diodes). <S> simulate this circuit <S> The device being powered is left "off" for a brief time while the relay switches (a few milliseconds typically). <S> If that's a problem then you might be able to add some capacitance to the output. <S> Many devices won't even notice. <A> The diodes prevent one source is charging/feeding the other. <S> And if you use the same diodes, the source that provides the highest voltage will "win", hence OR-ing. <S> Either that OR that diode will conduct. <S> Benefit of using ORing diodes is there will be no switching gap like when using relays/switches. <S> In some cases (with power sources having about equal voltages) you want the wall adapter to 'win', so you pick a Schottky diode (with lower forward voltage drop) connected to the wall adapter. <S> However, the come closer to the requested 4.5V <S> I'd suggest using a normal diode with 0.6V - 0.7V voltage drop. <S> Maybe you want even to use a Schottky diode connected to the batteries to not lower the battery voltage too much. <S> Note <S> the forward voltage of a (Schottky) diode is current dependent: if the device can draw less than 500mA you should check whether the configuration still works as expected. <S> Read the datasheets for typical voltage drops for the choosen (Schottky) diodes. <A> I would recommend just buying an AC adapter that outputs 4.5V ( Example ). <S> You can then use a barrel power connector with an internal switch to determine whether it should run off battery or AC.
| When you plug it in, it disconnects the batteries. Assuming the batteries will lower in voltage during use and the powered device will still accept it (so, the required 4.5V isn't a very strong demand) , I would use ORing diodes. If you cannot do this then you would need a voltage regulator to convert the 5.1V to 4.5V. Secondly, you could use logic of some kind.
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How to overcome over-current in a circuit? I have a relay switch that operates at 10 A and 120 V connected to a DC motor that runs at 18 A and 36 V. Will the relay switch burn out? If so, how can I protect the relay switch? <Q> You need a relay that is rated to work at the current that you are running it at. <S> Additionally you would want an inrush current limiter to prevent the inrush of current when the relay makes contact from destroying the relay. <A> Will the relay switch burn out? <S> Yes. <S> There is no good way to run 18A of current through a device that is only rated for 10A. <S> You should choose a relay so your load current does not exceed your current rating. <S> If so, how can I protect the relay switch? <S> You could use additional relays in parallel to increase the current capacity. <S> Ex: with two 10A relays in parallel, you could theoretically run up to 20A through the pair. <S> It's important to note that there may be a difference between recommended operating conditions and absolute max conditions. <S> If the absolute max rating of the relay is 10A, then you need to determine if 10% safety margin is sufficient for your application. <A> Yes, you'll likely burn up that relay. <S> There are plenty of solid state relays that can handle north of 20A with a heat-sink - I would look into those, or use a simple amplifier to drive a larger relay with your Arduino (although I'd recommend solid state). <S> Especially since we're talking about controlling motors, be advised that the larger the motor the worse the back EMF is. <S> You really want a snubber/ flyback diode across the motor leads to dissipate that energy and protect the relay from arcing. <A> In most cases you never want to go to motor current limits, except some stall cases. <S> Mostly even large engines run with much smaller current than max rated current, yours is probably around 3-4 A in normal running, starting could be more. <S> Your relays should be fine for the case, some relays have smaller low voltage DC current rated, so check it up for your relays, but mostly it stays the same. <S> Fuses are the most common way to protect against over-current. <S> For engines so called slow-blow fuses are used, because of starting spike in current. <S> Car fuses can be used as they are cheap and easy to access (every gas station), they also sell connectors to cables for these <S> and you could solder them in your circuit. <S> Nowadays most schematics use resetable fuses, so called PPTC. <S> The oldest way of limiting current is current limiting resistors and as wires have resistance, you don't want them to become fuses. <S> It may cause smoke and fire.
| Properly sizing the relay in the first place is likely your best bet, as trying to run relays in parallel can cause unexpected issues if one fails to operate. Whichever way of fusing or protecting you use, your fuses have to be rated for smaller currents than your cables.
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Can you replace a RJ45 connector with smaller connector? I'm designing a small board that needs an ethernet connection. However, the RJ45 jack takes considerable space. I would like to reduce the footprint of the connector to safe some space. Would it be possible to replace it by a (non-standard) smaller connector, like a micro usb connector or a JST PH connector? Which connectors would be suitable and why (or why not)? <Q> Yes, there are already many ways this is done, laptop dongles are one way. <S> HDMI has Ethernet routed in it. <S> Whatever cable\connector you use needs to be able to handle differential pairs with an impedance of 100Ω <A> While using Ethernet, then de facto standard is rj45. <S> The industrial Ethernet specifies some flat connectors as well, but you probably don't want to pay the higher price for the connectors and special cables, also your other end device requires the special connector then as well. <S> It is not possible to send Ethernet signal over JST, because of interference and electromagnetic behavior of Ethernet (it is really fast), it is just not up to specifications. <S> USB 2.0 doesn't have enough cables for that purpose, but you could set up Ethernet on USB (the other end would be USB as well).Lucky exploring. <A> You can use any connector or even solder directly wires on the PCB. <S> The drawback is that you may not be able to be compliant to Ethernet standard, have EMC issue and/or can't reach 100m. <S> But if it's for debug purpose, usually you don't need all the previous. <S> This little thing (35x25 mm) feature a Gigabit PHY (KSZ9031) and have a soldered RJ45 cable (2m, 5m or 10m). <S> It is used for debug purpose and have been developped due to very tight space constraints and especially the board needed to be as flat as possible. <S> We were able to stream 400Mbps on 5m cable without any issue.(RGMII is provided through a connector on bottom side) <A> IF you're not interested in following any particular standard, just achieving a smaller connector footprint within a system you have control over, then I suggest using a USB-C connector. <S> With careful allocation of the 4 x Superspeed Differential Pairs, combined with Auto MDI/MDIX switching available in many Ethernet Interfaces, it should be possible to use a standard USB-C cable. <S> The two-fold rotationally-symmetry is not then a problem. <S> Impedance is well controlled, and should be fine for short distances. <S> A further possibility is additionally using the power conductors in the USB-C Cable for power distribution.
| Note that the USB-C Cables are not quite rotationally symmetric pinout-wise.
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Contradiction:Maximum Power Transfer and High resistance of load Whenever we load a circuit, there's a drop from the open circuit voltage \$v_{oc}\$ due to the internal source resistance or Thevenin's resistance \$R_{th}\$ . So we need to have an \$R_{load}\$ that is very large compared to \$R_{th}\$ , (i.e) $$R_{load}>>R_{th}$$ so that almost all the voltage drop occurs across \$R_{load}\$ . But according to maximum power transfer theorem we need to have an \$R_{load}\$ that is equal to the \$R_{th}\$ to obtain maximum power transfer. Aren't these two statements contradictory, maybe contradictory isn't the right word, but you get it. So, which one do we need to follow while constructing a circuit. <Q> You are in good company. <S> The maximum power transfer concept frequently causes confusion. <S> Take note of the fact that maximum power transfer does not mean maximum power efficiency. <S> I think this may be the main point that trips people up. <S> In fact, when Rload = Rsource, then the maximum possible efficiency is 50%, since the load and source will be consuming equal power. <S> But often, it is more important to protect the source from getting hot, or to maximize power efficiency from the standpoint of DC input power to the amplifier. <S> In these cases, Rsource should be much lower than Rload. <S> You see this in audio amplifiers, for example. <S> Also, usually when you are discharging a battery, you make sure Rload is much greater than Rbattery to avoid over-heating the battery. <S> Another point is that when signals need to pass through transmission lines (such as coaxial cable or similar) the load must be matched to the source (and line) to prevent reflections or other signal degradation. <S> Key points <S> Maximum power transfer is not maximum efficiency. <S> When maximum energy efficiency is desired in an amplifier, source impedance should be kept low compared to load. <S> Sometimes, when transmission lines are involved, power concerns are secondary, and signal integrity dictates that load and source be matched (to avoid reflections). <A> You are confusing power transfer and voltage transfer. <S> To have maximum voltage transfer <S> Rload > <S> > <S> Rsource. <S> To have maximum power transfer Rload == <S> Rsource. <S> The reason is that as Rload goes up, more of the source voltage is dropped across Rload but the total current flowing in the circuit decreases. <A> The answer to the apparent contradiction relates to the inverse relation between resistance and power, mixed with a direct relation between resistance and voltage. <S> These thoughts can quickly become misleading. <S> After all, the power dissipated by a resistor is proportional to voltage <S> squared: $$P = <S> \frac{V_{load}^2}{R_{load}}$$ <S> So the first instinct is to assume to maximize \$P\$ , one should maximize \$V_{load}\$ . <S> Using a Thevenin voltage source, the resistor divisor means the maximal \$V_{load} = <S> V_{th}\$ <S> occurs when \$R_{load} > <S> > <S> R_{th}\$ . <S> However, remember another expression for power: $$P = <S> V_{load} <S> I_{load}$$ <S> you can quickly assess that, increasing \$R_{load}\$ all the way to \$+\infty\$ (an open circuit), no current flows and \$P=0\$ . <S> In the same manner, with \$R_{load} = 0\$ <S> (a short circuit) there is no voltage drop, and again \$P=0\$ . <S> This means the maximal power transferred is neither at an extremely high nor extremely low \$R_{load}\$ . <S> In fact, the maximal power is when both current and voltage meet "halfway". <S> This happens when \$R_{load}=R_{th}\$ .
| If the goal is to extract power from the source at the highest possible rate, then the load and source should be matched.
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IP Camera and USB data transmission over Ethernet Cable Currently I am working on a project that requires connection of the Arduino and IP Camera over single Ethernet cable to a laptop and additionally supply power to a robot. The goal is to stream Video from camera and communicate with Arduino board. 4 wires from Ethernet cable are required for the camera. 2 wires are needed for power supply. That leaves me with 2 spare wires for USB connection Rx/Tx but no wires left for USB ground. My question is: Do I need to run a separate GND wire for USB connection? Or will the GND be shared over RJ45 IP camera connector? And is it safe to operate? I am not sure how camera data transmission works, but it's logical for me to assume that camera and laptop share GND. simulate this circuit – Schematic created using CircuitLab <Q> A USB cable already contains a ground, so there could be a potential for a ground loop which can create noise. <S> If that is a problem, then you may want to consider isolating the 12V. USB should not be run over cables that do not have the proper tranmission line impedance. <S> USB cables should have an impedance of 90Ω and twisted pair. <S> Otherwise the error rate could go up significantly. <S> USB should be ran in it's own shielded cable and not with other signals that could introduce crosstalk. <A> You do not need a separate ground for the arduino and the IP camera. <S> You can share the single ground wire. <S> From a signal integrity standpoint, your setup may be suspect and could have USB data errors. <S> But I suspect that if your length is not too long, and you are using 12Mbit USB (as opposed to high-speed 480Mbit), it will probably work. <S> Be sure to use the twisted pairs in your cable appropriately <S> : use one pair for the power/ground, and another pair for the USB Dp/Dn <S> *. <S> *Just as a point of clarification, the USB lines are not transmit and receive as you describe, but are a differential pair (+/-) <S> that is used for bi-directional communication. <A> Switch to RS232 instead - that is what the original Arduino had, today you would need an add-on level shifter - and then use a USB serial adapter (ie, with a DE9 connector) at the computer. <S> Ethernet does not provide a ground; but if you are supplying DC power, you would be providing a ground there. <S> RS232 voltage levels should give you some noise margin. <S> Another option would be to put more of a "computer" at the robot end, which would multiplex the Arduino data over Ethernet; the downside is that many such solutions end up having delicate file system <S> state which is hard to maintain reliably if your robot can unexpectedly lose power. <S> Power consumption will also likely increase. <S> But something like an MT7688 board that brings out two distinct Ethernet ports (one for the IP camera, one for backhaul) could be an option. <S> Or a single Ethernet, and use a USB or dedicated camera on the embedded computer, rather than a complete IP camera.
| You really should not be trying to use USB over non-trivial distance or with alternate wiring - but fortunately you do not need to for an Arduino project. For an even better solution, use RS485 differential signaling with transceivers and a software protocol which permits both to device and from device data to share the same differential pair.
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Is it possible to convert 6V 4A (24W) DC to 12V 3A (36W) DC using step up booster? I have Yamaha RX100 1996 model which has 6 V power supply. This 6 V power supply is the main drawback for the headlight, where the circuit diagram is shown in these images: The official bulb requirements are given below: For further details about the circuit diagram open the link below (page - 63): Yamaha-RX-100-Workshop-Manual . I plan to convert 6 V 25 W to 12 V 36 W using either this boost converter: 6V 9V to 12V DC DC Boost Converter Step Up Module, 1A 2A 3A Power Supply , or; 6V to 12V Step Up Converter, DROK DC 12V Boost Converter DC 5V-11V to 12V Voltage Regulator Module 3A 36W Volt Transformer Power Supply for Car Stereo Radio LED Display Is it possible or not? <Q> No. <S> To put it quite simply, you cannot have more power out than you have going in. <S> You can increase the voltage available, you can increase the current available, but you cannot increase the available power. <S> 25W going in means a maximum of 25W out. <A> IF your question was about power supply conversion, the answer WOULD be a solid "No". <S> But... <S> 25 W is the power consumption of the bulb, not the power supply output. <S> You could step up the power as you asked IF you checked all the components involved in that bulb's circuit (including wiring, connectors, etc) making sure they can handle the extra power you push through them. <S> That sounds like a lot of pain, but possibly doable. <S> In case the above path isn't feaseable, you could try using the original circuit to control a relay powered directly from the battery (fused!). <S> Also, the increased bulb power output would put out more heat than the orignal one and could damage it's housing, reflector and connector. <S> Maybe it would be best to check for a compatible 6V bulb. <A> If you look at the connections in the light switch, only the park light is fed from the battery supply (brown wire), the other lighting is fed from the Y/R wire from the magneto coil, so the headlight supply is AC. <S> That precludes the use of a DC-DC converter, unless you rectify it first - and with half-wave rectification you won't get the same power out of the magneto. <S> The coils on these magnetos are often wound with a specific load in mind, so the terminal voltage is determined by that load - put a lower current bulb in the circuit and the terminal voltage will rise, though probably not by enough to successfully run a 12V bulb - but the total power available is likely limited by the magneto coil. <S> There's a separate winding (or maybe a tap) that goes through the rectifier/regulator to charge the battery. <S> The regulator is typically a shunt type - essentially a large Zener diode that holds the maximum voltage down by dissipating the excess generated power as heat. <S> Running a DC-DC converter from the battery to power a 36W bulb will likely result in a flat battery.
| One possibility might be to find an LED bulb you can power from this circuit, either directly or through a DC-DC converter if you're looking for more light output, but you're unlikely to be able to get more out of the magneto to get a brighter incandescent bulb.
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What code issues could persist past a reset? (STM32F103) I have a project running off a Bluepill board (STM32F103C8T6) that ceases to function only after it runs for ~6 hours. Up to that point it behaves totally normally, and UART logs don't indicate anything out of the ordinary until it just stops. This has made it very painful to debug since it's so difficult to gather information and attempt iterative solutions. The only lead I have is that when I find the device in the hang state, it cannot be reset properly . In the first few (correct) hours of operation, rebooting the device plays a speaker and blinks an LED. After the hang, resetting results in a solid LED - no sound, and no operation past that point. Only when I power cycle the device does it return to normal operation, and then everything works exactly as I want it to for another ~6h before it hangs again. Conventional wisdom would suggest this is a memory leak or a timer overflow, but I don't see how either of those could persist past a system reset! The Bluepill is attached to I2C and SPI devices, but even when those are removed from/returned to the circuit it does not reset properly, and there's nothing else in the circuit that can hold state. I've also heard that an unconnected VBAT could cause this but it's connected to VDD, and reading the datasheet shows no unexpected NRST behaviors. I'm reluctant to post all 250 lines of code for this project because I haven't been able to narrow down the cause of this at all, so I'm trying to stay general with this question. What kinds of code bugs are capable of surviving an MCU pin-based reset, but not a power cycle? EDIT: Here's the schematic. VBAT was manually connected to the 3.3V pins later. EDIT: This is terrible form for stack overflow but here is a pastebin of the code. I hope to condense this to only the problematic code ASAP but due to the massive repro time I've had issues narrowing it down. https://pastebin.com/nKDyzRKa <Q> Maybe something is being clobbered that you are not initializing in your code and that is not automatically initialized by the reset. <S> When the power is removed it may take on a random value that happens to be okay. <S> 250 lines of code is not very much, but maybe you can further reduce it to try to isolate the problem. <A> The microcontroller datasheet p. 59 mentions a few things that EMC hardened software might worry about: <S> Corrupted program counter Unexpected reset Critical Data corruption (control registers...) <S> It also says you can provoke some of these failure mechanism by manually pulling RSTN or Oscillator pins down for ~1 second. <S> The datasheet also mentions some failure monitoring mechanisms (supply, clock) that could trigger an interrupt. <S> Have you tried reading the interrupts after the failures or is the device non-responsive on all interfaces? <A> In your case you said there are I2C peripherals so the answer is easy. <S> For example what happens when the reset button is pushed while MCU is reading or writing data to/from slave chip? <S> The slave chip does not know that the CPU just went away and if the slave was pulling SDA pin low (data or ack bit), it will continue to do so forever. <S> If the MCU just initializes the I2C peripheral as usual and tries to start communication with the slave, it can't generate a start condition as it regards the bus being not idle as SDA is not high. <S> So instant jam situation after reset, until power cycling or resetting the slave. <S> Therefore the results of I2C operations should be checked after a timeout expires. <S> Then you can blink leds differently to signal an error. <S> Or perform a sequence to bring the slave out of that situation manually.
| Many kind of issues will persist past reset. If you have a debugger attached to the system when it hangs, the cause could be related to the external system.
