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Are datasheet min/max values "guaranteed" for the lifetime of the part? In electronic parts datasheets we always see min/max values specified for voltages, currents, etc. together with the temperature ranges. Some values are tested by the manufacturer, some are specified as "guaranteed by design". However, are these min/max values for the specified temperature supposed to hold for the lifetime of the part? For example, if a supply current is given as min 10 mA and max 20 mA, with a typical of 15 mA for 25C operation, is this current expected to be within those ranges for the part's lifetime? If so, what is assumed to be the part's lifetime for which the datasheet is guaranteed: 1 year, 10 years? How can I find that information without having to ask each individual manufacturer? EDIT: To lead the discussion in a better direction and get to the main point of my question:Let's suppose we set up a controlled environment where we guarantee the datasheet temperature is maintained (let's say room temp), with all the parameters meeting the spec at nominal values. How long am I supposed to assume that the datasheet min/max will be maintained? If parameters start drifting beyond min/max 1 year in, is that a violation of the datasheet? 10 years in? Where's the cutoff if there is any? Thanks! <Q> The datasheet describes what the manufacturer has to provide to you in a legally binding fashion. <S> So, unless it is noted otherwise in the datasheet, everything which is in it has to be correct under the conditions specified in the datasheet. <S> You have used the term lifetime of a part. <S> It is also often not well defined, if defined at all. <S> It is the designers responsibility to understand how the parts work, and which part is the most critical for the products lifetime, and do everything reasonable in his disposal to protect that part. <S> Of course it is virtually impossible to reliably determine whether the part was operated within the specified conditions, which makes this all promise thing a little vaguely controllable. <S> There are some failure processes which are exponential function of the temperature and stress, which property can give you some confidence if your product's operating conditions are less challenging. <A> You wouldn't want a mobile battery to burst into flames if the screen died, would you? <S> Reliability can never be guaranteed, so failure is a statistical measure... <S> usually mean time before failure (MTBF) for a reasonably large sample. <S> When a particular specification is important to be maintained even if the part dies, one would typically select components (even within an IC) which would have a greater MTBF, so the specification is not exceeded even if the part as a whole fails in some other determinate or indeterminate way. <S> At one time Intel sold off quad core processors as single/dual core when some cores were found to malfunction in a large part of the batches. <S> Obviously the malfunctioning cores went out of spec during the life of the entire processor. <S> Imagine two identical parts, one running at 25C and the other at 85C. <S> At some point one of them will fail before the other one does. <S> It is reasonable to assume that the part which fails will also go out of spec before it actually fails. <S> So to answer your question... <S> no, the datasheet values are not guaranteed for the life of the part. <S> I'm guessing the guarantee will be some temperature dependent mean time limit, possibly found by extrapolation of accelerated ageing at higher temperatures... <S> which may or may not be either accurate or a conservative estimate... <S> which is probably why we are amazed by the voyagers or even more so by the amsat-oscar-7 intermittently silent for 21 years between 1981 and 2002. <A> "How long am I supposed to assume that the datasheet min/max will be maintained? <S> If parameters start drifting beyond min/max 1 year in, is that a violation of the datasheet? <S> 10 years in? <S> Where's the cutoff if there is any?" <S> Good questions. <S> For a typical commercial-class part you buy, all bets are off once they sell you the part, unless something in the data sheet or purchase document indicates otherwise. <S> But that doesn't mean they won't last years, performing to their specs the whole time. <S> It's just that the manufacturer won't guarantee that kine of longevity. <S> For parts used in critical mil and space applications, most/many parts are procured to MIL/DESC drawings (the responsibility of the government), or something we call Source Control Drawings (SCDs). <S> These documents call out all of the stress screening and environmental test that an individual part or sample of parts (lot testing) have to go through. <S> As was mentioned by others, the place to start is the basic reliability of a component, specified as MTBF (Mean Time Between Failures), or more recently FIT (Failures in Time). <S> These two measures are inversely related to one another. <S> Mil-STD 217 is the a good reference for this kind of information. <S> It gives a basic failure rate for a general class of parts, such as SSI components (54XXX), power transistors, diodes, FPGA, etc. <S> This basic failure rate is usually specified under fairly benign environments, such as a lab. <S> Then this basic failure rate is modified by 1) the actual use environment and 2) <S> the quality level of the part. <S> The environmental factor almost always degrades the MTBF (increases the FIT) of the basic part. <S> For example, a part used in the nose of a fighter jet is going to have a shorter expected lifetime than that same part used in a lab. <S> Conversely, the higher quality level of a part increases the expected lifetime (that is, reduces the FIT rate). <S> But these higher quality parts are more expensive than "standard" parts bought off the shelf. <S> In some hi-rel products I have worked on, parts are still going strong after 25 or more years. <S> This time included the development, production, and delivery of a system (10 years or more), and 15+ more years in use. <S> You also might want to look over this discussion - How to design electronics to last 40 years or more?	It depends on the design of the part and if a particular specification was important enough to warrant failsafe mechanisms to be designed to remain in spec even if the part failed.
Are higher rated capacitors interchangeable with lower rated ones? I have the follow capacitors that I need to change: I can't find these specific ones on mouser or digikey and I can't buy in such huge quantities neither if they did have. I was told that I can use the following replacement for the Pink ones: https://www.amazon.com/gp/product/B07RBC8F62/ref=ox_sc_act_title_1?smid=A1BBEKKND9EC89&psc=1 EKZE250ELL821MJ25S, Cap Aluminum Lytic 820uF 25V 20%(10 X 25mm) Radial 5mm 2150mA 4000h 105°C Bulk (50 Items) But would I be able to use these same ones to replace the smaller blue ones too? These caps are both 16V, but the uF is different only, 470 vs 270. The replacement ones are 820uF and 25V much higher numbers. I'd like to know why would these high rated ones work aswell? What if I purchased or extracted caps from other boards that are greater than 470uF and 16V would those also work too? Even though they do not look the same? <Q> For the replacements, all you need to do is look for one that will fit on digikey that is <S> 16V 470uF. Digikey will sell you most parts in single quantity. <S> Try this list: https://www.digikey.com/short/z9q7qt <A> It depends on the application, if the application is a power filter capacitor (used on Vcc of a component) then bigger capacitors have the same function. <S> What is happening is we are forming an RLC filter. <S> The LC components are formed by wires or traces. <S> To prevent power dips sudden changes in Vcc we place a power capacitor to source power when the voltage drops (because the load changes), the value of C doesn't change the filter pole that much and bigger is usually better (but not if the ESR of the cap changes when we move to a bigger value). <S> When replacing power filter caps the voltage needs to be greater than that of the power rail, if you don't know what that is, then just make sure it's bigger than the cap. <S> Make sure the ESR is equal to or smaller, and the value of the cap is equal to or bigger. <S> If the application is an audio filter or some other filter where the filter pole needs to be exact, then it's best to replace the cap with the exact parameters (ESR and component value) to get the same result. <S> So you could probably replace all caps on that board with the 470uF's (if you could figure out a way to make them fit) but not the other way around. <A> Higher voltage ratings are okay in and of themselves. <S> But all things being equal, higher voltage caps have higher ESR and that might affect things if it's used in a switching circuit. <S> You're probably good though unless the ESR is much higher or the circuit was borderline designed. <A> Before you replace any capacitors, you need to determine what caused them to blow in the first place. <S> eg. <S> If the circuit was designed & intended to be used in the 90V -> 120 Volt range then using at 232V would explain the damage.	Higher voltage ratings are probably ok, but it's not great generally to use higher capacitance values unless you know what its for.
How to superimpose a synchronization pulse on a 12V DC power line I have a 12V DC to DC isolated converter powering a remote display. The DCDC is connected to a 12V supply. I would like to add a 1ms or so pulse on that every second that can be detected by the display unit that then has an exact tick. Crazy idea maybe, but is it feasible? Edit: I am concerned about the power supply side. How to get 12V plus say a 1V pulse on top of it? And then isolated. Or .5V or -1V. Total current is max 100mA. The remote side I will be able to do figure out later. I do not expect that to be too hard. <Q> The way that I would approach this is to have the output of your isolated DC-DC converter feed the end device via a diode. <S> There is a bulk storage capacitor after the diode that feeds the end device. <S> You also have something that detects when the input voltage to the diode drops. <S> Obviously, the diode and bulk storage capacitor are located at the end device. <S> It's far easier to detect a voltage drop than it is to try and force sufficient current into the DC line to cause a voltage rise. <S> You would have something like an opto-isolator at the supply end that either interrupts the output of the DC-DC converter for s specific period or causes a voltage drop. <S> Either method can work but detecting a complete absence of voltage at the end device is probably easier. <S> The above is a simple approach that is already in common use in several types of devices. <S> But you are going to have to provide many more details before anything concrete can be suggested. <A> Yes, this is feasible. <S> See, for example, the "parasite power" capability of some of the Maxim Integrated one-wire products. <S> The devil is in the details, of course, but you haven't given us any of the details. <A> One winding is connected to the pulse source, or audio-tone source, the other winding is a pass-through/series for the +12V line.	Perhaps a 90-deg rotated "modulation" transformer can be used to modulate a pulse onto the DC line, provided the transformer can handle the DC current.
How to intentionally overload almost any value resistor with a DC power supply? When attaching a resistor to my power supply (which is a Extech 80W 3-in-1 Switching DC Power Supply, capable of 36V 5A) and the amps are set to let’s say 5A at 16V the voltage stays the same but the amps will shoot down to something like .11A, not even close to half an amp. I know this is because of Ohm's law but how can I exceed and force the resistor to take in what the power supply wants to put out? Of course it's the same situation on my weaker power supply as well. Do I have to toggle something on the PSU itself? I've seen at least 5 videos of others frying resistors, but they never show the readings on the PSU's display. I know it's possible with my own PSU, I'm just clueless as to how to configure it. I've tried it on a 4 band 120 ohm and a 5 band 1k resistor. (I'm not a completely destructive mad man, I just have some useless components that no one wants. I want to stress test them out of curiosity instead of throwing them away.) <Q> You can never "force" a current higher than 0.11A through a 120 ohm resistance while maintaining the voltage at 16V. <S> You can either set the voltage or the current, not both. <S> Your power supply is (within limits) <S> a voltage supply, so it sets the voltage. <S> If you want to increase the current, you need to either increase the voltage, or decrease the resistance. <S> This means that the power supply will supply up to 5A, but the actual current is determined by the voltage. <S> In order to "blow up" the resistor, you would have to exceed its power rating (by quite a bit). <S> These are probably 1/4 watt resistors, so you would need to give them something like 5W if you want to see interesting things happen. <S> For the 120 Ohm one, you would need 25V (P=IV, <S> V=IR, P=(V^2)/R, V=sqrt(PR)=sqrt(5120)=24.5V). <A> When you set your power supply to 16 volts and 5 Amps, the supply will maintain 16 volts while delivering UP TO 5 Amps. <S> The actual current delivered will be determined by the resistance of the load. <S> If the supply delivered 0.11 amp at 16 volts, by Ohm's Law (V = I x R), the load resistance was about 145 Ohms, and the power dissipated in the load was about 1.76 watts. <S> A 1/4 Watt resistor would quickly go up in smoke, but a 5 watt resistor would just get fairly warm. <S> A 3.2 Ohm resistor would draw 5 Amps from your 16 volt supply. <S> If you connect a 1.6 Ohm resistor to your 16 volt 5 amp supply, the supply would reduce its voltage to about 8 volts, in order to limit the current to 5 amps. <A> The power (W) dissipated in a resistor (R) with voltage (V) across <S> it is... W = <S> V^2 / R Note that the power supply current limit is nowhere in that equation. <S> The current limit setting on the power supply is just a max allowed value. <S> The logic is like this. <S> if V / R < current_limit then the supply will output I = <S> V / R Amps. <S> otherwise the supply will output I = <S> current_limit Amps (and the voltage will drop to make it so). <S> You can see from the power equation, that since we are dividing by R, lower values of resistance yield more power for the same Voltage. <S> For example... <S> (10V)^2 / 1K ohms = 0.1W <S> (current is 10mA) <S> (10V)^2 / 100 ohms = <S> 1.0W (current is 100mA) <S> (10V)^2 / 10 ohms = <S> 10W (current is 1A) <S> (10V)^2 / 1 ohms = 100W (current is 10A) <S> So suppose you had a 1/4W resistor and you set the supply to 10V. <S> If you wanted to find the resistor value that would be at 1/4W you would write... 1/4W = <S> (10V)^2 / R <S> And then solve for R... <S> R = <S> (10V)^2 / (1/4W) <S> = 400 ohms <S> To burn up the resistor you might want to set the wattage to 4X that, so make it 100 ohms instead of 400.	To fry a resistor you need to put into it more power than its rating. When you set the "Amps" on your power supply, what you are really setting is the limit.
75 Ohm Coaxial Cable Used On a 600 Ohm Output I'm working on restoring an HP 211A square-wave generator and I'm looking to add a shielded cable to the 600 ohm output. I've got some spare 75 ohm RG59 cabling and I was wanting to know if using that would have any detrimental effects on the output such as signal loss due to the impedance mismatching. The length of cable will be 10" - 12", the output will be between 1Hz and 1MHz and mostly used on loads like oscilloscopes and maybe stereo receivers. To clarify, the RG59 cabling would run from the V10 and V11 plates to the wiper on R71, and then another cable from the non-grounded pin of R71 to the output screw post. I'd likely use 75 ohm cabling to run to any devices on the output though anyways. Thank you. <Q> If I searched right, the unit has a maximum output frequency of 1MHz. <S> A 1MHz wave has 300m wavelength in free space. <S> It will be slightly longer in a cable. <S> So the cable you will use, will be electrically short for sure, and only the capacitive loading of the cable and thus the bandwidth is to be assessed. <S> Such a coaxial cable should have roughly 100pF/m capacitance. <S> 600Ω output resistance and 100pF forms a pole at 2.67MHz, so you are good* up to 1m cable length for the full range of the instrument. <S> If you want a 1MHz frequency with a waveform including harmonics, then you should reduce your cable length, or the output impedance before the cable. <S> *Only 6% amplitude loss for a 1MHz sinusoid signal <A> At the frequencies your using there is little effect in using a lower impedance cable. <S> However you will want to terminate it with a 600 ohm load so the source can still drive it at full voltage. <S> Frequencies at or above 50MHZ are very fussy about impedance matching and standing waves <S> and you are nowhere near that high in frequency. <S> 600 ohms is typical for audio < 0.01HZ up to 1MHZ. <S> That is the range of my 600 ohm analog audio generator. <S> 1MHZ to 10MHZ for ultra-sound <S> is common today, so 10MHZ is nothing special anymore. <S> You do want to avoid cable with an impedance above 600 ohms, as this could deform the edges of frequencies approaching 1MHZ (for square waves). <S> Avoid 50 ohm and 25 ohm cables for this reason, and their higher cost. <S> For a "cleaner" square wave at 1MHZ you can load the end-point with less than 600 ohms, by going down a 100 ohms at a step, but the signal amplitude will drop. <S> You have to decide if signal amplitude is more important than signal quality. <A> RG59 Coax is 75 Ohms and as I recall 30pF/ft. <S> Twisted pair impedance and pF/ft depend on twists /ft. <S> The fewer twists, the higher the impedance and inductance must be considered as well. <S> It becomes a matter of tradeoffs whether you have rounded edges or ringing or attenuation with properly matched RC impedance divider like an 8:1 to 75 Ohms or unshielded wire prone to magnetic pickup or lightly shielded twisted pair with large twists. <S> degree of squareness at 1MHz length of twisted pair <S> overall <S> the impedance of load <A> 600 ohm is unrealistic; needs center conductor about the size of an atom, shield with diameter of the universe, from what I heard. <S> Use the 75 ohm; you are not expecting standing waves, so don't worry. <S> Regarding achieving 600 ohm COAX is not the same as achieving 600 ohm TWINLEAD. <S> Pole/crossbar mounted telephone cables are twinlead.	Cables lower than 75 ohm become a "bulk" problem as the center conductor can be thick, making the cable tough to bend.
Hardware to run a fixed neural network Suppose I have a feedforward neural network and I already have decided what its directed graph, weights and activation function should be. I want a device than runs this neural network as quickly as possible, with secondary considerations being size of the device, power consumption and cost. What is the best hardware design for this purpose? Have there been cases of say ASICs being built for a particular neural network? To give a rough idea my network has around one million neurons and ten billion connections. <Q> To give a rough idea my network has around one million neurons and ten billion connections. <S> That means ten billion weights. <S> From a NN perspective, this very much sounds like you want to apply L1 regularization to reduce the number of connections; unless you really have good reasons why this needs to be this large, I'm almost certain that you didn't sufficiently optimize your network. <S> Anyway, in its current form, you'd need something to store 10 billion coefficients. <S> The only realistic way of doing that is using SDRAM. <S> Then, you need an SDRAM controller. <S> And then, you need something to fetch the data from that controller and execute your inference. <S> In other word, you're describing a large GPU. <S> So, you need a large GPU. <S> A GPU these days essentially is an ASIC for NN applications. <S> The whole graphics aspects of it can be safely ignored. <A> If you want to implement a neural network in hardware, you need to prune <S> your neural network first. <S> It is not a good idea to implement a fixed neural network as an ASIC which has no option for reprogrammability as the requirements of your application <S> are likely to be modified in the future. <S> Deep Learning is well suited for parallel architectures. <S> 2 options for hardware which is inherently parallel are GPUs and FPGAs. <S> Read this paper to see which of the two are best suited for your application. <S> Since your application is computed at the edge, I would recommend the FPGA since it has lower latency when compared to a GPU. <S> FPGAs are reprogrammable and have computing, logic, and memory resources in a single device. <S> FPGAs have built in blocks for DSP, so they can efficiently perform convolution operations . <S> The reason why GPUs are preferred over FPGAs is because most AI scientists come from a software background and are familiar with GPU based parallel programming using CUDA and until recently FPGAs had to be programmed using Hardware Description Languages only. <S> But now, they don't have to be programmed using HDLs. <S> You can also use High Level Synthesis tools or OpenCL or other software like Vitis or OpenVino. <S> And you can reconfigure the hardware to efficiently execute algorithms if necessary <S> (in some cases it will be necessary). <S> Sources : <S> https://www.intel.in/content/dam/www/programmable/us/en/pdfs/literature/solution-sheets/efficient_neural_networks.pdf <S> https://www.xilinx.com/support/documentation/white_papers/wp504-accel-dnns.pdf <A> Edge TPU for hardware. <S> Notably, there is one from Google and another Chinese company called Sophon from Bitmain with several products like this one <S> Note that those products are fairly recent and not mature, so it might be difficult to implement. <S> They have either chips or modules, depending on if you intend to do your own design. <S> There are many different solutions, but depends a lot on your requirements, which are not very clear (size, power, etc..)	You can also consider using an accelerator which is based on FPGA (like Xilinx's Alveo.) There are some ASIC starting to go on the market for AI application, it is usually called TPU (Tensor Processing Unit) or
Why don't we use lenses for RF? Radio waves are a class of electromagnetic waves.Light also is a class of electromagnetic waves. By shaping a material in which the speed of light changes, we can bend the propagation direction of light, we call this a lens , and we call the speed change rate refractive index \$n\$ . Also for RF we can define a speed change rate as \$\frac{1}{\sqrt{LC}}\$ or \$\frac{1}{\sqrt{\epsilon_r\mu_r}}\$ . We also use the exact same destructive interference in the same way for radomes and Anti reflective coatings. So here comes the question: why don't we use lenses and ̶m̶i̶r̶r̶o̶r̶s̶ for RF, for example for focusing RF beams instead of using complicated directive antennas? EDIT: yes, we actually use mirrors <Q> In optical engineering, the choice between lenses and mirrors often comes down to aperture diameter: less than a few inches and lenses can be made cheaply and with high accuracy. <S> Larger and costs increase exponentially, so even 6" diameter systems usually work better reflective. <S> At RF frequencies, a 6 inch lens is on the order of a wavelength, and so not useful for focusing. <S> It isn't until you get towards the edge of the microwave spectrum that the wavelength gets short enough for lenses to start to become practical. <S> Of course if you don't care about cost, and you don't mind it being extremely heavy, you could build a lens to use with a WiFi antenna. <S> It just doesn't make much practical sense. <A> i don't know of any rf lens usage case <S> That can be changed. <S> They are quite common for ku-band and up. <S> Think satellite communications, radar, point-to-point links where you want high but can't use a dish, e.g. for weather reasons. <S> Look at <S> this nice lens antenna: <S> Or these nice insets to convert an open waveguide to an actual antenna <A> Luneburg Lenses are generally spherical, made of concentric shells of material with a stepped refractive index for practical purposes, but ideally the refractive index should be continously varying. <S> (image from www.rfwireless-world.com ) <S> Applications are radar reflectors, microwave antennas and laser collimation. <A> You can make lenses out of metamaterials - see this Phys.org story A three-dimensional self-supporting low loss microwave lens with anegative refractive index Journal of Applied Physics 112, 073114(2012); https://doi.org/10.1063/1.4757577 Isaac M. Ehrenberg,Sanjay E. Sarma, and Bae-Ian Wu <A> Heinrich Hertz used paraffin wedges, in the 1890s, as part of his 60MHz RF communication link. <S> The wedges were crucial to proving WAVES were being emitted and collected and detected. <A> RF mirrors are quite common. <S> RF lenses are possible but much less common. <S> The short explanation is cost. <S> Cost is almost always part of engineering. <S> If there are two ways to do something, and one costs less for adequate performance, then that is the better engineering solution to the problem. <S> An array of wires to form a directive or reflective antenna element is cheaper to build and maintain than a solid lens structure for most applications. <A> Another well-known example of RF lenses are Fresnel zone antennas . <S> They are based on the same principle as the optical Fresnel lenses: The focusing effect is achieved via the phase shifting property of its surface rather than volume, which allows for compact or arbitrarily sized antennas (e.g. built into a curved surface). <A> Lenses can be used in the millimeter wave region. <S> See, for example, https://doi.org/10.1007/BF01014036 <S> Which describes use of Rexolite lenses for 90GHz military applications.	There is the Luneburg Lens , which can be used for various applications, from optical to RF.
How to estimate the temperature of a filament? I would like to know how I can estimate/calculate the temperature of a heated filament.For example, the filament (rhenium) is 0.2mm in diameter and heated with a voltage of 6.4V and a current of 2.72A (i.e. power=17.4W.) My second question: How does the temperature of the filament depend on the filament current? <Q> The method you use will depend on the accuracy you want. <S> As you've used the word 'estimate' in your question title, I presume that you don't need standards lab accuracy. <S> Adding an accuracy specification, or adding the purpose of the equipment to your question, may get you more relevant and focused answers. <S> The simplest way to estimate is to measure the resistance when cold, and at temperature. <S> This depends on having a graph of resistance versus temperature. <S> This data is for pure rhenium from Kaye and Laby at the npl.co.uk via the wayback machine <S> temp K resistivity (10^-8 ohm.m) 78.2 <S> 2.62 273.2 <S> 17.2 <S> 373.2 <S> 24.9 573.2 <S> 39.7 973.2 <S> 63.51473.2 <S> 84.4 <S> There will be errors from the purity of the element, and the fact that the measured resistance is an average of the cool bits of element near the mounts, and the hot bits in the middle. <S> There may be a steady drift in the diameter and so resistance of the element due to degradation of the element, though this should be recoverable by a new cold measurement. <S> There may be a drift in the resistivity of the element material, due to reaction with the atmosphere it is operated in. <S> The main alternatives are infrared thermometry, and spectroscopy. <S> The latter may be possible fairly cheaply with a diffraction grating and a camera. <S> For the roughest and dirtiest way, look at the colour of the filament, and interpolate from colour description versus temperature. <S> You will find many tables of colour versus temperature online, generally from artisans who weld, operate kilns, heat-treat metals. <S> Spectroscopy is just a way to instrument this process and make it objective rather than subjective. <A> Adjust the current through the lamp until its filament disappears into the sample. <S> Approximations of the tungsten filament temperature may by accurate enough from calculations. <S> Calibration and temperature profile of a tungsten filament lamp suggests approx 2% accuracy. <S> The open access paper "Analytical expressions for thermophysical properties of solid and liquid tungsten relevant for fusion applications" https://doi.org/10.1016/j.nme.2017.08.002 is a goto source of information. <S> https://en.wikipedia.org/wiki/Disappearing-filament_pyrometer <A> It sounds like you want to estimate the temperature of the filament before you even build it. <S> That is, you want to calculate the temperature without making any actual measurements. <S> The difficulty we have with questions like this is that you really want to solve a heat transfer problem, not an electrical problem. <S> You know how much power you are dumping into the filament and you know that almost all of this power will be converted to heat. <S> To know the temperature of the filament you need to model the flow of heat away from the filament to the surroundings. <S> You need to know the thermal conductivity of the electrical connections to the filament and of any physical supports for the filament. <S> You need to know about the airflow and ambient air temperature. <S> At high temperatures, or in a vacuum, you would also need to calculate the power that is lost through radiation to the surroundings. <S> I can't help you with any of that...you might need a mechanical engineer or a physicist.	A simple way to measure the temperature of incandescent surface is to compare the colour of the filament of a known lamp.
What cause Diode bridge to short circuit? My circuit got 220V to 36V Transformer (Approximately 50VA) KBPC2510 bridge rectifier (1kV, 25A) relay to coltrol motor of/off 36V DC motor (100W, 4A), just drive rotary knife to cut fabric. At first place, primary side of transformer got short circuit winding and triped fuse. after replace transformer and fuse I found that diode bridge broken as well by 2 diode act like short circuit (negative pole to both AC pole)My question is. Is that normal failure mode of bridge rectifier? How it happen? Is my circuit lead to failure and how to prevent. -- Edit --Added schematic.Note: For record I draw this schematic as it's wire in actual mechine, not my design. and I still wonder why it's look so minimal. simulate this circuit – Schematic created using CircuitLab Edit 2Additional information Transformer has no datasheet or VA rate. I measure E core and got 3"x1.5"x2.5" . As I found similare size transformer It say 50VA. DC Motor nameplate say ZD Motor(Z3D100-36A2), 100W 36V 4.0A But it's used to cut fabric with rotary knife, shouldn't draw much current. <Q> The diode bridge is burnt due to the inrush current absorbs by the DC motor at the beginning. <S> To avoid this I can suggest you two methods. <S> Use a NTC thermistor( negative temperature coefficient resistor) in series with DC motor. <S> Use a soft starting circuit to control the DC motor. <A> Your setup as show, no filter cap, is a reasonably common situation (no filter cap) to run DC motors from an AC source. <S> A 1kV 25A bridge should not fail easily with a transformer that tiny, I suspect short circuit current is less than the bridge rated current, though without heatsink it might conceivably have eventually overheated and failed. <S> You could probably see some evidence of that (discolored markings, for example). <S> If that makes sense, I suggest you replace it with the same type and adjust the fuse to a slow-blow type that has a relatively low current rating. <S> Then, if the motor is stalled for many seconds the fuse will open. <A> KBPC2510 bridge rectifier, rectify 36VAC to 36VDC <S> No, way off - the peak voltage will be more like 50 volts and, if you are using a smoothing capacitor the DC level (ignoring ripple) will be about 50 volts also. <S> Draw a schematic - that is the language of EEs.	Assuming there is actually no (substantial) filter capacitor, it's possible that an external short circuit or a stalled motor caused the transformer to fail.
Do microcontrollers have a big resistance? Well, it maybe seems a very basic question, but I have this question because I'm trying to understand the 0 weak and 1 weak on pull-ups and pull-dows, for example: Here, when the button is not pressed, we have a weak 1 on the input pin, and when the button is pressed we have a strong 0 on the input pin (or, at least, this is what our professor taught us). But this strong 0 is because the MCU has a BIG resistance so all the current will go directly to the left GND instead of going through the MCU. Am I right? Can you give also typical values of pull-ups resistors and values of MCU resistance? <Q> That is correct, in the sense that it can be thought like that for the sake of simplicity, but in reality there is no resistance at all, as it's a CMOS input which has an extremely high input impedance. <S> However, nothing is perfect, and there can be leakage current flowing in or out of that pin, but in worst case it is in the order of 1 microampere at 5V voltage, so it can be thought of as resistance larger than 5 Megaohms. <S> Typical values for pull-ups is a more broad question, as there is no single answer. <S> Depends on supply voltage, switch current rating, level of coupled noise etc. <S> So from 1k to 100k maybe. <A> Can you give also typical values of pull-ups resistors and values ofMCU <S> resistance? <S> 'Typical' MCU input resistance is >100MΩ, except for those that have built in pull ups or pull downs. <S> But you shouldn't rely on a 'typical' value - read the datasheet to find the specs for your MCU (and remember that temperature can have a large effect). <S> A higher value draws less current when the input is low, but has lower EMI immunity. <S> Some switches may require a lower value just to get enough wetting current to keep the contacts clean. <A> A microcontroller input does not have a weak pull down as you have drawn it (R2). <S> It has a high impedance in the form of a MOSFET gate which should be many megaOhms. <S> There are other parasitic paths to the power and ground rails which would reduce that impedance <S> but I don't know typical values, or even if there are typical values. <S> Seems doubtful. <S> So there is an "R2" so to speak <S> but it's not necessarily tied to GND. <S> It could be tied to +V. <S> You don't know. <S> In reality there's probably one to both <S> and what voltage it sits at when disconnected at <S> is indeterminant. <S> The way you have drawn it implies that with nothing connected to the input, the input pin will read zero which is not true. <S> But if you are talking about inputs where an internal pull-down (or pull-up) is enabled your diagram is correct (for a pull-down). <S> I would expect the internal resistance to be no less than 20K typically. <S> In either case, you can say that all the current flows through the pushbutton <S> but it is because the pushbutton resistance is very low (basically just copper). <S> simulate this circuit – <S> Schematic created using CircuitLab <A> For compactness on silicon, the pullup is likely to be a long_channel narrow_width Pchannel Fet, or perhaps a very weak current source. <S> By manipulating the FET of the long_channel FET, you can enable or disable the pullup current. <S> Shutting off a current source, perhaps driven from a central bias generator, is only slightly more complicated. <S> My views on this are from the circuit designer level, using some understanding of FETs and how they are modeled. <S> I am not a skilled device physicist.	It can be thought as resistor so large in value that it's not even there, and theory no current flows in or out that pin. The pushbutton switch resistance is so low it almost doesn't matter what the input's pull resistance is. For external pullup resistors the 'typical' values I have seen are around 1-50kΩ, but it depends on the circuit requirements. So "all" the current would flow through the pushbutton if the input resistance was many megaOhms, 20K, or even 100 Ohms.
Can lightning strike into the phone line damage electronics in the house through the ADSL modem? I was always wondering about the following: If lightning strikes directly into the phone line outdoors, can the excess power coming from the phone line jump over the switching power supply of the ADSL modem into the mains in the house and damage any computers / TVs / washing machines? Let's assume that the ADSL modem is WiFi only and that there are no cables connected to it except for the phone line and mains. <Q> There are many 'entry points' to your house for electrical disturbances, once lightning has struck close by, and the most important of these is the mains. <S> A nearby strike will lift the ground potential, and induce a voltage into any wiring. <S> My anecdote is that I live in a row of houses. <S> A house 4 doors up was struck by lightning. <S> They needed to be totally rewired and replastered. <S> Cables had blown out of their walls. <S> Their neighbour's wiring survived, but they lost every item of electrical equipment, including simple motor-only things like fans. <S> Their neighbour's motor-based equipment survived, but they lost all electronics. <S> I lost only my ADSL modem, perhaps as it was connected to mains and the phone line, but all my other electronics survived. <A> Back in my hardware days, I was a technician repairing modems for a modem manufacturer. <S> I repaired many modems hit by lightning strikes. <S> They all had a couple large resistors connected to the phone line that would take the brunt of it. <S> The smell of a fried 2W resistor is quite distinctive. <S> I could tell it was a lightning strike when I took the case cover off. <S> At most it would char part of the circuit board. <S> I had to grind off the charred part, glue down traces from a scrap board if needed, the solder everything back up. <S> I never saw a power supply get harmed. <S> I would think other things would be more susceptible to power surges through the mains than the modem. <S> That's how we lose most of our electronics around here. <S> So, don't worry about it. <A> It's unlikely. <S> The wires of your telephone line are very thin and are close to ground at enough points to first vaporize before they can hit your washing mashine. <S> Let's assume that the ADSL modem is WiFi only and that there are no cables connected to it except for the phone line and mains. <S> If there is any vector for damage to your washing machine it's through mains, anyway. <S> Generally, lightning strikes into cables close to you are far rarer than people think they are <S> (no, anecdotes don't change that). <S> The likelihood of telephone lines being hit is smaller than that of power lines, because these typically hang higher for isolation purposes. <S> Multiply that with the small probability that damage to the phone system will propagate any further than the point at which the telephone line enters your house, and you'll find that you're worrying about something that is far less likely than your building collapsing onto your washing machine, computer and TV. <A> In my experience the failures have always been limited to power supply components. <A> At that time analog modems were in use and the computer the modem was connected to was sometimes destroyed. <S> However, only devices that were directly connected to the modem were destroyed. <S> WiFi did not exist at that time.	The probability of damage to electrical equipment through mains supply surges, after a nearby lightning strike, is much higher than that caused through telephone lines. In Austria I have seen such lightning damages years ago.
How does a pull-down prevent a "false 1" in a pin? yesterday I asked a question about pull-ups and pull-downs and resistance in an input pin, but I feel that the question that invaded me today is different from my yesterday's post so this is why I'm asking a new question instead of just editing yesterday's question. Maybe my question is similar to other questions in this site (like this: Why does a pull down resistor eliminate floating input , but I feel that they are not exactly answering what I want to know. Let's start with the question, we have this: I understand that, if the pull-down didn't exist, we would have a floating input in our microcontroller pin, because some interferences could give a logic '1' to our microcontroller. But what I don't understand is how the pull-down does to always get a logical 0 on that pin (when the button is not pressed). I mean: supose that interferences are giving 4V (logical 1), for example because we touch it with the finger. Here: So what I imagine that is happening is this: This is why I'm not able to understand why pull-down is avoiding a voltage on the microcontroller pin due to interference. I want to add this question too because I don't understand very well this concept: Why is it considered to be a weak 0 in a pull-down and not a strong 0? <Q> If you don't have a pull-down, and you push the button to give the MCU 4V which it understands as logic one, everything is fine. <S> Now, as the MCU input is a CMOS input, it has extremely high impedance, assume it's infinite, and assume there is some parasitic capacitance that is now charged to 4V as well. <S> When you stop pushing the button, the MCU input is disconnected from 4V supply, but since there is nothing to discharge that 4V from the parasitic capacitance, it will stay at 4V forever, and will never read as logic 0, no matter how long you wait. <S> When you have the pull-down resistor there, as soon as you stop pushing the button, it will be a DC path to discharge the 4V, so the voltage will end up to 0V and the MCU is able to read that as logic 0. <S> So if you pick a too high value as the pull-down resistor, like 100 Megaohms, it too high to prevent voltage fluctuations if you touch the pin with a finger. <S> If you pick too low value as the pull-down resistor, like 1 ohm, you are just wasting power with too high current, and the pushbutton may not be rated for it. <S> It is considered a weak pull-down because it is not strong. <S> It's weak in the sense that it allows the pushbutton to control the voltage without too much current, and it's strong enough to keep the voltage at 0V when pushbutton is not pushed. <S> For argument's sake assume it is 10 kilo-ohms. <A> There is a simple rule that comes from Ohm's Law: there is a current -> <S> there is voltage drop (difference). <S> No current - no voltage drop. <S> When signal is high, you obviously have current through the resistor, which means there MUST BE a voltage drop across it. <S> Voltage is not equal on the sides of the resistor. <S> But if you disconnect the switch and you have just a pull-down resistor, there is no power source attached to the pulled down line, thus no current across resistor, thus both sides of the resistor have the same voltage. <S> The only defined voltage is the ground, so the other side will automatically be zero too, preventing floating. <S> I tried to simplify the model for understanding a little <A> Think about what voltage is. <S> A voltage source has to supply some power at some potential. <S> A resistor can't supply power or potential as it's not a voltage source. <S> The voltage across a resistor with zero current flowing in it is zero. <S> Ohms law. <S> As the input has zero current flowing (ok in reality there IS a very tiny current) <S> but zero as near as matters. <S> Zero current in resistor, zero voltage across it. <S> A weak pull down (a large resistor) may exhibit a small voltage due to current leaking out of the input. <S> A strong pull down will see the same current but generate a smaller voltage.	The pull-down resistor has to be low enough in value, that any interference that might be there, will not be able to make the voltage rise above logic 1 threshold.
I need help with determining voltage wave form of LM317 Now I calculated Vout as being 5V, so is the Vout wave form the same as Vin? <Q> The LM317 is a linear voltage regulator. <S> It expects a DC input voltage at least 2 volts above the desired output voltage for proper operation. <S> If you give it a sine wave input as shown, it will produce a pulsed, varying voltage, with a maximum voltage of 5 volts. <A> There is a minimum input voltage that the LM317 needs to be able to regulate the 5V level on the output. <S> Anything below that level, the output level will fall. <S> Anything above that minimum input voltage level, the output should remain relatively close to 5V. <A> There is not really enough information on the data sheet to allow you to answer this question accurately. <S> I would expect it to look a bit like <S> the 7805 input-output characteristics: <S> Figure 5 in the datasheet shows you where the output voltage starts to decline linearly, but I see no information on what happens at the output as the input voltage approaches zero. <S> At some point it will likely drop to near zero, similar to the LM340 aka 7805. <S> So the output waveform would stay at zero for a spell, zoom up to some voltage, increase like a segment of a sine wave until the output hits 5.0V (at which point the input will be somewhat higher) and flat-line until the input falls again to the drop-out voltage at the output current (in your case, just the resistors form the load).	No, LM317 is a regulator and will try to hold the output voltage at the predefined value determined by your setpoint resistors.
Install grounding cable with eyelet to outlet, no center screw I've found an ESD mat that comes with a grounding cable with an eyelet that's supposed to be screwed into the center (grounded) screw of an electrical outlet. But in my home, the outlets do not have one center screw; they have two screws, a top one and a bottom one. Would either of these screws be grounded? If not, how would I connect the grounding cable? Edit: The outlets are three prong. I live in Canada. <Q> The round "D?" <S> outlets use 1 centre screw. <S> The square outlets use 2 screws for the cover. <S> Both are grounded. <S> You can file a 1mm slot for the wire exit and fasten to the metal box outlet then cover or use an external centre wire with a ringlug that is a threadtight fit. <S> Your choice. <S> You can buy or make these and somewhere <S> there is a 1Mohm R in series. <S> Just fasten any suitable wire to the box, file a relief slot for the wire to exit on the edge of the drywall and cover then connect to matt and use strain relief. <S> Voila! <A> We are all making assumptions about outlet grounding, and that is not a good path to follow. <S> Many homes over 70 years old may not have grounded outlets, so if you move into an old home as a fixer-upper, you may have to upgrade electric and plumbing to meet codes. <S> One of my brothers moved into a old house 20 years ago and bought it twice after upgrading many things. <S> Though the electrical outlets had a grounding hole, the Romex cable was only 2 wire with no ground. <S> So the refrigerator could shock you sometimes. <S> The key is to remove the plastic plate and check for a bare 14 to 12 gauge ground wire. <S> If it is there it will be wired to the ground lug on the power outlet. <S> But old bakelite boxes and todays plastic boxes are common, as code only requires that a ground wire of equal size to the power wires be run to each outlet. <S> If there is no ground wire in the box it cannot offer you a grounded connection, and you must contact a certified electrician to run ground wires. <S> If there is a ground wire then the metal screws that hold in the plate should be grounded as well. <S> The screws are normally 6-32 in the USA so a number 6 ring terminal with a wire crimp will get you a valid Earth ground. <S> Even more reason to take off a cover plate at least one time to verify the house / trailer / apartment has a ground wire. <S> If wires are stranded the ground wire is often colored green or green with a yellow stripe. <A> Every Decora-style duplex receptacle that I have seen for the past several decades has two screws that attach the cover plate to the receptacle. <S> Those two screws go into the metal frame that the receptacle is built around. <S> So long as the receptacle is properly grounded, either of those two screws are an adequate ground attach point for your ESD mat. <S> A great method of attaching to one of those screws is to use a #6 ring-tongue crimp terminal under the screw head.	Since it screws into what is sometimes a metal outlet box it is also grounded. Removing the cover should reveal the bare copper ground wire fastened to the box. It is understood that grounding codes in other countries may exceed or fall way short of USA NEC codes.
Choosing series resistor for Optocoupler LED How do you choose the series resistor value for the LED side of an optocoupler. My application uses this opto to interface a button that is about 25meters(50 meter wire roundtrip) away from the device. For choosing R44 or the collector pullup was pretty easy since the datasheet has provided a 1k. On the other side however there is no mention of a resistor. I would like for the led to be able to take inputs from at least 3.3 to 12volts. And having a look at the input characteristics of from the datasheet the maximum forward current is 60 mA(for a typical LED this is very very large). What would be the minimum current of the LED for so that the PIN can still safely register an "ON" state. Would a higher current(lower value resistor) on the led (30-40mA) be safe for repetitive use? Going further, to skip using a voltage regulator (it will come from mains anyway) is it possible to use the mains voltage 240 VAC instead since the optocoupler is capable of using AC anyway. How would I choose my resistors then ? <Q> Speed suffers approximately proportionally, but that's not so much a concern with a manual pushbutton. <S> The next simplest thing is to add a constant current circuit to the input. <S> For example: simulate this circuit – Schematic created using CircuitLab <S> By adding a couple pennies worth of parts, the current is typically maintained fairly constant from about 2V to 25V (at room temperature). <S> In either case you want to have a lot of margin <S> (maybe 2-3:1) to make sure the transistor saturates and to allow for normal aging. <S> The output current is 3.3mA so if your minimum CTR is 100% you might want 10mA nominal at the LED. <S> That would lead to excessive current with 25V in and a simple resistor R = <S> (3.3V-1.2V)/0.01A = 210 \$\Omega\$ <S> so at 25V you'd have 108mA (but not for long). <S> If you allowed only 15mA at 25V in that would require about a 1.5K resistor, so the current at 3.3V would be only 1.4mA <S> so with a CTR of 100% you'd want more like 7K at the output. <A> These devices usually spec a current transfer ratio which makes it hard to determine how the LED is actually working (it would be better if they gave a curve for the LED). <S> With LED's more current can cause lifetime issues, so it's best to stay at the nominal value. <S> In their testing diagrams they used 10mA, I would stick with that. <S> If you were running into issues on the transistor side with it not turning on, then I could see needing to get a better current transfer ratio, but it looks like you can also set R44, so that would be a better way to adjust the transistor current. <S> So if you use 10mA and a forward current of 1.15V with a 3.3V <S> Vcc <S> then you'd need a V drop on the resistor. <S> V/I = <S> R <S> so <S> 2.15V/0.01A = 215Ω <S> For (12V-1.15V)/0.01A= 1085Ω On the transistor side, it really depends on what the transistor of the opto is connected to. <S> The resistor needs to source more current than the load (whatever pin is connected to). <S> For most microprocessors 20k or 10k is plenty. <S> Source: <S> https://www.vishay.com/docs/83653/ild620.pdf <S> As far as connecting the LED directly to mains, it's not safe. <S> At minimum need an isolation transformer. <S> You could use a step down transformer. <S> The problem with connecting the LED side to an AC signal is if you push the button down and turn the LED on, it will cycle at the mains frequency and turn on and off rapidly, if you don't want this to happen the best way <S> would probably be to use an LDO. <A> There is no need for an Optocoupler here. <S> RS232 can go much further than this easily with higher impedance. <S> Just need UTP cable for balance on ingress EMI. <S> You show common grounds and there is probably supply crosstalk leakage. <S> Just use a low impedance switching circuit to shunt Vbe with ESD protection when the switch closes. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> All R values can be scaled up if 5mW power is a concern. <S> If there are high current noise impulses on wires nearby, you can add 10nF to 100nF Cap across Vbe. <S> Use UTP wire for CM rejection and STP wire for an aggressive noise environment. <S> Optoisolator will not block help on the above noises since there is no ground current in the open switch state to have crosstalk. <S> Thus no advantage. <S> But if there was capacitive coupled noise to the cable, the opto would pass that thru as current.	If you want to accommodate a wide range of voltages, the easiest thing to do is to increase the resistance at the phototransistor so it will work at lower currents. Maybe 2K and 10K would be acceptable, again assuming a nominal CTR of 100% minimum over the LED current range of 1~10mA. I would not go over about 15mA, LEDs age and will eventually drop in brightness to the point where the CTR is not high enough to work.
How to connect 5 kW electric heater to the weak wiring in house? The water heater has a 5kW heater and requires 20A. Unfortunately, the wiring in our house is old and this allows us to get only 10A. Is there an easy way to reduce energy/current consumption and connect such a powerful heater to the network with an energy consumption of about 2kW? I found, that "Triac power regulation" can be used to reduce the power consumption: I've tested schematic using LTspice and found, that peak current is the same (very big) - 20A: In this case, power consumption is about 2.5kW, but I am worrying about weak house wiring because of peak current is 20A. I'm not sure if it is safe. P.S. The voltage at the outlet is AC 220V (50Hz). <Q> Running double the current through your wiring causes 4 times the dissipation in it, and so more or less 4 times the temperature rise (assuming conduction and natural convection as the principal sources of cooling at these temperatures). <S> It's the temperature rise that causes the damage, softening insulation, starting fires etc. <S> Using the triac circuit is supplying power for 50% of the time. <S> Although that would cut your heater dissipation in half, it only cuts your '4 times too much' wiring dissipation in half as well, leaving the wiring temperature still double what it should be. <S> Given that the existing '10 A' wiring is rated for a significant temperature rise, even twice the rise is still far too hot for safe operation of the plastic insulation. <S> At double the rated temperature rise above ambient, it's likely to soften, and flow away from anywhere there's pressure on it. <S> This could cause conductors to contact each other. <S> In the best case, the short would open the circuit breaker. <S> You'd then have a dead circuit that would have to be repaired/rewired by an electrician. <S> In the worst case, an arc between the conductors could start a fire. <S> Your only two sensible options are to rewire, or to buy a much lower power heater. <S> Or three options, stop using the heater. <S> Or four options, continue using the heater, and have a good story ready for the fire investigators. <S> Or the fifth option, install a 2:1 step down transformer at the sending end of your 10 A feeder. <S> This will run the wiring at 10 A, so the wiring will run at rated, but the heater will run at 1.25 kW. <S> What about the 6th option of using the triac to turn on for less than 1/2 a cycle? <S> That's possible. <S> The heater and wiring will dissipate 25% of their 'full' power. <S> You don't need to worry about the peak current, just the rms aka heating effect, at least for modest peaks like this. <A> Peak current is not a problem. <S> Average current is. <S> As Neil_UK explained, the risk is from heating. <S> That's why the circuit breaker must never exceed the capacity of the wire. <S> First see what wire it is. <S> If it's 2.5mm you can safely use 20A. <S> If it's 1.5mm, only 16A. <S> Anyway, I would NOT recommend a regulator circuit on a 1.5kW installation because the regulator itself will heat. <S> If not the wires. <S> Components should be rated high. <S> You can do mistakes etc. <S> You will also waste power. <S> Maybe 10%. <S> The best is still taking a drill and buying a 4mm cable (or at least 2.5mm) and a dedicated 20A circuit breaker. <S> But you will need an engineer to design your triac circuit properly. <S> ;) <S> •1,5 mm² → 10 <S> A → 2 <S> 300 W <S> •2,5 <S> mm² <S> → 16 <S> A → 3 680 W (also good for 20A - 4 400 W) <S> •4 <S> mm² <S> → 25 <S> A → 5 750 W •6 <S> mm² <S> → 32 <S> A → 7 360 W •10 <S> mm² <S> → 40 <S> A → 9 200 W. <A> Water heater elements are replaceable. <S> They have to be because the elements die - often long before the heater tank does. <S> Just remember that your water will take significantly longer to heat up.	Wiring correctly the circuit must also be done carefully. It will also only work if this is an isolated feeder to only your heater, due to the non standard voltage. You don't need an electrician to do this. If you turn the triac on for the last 1/3rd of the cycle, then you'll deliver about half the rms current. Insulation relies on being solid. So the solution to your problem is easy - remove the 5kW element and replace it with a lower power one, somewhere in the 2kW - 2.5kW range. It might be that buying a suitably sized transformer is competitive with hiring an electrician to run a 20 A cable.
What is the ammount of current when using no resistors Ohm's law states that V=IR, thus if we have a 9V battery and one would use no resistance (R=0) then I= 9/0 which can't be done. Then what happens to the current? Does this mean there is an infinite amount of current? Or does a battery have a max amount of current it can provide (if so, what is this max amount of current)? Or do wires have a little resistance for example: R = 1 * 10^-6. This leads to an insane amount of current. <Q> so there would also be no voltage drop <S> so it's a very undefined situation. <S> However, if you look at V=I*R from another angle, having zero resistance means that there is no voltage drop anywhere, regardless of what the current is, which might make more sense. <S> In practice, wires have resistance, but the battery has even larger internal resistance due to it's chemistry and structure. <S> It is not likely to get more than few hundred milliamperes of short circuit current out of a 9V battery. <S> OK, I was wrong, I shorted a 9V battery with a multimeter in 10A range, and got a short circuit current of approximately 2.5A. <S> The open circuit voltage of the battey was 9.15V. <A> The 9v battery has internal resistance (see datasheet), so R is not actually equal to zero. <S> If you short out a very large power supply (R<<1) you get a lot of current, but R never actually reaches zero, so you can plug it into that formula and get the short circuit current. <A> I think the theoretical answer would be an infinite amount of current. <S> A more practical approach would be to do a simple experiment using different resistances and noting down the different current values. <S> Then you can extrapolate that result to a very small resistance value to see how the current behaves. <S> Might be interesting to check out the actual result.	In theory if it was an ideal power supply without internal resistance, and wires would not have resistance, you would have infinite current flowing, which is actually not possible as there is no resistance
Is this a safe way to measure mains power on an oscilloscope? I wanted to use my cheap battery powered single-channel oscilloscope or my Rigol 1054Z to see the 120V AC mains signal in my house. No particular reason, just curious. I at least knew enough to not just blindly stick the probe in the wall socket so I got on YouTube and found a lot of information, some of it contradictory. So taking what seemed like a reasonable approach, I built an "isolation transformer" from an old wall wart I had laying around. (Yes, this was an actual transformer and not just a switching power supply.) Now I'm wondering if this is actually a safe approach. Upon opening the case, I saw the transformer and a small circuit board with a full bridge rectifier. So I removed the circuit board, which left me with just an AC - AC transformer. Using my DMM, I measured 115 VAC in and 9.9 VAC out. I then soldered the two leads from the transformer to two banana jack ports. There was plenty of room in the case to drill holes and mount these so they can be accessed externally. See the attached picture. I tried this out with my battery powered oscilloscope and got the expected results. The 9.9V signal was a clean sine wave at 60Hz and I didn't fry anything or die, so I figured that was a success. My questions for you - Is this safe? Can I hook it up to my mains-powered Rigol? Or is there an important point I've missed? Thank you for your help. <Q> That looks like a good safe project to me. <S> You might consider checking the voltage between the 9.9 V terminals and earth. <S> You will probably find some random voltage perhaps even 115 volts. <S> Connect a 115 volt light incandescent bulb between the 9.9 V terminal and earth. <S> With that connection, the voltage across the bulb should drop to nearly zero. <S> That should demonstrate that there is some small leakage current through the transformer due to capacitance between the primary and secondary winding. <A> From your description, it sounds like you've made a safe isolated source of 9 VAC. <S> To verify that the output is isolated from the AC Mains, you could measure (with the thing unplugged!) <S> the resistance between the mains terminals and your output terminals - there should be no continuity (inifinite resistance) between the mains pins and the 9 VAC jacks. <A> and found a lot of information, some of it contradictory. <S> Some people take a more paranoic approach to safety than others. <S> In general there are three main issues when using a scope directly on the mains. <S> You can get around this by just measuring with respect to ground. <S> It's probably close enough and you can always measure the neutral and subtract. <S> Voltage ratings, many scopes have too low a voltage rating for mains power. <S> You can work around this with an attenuated probe. <S> The thing that many people forget, overvoltage categories. <S> Many scopes are only "Cat 1" which are not suitable for direct connection to the mains. <S> For connection to mains power from a domestic outlet you want at least "Cat II". <S> To fix this you really want an isolated probing soloution. <S> Your transformer solves all of these issues, and it is safe to connect your scope to it's output <S> (secondary circuits of normal mains transformers are considered CAT 1). <S> However it may well distort your results. <S> The proper solution is a high-voltage differential probe, but such a device is out of the price range of the casual tinkerer. <A> You are on the right track. <S> There one thing to watch is the ground on your scope is not "ground" once you isolate it. <S> You can easily get line voltage on the case of your scope since it is the point your probe ground is attached to.	Also some scopes are more suitable for mains measurements than others. Scopes are normally referenced to mains ground, so if you hook your probe ground lead to mains neutral you will create a neutral to earth fault which will trip RCDs/GFCIs and could potentially melt wires.