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Is there a preferred order in which to apply (analog) filters in a chain? I'm working on a project where a very weak signal need to pass through a bunch of filters. However, I am seeing quite a lot of different ways in which the order of the various filter stages are applied. So my question is: Is there a preferred order in which to apply different filters in an analog signal chain? For example: [weak signal sensor] (input)--> [10 kHz LPF] --> InAmp --> [50Hz Notch (active)] --> [500 Hz LPF-2] --> [20 Hz HPF] --> ADC (output) There is a similar question here but clear answers are lacking and is not related to very weak signals. I'm looking to understand why, for example (above), one want to pre-filter the input before the IA, or in other cases where people decide to place a HPF before LPF. The only thing I have read (from Texas Instruments docs) is that one should apply the greatest gain as late as possible, but that also seem to vary. <Q> There is no 'one size fits all' solution. <S> It depends on the detail of your individual amplifiers, and your input signal. <S> First, you have to see whether it's even theoretically possible to build a signal processing chain from your chosen amplifiers, for your chosen specification. <S> Your specification should allow you to derive a dynamic range for the system. <S> This is the difference between the minimum signal (usually defined by noise and signal offsets) and the maximum signal (usually defined by some measure of distortion (harmonic, intermodulation, spectral growth) slew rate, or even clipping against the rails). <S> Each of your amplifiers will have a dynamic range. <S> All amplifier dynamic ranges should be greater than your specification dynamic range, the greater the margin the better. <S> The amplifier with the smallest margin is the one to worry about first. <S> If your signal, or your specification has a small dynamic range, then your job is easy. <S> The amplifier dynamic ranges will apply at an optimum signal level. <S> As you move lower than this level, the effect of noise will increase, higher and you get more distortion, both decreasing the dynamic range. <S> This defines a signal level profile through the signal processing chain. <S> Does your input signal have any defects? <S> Much HF noise? <S> Much DC offset? <S> Put a HPF first, or a differential stage, to remove the DC and centre-up the signal for subsequent stages. <S> Finally, arrange your stages in an order that meets the signal level profile. <S> If there are lots of ways to do this, lucky you, your job is easy. <S> If there is only one way (let's say it's an off-air signal, so you need a bandpass filter and LNA first) with subsequent stages of gain with increasing signal level handling, then it's straightforward. <S> If there are no ways, then find the limiting element, the one with the lowest dynamic range, build or buy a better one, rinse and repeat. <A> The instrumentation amp generally converts a differential signal to a single-ended signal. <S> The input impedance is high, and equal on both inputs. <S> It has great CMRR. <S> It's generally first in my signal path. <S> The high-pass filter removes any offsets. <S> I generally put this next, as amplification at my InAmp will amplify offsets as well. <S> Any further amplification may saturate any subsequent amplification. <S> The caveat of this approach is that if the input has substantial offsets, the InAmp needs to be capped at modest gain. <S> I tend to not prefilter before the InAmp (other than RF filters usually recommended in data sheets). <S> I don't want to do anything that might dork around with CMRR. <S> If the inputs are people-mounted electrodes, I do put current limiting resistors in place to comply with NFPA99. <S> ADC, of course is last, and is preceded by low pass filters. <S> There may be more LPF's in other stages, as well, and other HPF's, if a large gain stage generates large offsets. <A> As @Neil_UK said, the answer will depend on the details of your application. <S> In your particular case, it seems like you're trying to capture a very weak signal of low frequency (20Hz~500Hz) that may be susceptible to AC power line coupling (50Hz). <S> I would also assume that you don't care about the DC component. <S> In this case I would use a simple AC coupling first that could be as simple as a series capacitor (that may give you the 20Hz HPF you want) followed by a very low noise audio pre-amp <S> (probably using discrete transistors) <S> so you get the best system noise figure you can get. <S> Then I would follow with the LPFs (I'm not sure why you need two LPFs with two different cutoff frequencies) and the notch filter. <A> Sometimes there is. <S> One audio system I remember generated quite a lot of broadband white noise from the thermal noise of some unavoidable high value resistors in one filter stage. <S> The simple solution was to move that stage ahead of a low pass filter like your 500Hz LPF, which then attenuated almost all of that broadband noise.
| I absolutely agree with Neil, there is no one-size-fits-all solution. Put a LPF first, to protect the slew rate of later stages.
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How do you know if impedance is the same for 2 ICs? in my last question I got some very interesting answers that made rethink a lot about my design as to whether or not I need termination resistors at all. Additionally, the design proposed in this answer as well as the reply in this particular question made me rethink about my approach in general. Now, this is probably going to be a lengthy question, but I hope someone will have the patience to read and clarify things. :) Background I am working on a project for sometime now which, among other things, involves the design and implementation of the communication between an FPGA and a USB 3.0 bridge, the MachXO2 and FT601 respectively. It is my first high speed design and I aim to gain experience and insight on the principles of this kind of design. I have spent last month studying a great amount of literature including resources from here , various other articles from credible sources (like T.I. or questions asked here) as well as Mr. Bogatin's book, "Signal and Power Integrity - Simplified". Due to time contraints, I have not studied in as much depth as I would like, but I think I have got the fundamental ideas. My problem and confusion lies in the application of those. First Approach My first approach on the subject was by examining the data I have for both ICs in order to determine whether or not I can get away integrity issued by keeping the traces short enough. FT clocks at 100 MHz but does not provide rise time. On the other side, the MXO2 provides an IBIS file for the design, which reports a rise time of 0.8 ns, which seems rational to use as a rise time for FT too. For 0.8 ns PCB Toolkit reports a max trace length of 1.25 inches. Although I achieved such length, the traces where tightly packed which may introduce significant crosstalk. Additionally, simulation using MXO2 IBIS for my design in Altium revealed significant overshoot and undershoot. Now I understand that this may not be entirely correct since I don't have a proper IBIS for FT, but it got me alarmed anyway. Terminating R's In order to be on the safe-side, I decided to use series resistors and terminate the traces with 50Ω. The PCB, of course, is designed in such way that the particular traces have an impedance of 50Ω. I chose that value because FT Configuration Utility lets you config the pin drive strength as 50Ω so we are fine by this driver, and MXO2 drives the lines as 3.3V, 8 mA which roughly translates as 50Ω too by what I have read. Also, the fact that many of the lines are bidirectional perplexes the situation even more. By what I have read, bidir lines must terminate both ways. So, I placed R's in my design both in start and end of the lines, run simulation again and I got some smooth curves. Then I made my previous question. My Source of Confusion Now, you may wonder why since both drivers seem to output at 50Ω, did I use the termination scheme. And my answer is because I don't know each receiver's impedance. And this is where the confusion starts for me. Theory and stuff is all good, but I am stuck in the practical application. According to theory, if Zt, Zr and Z0 are the same (or close) then reflections will not occur. I know Z0 since I designed it, but what about Zt and Zr? And to further complicate things, those lines are bidirectional so they switch functionality. For the FT, I know it can be 50Ω since you can configure the chip to do so. But although MXO2 at LVCMOS33, 8 mA should have 50Ω too, and IBIS viewer reported that line have actually a lot less like 25-30Ω or so, which complicated matters more for me. But regardless of the IBIS file (many vendors don't provide one anyway), my confusion sum up to this simple and probably naive questions: I have the datasheets of both ICs. Where is it written, implied or can be calculated/extracted the input/output impedance of the pins? Is the output impedance the same as input impedance for I/O pins? Does the specification of the protocol (in this case LVCMOS33) specifies the I/O impedance? How a designer is supposed to know that impedances match between those 2 ICs so a controlled line is all that is required? Since the design mentioned in my other question does not use R's, then the only reasonable answer I could think of is that probably the whole design is indeed a specific impedance as well as the FPGAs attached to. All that probably seem quite trivial for an experienced designer, but have really puzzled me for quite some time. If my reputation permitted I would gladly offer a bounty for this question. I did search but I could land to any information. <Q> I didn't read your entire wall of text, but I don't think the answers to your questions depend on much of what you wrote. <S> I have the datasheets of both ICs. <S> Where is it written, implied or can be calculated/extracted the input/output impedance of the pins? <S> Different vendors have different conventions for how they report it. <S> One might report an output impedance. <S> One might report an \$|S_{22}|\$ limit. <S> One might report a typical \$S_{22}(f)\$ curve. <S> For some kinds of ICs, input and output impedances aren't specified at all. <S> Is the output impedance the same as input impedance for I/O pins? <S> Not necessarily. <S> For a digital buffer, you'd expect input impedance to be very high and output impedance to be very low, for example. <S> Does the specification of the protocol (in this case LVCMOS33) specifies the I/ <S> O impedance? <S> For CMOS, yes, you can expect a "very low" output impedance, maybe a few ohms. <S> And input impedance will be dominated by capacitance, which will very often be specified. <S> How a designer is supposed to know that impedances match between those 2 ICs <S> so a controlled line is all that is required? <S> For CMOS designs, generally you don't match impedances, you just keep the lines short enough that it doesn't matter. <S> A few ohms series resistance on drivers may be used to reduce ringing, but still an actual matched termination is not expected. <S> One thing I did pick out of your text, <S> For 0.8 ns PCB Toolkit reports a max trace length of 1.25 inches. <S> One option is to use a series resistor at the source to increase the rise and fall times. <S> If you increase the edge time to a more reasonable 2 ns (for a 10 ns clock period), you will get a maximum unmatched trace length that's much easier to work with. <A> This is a point to point link? <S> Then source termination by the drivers output impedance is probably all you need if you can configure the driver to match the line impedance (which it sounds like you can). <S> If for example, the driver is 50 ohms and the line impedance is 50 ohms driving a cmos input (so essentially a slightly capacitive open circuit) <S> then you will get almost a 100% reflection at the receiver, which is fine because it will be correctly terminated by the driver output impedance. <S> Note that the double voltage at the receiver compared to the voltage propagating in the line is correct because of the divider formed by the line impedance and the output impedance of the driver, obviously once the reflection arrives at the driver and is terminated the steady state condition applies and line impedance is no longer relevant. <S> Where this sort of thing goes pearshaped is if you have either a multi drop bus topology or very stiff drivers, either of which will wind up needing some form of termination to get good behaviour, but it does not sound like that <S> is the case here. <A> There are many ways to interconnect two devices. <S> Depending on timing and speed of connection, you can use a spectrum of connection, starting from single ended unterminated lines, to many sorts of impedance-controlled differential transceivers. <S> I am not sure about Altera, but just get almost any of Xilinx datasheets or I/O design guides, like this one , and you will see whole variety and many available options.
| You'll have to read each individual datasheet and see how they spec their chip.
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How to determine total tolerance of a ceramic capacitor? For the sake of this question (as an example only) lets consider MLCC X7R ceramic capacitor. Such a component among others is characterized by a number of tolerances. Temperature dependeant tolerance, DC bias voltage tolerance or aging tolerance as a few of them might be listed. Ommitting the problem of determining tolerances themselves (assuming knowledge of particular tolerances values), how to calculate total tolerance of a capacitor? Is it just a sum of all tolerances or it is more complicated (e.g. there exists some sort of tolerance "propagation")? <Q> For a first order worst case analysis (WCA), you can assume that all tolerances are summed (stacked). <S> The real complication occurs when you have multiple components. <S> Since it often isn't obvious if positive or negative tolerance is worst case for each component, 10 components would require 1024 different analyses. <S> Some automated simulations can be setup to calculate all possibilities. <S> If absolute worst case doesn't pass, and redesigning is expensive, there are other techniques that assume that all parts won't have worst case errors at the same time. <S> RSS (Root Sum Square) and Monte Carlo are 2 methods. <S> Certainly for errors caused by temperature, parts on the same PWB will have similar temperatures, so it is unfair to assume that all will have worst case errors at the same time. <S> https://pdnotebook.com/statistical-tolerance-analysis-basics-root-sum-square-rss-632f2eb274ba <A> MLCC and for example the classification X7R is just a classification not anything more, it's not a material group or determined structure. <S> There for the properties varies between X7R capacitor from manufacture 1 and 2 and different models. <S> It just says that this MLCC full files the X7R classification. <S> Therefor the attributes depending on temperature and voltage is different, between different models. <S> The easiest way to determine the total tolerance of operation for the chooses capacitor would therefor in my opinion be to go to the specific manufacturer that hopefully has enough data publicly available so you can take out the tolerances for the operations of the capacitor of your use <S> and then the ageing deviation from there. <A> Capacitors possess a capacitance tolerance around the nominal value but temperature coefficients, DC bias and ageing are distributions (not tolerances) that cannot be summed.
| I don't know that there is any such thing as "total tolerance" of a capacitor.
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How to remove these lines in Altium Design In my design, there are some lines with "=" on them. Can someone tell me why these lines appear and how to delete them? I'm using Altium18. Thank you. <Q> From the Tools Menu, select "Reset Error Markers". <S> That should solve your issue <A> Quick trick: if you want to check for un connected net, just press ALT and select over your whole board. <S> This selects connections (unconnected) <A> This is one of the more annoying misfeatures of altium. <S> Error markers for un-routed nets that are redundent with the rats nest lines and don't dissapear as you route the nets or move around as you move the components, they just stick where they are until you run the next design rule check. <S> As Elmesito says you can do a "reset error markers" but that will remove all violation markers. <S> You can also run a design rule check with "un-routed nets" unticked <S> but then you won't get them in the report either. <S> I'm not sure if there is a way to get it to include un-routed nets in the report but not to add those stupid redundent violation markers for them.
| Those lines are error markers that are there to show that the net is not connected with a trace in copper (open circuit).
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bldc motor, esc and battery draw, nominal vs peak I want to better understand how Electronic Speed Controllers (ESC) work for BLDC motors. Specifically, I am trying to understand battery current draw. Question 1:When an ESC is rated at 200 amps nominal and 500 amps peak is it safe to assume that the draw on the battery pack will not exceed 200 amps? I'm rationalizing this in my head by assuming the capacitors in the ESC are just discharging more current than normal for short durations when this 'peak' value is experienced. I'm also assuming it will never try to pull 500 amps from the battery pack. If this isn't right, please explain. Question 2:If the motor is run indefinitely at max throttle (theoretically) will the ESC draw 200 amps from the battery continuously or less than that? I'm now thinking the draw will be more along the RMS value of the output pulse wave. This would explain why the ESC input wires from the battery pack are relatively small in relation to the output current rating. When I see 200 amps I think of thick 2/0 awg gauge wire, not the 6 awg or 8 awg gauge wires I see on my ESC. Question 3: (slightly off topic)To modulate motor speed does the ESC restrict current flow or voltage? I assume it is current, I just want verification. Let's say the throttle is at 10% and assume the throttle curve is linear. I just want to understand the theory of how ESC's work at a high level. Much thanks! <Q> Question 1: When an ESC is rated at 200 amps nominal and 500 amps peak is it safe to assume that the draw on the battery pack will not exceed 200 amps? <S> I'm rationalizing this in my head by assuming the capacitors in the ESC are just discharging more current than normal for short durations when this 'peak' value is experienced. <S> I'm also assuming it will never try to pull 500 amps from the battery pack. <S> If this isn't right, please explain. <S> Without a datasheet we're guessing but the 500 A peak current is likely to occur during start-up and this may last one or more tenths of a second. <S> A capacitor to supply 500 A for 0.1 s at a useful voltage will be very large. <S> It's much more likely that any capacitors on the DC input stage are acting as short-term filters to pass high-frequency noise to ground. <S> Question 2: <S> If the motor is run indefinitely at max throttle (theoretically) will the ESC draw 200 amps from the battery continuously or less than that? <S> It will need to draw enough to power the motor. <S> It completely depends on the load the motor is given. <S> This would explain why the ESC input wires from the battery pack are relatively small in relation to the output current rating. <S> When I see 200 amps I think of thick 2/0 awg gauge wire, not the 6 awg or 8 awg gauge wires I see on my ESC. <S> This may be a warning sign that you have a low quality product. <S> Question 3: (slightly off topic) <S> To modulate motor speed does the ESC restrict current flow or voltage? <S> I assume it is current, I just want verification. <S> Let's say the throttle is at 10% and assume the throttle curve is linear. <S> The ESC will operate using pulse-width modulation (PWM) as it results in very low losses in the switching transistors. <S> PWM applies pulses of 100% voltage to the motor for a short period, switches off, waits and then repeats the process. <S> Figure 1. <S> PWM signal transitioning from high pulse width (75%) to low (25%) and back again. <S> Note amplitude remains constant. <A> All answers assume you're talking about "hobby" ESCs, as used in planes, boats, cars, robots, etc. <S> Question 1: When an ESC is rated at 200 amps nominal and 500 amps peak is it safe to assume that the draw on the battery pack will not exceed 200 amps? <S> No. <S> For the most part, the ESC draws what it draws given the motor, battery and load on the motor. <S> An ESC could be designed to limit current in such a way (I've designed industrial motor controllers that do), but AFAIK ESC's don't. <S> Question 2: <S> If the motor is run indefinitely at max throttle (theoretically) will the ESC draw 200 amps from the battery continuously or less than that? <S> See answer to question 1. <S> Question 3: (slightly off topic) <S> To modulate motor speed does the ESC restrict current flow or voltage? <S> Neither. <S> At the level of the throttle command to motor, it supplies a fixed PWM to the motor -- essentially it chops the battery voltage and hands it to the motor. <S> This is complicated by the fact that some ESC's have a governor function, that servos the motor speed to the signal coming in -- but they still have an internal throttle command that works as above. <A> The max operating temperature or heat rise above 25'C is what limits the nominal continuous current rating . <S> The peak is an absolute maximum which may or may not be protected and should never be exceeded, such as starting a couple of these motors below at full throttle. <S> One must mind the rated peak currents to prevent damage to the electronics. <S> Similarily, a motor full load <S> current rating is defined by the same method of temperature rise above 25'C for the winding hot spots. <S> However every market does not use the same hot spot temperature for a long useful life and if no standards are given or not temp rise is specified, you might guess they don't want you to know. <S> However it is quite common to give the peak motor current as surge, start or locked rotor current. <S> These are all the same as is simply Ohm's Law applied to the DC resistance or DCR, so Imax= V/DCR . <S> This is typically Imax = <S> 8 to 12 times the rated current in modern BLDC, high-efficiency DC motors. <S> Here they generously rate the motor current with a 180 second limit. <S> This would be at max internal temperature.
| Some of the better ESC's do have current limiting, but experience in the flying field is that if you dork the thing into the dirt at full throttle, chances are that you'll damage the electronics due to overcurrent.
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Converting a sprinkler system's 24V AC outputs to 3.3V DC logic inputs I am working on a system that can detect which of the AC outputs on a sprinkler box is on and for how long. The sprinkler box has one common wire and then 18 connections that output 24V to open sprinkler valves. My plan is to use several full bridge rectifiers which connect to some transistors. The Base of the transistor connects to the positive end of the bridge rectifier, the collector connects to a 3.3V line, and then the emitter is connected to ground so that when one of the sprinkler leads turns on it grounds the corresponding transistor. I then have a micro controller looking at the voltage of the collector side of the transistor to see which lead is on. The problem I am having is that when one of leads from the sprinkler box turns on it flip on all of the transistors and grounds all of the logical leads to the micro controller. I did some tests with a multi-meter and when one of the leads of from the sprinkler leads turns on it is at 24Vrms and all of the other outputs are at about 1.4Vrms. The Voltage after being rectified is 35V on the input which is currently on, and then 34V across the all of the other rectifiers. I don't know much about rectifying AC current to DC current but I am thinking the issue is with the common wire. I think the common wire may be creating essentially a half bridge rectifier and outputting a lower voltage that is still causing the transistor to ground the input. Anyone have any ideas of either how to fix this problem, or of what is actually happening? <Q> the problem with your circuit is that the input can cross-talk via the bridge diodes and the capacitors. <S> in1-d1-c1-ground-d2-in2 and in1-d1-ground-c2-d2-in2 etc. <S> try this, it should give better isolation between inputs. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Optoisolators (AC input) would be an easy solution. <S> but it will be slow, which probably doesn't matter in this case). <A> I expect your problem is the Common is high side with low-side switching , and meanwhile you have low-side detection common, which will result in 2 diode drops at low current and thus 1Vdc lower on all the other bridges. <S> You can easily verify this and if so a better detection design be implemented.
| Or half-wave rectify the inputs (just a diode rather than a bridge and increase the capacitor values..