Even though the signal gets low several times a second why our eyes are only able to see the bulb glowing all the time I was learning Pulse Width Modulation and it told me that it just the technique to change the amount of time the signal gets high in one wave, but why am I able to see the bulb glowing at all the time even if the duty cycle is set to 40%. <Q> Persistence of vision traditionally refers to the optical illusion that occurs when visual perception of an object does not cease for some time after the rays of light proceeding from it have ceased to enter the eye. <S> The illusion has also been described as "retinal persistence", "persistence of impressions", simply "persistence" and other variations. <S> According to this definition, the illusion would be the same as, or very similar to positive afterimages. <S> Source: Persistance of vision . <S> There is more in the linked article. <S> If you move the PWM display across your field of vision while looking straight ahead you should be able to see the strobing effect. <S> I've also managed to see it during humming while looking at a multiplexed LED display. <S> You need to vary the pitch of your humming until your eyeballs vibrate at a multiple of the strobe frequency. <S> It's quite weird! <A> Depends on PWM rate (frequency) and what light source you use. <S> It also depends on the ability of the person to detect the blinking as it varies. <S> If you apply 10 Hz PWM to a LED to turn it simply on and off at 40% duty rate, you will see it being on and off, at least when moving it. <S> People don't have a fixed "flicker fusion threshold" <S> so the point where it can or cannot be seen varies between people. <S> If you apply high enough PWM rate like 10 kHz, nobody will be able to see it blinking any more. <S> If the light source is an incandescent bulb, the filament stays glowing hot even when it does not receive power, so 10 Hz PWM rate will be fine. <A> At frequencies above the flicker fusion frequency, pulsing light is perceived as steady light at an average intensity. <S> Note that the minimum frequency varies with light intensity and where the light is in your field of vision <S> (peripheral vision is more sensitive to flicker) and other factors, as described in the linked Wiki article. <S> If there is relative motion between the light source and your eyeball you can perceive something like the a multiplexed LED display or indicator "breaking up" into dots characters because the light hits different receptors in your eye. <S> For this reason it's good to keep the frequency well above the minimum, especially where vibration is involved. <S> More like 1kHz than a few tens of Hz. <S> An interesting parlor trick (courtesy of John Larkin) is that if you look at the screen of an old-fashioned analog oscilloscope with the trace sweeping at an appropriate rate, and whack yourself on the top of the head, you will see a trace representing the control system tracking response of your eyeball.	It's due to the response of the eye and brain to rapid variations of light intensity.
Is building a long range wifi antenna to connect to the up and coming starlink internet possible? I want to try to convert a residential satellite dish into a long range wifi antenna to theoretically see if I can connect to the up & coming starlink internet that is being released in Canada and United States from virtually wherever.It's just an experiment, and I don't intend to use it for illegal purposes. I imagine there must be some sort of security feature that would stop me from intercepting starlink wifi signals. But this is my first time ever doing anything of this nature, does anyone have any experience with intercepting or capturing wi-fi signals or could maybe tell me why my line of thinking is incorrect, or maybe if I'm possibly going in a right direction? <Q> Starlink doesn't use WiFi, so you won't be able to capture WiFi from it. <S> If you had the right antenna and electronics (which you certainly do not given the cost), you could see the RF signal. <S> You wouldn't be able to decode it however. <S> If you want to do amateur radio, there are lots of lower frequencies being used on ground and in orbit. <S> These are much more accessible with affordable electronics. <A> You can intercept the downlink signals if you like, but they won't be of any use to you since they will either be for other subscribers or encrypted or both. <S> Even ordinary Wifi is encrypted, so it is likely that Starlink will be. <S> Details on how it works are sparse, but I imagine they won't accept transmissions from you without a paying subscription, and a one-way internet is of very little use. <A> However, what makes you think the satellite system even uses WiFi signaling? <S> Most likely it does not, <S> so by combining a WiFi base station with a satellite dish, it will not be possible to communicate with the system in any level. <S> So without specification how it works and proper tools to make it work, you won't be able to build one. <S> And by looking at Wikipedia, it seems to use frequency bands above 10GHz, <S> so no, it won't be compatible with your WiFi equipment. <S> It also does not use a single dish but a phased array antenna. <A> Your idea has technical, legal and possibly other important aspects. <S> First, Starlink is not WiFi. <S> I am yet to see their technical specs, but given the completely different task, spatial arrangement and business model, they use different frequency bands, probably different modulation schemes and probably different access protocol as well. <S> Your high-gain antenna will have rather narrow beam width, so it will have to track some particular satellite as it orbits across the sky. <S> They are on low orbit so each one of them is visible for less than an hour <S> and then you have to communicate to another one. <S> As I am thinking about the whole task, antenna is the easiest. <S> Legal: <S> Frequency bands are a finite resource, so they are regulated by governments and international organizations (see ITU). <S> In most jurisdictions, you are free to listen whatever you want (some restrictions may still apply, depending on where you live you may or may not be allowed to listen to police channels). <S> In order to transmit, in most frequency bands you need an explicit permission and a transmitting device certified according to some standard. <S> That's how mobile networks work - an operator pays a great deal of money to the government in order to use some frequency band and both the operator and the subscribers have to use compliant devices. <S> Starlink will not be any different. <S> Business: the Starlink network will be open only for subscribers. <S> Even if you get a subscription, you will be obligated (probably) to use equipment that only the operator sells or rents. <S> They will rather not allow subscribers to use third-party devices, until some common standard for these types of communication is developed. <S> In short, find a better target for your curiosity.	Surely it is possible to build an antenna for it, as they must have built antennas if they are selling them.
Something like a flip flop I have a pneumatic chuck that needs a pulse that lasts a bit to open and close. Currently I have 2 foot pedals and each works in one direction. I would like to have just one pedal and a relay circuit to make it a flip-flop, but I only know how to do it digitaly. When the pedal is pressed first time the signal should go in one direction as long as the pedal is pressed. When it's released the direction should change, and by pressing it again it should start again. ____ ____ ____FOOTSW ____\| \|______________\| \|____________\| \|_______ ____ SOLN-O ________________________\| \|_________________________ ____ ____SOLN-C ____\| \|________________________________\| \|_______ Basicaly to get both a push button and a latching switch with the same pedal that in hadrware only has push buttons. Is it possible to have this kind of functionality with just relays, and how? I can solve the problem in general, I would just like to know if there's a solution to do it without other elements like timers, capacitors and microcontrollers. <Q> Solution 1: A latching relay. <S> Figure 1. <S> There's an animation of one on [HomoFaciens.de]( https://www.homofaciens.de/technics-base-circuits-relay_en.htm ) <S> that explains the operation very well. <S> (Click the "animate" link in the article to see it in action.) <S> Solution 2. <S> A relay flip-flop. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> A weird relay toggle circuit. <S> How it works: <S> On power-up both relays are off. <S> The chuck solenoid, SOLN, is de-energised. <S> (This is good as it gives you a predictable power-up state.) <S> When FOOTSW is pressed RLY2 is powered on. <S> SOLN will then be powered on. <S> While FOOTSW is pressed RLY1 is shorted out and cannot energise. <S> When FOOTSW is released RLY1 is connected in series with RLY2 and will energise. <S> Both relays now receive half of the supply voltage. <S> (See Note 1 below.) <S> When RLY1 energises contacts RLY1a and RLY1b switch over setting it up for the next FOOTSW operation. <S> On the next press of FOOTSW the bottom of RLY1 is grounded (giving full voltage to RLY1 as long as FOOTSW is held) and RLY2 is shorted out. <S> It will switch off. <S> When FOOTSW is released the circuit resets to the original state. <S> Note 1 <S> : The circuit relies on the relays energising in series and on their own. <S> That means operating on half of supply voltage and full supply voltage. <S> You'll need to rummage through your selection of relays and find a pair that will operate when the coils are connected in series. <S> Figure 3. <S> Timing diagram. <S> ____ ____ ____FOOTSW <S> ____\| <S> \|______________\| <S> \|____________\| <S> \|_______ ______________ _______RLY1 <S> _________\| \|______________________\| <S> ___________________ ____________RLY2 <S> ____\| <S> \|_________________\| <S> From the comments: <S> ... <S> but I need a pulse in different directions for the duration of holding the pedal. <S> Figure 4. <S> Updated timing diagram. <S> ____ ____ ____FOOTSW <S> ____\| <S> \|______________\| <S> \|____________\| <S> \|_______ ____ SOLN-O <S> ________________________\| <S> \|_________________________ <S> ____ ____SOLN-C <S> ____\| <S> \|________________________________\| \|_______ <S> So the latching would chose the direction and second contacts on pedal would just activate it. <S> Correct. <S> See Figure 5. <S> simulate this circuit Figure 5. <S> Schematic updated for timing diagram of Figure 4. <A> The latching relay referred to in 'Solution 1' of the accepted answer is also known as a ratchet relay. <S> It gives an optimal solution with a part count of only one. <S> Here's the schematic using a readily available ratchet relay with 2NO + 2NC contacts. <A> Maybe I have't understood the qestion, but:If you are looking for a Simplest Soft Latching Power Switch Circuit, here you can see detailed review of <S> it's schematic: in this video from EEVBLOG <S> To answer first qestion I can advice you some kind of relay IC: for example ADG736 ADG736 or similar. <S> If you now looking to use it with high current, you can use relay connected to mentioned schematic.	* Have a look at latching relays.
Search for "pinhole" SMT switch (like "reset" switch operated with paperclip) I search for a small switch, momentary on, SMT mounting type, I can put on my STM32 board for entering DFU. I search for a small switch with only a small hole to put a pin or a paperclip. I thing I don't get the right name of this kind of switch to search. Can someone can tell me the name or have an entry point? EDIT: <Q> If you don't have an enclosure (just a PCB) <S> you can't do that, but you could use a reverse-mounted switch with a hole in the PCB instead... <S> This is a pretty uncommon and hard to find component though. <A> Most devices use a right angle toggle switch that is mounted to the PCB. <S> Plastic injection molding from the case cover the switch and PCB from the user. <S> You can make the hole as small as you want. <A> Edit: After clarifying what you mean I am unsure something like that exists. <S> Most users would not randomly do that I would guess.	All the devices I've opened that had a "paperclip pinhole reset" had a hole in the enclosure and a standard pcb switch like this one behind it. What I personally would do in this case is use a 2 pin header instead, which you can short if you need a reset.
What is the impact of unhandled exception on microcontroller peripherals? I always assumed that when an unhandled ARM exception occurs (i.e segfault), the microcontroller just stops executing code. This assumption comes from debugging observation: most MCU I worked with stopped in an assembly routine when segfaulting. As I'm building more safety-critical applications, I want to better understand what is the effect and impact of those errors. Can I assume the peripheral will stay configured as is? Can I assume the GPIOs will stay in their current state? Is the unhandled exception behaviour defined? Tips and resources to learn more about the fundamentals are appreciated. Thanks for your time. <Q> You should handle all exceptions/fault interrupts, so there should not be any "unhandled" ones. <S> Unhandled means that the exception and/or interrupt table has an instance which sends the program counter nowhere/to the wrong address. <S> Please note that in circuit debuggers might be designed to halt upon encountering such problems, to give you a chance to spot one. <S> While the real program will hopefully just reset the MCU and start over - at watchdog reset if not sooner. <S> Can I assume the peripheral will stay configured as is? <S> Can I assume the GPIOs will stay in their current state? <S> No. <S> Hardware peripheral registers are to be regarded as RAM, which in turn is traditionally regarded as less reliable than NVM. <S> In systems where very long up-times are expected, it is custom to initialize those registers on regular basis, by copying down the default settings from flash. <S> Additionally, in safety-critical applications, I usually recommend to implement a vector table where every unhandled interrupt that you shouldn't be getting results in a defensive ISR. <S> This ISR does nothing but shutting off the source of its interrupt and then returns. <S> (This of course depending on the nature of the application: is the safe state a stop or is it "keep running as well as you can".) <S> That is, if you aren't using UART in your application, still write a UART ISR which does nothing but shutting off the UART interrupt, that you shouldn't be getting. <A> It is all up to the software developer. <S> Has nothing to do with the hardware. <S> The hardware has known locations for where handler code should go (be it an exception table or vector table depending on which ARM core you are talking about, or in general for any core). <S> The software developer chooses how to handle each. <S> Worst case is that no consideration is made for handlers and where the entry points are may instead be code, so in that case the software may take a partial reboot path in that it starts executing near the beginning but not at the beginning, but the chip was not in reset so how that code is written and where you enter determines if you crash or not and how. <S> If nothing else the developer should have a handler even if it is an infinite loop which to your question would mean nothing changes, the peripherals stay in their as-was state. <S> And then there is some sense of full support, but that is part of an overall system design, what, from a system perspective do you want the product to do for those exceptions, or within an exception if more info can be determined what to do for each sub category of exception. <S> Is this an rtos <S> do you attempt to abandon an application and keep the system going, etc. <S> Takes a lot of design to determine who caused the fault in an example like that and what keep going means. <S> The long and short of it is <S> the answer is 100% in the developers hands, the hardware has little to do with it. <S> You do the system design and from that the hardware and software design, and within that the handlers. <S> ARM is pretty good with their documentation (should have read that first before asking here) so the various exceptions are defined. <A> Most likely the startup code you are using will have default handlers to exceptions you don't explicitly say you want to handle yourself in your code. <S> These handlers do get executed, and usually by default they just go into while(1); loop or equivalent. <S> Some things are not enabled by default by the CPU, like division by zero won't raise an Usage Fault handler to execute, but it all depends on your compiler startup code if it provides that. <S> Same thing happens if for example you enable UART receiver interrupt, but forget to install the handler, so again the default code catches it, even though it is a perfectly valid interrupt. <S> So yes, always install specific or generic handler to all vectors, even if these are not used. <S> And make sure the code makes known something unexpected happens, because it is a bug anyway. <S> During developement, blink a LED or something so it can be debugged why it happened. <S> In release code, restart the MCU, if it can be done safely. <S> In general, why a fault happens, depends of course what caused it, and how much you can trust a system state after it. <S> If you divide by zero, and have the fault generation enabled for that, it won't of course destroy any GPIO or peripheral state. <S> If there is a fault from reading or writing to a memory address that is non-existent (could include the NULL address), you can't really know what has gone wrong, if a pointer has already become invalid, and you can't really continue safely from that any more.	When debugging, the debugger and the UI do know that now a fault happened, so most likely they stop execution and break at the fault entry point, because that is most convenient for the person debugging, and the code can't continue anyway.
Is is possible to create modular logic gates from transistors for teaching purposes? I'd like to work through some basic logic circuits with some youngsters and wanted to build them up from scratch (transistors), not logic gates if possible. We've done the basics: buffer, NOT, AND, OR, NAND, XOR just using NPN mosfets and resistors. We've even been able to get to D-latches. As an example, the AND gates we're playing with have been build like this: simulate this circuit – Schematic created using CircuitLab The problem is that as the circuits have grown in complexity, with transistor emitters feeding into other transistor bases, I've only been able to get things to work by very carefully balancing resistor values to make sure transistors are triggered when they should be. I would like to be able to build modular logic gates, which are identical and can just be plugged into one another without needing to carefully calculate resistor values. Another restrictions is that we currently have 200 BC547ATA NPN transistors and a big bunch of resistors, so if its achievable without buying any more components, that would be ideal. <Q> If all you have is NPNs and resistors, you'll want to use some form of RTL . <S> Here, the basic logic element is the NOR gate. <S> The one-transistor gate uses fewer transistors (obviously), but the multi-transistor gate is more robust in several ways. <S> And there's a teaching opportunity here, to show how the NOR gate is a "universal" logic element <S> — all other functions can be created by combinations of NOR gates (including the degenerate 1-input NOR gate, or inverter). <S> Even large systems have been built this way. <S> The original Cray-1's logic was entirely implemented using 4- and 5-input ECL NOR gates! <S> One implementation strategy would be to build up individual 3-, 4- or 5-input gates on single-inline modules like these: <S> ( source ) <S> These are easy to plug into a breadboard socket, allowing the students to focus on the logic they're building. <S> If you need a large number of them, have a custom PCB made. <S> And if you're really ambitious, you could plug them into a universal wirewrap card for larger projects and a more permanent implenentation. <A> Building discrete logic with MOSFETs is much easier than bipolar transistors, because the high gate impedance prevents the issue you describe. <S> NMOS <S> circuits are easier than CMOS (which is more commonly used because it has lower power). <A> I think you want to follow TTL logic. <S> I haven't done it myself, but the problem with this simple wiring up of transistors that you are showing is prone to fail when you put more of them together. <S> So here is the TTL way of doing things . <S> This is a NAND gate what I find a good instruction here -- and you might want to sprinkle in a littler diode-resistor logic too -- is that you are entering on the emitters. <S> This keeps the students appreciating the difference between how the logic bits "flow" as opposed to how currents flow, you know, appreciating the fact that a logic 0 state means the output element needs to sink current (that's the same lesson with the diode AND gate). <S> And you don't need dual emitter transistors <S> PS: if you can use my approach (please report back) and you find that you're now in need of more transistors, I will donate another 200 to your project. <S> The totem-pole output stage will also be interesting to experiment with. <S> The whole nine-yards: open collector, normal always on, and three-state. <S> Cool project. <A> The modules below are liable to be a better starting point. <S> Resistor value may need adapting to suit. <S> I've shown BC337-40 because they are cheap and superb. <S> The BC547 is acceptable also. <S> Smaller base resistor values MAY be needed for lower beta transistors but a high ratio between collector and base resistors minimises loading. <S> Note how the AND works. <S> The inputs via diodes stop R1 pulling the base high. <S> So an open input is equivalent to a high input. <S> The inverter input is undefined without an input. <S> If liable to be left floating add and eg 100k to ground AT the input. <S> The buffer has a Vbe voltage drop but should drive an LED to ground with series resistor acceptably. <S> Reduce base resistor value for more LED drive. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> The simplest BJT gate circuit is the DCTL (Direct Coupled Transistor Logic) <S> 2-input <S> NOR. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This is a universal component: any digital system may be reduced to these. <S> To get it to fan out, you need transistors with closely matched Vbe. <S> If your transistors are all from the same manufacturing batch, you're probably OK. <S> Otherwise, you may add a ~100k resistor in series with each base to prevent any input hogging the current from the output it's connected to.	, I have read you can just wire two transistors in parallel, base and collector together and separate emitters.
Why can't I find a 2.2ohm 0.25W 10% resistor I am following the guide to create the Colour Maximite 2 computer (here: https://ufile.io/wv2xo4wg ). It is asking for several resistors. Here are some of them: Carbon Film 2.2Ω 0.25W 10% ResistorCarbon Film 10Ω 0.25W 10% ResistorCarbon Film 1KΩ 0.25W 10% Resistor It is hard to find the exact resistors. Do I have to have the exact one? <Q> The 0.25W is the wattage that the resistor should be able to withstand. <S> E.g. it can do 10mA (Milliampere) at 25V (Volts) because 0,25W = <S> 10mA <S> * 25V (from P = <S> U <S> * I).Generally <S> : More wattage is better. <S> E.g. that allowes for resistors that are labeld "1kΩ" to be between 900Ω and 1100Ω. <S> So a 1kΩ resistor with 5% tolerance (or even less tolerance) will be well within specs. <S> Generally: Less tolerance is better Conclusion: You can use any 2.2Ω / 10Ω / 1kΩ resistor with at least 0.25W and a maximum of 10%. <S> Adittionally you should look out for the formfactor of the resistor. <S> SMD or THT? <S> How big (as in "5mm long")? <S> You need to know these values if you want to solder it to a pcb. <A> Here's what you can do: Visit the web page some component vendor. <S> In this case I used Digikey. <S> Click "passives" -> "resistors" -> "chip resistors", assuming SMD. <S> Enter filters: resistance 2.2 ohm, power 0.25W, tolerance 1%. <S> 63 suggestions, such as this: https://www.digikey.com/product-detail/en/panasonic-electronic-components/ERJ-8RQF2R2V/P2-2RTR-ND/250222 <S> Very easy. <S> The only actual electronics knowledge you need in order to do this, is that 250mW is usually 1210 or larger SMD packages. <S> The above one was a 1206, but traditionally 1206 only goes up to 125mW. <A> After checking out the schematic and gerber(pcb) files provided by the manufacturer. <S> Following resistors are what is causing you the confusion. <S> Through Hole, 0.25W Green Box: 10k, Red Box: 2.2ohm, Cyan Box: <S> 10ohm, Yellow Box: 1k <S> For knowing these resistors relevance, it is good to check the schematic and find out their usage, if particular resistor values are not available, could something else be used or not. <S> Following schematic file shows the colored resistors. <S> Following are the points: <S> 1k resistor: Used for LED, <S> so within 1-2 kohm range, any resistor could be used. <S> Only brightness intensity would decrease. <S> 10k resistor: As they are behaving for pull-up functions, could be used in the range of 10-12 kohm , it will not impact your circuit. <S> 2.2ohm,10ohm resistors are used for 3.3V supply protection. <S> Higher value may not be recommended as they would be limiting the current required. <S> Slightly smaller value than specified could be used but should not go beyond 10%. <S> Commenting about the power rating, it is good to use higher wattage rating resistor than specified, if space is available on PCB. <S> It looks like <S> 0.5W resistors also it can support , if 0.25W resistors are not available.	The 10% is the maximum the tolerance that is allowed.
Is SPI clock always active? sorry it could be a basic question but still, I didn't find the answer on the web is the SPI clock always active or active only when CS is low (during data exchange)? <Q> SPI is not some rigorously defined standard, but more of a de-facto thing. <S> In general a SPI transaction looks like. <S> The master asserts chip select The master clocks through the desired number of data bits. <S> Normally data is clocked out on one edge and clocked in on the other edge avoiding the need for highly precise timing between clock and data. <S> Exactly which edges are used depends on the SPI "mode" The master deassets chip select to end the transaction. <S> If you are designing a slave you should be aware that it's common for one master to drive multiple slaves with separate chip selects. <S> So your slave must ignore any clock transitions that happen while it's chip select is not asserted. <S> This of course in-turn means that a master could run the clock continuously (as long as it took care to observe the required timings between clock transitions and chip-select transitions). <A> The SPI clock is only active while the chip select is low, yes. <S> As correctly stated in the comment, if there's no transmission active, the clock will stay idle even if the chip select is low. <S> The idle state of the clock (high or low) depends on the chosen SPI mode https://www.analog.com/media/en/analog-dialogue/volume-52/number-3/introduction-to-spi-interface.pdf <A> Data is clocked out of MOSI and into MISO when the clock is active, it's only driven when CS is low and there's a transaction happening. <A> While the other answers describe a common implementation, there is no requirement that the controller cease clocking when no controlled device is selected. <S> Controlled devices must ignore the clock when not selected. <S> The controller may continue to drive the clock but it need not do so. <A> No. <S> It's active only when data is being transmitted. <A> The SPI clock may or may not be always active . <S> This is application dependent . <S> Example: <S> http://ww1.microchip.com/downloads/en/devicedoc/39699b.pdf <S> In the PIC24F Family Reference Manual 'SPI' chapter, it explains that the PIC24 series SPI channels support both a mode where the clock toggles only during data transmission, and a mode where the clock is continuously free-running. <S> 1 <S> In Standard Master mode, the system clock is prescaled and then used as the serial clock. <S> The prescaling is based on the settings in the PPRE1:PPRE0 (SPIxCON1<1:0>) and SPRE2:SPRE0 (SPIxCON1<4:2>) bits. <S> The serial clock is output via the SCKx pin to slave devices. <S> Clock pulses are only generated when there is data to be transmitted. <S> SCKx PIN IN FRAMED <S> SPI MODES: <S> When FRMEN = 1 and MSTEN = <S> 1, the SCKx pin becomes an output and the SPIx clock at SCKx becomes a free-running clock. <S> When FRMEN = 1 and MSTEN = 0, the SCKx pin becomes an input. <S> The source clock provided to the SCKx pin is assumed to be a free-running clock. <S> -- <S> 1 <S> In either mode, the PIC can produce the clock itself (Master mode) or it can receive the clock from the peripheral (Slave mode). <S> In either case, the !SSx pin determines whether any action is taken or not.	Normally the SPI clock is only active when the master wants to send/receive data. It is not active always, and it is not active when CS is low.
Implications of having duplicate case statement in verilog/system verilog design module Are there any implications if you have a design module as shown below where you have a duplicate case statement (duplicate case 2'b01 : )? As far as I known the execution is via precedence, so the second duplicate statement doesn't get executed. In terms of fabrication and any other reasons why this may not be ideal or it doesn't matter. module jk_ff (input j, input k, input clk, output logic q); always @(posedge clk) case ({j,k}) 2'b10 : q <= 1; 2'b11 : q <= ~q; 2'b01 : q <= 0; 2'b01 : q <= ~q; default : q <= q; endcase endmodule <Q> The implications will show up when you get to a real project. <S> It will show up when you get into code coverage. <S> It's one of many ways you can create unreachable statements. <S> SystemVerilog has a unique case that flags multiple matching case items during simulation. <S> BTW, It's also a bad idea to have q <= <S> q <S> ; statements—leave that behavior implicit. <S> You try to set a register when you don't think it's being used, and your setting gets overwritten with the previous value. <A> A good simulator should give you an error message, but I wouldn't bet my career on that. <S> A synthesis tool will certainly choke on this, because the different cases must be mutually exclusive. <S> The logic you describe is not synthesizable. <A> In QuartusII this will be synthesized by the following: <S> so the first case 2'b01 : q <= 0; will only be considered. <S> However, it will signal the following warning ( case item expression covers a value already covered by a previous case item ): Warning (10272): <S> Verilog HDL Case Statement warning at question509393.v(7): case item expression covers a value already covered by a previous case item	It gets in the way of debugging or what called backdoor access . Since you are intentionally inferring sequential logic there is no need for the default case...the flip flop will just keep its current value.
Reduce voltage before LDO I am working on a remote gate opener setup. It will be powered via a 12v battery (12-13.5v) and charged via pwm solar charger at up to 14.5v. I would like to use the HT7333 LDO datasheet here It has a max input voltage of 12v and I would like to avoid the magic smoke :) I am looking for a way of reducing the voltage to a safe range so the LDO will work while not draining the 7 Ah battery much when on standby. Background: The small micro controler will sleep for 1000ms then wake, turn on radio and listen for a signal for a few ms then back to sleep so the bulk of the time the current will be < 20 uA however when awake and opening gates its CPU:10 mA + 2xrelays 70~150 mA. My thoughts so far: LDO vs Switching Regulator - LDO as its sleep current is only 4 uA most SR are in the mA range and will empty my battery in winter. Also I have long range radio and the switching of the regulator will cause issues or make my PCB design more complex. Use another LDO with voltage range i need - There are many designed for greater voltages but nothing in that class comes close to 4 uA current while CPU sleeps in fact most are 2-3 mA. Use diodes to reduce voltage before the LDO - This seems the most promising as they will only consume power as it is drawn and not much if anything when not. However from the datasheets I have looked at it appears voltage drops on diodes are not static and change significantly based on the current load, so keeping it in a safe range could be tricky and I don't want so many that I have to extend the PCB (could fit in 4) So I thought I would ask smarter people than I :) thanks in advance for your help. <Q> This isn't meant to be a shopping site, but AP7370 <S> and TPS7A25 should do what you want. <S> In very low duty cycle applications, it's common to use an LDO instead of a buck regulator. <S> You can also use both in parallel, turning on the buck when you need more current. <S> See here for more info on LDO+buck <A> <A> Use a 3.9 V zener diode. <S> At the 4 uA quiescent current it may only drop about 3 V, but that's still safe for you. <S> At high load (10 mA), it will be very close to 3.9 V. <S> A lower V zener won't give you as much margin; a higher voltage one will limit you as the battery discharges. <S> I presume the MCU doesn't (can't) directly drive the relays -- likely you use a transistor (e.g. 2N2222). <S> The coil of the relay doesn't have to connect to the LDO -- it can be connected to the full input.	Zener diode in series with the input could work, if you select one with large enough voltage drop at the rated 4 uA. Or, change the LDO.
What is the use of this diode in the circuit? What is the use of this diode in the circuit? I guess it is used for short circuit protection and turning off the LM2576. <Q> This circuit is for a buck converter. <S> If you look into the working of the buck converter, this diode is essential for the purpose of voltage conversion. <S> What happens is that, during Toff duration, inductor current cannot change instantaneously, hence, a freewheeling path is required when the transistor inside the IC is switched off. <S> That is why this diode is called freewheeling diode. <S> TI LM2576 Datasheet <A> It is explained in the/a datasheet of the LM2576: https://www.onsemi.com/pub/Collateral/LM2576-D.PDF <S> Below the excerpt: <A> That's a switch mode regulator, so that is the catch diode that conducts the current from inductor to output during the time when the switch is off. <S> It is not for short circuit protection or turning off the chip.	In short, the circuit you mention is a buck converter, and the diode is called a catch diode.
Possible to buy Atmega328p with programmed Flash and EEPROM? I've recently upgraded my old project that goes to mass production (a pair of thousand pieces, pretty niche). Previously I used DIP atmega328p, which were manually programmed in China. I sent them a detailed instruction, HEX of Flash and HEX of EEPROM, all was OK, I made the process as quick as I could, given limits. But I've decided to switch to SMD version of the MCU (hardware/spacing related reasons). So it will make assembly process quicker and easier. So I thought why not to optimize MCU programming step as well, now that I have a little more experience and knowledge (the original project was one of my very first projects ever). So if I can also get rid of manual programming, it's a double win. I'm aware of zif sockets for qfn-32 which can be used to program the MCUs manually, it's a backup option. I want pick-and-place machine to solder the MCU onto the board and be done with it. Manual programming means manual soldering of the MCU, which, in case of QFN-32, sounds like a really bad idea. Unfortunately, fierce and fingerbreaking googling didn't yield expected (or desired) results. The best I saw was purchasing Atmegas with a bootloader, which doesn't really solve the problem of course. So is there a way to buy Atmega328p SMD with Flash and EEPROM programmed? The rest of the PCB will be assembled in China, so if there is any way to do that in China too (LCSC?), that would be great, but I just want to find out if/how this works in general, who could I contact? Should I contact, say, LCSC? Or someone else? It's critical to have BOTH FLASH and EEPROM. In case it somehow matters: in my project the Atmega328p runs from the internal 8mhz oscillator <Q> In my experience, this has only economical for gigantic production runs; the cost of this value-add service can be high (I've been quoted a multiple of the price of each controller.) <S> Consider the cost of the service vs. how much it costs for 30-60 seconds of labour on the production line to program the board and go for what makes the most economic sense. <S> It's often easy enough to add some circular copper pads on the board and use a pogo-pin fixture to program the micro as part of the board's assembly process. <S> This allows for much more flexibility - imagine what might happen if you need a firmware update and <S> you have 100000 pre-programmed parts in stock. <S> If you don't have programming pins on your board, you'd need to ship'em all back and get them unpackaged/reprogrammed/repackaged. <S> (Also, every time a part is handled there's a risk of ESD or other damage <S> -yields may be affected) <S> This still allows you to control and update firmware as needed. <S> Once the bootloader is working, it's rare to ever need to update it <S> so it's much safer to pre-program <S> that part of the software. <S> At my workplace we produce many controller boards with multiple microcontrollers and we always go with the post-assembly pin-programming option. <S> Any EEPROM programming is taken care of during functional test. <S> The cost of doing it at our CM is much cheaper than pre-programming; also, because we use the microcontrollers in multiple projects we would need a unique P/N for each micro + firmware combination in order for the CM to not get mixed up - that would be a logistical nightmare. <A> You can order them directly from Microchip <S> pre-programmed <S> and they can ship to your assembly house. <S> It might be more economical to have the distributor do it, or to incorporate it as part of the testing procedure. <S> Sometimes you might want to load a self-test program and then load the user program when it passes. <S> Unless your code is extremely simple and stable and your product is disposable, I would strongly suggest retaining some means of reprogramming the MCUs. <A> Chip preprogramming is a so-called "Value Add" Service offered by many IC distributors. <S> You will receive your programmed chips in a tape, just like you would an unprogrammed SM IC. <S> Arrow.com has IC programming-as-a-service <S> Future Electronics has value-add-programming or, in this case, you can order programmed chips directly from microchip: https://www.microchipdirect.com/programming/price-lookup Note, when you use a preprogrammed chip and a contract manufacturer (someone assembling your boards), Definitely use your own part number, do not use the normal MFG pn. <S> The CM will tend to just order the "normal" part and skip ordering your special preprogrammed variant. <A> Most projects I've seen avoided pre-programmed chips. <S> One picture is worth a thousand words: <S> It doesn't have to be a commercial solution like the Tag Connect in the picture, we are using similar in-house connectors. <S> They take very little PCB area (and zero vertical space), and it's possible to make the mating completely automated for high volumes. <S> Guide <S> holes on the PCB are optional: depending on your product it may be easier to align the connector to parts of the case instead.	If you really must go with pre-programmed, consider having the supplier only program a bootloader into the controller, and do the programming as part of your product test suite.
Generating a delay on the order of microseconds (timer vs HAL_Delay()) So I am using an nRF24 radio module and I probably need a 10us delay after enabling chip select. But the thing is I can't use HAL_Delay() since the least it provides is 1ms delay ( SCK/1000 is hardcoded in the HAL function). I was suggested to not use SysTick timer for too small of a delay anyways and rather go for hardware timers. I read up on Timers and to me, it looks more like they're mainly used for tasks where you need to do a function at a specific interval i.e blinking LED every 4 seconds or something by changing the prescaler and counter period and I did try it out as well -- and it's nonblocking as opposed to a delay. So every time a counter hits the specific counter period, it generates an event/IRQ Am I interpreting it incorrectly? Is there a way around it? I'm using STM32F401RE. <Q> You can start a timer and read the count until it is large enough. <S> Or you can poll the overflow flag when it rolls over. <S> No need for interrupts. <S> You can also sit in a while or for loop for enough counts to reach 10us. <A> For a delay of at least some amount, You could set a flag in the main loop and enable a timer interrupt that resets the flag. <S> The main loop polls the flag in sequence while running other tasks and skips the function that executes the second half while it is set. <S> You may need another flag to indicate that the second half is armed to run. <S> (Or just poll the timer counter directly like what justme said. <S> Much simpler if nothing else needs to happen right when the delay is over). <S> An interrupt of exactly some amount requires running things in the interrupt which is bad practice if not absolutely necessary. <A> Timers are great at making accurate time intervals. <S> Add a pragma to it so that it doesn't change with optimization levels and you are good to go. <A> I was suggested to not use SysTick timer for too small of a delay anyways and rather go for hardware timers. <S> Yes, this is the sensible approach. <S> 10us is a pretty "hard" real-time constraint. <S> I read up on Timers and to me <S> , it looks more like they're mainly used for tasks where you need to do a function at a specific interval Not necessarily. <S> They are supposed to be general enough to be used in a lots of different ways. <S> Periodic interrupts, one-time interrupts, PWM generation, input capture, output generation etc etc. <S> and it's nonblocking as opposed to a delay <S> Indeed, this is often a big advantage over fishy busy-delays. <S> Is there a way around it? <S> Set up the pre-scaler, enable the timer interrupt, start the timer, from the ISR that triggers 10us later simply disable the timer interrupt, then set a application-specific flag. <S> Preferably wrap this whole functionality including the ISR inside some manner of HAL. <S> You could have something like this in the caller code: tim_init(10);...tim_enable(timer_n);while(!tim_done(timer_n)) <S> { /* optionally do other stuff here while you wait <S> */} <A> Or you might use the 74LS121 one-shot delay IC. <A> Here is the code on how ST implements a delay in one of their example projects (also as another user has suggested here): for (volatile uint8_t i=0; i!=0xFF; i++); Change the value of 0xFF to the max count representing the max time delay you would like based on the system clock of your device. <S> Assuming that you use STM32CubeMX to generate your project template, this system clock is the same SYSCLK (MHz) clock under 'Clock Configuration'. <S> Example: if you use a max count of 8000 on a system clock of 8MHz, the delay would be 1ms.	But if you just need a one time delay of some period longer than 10us, when the chip is enabled and don't want to hassle with configuring a timer, then a simple for loop with a NOP might be the most expeditious path.
How to create a smart 3-way positional switch? Looking at this switch as an example: https://www.directindustry.com/prod/tacthill-kynoppe/product-192123-1963276.html I have a similar 3-position switch in my garage that controls a water pump. Central position = off, down = permanently on, up = timer. My goal is to replace this switch with a smart switch that I can incorporate into my home automation. Ideally I'd love a z-wave compatible one, but I'll just be happy with any if it stops me having to visit this switch regularly. Does anyone know if there are such a smart switch, if not, maybe I could build one even with an arduino if need be, but it'd be nice to not have to buy all those parts, particularly the z-wave transmitter. Any guidance at all on how I could achieve this goal would be great. Thanks <Q> A 3-position switch would be implemented in z-wave by independent 2 on/off switches, and the "3-position" function is programmed as a "radio-button" activating one or none (but not both) of the 2 on/off switches. <S> Have a look at this: https://www.homecontrols.com/Qubino-Flush-Relay-QUZMNHBD3 <S> If you do not want to use a smart-hub, then you could try it with a "remote switch". <S> There are models that provide one remote to control several outlets, e.g. for outdoor lighting. <S> https://www.amazon.com/Etekcity-Weatherproof-Electrical-Unlimited-Connections/dp/B078YL37MM?ref_=fsclp_pl_dp_12 <S> If you are somewhat handy <S> (and I think you are, if you are thinking about the Arduino), you can wire the double-relay unit so that it overrides or replaces your wall switch. <S> Since you are using an independent timer as one connection option, I don't think you can use these remote receivers straight out of the box. <S> You have to open up the case and modify the wiring at the switching poles of the relay. <S> Basically you reverse wire <S> the switch: the relays have to select between two input plugs (one powered permanently, one powered through the timer), and supply one socket. <S> Simply reversing the input plug and output socket of the entire unit would not work because the unit derives its own power from the plug side. <S> I can update this answer with a sketch if you need one. <A> simulate this circuit – Schematic created using CircuitLab <S> The OR arrangement requires you choose XOR or exclusive OR in the remote operation as one over-rides the other when active. <S> The AND series arrangement gives you exclusive OR using a SPDT. <S> There are over 3000 interoperable Z-Wave products throughout the world. <S> But perhaps you can find a Up/Stop/Dn <S> Blind Z wave Remote that has <S> the single pole or double pole function your prefer to drive your AC power Relays using either DC coil power (more efficient) oR AC coil power. <A> Here's how, using 2 DPDT relays <S> K1 & K2 (could be a 2 channel RF relay board). <S> The above network would replace the toggle switch. <S> The bottom terminal would get connected to the point where the centre terminal of the toggle switch was and the other two where the end terminals were. <S> The logic would be 'K1 on' or 'K2 on' or 'both off'.	If it's ok to switch one or the other mode, and no hazard exists if both are on, as in your case (no harm if timer is on and switch is also on), there are two-relay modules that could fit inside the electrical box of your current switch.
What would happen to the magnetic fields of a solenoid if it were wrapped around a hollow ferromagnetic core? I'm well aware that a solenoid wrapped around a solid ferromagnetic core produces a stronger magnetic field, but what about a hollow ferromagnetic core? Would the fields farther from the center near the hollow ferromagnetic core become more powerful, or would the middle of the field become more powerful? If something else happens do say so! I'm seeking to understand the fields produced by such a setup. <Q> The magnetic field would be concentrated in the areas occupied by the mass of the ferromagnetic core. <S> The empty centre of the core would be largely devoid of magnetic lines of flux. <S> This usually means that there will be a higher level of flux density in the material mass of the core compared to a solid core. <S> However, at the ends of the core (where air dominates), the flux lines would rapidly become very similar the those lines of a solid ferromagnetic core. <A> I'm adding this in an answer just so i can show the graphic - the results seem to agree with how Andy explains it in his answer and subsequent comments (Charles answer as well). <S> I did this model in the free 2-d FEMM software. <S> I started with the .fem file from this neat page and watched this video to learn how to use the simple program. <S> The clear rectangle is one side of the current carrying coil that wraps around the inner hollow cylinder. <S> The inner hollow cylinder is M-19 steel. <S> Inside the hollow cylinder (far left) is air. <S> If you take the mirror image of this and place it on the left you can imagine a 2-d view of the full thing. <S> The magnetic field lines really pack into the low reluctance cylinder, much less in the air in the hollow. <S> My first look at FEMM, seems like great tool. <S> Also, all the blue is air. <S> EDIT : <S> In the image below i created a mirror image and stuck them together. <S> If you imagine a vertical field line in the center i think you would have a fair understanding of the field (at least this 2-d slice). <A> Iron provides the easiest path for magnetic flux. <S> The flux will be concentrated in that path the extent that is doesn't need to go too far in the air to get to that path. <S> The magnetic field will try to move whatever iron is nearby to become a part of that path. <S> Without the iron, the magnetic field would tend to go through the center and around the outside to the other end. <S> Whatever you put in that path will cause a stronger force to close the part that you leave open. <S> The strongest force would result from a layer of iron in the inside, connecting on one end to another layer on the outside of the coil. <S> That would give the maximum force to pull something to close the open end. <A> If the core of the solenoid is not homogenous, and composed of a hollow cylinder, as in your case, then the H field is still homogenous all around, since it only depends on current that generates it (current density specifically, but that gets into the details of Maxwell's equations). <S> The H field depends on current only, and not material. <S> The B field depends on H and the permeability plus any magnetization, and so where the permeability varies so does the B field. <S> The flux is the integral of the B field at a surface. <S> By summing the B field across the inner of the solenoid we can calculate the total flux inside the solenoid. <S> The flux is increased if measured through materials with higher permeability. <S> The flux determines the inductance L of a coil inductor, even if it is an air inductor, or a straight wire. <S> That's why the "flux is concentrated" in a metal or ferro-metallic core. <S> The B field, -and not H- is the bases for flux, and it is much stronger there, and relative to the surrounding air it appears "concentrated". <S> Where the B field is stronger, the "B lines" are drawn closer together. <S> Each line represents the direction of the B field, and their proximity represents the strength. <S> The total flux in a solenoid core that is part ferromagnetic and part air (as in the cross section of your cylinder) will be less than if it is all-ferromagnetic. <S> But the magnetic H field and the B field are not zero in the air inside the cylinder, and the H field is homogenous.	The magnetic field inside a long solenoid of homogenous material (all air, all ferro rod etc...) is homogenous: the field strength (density of field lines) is the same near/at the center of the core as it is near the edge of the core where the windings are.
Protecting reed switches operating 240V AC relay from welding shut I'd be grateful for some advice about protecting the reed switches in this pump-down setup I am installing. I naïvely thought that the relay alone, which is rated as drawing only 5.3mA, would be adequate to protect the reed switch contacts from welding shut. I now know that this is not the case and that some sort of additional protection is required. Research led to this post that recommended using a TVS (specifically a P6KE300CA) across the coil. However, other posts put snubber circuitry across the switches themselves and I'm not confident about what configuration will produce the most reliable results. Advice and suggestions would be welcome. To give some more details of the application, it is a condensate pump for a gas-fired ducted heating system. The unit was installed with a pump used for split air conditioners. With a very small reservoir, it ran almost continually and was been very unreliable as, I believe, is often the case with these things. There is plenty of space for a bigger tank to store the litre or two of water produced in day, so I sought to design a latching circuit that would fully pump out the tank only when full and operate just once or twice a day. The float switches used are on Amazon and were sold as having a contact rating of 1A at 220V AC so I didn't think powering a relay would be a problem. <Q> There is an error in the schematic that requires correction. <S> The lower and upper float switches, being simultaneously on, would carry the high pump motor current till the relay switches on, and get damaged. <S> It could also be a VDR across it or across the relay coil. <A> Figure 1. <S> The original circuit. <S> When both switches turn on the pump is powered directly through the reed switches. <S> As @vu2nan has pointed out the likely problem is switch on rather than switch off. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> The solution is to separate the control circuit from the "power" circuit. <S> For this you need a 2-pole relay. <A> As others have said trhe problem is that the reed switches are passing the motor start current. <S> if you can use a dual-throw reed switch in the upper position this can be prevented. <S> AC relays are slow enough that the interruption while the upper switch changes will not cause the relay to disconnect.	In an AC circuit, reed switch protection could be a resistor and capacitor (in series) wired across it.
Can we drive an led from output port of TCA6424ARGJR IC? I am using TCA6424ARGJR IC so from its output port 24 I need to keep an led, which glow when there is high. Can we drive an led from output port of TCA6424ARGJR IC? I have connected as shown in IMG. <Q> Yes, you can light a LED, but not at 30mA, because it exceeds the recommended value. <A> Each (P) Port can source 25mA from Vcc, which may not be enough for the 30mA that you want in the diagram. <S> Make sure you size the resistor <S> so no more that 25mA (maybe a little lower also, like 24mA) so that the ratings in the datasheet are not violated. <S> Also make sure that the sum of the current from all the P ports does not exceed 160mA <S> Source: <S> https://www.ti.com/lit/gpn/TCA6424A <A> You can drive up to 160mA as a high side LED driver with Current limiting R. <S> But that doesn't mean you should. <S> Nor does it mean you should use 30mA the common 5mm Absolute max for an indicator. <S> With LED indicators up to 10kmcd, all you need is about 2 mA. 30mA would be blinding. <S> Ignore <S> the Port Out max current. <S> That is just for Logic levels. <S> 0.8V/25mA <S> max indicates the minimum RdsOn is 32 Ohms for the driver.	But if you chose an old 100 mcd RED LED then I can see 20mA is OK, but 30mA is not recommended unless you are just playing around as the junction will be burning hot then.
Using thermistor vs. soft starter for motor What is the advantage that complex soft starters give over NTC thermistors if one is only trying to get rid of inrush current spikes? <Q> A VFD is a complex soft starter for Motors that limits acceleration and current and thus RPM rate of change with source frequency with constant V/F with high efficiency . <S> A huge NTC must get hot to operate and thus start at reduced current. <S> The amount of power determines the best method. <S> There are many more examples. <A> The advantage a "complex soft starter" provides like the one describe here <S> Does Thyristor (SCR ) works for Inrush current protection? <S> is when a unit is subjected to repeated inrush events in a short period of time. <S> If you were to use an NTC and there was a blackout on the supply, you are restricted to the thermal time constant of the NTC to realise a repeatable management of the inrush current. <A> NTC : <S> Pros: cheap, only one component in the BOM, a few tests to find the correct value for you application. <S> Cons: Waste of energy because of small residual resistance <S> when hot, Get hot. <S> In a nutshell resistor is always "there". <S> Should be cold to be efficient, so it will not always have the same effect on inrush currents depending on its temperature making it not a robust solution. <S> Not ideal for high power solution. <S> Inrush current limiter chip : <S> Pros: Behave always the same way thanks to current limitation (through shunt resistor or using the Rds,on of the MOSFET) <S> so it's a robust way to do it. <S> The chip complexity often allow more functions like timed overcurrent detection, UVLO, short circuit protection (act like a very fast fuse. <S> Example: the LTC7000 can cut power in 70ns if current is above a threshold). <S> Cons <S> : Chip could be expensive (several dollars), a lot of components in the BOM, some chips are complex to understand and thus you will have to fully test it for industrial purpose. <S> My opinion: <S> all the functions offered by those chips make me prefer them compared to NTC. <S> But it is up to you to design a solution without the need of reliability and high level of safety (for a personal project or a low power product for example).	An NTC is good if the likelihood of "fast" restart or intermittent power outage is low.
Why does my ceiling fan spin the wrong way? My ceiling fan blows are at the ceiling (uselessly). At first, I wondered whether I could fix it by maybe removing it and switching some kind of polarity. But is there a polarity? Aren't the residential power wires live and ground? How does the fan know which way to spin? <Q> The fan would have a capacitor to put phase shifted voltage to additional coil in the motor. <S> Residential power depends on where you live. <S> Some have single phase only, some have two phase, and some have three phase. <S> Nevertheless, a ceiling fan would usually have single phase and neutral return, and protective earth ground. <A> "Uselessly" is wrong. <S> If you want to circulate air, breaking up the hot layer near the ceiling, then either way will work. <S> If you want to sit underneath the fan and feel a breeze, then yes, it must direct air downwards. <S> But if you want the cooling, but don't want to blow papers off your desk, it needs to point upwards. <S> (When pointed upwards, air descends all around the walls of the room.) <S> Look up the part number? <S> These fans almost always have wiring for a 2-direction switch. <S> Sometimes there's a pull-chain for doing this. <S> With a very old fan, the pull-chain may be broken off. <S> That, or the 2-direction wires were never hooked up, and are buried in a junction-box in the ceiling (attic?) <S> directly above the fan. <A> A ceiling fan has a split-phase motor with a capacitor. <S> The required direction of rotation is obtained by interchanging the 'Line' connection as shown.	Sometimes there is a switch to change direction and it changes which way the motor runs, by changing which way the phase is shifted.
Why do layout footprints for Crystals often define via keepouts? For example, one of the EagleCAD standard footprints for an HC-49U package through-hole crystal (mounted horizontally, so that length of the "can" lies against the PCB) looks like this: The green "shaded" area is the vRestrict area, which restricts the placement of vias in the designated area for design-rule checking (DRC). Why is this an important consideration? A follow-on question that is begging to be asked in follow up, is why not also a bRestrict and tRestrict to prevent any copper (e.g. a ground plane or a signal), and is that something I should do anyway? <Q> The green "shaded" area is the vRestrict area, which restricts theplacement of vias in the designated area for design-rule checking(DRC). <S> Why is this an important consideration? <S> Because the metallic (and therefore conducting) crystal can is laid flat and could make contact to exposed PCB traces that are not covered with solder resist. <S> Typically, vias are not covered with solder resist. <S> However, it's not a good idea to put any copper traces (covered in solder resist or not) in the area directly under the metal can. <S> You can use an insulating pad of course <S> then it's OK. <A> The metallic crystal casing might short vias. <S> tRestrict is the only way of preventing tracks from being laid under the crystal, unfortunately it also prevents copper pour, even though it's considered a good practice to lay the copper. <S> In theory it adds capacitance, but that shouldn't have an effect for a mechanical resonator and it doesn't load the resonator circuit too much. <S> Having bRestrict sounds overcautious to me. <A> EMC: <S> the signal from the crystal to the Xtal_in pin is often relatively weak; if there are any strongly-driven digital traces near/under the crystal, they might interfere with this signal and cause jitter / frequency shift. <S> short circuits: While vias are often covered with solder resist ("tented"), sometimes the tenting is incomplete (partially exposing bare copper); sometimes boards are manufactured "without tenting", completely exposing the copper at the via. <S> The metal case of the crystal would short out those signals. <S> Most design guides for oscillatorsrecommend a solid ground pour underneath the oscillator as part of a general ground pour over the entire PCB, for EMI/EMC reasons: <S> AVR186: <S> Best Practices for the PCB Layout ofOscillators TPS65950/30/20 32-kHz Oscillator Schematic and PCBLayout Guide <S> Oscillator <S> So we don't want any signal anywhere under/near the crystal or the <S> Xtal_out / Xtal_in lines connecting it to the IC.But we do want a solid ground pour under it. <S> we would put a pad of ground under the crystal witha couple of through holes (or a bare spot) and either wrap a wire overthe can soldered down or solder the can directly to the PCB.-- mtripoli	Some of the things a designer might think about: EMI: the signal from the Xtal_out pin to the crystal has a strong, high-frequency signal; if there are any other traces near/under the crystal, they could pick up some of this signal, which might directly interfere with their function -- and even if they are digital lines and it doesn't interfere with their function, they might re-radiate that signal and cause the device to fail radio EMI emissions test.
12 v DC relay coil shows resistance in both ways 86 and 85 or 85 and 86. Why? My relay coil 12v measure 100 ohm of resistance between pin 86 and 85 or between 85 and 86. A diode IN4001 is added in parallel to the coil. However, after adding the diode, the coil has the same reading 100 ohm when measured between 86 and 85, and again vice versa, 85 and 86. Why? Is not the relay supposed to show resistance only in one direction???? Why the relay coil resistance still stays at 100 ohm after adding a diode in parallel with the coil?? <Q> A diode 1N4001 is added in parallel to the coil. <S> Your meter (on a low ohm scale) might inject 1 mA into the "thing" it connects to in order to determine resistance. <S> 100 ohms (the relay coil) and 1 mA produces a potential difference of 0.1 volts and this is way lower than what any normal silicon diode needs to start any meaningful conduction. <S> So the diode appears open circuit. <S> Is not the relay supposed to show resistance only in one direction? <S> Edit - diode current at low voltages <S> The above is for a 1N4148 diode (rather than a 1N4001) because the data is available and therefore it can be extrapolated (as I have done in red). <S> So, 0.1 volts divided by 40 nA = 2.5 MΩ and, compared to 100 ohms, it conducts 0.004 % of the current. <A> The problem would be unique to digital multimeters, in which the voltage applied by the prods for resistance measurement would be quite low (less than 0.6V) for the silicon diode to conduct. <S> With an analogue multimeter the applied voltage would be high enough (equal to multimeter battery voltage) and the silicon diode would conduct. <S> Measurements across a 1N4001 in parallel with a 100Ω resistor, made with an anlogue multimeter (battery voltage 3V), gave a reading of 100Ω in one direction and 18Ω in the other. <A> A DC relay coil requires DC current, but that doesn't mean that it passes current in only one direction. <S> After all, electrically, it's just a long thin wire. <S> In fact, until you add the diode, the relay works equally well with either terminal positive with respect to the other. <S> Relays with AC coils (still just wire) are constructed differently in the magnetic circuit. <S> First of all, they depend on reactance as well as resistance to limit the current. <S> And they often have shaded poles in order to avoid any buzz or chatter from the power line frequency.	If the diode were wired in series then this would be the case but, you have the diode wired in parallel.