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Why output is high even though trigger pin is high in Proteus simulation of IC555? In the below attached circuit of IC NE555 Proteus 8.0 simulation, TRIGGER pin (Pin 2)is set at Vcc. So why the output is high? It should be low since voltage at TRIGGER pin (Pin 2) is more than 1/3 of Vcc. Please let me now if there is anything wrong in the circuit or is it a problem in the simulator? <Q> The trigger being low (less than 1/3 Vcc) forces the output high. <S> But the trigger being high does not do anything to the output. <S> Internally, the 555 is a set-reset flip-flop driven by a pair of voltage comparators. <A> The trigger, TR, is active low. <S> Try connecting it to GND. <S> Add a series resistor - 1k should do - between Q and D1. <A> The highest priority input is RESET. <S> But you have that tied high. <S> The next highest priority input is TRIG. <S> But only if it is low (below it's \$\frac13\$ \$^\text{rd}\$ <S> threshold.) <S> You have that tied high, as well. <S> So TRIG doesn't dominate, either. <S> The next and last priority input is THRES. <S> But only when it is high (above it's \$\frac23\$ \$^\text{rd}\$ <S> threshold.) <S> Here, you have it tied low. <S> So none of these inputs dominate -- the reset isn't actively resetting the FF state; the TRIG isn't actively setting the FF state; and the THRES isn't actively resetting the FF state. <S> This probably means (I don't know for sure) that the internal FF comes up in a default state that yields the result you see. <S> I don't recall seeing specific documentation about the power-up state of the FF in the conditions you are using. <S> But that's what you need to look for, I think. <S> It will probably explain the results you observe, once found. <S> (And which the simulator simulated.) <S> I know it is just a simulation <S> and I'd guess that you are just exploring and that this isn't an actual circuit you intend building, but as Transistor mentioned you will need to use a current-limiting resistor in series with the 555 output and the LED (unless you buy an LED with a resistor or current limit built into it, I suppose.)
| To force the output low, you must either drive the reset pin low or take the threshold pin (pin 6) above 2/3 Vcc.
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Spurious Ouputs from Nexperia 74HC154 4 to 16 Decoder/Demultiplexer Problem: Driving 74HC154 Decode with Atmega328 using Arduino software. Decoder Outputs Y0 – Y10 (pins 1 – 11) seem to be picking up signals from each other. The outputs are not random noise. Changing which output pins are being turned on changes which output pins have the spurious output signal. This is occurring with the 74HC154 outputs not connected to a circuit and also when connected to an LED array. Since 74HC154 is being used to turn PNP transistors on and off, the spurious signals are incorrectly turning on PNP transistors. Test Setup: Breadboard with Atmega328 running Arduino software. Nexperia 74HC154 Pinout and 74HC154 Truth Table Example 1: Arduino Code: //Inputs into 74HC154 pins 20-23 int InputA0 = 4; int InputA1 = 5; int InputA2 = 2; int InputA3 = 3; void setup() { pinMode(InputA0, OUTPUT); pinMode(InputA1, OUTPUT); pinMode(InputA2, OUTPUT); pinMode(InputA3, OUTPUT); } void loop() { //Output Y0 digitalWrite(InputA0, LOW); digitalWrite(InputA1, LOW); digitalWrite(InputA2, LOW);digitalWrite(InputA3, LOW); delay(1); //Output Y7 digitalWrite(InputA0, HIGH);digitalWrite(InputA1, HIGH);digitalWrite(InputA2, HIGH); digitalWrite(InputA3, LOW); delay(1); } Outputs Y0 and Y7 look great. Output Y6 should be off, but has 5 volt short spikes/bursts. These spikes seem to line up with Y0 and Y7 transitions from High to Low. Same with Y3 and Y4 - Spikes..why? Outputs Y2 and Y5 are off as they should be with no spikes. Outputs with spikes: Y1, Y3, Y4, Y6, Outputs without spikes: Y2, Y5, Y8, Y9, Y10 Now the interesting part: Example 2: Turn off Y0 and turn on only Y1 and Y7 //Output Y1 digitalWrite(InputA0, HIGH); digitalWrite(InputA1, LOW); digitalWrite(InputA2, LOW);digitalWrite(InputA3, LOW); delay(1); //Output Y7 digitalWrite(InputA0, HIGH);digitalWrite(InputA1, HIGH);digitalWrite(InputA2, HIGH); digitalWrite(InputA3, LOW); delay(1); Spikes now on: Y3, Y5 No spikes on: Y0, Y2, Y4, Y6, Y8, Y9, Y10 So Y3 kept the spikes, and spikes went away on Y4 and Y6.Y5 which had no spikes, now has spikes. Finally, this one may have a clue in it. Example 3: Turn on only Y4, Y7 and Y8 //Output Y4digitalWrite(InputA0, LOW); digitalWrite(InputA1, LOW); digitalWrite(InputA2, HIGH);digitalWrite(InputA3, LOW); delay(1); //Output Y7 digitalWrite(InputA0, HIGH); digitalWrite(InputA1, HIGH); digitalWrite(InputA2, HIGH);digitalWrite(InputA3, LOW); delay(1); //Output Y8 digitalWrite(InputA0, LOW);digitalWrite(InputA1, LOW);digitalWrite(InputA2, LOW); digitalWrite(InputA3, HIGH); delay(1); The clue is that spikes appear in Y4 only when BOTH Y7 and Y8 are also on. Y4 does not have spikes if only Y7 OR Y8 are on: Spikes appear to be related to internal switching within 74HC154. Question: Are these spikes internal to the chip? Is there any way to eliminate them? Is this an issue with the Nexperia 74HC154? <Q> You have encountered the well known phenomenon of glitching in multi-bit logic transitions. <S> Your problem is especially severe, because your Arduino code changes the controlling word one bit at a time, using the very slow digitalWrite() <S> function for each bit in turn. <S> You could substantially reduce the glitching if you used bits from the same ATmega GPIO port, and wrote them all at once by writing directly to the port register. <S> However, even changing all the input bits to a logic function at exactly the same time does not mean that the output will change cleanly from one state to another. <S> For this reason, systems which must avoid spurious outputs rely on synchronous logic whereby the output of a combinatorial function is held in a register, and the time when the inputs change is purposefully separated from the time when the register is clocked to latch a new result, permitting the glitchy transition to complete and settle before "anything looks at" the result. <S> So adding a register to the output could be a complete solution. <S> If your application does not require that some output always be active, this can be a simple fix. <S> Another solution for output demultiplexing is to use something like an I2C <S> I/O expander, where you clock data into the peripheral chip's registers, but the new value only appears on that chip's output pins when the transfer operation is fully completed. <A> You should get a lot from Chris' answer, but just to add a suggestion you could use a 74HC374 latch which has both the synchronous nature and the output enabling ability. <S> You connect the inputs of the 374 to the outputs from the micro, then once all the bits are written to the ports you clock the 374 to transfer the whole word to it's outputs which you then connect to the decoder. <S> Depending on how many LEDs you need you could eliminate the decoder and use the 374 outputs directly or add another 374 with it's own clock to get the 16 outputs. <S> This scheme would allow any combination of the LEDs to be on at once if that would be nice for your application. <A> You write the chip input one bit at a time so the output just follows the input. <S> In fact it does exactly what your program tells it to do <S> and that way causes pulses. <S> But even if you changed all inputs at the same time, it might have short spikes because of how signals have propagation delays inside the chip. <S> For this reason there are the enable pins that can disable the chip to stable state when changing the inputs.
| Another option workable in some cases is to use an explicit enable pin to disable all outputs, then change the control word, wait an instant, and re-enable the outputs.
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Is DC-to-DC (24 V to 12 V) buck conversion typically more efficient than AC-to-DC (110 V to 12 V) conversion? While building an online UPS system, I'm basing the choice to eliminate as many AC inverters from the system as possible because of the assumption that DC-to-DC conversion is more efficient than AC-to-DC conversion. I'm not totally sure that this is the case though. I've read some articles stating that buck conversion might only be 85% efficient, yet many product offerings tout 97% efficiency. Also, one of the big selling points on AC is the efficiency of conversion. I am hoping to get some solid information on this specific efficiency comparison. <Q> No general statements can be given. <S> Generally, the larger the input/output voltage ratio is, the smaller the efficiency for basically all converter topologies. <S> However, whether you end up with 70%, 85% or 99% efficiency depends on the actual converter and the load. <S> The reason installation wiring uses higher voltages like 120V or 230V <S> is that the energy losses during transportation will be lower if you need less current to deliver the same power – <S> so, the very standard way of having 120V to racks of equipment and then individual converters where you need your power makes a lot of sense. <S> AC/DC converters are technically so close to DC/DC converters that the difference between those is shadowed by other technical considerations, such as how much you need to "overdimension" your power supply, how much money you want to invest in copper, and whether you're going for constrained size etc. <A> There will be almost no difference in the conversion efficiency of an AC-DC versus a DC-DC convertor. <S> For example if you used a PC power supply to convert 120/240 to 12VDC (I'll simply ignore the other voltages produced), most modern SMPS supplies are at least 85% efficient: Here you can clearly see that the efficiency at some reasonable load range is much better than 85%, and up to 90% for some portion of the load managed. <S> A DC-DC convertor specifically designed for 24-12VDC conversion could be expected to achieve similar efficiencies at reasonable loads. <S> However, there are other differences that may impact your decision. <S> A task designed 24-12V convertor may run at a much higher frequency than a PC power supply and be more space efficient for the same power output. <S> For any given 12VDC load current the 24-12V convertor will have more input current flowing than the PC power supply. <S> This may significantly impact the I^2R losses on the primary side. <S> The device cost (FETs voltage rating, isolation components, inductors) will be significantly higher in the PC supply. <S> You need to decide where your cost vs performance decisions are made. <A> If you need kilowatts, you'd better go AC to DC. <S> In between the efficiency is unlikely to be regulated by this particular choice as pointed out by Marcus. <S> I have used buck convertors with 90% or greater efficiency, and they are pretty well developed nowadays. <A> You will be hard pressed to find any 120Vac to 12Vdc converter as efficient as the best 24V to 12V DC-DC converter in the mid 90% for a 120W range and up. <S> But then the energy must come from somewhere else besides 24V which will also have efficiency losses just as the AC to DC conversion does in the 1st stage.
| I would suggest, based on my own experience, the DC-to-DC route is better for low power situations like just a few watts.
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In automobiles, are we using the car chassis as a ground terminal of battery? Then why we don't get shocked while touching it? My concern is to know if we can use the metallic chassis for a replacement of ground wire. In automobiles, the positive wire is taken to the headlamps and ground is provided by connecting the neutral wire to chassis of car. <Q> To get a shock requires a sufficient current through the body. <S> That requires a few things. <S> Firstly under normal circumstances you need enough voltage to break through the skin barrier. <S> How much voltage this is does vary depending on circumstances but the normal rule is under dry conditions voltages up to about 50 volts are considered safe to touch. <S> Secondly (ignoring static electricity which does lead to shocks but they are too short in duration to be dangerous to the body and ignoring capacitie coupling which is not relevant at DC) <S> current normally flows in loops. <S> No part of the cars electrical system is deliberately connected to the earth, so when you touch the cars body and the earth at the same time there is really nowhere for current to go. <A> Most cars built with metal chassis or metal bodies use the chassis / body for the return to the battery. <S> This saves on wire and weight and of course, cost. <A> If you tied the +12 volts to the earth, you might feel 12 volts across your body by standing with bare feet on the dirt and then licking (with your tongue) <S> a bare part of the car's chassis.
| The voltage between different parts of the cars chassis will normally be a small fraction of the (already low enough to be safe) battery voltage.
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Power factor and cos φ I am confused about the difference between "power factor" and "cos φ". Some textbooks state that those two quantities are equal. Other textbooks state that those two quantities are not equal. And only in one internet resource, I have read that power factor equals cos φ plus non-linear distortion factor. As far as I know, those two quantities are equal when we have ideal sinusoidal current. Please help me, as I cannot continue my education without understanding. Also I will be very happy if you can give me a link to some textbook, in which these topics are covered in more detail. <Q> The power factor is the ratio between real power and apparent power. <S> It is a generalization of the concept of cos φ. <S> In case of a sinusoidal current, the power factor is just plain cos φ, but in case of non-linear current consumption (which is typical for phase-angle control and rectifiers, so a whole lot of electronic devices today), the power factor is affected by the current waveform as well. <S> Power grid operators prefer power factors close to one, because power is (by residential households) paid for real power, whereas the energy losses in distribution depend mostly on apparent power, so power factor compensation , the act of getting the power factor close to unity, is a great deal. <S> In the case of phase shift, the power factor can be brought to nearly one by just adding a parallel inductor or capacitor to the load, so that their reactive powers cancel out and just the true power remains as apparent power. <S> In case of non-sinusoidal current consumption, adding inductors or capacitors still is able to change the apparent power (and thus the power factor), but no amount of parallel inductors or capacitors can bring the power factor to one. <S> So you can split the power factor into two parts: <S> The displacement power factor is introduced by phase shift (called φ) and can be compensated using suitable reactance, while the distortion power factor is introduced by distortion and can not be compensated that way. <A> Power factor (PF) is defined as: \$\dfrac{\text{RealPower}}{\text{ApparentPower}}\$ <S> where ApparentPower is simply the RMS voltage multiplied by the RMS current. <S> RealPower can be more complicated to calculate if the voltage and current are not perfect sine waves of the same frequency (Image source: Envirotec Magazine - Monitoring power factor for effective energy management) <S> In the case where they are however from the diagram above we can see \$\text{RealPower} = <S> \text{ApparentPower}\cdot <S> \cos(\varphi)\$ <S> and \$\text{PF} = cos(\varphi)\$ <S> Where \$\varphi\$ <S> is the phase difference between the voltage and current. <S> For the more general case \$\text{RealPower} = \dfrac{1}{T} \cdot <S> \int_0^T v(t) <S> \cdot i(t) <S> \text{ d}t \$ <S> where \$v(t)\$ and \$i(t)\$ are the instantaneous voltage and current values, with respect to time, \$T\$ <S> is the time for any whole number of cycles. <S> There is more information on calculation here on Wikipedia <A> Best way to understand this is to think of Power Triangle . <S> (Image source: Envirotec Magazine - Monitoring power factor for effective energy management ) <S> Because true power and apparent power form the adjacent and hypotenuse sides of a right triangle, respectively, the power factor ratio is also equal to the cosine of that phase angle. <S> Power Factor of 1 implies that all power is utilized and there is no reactive power. <S> This can be seen by realizing simple values: PF = <S> True Power/Apparent Power PF = 150W/165VA <S> PF = 0.909, which can also be realised as <S> cos(24.6) = 0.909 <S> Looking from the graph above, you can see that as the reactive power is decreased in magnitude - the phase angle becomes smaller. <S> With no reactive power component the angle becomes 0. <S> PF of <S> cos(0) = 1.
| The total power factor is the product of the displacement power factor and the distortion power factor.
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How to make a motor capable of resisting physical movement? Is it possible to make a motor capable of resisting physical movement? For example, at the gym you control the resistance to movement by putting more load, more weight. Is there a way to control that resistance electronically? Maybe by using a motor controller adjusting the PWM signal? <Q> You could use a generator as a variable load. <S> Use a PWM signal to short circuit the output of the generator. <S> Shorter duty cycle is less load, longer duty cycle is more load. <S> A generator is just a motor whose shaft is turned by an external force. <A> Somethink like that. <S> If you wan't a constant torque controller then you should measure the motor current and adjust PWM duty ratio so that the motor current is equal or less the current setpoint (torque setpoint). simulate this circuit – <S> Schematic created using CircuitLab <A> The simplest method, which you will find in a number of stair-stepper, elliptical and stationary bicycles at the gym is to use an automotive alternator. <S> These have the advantage of a wound field, so it's possible to vary the field current, which will be much lower than the output current, by either changing the voltage applied, or use of PWM from a fixed supply, and dissipating the generated output in a load resistor. <S> This also has the advantage of being able to generate the supply for the control electronics and display. <S> With a permanent magnet machine, you would have to control the generated current, with what would effectively be a controllable constant current load. <S> This works for rotary motion, and for reciprocating or linear motion you can arrange some kind of linkage to translate that to rotary - the stair steppers use a chain drive. <S> Transformer E laminations work well for making these.
| Alternatively, you could use a eddy current or hysteresis brake, where the moving element (aluminum or copper for eddy current, steel for hysteresis) passes through the magnetic field generated by an electromagnet.
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How to interpret this timing diagram? I am very confused on how to interpret this timing diagram? <Q> It looks like the colors of the wires in the schematic are supposed to indicate the logic level of the signals. <S> Specifically, the schematic is showing the state of the circuit at the point indicated by the vertical yellow line in the timing diagram, just after the falling edge of S. Q and E are high so they are red; S, R, and Q-bar are low so they are blue. <S> Having said that, using red and blue as colors of signals in the timing diagram seems like a very poor choice to me. <S> I suggest you ignore the colors of the signals in the timing diagram; they don't seem to have any significance. <A> Timing diagrams represent timing to reach valid logic states. <S> Truth tables represent all possible states assuming all timing is valid. <A> The color in the schematic will represent whether the signal is high or low at that particular point. <S> So blue would be '0' (low) and <S> red would be '1' (high). <S> There is however no correlation between the colors in the timing diagram and the schematic . <S> I think this may be confusing to someone that is looking at first without any background on what the colors mean.
| The timing is usually the same to reach all states.
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Can I use 1000v rectifier diodes instead of 600v rectifier diodes? I plan on doing some wiring but the directions are asking for 1A 600v rectifier diodes. I could only get my hand on 1000v rectifier diodes and before I continue I want to make sure this is safe as if I wont cause a fire or a short or make something blow up. <Q> In general, it is safe to be overly conservative with a component specification. <S> If your design calls for 600V rated diodes, then using diodes rated for 1000V shouldn't hurt anything, all else begin equal. <S> However, you have only provided information for the voltage rating. <S> If the 1000V diodes can't handle 1A of current, or are insufficient in some other area, you could still have issues. <A> No* properly designed circuit, however, will rely on the reverse breakdown behaviour of a conventional diode. <S> *Note: there may be exceptions, but they will be exceedingly rare, and I can't think of any case where a Zener or avalanche diode wouldn't be a superior choice. <A> That rating is just the maximum reverse voltage they can take. <S> More is better. <S> If they are recommending a specific part, you may want to look at some of the specs like reverse recovery time or forward voltage drop, but it almost certainly doesn't matter.
| The only situation in which you would not be able to increase the voltage rating (all other performance metrics being equal) would be if the circuit depends on the reverse breakdown behaviour of the diode. There is unlikely to be any negative consequences to doing this, just make sure they are big enough to handle the current.