Energy Transformation in Flyback Transformers I am trying to understand how is energy transferred as a current to secondary of flyback transformer. First due to the dot convention and secondary side diode, current will not flow in secondary and energy is going to be stored in the core. But after opening the switch, how polarity changes? I mean the current in the primary wont change suddenly and will try to flow in the same direction, as well as the flux in the core. The created flux is going to force secondary current to flow in the same direction as it was closed. What causes the change in the secondary polarity? Can someone explain it? <Q> and energy is going to be stored in the core actually, in the magnetic field. <S> When the switch closes, the secondary diode will be reverse biased (remember the dots - winding polarities) and does not let the current flow. <S> But after opening the switch, how polarity changes? <S> Remember the basic inductor behavior: Inductors always generate an emf to counteract any change on the applied voltage. <S> For example, if the applied voltage is suddenly removed then the inductor will generate a very high voltage with reversed polarity across its terminals. <S> Let's put this behavior in a flyback converter: <S> When the switch opens, the primary inductor will not let the current stop suddenly and will generate a reversed and very high voltage across its terminals. <S> So the voltage at the switch side of the inductor will be greater than the voltage at the supply side of the inductor. <S> That's how the polarity changes. <S> The stored energy will be partly (or completely, hence the name DCM or CCM) transferred to the secondary. <S> This will induce a voltage across the secondary winding. <S> If the stored energy will not be transferred to the secondary then the core will never reset and will saturate eventually (Remember the B-H Loop). <A> I mean the current in the primary wont change suddenly and will try toflow in the same direction, as well as the flux in the core. <S> That is correct except the current will now be flowing in the secondary because, in an ideal flyback transformer there is no path for current to flow in the primary. <S> Maybe consider simplifying the flyback transformer to a basic inductor-switch circuit like this: <S> - Notice that I've re-positioned the secondary circuit in the left image to be inverted to help make the transition easier to the simplified inductor-switch model (on the right). <S> S1 is the MOSFET <S> S2 is the diode <S> The created flux is going to force secondary current to flow in thesame direction as it was closed. <S> That's basically where your error of thinking is. <S> Just think about a normal transformer with two identically wound and 100% coupled windings: <S> - <S> \$I_P\$ is naturally in the opposite direction to \$I_S\$ . <S> If it were the other way round then the output voltage would have to be inverted and, of course, that is not what the dots associated with each winding is about. <S> Picture from here . <S> What causes the change in the secondary polarity? <S> Can someone explainit? <S> If there is no voltage polarity change then there has to be a polarity change in current . <S> If there is no current polarity change then there has to be a change in polarity for voltage . <A> Understanding transformers - as well the usually used ones as the ones in flyback circuits - can happen after a person understands that induction happens in the magnetic field around the windings. <S> High permeability transformer core helps to keep the magnetic field inside a certain area and direct the magnetic flux through the windings. <S> The induction law states that as soon as for any reason a magnetic field changes there exists a circular electric field around the magnetic flux lines. <S> In transformers the induced electric field has the same direction as the wire in the windings, so having plenty of turns collects a substantial voltage. <S> The induction law (in the vector field differential equation form) states also that the generated electric field has a polarity which tries to cause such current in the windings that the change of the magnetic field is reduced. <S> Thus if you break the current in the primary of a flyback transformer both windings generate a voltage. <S> In the secondary there's a diode which allows the secondary output current. <S> That current has direction and strength which generates at first just the same magnetic field in the core that you just tried to break by stopping the primary current. <S> The current decays gradually as the magnetic field energy decreases; it goes to the output. <S> When you break the primary current the voltage over both windings jumps as high as needed to make the current possible. <S> The current bulldozes its way through the insulations if needed. <S> It's capacitance is so high that the voltage rises only very little, say less than 100 millivolts with one induced pulse. <S> Induction lifts the voltage in the winding just as high as it's needed to charge the capacitor.	In flyback power supplies the diode in the secondary allows the current from the secondary charge the output capacitor. Changing the polarities on the windings is the only thing that the flyback transformer can do because it has to transfer the stored energy.
12V 20A: What to do for high heat I have 4 DC motors. they each require about 12V 9A DC (108W each), and have a potential draw of up to 12V 20A each (240W each) stall current. I am using 22 AWG insulated wire. The length of the wire, per motor terminal is about 1.2 ft. This is all on a small(ish) sized robot. I need to know what kind of wire I need (per motor terminal) in order to prevent wire insulation melting or fires, with wattage like that in mind <Q> You must be sure to select an adequate wire size for the current considering the temperature in the area where the wires are located. <S> Also make sure that the wires are adequately protected from mechanical damage. <S> The connections must be tight and connected from corrosion. <S> There must be protection from excess current due to motor overloading or short circuit. <S> You should probably use 14 AWG or larger depending on the insulation. <S> For a vehicle, you should probably use finely stranded wire. <S> If you use a high temperature insulation, you can use a little smaller wire, but you may need to be concerned about the mechanical strength of the wire and insulation. <S> I would be inclined to use 16 AWG with 27/30 stranding in silicone insulation covered with fiberglas braid. <S> Or perhaps this: https://www.amazon.com/BNTECHGO-Silicone-Flexible-Strands-Stranded/dp/B00TG1TRL2/ref=sr_1_12?dchild=1&keywords=silicone+insulated+wire&qid=1594775032&sr=8-12 <S> That offers the advantage of leads that are easier to work with, lighter weight and more flexible. <S> The disadvantage is that any material that comes in contact with the wire must be capable of withstanding that same temperature without damage. <S> Silicon insulation is rated 150C to 200C. SEW seems designate 200C silicon with a braided fiberglass outer protecting layer to abrasion resistance. <S> Teflon insulation is rated 200C to 250C. TGGT seems to designate 250C insulation with a fiberglass cover. <S> SRMC seems to designate 150C silicon insulation with no outer cover. <S> Silicone insulation with no fiberglas cover would have better abrasion protection. <S> The Amazon product linked above has 200C silicon insulation with no outer cover. <S> It has very fine stranding. <S> The combination of very fine stranding and no outer cover makes it very flexible. <S> I mentioned it because it seems to be easily available in sort lengths. <S> I would think that the motor controllers could not withstand the stall current for more that a minute or so. <S> You might want to use 18 AWG wire for mechanical strength, but you will probably not get it above 150C. You probably don't need to select the wire based on heat. <S> You will need to decide if you need the best abrasion resistance and how difficult it is to get the total length wire that you want to buy. <A> The NEC requires at least #3 AWG wire for 80 amps. <S> You may need larger wires depending on the length of cable and allowable voltage loss. <S> With 80 amps in #3 cable, you will have a 3% voltage loss in a 12 ft two conductor cable on a 12 volt circuit. <S> Edit <S> If each motor only draws 36 amp, you can use #10 AWG wire to each motor. <S> #12 or even #14 may be adequate if the wires are in open air, or for low duty cycle - check the wire temperature while the motors are operating. <A> You might be able to get away (temperature wise) with as small as AWG20 with PTFE insulation if you are willing to allow the wire temperature to go to a rather toasty 200°C. <S> That's for 20A or so, assuming no current path is shared between motors. <S> Voltage drop will be large (about 1 volt for a 2.5 foot round trip), you'll lose some torque, so use a higher gauge if possible, and don't bundle even two of the wires together if they're on the smaller end of the sizes. <S> If you're using low performance wire with PVC etc. <S> insulation rated at lower temperatures as suggested above, the insulation will drip off the wire potentially causing shorts, and nasty chlorine gas may be emitted.	Make sure that the wire connections are enclosed or otherwise protected from coming in contact with anything that would cause a short circuit. Using high temperature insulation allows a smaller wire gauge to be used. For a 30 ft cable run, you'd need #00 wire for a 3% voltage loss.
How to protect Relay's NC contact tip from wearing out? I made a rev limiter by using a 5V SPDT relay on motorcycle between the CDI and ignition coil. The relay works fine. However, the NC contact tip wears out very fast. I added a 1N4007 at NC and COM to protect the contact pin, but the voltage still passing through from NC to COM although the COM already switch to NO pin. So this method is not working. Am thinking to add an RC snubber to the relay, but wondering whether it will have the same problem as above. Should I replace the mechanical relay with an SSR instead? This is the relay that I'm using now: Edit: This is my current schematic. <Q> Where you add the diode matters and it should normally be in parallel with the inductance that's producing the arc in the relay, not the NC-COM relay contact. <S> Even then, a 1N4007 is too slow. <S> Silver cadmium oxide contacts help and are more resistant to erosion. <S> You might want to look at a thyristor which is unable to interrupt the spark while it is in progress (so none of the harshness invovled with interrupting something mid-arc), but will block any future sparks. <S> This is because a thyristor actually latches on when triggered and can't actually interrupt current itself. <S> However, if the current falls below a certain level (ideally zero) due to the external circuit while the trigger is removed then thyristor will turn off and block future current from flowing <A> The CDI module puts out pulsed DC, in the few hundred volts range, but nowherenear 10A; if you can switch the INPUT power to that CDI module instead of theoutput, you'd be switching higher current (maybe several amps) butlower voltages. <S> The relay is rated for that higher current at low voltages. <S> Much of the contact 'wear' is in the high voltage causing a brief, hot, arc. <A> Relay damage may be avoided by using it to switch a signal instead of power. <S> The NC contact of the Rev Limiter relay is to be used to disconnect the CDI thyristor gate, when the RPM limit is crossed.	The problem is since your circuit is designed to make a spark that can jump across an air gap so adding any snubber, either diode or RC, in the proper place to suppress that spark across your relay could also suppress same ignition spark.
Why does this datasheet specify 12 100nF capacitors? I am learning to make my first PCB with an MCU and looking at the Power Scheme, it asks for 12 x 100nF capacitors. My question is why have 12 of them rather than 1 x 1uF, and 2 x 100nF (I see that there is limited value selection, but these are popular values too)? This would save space, and manufacturing costs. I also assume these are in parallel to add up the capacitance, is this a wrong assumption?Thank you for your help. EDIT:Thank you for all the good advice, due to request, here is the datasheet: STM32F446xC Datasheet <Q> Because the chip has 12 (maybe depending on the package) pairs of Vdd and Vss pins. <S> Each pair should get its own decoupling cap. <A> Each of your decoupling capacitors exists to provide current when the load (your MCU) switches and needs more current very quickly. <S> The power supply takes time to respond to this transient demand and increase its current output to stop the supply rail falling and the connections to the power supply have an impedance that also slows its response to the load. <S> So you have decoupling capacitors that supply these fast transient currents as needed when MCU switching occurs and charge back up when during steady load times. <S> A capacitor has a capacitance value and, among other characteristics, an Equivalent Series Resistance (ESR) value which is specified at some frequency of operation. <S> The ESR can be seen as a resistance in series with the capacitor. <S> Using 12 x 100 nF capacitors in parallel and placed closely together instead of <S> 1 x '1200 nF' capacitor would give the same capacitance but with a lower ESR because all the 12 capacitor ESRs are in parallel with each other. <S> It doesn't give an overall 12th of a single capacitor's ESR, because the interconnecting PCB tracks have an impedance, but it's much lower than the ESR of a single capacitor. <S> Here, your MCU recommends placing a capacitor near to each of many (12?) <S> MCU supply pins. <S> Each supply pin now sees a nearby capacitor with a lower ESR, and therefore a faster transient current response, than if it was one larger capacitor shared between them. <A> At high frequencies, the inductance of wires in particular becomes harder to neglect, so you cannot do that. <A> Beware though of the parasitic R and L of capacitors. <S> Here we have 12 X 100n in parallel with 1X 4.7uIf <S> the capacitors are ideal there would be no need for the 12 X 100n <S> In fact the 4.7uF will have significant self inductance and so have high impedance above a few MHz, The 100nF are there to provide low impedance up to higher frequencies. <S> A 1200nF will be worse at those frequencies. <S> It's not uncommon to put a 10nF in parallel with a 100nF to provide good decoupling up to the GHz range. <S> 10nF caps are available with better dielectrics so they have lower parasitics, but you can't fit much capacitance in the same size package. <S> Also as others have said keep the track to the power pin and ground short or the cap has no effect. <S> Alway check the datasheets; https://www.farnell.com/datasheets/2167237.pdf Building circuits is always a bit harder than just the schematic, but it does make for an interesting job.	To add to the good answers already here: you can replace multiple capacitors in parallel by one with the sum of their capacitance only if you can neglect the resistance and inductance of the wires that connect them.
Is this cable meant for metal case grounding? I opened a heat pump cover and saw this little cable screwed onto the case but it appears to still be insulated from case and is supposed to be used by an electrician. Is this meant to ground the metal case of the machine? Since there is no other external pin on the case, is this internal wire meant to be in contact with the case itself and in case of a malfunction that would put case under current, it would provide the current a path to the ground by going (I assume) to the same place the power supply cable ground cable connects to internally? Or could it potentially be connected to the case on the other side (not visible on image) and is expected from electrician to pull this cable out through the case and connect it to some external grounding wiring? Or could it also be used just for bonding with other conductive items in the area? I understand this requires professional. I am just curious to know if this is a rough explanation of what it does. The instruction manual provides no explanation and both seller and manufacturer failed to provide any response so far. <Q> If it has two wires on the end that plugs into a PCB, then it is a thermistor. <S> By the looks of it, it is there to measure the temperature of the air (as it isn't thermally connected to the case). <S> A lot of thermistors I use look like this in my designs to measure ambient temperature. <A> These devices can have resistance varying from 1K Ohms to 100K Ohms. <S> I remember finding something similar <S> and it was as discrete thermistor, which has uni curve with bare leads and epoxy on the top. <S> The one that I found was this honeywell 192 series thermistor ( https://sensing.honeywell.com/sensors/thermistors/192-series ) <S> How to Test Measure the initial resistance Bring the tip of the sensor close to the heating element and measure the resistance and compare to the initial measurement, the change observer must be high and must be sensitive to touch <A> No, the wire in that picture is not used for grounding. <S> There are two wires coming into the sensor which has a protective coating, it looks like it is dipped to black epoxy resin. <S> The sensor is mounted with the plastic part to the case with a screw, to place the sensor element to monitor temperature from the correct place inside the unit. <S> So leave temperature sensor where it originally was measuring the temperature.	That is a temperature sensor. There are high chances that this is a thermistor very similar to Honeywell 192-series discrete thermistor. Measure the resistance, and it should come in at a value that is 10KΩ's (could be more, but this is the usual value.)
How can these two motors produce the same torque if their volt/amp ratings are different and their watts/RPM are the same My rudimentary understanding of a DC brushed motor is speed is proportional to voltage and torque is proportional current. Therefore a 36 volt 500 watt motor should have a higher RPM rating and a lower torque rating than a 24 volt 500 watt motor which should have more torque but less RPMs. However, the datasheet suggests that even though the current ratings vary between the two voltages (as expected), the speed and the torque are the same, implying that torque is a function of watts rather than current. Question 1. How can the voltage / current be different and the speed / torque be the same? Question 2 Practically speaking, will the 36 volt motor be preferable to purchase since I would get the same torque as the 24V motor but less current (and the heat this produces) going through my circuit? I know I have misunderstood something. All help appreciated. <Q> P=IV and that gives 640W for both. <S> If the designs are different ie number of windings etc then the voltage can be changed for a different current while delivering the same power at a given speed. <S> Now power = <S> Force <S> * distance <S> * time <S> So, the torque is related to power. <S> For a long explanation, see: http://www.epi-eng.com/piston_engine_technology/power_and_torque.htm especially at the bottom of the page. <A> My rudimentary understanding of a DC brushed motor is <S> speed is proportional to voltage and torque is proportional current... <S> correct as far as it goes for any particular motor <S> Therefore a 36 volt 500 watt motor should have a higher RPM rating and a lower torque rating than a 24 volt 500 watt motor which should have more torque but less RPMs. <S> Any given motor has a constant ratio of speed to voltage. <S> A different motor may have a different ratio. <S> That ratio is dependent on how the motor is built. <S> It depends on armature turns, armature pole area, and magnetic field. <S> Typically motors of a given power will be offered in different voltages, where the difference between them is the number of turns on the armature, and the wire gauge, because some users will find one voltage more convenient than another. <S> I will wager that these motors are exactly the same mechanically, and the 36 V one has 50% more turns of thinner wire than the 24 V one. <A> These two are actually the same physical motor - but the 36V one is wound with more turns of thinner wire. <S> Therefore it requires more voltage to reach the same speed, and produces the same torque with less current, by generating the same flux (Ampere-turns) with more turns and fewer amps. <S> You can either think of the higher voltage motor as having a built-in step down transformer to match the lower voltage one, or having a different (electrical) gearing to the shaft.	You'll notice that the two motors you've highlighted are rated to produce exactly the same torque at the same speed, for the same electrical power input.
What polyfuses protect USB ports in a typical modern PC? I would like to know what is the typical rating of a polyfuse used on modern PCs (desktop or laptop) with USB 2.0 and 3.0. TE Connectivity apparently recommended in 2011 that a single USB 2.0 port (non-ganged) be protected by a polyfuse with hold current 750 mA and trip current 1.5 A, and double that for USB 3.0 that is not charging-enabled, and higher still for a charging-enabled port. However, I have been unable to find the actual specification of the USB polyfuse ratings for desktop PCs and personal laptops. Many technical manuals I could find for various PC models mention that they do have over-current protection, but for whatever reason invariably neglect to give any further detail. (I recall some of them explicitly stated "polyfuse" but do not have them at hand right now.) I can find Littelfuse specifications for the above-recommended polyfuses, and the Littelfuse 2016 selection guide recommends using one of its 0805L, 1206L, 1210L, 1812L or USBR polyfuses for USB, and all of these have hold current at most 2.6 A, but I have no idea what PC manufacturers actually use. If "modern PC" is too broad, I am happy to know the answer just for the Dell Optiplex 780/790, the HP Pavilion p7-1030/1154 and the Lenovo Thinkpad X220/X230, which is around the time that integrated USB 3.0 ports seemed to start appearing (2011). I understand that later PCs can have BC (Battery Charging) 1.2 ports or PD (Power Delivery) ports, which are rated to provide up to 5 A, and so would have much higher overcurrent protection limits, which I am less interested in. My main motivation for asking this question is to know whether plugging a faulty device into an older USB port can ever result in that device drawing a much larger current (but not short-circuit) than the USB specifications provide for, and how much more (before the polyfuse trips). I chose these relatively popular PC models, but if you can only find information on other Dell/HP/Lenovo models, that would also be helpful. <Q> Sorry, but all answers are incorrect. <S> Vast majority of PC mainboards do use polyfuses (aka resettable fuses) to protect USB ports from overcurrent and prevent damage from melting wires and connector's pins. <S> Below are two examples, for ASUS Z170-A mainboard <S> And for ASUS P9X79LE USB specifications define that "high-powered hosts" must supply AT LEAST 500 mA for USB2, and 900 mA for USB3 ports (and more for Type-C). <S> At least . <S> General rule of design is consumer safety and to reduce OEM liability for setting fires, so the designers tend to limit the available current. <S> High-side switches are expensive, and require even more expensive wire and extra pin on CPU or EC (or whatever). <S> So simple polyfuses do their job quite economically. <S> Also, 95% (or (99?) of all usb hubs do use polyfuses as well. <S> As you rightfully found, maindoard use polyfuses at about double the minimum, some can supply up to the level of entire +5VSTBy limit (2.5 A and above). <S> If someone still has a concern about newer mainboards, here is a snippet of Gigabyte X570 AORUS MASTER (released sometime in 2019), with polyfuses circled: <A> They may be using a dedicated IC known generally as a USB power switch. <S> If you know the specific USB power switch used, you can look up the datasheet and find the typical trip current. <S> Finding this is tricky, as you'll need the schematic which is typically proprietary and unpublished. <S> Otherwise, you can look at the PCB and figure out what power switch is used. <S> To give a specific example, I searched for a ThinkPad X220 schematic and found an alleged copy on a forum. <S> From there, I found that the power switch used is the TI TPS2069 . <S> If you take a look at the electrical characteristics in the linked datasheet, you will find that the typical overcurrent trip threshold is 2.85A, with a minimum of 2.3A and a maximum of 3.4A. <A> There are no fuses in the general USB 3.0 controller (unless you count an electronic fuse circuit as a fuse). <S> The fusing is provided by a mosfet that is built into an IC (host controller or external mosfet) with a current level detection circuit. <S> Source: <S> https://www.usb.org/sites/default/files/usb_20_20190524.zip <S> (USB 2.0 ECN VBUS Max Limit.pdf) <S> A device cannot sink more current than is sourced, and it is my understanding (based on observation of shorted USB ports that I have shorted) that all ports have some kind of over current detection (and will flag the port in the OS when the current is overdrawn and shut down power to the port). <S> Here is a good question on the current outputs of each spec: https://superuser.com/questions/1131515/usb-c-to-usb-a-cable-usb-3-0-with-high-current-at-2m-length <S> If the manufacturer does not report which port belongs to each spec, then there isn't a good way to find out unless the circuit is reverse engineered. <S> Many manufacturers produce USB ports that are out of spec. <A> I would like to know what is the typical rating of a polyfuse usedon modern PCs (desktop or laptop) with USB 2.0 and 3.0. <S> My main motivation for asking this question is to know whetherplugging a faulty device into an older USB port <S> can ever result inthat device drawing a much larger current (but not short-circuit) <S> thanthe USB specifications provide for, and how much more (before thepolyfuse trips) <S> These are conflicting requirements. ' <S> Typical' ratings tell you nothing about the maximum possible. <S> Even having information on a specific laptop model doesn't tell you, because you don't know that the manufacturer didn't change the circuit at some time (which is very common). <S> If you are designing a product then you surely don't want to limit its use to particular computer models and productions runs, and you don't even know that the user won't plug your device into something else with a USB port. <S> So you should not assume that the host will provide any protection. <S> One thing is for sure though. <S> If the USB port is protected by a polyfuse (which I know some are) <S> then the maximum possible current must be a lot higher than the hold current, which should be similarly higher than the USB port's rating. <S> If you need actual numbers then I suggest testing as many computers (and other host devices) as you can, using a USB load tester. <S> Alternatively you could start an online survey, or look for existing surveys asking people to report the maximum current their USB ports can provide. <S> This will provide more accurate data than polyfuse specs, since the actual trip current and time is dependent on its thermal characteristics when installed in the device. <A> No modern PC would use a polyfuse. <S> They use a high-side switch with built-in current limiting. <S> The limiter serves three functions: power domain control, overcurrent protection, and signalling the host when the overcurrent condition occurs (OC trip.) <S> A polyfuse can only do one of those things, and not very accurately or cost-effectively. <S> The PC port must deliver at least what is required by version of the USB spec it claims to support. <S> For USB2 and earlier, this was 500mA. <S> For <S> USB3.0 this is at least 900mA, and can be up to 2.1A for a battery charging port. <S> USB3 Type C introduces a more sophisticated approach with options for higher voltage. <S> The takeaway is, no polyfuse in a PC. <S> And even for non-PC applications, based on my direct experience (set-top boxes) <S> USB power switches are very, very cheap (cheaper than a polyfuse) and more predictable.	Not all computers necessarily use a polyfuse to provide overcurrent protection for USB. With USB 2.0 and earlier the current limit is provided by the host controller and that limit is 500mA . If the current goes beyond the spec, then the circuit is shut down.
Measure distance using an ultrasonic sensor in a pipe I am trying to find the distance of an object in a pipe, but I am getting wrong values using the ultrasonic sensor with an Arduino as the sound is bounced back from the surface of the pipe. What do I need to do to get the right values? This is the picture of what I am trying to do. <Q> I assume the pipe is only to protect the sensor from environment noise. <S> If the pipe is not filled with anything, then just line the inside of the pipe with sheets of sponge or foam to eliminate or reduce internal reflections. <S> Searching the net yields results for sound proofing foam sheet Images from amazon.com and efoam.co.uk (no affiliation, just top google results for me) <S> The principle is similar to the black paint used inside telescopes and camera lenses. <A> Ultrasound Sensors measure the distance by sending an ultrasound signal and measuring the time it takes to be reflected from the object and come back. <S> The biggest problem are reflections inside the pipe, which can be solved using sound absorbing material inside the pipe. <S> Another problem is, that the speed of sound is dependent on air pressure. <S> Even small changes, like opening a door or moving an object quickly in front of the sensor completely ruins the measurement. <S> A much better solution is to use a time-of-flight laser-ranging sensor. <S> They are really inexpensive these days (the VL53L0X is around $10) and give vastly better accuracy and performance. <A> You may very well need a completely different technology than the ultrasonic. <S> It may be better to step back away from the ultrasonic problems and restate the problem in a more general way as to what you are trying to measure. <S> When doing that you will also have to provide more detail about the environment and what is in the pipe. <A> If your sensor is like the HCSR04 <S> it generates an 8 cycle burst of a 40 kHz tone. <S> In free air at 330 m/s the wavelength is 8 millimeters and the burst is roughly 6 centimeters long. <S> The signal from the receiver is amplified and some simple discriminator circuit tries to decide either when the leading edge or the maximum of the delayed and terribly distorted return burst happens. <S> This works fairly well in relatively free space where mostly a single reflection from an oddly shaped object is detected, but assuming that your pipe is larger than 8 millimeters in diameter it will reflect the sound many many times and each angle will produce a different delay due to it's zig-zag pattern. <S> A good analogy would be comparing single-mode optical fiber which works like a waveguide and can maintain absurdly high modulation frequencies over tens to hundreds of kilometers with amplification but without regeneration, to the earliest kinds of step-index multi-mode optical fiber which only worked at much lower bandwidth and that strongly depends on the distance. <S> (The graded index MMF has much less modal dispersion than the old step-index fiber.) <S> What you have is an analogy of the step-index fiber with huge modal dispersion; the higher the angles of the zig-zag reflections the longer the path length and the slower the reflection. <S> Your sharp "ping" input gets smeared out terribly with distance. <S> That means that the simple discriminator in your budget device is totally unprepared to receive this mess, and gives a poor result. <S> The other answers all provide good advice. <S> If you can line your pipe with material that blocks acoustic reflections as recommended in <S> @AJN's answer so that the signal propagation is more like that of a free-space situation, you may get much better results! <S> All is not lost! <S> From Graded Index or Step Index Multimode Fiber :	Some type of focused and targeted sensor beam reflected back from a target on the distant object may be a possibility but much depends upon the environment, what is in the pipe and just what you are trying to achieve.