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In a vacuum triode, what prevents the grid from acting as another anode? A triode vacuum tube has three electrodes, namely an anode, a cathode and a control grid in between of those.Assuming that the filament is hot, current would readily flow from the cathode into the anode when no bias voltage is applied to the grid, like a diode valve. But doesn't the grid also have the ability to capture electrons? Would there be a non-negligible current path between the cathode and the grid? <Q> The grid is maintained at a negative voltage with respect to the cathode (similar to the operation of a N-channel JFET, or a depletion mode N-channel MOSFET), so electrons will be repelled by it. <S> A result is that fewer electrons reach the anode, which has a positive voltage with respect to the cathode in order to attract electrons. <S> Whatever small current flows in the grid is considered leakage. <A> But doesn't the grid also have the ability to capture electrons? <S> You are correct, it does and some non-negligible current can flow. <S> However, to get a current the electrons need to actually enter the grid. <S> If we look at the construction of a triode: we can see that the grid is just a thin wire, so the chance of an electron hitting (touching) <S> it is small. <S> Most electrons will "miss", travel between the grid's wires and reach the anode. <S> Depending on the voltages at grid and anode, the "pull" (due to the electric field) on the electrons from the anode might be stronger preventing the electrons from entering the grid. <A> Apart from tube design specifics (see what is said about grid shapes in other answers), circuit design keeping the grid sufficiently biased negative does. <S> "Grid current" is a well known phenomenon with vacuum tubes biased well into the positive. <S> Considered part of some power amplifier designs (especially transmitters); obviously, the previous stages need to be able to deal with having to feed that kind of nonlinear load. <S> Considered an overload condition in other designs. <A> If the grid goes positive, it does act as an anode & you get grid current. <S> In transmitter class C output stages, they used to use the grid as a rectifier to generate its own bias, which was stored on the coupling capacitor.
| If you run a valve with the grid floating, it will pick up some electrons due to the Edison Cold Plate Effect and bias itself partly off until there is enough negative voltage on the grid to repel any more electrons.
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Should I re-apply conformal coating after hot air soldering? I'm working on a PCB of a drone. It had some failed MOSFETs that I've replaced using hot air soldering. It works fine again, but parts of the PCB were covered in a glue-like substance which I think is conformal coating. I figured the coating would probably interfere with the hot-air soldering, so prior to applying the hot air, I removed parts of it using paint thinner. Now I have a few questions: What is the purpose of this coating? I suspect it's to prevent malfunction of the PCB when it gets moist. Maybe when the drone would fly in moist air or fog. The drone really isn't made to fly in the rain. I'm confused about it because it's only partial - only at the places where components are. I'd expect a coating like that to be applied over the whole surface of the PCB. Or are traces and vias not prone to getting moist because of the solder mask? Could there be another reason for a coating like this? Is conformal coating ever used to prevent service to a board? Like when a manufacturer would rather sell new boards or new devices than having a technician service a broken PCB? Removing the coating before service is a lot of work and might discourage the technician. I've taken off the coating at the place I needed to do the rework. But it takes quite some scrubbing with paint thinner because the surface is very uneven (mounted SMD parts.) The coating becomes very sticky when in contact with the thinner, but needs scrubbing to come off. So where the coating is not completely removed, there is now a mix of re-hardened coating and fibers from the cloth that I was using. (When the thinner evaporates, the coating becomes hard again.) The fibers are very small, and best seen with a magnification glass (hence no photo), but are also in-between the legs of the ICs (see picture 2.) Is it likely that this will lead to problems? E.g. if moisture would be trapped in those fibers? Is it advisable to re-apply the coating? If so: Are there any household products I could use for that? I was thinking about hairspray. Or is that a really bad idea? Picture 1: Partially applied coating. It has a blue-ish/transparent color to it. Picture 2: Closeup of the part I did rework on, prior to the rework. The coating on the bottom of the PCB (picture 1) was clearly applied nicer than the top (this picture). On the top, there is quite a thick layer at some places. I removed the coating near the MOSFETs, but also around the two high pin count ICs (MOSFET driver and microcontroller). <Q> Conformal coating is used to increase mechanical toughness (like abrasion), shock dampening, and to protect against moisture, dirt, grime, and arcing. <S> Certain types of conformal coating are very nasty to get off if you need <S> re work so they can be used to prevent service. <S> But normally you would just use epoxy for that. <S> Note <S> Silicone, for example does prevent moisture from getting to the PCB, but it is not completely impermeable. <S> Therefore, if air pockets exist, the moisture can permeate through and accumulate against the PCB where they it will never leave. <S> Do <S> NOT coat a board with anything that it is not meant to be coated with. <S> Hairspray, for example, is meant to stick and stiffen things, not seal them off. <S> Furthermore, isn't hairspray flammable? <S> The most common types of conformal coating that can be applied by a normal person is acrylic, polyurethane, and silicone. <S> I prefer silicone for reworkability. <S> It's feels softish and pliable. <S> The others are nasty get off. <S> Paraylene is the really nice one but requires vacuum deposition equipment <S> so it's NASA/military stuff. <A> 1) <S> What is the purpose of this coating? <S> Generally it's to keep humidity or dirt away from components. <S> This site has a good write up on it. <S> 2) Is conformal coating ever used to prevent service to a board? <S> Not that I know of <S> , if people want to hamper efforts to service a product, usually potting is used, which is an epoxy that coats the board and all the components usually more than 1/4" thick. <S> Very difficult to remove without collateral damage. <S> Conformal coating can make it more difficult to service, but people that don't want parts serviced usually start by grinding off component ID's, then you have to etch off the package and find a nice microscope. <S> Is [the fibers in the conformal coating] likely that this will lead to problems? <S> Yes, if the fibers are not covered with conformal coating they could provide a pathway that penetrates the conformal coating. <S> Ideally they should be removed before recoating, but this might not be possible. <S> Acetone might be a better way to get the coating off, but whatever you use, be careful as some solvents can damage components. <S> 4) Is it advisable to re-apply the coating? <S> If so: Are there any household products I could use for that? <S> I was thinking about hairspray? <S> Or is that a really bad idea? <S> Ha! <S> Hairspray is not a conformal coating, I suppose they have the same consistency but are not even close to being chemically similar. <S> There are many places that sell conformal coatings, including digikey <A> Please do not use hairspray! <S> I highly recommend re-applying coating. <S> I have used conformal coating, specifically silicone in the past with great success. <S> It's also a lot easier to rework if needed. <S> Great source of information about conformal coating and the different types on this site if your just looking for information on it and still deciding what to use: https://www.paryleneconformalcoating.com/conformal-coating/
| if conformal coating is applied improperly it can do more harm than good. Silicone conformal coatings can be soldered through for rework.
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Isolated Power Supplies for Inverter Gate Drivers I am using a 3-Phase IPM IGBT Inverter Module in motor driver. This module has built-in gate drivers which require 4 Isolated power supplies for their operation. In all I have to design 5 Isolated Power Supplies as shown in the attached drawing. My input is 2-Phases (380V AC) and the output will be x5 regulators 7815. All 5 outputs will have isolated GND from each other. Max current from each output is less than 500 mA. If cost is the main factor then how can this be achieved? Should I wound 1 big 50 Hz step-down transformer (380 VAC to 24 VAC) with 5 separate secondary windings or is there any better design technique than this? The IGBT module is following: <Q> Should I wound 1 big 50 Hz step-down transformer (380 VAC to 24 VAC) with 5 separate secondary windings...? <S> No. <S> There are at least two alternatives that will likely be less expensive. <S> One possibility would be to convert 380 V, 50 Hz to a higher frequency and use that for the primary of a single transformer. <S> Another possibility would be to use 5 individually isolated, 380:?? <S> V SMPS's. <S> In order to make a valid decision, you need to estimate the costs based on your anticipated manufacturing location, volume etc. <A> Can you supply the thing with power for the lower Vcc of the low side drivers? <S> Royer oscilator driving a pot core with 5 secondaries, 5 * {diode, cap, 7815}, job done? <S> Shunt regulated flyback? <S> Lots of ways to skin this. <S> Alternatively, and it might be cheaper, just buy 5 isolated DC/DC modules and run them off the supply for your control circuitry, 15V, 7.5W DC/DC is not expensive in terms of a system with the sort of 380V IGBTs that need 500mA up em. <A> Look into a multiwinding flyback converter. <S> That is likely to be least cost option. <S> You should further consider isolation voltage requirements, isolation capacitance requirement, and required power ratings also. <A> Your Bridge rectified 3 phase DC bus volts will be around 500 VDC .This <S> is too much for an isolated flyback SMPS using orthodox parts .Some <S> countries use 480 VAC meaning even higher DC bus voltages .A <S> two transister Flyback will be reliable if 1200V switching devices are used .Efficiency wont be brilliant but it will cope with wide mains variations .
| That would make this a low voltage problem with an isolation requirement which is easier to deal with.
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Viable alternative for expensive MLCC capacitor? I'm thinking of building and selling a product which uses a DC-DC isolator. I've been prototyping with Recom RI3 3W Isolated DC-DC Converter Through Hole, Vin 5 V dc, Vout 9V dc, I/O isolation 1kV . The latest datasheet suggests an EMC filter using a 10uF 100V MLCC capacitor to meet EN55032. These capacitors are expensive, and so might affect the viability of my product. For my own purposes, the prototype works fine both without a filter, and with an aluminium capacitor in place of the MLCC one. Are there other types of capacitor I could use which might enable me to achieve FCC and CE compliance while keeping down the cost of components? <Q> Probably film caps with MLCC in parallel too, but they're not so viable these days. <S> Keep in mind that the voltage coefficient of capacitance is quite large (and negative) with MLCC caps so using them near the rated voltage will result in their capacitance value typically being much less than the nominal value with a small DC bias. <S> Similarly, the inductor has to be chosen so that it's not close to saturation. <S> As an example, suppose you tried to use two CL31B475KBHNNNE <S> Samsung 4.7uF/50V caps in parallel (about 7.5 cents USD each). <S> The capacitance at 30V will be nowhere near 10uF.... <S> That's probably why the datasheet specifies 100V rating. <S> There are DIY LISN designs out there if you're concerned about pre-compliance testing for conducted noise. <S> Note that particular DC-DC has an unusually low switching frequency, which may be of benefit, depending on which standards you are trying to meet. <S> Testing is a must in any case. <A> You could consider aluminum solid polymer e.g. Vishay 184 CPNS series has 10µF 100V version and Panasonic SXV series even has a 15µF 100V version. <S> I don't know the cost but would think they are considerably cheaper than a big MLCC (and probably easier to source). <A> You may have to learn the hard way about EMI, noise, RF filters and SMPS... <S> Because doing it well is hard and we don't have near enough information to know what your layout and effect of cables between DC and boost regulator and other design info will be. <S> Or you may be lucky , and never have any problems because you probably are only going to use < 10% of the 333 mA <S> rated boost current on the 5 to 9V converter for Op Amps. <S> The 10uF Alum Electrolytic Caps on the LC "Pi" filter act as current buffer for the 20 to 50kHz switching boost regulator as well as attenuating noise going out to your 5V source and radiating on all your cables like an antenna. <S> So it is bi-directional dual purpose. <S> REF <S> https://www.tme.eu/en/details/ce10_35-smd/85degc-smd-electrolytic-capacitors/ <S> There are combinations of impedance and reactance that can work well and others than amplify the noise DC input cable inductance shown above. <S> ( I picked a length random.)
| You could use a solid polymer (low ESR) capacitor, a low-ESR aluminum capacitor or parallel MLCC capacitors of lower voltage and/or capacitance. These Korean electrolytic caps might do the job as well.
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How can I add "standoffs" to elevate a SMD component above the PCB? I made a mistake when creating a symbol/footprint for a small (2.0 x 1.6 mm) SMD crystal and didn't notice until the prototypes didn't work. Unfortunately, the boards are tiny and expensive (0.4 mm WLCSPs, 0201 passives, etc). I need to use these existing boards for development and testing. To fix it, I can rotate the crystal 90-degrees and solder it to the existing pads. The problem is that the metal case could now contact some vias and perhaps a nearby guard trace. These features are covered by solder mask, but I don't want to rely on the mask for electrical isolation. So I want to raise the package up off of the board by 4 mils or so (~100 um) and solder it in place. As of now, I'm envisioning cutting pieces of copper foil and placing them on the pads with solder paste, then placing the component on these risers (with additional paste). Although I have a (cheap) rework station, this will be difficult: the package covers the pads, the pads are only 0.55 x 0.65 mm, and they won't line up perfectly: Seems like there might be a better way! Anyone have any good ideas for me? I'll need to modify 4-5 boards. <Q> Adding some carefully-cut strips of Kapton tape to protect the vias you're concerned about is an option, but personally I'd just rely on the solder mask for insulation in this sort of prototype situation. <S> You can probe it after the rework to check for shorts, I bet there won't be any. <S> Additionally, isn't the actual crystal package smaller than the footprint? <S> In your image you've overlaid the same footprint on itself rotated, but the actual component ought to be smaller and might fit entirely within the footprint despite the rotation. <S> Or if it doesn't, perhaps you can find a smaller crystal with the same frequency and similar capacitive loading requirements. <A> In the past I've done something similar with crystals. <S> The solution I used for lack of any kapton tape, was to place a big blob of solder on the pads to form essentially the standoffs you describe. <S> Then also tin the pads of the crystal with solder. <S> Then place the crystal on top of the solder blobs (which have by now cooled solidified) in the correct orientation - it will be sitting a short distance off the board now. <S> Re-heat the solder blob for one corner to allow the solder to flow and connect the tinned crystal pads to the board. <S> Make sure to hold the crystal steady for the first join so that it sits level. <S> Now that the crystal is held in place proud of the board, you can reheat the remaining corners in turn to complete the connections. <S> I wouldn't do this on a production board, but it is easy enough to do for a few test boards. <A> Ok, this maybe weird but how about soldering some other SMD resistors (very high value) or capacitors(very low value) underneath? <S> It greatly depends on which pads are active and which are passive. <A> I think you may be overthinking a bit; Solder mask <S> /is/ insulation, it's thin and doesn't hold up well to mechanical abrasion, but it does insulate. <S> Probe out the case and the pads on the crystal, <S> I suspect it may actually be isolated from the workings anyway <S> and just there to provide an RF shield. <A> Tom Carpenter has a pretty good solution, I've done that in the past too, but getting those tiny pads individually soldered with a crystal that small is going to be really difficult without reflow, even if you have some extra exposed pad around the crystal. <S> I have two suggestions you could try - <S> If the balls are big enough, they'll make contact anyway <S> If you don't have kapton, the crystal is low frequency and you have 0ohm 0201 jumpers <S> , you could solder 4 of them vertically and very carefully hand solder the crystal on top to form a table with resistor jumpers for legs. <S> I've done this with 10k resistors when I've forgotten to put in a pull up/ pull down resistor and soldered magnet wires to the tops of the resistors. <S> It's easier than you'd think. <A> Liquid tape is magical stuff, you do have to apply several coats, but is should be able to provide an insulation from what you don't want the crystal case to touch. <A> Buy a 2 pin THT crystal, bend the pins and solder them on to the 2 relevant pads <A> Make a bump stickers from kapton and locate them between crystal's pads. <S> Place crystal on these bumps and solder.
| Do the solder balls on the pads, but put a couple layers of kapton in the middle to hold the component off the board before reflowing the balls with a hot air gun.
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PN junction band gap - equal across all devices? With a given material eg silicon, is the band gap constant across devices? For example, at a constant temperature is the band gap of a diode ("diode drop") exactly the same as that of (say) an NPN transistor? If not, what kind of variation would one expect between devices? <Q> For a 100% pure element (no doping) it will be the same. <S> For a practical diode or transistor, the dopant levels and purity have quite a large effect on both the forward voltage and temperature coefficient. <S> For pure silicon the tempco is -2.1mV / degree C, but the venerable 2N3904 usually has -2mV / degree C (depends on manufacturer). <S> If you read the SPICE files for some devices, this can be inferred from ideality factor as seen in the Schockley diode equation. <S> For a diode, this is N in the model. <S> For a transistor <S> NF is the forward mode ideality factor and NR <S> is the reverse ideality factor. <S> A perfect device will have N=1.Here is the model for the MMBD4148 <S> small signal / switching diode <S> *SRC= <S> MMBD4148;DI_MMBD4148;Diodes;Si; 75.0V 0.250A <S> 4.00ns <S> Diodes Inc. Switching Diode .MODEL <S> DI_MMBD4148 <S> D <S> ( IS=300n RS=0.422 BV=75.0 <S> IBV=2.50u <S> + CJO=1.99p <S> M=0.333 <S> N=2.77 TT=5.76n ) <S> Here N is 2.77 (the higher this value, the less ideal it is) <S> Looking at the 2N3904 model Model Generated by MODPEX <S> * <S> Copyright(c) <S> Symmetry Design Systems <S> All Rights Reserved <S> * <S> UNPUBLISHED LICENSED SOFTWARE <S> * <S> Contains Proprietary Information <S> * <S> Which is The Property of <S> * SYMMETRY OR <S> ITS LICENSORS <S> * <S> *Commercial Use or Resale Restricted <S> * by Symmetry License Agreement <S> * Model generated on Aug 7, 01 MODEL FORMAT: <S> PSpice <S> *$ .MODEL <S> Q2n3904 npn +IS=1.26532e-10 <S> BF=206.302 <S> NF=1.5 <S> VAF=1000 <S> +IKF=0.0272221 ISE=2.30771e-09 <S> NE=3.31052 BR=20.6302 + <S> NR=2.89609 VAR=9.39809 <S> IKR=0.272221 ISC=2.30771e-09 <S> + <S> NC=1.9876 RB=5.8376 <S> IRB=50.3624 RBM=0.634251 +RE=0.0001 RC=2.65711 XTB=0.1 XTI=1 <S> +EG=1.05 CJE=4.64214e-12 <S> VJE=0.4 <S> MJE=0.256227 +TF=4.19578e-10 <S> XTF=0.906167 <S> VTF=8.75418 <S> ITF=0.0105823 <S> +CJC=3.76961e-12 <S> VJC=0.4 MJC=0.238109 XCJC=0.8 + <S> FC=0.512134 CJS=0 VJS=0.75 MJS=0.5 <S> +TR=6.82023e-08 <S> PTF=0 KF=0 AF=1 <S> Here the forward ideality factor is 1.5 (much closer to an ideal diode). <S> This is one of the reasons that temperature sensors often use a diode connected transistor. <S> So the forward voltage is process dependent (as is reverse leakage). <A> No, it certainly isn't equal for all devices. <S> Semiconductors are not made of pure silicon, but silicon that is carefully mixed ('doped') with other materials. <S> This is done to influence the bandgap voltage, creating areas that are more positive or more negative than other areas on the device. <S> This is how the P and N regions in a PNP transistor, diode etc. are made. <S> The specifications of the device will usually show what kind of spread can be expected in forward voltage for a given current and temperature. <A> You said ... <S> is the band gap of a diode (" diode drop ")... suggesting that both things are the same, which is not the case. <S> Band gap and diode drop are two completely different things. <S> Common for both is only that both are measured in Volts. <S> Band gap is a constant specific to a semiconductor material (a material constant). <S> It is the difference between the potential of the lower limit of the conduction band and upper limit of the valence band. <S> For Si it's about 1.1V (with only small dependency on temperature and doping). <S> Diode drop <S> is a constant specific to a PN junction, i.e. a specific semiconductor device . <S> For Si PN junctions it is about 0.6V; For Si Schottky junctions it is only about 0.3V; at room temperature (with substantial dependency on temperature).
| The bandgap is very sensitive to the dopants, and will usually differ slightly even between devices of the same type.