Do I use a relay, MOSFET, or just a switch? (5VDC 15A load) I have a 5VDC 15A power supply that will be powering both an Arduino and 288 Addressable LEDs. How do I switch the circuit on and off? First I looked at relays. The only relays that I could find that support a 15A load are ones that use High Voltages around 250V. I think they had a minimum recommended load voltage of 12V.I also looked at mosfets, but I found that many would burn up if I put 15A of current through them. I have also considered just using a switch, as everything is at the same voltage. In that case, I would need to make sure the switch I use can handle the 15 A of current. Hopefully a switch like that wouldn't be too bulky. Basically, My load voltage is too low to use a relay, but I'm afraid the current is too high to use a mosfet or normal switch. How should I turn this thing on and off? <Q> simulate this circuit – <S> Schematic created using CircuitLab <S> The particular part listed above is cheap and has an Rds(on) of less than 2.5m \$\Omega\$ @25°C with 5V drive, so it will drop less than 38mV at 15A, so it will dissipate about 0.8W when hot, requiring a few square cm of copper to run reasonably cool. <S> A slightly more expensive part (still less than $1.50) is <S> more like 2m \$\Omega\$ @25°C <S> so it will run cooler. <A> You should be able to use a simple relay, like this one: https://ar.mouser.com/datasheet/2/307/en-g2rl-1670825.pdf <S> Don’t confuse the coil voltage with the max voltage the relay can handle. <A> It depends if you are looking to control the system manually (ie flip the switch by hand) or make another electronic system to flip the switch for you. <S> In the latter case, I would look for an automotive N-channel mosfet to install on the low side of your circuitry. <S> The mosfet would also allow you to use a much smaller switch, if you are going for the manual solution. <S> There are plenty to handle 5V 15A. <S> For instance IRLZ34NPBF (the first to appear in my google search). <S> A relay would also work, but I would not recommend using one. <S> You should use a mosfet since you are dealing with direct current. <S> Relays are more expensive and less reliable. <A> Be careful using relays intended for AC loads. <S> These are designed so that any any arcing that occurs as the contacts open is quenched by the AC signal becoming zero 120 times per second. <A> So lots of these answers detail very well that you can use a MOSFET or a relay or even a switch. <S> But they all miss an easier solution. <S> You say these are addressable LEDs. <S> So before you go whacking the power off to a 15A load, first command them off using the arduino (which is presumably generating the commands to the LEDs). <S> Bam. <S> Your load just went from 15amps to like 10mA (arduino current). <S> Now if you want to kill power using an external switch it becomes trivial. <S> Now as to why the mechanical solutions can potentially have arcing issues: while the system voltage is indeed 5v, the moment you try to pull away the pieces of metal conducting those 5v @ <S> 15amps there is a rather large change in current per unit time. <S> Any inductance in the circuit (either stray or intended) will resist that rapid change in current and cause a voltage spike that will likely arc per V = L*di/dt. <S> Mechanical switch/relay designers know all this, the plating on the switches and relays are designed to handle this presuming you stay within the DC limits of the switch/relay current. <S> Not doing so will eat away at the lifetime of the switch/relay and would require you re-rate the switch lifetime (re-calculate the expected life of the switch based on the new info). <S> Like somebody else said, automative relays are a great source for higher-current relays, and plenty, plenty of MOSFETs are more than capable of handling 15amps.	You can use a simple p-channel MOSFET as a high-side switch. An automotive relay would be a better choice, although its coil may require 9 V to operate. The 250V refers to the maximum voltage the relay can switch on/off because of the electric arc that's formed during commutation. All the mechanical solutions (relay and switch) will have the problem of arcing at these relatively high currents (more on that in a bit).
how to calculate maximum temperature of wattage on wire? how do a calculate the maximum temperature that a certain wattage can get on a wire?I know how to convert wattage to "Celsius heat units (IT) per hour", but eventually the wire has to stop heating up, right? EXAMPLE: if i have 144 watts, converting it to Celsius heat units (IT) per hour would be 272 degrees Celsius. But i still don't know, if that is the hottest it will get. <Q> I found a definition of the "Celsius heat unit" but I have no idea if it is true or correct: <S> The Celsius heat unit is the energy required to increase the temperature of one pound of water by one degree Celsius (or 1.8 degrees Fahrenheit) at a constant pressure of one atmosphere. <S> One Celsius heat unit is equal to 1.8 British thermal units. <S> Source: <S> Conversion Website <S> The definition seems to be a confused mess of metric and imperial units and you will find it far better to work in SI (metric) units. <S> The Celsius scale measures temperature . <S> Use watts <S> (W) to measure power . <S> Use joules <S> (J) to measure energy . <S> if i have 144 watts, converting it to Celsius heat units (IT) per hour would be 272 degrees Celsius. <S> But i still don't know, if that is the hottest it will get. <S> The temperature will stabilise when the heat lost to the surroundings of the wire is equal to the electrical power input to the wire. <S> To calculate this you would need to know the length, surface area and emissivity of the wire to calculate radiated energy and the temperature and flow rate of the surrounding fluid (e.g., air) for convective losses. <S> You might be able to ignore conduction losses depending on the geometry of the setup. <S> Forget about "Celsius heat units (IT) per hour". <S> It's not used in general or electrical engineering. <A> You'll need to consider the thermal resistance from the wire to ambient surroundings, call it \$\theta_{wa}\$ <S> (units of degrees Kelvin per Watt). <S> This will need to take into account any forced airflow, convection, contact with supports/substrate, and direct radiation as applicable. <S> This may be hard to calculate and thus might require experimental techniques to determine. <S> The wire will reach a temperature of <S> \$P \cdot \theta_{wa} + T_{\text{ambient}}\$ under the assumption that the power is constant as temperature changes. <S> This is another approximation that may not hold; as the wire heats up its resistance typically increases, which may lead to an increase or decrease in dissipated heat depending on how the wire is connected to the circuit. <S> If driven with a nearly constant voltage, power will decrease as the wire heats up; if driven with a nearly constant current, the dissipated power will increase. <A> Probably about as many people calculate the maximum temperature of wire as use units like CHU per hour or furlongs per fortnight. <S> Most people find tables in electrical codes for building wiring or published by wire manufacturers and others for equipment wiring. <S> Tables give ampacity for various size wires and conditions. <S> Some conditions are handled by the use of adjustment factors. <S> Here are some examples: IEWC Global Solutions Cooner Wire Belden <A> According to this , the "celsius heat unit" is an actual unit of energy, roughly equal to the amount of energy required to raise the temperature of one pound of water by 1 C. <S> I have never previously heard of this unit. <S> It is not used regularly in academic physics or in electrical engineering. <S> Perhaps there is some industry where it is used or has been used historically; refrigeration and building heating or air conditioning seem likely, given the scale of the unit. <S> But given that the CHU is a measure of energy, it is not a measure of temperature. <S> You cannot compare CHU and degrees celsius directly. <S> As the other answers have said, the temperature achieved by a wire (or any other object) being heated by a constant power depends entirely on how that wire is cooled (by conduction, convection, and radiation) and not at all on the conversion factor between watts and CHU per hour.	You need to know the wire material, the ambient temperature, the current that the wire carries and whether the wire is mostly surrounded by air or bundled with other wires.
Should a ferrite bead be placed close to the MCU or close to the power supply? I am trying to design a PCB based around the STM32F446RE microcontroller. I looked at the schematic provided by ST for their Nucleo-64 board, and physical placement of the ferrite bead on the board. The ferrite bead they use is a Taitec FCM1608KF-601T05 . Which is placed on the VDDA line (pin 13). They do not have a ferrite bead anywhere else. I have seen a few designs done by others, and they seem to place the ferrite bead at the input side of the their 5V linear regulator . In other words, ST seems to concerned primarily with noise coupling on to analogue measurements and have placed the bead after the 3.3V regulator to the MCU. Other designs seem to be concerned about noise getting into the overall circuit. What are the pros and cons of each approach? I am considering putting my ferrite bead at the input stage to the 5V regulator, with the thinking I can filter out at everything right at the beginning - but get a feeling that this may introduce other issues. The ferrite bead I would be using is Würth Elektronik 74279265 which has better DC resistance and impedance characteristics over a wider frequency range. <Q> You should generally place filtering components as close as possible to the source of the noise you want to filter out. <S> If you're concerned about noise from your power supply, put your filter near the PSU. <S> For sensitive analog inputs, though, you want to put a filter close to that input. <S> This can be thought of as because the entire trace leading to it is the noise source you're concerned about here; you just want to isolate that antenna of a trace from the input. <A> Place it as close as possible to the source, e.g. switching circuit of the power supply. <S> This will attenuate the noise before it can travel through wiring, to avoid emissions by the wiring. <S> These emissions (egress) can cause ingress or interference within the device itself and into other devices. <S> as close as possible to any sensitive analog inputs. <S> This will reject ingress captured by any wiring and connectors that lead to these sensitive inputs. <A> Filters need TWO components: a series (here the FB) and a shunt (which is missing here). <S> For that bead to be useful, add 1uF at AVDD. <A> There is no "best" position to place a ferrite choke. <S> Placing it on the input or output of your switching PSU, or on a pin input of another IC achieves different things. <S> You should assess the need for any of these chokes individually, and implement as many of them as you believe will benefit your board. <S> Input of PSU: <S> Reduces noise generated by the PSU onto the power input net, for example a 24V rail. <S> Can be useful if for example a device is using the same net directly without a dc/dc, and it does not have a high ripple tolerance. <S> Output of PSU:Will reduce the overall ripple on the output net, which might be required for some components and reduces EMI. <S> Input of IC:Will both smooth out the net for the specific pin, which is often necessary for many MCU VDD pins (usually require less than like 20-50mV ripple), and can also prevent ripple from the pin to the incoming net in cases where the pin belongs to a very fast switching mechanism like an internal PSU, oscilator etc. <S> In your example I believe the placement near the MCU pin is done because that specific pin has a much stricter ripple requirement than the other pins using the same net. <S> Placing the ferrite there will give a better ripple reduction than placing it on the PSU output, because placing it on the PSU output would not attenuate ripple generated by other MCU pins nor EMI emitted onto the net between the PSU and the MCU pin.	If you're concerned about noise from the microcontroller, put the filter near the MCU. Keep it close to the inputs, to cover as much input wiring as possible.
Need help identifying wires on 3-phase single-volt AC motor (6 leads) I'm far from an electrician. Just a guy trying to get some new life out of an existing piece of machinery. See the image below, this 5HP motor comes from a heavy duty commercial treadmill. Based on the fact that it has no start capacitor, 6 leads, and that heavy flywheel on the right this is a 3-phase single-volt motor. The labels originally were W,V,U but some of them were scratched off and unreadable. But there's two additional wires both black and missing labels, and the green for total of 6 wires. I've labeled the black wires A,B,C,D,E and measured them using ohms setting on a multimeter: * A<->B: nothing* A<->C: nothing* A<->D: 0.6* A<->E: nothing* B<->C: 2.2* B<->D: nothing* B<->E: 2.5* C<->D: nothing* C<->E: 2.5* D<->E: nothing My questions are: how do I restore the original labels of W,V,U from the information above? if 3 wires were W,V,U and the other two had no labels remaining, what purpose are the other two wires? assuming i identify all the wires are correctly labeled, any quick way to test the functionality of this motor before wiring it to a control board? <Q> You have a three phase motor with thermal protection. <S> From True Fitness Max Drive Motor AC 5 HP CS6.0 <S> CS8.0 OC460000 Works with Treadmill by True Fitness. <S> So if you look at your results. <S> A is only connected to D. <S> This corresponds to P1 and P2 Thermostat leads. <S> BC <S> = 2.2CE = <S> 2.5BE <S> = 2.5 <S> These correspond to UVW <S> (Line 1 to 3). <S> The order does not mater, aside from directional. <S> Reverse any two and the motor should change direction. <S> Only true way to test it is to wire it up and apply power. <S> Usually, threadmills are discarded when the motor dies. <A> The resistances between leads B, C and E seems high for stator resistance, but the fact that they are equal would indicate they are U, V and W. Which is which doesn't matter much. <S> If the motor runs backwards when powered, swap two of them. <S> The resistance between A and D would be nearly zero if it is an over temperature switch. <S> Both this and the B, C, E resistances could be just an indication that the meter doesn't read low resistances accurately. <S> I doubt that the motor will harm the controller board. <S> I would be inclined to test the motor and controller board together rather than try to test the motor my itself. <S> The current on the nameplate posted by @StainlessSteelRat indicates that the "SPL" marked after 5 <S> Hp stands for "sales person lie. <S> " That is a "peak Hp" rating that means it can produce the torque of a 5 Hp motor for a very short time prior to failing. <S> It is really something like a 1.5 Hp motor. <S> The measured resistances are more reasonable for 1.5 Hp. <A> I definitely agree with Stainless steel Rat. <S> I'm an Winder and work with Electric Motors for 4 years now. <S> You'll get an open circuit between A&D if the motor heats up beyond 150° <S> C most likely but that won't open the winding, the contactor or starter will kick the overload. <S> But to be 100% sure <S> if you post a pic of the winding I can definitely tell you if it's single phase or 3 phase. <A> A-D thermal trip, goes open circuit when internal winding temperature goes too high(usually 80-105C)B,C,E 3 phase supply <S> Above assumes you are in 110v/60Hz country. <S> When connected correctly motor should spin clockwise when looking from the back, swap any 2 phases if going backwards.	It certainly appears to be a three-phase induction motor.
Why is my fan circuit with SSRs short circuiting on power down? I am attempting to build a "smart" ceiling fan for a home automation system I have created and added to over the years. However, this is the first time I've tried controlling a fan motor and the first time I've used solid state relays (SSRs). My research (and now experience) shows me these two elements can be a bit tricky. Note: I am a software developer with no formal electrical engineering education so you will forgive my ignorance when it shows. In brief, the system powers on fine and can be controlled by my software running on a NodeMCU successfully. It can do this for several days. However, at some point, during a power-off cycle, the circuit will short and blow the fuse. Once this occurs, 4 of my 8 SSRs will be permanently closed. I am very confident the short is occurring during power down (i.e,. while turning off the wall switch that provides 120v to the black (#6) wire as seen in the attached SmartCeilingFan schematic). This is true because when I had it on my bench testing it, I saw a blue flash from the fuse when I turned off the system after a successful test run. After that, the SSRs mentioned below were permanently closed. (This has occurred on three different prototypes I have done so it is clearly repeatable.) Troubleshooting I have done: I have placed an oscilloscope on all of the used data pins on the NodeMCU and confirmed their behavior on start up to be as I state in the attached SmartFanWifi image. I have placed an oscilloscope on the data pins for both of the SSRs I suspect to be coming on at the same time and causing a short and confirmed that neither bounces HIGH during power down. I have gone over the G3MB-202P datasheet and the SSR's Common Precautions sheet in detail. However, this is where my lack of formal electrical engineering training will show... while I believe I have followed the guidelines and am in spec for the SSRs, I somehow must not be. I have bench tested the max amp draw of the ceiling fan by placing an amp meter in series on the load side of the fan and then held the fan from spinning (not a perfect way to do it I am sure, I'm willing to be taught). The max amp reading using the method was around 500 mA. Additional information: The ceiling fan is a Hunter fan with this wiring diagram out of the box. Here is the datasheet for the L293D ICs. SmartCeilingFan Schematic Image SmartFanWifi Schematic Image <Q> I notice your hand-written notes on the schematic. <S> Figure 1. <S> Details from the schematic. <S> I suspect that on switch-off that you might be getting a high \$ \frac {dV}{dt} \$ (very sudden rise in voltage). <S> Triacs can be turned on when the voltage across their MT1 and MT2 terminals rise suddenly. <S> It may be that this is turning on A and B simultaneously and shorting the motor windings as indicated in your note. <S> This in turn may cause high current to flow and destroy your triacs. <S> You'll need the appropriate snubber for this. <S> Links Phase Control <S> Using Thyristors by Littlefuse. <S> Thyristor Theory and Design Considerations Handbook by ON Semiconductor. <A> This means that the relays can be on for almost half a mains cycle after you remove the control signal. <S> Introduce a delay of at least 10ms-20ms between turning off one relay and turning on another if having both on causes a short. <A> Only 2 answers come to mind: <S> You missed rogue On signals on power down. <S> Possible if you searched for them with motor disconnected and back EMF spikes are source of rogue on signals <S> Voltage spike on mains disconnect are not dampened enough by varistors.	A thyristor or triac will not turn off until the current through it drops to zero.
Analog circuitry for measuring capacitive touch I'm interested in adding capacitive touch input to a project I'm working on. Ideally I would like a tapered shape so that a finger can slide up as if it's a mixing desk fader or something like that. However, everything I read about capacitive touch input online involves using an ADC/microcontroller to digitise the value, then use software to determine the output. I am wondering if there is any circuit which can achieve this without any code: a capacitive touch input, which creates a voltage proportional to how much capacitance* the finger is exerting on it. So, is there an analog circuit which can achieve this without any microcontroller? The capacitance varying through either the distance of the finger from the sensor, or some kind of tapered sensor as I mentioned. I'm aware that external factors could affect the capacitance too. <Q> So, is there an analog circuit which can achieve this without any microcontroller? <S> Sure. <S> The classical way of measuring capacitance is building an oscillator whose frequency depends on the capacitance. <S> From there, multiple ways: if you waveform is already stable in amplitude, i.e. the amplitude doesn't depend on any factors and especially not the capacitance, a simple low-pass filter can be used to dampen the wave proportionally to its frequency; a rectifier following that (and another, but higher-frequency low pass, to get rid of the harmonics) can convert that then into an inversely frequency-proportional voltage. <S> If you can't make reliable statements on the oscillation's amplitude, a number of ways, easiest probably being strong amplification sufficient for converting it into a square wave, can be walked to get a constant-envelope waveform. <S> If you have complexity to spare: use a reference oscillator, and a phase error detector as in a PLL. <S> The error signal's average is proportional to the frequency deviation. <A> So, is there an analog circuit which can achieve this without anymicrocontroller? <S> A Theremin springs to mind: - Not just outdated since black and white days but still made: - Basic Theremin kit including schematics. <S> Here's a nice tube/valve schematic from here <S> (pity about the lack of values) but I'm sure if you dig around they will turn up: - <A> I mentioned this in a comment, but one simple implementation (of Marcus Millers post) is to use a relaxation oscillator, and then filter out everything except the edges (with the 1nf capacitor + 100 ohm resistor + diode), then smooth this out and amplify it with another opamp. <S> You can play around with this here . <S> The voltage will go down when a finger is placed on the touch sensor (a typical range is 100-200pF with finger present, <10pF with finger not present), and will vary (linearly?) with the amount of capacitance. <A> you could simply rectify what 60Hz energy is coupled into a tapering metal stripe udre som ethin insulation. <S> By peak_detecting the energy, you have a linear measure of capacitance; this will change dramatically, as your body moves around in the Electric Field in your workspace; an alternative is to rest your forearm on a plate driven by 1,000Hz, and you simply filter for the 1,000 Hz, now having your entire body moving slightly with 1,000Hz sin. <S> WHen your finger is on the very narrow end, most of the 60Hz Efield moves to the underlying metal plate. <S> As your finger slides to the wider part of the tapering metal strip, more and more of the 60Hz displacement current couples onto the strip instead of moving a millimeter further to the underlying metal plate. <S> And at the very widest part of the tapering metal strip, essentially ALL of the 60Hz Efield from your finger has coupled onoh the tapering sensor strip. <S> As you press your finger harder, you will get more displacement current, so you need to anticipate that effect. <S> ========================================== <S> You need two layers of insulation, and a metallic base plate, and the tapered metal strip which is your sensor. <S> How much current? <S> Assume 1cm by 1cm area, and 1mm distance form your finger to the taperer sensor. <S> Assume the Relative Dielectric Constant of the insulation is 5. <S> Why not? <S> C = <S> E0 <S> <S> Er <S> * Area/ Disrance C = <S> 9e-12 farad/meter * 5 <S> * (1cm <S> * 1cm) <S> * 1mm <S> C = <S> 45e-12 <S> * 0.0001/0.001 = <S> 45e-12 <S> * 0.1 <S> = 45.e-12 = 4.5 picoFarad plus/mimus <S> At 160 volts at 60Hz, the SlewRate is 160 * 377 ~~~ <S> 60,000 volts/ <S> second <S> assuming there are no SPIKES from motors or fluorescent lights. <S> The current (Maxwell's Displacement Current) is using Q = <S> C * V, differentiate with <S> dC/dV being constant, get dQ/dT = <S> C <S> * dV/ <S> dT whihci becomes I = <S> C <S> * dV/dT <S> And substrituting, <S> I = <S> 4.5pF <S> * 60,000 volts resone sinusoid <S> I = <S> 4.5e-12 <S> * 6e+4 = 27e-8 = 270 nanoAmperes (with maybe 50% accuracy on this math) <S> NOte 1 MegOhm shunt, sensed by UNity Gain BUffer, produces 0.27 volts. <S> For 1,000 Hz at 10 volts, (charging your body thru 1Kohm resistor! <S> why not), yo uget approximatey the same 0.27 volts.	Other ways include delta-sigma conversion of the capacitance through measuring the time it takes to charge it; but now we're getting dangerously close to digital domain again, and you wanted to do it in analog.
Why is the reverse biased zener diode used for voltage regulation? I would expect a forward biased regular diode to be used. Especially in applications in which you don't expect a change in current direction. Is the curve more predictable in the reverse biased state? <Q> The curve is much more useful in the reverse breakdown mode. <S> Especially for avalanche diodes, which every "Zener" diode over above 5V actually is. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Here I have swept the current from 100uA to 100mA on each type of "regulator" diode. <S> Compare the regulation of each: <S> Temperature coefficient is also ~4x better, about +6.5mV/ <S> °C vs. roughly -25mV/°C <A> A forward-biased silicon diode (zener or "normal") will have a voltage drop of about 0.7 volts - not too useful for voltage regulation. <A> The reason we use zener diodes in voltage regulators is because zener diodes exhibit almost identical behavior to a forward biased diode above the zener breakdown voltage. <S> It is like having a diode but with a forward voltage of any value you want.	Zener diodes can be made with many diffrerent reverse breakdown voltages, so you can use them in suitable circuits to produce many different regulated voltages.
Different outputs of a sensor in its recommended load range I want to use this sensor by using 5V supply voltage and I do the connections as shown in its datasheet. According to he data-sheet the minimum load is 80k which means I should use less than 80k.But when I use different resistor values I get very different outputs and Vsupply is 4.96V in both cases: For instance, using a 15k load I get 1.08V output; and if I use a 22k load I get 1.48V output. Both loads are less than 80k but I get very different results. What could be the reason? <Q> Probably they mean minimum load resistance. <S> It’s not 100% clear from the datasheet, but they don’t specify a maximum output current or equivalent drive capability so it has to be that. <A> Duplicating my own answer from an earlier question on the same subject: I have no experience with the HIH 4000 (or other humidity sensors,) <S> but I thought I'd have a look and see what I could find. <S> From what I have found, the datasheet isn't telling you that you need to put an 80k resistor from the output to ground. <S> What the datasheet trying to say is that the load impedance connected to its output must be higher than 80k ohms. <S> This thread on the Velleman forums refers to an ADC module with a 20k ohm input imopedance. <S> An HIH 4000 connected to that module (the Velleman K8061 ) <S> would consistently show a wrong output (as measured with a voltmeter) <S> when connected to the K8061, but would show a proper voltage when disconnected from the K8061. <S> The recommended solution was to remove the 20k resistor in parallel with the input on the K8061. <S> Alternatively, an op amp could be used to buffer the HIH4000 output. <S> There's a similar problem and solution with the Velleman K8055 mentioned in this forum post. <S> Something similar happened with an ADC for a Raspberry Pi. <S> Again, the solution was to increase the input impedance of the ADC module. <S> Summary: <S> Given that you have a high impedance output coupled to a high impedance input, you have a very good chance of picking up stay noise over the wires from the sensor to the ADC. <S> The solution to that is to use the minimum load allowed in all cases (80k in parallel with the ADC input) or to install a buffer right at the sensor output so that you have a lower impedance line going to the ADC. <A> It is apparent from your test that, the lesser the load resistance relative to 80kΩ, lower the output voltage. <S> Hence it would not be incorrect to presume that the value of the load resistor should be 80kΩ or greater. <S> You may confirm that by checking the output voltage at 68kΩ, 82kΩ and 100kΩ.	The 80k is the minimum input impedance of the ADC or buffer amplifier you connect to the sensor.
Why does a spark occur when an SMPS is connected to the load circuit? I am using a 36VDc SMPS as a power supply. A spark is generated when I connect 36V with my load circuit. Specification: Voltage source: 36VDc Current consumption: more than 3A I want to know why a spark is generated when I attach my load circuit, and how can I reduce the spark. I have tried some RC circuit solutions given for arc suppression circuits but I can not identify how can I use those solutions. Is there another way to reduce the sparks? I tested with a 1545CT Schottky diode to put at the beginning of my circuit as mentioned in this answer, but the solution isn't useful in my case. @Manumerous, can you guide me for which type of MOSFET I have to use in my circuit with configurations to avoid a spark? <Q> This is a very common occurrence if your load contains a large capacitor bank. <S> The problem is that if the voltage over a capacitor changes fast (as when you connect the load to the power supply), this results in very large currents. <S> This can be seen by the following differential equation for the capacitor current: $$I_c(t) = <S> C \frac{dV(t)}{dt}$$ <S> Unfortunately this is exactly the case when you connect your load to the power supply. <S> In this case your capacitors will draw a lot of current (which is often called the inrush current) <S> You can reduce this by limiting the current with a large enough resistor. <S> This approach is the simplest, but adds additional losses through the resistor and increases the input impedance. <S> A better Option would be to use an NTC Thermistor in that case. <S> These resistors have a negative temperature coefficient, which means that there resistance decreases when they heat up. <S> Therefore, the resistance of the NTC Thermistor becomes very low after it is warmed up by the current passing through it. <S> This achieves a power loss that is lower than when a fixed resistance is used. <S> You can find some information about NTC Thermistors here: How to Use NTC Thermistors for Inrush Current Limiting <S> Alternatively you could add an electric switch (like a MOSFET for example), which ramps up the voltage and therefore eliminates the spark. <S> An in depth discussion about that topic as well as schematics for how to connect the MOSFET can be found here: <S> StackExchange Electrical Engineering: <S> P-Channel MOSFET Inrush Current <S> Limiting <S> Furthermore also the geometry of the connector can influence the stark. <S> There are dedicated anti-spark connectors which help reduce it or ensure that the spark can not come in contact with the skin. <A> An electric spark is an abrupt electrical discharge that occurs when a sufficiently high electric field creates an ionized, electrically conductive channel through a normally-insulating medium, often air or other gases or gas mixtures. <S> - Courtesy Wikipedia. <S> A spark could be caused by breaking current flow through an inductor (resulting in a high back emf generated by a collapsing magnetic field to return its stored energy to the source). <S> Other causes could be high surge currents through incandescent lamps etc. <S> Identical scenarios would be possible, when a contact is made using a switch or a relay, due to 'contact bounce' or 'chatter'. <S> The solution would be to use RC snubbers or VDR's across the contact or across the load. <A> Your design has excessive inrush current because it has no protection for a bulk capacitive load. <S> Thus the surge current is I= <S> V/ESR for ESR includes cap and contact resistance. <S> You could consider a PTC rated for the max Joules of charge energy but that adds a minor cost and heat. <S> Even 2.4A USB power to Apple iPad products creates a black carbon strip in the middle for +5V and burns out one connector pad. <S> With an extremely thin flash coating of gold, the plugs eventually wear out in a year or 2 from a daily connection. <A> You can get a spark hooking up 12V jumper cables between cars because you are making and breaking a circuit in the many-millisecond time it takes to connect. <S> The inductance in the cables or receiving circuit is surprising in its ability to keep current flowing in a collapsing magnetic field when a circuit is broken. <S> It's nothing to worry about. <S> If you want to reduce it, put a switch in the line and throw the switch after you connect. <S> There will be little or no "bounce" that way. <A> My suggestions, MOSFET based inrush current limiter: https://github.com/msglazer/Anti-Spark_Switch (Generally used in octocopter) <S> Anti-Spark XT60, XT90 connectors explanation: <S> https://youtu.be/X71Suakve6A <S> (Video)In Anti-spark connectors, they are using small value high power resistors approx 6 ohms. <S> This Resistor helps in reduce inrush current due to higher capacitance in your circuit. <S> But this resistor only contacts in a small period of time. <S> If you connect this loose connector maybe meltdown. <S> Thank you for reading. <A> I am using IRF3205 to switching GND, and the gate is controlled by the microcontroller for the delay, so when the connector attaches to the load circuit, the GND is already disconnected, so the spark not happens. <S> Solution: - <S> When the power In signal detected by the microcontroller, it turns on the gate after a few seconds to avoid spark.(Note:- Here I can use a relay for the same purpose, but again relay is the one kind of physical switch, so spark happens inside a relay <S> is again cause a problem after a long time so that kind of solutions are not good <S> , so I preferred this type of solution) <S> Thank you all for your help.	It could also be caused by breaking a high charge/discharge capacitor current.