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Decoupling cap routing on a 4 layer PCB I would like to place some decoupling caps for an IC. The IC is on the top while the caps will be on the bottom, opposite side. How should I route them: A single VIA that connects the IC pin, power plane and cap Two VIAs, one that connect cap with power plane and another that connects cap with IC pin Please excuse me if this has been asked previously, but I read a load of resourses and could not find an answer to this. Thanks in advance. Edit: More detailed picture of the 2 scenarios (But first, I want to thank everyone for the great answers and comments!) I have decoupling cap. One pin of the cap is connected to the ground plane by a VIA. That is for both scenarios. The other pin of the cap will be connect a) to power, b) to IC input. Thus, three things have to be connected together ultimately: IC pin, power, cap pin. Scenario 1: A VIA connects the 3 of them, that is a trace from IC pin goes to the VIA, the VIA is goes through power plane and actually connects with power plane, a trace connects this very same VIA with the cap pin. Scenario 2: A VIA is connected to power plane and a trace connects cap pin with this VIA. Then another VIA connects the IC pin with the that trace, but the second VIA although it goes through the power plane does not connects with it. So you have two VIAs: one that connects the cap with the power plane and another that connects the cap with the IC pin. <Q> It actually does not matter too much, according to Henry Ott's book on EMC. <S> Proximity, trace length, and loop size matter more. <S> Multiple vias also helps. <S> Note this also applies to connecting IC power pins to the power and ground planes. <S> Those are part of the loop too. <S> From Electromagnetic Compatibility, Henry Ott 2009 <S> Similarly, placing two caps on the same side on opposite sides of the IC does something similar since the loop currents flow inside each loop in opposite directions and the magnetic overlap a fair amount if the loops are close to each other. <S> From Electromagnetic Compatibility, Henry Ott 2009 <S> So you can see that the first requires a pair vias for both the cap and pins, but the vias can be tightly coupled, while the second requires longer traces/loops but no vias. <S> It's twelve of one or a dozen of the other. <A> Thus placing caps on the back, UNDER <S> the IC's VDD/GROUND pin-pairs, is a good mindset. <S> Think in 3_D, draw sketches in 3_D, and minimize the total area and total volume. <A> Important is that VCC first connects to the Cap, then from there you go to the IC pin. <S> If you need a second via to get VCC to the bottom layer at this place, then that's the way to go. <S> The bypassing effect exists, because every trace on the PCB has a certain inductance. <S> If there is a change in current draw, the inductance allows the current flow to change only slowly. <S> The longer the trace, the stronger is this effect (I guess you knew that already). <S> Now it is pretty easy to imagine, that the capacitor can act way faster, when the supply current is typically already flowing along its pin. <S> If now the supply voltage collapses (or has some ripple) <S> the current delivered by the capacitor is producing a dI/dt only at the leads of the cap. <S> The rest of the trace does not see a change in current (in an ideal scenario). <S> Nevertheless the bypassing would be significantly better, if the cap was also on the PCB side of the IC, so that you need no via at all (at least not between cap and IC), because a change in current draw (e.g. due to some clock of the chip) IS still producing a dI/dt on the trace to the capacitor.
| Minimize the total area, and the enclosed volume, of the current-loop. If the IC and cap are on different sides of the PCB, placing both vias along the edge of the capacitor pad side (rather than the ends) so they are closer also reduces inductance since magnetic fields of opposite currents in the via cancels out.
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Transmission from atmega32 to PC using max232 not working Transmission from atmega32 to PC using max232 not working. When I checked the data from the Tx pin (in the realterm software) of atmega32 it was fine. But there was no output data from max232, when I was trying to read data in the realterm software. I have attached the circuit that I used.Where am I going wrong? <Q> <A> Check if the voltages at output of level converter are correct by giving logic 0/logic 1 at its input <S> (use a multi meter to check output voltage of level converter). <S> You haven't mentioned about the software used at host side to send/receive data. <S> If you are using any terminal emulator like putty/minicom, check their settings for correct port settings, baud rate settings and make sure that flow control lines (RTS, CTS) are disabled. <S> If all of these are not working, it would be a good idea to check the output of TX line of microcontroller using a logic analyser/oscilloscope. <S> If such a setup is not available, use a loopback connection to ensure that your UART driver can actually send and receive bytes. <S> Also check if your microcontroller is properly decoupled from power supply. <A> I am grateful to all of you for your valuable suggestions. <S> My fault was in the RS-232 to USB converter. <S> I changed it and now I am able to read data in my Realterm terminal software.
| As is mentioned (above) the specific MAX232 chip may need larger capacitors and the polarity of them (usually) is important.
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Can a Variable Frequency Drive convert standard wall outlet (120V/15A/single phase) to ~220V 3 phase? In my workshop I have a device that requires 220V 3 phase power, with a requirement of up to 1.5kVA. In North America, my wall outlet provides ~120V/15A single phases. (EDIT - my device can take +/- 10% voltage, so anywhere from 220V to 240V 3 phase is fine.) Doing some research, I found devices like these, called 'Variable Frequency Drives' that seem to take this type of single phase input and give the 3 phase output desired: https://www.mcmaster.com/variable-frequency-drives I am a noob at electronics, and I just want to confirm if I could buy one of these VFDs to solve my problem as described. I guess I would get the 2 HP version to match approximately 1.5kVA. Thank you for any help, and open to other suggestions if this won't work. <Q> There are VFDs that will accept 115V single phase input and give you a 230V 3 phase output. <S> They use what is called a "voltage doubler" front-end rectifier, meaning they rectify the 115VAC to approx. <S> 165VDC, then put it through a circuit that doubles the DC to 330VDC, then the inverter section uses PWM to recreates an output that MOTORS react to as if it is 230VAC 3 phase. <S> The key point however is MOTORS; this can ONLY be used for 3 phase AC induction motors, you cannot use the output as a general 3 phase power supply source. <S> it is the inductive nature of the motor that allows the PWM to become a pseudo-sine wave. <S> Without the motor, it's just pulsating DC that changes direction. <S> Then there is the size issue. <S> On a 15A 115V circuit, you can actually only load it to 80% of that continuously, so 12A. The largest 115V input VFD is for a 1-1/2HP motor, but <S> the INPUT CURRENT on the 115VAC side is actually 20A; too much for a standard 15A circuit and actually, too much for even a 20A circuit (80% rule means a 20A circuit is only good for 16A continuous). <S> For that reason, MOST of the companies that sell a 115V input VFD limit it to 1HP, and even those require a 20A circuit (the NEC allows you to use a 20A breaker for a 15A outlet if you use 12ga wire). <S> Bottom line, if you have only a 15A 115V outlet, you cannot (legally) connect a device to it that requires 1.5kVA continuously, that is too much for that circuit regardless of what you put on it. <A> Knowing the cost of doubling line voltages for 3 phase from 120 to 240V, I think it would be better to install a 2nd line to get 120+120=240V service wire to hard wire <S> the VFD and the variety of choices is better, at a much lower cost. <S> Selection <S> my pick $275 <S> The come with a fancy set of interfaces that don't have to be used but can add value to some operations. <S> There is also a variable torque version can drive up to 3Hp for the same size/cost unit as this 2Hp constant torque version. <A> With the luxury of not attempting to answer the question while it was so incomplete and misleading, I would say that you should not use a VFD. <S> Using a VFD to power something that is essentially another VFD is asking for trouble. <S> I recommend that you consider: <S> Determine if the servo drive will accept single phase input. <S> If it will, see if you can get a 240 V outlet installed. <S> Consider using a 120:240 V step-up transformer. <S> If the servo drive will not accept single phase, there is probably a way that is can be hacked to make it work. <S> However that will probably require a level of assistance that you can not get here. <S> If you absolutely need 3-phase: <S> There are both electronic and rotary phase converters on the market.
| If installing a new line for 240 V is not attractive, investigate the methods for using two existing 120 V circuits. There are hobby groups for metal working and perhaps other areas that have online plans for making phase converters.
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Would a small hole in a Faraday cage drastically reduce its effectiveness at blocking interference? I'm shielding the pickup cavity of a guitar to minimize interference. I'm doing this with copper tape that has conductive adhesive, and I'm connecting this tape to ground. Without getting into too much detail, let's say there's a small spot I miss or purposely leave uncovered because it's difficult to reach, so that the Faraday cage is not completely "sealed" all the way around. Would this drastically reduce the effectiveness of the cage, or would it simply reduce it proportionally to the size of the hole? I ask because I'm fine if it's just slightly less effective, but if it ruins the whole thing, then I'll put in the extra effort. <Q> It's fairly common to make Faraday cages out of mesh rather than sheet copper, so you can imagine that a single small round hole is not going to degrade the effectiveness enormously. <S> But the holes in the mesh must be much smaller than the wavelength you're trying to screen. <S> In particular, it's the largest dimension of the hole, not its area, that matters. <S> A 1-mm round hole will allow much less leakage than a seam 10 mm long but only 1 um wide. <A> Regarding the actual use case in question: Having built musical instruments and experimented with this tape in the past, I can say it's a complete waste of time if you're using humbuckers, since those pickups are designed to cancel hum anyway, and almost a complete waste of time if you've got single-coil pickups, since most pickups nowadays come with shielded leads. <S> If you do a good job grounding the pots, bridge, jack, etc., you'll be fine with no extra shielding. <S> (How's your soldering?) <S> If you absolutely must shield the electronics cavity, use conductive paint, since you can paint it into every nook and cranny. <S> With paint, there are no tape overlap regions that may or may not be in good electrical contact, and you don't have to worry about the tape adhesive losing its grip when you leave your instrument in a case in a hot car, causing the tape to fall off the cavity wall and short out your wiring <S> (unbeknownst to you!). <S> If you're using vintage pickups with unshielded leads, you may consider just sleeving your leads between the pickup and the electronics cavity in a tube of conductive tape (just make this on your own from a sufficiently long piece of tape), and grounding that tape to the pot body to mimic shielded modern pickup leads. <A> The effectiveness of the shield (with and without holes) will depend on the frequencies you're concerned about, since the maximum size of holes in Faraday cage is supposed to be 1/10 of the wavelength or less. <S> Reality check: <S> a domestic microwave operates at 2.4 GHz (12.2 cm wavelength) and has a shielded window with holes of 5 mm or less. <S> If we're talking about audio frequencies, your biggest concern will be the skin depth of copper which is about 8mm at 60 Hz, so a copper tape (which is often 35μm) <S> thick is essentially transparent to such waves. <S> At 1 MHz the skin depth will be about 60μm so several layers of copper tape may have an effect. <S> A wavelength at that frequency is still around 300m, so small holes will not matter. <S> Note that if you're in an environment where an object less than 1 meter in size picks up significant audio interference at 1 MHz, nearby objects about a quarter-wave length (75m) should noticeably resonate (as in, long metal cables would "sing" loud enough for you to hear). <S> At 100 MHz the copper foil is really effective (with skin depth of only 6μm). <S> The wavelengh is around 3m, so holes of reasonable size will not be of your concern. <S> Only if you're expecting radiation in GHz range interfere with your guitar, holes in your shield can become problematic. <A> <A> Faraday cages aren't entirely closed boxes. <S> Like cages, they have gaps. <S> The size of the gap mainly influences what wavelengths will be able to penetrate, and not so much the amount. <S> For example: a microwave oven always has a mesh in front of the window. <S> The holes in it are small enough to prevent the microwaves from escaping, but large enough for light to pass through. <S> Another example: your car can be considered a Faraday cage in the event of lightning. <S> It will protect you from the strikes, because the wavelength is way too large. <S> However... because of the huge gaps in the cage (glass windows) <S> we can still receive cell phone signals through it. <S> I'm not sure what kind of signals you're trying to block, but since we're talking about audio I'm guessing fairly low frequencies (large wavelengths). <S> So as long as the gap isn't too big, I don't think it'll be too much of an issue.
| As long as the size of the hole is less than one wavelength of the frequency you are concerned about, AND your circuit is at least a wavelength inside the shield, you will be fine (6.28 nepers of attenuation will be yours).
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Voltage divider does not work with LM393 sound sensor I have a sound sensor which I want to connect to a RaspberryPi. RaspberryPi input voltage on GPIO is 3.3V and this sound sensor output gives 5V. I've made voltage divider that looks like this: It works fine when I connect it with power supply. It gaves me ~3.1V on V out. However when I connect it to sound sensor it does not give me the expected voltage. My circuit looks like this: Output voltage is ~2V and voltage between sensor output and resistor R4 is ~3.3V. I've also noticed that the small diodeon sensor (which lights when sound is detected) lights a bit when circuit is connected that way. What might be causing this problem? Should I add a diode between R4 and sensor output? Will this solve the problem? This is the circuit of the sound sensor: <Q> The LM393 is a comparator with open collector outputs. <S> So, based on your measurement your sensor output circuit will look like this: simulate this circuit – <S> Schematic created using CircuitLab <S> And this is why you are getting 2V at the output. <S> simulate this circuit <S> Have you tried to supply your sound sensor from the 3.3V rail? <S> There is a chance it will work. <S> Also, this 2V at the output will be interpreted by a RaspberryPi as a high state. <S> So no problem either. <A> You can find the schematic for the module by clicking the download in the link you supplied. <S> It looks like this: <S> It looks like when the threshold is reached, your output will go LOW, which enables the LED to turn on. <S> When there is no noise, the output is high, and LED is off. <S> With a 10k pullup, you are not going to get the correct voltage on your divider. <S> It may be better to use a MOSFET or transistor as a switch. <S> Moral of the story is always find a schematic if you are working with a breakout board. <A> If you track down the circuit on the seller's site, you will see that it already has a resistor between the 5V supply and the output. <S> There's a 10k resistor on the board, and you have another 6k on your voltage divider. <S> What have therefore is a voltage divider of 16k to 10k. <S> That ratio delivers 1.92V at the output. <S> Almost exactly your measured value. <S> The circuit diagram also explains why the LED lights up (weakly.) <S> The lower resistor of your divider provides a path to ground for the LED. <S> It is 10k, so not much current flows - but it is enough. <S> The simplest solution is to use a diode instead of a voltage divider. <S> Put a pullup on the GPIO pin on the Pi, and connect it to the sound sensor with a diode. <S> Cathode (banded end) towards the sound sensor. <S> The Pi has a proper high input when there's no sound. <S> No current flows through the LED when it's quiet. <S> And, the LED will light up correctly. <A> If so, you will have to recalculate your voltage divider. <S> It's better to use a transistor in your case.
| When there's a noise, the sound sensor pulls the output low and the Pi "sees" that through the diode - it has a proper low signal when sound is detected. You should measure the resistance between Vout and Gnd on the module to see if there is a pull down present. Furthermore, a voltage divider is usually only a good idea to use with high impedance IO since any for of parallel resistive load will screw with your ratio.
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PIC 16f628a succesfully programmed and simulated but not working on circuit I simulated this circuit successfully in proteus but it doesn't work on breadboard. The motor is unipolar stepper motor I checked the motor and ULN2003A darlington transistor IC. They work perfectly. Only problem is with the pic. I used a 16f628a edit 1: programming device detects and programs the PIC. but when i put the PIC on the breadboard it doesnt do anything. edit 2: programming device is "brenner 8" and software is "USburn" edit 3: fixed circuit and codes after answers and comments edit 4: after fixing; only 2 leds are constantly lit and nothing else happened. edit 5: (May 10th 2019 11:15 hours GMT) it doesnt work no matter what i tried (i tried all answers to this date). Fortunately the teacher gave me a decent grade. I will try with a pickit 3 if i want to try again(currently using a brenner 8). // CONFIG#pragma config FOSC = INTOSCIO // Oscillator Selection bits (INTOSC oscillator: I/O function on RA6/OSC2/CLKOUT pin, I/O function on RA7/OSC1/CLKIN)#pragma config WDTE = OFF // Watchdog Timer Enable bit (WDT disabled)#pragma config PWRTE = OFF // Power-up Timer Enable bit (PWRT disabled)#pragma config MCLRE = OFF // RA5/MCLR/VPP Pin Function Select bit (RA5/MCLR/VPP pin function is digital input, MCLR internally tied to VDD)#pragma config BOREN = OFF // Brown-out Detect Enable bit (BOD disabled)#pragma config LVP = OFF // Low-Voltage Programming Enable bit (RB4/PGM pin has digital I/O function, HV on MCLR must be used for programming)#pragma config CPD = OFF // Data EE Memory Code Protection bit (Data memory code protection off)#pragma config CP = OFF // Flash Program Memory Code Protection bit (Code protection off)#include <xc.h>void wait();void main(void){ CMCON = 7; TRISA = 255; TRISB = 0; PORTB = 1; PORTA = 0; int portb_value = 1; int minimum_step_count = 3; int counter = 0; wait(); while(1) { if(PORTA == 1) { while( counter < minimum_step_count ) { counter++; if(portb_value != 16) portb_value=2*portb_value; if(portb_value == 16) portb_value=1; PORTB = portb_value; wait(); } counter=0; } if(PORTA == 2) { while( counter < minimum_step_count ) { counter++; if(portb_value == 1) portb_value=16; if(portb_value != 1) portb_value=portb_value/2; PORTB = portb_value; wait(); } counter=0; } PORTB = 0; }}void wait(){ int time = 0; while( time < 30000 ) { time++; }} <Q> You need to set the configuration. <S> See the datasheet 14-1 . <S> In particular, set /MCLR to an input or tie it to Vdd with a 10K resistor and set the clock up to use the internal RC clock and disable the WDT. <S> While you can typically set this in the IDE, you should insert pragma statements to set it in your code. <S> Here is an example ( not suitable for your situation) from this tutorial. <S> If the /MCLR pin is enabled for use as /MCLR function and you decide to add a reset switch, be sure to follow the datasheet recommendation for a series resistor of 50-100 \$\Omega\$ . <A> Do you have the program configured to use the internal oscillator? <S> ( Page 99 of the datasheet ) You should also try tying the MCLR pin to Vdd or use one of the circuits described on page 108 EDIT: Be sure to add a 0.1uF decoupling capacitor near the power input of the microcontroller. <S> There are some additional problems I noticed with your code. <S> You should increase the time for <S> the wait() function since 30,000 ticks is going to be a short period of time. <S> I would use an unsigned int and take it to 150,000. <S> The other thing is that you need to change the second if statement in your while loops to an else if. <S> Currently when portb_value == 1 then it gets changed to 16, it makes portb_value ! <S> = <S> 1 for the next if statement, <S> and so it becomes 8 by the time it gets written to PORTB. <S> Can you also confirm electrically that the switches are working? <A> These pics can be programmed in low voltage mode or high voltage mode. <S> You have chosen low voltage with #pragma <S> config LVP = <S> ON <S> In this mode, pin RB4 is not a programmable <S> I/O and becomes a program mode select input. <S> As it's not connected, the PIC might be in program mode - Unless you have a good reason, always use high voltage programming...... <S> so #pragma config <S> LVP = <S> OFF <S> Also, int minimum_step_count = 3; should probably be 4 <S> Also maybe the higher bits set in port A are set. <S> Try just looking at 2 bits if((PORTA & 3) = <S> = 1) ... <S> if((PORTA & 3) = <S> = 2) ... or better while((PORTA & 3) == 1) ... <S> while((PORTA & 3) == 2) ... <A> I am unable to comment because i have less than 50 rep <S> so I am writing here. <S> You are writing code in MPLAB, so one of the thing that could lead to a problem you're having is that when you create a project in MPALB you have to select a hardware tool you are going to use, and your code is configured for that kind of tool i guess. <S> You can check if your tool is present in selection menu. <S> Did you checked that all wires are connected properly, if not it could create a problem. <S> Check if your power source is good enough (voltage, amperage). <S> Last <S> but not to be ignored is your pic functional? <S> Sometimes hardware tool can detect your controller but that doesn't mean it's working propertly. <A> To be honest, I wont suggest you to use simulator software to predict the behaviour of your circuits, esp the ones revolving around microcontrollers. <S> Reason being that, the model of MCU fed into the sim software is too "forgiving" meaning that it overlooks most of the mistakes you make. <S> These mistakes would, however, produce a completely output different in hardware implementation. <S> Now coming to why your circuit isnt working? <S> Here are a few pointers <S> - double check the connections - particularly Vcc and Gnd. <S> If ground isnt connected, control signals will have a floating voltage (undefined voltage state), hence the motor wont function as expected. <S> Instead of driving a stepper directly, check the circuit (hardware implementation) by controlling a normal dc motor first. <S> Simply send a constant HIGH or a variable voltage signal to ULN2003. <S> If the motor shaft rotates, then be happy there is no issue with your hardware implementation. <S> Simply <S> Once hardware is okay, focus on the software. <S> Set the port pins as output. <S> Check the clock pulse, reset pulse and then phase voltage pattern (generated by the uC) on a DSO (you can use the simulator for the last one). <S> Check if the motor phases are powered in the proper sequence. <S> It is always nice and convenient to have a LED (or LEDs) connected to the uC. Use this LED to indicate the current status of the code. <S> This can give you a visual indication of the code flow. <S> After every phase change, blink the LED once to check whether the phase signals are generated properly or not.
| As mentioned in the comments, you should either tie inputs to a known voltage or enable the internal pull-up resistors (see the datasheet for details).
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Incorrect operation of the Allegro A3941 driver I use the Allegro A3941 driver in these modes: At very low duty cycle (<10%)PWM motor consumes 0.05 A and produces a squeak. If you increase the PWM duty cycle of the PWM to 10% the squeak disappears the motor to spin at the same speed and consumes a current of 0.01 A. The PWM frequency is 84 kHz. My circuit: My PCB: Sometimes the driver burns out for no apparent reason. Where could I have made a mistake that causes the driver to burn out? I do not understand this, because I believe that all the large current flows through the transistors, not through the driver. And, what causes the squeak? <Q> The FETs you are using have diodes to catch the voltage spikes from switching the motor on and off. <S> These diodes short the spikes to the power rail (12V.) <S> If that were a battery, then it would absord the spikes and all is good. <S> Many powersupplies can't do that. <S> The spike hits, and the voltage on the rail jumps. <S> Most of the time, it jumps a little and your driver gets to go on living. <S> Sometimes it jumps a lot, and exceeds the input voltage limts of the A3941. <S> You can check it out by attaching a scope to the 12V while the controller is running. <S> Watch for spikes, or even just excessive noise on the 12V rail. <S> Maybe someone else can explain that. <A> The driver burns out sometimes due to damage of bootstrap driver. <S> Please refer to the Bootstrap issues article, eg. by ON Semiconductor. <S> It's important to protect some circuits of A3941 from spikes. <A> C3 , at 22uf, is way way way too small. <S> And you have no ground plane. <S> Get copper tape, place that on backside of your PCB, insulated in some way, and solder the copper tape to a number of GND nodes.
| I think your A3941 drivers keep burning out because you are using a power supply instead of a battery. Given that your driver is operating at 84kHz, I don't know why you would hear an audible squeal from the motor at low speed.
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Why are oscilloscope input impedances so low? My question is two-fold: Where does the input impedance come from? I'm wondering where the input impedance of your average multimeter or oscilloscope comes from? Is it just the input impedance to the device's input stage (such as an amplifier or ADC input stage), or is it the impedance of an actual resistor? If it is the impedance of an actual resistor, then why is there a resistor at all? Why not just the input circuitry? I measured the input impedance of my oscilloscope with a DMM. When the scope was turned off, the DMM measured about \$1.2\mathrm{M\Omega}\$ . However, when the scope was turned on, the DMM measured pretty much exactly \$1\mathrm{M\Omega}\$ (I could even see the 1V test input applied by the DMM on the oscilloscope screen!). This suggests to me that there is active circuitry involved in the scope's input impedance. If this is true, how can the input impedance be so precisely controlled? Based on my understanding, the input impedance to active circuitry will depend somewhat on the exact transistor characteristics. Why can't the input impedance be much higher? Why is the input impedance of an oscilloscope a standard \$1\mathrm{M\Omega}\$ ? Why can't it be higher than that? FET input stages can achieve input impedances on the order of teraohms! Why have such a low input impedance? I suppose one benefit of a precise standard \$1\mathrm{M\Omega}\$ is it allows 10X probes and the like, which would only work if the scope had a precise input impedance that wasn't unreasonably large (like that of a FET input stage). However, even if the scope had a really high input impedance (e.g., teraohms), it seems to me that you could still have 10X probes just by having a 10:1 voltage divider inside the probe itself, with the scope measuring across a \$1\mathrm{M\Omega}\$ resistor inside the probe. If it had an input impedance on the order of teraohms, this would seem to be feasible. Am I misunderstanding the input circuitry of a scope? Is it more complicated than I'm making it out to be? What are your thoughts on this? The reason I thought of this is that I've recently been trying to measure the common-mode input impedance of an emitter-coupled differential pair, which is much larger than the scope input impedance, so it made me wonder why the input impedance can't be larger. <Q> A lot of things are the way they are because of history, and de facto standardisation. <S> A general purpose oscilloscope input is a difficult compromise between not loading the circuit, not being damaged by high voltage, having reasonably low noise, and being able to maintain a decent bandwidth. <S> 1Mohm in parallel with 15pF to 30pF satisfies a lot of people for a lot of applications. <S> There's little incentive for manufacturers to build a general purpose oscilloscope with a different input, to address tiny parts of the market. <S> When you do need better noise, or a differential input, or a higher input impedance, then you use a custom pre-amp. <S> When you need wider bandwidth, you switch to a 50 ohm input impedance. <S> There are special purpose oscilloscopes made at high prices that do address niche applications. <A> I would say a combination of a few factors. <S> The input stages of an osciloscope are a difficult compromise. <S> They need to be have a wide range of gains/attenutations, they need to be tolerant of user errors, and they need to pass high bandwidths. <S> Adding a requirement for a very high DC resistance would just further complicate matters. <S> In particular attenuators needed to handle the higher end of the scopes input level range would get much more complex/sensitive if they needed to have a very high DC resistance. <S> It's a de-facto standard, changing to something else would lead to incompatibilities with existing probes etc. <S> There wouldn't be much benefit anyway. <S> To further explain point 3, at moderate frequencies (from a few kilohertz upwards) the 1 megohm DC resistance of the scope input is not the dominant factor in the overall input impedance. <S> The dominant factor is the capacitance, with the cable making probably the largest contribution. <S> (in fact at UHF/microwave frequencies <S> it's common to reduce the scope input impedance to 50 ohm, so <S> the inductance in the cable can balance out the capacitance and the cable becomes a properly matched transmission line) <S> What this means is if high input impedances are desirable <S> then it's much better to deal with that at the point of probing than at the scope. <S> The typical compromise of cost/flexibility/input impedance for general use is an x10 passive probe. <S> If you need a really high DC resistance then the solution is to add a FET based amplifier in front of the scope, preferably as close to the point of measurement as possible. <A> Actually, it is ridiculously high for a wideband input. <S> There is no practical connector or cable that actually has an impedance (from a transmission line view. <S> Resistance, but for coaxial cablers, gold platers, and waveguide plumbers. <S> RF dudes.) <S> of 1 megaohms, leaving the input utterly mismatched - even worse, a 15-45pf capacitor across an 1 megaohm (transmission line impedance) input would mismatch it to oblivion. <S> However, once you are dealing with RF or fast digital circuitry, the parallel capacitance of the scope input (which you can't make too small, again because of probes, cables, connectors) will dominate ... and bring the actual input resistance of that input down to 5 to 10 kiloohms once you reach one megahertz, 500 to 1000 ohms once you reach 10 megahertz. <S> Reach VHF (hint: ACMOS or F-TTL circuitry is VHF stuff even if you don't clock it at VHF) and you would be better off with a matched 50 Ohm input, since you could connect a (within reason) long 50 Ohm cable and still have a 50 Ohm input on the circuit end, instead of an even bigger capacitive burden. <S> With the conventional kind of probe and input, you will overload RF circuitry easily. <S> RF optimized oscilloscopes tend to have inputs that can be switched to 50 Ohm input impedance (any oscilloscope input can, with a parallel/through terminator) - which is, interestingly, BETTER suited, since now you can use probes (eg Z0 probes or active FET probes) that actually can be made to present much higher effective input impedances at the probe point. <S> Or just provide a reliable 50 Ohm connection to your circuit with any old RG58 cable.
| The reason it is 1 megaohm is for supporting standard 10:1 probes, which you indeed need to not overload the kind of circuit carrying audio frequency signals at high impedance and with high DC offset (think audio vacuum tube circuits, the probe designs are from just that era).
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Supply white/blue led with 3V3 at constant current I need to drive a white/blue led at low constant current (about 10mA.) To do that I usually use the schematic below, but in my case I can only supply the LED with 3V3. Since the typical forward voltage of white leds @10mA ranges from 2.7V to 3V, the voltage drop across R2 could range from about 0.5V to 0.3V. For tolerance reasons I'm afraid that with this variability I cannot guaranteed a repeatable luminous flux with different leds. Can you suggest to me a different way to drive a led that can control the resistance? simulate this circuit – Schematic created using CircuitLab <Q> The following might work with the compensation cap and resistors set to zero (i.e., no C1, wires for R3 and R4), but only if you choose an op-amp that is designed to drive capacitive loads, and a FET that has a particularly low-capacitance gate. <S> In addition to gate capacitance, your FET also needs to be "logic level", and should probably be rated for a 3.3V gate voltage. <S> The component count is distressingly high, but you should be able to get away with cheap parts. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> To use a low voltage (3.3V) source, and attain a given current, a simple high-value resistor does notfill the need. <S> This is one way to do it (a current mirror): simulate this circuit – <S> Schematic created using CircuitLab <S> The nominal 0.6V base-emitter drop, and R1, make a circa 10mA current in Q2,and <S> the (matched) Q1 has a relatively high collector impedance, withsimilar current. <S> Matched-pair transistors, packaged together to have the same temperature of operation, are available for this purpose. <A> There's slightly more advanced constant current sinks (or sources) than your single BJT plus resistor could be; these could be built with lower effective output resistance (and thus, a lower output voltage drop than the inevitable 0.7 V that you'll get here, at least). <S> For example, a self-biased JFET might do that – and they even come in handy small packages, e.g. for 15 mA . <S> However, even that device would require a headroom of about 0.5 V. <S> So, if you really need full forward voltage headroom, and have a lot of LEDs in parallel to drive, anyways, designing a boost (or inverting) power supply interestingly becomes a viable option – in the simplest case, one supply for all your LEDs to give you e.g. 4.2 V, so that a simple biased transistor regulator works. <S> In a more advanced case, if you really need the reliable luminosity: there's plenty of multi-channel LED drivers; and they support things like telling them brightness correction values for individual LED channels. <S> Look at TI's and Maxim's and ON Semi's websites. <A> I highly doubt the human eye could distinguish the difference in illuminance. <S> The eye's pupil will moderate the perceived brightness much more than the LED's V f . <S> I would suggest using a newer Cree LED (e.g. XP-G3) which will have a low enough V f (i.e. less than 2.7 V) to regulate the current with an FET. <S> When a datasheet specifies 2.7-3.0 V f , that is typically over the full current range of the LED. <S> At 10 mA you should be able to easily find an LED with a V f less than 2.7 V. <S> There may be an LED with the required illuminance at a much lower current (e.g. less than 1 mA), and consequently a lower V f .
| To drive an LED with constant current, given that the LED may have bothmanufacturing and temperature dependence in its terminal voltage, youwant a very high impedance drive. Doing such a design, e.g. with TI's webbench power designer (ti.com) is actually not very hard, and there's probably even a module that you can simply buy that does what you need.
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What happens if we force a voltage on Vbe of a bjt transistor I'm wondering what happens in a BJT transistor if we force a voltage bigger than 0.7V on Vbe. A colleague of mine said that if we put on Vbe, let's say, 5V the transistor would be saturated. I haven't seen the case in question but I suppose that happens if Vbe > Vce. Still is there any harm to the transistor if we force a bigger voltage than 0.7 there? <Q> As long as the current is low enough to not damage the transistor, the voltage drop on Vbe will be 0.7V. <S> If you measure more than 0.7V from the base to the emitter of a bipolar junction transistor (BJT,) then you have destroyed it. <S> Between the base and the emitter of a BJT is the equivalent of a diode. <S> That's why Vbe is 0.7V <S> - it is the foward voltage of a silicon diode. <S> Just like any diode, the forward voltage stays around 0.7V unless it is damaged. <S> So, no, you can't "force" Vbe to be higher than 0.7V if you want to actually use it as a transistor afterwards. <S> I've been reminded that not everyone will realize that "0.7V" is a sort of shorthand for "the rated Vbe of your transistor." <S> Depending on how the transistor is made, Vbe can be higher or lower than 0.7V. <S> It also varies depending on forward current and temperature just like in any other diode. <S> In any case, the actual Vbe is inherent to the diode and the current through it. <S> If you try to force any higher voltage onto it then you will destroy it. <A> As 'Transistor' says - a BJT base-emitter junction IS a diode. <S> The normally seen voltage of around 0.6 - 0.7 <S> V is what is adequate to provide a base current that causes the transistor to operate in a normally desired manner. <S> If you increase Vbe the Ib also increases - up to some maximum allowable Vbe limit. <S> At some point above that the transistor will be damaged or destroyed. <S> Some transistor data sheets provide Vbe/Ib curves. <S> This is reasonably uncommon. <S> In most cases Vbe of <= <S> 1V would meet all needs for small signal transistors. <S> In some power applications, with Ibe in the amps range, Vbe MAY be more than 1 Volt. <S> Some data sheets specify Veb REVERSE bias breakdown limits of the base-emitter junction where it acts as a zener diode. <S> This datasheet for a BC337 transistor specifies Vbe_max of 5 Volts. <A> Isn't that like asking what happens if you put 10 Amps thru a 100mA diode? <S> You can ignore the Shockley equation above the rated max current, because then it behaves like a heater resistor and it will never come close to the giga-amp calculated current at 5V. <S> For BJT's I call it saturated when the bulk resistance dominates the VI curve when it becomes linear. <S> For the Vce(sat) it is similar except what happens is the hFE drops towards 10% of its linear hFE as the junction saturates. <S> Then increasing load current follows the Rce load line and not the hFE current ratio. <S> I have a rule of thumb for Rbe bulk resistance R=5/ <S> Pd which is the device max power rating Pd at 85'C. <S> THe Rce saturation resistor is typically Rce = 0.5/Pd with a 50% <S> but there are special device designs much lower. <S> There are some transistors however which are rated for 1A Max. <S> by design and IC is rated for 3A max or 6A pulse. <S> But attention to Joules of heating or the Safe Operating Area (SOA) are necessary. <S> So the bottom line is damage to the junction by overtemperature will result if the Safe Operating Area or the Absolute max current is exceeded.
| You can't make Vbe larger than 0.7V without destroying the transistor.
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Does the neutral wire have any voltage? In this article , the author talks about 3 phase and the magic of a missing neutral wire. And there's the following paragraph: In a 3-phase system the voltage between any two phases is 3 times higher than the voltage of an individual phase by a factor of 1.73 (square root of 3 to be exact). If your X-N (and Y-N and Z-N) voltage is 120V (common in the US), the X-Y (and Y-Z and Z-X) voltages (a.k.a. “cross-phase” voltages) will be 120V * 1.73 = 208V. When he says the X-N voltage is 120V, with N being the reference voltage, is the reference voltage 0V or does it have some value? From the diagram (in the link,) I realise that the neutral wire is grounded, but is this the same as earth grounding? Doesn't the neutral wire go back to the transformer so as to provide a closed circuit? <Q> The original article you've quoted is garbled. <S> ... <S> the voltage between any two phases is 3 times higher than the voltage of an individual phase by a factor of 1.73 (square root of 3 to be exact). <S> It can't be three times higher and <S> \$ <S> \sqrt <S> 3\$ <S> times higher simultaneously. <S> The correct value is \$ \sqrt 3\$ times higher. <S> Figure 1. <S> The diagram in question. <S> Source: PacketPower . <S> When he says the X-N voltage is 120 V, with N being the reference voltage, is the reference voltage 0 V or does it have some value? <S> The reference is the wye (star) point where the three windings share a common node. <S> In this diagram it is neutralised by connecting to Earth. <S> (That means that this conductor should not see any significant voltage on it with respect to earth.) <S> There will be 120 V between each of the phases and the wye point whether it is earthed or not. <S> From the diagram (in the link,) I realise that the neutral wire is grounded, but is this the same as earth grounding? <S> Doesn't the neutral wire go back to the transformer so as to provide a closed circuit? <S> It depends on local regulations. <S> If we're dealing with a transformer secondary here then typically the primary will be delta powered <S> so there may be no neutral on the incoming supply. <S> The solution is to ground the wye to the building's earth bonding in which case it would be the same as earth grounding. <S> In the case of a ship, for example, the "earth" would be the ship's hull. <S> Figure 2. <S> A delta-wye (delta-star) transformer connection. <S> Source: <S> Gamatronic . <S> Here we can see that there is no neutral connection on the incoming supply. <S> Using a wye configuration on the secondary allows us to create one for internal use. <S> The wye point can be left floating or can be grounded. <A> The neutral wire is defined to be 0V. <S> And each phase individually has a voltage relative to neutral of 115V if you are in the US, and 230V if you are in most of Europe. <S> But because the three phases are 120 deg out of phase compared to each other this results in a voltage difference of 115V * sqrt(3) = <S> 200V (or 230V <S> * sqrt(3) = 400V) between any two phases. <S> Depending on the type of mains distribution system you've got in the country you live <S> (I don't know how it is in the US) <S> the neutral wire might be connected to earth at the power company. <S> Keep in mind however, that even though the neutral wire might be connected to earth, this doesn't mean that it is actually at earth potential, because a return current might be flowing in the neutral wire, and due to the resistance in the wire this will cause a voltage at your end of the neutral wire. <S> It is also important to keep in mind that in many countries the mains plugs we use can actually be turned 180 deg putting line where neutral was supposed to be, meaning that with these kinds of systems you can never be sure if your neutral is actually neutral or line (BE CAREFUL!) <S> So to sum it up: The neutral wire is defined to be 0V. <S> But will typically be at some voltage potential (relative to earth). <A> Please note, this is for a healthy working balanced system. <S> So yes, consider the neutral wire 0V if everything is in working order. <A> "Here we can see that there is no neutral connection on the incoming supply." <S> So what provides the return path for flow of current? <S> This was asked as a comment, but to answer it I need far more space than a comment allows. <S> With a simple two wire AC connection, the outgoing current on the hot wire is has the exact same absolute magnitude as the return current in the neutral wire, but they have opposite signs. <S> In addition to this, the current is actually a sine wave. <S> Now, lets take three copies of this, but adjust the phase of two of the circuits so that all three are 120 degrees out of phase with respect to each other. <S> Next, lets replace the three neutral wires with one big wire three times the capacity. <S> Up to this point, everything is working just fine. <S> However, lets now take a close look at the current flowing in the neutral wire. <S> It'll be the sum of the three individual component flows, i.e. the sum of three sine waves, each 120 degrees out of phase with respect to the others. <S> If you either draw it out, or sum it up mathematically, those three sine waves always add up to zero, at any point in time. <S> This in turn means that there's actually no current flowing in that huge return wire. <S> So why not just remove it, and let the three hot wires carry on doing what they're doing. <S> In practice, this works well enough that we can transmit power over large distances without an explicit return wire. <S> Due to imbalances in the loads, there will be some return current, but it's generally small enough that using Earth as a return is sufficient.
| The neutral wire is grounded, so if you literally took a multi-meter and tried to read voltage between the neutral wire and ground, you will read 0 voltage.