What features should/must be present inside a prototype FPGA board to aid in debug? When making any prototype PCB it is always a great idea to have features within it that can be used to aid in debug of the design. A few examples are: Use the largest FPGA available that has migration path to the one we expect to use in the final product. This means we can have a lot of logic to implement debug modes and internal logic analyzer like signal-tap. This does mean that our power dissipation measurements won't be accurate since a larger FPGA will have higher static power dissipation. Embed features into the board that makes it possible to connect certain signal paths to a scope. These could be analogue signals or digital signals. I am not talking about the signals that directly connect to the FPGA. There are different ways to do this, and I am not sure about the best way to do this. Have 8 or 16 pins exposed via header, that connect to the extra general purpose I/O on the FPGA. These can be used to output internal FPGA signals to the pins which can then be connected to a scope or logic analyzer. The question that arises here is, what type of connector to use since high speed signals require signal integrity to be kept in mind. Certainly we must choose a connector that a scope or logic analyzer can easily connect to. Have 1ohm resistors on the power supply rails that can be used to accurately measure the power dissipation on all the supply rails, especially the ones connected to the FPGA. Put in an FTDI chip that can be used to implement USB-UART. Have an easy method for the prototype to be connected to another FPGA or microcontroller board in case the need arises. One way to do this is to have Arduino shield connectors on the board but then we must be sure that what we intend to connect is compatible before we make the PCB. I am sure that people that have made many prototype PCBs and then gone to production can give their opinion on this list and suggest some improvements. It is true that the features also depend on what the board actually has been designed to do, so the same techniques can be applicable to every project. So now my questions are: Is there any improvement over what I have recommended above? Are there some features further to what I have mentioned that should be utilized? Having lots of capability to aid in debug is a great idea but, it creates an issue when we come to production level tests. The path to go from the prototype board to the production level PCB for alpha testing e.t.c must be simple. I would be grateful if someone can put in their two cents. This question is VERY specific and not broad or off topic. I think I have covered most things already. Thanks. <Q> This is especially useful when you want to sample a large chunk of data, like a frame or a package, which is larger than signaltap's sample capacity. <S> Some more advanced features can also be designed based on this sample memory like sampling only the packages whose header match a specific value. <A> I would actually advise against using the largest FPGA available simply due to the extraordinary device cost and implementation challenge of the largest FPGAs. <S> Don't forget adding some LEDs for debug output indicators. <S> You might also want to add jumpers to various places (like disconnecting power from certain sections) for debugging in case something goes wrong. <S> I see little point in an exposed GPIO header for a board that already has a set purpose. <A> Your prototype should have these features too 8in no order): embedded USB to JTAG module like SMT3 from Digilent 2 port or 4 port <S> USB-UART <S> (I also like FTDI like you mentioned) <S> note: <S> a modern FPGA or FPGA+SoC needs one UARTs per operating system and FPGA part <S> 1 = <S> > <S> Linux <S> 2 = <S> > <S> RTOS <S> 3 = <S> > <S> FPGA tiny USB 2.0 hub and a USB-C connector used as USB 2 as service port (mechanically more stable then micro-USB) more complex boards might need a board management controller like MSP430 or similar.this IC can have USB (see USB hub) and 2x I²C to control PMBus to control temperature, voltages, fan speeds, ... it can also serve to configure the board before powering the FPGA control power with PMBus controllers like from Linear. <S> Measure voltage, current, power, temperature and control power staging or savings. <S> EEPROM or better F-RAM <S> so save versions (at best Git hashes, production dates, serial numbers, board configurations, modes, last power state, ...) <S> 1 yellow LED (low-active) + one RGB led that can drive any color and blinking code :) <S> 2 (micro) push buttons 3-pin header for I²C bus (PMBus) 3-pin header for main UART test points for all power rails (at best on top-side) ground pin or place for a ground clamp - used by osziloscope Fan connector, while you board is not in the final chassis while testing temperature sensors distributed over the board to get heat distribution like you want to upsize/downsize your FPGA in the same footprint, you can do the same with DC/DC converters. <S> design your additional pin headers as Pmod 2x6 headers (8-bit data + 3v3 + GND) <S> so you can use extension boards with OLED display, additional puttons, WiFi, Bluetooth, Flash, EEPROM, ... <A> Always LEDs. <S> A 3 pin UART header for TTL-232R-RPI. <S> I usually adds an Ethernet port for debug and programming. <S> FPGA Cores provides an easy way for the customer to update the firmware with the remote programmer tool .If <S> you want there are also an embedded scope that you can use. <S> I really appreciate these cores. <S> You can find the cores here: https://www.fpga-cores.com/cores/	If the memory chip is connected with FPGA, like SDRAM or DDR or others, I'd like to allocate some extra sampling memory space for debugging.
Why do we need array antennas? If we can increase the directivity of the single element antenna by increasing its electrical size then why do we need array antennas? <Q> Array antennas are the basis for beam forming. <A> If we can increase the directivity of the single element antenna byincreasing its electrical size then why do we need array antennas? <S> Increasing the size of a single element means it will certainly receive unwanted lower frequencies and this might make the electronics more complex. <S> It also means that the impedance of the antenna at the "wanted" frequency changes from the straightforward near-resistive impedance of the half-wave dipole antenna (for instance) to a less useful higher impedance as shown below: <S> - So, when the dipole is dealing with a frequency corresponding to half a wavelength, it's impedance is around 74 ohms resistive <S> but, if the length doubled the impedance sky-rockets and this becomes a bigger problem to receive. <A> Long wire antennas can do that, but only in two of 3 axis. <S> And the usual narrow band response of a dipole has been ruined.	With array antennas, you can control the directivity by driving each antenna of the array with a different phase.
3.7 V to 12 V boost converter - effect on cell If I boost a 3.7 V 18650 battery to 12 V with a boost converter, would it affect the durability of the battery? For example, if I have a 2100 mAh Li-ion boosted to 12 V, would it affect the 2100 mAh capacity of the battery? <Q> You won't get as many mAh from the 12V supply as you do from the battery, but you'll get almost as many mWh. <S> A boost supply can't provide free energy, so when it generates a higher voltage than at the input, it must draw more current from the input than it puts out. <S> In the ideal case, Po= <S> Pi, meaning Vo <S> * Io = <S> Vi <S> * Ii. <S> In real life, the power supply will have an efficiency lower than 100%, so Vo <S> * Io = <S> Vi <S> * <S> Ii <S> * Eff. <A> Basically your capacity will reduce in accordance with the efficiency of the boost converter and the voltage transformation ratio of the boost converter (if you are measuring the current at the 12V output. <S> Assuming the efficiency of the converter is 80%, then 2100mAh will be effectively reduced to: $$ 2100 \text{mAh}\times 0.8 <S> \times \frac{3.7}{12} <S> = 518 \text{mAh}$$ <A> Using a booster can have an effect on the battery. <S> Say you need to get 1 ampere at 12 volts from your 3.7 volt battery.. <S> Your booster has to put out 12 watts continuosly <S> (1A <S> x 12V = 12 watts.) <S> To put out those 12 watts continuosly, the converter has to take in 12 watts continuosly. <S> Given 12 watts and 3.7 volts, the converter will draw 3.24 amperes (12 watts/3.7 volts = <S> 3.24 amperes.) <S> Your battery must be capable of supplying over 3 amperes of current to the converter so it can put out 12 volts at 1 ampere. <S> Batteries have a capacity rating (ampere hours) and a current rating (amperes.) <S> That damage may show as reduced capacity, or it may out <S> right destroy the battery (effectively no capacity at all.) <S> If you are using cells with over current protection, the protection might kick in and shut things off before <S> (much) damage occurs. <S> Summary: <S> You have to consider the current rating of the battery along with the current the booster will need to deliver the required power to the load.	If you draw more than the rated current from your battery, you may permanently damage it.
What is this component and what is it used for? Today I opened an old adapter to salvage the transformer and found this component which I also desoldered. What is this component and what is it used for? It seems to have a resistance of 10 Ohms as shown and it's labeled as R1, but the joule is confusing me, so I don't think it's a regular resistor. I haven't been able to find anything online. <Q> 10 ohm +/-5% wirewound resistor (probably inductive, perhaps fusible <S> but it's hard to tell). <S> Called a "cement" resistor. <S> Made by KOA (Nagano, Japan-based company). <S> If it's 13mm long, then 1W rating. <S> J is the JIS or EIAJ standard code for +/-5% tolerance. <S> Here is a datasheet . <A> The "J" stands for 5%. <S> what is it used for? <S> so I would urge you to read Wikipedia's page on resistors <S> and hopefully it might enlighten <S> you given that you have the device in your hand <S> whilst all I have is a picture of it. <A> That's a 10Ohm fusible cement resistor with a 5% tolerance --->(J).	It's a power resistor and may be used for many things It's a 10 ohm resistor hence the label on the PCB called "R1".
Why does GPU consume so much power? I am trying to understand why GPUs consume so much power. For example, the max power consumption of the P100 GPU is 250W. As I understand power is measures in watts as current x voltage . Given a fixed power source (i.e. voltage), I would assume that GPUs draw a lot of current. If I understand this correctly, why does a GPU draw so much current? I think GPUs mostly consist of transistors? So why does the configuration of transistors in a GPU lead to more power being consumed than in a CPU? Thanks! <Q> A GPU is basically a LOT of simplified CPUs in parallel. <S> Each of them is not as capable and flexible as a real CPU, but there are thousands of them to give this massive parallel computing performance. <S> But this also means that it takes many billions of transistors to build a modern GPU. <S> And for logic chips we use FETs, so with every clock cycle all the billions of gate capacitances have to be charged and discharged. <S> This is where the large amount of power is going. <A> This question asks why a faster clock in a given CPU requires more power, and has a number of really good answers. <S> Take those answers, and add to them the fact that the power consumption is a consequence of not just how fast the logic involved is switching, but how much of it there is -- and <S> a GPU has lots and lots of logic, all of it switching madly when it's hard at work. <A> Relative to a high end <S> GPU, CPUs usually have many fewer transistors switching at any time and so do not require as much power. <S> This is not always true, you can get low power GPUs and enormous server CPUs with very high power requirements. <A> A GPU is a many, smaller processors operating simultaneously in parallel. <S> Each processor contains a lot of transistors. <S> Many processors in parallel means even more transistors. <S> Every time each transistor switches, it must charge or discharge a parasitic capacitance inside the transistor. <S> So the more often a transistor switches, the more often this capacitance is charged/discharged which means the more power is consumed. <S> GPUs run fast <S> so they switch very often. <S> So the fact that GPUs contain many transistors switching at the same time (because it's not just one processor operating at any one instant, its many processors operating at any one instant) and switching very often means very high currents to charge and discharge all those parasitic capacitors simultaneously. <S> A normal processor is more complex and has more transistors than than any single processor inside a GPU since it is has higher functionality, but it is also doing only one or two things at a time so only a small fraction of the transistors are switching at any point in time. <A> Note: This is intended as a short comment on the answer posted by @user1850479 <S> ( Why does GPU consume so much power? ); the only reason it exists as an "answer" is that posting a comment requires 50 reputation, which I do not have. <S> If anyone wants to help out by copying its contents to a (true) comment in the appropriate place, it's perfectly fine by me and this post can thus be deleted. <S> No attribution nor acknowledgement is necessary. <S> License: CC0 <S> Thanks. <S> :) <S> It's also worth noting that CPUs executing vector instructions (SSE, AVX variants) can draw more power than they otherwise would just executing standard instructions. <S> This can be observed by running 3D programs and forcing them to use software rasterizers -- in essence, forcing the CPU to take on a GPU-like workload -- such as <S> WARP or SwiftShader , which usually make heavy use of vector instructions. <S> Intel's processors even have AVX-specific power throttling functionality <S> that kicks in to keep the processor within its rated TDP .	GPUs consume a lot of power because they have a large number of transistors switching at high frequency.
How to light LEDs with Arduino without having any resistors I've got a silly situation. I have an Arduino board (2009). There's just one inbuilt LED but can't do much with it beyond blinking. I have five LEDs (3 yellow, 2 green), breadboards, jumper wires but not resistors. (And there is a total lockdown here) I wish to play with these LEDs but can't risk it putting it directly. I tried using them in series but that did not light them up.I know each LED needs a 330ohm resistor but that I don't have. I even have a 2x 7 segment display too but the same problem. Any way to use them?Thanks. Update: On suggestions, we have got two resistors from a faulty solder gun (blue-grey-orange-gold = 68k) and a (brown-black-brown-gold = 100) and 1 IN4000 diode. <Q> #include<avr/io.h>#include <S> <util/delay.h <S> > <S> int <S> main(void){ <S> SFIOR &= <S> ~ <S> (0<<PUD) <S> ; PORTC \|= 0x01; while(1) { DDRC ^= (1<<0); _delay_ms(10); }} AVR microcontroller has a built-in 10K resistor with every output pins. <S> Arduino should have libraries related to it. <S> They are called pull up resistors. <S> I checked it using the code I provided. <S> This screenshot is taken from the AVR datasheet. <S> You can see that there is a Resistor inside that can be operated by switching the internal MOSFET. <S> So, ask in the Arduino forum about the library associated to it. <A> Try this as the loop function of your Arduino sketch: void loop() {pinMode(2, INPUT_PULLUP); //Pull up is built in to the Atmega chip. <S> //It <S> will be at least 20k - <S> very little current will flow through the LED.// LED should light dimly.//May need to turn off room lights or cover with hands to see.delay(1000); // <S> Wait one second.pinMode(2, INPUT); //Input mode turns off the pull up. <S> No current to LED.delay(1000); } Connect an LED to GPIO 2 (or use another pin and change the code to match) with the LED cathode connected to the Arduino ground. <S> It may be very dim. <S> The built in pull up will be at least 20k, and may be as high as 200k depending on which processor is built in to your Arduino. <S> Information on "pinMode". <S> Information on Arduino pull ups. <S> The above is a trick, and may not work. <S> I haven't tried it out. <S> Whether it works or not will depend on how efficient your LEDs are and which processor you use. <S> Try it out. <S> It <S> cannot hurt your processor or your LED. <S> Do not change the pin mode to output. <S> That may damage your processor (though probably not.) <S> This is the same idea that Sadat Rafi had except using the Arduino library functions rather than generic Atmega code. <S> The same cautions about LED efficiency that I've given here apply to that answer as well. <S> Pin 13 was a poor choice on my part. <S> It is used for other things on the Arduino. <S> I've changed the code to use GPIO 2, which shouldn't have any conflict. <S> Do <S> NOT use GPIO 0, 1, or 13 with this trick. <S> Those pins are connected to other things on the Arduino board. <S> Those other things could cause damage to the Arduino if there are LEDs attached, and will almost certainly interfere with normal operation of the Arduino. <A> Even if you try to connect them in series of 2 (2.5V per LED) or 3 LED (1.6 v per LED), it is not advisable. <S> Don't you have a broken electronic device that you could scavenge for resistors? <S> Even a burnt CFL or LED lamp can have some resistors that you could use. <S> 330 ohm is just a minimum value, but as you want only an initial learning experience with Arduino, LED's values from 330 ohm to even 10k ohm can permit Arduino to light a LED safely. <S> What are the LED's colours? <S> Their voltage depends on colour. <A> If you have higher valued resistors, try those. <S> I've had up to 20K series resistance with a red LED (high luminosity, admittedly) <S> and it was still clearly visible in office/lab lighting. <S> You can go for more intensity later, but this may be sufficiently visible for bench debug. <S> How high you can go on resistance is probably different for each device, and I've tried this with exactly one sample, so YMMV. <A> Anything between about 250 ohms and 10K will work fine. <S> Surely you can sacrifice something and pull resistors out of it and maybe extend the leads a bit.	I'm sorry for you, but no, there's no way of connecting LED's to Arduino without resistors, without risk of damages to your Arduino's ports or the whole chip itself. CFL lamps have diodes, you could connect a series of 4 diodes to make a 2.4/2.6 V reductor, this series of diodes could permit using a lower value resistor (CFL lamps usually has a low value resistor, under 100 ohm).
Effect of length on a thermocouple's accuracy I'm a mechanical engineer working on acquiring some temperature data using thermocouples. Q. I wanted to know the effect of wire length on a thermocouple's accuracy. I don't have much of electrical engineering knowledge. I did some amount of search and people have mentioned that thermocouples don't experience much of voltage drop even when they are really long (100-150m, infact long thermocouples are used in powerplants for temperature measurements. I wasn't able to figure out the physical reason behind it. I want to particularly understand effect of voltage drop in particular, I'm not interested noise due to interference. Any leads would be really helpful! <Q> For example, the LT1052 (which would be great for a thermocouple) has a 30pA input bias current, which corresponds to over 100MΩ of resistance. <S> Wire cable even in the 100Ω range with a 30pA current would be about 3nV of error. <S> What does cause greater problems are temperature effects from other thermocouples, such as the connectors to the board and the thermocouple junctions formed between the pins of the opamp IC and the PCB. <S> It is necessary to make sure that the connector is thermocouple effects are matched to reduce error. <S> Another problem is the voltage offset of the op amp, because themocouple voltages are in the mV range, voltage offsets contribute to error, so look for a low voltage offset (and one that doesn't change with the temperature of the op amp much) <A> There's no significant voltage drop because there's no significant current flowing - thermocouple amplifiers have a very high input impedance. <S> You also need to remember that a thermocouple sensor does not generate a voltage at the sensing junction - the voltage s are generated along the length of the wire <S> s and the voltage generated along each wire depends on the temperature difference from one end of the wire to the other (from sense junction to thermocouple amp). <S> Because each wire is made from a different material, the voltage generated is different even though the length is the same, and the thermocouple amp measure the difference between these 2 voltages. <A> Thermocouples used for temperature measurement don't suffer much voltage drop because the thermocouple interface electronics don't draw any current from the thermocouple (ideally, they don't draw any at all). <S> You may have problems with other effects, but voltage drop should be negligible.	Wire cable resistance is insignificant when compared to the input impedance of the amplifier which is so high that the wire doesn't affect the measurement.
Failure of a C cell in a 4x battery pack, potential causes? I am trying to understand failure of cells in battery packs of 4 x alkaline C cells (in series, so ~6V output). The failed cells, from 2 separate battery packs, have bulged and leaked out, you can see the photo below: Unfortunately I didn't get a chance to see the battery pack as whole unit before I started investigating but I've been told that at least 1 cell (not clear if it is the affected one) might have been placed in reverse polarity in the battery pack. Would I be right in thinking that given that a cell might have been placed in reverse polarity, it is possible perhaps that somehow other batteries around it (with more charge than the affected one, possibly) could have caused a "reverse flow" of sorts and heated up the cell to the point that it vented out its innards? These battery packs were assembled on the factory floor so it is also likely that some of them were dropped on concrete floors before they went in the packs, seems unlikely but do you reckon the combination of the reverse polarity plus potential damage from a fall could have contributed to the failure? Difficult to get to the bottom of this with unknowns but I guess the jist of it is: Can a reverse polarity cell, in a series arrangement, be damaged by other cells around it? <Q> And it says very clearly on the battery <S> "Do not charge - may leak or explode" <S> ;) <S> Take a look inside a young-age-grade toy's battery box <S> and you'll find there's extra plastic or battery contacts added specifically to make it impossible to reverse-insert a cell, thus preventing parents from burning their kids this way. <A> Difficult to get to the bottom of this with unknowns <S> but I guess the jist of it is: Can <S> a reverse polarity cell, in a series arrangement, be damaged by other cells around it? <S> Most definitely if there is a load which will pass more than trivial current completing the circuit. <S> Consider something simple like a 4-cell flashlight. <S> You basically have 4 cells in series with a small resistor. <S> But if you turn one of the cells around, now you have three cells trying to charge one cell through a small resistor. <S> Given the high voltage and the low resistance, this will probably exceed the charging limits... especially for a cell that is not designed to be rechargeable to begin with. <S> But that doesn't mean you can rule out other damage mechanisms. <S> Or fakes. <A> Backwards could cause it, but it can also just happen. <S> I had one cell from a pack of four loose voltage well before the others. <S> this triggered an alarm so replaced them all. <S> If I had continued with them it could have reverse charged the weak cell and caused a leak. <S> after a second duracell destroyed my maglite <S> I stopped using that brand.	A reverse-inserted cell in a series chain of cells effectively is being charged by the other cells.
How does this surge protection work? This circuit signals an MCU via OU lead. I am not interested in the MCU side. I just show the concept. My question is very simple: How does the surge protection(actually ı meant detection) part work? By the way, I have basic knowledge of TVS, MOVs, opto-couplers and RC theory (phase etc.) <Q> It starts with an MOV at the L01 inductor input, which is sensitive to surges and outputs a voltage based on surge current through a nearby conductor. <S> But it only works with a minimum surge. <S> The MOV is to clamp over-voltage before it is fed into the bridge rectifier. <S> Regardless of surge polarity the bridge rectifier gives it a single peak value. <S> This peak is clamped by TVS01, then current limited by R011 before sending the current to an opto-coupler, which prevents violent surges from directly entering the MPU. <S> The MPU can be used to count surges over a certain current level. <S> With flash storage it can show surge activity per hour/day/week/month/year. <S> This can be tied in to storms or heavy motor activity. <S> To the bottom right another TVS device protects a serial port, either USB or RS-485. <S> This design has adequate surge protection, not including external SPD. <A> I can read and write Chinese. <S> So let me translate. <S> The schematic has 4 blocks on the top row, another at the bottom. <S> The labels are: Top: Sensor , Voltage limiter, Surge processing circuit, Optocoupler Bottom: <S> Remote input, alarm circuit, core MCU, long distance transmission circuit. <S> I think it is the top left block's L01 inductor that picks up the surge and kick starts everything. <A> How does the surge protection part work? <S> It may not work at <S> all - surge suppression is a "numbers game". <S> This means that any surge suppression method is reliant on assuming that the incoming surge energy is below a certain value. <S> If the surge energy is above that which the circuit was designed to withstand then the circuit fails to suppress and may sometimes become a hazard in its own right. <S> This is because devices can go short circuit and blow fuses. <S> For instance, the MOV in the top left circuit box will go short circuit if over-exposed to surge energy - so it needs a fuse else it might short-out a bona fide AC supply. <S> The selection of a fuse is a numbers game; surge suppression is a numbers game <S> so, where are the surge limit numbers? <S> If the surge is undefined then any circuit is meaningless. <S> If the devices are not linked with data sheets then analysis is also meaningless. <S> A surge suppression circuit is only as good as the specification it was designed to.	This is actually a surge detection device, so to work dependably it must have its own surge protection.
Does EC11 Incremental Encoder need hardware debouncing and VCC for encoder? I'm implementing an EC11E rotary encoder which has 3 pins + 2 pins for a switch. It seems like most of the projects I've seen online wire A/B and GND to the microcontroller, which to me, makes sense. However, at the same time, I've seen a few references to schematics which include debouncing capacitors and resistors like below. However, I don't understand why this is actually necessary. From what I understand, since the quadrature encoding effectively presents itself as a square wave, it shouldn't need debouncing, given the contacts either touch or don't touch. Do I actually need these debouncing capacitors and resistors in my circuit? As well, it seems like most breakout boards for encoders include a VCC line for pull-up. However, the MCU I'm using [AT90USB646] already includes internal pull-ups on all its GPIO pins. WIll I still need a pull up line then? <Q> Those rotary switches are nothing more than mechanical switches. <S> And like all mechanical switches, these will have contact bounce. <S> The contacts either touch, or don't, and they do that many times rapidly during the transition, and due to oxidation there always a state where the contacts have a bad connection so it can be indeterminate if the signal is high or low. <S> Some other rotary encoder manufacturers (Bourns) clearly state a default application circuit which includes pull-up resistor, a capacitor, and a series resistor, and also include default values for them. <S> The pull-up resistance is selected to provide at least the minimum specified wetting current to keep the contacts clean. <S> The series resistor will limit the current below maximum rating so that the contacts do not just short the fully charged capacitor directly. <S> And then the capacitor is there to act as a low pass filter for the contact bounce. <S> The pull-up resistance must be low enough value to make sharp enough rising edges. <S> Internal pull-ups in MCUs are quite high in value and thus can result in slow edges, especially if you have the capacitor there. <S> As the capacitor makes the edge rate slower, in some applications a Schmitt trigger buffer is useful to add hysteresis and square up the signal again to have sharp edges as in some applications the slow edges can be a problem. <S> The software just needs to cope with whatever garbage happens on the encoder lines, which means more complex software debouncing, dead time between detecting interrupts etc. <S> And if the pull-ups don't provide enough current, the encoder contacts may oxidize and bounce even more over time. <A> No, you don't need them. <S> You do need to cover the bounce period as specified in the datasheet using firmware methods (or equivalent if you are using an FPGA). <S> Fast (enough) sampling using a periodic ISR works well. <S> They do not actually debounce reliably unless you have Schmidt trigger inputs. <S> They help with possible ESD through the metal shaft. <S> Personally, I like to see some series resistance on any inputs, and the internal pullups on most MCUs are a bit high resistance for my preference (for noise immunity with the switch in the open state, especially on a one or two layer board), but you can certainly use the internal pullups if you want. <A> I have been currently using a EC11 rotary encoder, which I have assembled just like described in the datasheet with no debouncing methods, and it works just fine as stated in the datasheet. <S> No matter how fast I would switch through states, they've never bounced and it works very well. <S> You can check that on an oscilloscope, as you move it in the clockwise direction you'll see one of the pins switching from either 0 to 1 or 1 to 0 (depending on the initial state) in \$\approx <S> 40\,ms\$ <S> ahead of the other. <S> When moving counterclockwise, you'll face the same event with the moving pin order reversed though.	So, the parts are not necessary, and you may not need any of these external components.