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What is the difference between a PLL and a frequency-synthesizer？ It seems that a LPF is contained in PLL and not in frequency-synthesizer. Maybe I had made a wrong judgement. <Q> A block schematic could look like: <S> And I see a Lowpass Filter in there <S> so your statement <S> LPF is contained in PLL and not in frequency-synthesizer. <S> is untrue . <S> Also a Phase Locked Loop (PLL) has a block schematic which is a copy of part of the frequency synthesizer: <S> So basically a frequency synthesizer could be a PLL but with some blocks (mainly frequency dividers / counters) added to increase the range of supported frequencies. <S> A frequency synthesizer can also be made using a DDS which has no PLL. <A> A PLL is a specific type of building block, that will control a variable oscillator (usually a VCO, though it doesn't have to be <S> voltage-controlled - YIG-based PLLs use current instead) <S> such that it's phase is locked to that of a reference (not frequency! <S> It just so happens that if you lock to a phase, you automatically lock to the frequency too). <S> Of course, they can put multipliers and dividers in between to have the oscillator frequency and reference frequency be different, but they are always 'locked' to one another. <S> A frequency synthesizer is the name we give to a block that synthesizes frequencies. <S> This can be done internally through a PLL, but this could also be an FLL (frequency-locked loop), DDS (direct digital synthesis, basically a DAC that just outputs a signal at the frequency we want), DLLs (delay-locked loops), etc... <S> In modern electronics the entire thing is not always as simple as it may seem, and many high-performance synthesizers may use hybrids of different types. <S> So this question is kinda like asking 'What is the difference between a Car and a Vehicle?' <S> - One is a subset of the other, so you can't really speak of 'differences'. <A> "Frequency synthesizer" is just a more general term than "PLL". <S> Every PLL is a frequency synthesizer but not vice versa. <S> Often frequency synthesizers are PLLs but not every frequency synthesizer is a PLL. <S> An example for a frequency synthesizer that is not a PLL would be a Direct Digital Synthesizer (DDS) .
| A PLL is one common type of several possible types of frequency synthesizers. According to Wikipedia a frequency synthesizer is an electronic circuit that generates a range of frequencies from a single reference frequency.
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Fastest temperature sensor for a bimetallic strip What is the best approach or sensor to analyze the temperature of a bimetallic strip? From experimental data, the temperature swings due to a current flowing into the bimetal, are up to 80°C in a period of 20s which means 4°C/s.I want to have a high sampling frequency (10Hz minimum, best would be 20-30Hz I guess).Ultimately I must have high accuracy as well since I want to model the transient dynamics. Edit: my goal is to have a batch of samples which contains the temperature transient (when the current is flowing) to make a model (for simulation/prediction). My need is to detect the temperature rising as soon as possible when the current is flowing (with no delays due to the kind of sensor being used, or at least to minimize it, in order to capture the dynamics of the physics as close as possible) The current solution (an unknown thermocouple) has a sampling frequency of 10Hz and it has a variable delay up to 15samples, meaning it takes around 1.5s before detecting a rise in temperature when the current is flowing. My main problem is that I would like to detect the rise in temperature due to the current as soon as possible (closest to the physics). <Q> It shouldn't be too hard to get fast electronics. <S> ADCs tend to have the speed advertised fairly clearly, and unless you mis-configure some kind of internal digital filter will not normally delay the data by more than one sample or so, so I'll talk about fast sensors. <S> A fast sensor needs: A low heat capacity A low thermal resistance link to the sample Low heat capacity tends to mean small. <S> A thermocouple would usually be a good choice. <S> Make sure it is a bare junction, it should just look like two thin wires welded together. <S> When I wanted good, small thermocouples I used to make my own from suitable wire, as commercial ones can have big blobs of metal or plastic covers which slow down the response time. <S> Low thermal resistance to the sample means short distance, and that distance filled with something with a high thermal conductivity. <S> Welding directly to the bimetal strip would be best, solder would be next, followed by high quality conductive epoxy (e.g. Epotek H20E, not the stuff used for PCB repair). <S> A non-electrically-conductive epoxy is not as good, so if you can set up your ADC so that it does not need to be electrically isolated from the bimetal strip that will help. <S> With a thermocouple formed by welding <100um diameter wire, welded onto a sample, I have been able to measure temperature changes into the low hundreds of Hz. <S> There is a technique in Physics called "AC calorimetry" where this is commonplace, and you might find more useful information in papers on the subject. <A> A bimetallic strip is a temperature sensor. <S> It changes geometry with a change in temperature. <S> If you need to know the temperature of the strip, you need a way to measure that geometry. <S> For a more specific answer, you need a more specific question, with much more information about what you're trying to accomplish <A> The thermal timeconstant of a cubic meter of silicon is 11,400 seconds, measured from one face of the cube to the opposite face. <S> The thermal timeconstant of 0.1meter cube of silicon is 100X faster, at 114 seconds. <S> The thermal timeconstantof 1cm cube of silicon is 100X faster, at 1.14 seconds. <S> The thermal timeconstant of 1mm cube of silicon is 100X faster, at 0.0114 seconds. <S> The thermal timeconstant of 100 micron cube of silicon is 100X faster, 0.000114 seconds or 114 microSeconds. <S> If you can get back-grinded diodes (which are about 100 micron thick) and install them on your bi-metallic strip, you'll have 114uS response, degraded by whatever thermal resistance the "installation material" interposes. <S> Or you can get a diode in SOT-23, with the die installed on the HEAT_REMOVAL metal tab, and just solder that to the bi-metallic strip. <S> Given the thermal timeconstant of copper is a bit faster at 9,600 seconds per cubic meter, if you can achieve 1mm distance from the strip to the diode, you'll have 0.0114 second timeconstant.
| To get fast temperature measurements, you need two things: A fast temperature sensor Fast electronics
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What are these round pads on the bottom of a PCB? What are these round pads on the bottom of a PCB marked ScX? I don't think these are test points - this board has some (not in the picture) and they're labelled with TP and the pad size and shape are different. My best guess is that they're termination pads, but googling it didn't turn up many results so I can't verify that. <Q> These fixtures have a specially-shaped array of pogo-pins which press down and make contact with the test pads on the board under test: These test pads and the test points are probably used in different parts of the manufacturing process. <S> For example, the test points may be used for board-level testing and the test pads may be used for system-level testing. <S> Or, perhaps, the test pads are used for programming and/or debugging firmware, when the device under test is clamped into the bed-of-nails fixture. <A> They're called test points. <S> They're places to stick your multimeter probes, or, more commonly, an industrial test jig. <A> I think SC could be short for "spring contact" (spring loaded contact), in which case the PCB will be mounted together with some other PCB, similar to a Arduino header, but a different connector type not using male and female pins.
| As mentioned in some of the comments, these are test pads designed for use with a pogo-pin test fixture, also known as a bed-of-nails test fixture.
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Should I include over-voltage protection for a 3.63V max device using a Lithium-SOCl2 3.65V battery? When designing an ultra-low-power device that is powered by a primary lithium battery, is it necessary to include over-voltage protection? Specifically, I am using a Microchip (Atmel) SAML21J which specifies a supply voltage of 1.62-3.63V. I am using Lithium-Thionyl Chloride (Li-SOCl 2 ) batteries with an open circuit voltage of 3.65V. In practice, I've never worried about the small potential over-voltage in testing on the bench, both because the voltage immediately drops slightly with load, and presumably the microcontroller can tolerate short-duration small over-voltage. However, as I look at other components with a max voltage rating of 3.6, I wonder if there is value in adding, say, a Zener diode to clamp the voltage. The factors against adding protection in this case are 1) cost, and 2) potential battery life decrease. I'm mainly concerned that brand-new batteries installed during production could cause damage to some percentage of finished units. <Q> I'd suggest you <S> DO need some form of regulation. <S> Looking at the datasheets for the batteries, the nominal voltage is 3.6V <S> BUT it is temperature dependent. <S> This datasheet from Minamoto shows the terminal voltage rising to about +3.8V at 70°C. <S> If you can guarantee that everything on your board will tolerate 3.5V, then you could use a low dropout linear regulator such as this Ablic LDO . <S> This has only 53mV dropout voltage and should follow the battery down below that point. <S> It's available up to 3.5V. <A> What datasheets mean: "operating range" translates to " <S> if you're inside this range, we guarantee it'll work like we say". <S> "absolute maximum" translates to "if you're inside this range, it won't be permanently damaged". <S> If you're between the operating range and the absolute maximum, then you can expect nasties like more than advertised currents, parts not working right at the extremes of temperature <S> , parts maybe not meeting timing (although usually CMOS is faster at higher voltages). <S> If you're not above the absolute maximum ratings, and you're buying a whole lot of parts, you probably have the leverage to ask an applications engineer for the company just what you can expect. <A> First, you can't design a cheap and low-consuming limiter that will be accurate to 20 mV difference. <S> Second, you need to check manufacturer's QA methods and specifications on their definitions of "absolute max" ratings. <S> All max ratings are derived from accelerated tests at elevated temperatures, and then extrapolated on normal operating conditions using certain assumptions and theoretical formulas. <S> And the assumptions usually are on quite conservative side. <S> More, the max ratings usually imply "prolonged" exposure to max conditions, so episodic fresh battery overexposure for few seconds may have no effect. <S> It could be that the rating is defined for 10,000 years of service, and exceeding the rating by 5% would lead only to 1,000 years according to their aging models (hypothetically speaking). <S> Lastly, if you are really concerned with reliability and making a high-volume device, you might be able to negotiate special sorting batches from the manufacturer, since there is always a statistical spread. <S> Or if you are happy with reduced lifespan ( like 1,000 years instead of 10,000), you can get corresponding assurance from them. <A> Table 46-1. <S> Absolute Maximum Ratings VDD Power supply voltage 3.8 V max <S> p 1008 of 1155 ref
| If you're just a hair over, then you may be OK, particularly if you don't have to push it really hard.
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How to measure current in very low duty cycle circuit with DMM? My circuit has 4Hz 50% duty cycle. How do I measure the average current using a DMM? I have a Fluke 87 DMM. It has an average setting but for 100 ms which less than one complete duty cycle. My circuit takes abouts 13 uA in sleep and 20uA in active. <Q> If you know the duty cycle, just measure the current in both states and do the average by hand. <S> If you don't know the duty cylce, get a tool with which you can figure out: an oscilloscope in high input impedance mode, attached across a shunt resistor would be a good tool. <A> 50% is not a "very low duty cycle", it's a very low frequency. <S> A low duty cycle would be something like 1mA for 10ms and 100uA for 240ms. <S> Say you use a 5K shunt resistor with an RC filter of 10uF ceramic and 200K, which should give you 40uA full scale on your meter 200mV DC voltage range. <A> My circuit has 4Hz 50% duty cycle. <S> How do I measure the average current using a DMM? <S> Easier to do than to describe :-). <S> Brief: <S> Provide 2 large filter capacitors. <S> "Large" = will change in voltage by a minimal amount during a cycle. <S> Connect a resistor Rm between them. <S> Connect Vin to C1, Vout to C2. <S> Measure V across Rm to determine I <S> m <S> ____________________________________ <S> Longer: <S> Ground capacitor negatives, connect C1 +ve to power supply, connect C2 +ve to load. <S> A capacitor Cm will change in voltage by ABOUT Vm in period Tm for a current of I <S> m <S> where Cm = <S> Tm <S> x <S> Im <S> / <S> Vm Set - I <S> m as max current <S> liable to be drawn when circuit is active. <S> - Vm as allowable change in capacitor voltage <S> - Tm = period of high drain - or some suitable period if unknown. <S> eg <S> Vm = 0.01V, Tm = <S> 1 second, Im = 20 <S> mA ie Vcap must not sag more than 10 mV in 1 second at 20 mA drain <S> So C = <S> Tm <S> x Im <S> / Vm = = <S> 1 x 0.020 / 0.010 <S> V = 2 F. <S> If this is larger than acceptable, revisit allowed values. <S> C2 is more important than C1. <S> C2 provides a reservoir into which Im flows. <S> C1 provides a stable source . <S> Connect a resistor Rm between the two capacitors. <S> Size Rm such that mean current drain will cause a voltage drop that is as large as possible BUT not so large as to significantly alter load operation. <S> eg <S> if 0.1V is an acceptable drop and mean current is expected to be say 25 uA then. <S> Rm = <S> Vdrop <S> /Imean = <S> 0.1 <S> V / 25 <S> uA = <S> 4 <S> K -> say 3K9 or similar. <S> If a larger dV drop is acceptable make Rm larger. <S> eg here <S> maybe 10K. <S> Measure: <S> Connect load to circuit. <S> Temporarily short out Rm to allow C2 to charge to Vc1. <S> Remove short. <S> Measure voltage across Rm. <S> This allow I average to be determined. <S> Iavg = <S> V/R = <S> V_rm/ <S> Rm simulate this circuit – <S> Schematic created using CircuitLab
| If your meter is not giving you a consistent reading because of sampling, you can make your own current shunt with a resistor and an RC filter.
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stm32 - is there a factory reset? for the past couple of months I have been developing software for a custom build board. Due to a non disclosure agreement i am not allowed to share the schematics , PCB layout or software of the board. What i am allowed to share is that the board uses a Murata cmwx1zzabz module as proccesor and for lora communication. The proccesor in this module is a Stm32l0 which i was able to program for the past couple of months. As i had a problem with the stability of the program (chrashed randomly after sometimes running perfectly for a couple of hours) i decided to let 3 of my 5 prototype boards run a program in a mode that uses one paticular sensor which seemed to cause problems. As a refference i picked another prototype board and let it run the same software version as the other three prototypes but in a different mode. The idea was to let the code run on the four prototypes and let it run the whole weekend as a stability check. The last prototype board i wanted to use for an energy consumption test. I loaded the same software as the other boards onto the board and did some measurements. It was a terrible idea to let all of my prototypes run the same software because keil uvision 5 cant program any of the boards anymore. I get this error: What i have tried to solve the problem is: use a different ST-link use different cables pulled the boot0 pin to '1' and tried to program it (normally pulled down) tried to load the program onto a Discovery board (It seems to work but i did stop the programming halfway through the procces because i am afraid to brick the disco board too) Created a "fresh" CubeMX project and tried to load the program on to the prototypes without succes Does anyone know a way to do a factory reset or has a sollution for this problem? My internship is almost finished and i dont really have the time to wait +- 4 weeks to wait on new prototypes. On the plus side the boards are still running but have some inconsistent readings. The only change i made to this version of the software and an older variant was to change return uwTick; to return HW_RTC_GetTimerValue(); in the HAL_gettick() function. some additional info of when i connect the stlink in my PC with the prototye connected. <Q> It was a terrible idea to let all of my prototypes run the same software because keil uvision 5 cant program any of the boards anymore. <S> It wasn't, you've found a problem. <S> A test where nothing fails is a poor test. <S> Try the ST Link Utility , and verify if the RESET pin is actually connected to the programmer. <S> Either because the lock bits of the manufacturer are set, or pin function was changed by your code. <S> Did you use the GPIO port configuration lock register? <S> There is no such thing as a factory reset if you cannot get into the chip. <S> If mass flash erase is not available over SWD/JTAG. <S> Then maybe over a bootloader (uart/usb). <S> Otherwise you have to replace the chip since you might have accidentally activated read out protection level 2. <S> (RDP level 2, see reference manual) <S> Also sometimes a lower clock on the SW interface can help if you have a poor board design. <A> Thanks everybody for the help. <S> Somehow holding the reset button and releasing it just before loading new code onto the board worked and i can program it again ( <S> Not like i used to <S> but it works). <S> I havent figured out how this is possible <S> but if i do i will share it here. <A> What is the VDD on those MCUs? <S> I've run into this exact message if ST-Link did not get correct voltage on the TVCC pin.
| Right now it says " No Target Connected ", meaning the debug core inside the ARM Cortex M0 is not reachable.
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How to avoid input floating ports on a microcontroller? I have 16-bit Thunderbird12 Microcontroller , similar to the Freescale 68HC12. I'm currently working on a project to monitor and control the level of water. Everything is working fine except for the input ports which have floating pins and they are randomly reading low and high. I did some research here and found that pull-up or pull-down resistors are the solution, but my set up is different. There are 5 "sensors" that feed into the input ports. Each sensor is connected with 2 wires; one wire goes into the pin, and the other into VCC, then both are inserted into the container with both ends exposed. When the water level goes up, it touches both ends and completes the circuit and setting the input to 1. The problem I have is that the input pins still read High even if the wires are not connected yet. What can I do do prevent this high impedance? I can't ground the wires or set them to 1, because the latter is supposed to happen only when the circuit is completed (when water touches them). Any help would be appreciated, thank you! <Q> You still need a pull-up or pull-down resistor as appropriate. <S> If the sensor connects between ground and the input pin, you need a pull-up resistor. <S> The resistor needs to be chosen so that it has a substantially higher resistance than the sensor. <A> There are 5 "sensors" that feed into the input ports. <S> Each sensor is connected with 2 wires; one wire goes into the pin, and the other into VCC, then both are inserted into the container with both ends exposed. <S> When the water level goes up, it touches both ends and completes the circuit and setting the input to 1. <S> ok, there is some challenges here <S> , first we have a simple sensor that completes an electrical circuit for a water level sensor. <S> I see this part of the circuit is going to be weak for switching cleanly the micro-controller. <S> I would normally solve it by low power Schottkey inverters (74ls04) in series to clean up the signal going in. <S> But since you need to trigger on a voltage, instead of completing a ground, then a nor gate (74ls02) circuit like the one below would work if you just use the "B" input for the sensor, nothing connected to "A" but the resistor to ground. <S> https://www.electronics-tutorials.ws/wp-content/uploads/2018/05/io-io81.gif <S> You could try high gain transistors, trasistor arrays (uln2003) or even a darlington comparator (that you can include a threshold adjustment) <A> Since your GPIO pin has extremely high impedance, you need a pull-up or pull-down (as most people have said). <S> I am working with a TI microcontroller right now, so I quickly wrote code for a quick test. <S> I also didn't make any assumptions on the resistance of water, I used my tap water in a tiny cup for a test. <S> It works with a 100k, so a 1M would provide good margin (for my water). <S> Your water may vary. <S> The resistance of water is dependent on the impurities, really pure water has a very high resistance. <S> Finally, if your water container is really large and the water may already be grounded, you should use a pull-up resistor, not a pull-down. <S> simulate this circuit – <S> Schematic created using CircuitLab
| If the sensor connects between Vcc and the input pin, and if the sensor is open-circuit when off, then it would be a pull-down resistor.
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Why does my circuit work on a breadboard, but not on a perfboard? I am new to soldering I'm a total beginner when it comes to soldering, and recently I've been trying (and failing) to solder together a simple circuit I put together for a Raspberry Pi sensor. Although it works fine on the breadboard, when I solder it onto one of my perfboards , the sensor no longer turns on. Here is a topdown view of my soldered circuit: Here is a view of the connections (the red block is just covering up old connections from past attempts): More pictures of the connections. What might I be doing wrong? <Q> Everyone here is right. <S> If you got rid of the solder mask you would see something like this: <S> You have to make the connections manually or buy this type of perf board. <S> Notice how it has the connections made in copper? <A> You actually did a good job on the soldering <S> The problem is that the board you are using, unlike the breadboard, has no connection for a given row of pads. <S> You have to add wires or solder shorts on the back to make the connections you want. <A> A perfboard is not like a breadboard. <S> A perfboard is called so, because it has holes in it, it is perforated! <S> So the whole perfboard contains only holes and no connections between any holes (unlike the breadboard). <S> You have to interconnect the holes yourself. <S> In this case, you have to connect the two leads of the resistor to two jumpers. <S> The first step is to solder every individual component on the perfboard. <S> You did this step correctly! <S> Second step is to make connections between the soldered leads. <S> In this case, you have soldered two resistor leads and two jumper leads. <S> To connect leads together, you have to solder another wire between them, or you can just use a solder joint between them, i.e., connect the two leads only using solder. <S> The purple lines represent the connections you should make, i.e., the wires you should place externally to connect the required perfboard pads: <S> This is how you can connect adjacent holes using solder bridges. <S> Source: How to make traces on an universal PCB? . <S> Look at the answer by JYelton. <S> Also, you can use wires to solder holes together like this - Source: How to make traces on an universal PCB? . <S> Look at the answer by Passerby. <A> The board you are using has no connections between the pads - you have to add wires between the pads to complete your circuit. <A> I think you are treating the perfboard as a breadboard. <S> There is no connections between the points you have soldered. <S> This is how a breadboard looks. <S> There are connections inside and you just have to pin in your wires. <S> When you want to solder a perfboard you will have to make a connection using a wire like this: Image credits: <S> Lab: <S> Setting Up A Breadboard Soldering A Perf Board <A> Ya <S> like everyone is saying you need to connect the components since it is not a bread board. <S> I've found solar panel bus wire works really well as traces on these boards since it can get soldered directly onto it.
| The perf board you are using does not contain the connections between pads like the bread board. Also, you have excessively long leads sticking out of the pads on the solder side of the board - this could lead to unwanted connections (short circuits) between points in your circuit.
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Does data rate increase with channel frequency? It is commonly said that a higher channel frequency implies a higher data rate. For example, in https://www.howtogeek.com/222249/whats-the-difference-between-2.4-ghz-and-5-ghz-wi-fi-and-which-should-you-use/ they say that 5 GHz wifi connection is simply faster than 2.4 GHz. However, Shannon–Hartley theorem states that the maximum channel capacity in bits/s depends only on the channel bandwidth and the SNR. Therefore, the data rate shouldn't really depend on the center frequency of the channel, only on the available bandwidth. Is it a misconception that higher frequency channels support a higher bit rate or is it true for some reason? <Q> Is it a misconception that higher frequency channels support a higher bit rate or is it true for some reason? <S> It is not the (carrier) <S> frequency itself that determines the supported bit rates but the available Bandwidth of the channel <S> Suppose I have 10 MHz of bandwith available at 100 MHz, for example 100 MHz to 110 MHz <S> or I have 10 MHz of bandwith available at 1000 MHz, for example 1000 MHz to 1010 MHz <S> Then the highest bit rate I can achieve will be the same as there is 1 MHz available in both cases. <S> However note how at 100 MHz that channel bandwith is 10% of the carrier frequency but at 1000 MHz it is only 1 %. <S> At that 1000 MHz I could fit 10 of those 10 MHz channels to come to the same 10% (or use a wider channel of 100 MHz.) <S> If we want divide a certain frequency band between service providers, that's much easier to do at the higher frequencies. <S> It can be done also at lower frequencies but that would result in narrow channels (small bandwidth) and therefore lower bit rates. <S> Suppose I go to 10 GHz <S> , that will give me another factor 10 more space. <S> The 2.5 GHz Wifi vs 5 GHz <S> Wifi isn't completely fair because the 2.5 GHz band is about 100 MHz wide (that's all channels together) but the 5 GHz band has about 900 MHz available (depending on your country <S> it might not be one continuous 900 MHz range though). <S> See here . <S> So there's simply lots more space (bandwidth) assigned to the 5 GHz standard. <A> I agree that sending the same bandwdith on different Carrier signal frequencies doesn't necessarily mean higher data throughput. <S> I guess what everyone means when they compare lower frequency applications like GSM900 with higher frequency applications like UMTS2100 or 2,4Ghz <S> Wifi with 5Ghz Wifi is, that the available bandwidth in that particular band is just larger. <S> E.g. the bandwidth in GSM900 is very limited, it is a scarce resource which is usually regulated by a national authority. <S> Same for any commercial mobile technology. <S> Wifi uses 2,4Ghz and 5Ghz non-regulated/licensed band. <S> But the boundaries of the bands are specified, and the 5Ghz seems to be much "larger". <S> The 2,4Ghz WiFi band has 3 non overlapping channels. <S> Whereas the 5Hgz band has 23! <S> (see https://www.electronics-notes.com/articles/connectivity/wifi-ieee-802-11/channels-frequencies-bands-bandwidth.php e.g.) <A> As mentioned in the other answer, the band around 5 GHz is less used in general which is advantageous. <S> In addition to this, it is often only practical to achieve very large bandwidths at higher frequencies. <S> If you look at some basic texts in microwave engineering, such as "Microwave Engineering" by Pozar you find many of the building block circuits may have bandwidth listed as 30%-40% of the central frequency. <S> At higher central frequencies this is obviously a larger bandwidth. <S> Intuitively, I think of this as being the case because the operation of say, a quarter-wave transformer, is dependent on the transmission line being exactly a quarter wavelength long at the central frequency. <S> Frequencies that have wavelengths near this central frequency will also behave acceptably. <S> At higher frequencies the wavelengths are spaced more closely together as: \$c= <S> f\lambda <S> \\\lambda=c/f \\\frac{\Delta\lambda}{\Delta f} = <S> -c <S> /f^2 \\\Delta \lambda = <S> \frac{-c}{f^2}\Delta f \$ <S> Decreases with \$f^2\$ <S> Therefore, practically it is often easier to achieve larger bandwidths at higher frequencies. <S> Ultimately, though, as you said the channel capacity is dependent only upon the bandwidth and SNR.
| To put it more simply: there's more "space" at higher frequencies so it "costs less space" to implement higher bit rates.
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Is it a bad idea to replace pull-up resistors with hard pull-ups? On many of my designs, there are ICs which have mode selection or similar inputs that are permanently pulled up or down using resistors. If I replaced all these with simple hard pullups or pulldowns I would probably save 10 placements per board on average, which is not nothing. Is this a bad idea? And if so why? <Q> The idea of pull-up / pull-down a signal is that the signal is being pulled high or pulled down low most of the time , but can sometimes be pulled down or pulled high respectively. <S> In the case you want to pull down a signal incidentally that is pulled up most of the time, you want to use a pull-up resistor to prevent huge current drawn from the supply. <S> So, if your circuit really uses pull-up or pull-down resistors as described above, don't remove them. <S> (For example, when a pin is open collector.) <S> If in your circuit a signal needs to high or low forever (when the supply is present) and no other component can change the state of this signal, then you can hard-wire it. <S> Those signal are not called pulled-up, but high, and, respectively not called pulled-down, but low. <A> I asked a similar question on the EEVBlog forum some time ago. <S> I had this idea in my head that any signal I needed permanently low <S> I would tie hard to ground, and any signal that I needed permanently high I would tie up via a resistor. <S> I didn't really know from where or why I used this scheme, so I asked about it. <S> I think it may have been something that I picked up somewhere <S> that was more applicable in the TTL days. <S> EEVBlog - Pull-up resistors - technically necessary vs preference? <S> The general consensus seemed to be, and as Huisman suggests, unless you need to be able to pull the signal in the opposite direction, you can simply pull it hard up/down. <S> It is worth noting that my question was in relation to CMOS-based devices - it may still be applicable/necessary if you are playing with TTL families. <S> More technical reasons revolved around noise immunity, especially if your pull-up resistor values were of a high value. <S> As I am writing this, other reasons I can think of for wanting to use pull-up/pull-down resistors might include troubleshooting abilities, "hidden features", or debug/service mode selection for example. <S> Since that thread I am no longer using pull-up resistors unless I need them functionally. <A> The answer will be in the datasheet. <S> If the logic input operating voltage specification includes V CC or V+ then connecting directly to positive supply is OK. <A> Let’s consider a pull-up resistor. <S> The job of a pull-up resistor is to pull a particular pin to the HIGH state. <S> However, the pin won't always be in the HIGH state because some circuitry can pull it down to ground. <S> Consider I²C lines. <S> They are pulled up via pull-up resistors and the microcontroller pulls them down as and when needed. <S> Had these lines been permanently pulled up AKA "hard pulled up", I²C communication wouldn't have happened. <S> The SDA line will see a permanent HIGH state. <S> Your scenario <S> In your case, if there's a resistor between the pin and GND/V cc , don't remove it. <S> If the datasheet says to put a resistor, do it. <S> However if you want to delve deeper and understand the functionality of the pull-up /pull-down, look for the IC block diagram in the datasheet. <S> Sometimes you may even find a circuit diagram of the internal blocks. <S> Try to understand the function of the particular resistor (if you find it difficult to understand the circuit, you can post the schematic here). <S> Like Huisman said, if a pin is permanently pulled up or pulled down, the pin's state isn't called pulled up or pulled down. <S> Instead it is HIGH or LOW. <S> Some info about pull-up /pull down Pull-up (and <S> pull-down) resistors are generally high, about 10 kΩ <S> generally and they keep the pin in a defined state - HIGH (or LOW) . <S> When an external circuit pulls down a pulled up pin, it provides a path of lower resistance to the ground for that pin. <S> Hence, the pull-up resistor value depends on the resistance offered by the external circuitry to the pin, to GND. <S> The pull-up value must be significantly greater than the resistance of pull-down path. <S> Some circuit designs with ASICs may use even a higher value of pull-up or pull-down. <S> If the pin state won't be changed anytime in future, you can hardwire it to either V CC or GND. <A> I am surprised to see that nobody mentioned DFT here. <S> In some case, using pull up/down resistor leaves room for a test fixture to inject a signal and put the input in a different state for the time of the test. <S> Let's use the simple example of a Chip Enable signal that you want to be always at "enable". <S> While performing an ICT test, you may want to disable the Chip Enable pin to put the output of the IC in High impedance mode. <S> Doing so, allows the test fixture to inject arbitrary signal at the output of the disabled IC, which would otherwise be impossible if the CE pin would be "hard driven". <S> This is an additional use-case. <S> Other answers in this threads are valids. <A> It really depends on why you are pulling it up. <S> Sometimes unused functions may be toggled by the built-in boot logic of a chip during startup. <S> If a processor can boot from multiple sources, it may have to auto-discover which source is attached during power on. <S> That can lead to some lines being toggled prior to code execution (prior to your code executing). <S> So if the datasheet says "pull up if not used," then you would want to double-check with the manufacturer before tying it high. <S> Or, if possible, maybe you can monitor the behavior of the line during start-up to make sure it is not driven low ever. <A> Hmmm. <S> Have not seen it mentioned yet, but a reason, especially for pins tied high, is to use a resistor for reduced power consumption. <S> Consult the datasheet for the device in question.
| So to summarise the answer, a pull-up or pull-down resistor is used when the state of the pin has to be changed via some circuitry.
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Why is it harder to turn a motor/generator with shorted terminals? The shaft of an unconnected motor is easy to rotate relative to a motor with shorted terminals. If a resistive load is connected to the terminals, the turning difficulty is somewhere in between. Why is this? (I'm using a BLDC motor.) <Q> I have to start with some terminology -- <S> sorry if it's esoteric, but this will bring things into line with how folks talk about this subject. <S> When you turn a permanent-magnet DC machine <S> *, the armature generates a voltage internally. <S> This is called the "EMF" <S> ** of the armature, or the "back EMF" if the machine is running as a motor. <S> This EMF is always generated when the machine turns. <S> When you run current through a DC machine, it generates a torque. <S> This torque is always generated when the machine turns, regardless of whether it's a motor or a generator. <S> When you put a resistance on the terminals of a machine and turn its shaft, it generates that EMF. <S> With the resistance connected, this EMF causes a current to flow that's proportional to the EMF <S> divided by the external resistance plus the machine's armature resistance. <S> This current, in turn, generates a torque that resists motion (due to conservation of energy, it must be in a direction to resist motion). <S> Shorting the machine puts the smallest possible resistance on it -- you can't get lower than 0 without resorting to active circuitry. <S> The back torque in this case is purely a product of the EMF and the armature resistance. <S> Increasing the resistance by putting a resistor on there means less current for the same machine speed, which means less back torque. <S> In the extreme, you have no resistor at all, which means infinite electrical resistance -- this means that the back torque will be from mechanical effects such as friction (and windage, if you're turning it that fast), and possibly mechanical and electromechanical effects as the field magnets work against the iron in the armature. <S> * <S> I'm calling it a "machine" instead of a "motor" because it can be a motor or a generator, depending on how you use it. <S> But you don't have to change anything internally to change how it's used -- hence, "machine". <S> ** <S> EMF stands for "electromotive force", which is just and older term for "voltage". <S> It seems silly to have two terms, but sometimes it's useful. <A> "applying a resistive load" to a running motor is essentially how an electric brake works . <S> As a first approximation, the torque produced by the motor is proportional to the current, that's turning the motor is harder as the load resistance gets smaller. <A> As I read the accepted answer my brain came up with the following simplification, which I think is loosely accurate (?) <S> : Motors are both dynamos and electromagnets. <S> Turning a motor invokes its properties as a dynamo. <S> Because the motor's terminals are shorted together, the generated voltage is applied to the motor coil windings, invoking the motor's properties as an electromagnet on its own axle.
| When you short the terminals, there's only the internal resistance of the motor which limits the current.
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Can only the master initiate communication in SPI whereas in I2C the slave can also initiate the communication? Is it possible for a slave to initiate communication in SPI? I thought due to the chip select i.e. NSS feature only master can communicate the communication. However, in I2C, the slave can also initiate communication by changing the RW flag in the i2c communication. Is this correct? <Q> With both only master can initiate the communication. <S> I²C can however have multiple masters and the nodes can change the roles, so it is a bit more flexible. <S> But saying that slave could initiate communication is still not correct. <S> A common way for a slave to indicate that it wants to communicate to the master is to use an interrupt signal. <S> On many sensors and ADCs they are called "data ready" or something similar. <S> After a slave has asserted the signal the master knows that the slave has some new data available. <A> Both SPI and I2C only allow the master to initial communication. <S> However, it is possible for slave devices to indicate that something has happened which merits communicating with them via an interrupt pin. <S> Interrupt pins are fairly common on ICs with serial busses. <S> The pin changes state to indicate that some event has occurred, and the master can then watch to pin to know when it should communicate with the slave to find out what happened. <S> Even in that case though, it's up to the master to start the communication and it can simply ignore the slave if it wants to. <A> No, in I2C only the master can start communicating on the bus, and all slaves must follow the master. <S> A slave can't initiate communication and specifically the slave can't change the RW flag in any way. <A> With SPI, the MOSI line always transmits data from the master to the slave, and the MISO line always transmits data from the slave to the master. <S> With I²C, there is only a single data line for both direction. <S> The R/W bit controls which device transmits data bytes and with device transmits the ACK bits, but the R/W bit itself is always controlled by the master. <S> (The R/W bit can be changed with a repeated start condition, but only by the master.) <S> And the clock is always controlled by the master. <S> (An I²C slave can delay clock cycles with clock stretching, but it cannot generate new clock cycles.) <A> In SPI, each device has a fixed role as master or slave. <S> In I2C, the roles of devices can change dynamically. <S> Except for a couple of details, I2C is predicated on there being exactly one master and one slave for any particular transaction: <S> When using longer addressing formats, multiple devices can act tentatively as slaves while awaiting their full address. <S> In that state, the only thing slaves can do is indicate to the master that the slave the master is interested in <S> might be ready to respond. <S> Multiple devices may act simultaneously as masters if all want to send the same data at the same time. <S> This may potentially cause trouble if e.g. two devices each request that a device increment a counter and then simultaneously request that the device decrement it. <S> If the two masters send bit-for-bit identical requests, neither one would lose arbitration, and thus each would think that its request was honored. <S> Most I2C systems contain only one device that can ever act as master, but the protocol would allow for multiple devices. <S> Unless a bus is expected to have multiple masters, however, it's easier to implement a device that expects to be the only master than one which can co-exist with other masters on the bus.
| Any device which initiates a transaction on the bus (as opposed to soliciting a transaction via some means outside the bus) must behave as master for that transaction, but that device could behave as slave for transactions initiated by other devices.
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Implementing digital control on analog potentiometer I have an analog sound processor unit, which uses either ICs or spare components. There are four controls, which are simple potentiometers.I would like to control their levels digitally, storing the settings in an EEPROM. Which are the architectural most advisable solutions? The idea is to transform an analog device into a MIDI controlled device, I am available in doing it with more elementary components or, if possibile, implementing ICs which can do the job. <Q> I've done a nearly identical project as a one-off for a customer in the past. <S> Motorised Potentiometers exist (ALPS is one manufacturer), but the response speed is slow, so that may not be suitable for your application. <S> Digital Potentiometers are available from multiple manufacturers and are generally SPI/I2C controllable, which are both easy to interface to your controller of choice. <S> Digipots do have a couple of serious limitations that need to be considered for this sort of project. <S> They have a maximum voltage window between different terminals (usually, but not always, the device supply voltage) and requirements for the ground reference. <S> This may or may not be a problem depending on how your existing device uses the potentiometers. <S> For my device the maximum voltage differential was within the acceptable range, but the ground reference was not suitable. <S> So, I used a small isolated DC-DC converter (e.g. XPPower), and a isolating-coupler (e.g. TI iCoupler) to allow a Digital Potentiometer (e.g. Microchip) to be placed on each channel while staying within the device limitations. <S> A low-cost microcontroller drove each channel via SPI. <A> You'll need a microcontroller to "speak" to your digital interface (e.g. MIDI). <S> Such microcontrollers typically come with internal non-volatile memory, so that solves your "how do I store the settings" problem. <S> If these potentiometers are used to set the gain in an amplifier (that's their most common use): There's also Programmable Gain Amplifiers <S> that can be had for cheap as drop-in replacement for an amplifier circuit. <A> You probably want an audio-taper digital pot like the ds1882
| There's digital potentiometers, which you can talk to from your microcontroller. If you don't want to modify the original analog circuit, you could use a servo/motor to turn the knob of a potentiometer, albeit that would be a pretty complex task and pretty expensive.
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