PNP Transistor Not Working as Expected with Higher Voltage I am a novice with electronics, I may know what my issue is but wanted to put this out and make sure my thinking is correct. The below circuit is one I am using to turn on an LED when charging a battery and turn off when the battery is complete for an amplifier I've designed. What is happening is the LED never turns off. 0.74V is coming from a pin on the charging module's IC. When it turns off it is about 10V. This was working fine with a 12v AC adapter as I was waiting on my 15V adapter to come from China. Foolish me, I completed my PCB design and placed the order before I was aware of this issue. Sure I could just continue to use a 12V adapter but the circuitry is more efficient with the 15V when charging and listening at the same time. I am hoping I just need a different transistor, my current one is a 2N3906 . From looking at the datasheet I am thinking my Emitter−Base Breakdown Voltage of 5 is not high enough and if I switched to something like a KSA708 with an Emitter−Base Breakdown Voltage of 8 I may get the result I am looking for with this project. Any insight would be greatly appreciated, thanks. <Q> Your issue is that you have Q1 configured as an emitter follower. <S> So the voltage on the emitter will be about 0.7V higher than the voltage on the base. <S> With your 12V adapter, when the charge ends you didn't have enough voltage across the diode to turn it on. <S> With the 15V adapter you'll have about 10.5V on the base when charging is done, and 11.2V <S> on the emitter. <S> If your diode drop is say 2V you will have 15-2-11.2 = 1.8V across R1, and 1.8mA in the diode. <S> Probably enough that it still glows. <S> No easy solution by changing the transistor, you'll have the same result. <S> It has nothing to do with breakdown. <S> If you don't want to change your PCB you could try a bigger resistor for R1. <S> Increase it until the LED no longer glows and see if you get acceptable brightness during charging. <S> The better approach would be to use the transistor as a switch, but that's a PCB layout change or lots of cuts and jumpers. <S> JRE's suggestion of putting some LEDs in series would also work. <A> Any insight would be greatly appreciated Try adding a 10 volt zener diode (rough and ready level shifter) like this: - <A> use a FET instead. <S> a p-channel depeltion MOSFET or JFET will turn off when it sees 10V on the gate, and turn on for a low voltage there. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Even an NPN would always be only with 1K to 10K to 15V. Maybe 33K is OK or no 10k pull-up needed if using logic levels. <S> REF Summary: <S> Change PNP to NPN Remove 10K for any logic level 1V to 5V ON. <S> Invert output in software if possible. <S> Define input signal & state when ON/OFF with Voltage and Z. <S> Let <S> Ic/ <S> Ib=20 <S> roughly. <S> The left side works. <S> Right side (no Good)PNP will always be ON. <A> So I think I am just going to bite the bullet and amend my PCB design. <S> I tested this design last night but welcome all feedback.	Making this an appropriate high side switch with a 20K resistor on the gate and 1k as a pull-up seems to work.
Are external data connections considered "unintentional radiation" by the FCC? Does a device with an external communication port of some kind fall under "intentional radiation" or "unintentional radiation" according to the FCC for testing and certification requirements? For the purpose of this question, the external connection would be outside of the device's enclosure and must be connected via wire to communicate; it does not use the air as a communications medium. For example: Lower frequency ports: CAN Bus (which is the reason I'm writing) RS232 I2C SPI Higher frequency ports (Are there others of note for the purpose of this discussion?): USB Ethernet MoCA (multimedia over coax) Are there some above that would be considered intentional and others that are unintentional radiation? <Q> An unintentional radiator (defined in Section 15.3 (z)) is a device that by design uses digital logic, or electrical signals operating at radio frequencies for use within the product, or sends radio frequency signals by conduction to associated equipment via connecting wiring, but is not intended to emit RF energy wirelessly by radiation or induction. <S> Today the majority of electronic-electrical products use digital logic, operating between 9 kHz to 3000 GHz and are regulated under <S> 47 CFR Part 15 Subpart B. Examples include: coffee pots, wrist watches, cash registers, personal computers, printers, telephones, garage door receivers, wireless temperature probe receiver, RF universal remote control and thousands of other types of common electronic-electrical equipment that rely on digital technology. <S> This also includes many traditional products that were once classified as incidental radiators – like motors and basic electrical power tools that now use digital logic. <S> From https://www.fcc.gov/oet/ea/rfdevice <S> So all of these fall under sends rf signals via connecting wiring <A> Intentional radiation means rf radiation, i.e., a wireless transmitter. <S> None of the examples you mentioned are intentional radiators. <A> The answer is "Yes. <S> " None of your examples constitute intentional radiation. <S> If they have a transmitter, then they are intentional radiators. <S> They still need to satisfy all requirements of unintentional radiators except in the band where the transmitter is authorized to operate. <S> In that band, the intentional radiator band, they would be required to meet band-specific FCC regulations. <S> You may wish to enter the following into your favorite search engine: "title 47 part 15". <S> In many cases, compliance testing can be greatly simplified if you use a pre-certified module for things like bluetooth and wifi and gsm (or other mobile). <S> Outside the USA, different rules apply, but they are somewhat similar in regards to intentional and unintentional radiators.	Pretty much all modern digital devices would be classified by the FCC as unintentional radiators unless they have a transmitter.
Construction for a homogeneous field using permanent magnets Please let me describe what I am trying to achieve and I will pose my question in the end. I am trying to design a small chamber which can have a uniform magnetic field inside. I want to see if a design using permanent block magnets is feasible in terms of cost and construction. The volume inside the chamber is approximately 13cm x 13cm x 13cm. I am thinking of using an iron or steel yoke as shown in the picture below. I got the idea from this article in supermagnete.de . The picture below shows a top-view of my current design, where the green is a block magnet and the gray is iron or steel: To reduce construction costs, I am planning to laser-cut many pieces and stack them as shown below: This is an experimental prototype design, in which I care more about the field homogeneity rather than the field strength itself. My question is the following: Would such a construction produce a fairly homogeneous field? I am concerned that the size of the magnet and the geometry of certain areas of the sheet matter, but I am not entirely sure how to design for them. Given the technical details of the magnets, for example this datasheet , how can I estimate the field strength inside the chamber? With that in mind, is there any free software that can easily simulate the fields in such a geometry? I could use that to see the effect of different geometries and magnets. Thanks for all the help in advance! <Q> The magnetic field in the region between the arms would be relatively uniform in direction , but non-uniform in intensity. <S> Below, I've attached an image of the field that I made using very simple simulation software (note: <S> this is extremely simple software meant for didactic use only. <S> its results are not necessarily representative of reality); notice how the magnetic field is more intense (whiter) on the left side, near the permanent magnet, and less intense (greener) as you move to the right. <S> If I adjust the intensity a bit to better show it, you can also see that the field between the arms is far lower in magnitude than the field nearer the permanent magnet: <S> This may be good enough for you. <S> Or it may not. <S> Personally, I'd use a Helmholtz coil and just deal with having to wind the thing manually; it's tedious, but at least you only have to do it once. <A> Looking at the following three bits: <S> I want to see if a design using permanent block magnets is feasible in terms of cost and construction. <S> The volume inside the chamber is approximately 13cm x 13cm x 13cm. <S> To reduce construction costs, I am planning to laser-cut many pieces and stack them <S> This is an experimental prototype design, in which I care more about the field homogeneity rather than the field strength itself. <S> and this answer <S> then it seems you can have some fun adjusting the shapes of your pole <S> edges to try to get more uniformity of field strength <S> and, if you need it, of field direction . <S> There is another approach that you should consider <S> , it's usually the solution people adopt when they need to produce a volume of uniform magnetic field. <S> You could build a Helmholtz coil . <S> In the first image the field is in the vertical direction and the upper and lower coils have current in the same direction. <S> Sources: 1 , 2 , 3 <S> You can get a uniformity of about 1% over a significant volume this way, and with a single knob you can adjust the strength <S> If you need to cancel the Earth's field, or if you want to be able to point the field in any direction you can build three sets of interlocking Helmoltz coils and with three knobs you can point the field any way you like. <S> Here is a higher order three-axis set of coils that uses a second set in each axis to improve the uniformity even further and over a larger volume. <S> It is discussed in the Physics SE question <S> How are these “supplementary” or “satellite” Helmholtz coils used? : <S> Annotated screen shot from the NASA video MAVEN <S> Magnetometer <S> Here is a wooden form on which you can see three pairs of interlocking grooves for a three-axis configuration for applying varying but uniform field strengths and directions to the magnetometer on the deep space probe Mariner 3. <S> It is discussed in the Space SE question <S> Why is there a large wooden ball on Mariner 3's magnetometer? . <S> Source: <S> Photograph Number 293-6619Ac . <S> Image credit: NASA/JPL-Caltech. <A> You'll get a field. <S> It will produce a fairly uniform field, for some values of 'fairly' that you may not recognise. <S> There are FEMM solvers, some free, that will show you how horribly non-uniform a field you'll get. <S> It all depends on your specification, does uniform mean +/- <S> 10%, or +/- <S> <1ppm (as required in MRI)? <S> You have a strong axis of asymmetry, with strong magnets and a lump of iron off to one side. <S> This will put an overall slope on your field. <S> You have flat faces, so the field will be uniform over a tiny volume dead-centre, and fall off towards the edges. <S> You'll be able to counter this effect by reducing the distance between the faces as the distance from the centre increases, your laser cutting flexibility could come in handy. <S> With modern high field magnets, you may be better off abandoning the return field limb, and simply using two arrays of small strong magnets held onto two separate, facing, flat iron sheets by their own attraction. <S> The field could then be tuned to uniformity by adjusting the density of the magnets, or by adding iron to the ends to change the spacing. <A> Even the Helmholz coil standard for uniform magnetic field fails at points other than the exact center. <S> How uniform must it be? <S> Uniformity will be perturbed by magnetic substances like water or liquid oxygen. <S> The bottom line: You can’t get a uniform magnetic field over a non-zero volume, just like you can’t get absolute zero temperature.	In a hybrid solution, you could use permanent magnets to produce the bulk of the field, then tune for uniformity using 'shim coils' (as they're called in MRI imaging) to tweak up the field shape.
How can I prevent my IR sensor from sensing the sunlight? I want to use an IR sensor for object detection. An LED will be turned on if an object is detected. But because of sunlight, it's malfunctioning (i.e. always getting logic 1). So, how do I omit the interference due to sunlight? <Q> I'd like to expand on analogsystemsrf's answer a bit: <S> Exactly. <S> Sunlight doesn't change quickly. <S> If your illuminating IR LED is pulsed on/off, e.g. at 25 kHz (easy with a microcontroller), then you can simply measure the photocurrent at every pulse: Add all measurements when the LED is on, subtract all measurements where the LED is off. <S> When you write that down, you'll see that you're simply subtracting the background light, which stays constant. <A> WARNING --- if your circuit is just a photodiode and a resistor, then high amounts of DC_photon flux (the sunlight) will push the output voltage close to rail, and the circuit will become VERY UNRESPONSIVE to changes. <S> Thus you need to find a way to AVOID pushing close to rail/VDD/GROUND. <S> Thus I offer this: <S> If you run the photodiode current thru a normal silicon diode, you can get a LOGARITHM of the current. <S> This --- the LOG --- greatly extends your dynamic range. <S> As the reflectance of your "object" changes with different objects, you may need to calibrate any thresholding or detecting activity in the circuits. <S> And there will be temperature effects. <S> Thus pulses, or squarewaves, of photons may be your best bet. <S> However, you need to bias the photodiode, despite the sunlight; Or use a black_looking IR filter, taken from a discarded VCR. <S> To handle sunlight, I've very successfully used a parallel_load for the photodiode: a transistor to pull DC (sunlight) <S> current from the photodiode a resistor, perhaps 10,000 Ohms, to convert small fast photocurrents into voltages; connect this resistor from the output node (photodiode, resistor, transistor_collector) to ground. <S> Operate the transistor base from a low_pass version of its collector. <A> You will need a filter. <S> Sunlight doesn't vary rapidly. <S> All you have to do is to use high-frequency pulse for driving the LED. <S> And design a bandpass or high pass filter to receive the desired signal. <S> And if you are using a high-frequency microcontroller (STM32, ATSAM, etc.) <S> then you can implement digital signal processing (FFT/ IIR/ FIR). <S> You will find library function related to FFT/ IIR /FIR <S> on the internet. <A> In addition to pulsing, you can also pick a wavelength where the sun is less bright. <S> 1.5 <S> um for example is often used for that purpose, although it does not work with cheap silicon diodes. <S> 930nm is also sometimes used, since water absorption attenuates sunlight somewhat at that wavelength. <S> Add a bandpass filter at one of these wavelengths and you will detect a lot less solar radiation.	Thus pulses, or squarewaves, of photons may be your best bet. You can find an IR sensor with built-in bandpass filter in the market.
How can I make a current source that can supply 1uA? Is there anyway I can make a constant current source which has an output of 1uA or can I get it from the market? The load is around 200kOhms, and I need accuracy to 1%. <Q> D2 generates a stable voltage, while the forward voltage of Q1 varies with temperature (by about -2 mV <S> /deg. <S> C). <S> You can compensate for this by putting a standard diode in series with D2 -- e.g. a 1N4148. <S> Another transistor like Q2 with its base-collector tied together would also work well. <S> Then you need to change R1 as the current is then 2.5 V (from D2) divided by R2. <S> D2 provides a voltage of about 2.495V below the positive rail with 1mA+ flowing through it via R3. <S> That voltage minus Vbe (about 0.6V) appears across R1, so a collector current of almost that much flows, provide RL is not more than about 2.5M. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Of course if you need 0.1% precision you'd use an op-amp etc. <S> but that's kind of boring. <S> And, if you have budget and need something for the lab, something like the Keysight <S> B2901A/2902A will give you 1uA ±(0.025 % + 500 pA) at up to 200V+ for a small fraction of the price of a new car. <A> Or you may look for Linear Technology, Analog Devices, Maxim, ST,... their portfolio for current sources and applications. <S> It also depends if your current source has a common ground or common positive, and how precise it should be. <S> See, 1uA source <S> /current example <A> A cheap and dirty way is to use a voltage source and place a really high series resistance on the output that dominates and the load resistance even as it changes. <S> That way, even as the load resistance changes, it doesn't really affect the overall resistance seen by the voltage source so the current remains more or less the same. <S> 10V in series with 10MegaOhms would give you that, as long as your load resistance never varies by a significant fraction of 10MegaOhms (1/100th or 1/1000th, for example). <S> Or you could go 100V and 100MegaOhms, but around resistances that high you have to start worrying about leakage currents since your resistance is approaching the resistance nearby insulating materials. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Not that large resistors are hard to come by, but in this case the advantage conferred is a wide range of acceptable load values (good compliance voltage). <S> That quality could be even more important if you are trying to do this with supply voltages less than 5V. <S> This circuit lets you source loads up to 4.5 \$M\Omega\$ from a 5V supply with modest impact on the current(about 3.5% reduction relative to 0 \$\Omega\$ load). <S> The current is set with R1 and R2, but is much more sensitive to R2. <S> The value of 1 \$\mu A\$ turns out to be a fortuitous value for this circuit, as standard resistor values for R1 and R2 get you very close. <S> If you wanted to tweak it a bit, you could replace R2 with two resistors to have finer control. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> An approximate design equation for the current through RL, in terms of the current \$I_1\$ (set by R1) is \$I_2 = <S> \frac{V_T}{R2}ln(\frac{I_1}{I_2})\$ <S> This approximate equation ignores differences in the reverse saturation current between Q1 and Q2, and is a high- \$\beta\$ approximation. <S> \$V_T\$ <S> is the thermal voltage, about 26 mV at 300K. <S> The equation is a transcendental equation, though, so you might find it easier to just tweak a simulation. <S> It should be noted that the output current varies approximately in proportion to absolute temperature in Kelvin, so if high temperature stability is required, this solution may not be the first choice.	You can use a precision current source like TI REF200, then use current mirror to multiply the reference current. The Widlar Current Source should be mentioned, famous for making small driving currents with modest-value resistors.
What's wrong with this audio amplifier and stereo to mono circuit? The image below presents part of a schematic of an MP3 decoder ( VS1003 ) with stereo -> mono conversion and audio amplifier to connect a speaker. The problem is that after assembling the PCB barely anything could have been heard. I (kinda randomly) shorted R17 and C28 node to AGND and after this I can hear the music (this basically removes C28 and LEFT audio channel), but the sound quality is still terrible with a lot of noise. I guess I did something wrong on the schematic. Can someone find a mistake here? What could be corrected to get this right? I would be happy to have even 1 audio channel. <Q> The circuit is designed to operate from a positive only rail so lack of a negative rail shouldn’t be an issue. <S> I think your problem is the DC biasing between the VS1003 and TPA2005. <S> The VS1003 outputs are biased around GBUF which is about 1.2V. <S> The TPA2005 wants its inputs biased around VCC/2 = 2.5V. <S> Since these are not matched you must put AC coupling capacitors in series to block the DC component. <S> The TDA2005 app note shows the preferred circuit for a single ended input: <S> The RI in this schematic is your R25. <S> The gain of the amp is 150K/RI <S> so you may want to lower your value way down (from 150K) to get more gain. <S> Make sure you change RI in both legs. <S> Choose RI low enough for desired gain, but CI high enough to maintain low frequency response fc at least 20Hz, as per formula in same datasheet: If you end up using a polarized cap for CI make sure the "+" side faces the TDA2005, because DC level is higher on that side. <A> There are two ways that the sound may become muted. <S> Firstly, thepower supplies to the VS1003 don't include a negative voltagesource. <S> That usually means that the 'ground' is not in themidrange of the AC output signals, and only a coupling capacitor(to block DC) will make a suitable connection to an externaldevice (amplifier input or speaker). <S> That DC-bias problem is not consistent with the observation thatthe sound can be heard after removing one channel (and <S> if theamplifier/speaker on the output has GBUF biased correctly,there is no need to modify the circuit as shown). <S> GBUF shouldcorrect this problem, but that depends on the common-moderange of the output amplifier, so a DC measure of both theabsolute voltage, and GBUF voltage, is advised. <S> Theother way that combining stereo signals can result in zeroingthe output, is if the 'stereo' is in fact monaural, andthe two channels are in opposite phases (basically, differentpolarity copies of the SAME signal). <S> Check the outputs for DC content, and if you have an oscilloscope,put left channel on X axis, and right channel on Y axis, and see ifthe display is a diagonal line, rather than the more nuanced squigglesof a known stereo sound source. <A> I've found the problem on my own - in the end it was very simple, rookie mistake. <S> Along with this C31 capacitor of 1uF, I added in parallel 100nF and 1000uF and all the problems went away. <S> It was just not filtered enough...	The DC bias can causean amplifier to saturate (hit its voltage rail, outputting onlya constant voltage with negligible signal content).
Equipment required to measure audio line-out voltage peak-to-peak? I want to measure the output voltage coming from the line-out of my tape recorder such that the output is 5 to 6 volts peak-to-peak by adjusting the volume control. I am new to electronics. Can this be done with a basic multimeter or will I need an oscilloscope to view the waveform? I am trying to follow a guide on loading a program stored on cassette tape onto an old 8-bit microcomputer and this is the recommended setting. <Q> If all you want is to find the peak value, one can potentially connect a diode and capacitor at the output of the tape recorder, and measure the DC voltage (something like a sample and hold). <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Also, an arduino can be set up to repeatedly sample the voltage and display the maximum. <S> Also, you can calibrate your multimeter with a square wave signal (because you have digital signal on your particular cassette) of known amplitude, then multiply tape recorder output with that multiplier. <S> Also, you can feed in 50-60 Hz pure sine wave of known amplitude into your tape recorder and measure the AC output. <S> Most multimeters are fairly accurate at least in the 50-60 Hz range, if the sine wave is not distorted (even more true if multimeter is legally certified). <A> Edit: With this method, ensure that you play a sinusoidal tone, e.g. 1kHz, but it will save you an oscilloscope... <S> There are numerous tone generators online such as this . <S> Keep things simple. <S> Then multiply the RMS value by 1.414 to get the peak value. <S> With this method you won't need an oscilloscope ;) <S> You can see more about the relationship between RMS and peak here . <A> I am trying to follow a guide on loading a program stored on cassette tape onto an old 8-bit microcomputer and this is the recommended setting. <S> While a scope would be wonderful if you have one, the cheapest solution here will be trial and error . <S> Adjust the tape recorder to different settings and see which give you an accurate result. <S> It may also be worth using a computer sound card to capture these recordings and digitally archive them, using PC audio software to get a nice consistent level, and then learn what PC audio settings product a good playback for interpretation by the target device. <S> (If you are going to compress the files, use lossless data compression, not a perceptual audio codec, though chances are the signal is simple enough that even something designed around human hearing rather than data storage would work to a degree if the bitrate were sufficiently high) <S> But really the best "compression" would be to extract the original digital data from the recording. <S> With a little care you could probably software demodulate the signals, store them as digital data (rather than digital audio samples) and re-synthesize them on need, eliminating all analog degredation, tape speed variation, etc. <S> If you then know any tokenizing employed by the basic interpeter you could probably extract textual source files. <S> Most likely some retro computing hobbyist has already done this... if you look around you may find a python script or something. <S> You are of course welcome to treat this need as justification to purchase an oscilloscope, but if you do so, get a "real" one - eg, something in the conventional form of a small bench instrument with its own display, control knobs, and at least two channels. <S> The less expensive alternatives all but invariably turn out to have too many limitations and quirks to make good purchases as a first oscilloscope - to the extent that they are useful, it is as compact portable devices, devices with data capture capability, etc for experienced users with a good understanding of the compromises made by the tool and how to work around those and interpret a display which may not be showing the real situation. <S> Traditional CRT-based analog scopes may occasionally be available on the used market and if operating and truly inexpensive (eg, $20 - $50), those might be a budget alternative. <A> You will need an oscilloscope. <S> However not a very complex one. <S> 10MHz should do it, and you don't even need a digital storage oscilloscope to measure amplitude. <S> i.e., analog scope works fine for this. <S> I'd suggest two paths. <S> Either go on eBay and find a used scope. <S> Or buy the infamous Rigol DS1054 or similar if you intend to pursue electronics further. <S> Anyway, old scopes are also good shelf-filler since they look cool.	Use a multimeter with a true RMS function, which measures the effective dc equivalent of the waveform.
Why don't we make CPUs with 1000s of layers to make use of space in the third dimension? I am wondering why don't we make processors such as CPUs with 1000s of stacked layers to make use of space in the third dimension now that we have three-dimensional transistors. To be clear I'm referencing making something of a rectangular-prism shaped processor. To be clear there's a vast amount I am not aware of when it comes to processor manufacturing, I am not an electrical or computer engineer, but I am very curious. I am aware of the heating problems this would cause considering an even more dense packing of transistors and the manufacturing problems considering you would have to laser etch through so much silicon, but unlike enlarging the size in width and height there wouldn't be problems with making the most of the circular silicon wafers, and unlike with that you would be keeping the parts very close which means it wouldn't slow down the speed of the electricity getting from one part of the processor to the other because the processor already has 1000s of transistors stacked horizontally and vertically. I am curious if you could solve the heating problems by laying down thin intermittent heatsinking layers, while keeping the vertical throughput. And fix the manufacturing problems partially by using separately etched wafers every 10 layers or so. Could this be possible or are there lots of problems I'm not thinking of (and I'm sure there are)? Thanks. <Q> Yield. <S> Every time you do a process step, you get less than 100% perfection. <S> Let's say you get 99% perfection per step. <S> In a process with 20 steps, you would be down to 82%. <S> In a process with 1000 steps, you would be down to 43 ppm, 43 successful builds for every million wafers started. <S> Heat. <S> Our existing designs are already limited by how fast we can extract heat from the bottom of the die. <S> So neither the opportunity to generate more heat, nor the opportunity to generate that heat further away from where it can be dissipated, is of any real use to us. <S> Those said, there are some devices building up into the 3rd dimension, bonding several finished wafers together, which mitigates the yield issue. <S> Those stacked wafers tend to be memory, which don't use anything like the power of a CPU, which mitigates the heat issue. <A> Heat removal is the issue. <S> Already some chips have higher energy density than a nuclear reactor. <S> Consider a hair_drier ---- 1,500 watts with a air_blast fan to cool the tungsten coils. <S> And the coils glow dull red. <A> But what would you get from that? <S> The number of transistors per mm² of mask you get would still be the same, you'd just have more masks <S> Alignment of multiple masks is way harder, the more masks need to be aligned. <S> You'd probably need multiple extra interconnect layers for each extra transistor layer Making a connection between layers is more effort than making a connection within a layer. <S> Heat dissipation would be worse <A> Transistors are far easier to make on the bottom layer, because the traditional structure involves "n-well" or "p-well" structures . <S> Also: Planarisation . <S> The bottom "substrate" layer is mechanically polished to a very high degree of flatness. <S> Subsequent layers on top are etched and deposited, but each time is less than perfect. <S> There is a risk of errors adding up causing features not to align properly on a "lump" in the surface. <A> There are 2 main reasons: Heat dissipation - this requires contact surface. <S> That surface transfer heat from the CPU to the cooling system. <S> If it's 3D, it becomes extremely hard to evacuate heat from the under-layers as the surface-of contact would be significantly less than needed to sustain the thermal transfer. <S> Yields - they are low in many cases anyway. <S> When nVidia was trying the GTX 285/295s they had initial yields of under 2% and after process stabilization <S> they were still under 10%. <S> That meant they had to cut-off parts of the chip to make lower class cards out of the remaining good portion. <S> And that was with a standard 2D process that was just over-sized. <S> Trying to put anything in 3D would have even lower yields if we were to ignore the heating part. <S> In addition, adapting the fab process completely (assuming everything else is fine) is not something many would be willing to just go forward with while there's more to be had from the current tech. <S> HBM Memory managed to do things in stacks. <S> It's not really 3D, it was called 2.5D due to having only a few layers <S> and it is an expensive solution. <S> The package size is large, and it comes with thermal management challenges (even if the heat generated is significantly lower compared to CPUs). <S> The advanced chip packaging technology that vertically connects DRAM chip dies using electrodes that penetrate the microns-thick dies through microscopic holes came to the rescue in this case.	The two killer reasons are yield, and heat.
How should I design a heat sink for an SMD power mosfet? I really like the specs for the Infineon OptiMOSPower-Transistor IPT007N06N, but how would I keep this beast cool? Usually I use regular aluminium heat sinks (with water cooling or a fan) for MOSFETs in this power category, but this has to be soldered onto a PCB. Would I create a larger copper area and then attach a heatsink to this, or is there a way to keep the PCB itself cool? What is the usual the strategy here and are there any sample designs? Datasheet of IPT007N06N (pdf 1.2 MB): https://datasheet.octopart.com/IPT007N06NATMA1-Infineon-datasheet-43351903.pdf <Q> Depends on your power levels. <S> From the datasheet you'll get a temperature rise of 62C/W on a minimal copper area, but this goes down to 40C/W with a larger cooling area. <S> A multilayer plane (if you can fit it) with plenty of vias for extra area and copper makes 6cm^2 not all that much. <S> If you look through datasheets for other TOLL-88 packages (or similar, Infineon has weird packages), you'll usually find recommended PCB layouts. <S> At 0.75mOhm rdson, 25C ambient, and Mercury in retrograde you could put 56A through that part before hitting 120C on your junction. <S> The datasheet says you can go as high as 175, but I never take power semis even close to that. <S> (With the minimal copper area you're looking at ~45A to hit 120C). <S> If this still isn't enough for you, then start looking into heatsinks. <S> The whole point of these packages is to save space, and heatsinks do not do that... <S> Of course, my analysis only works if you are not switching frequently as it does not take turn-on and turn-off losses into account. <S> It assumes your max ambient is 25C, etc, etc. <S> I'll leave calculating your actual power consumption to you. <A> You can buy heatsinks that can solder to the PCB copper but, for this device, if you look at the thermal characteristics in the data sheet on page 4, you will see that directly attaching a heatsink to the case is probably you best bet. <S> The typical value of thermal resistance from junction to case is 0.2 degC per watt. <S> Something like this ought to be an option: <S> - Available from Farnell . <S> There are plenty of options to consider and plenty of offerings from the usual heatsink suppliers. <A> Cooling from the bottom: I know from some "Gansystems" applications, that they attach heatsinks on the other side of the PCB. <S> This will require multiple vias in the component footprint to get the heat down to the other side of the PCB. <S> But this strategy is viable, and maybe you have a lot of space on the back. <S> Cooling from the top: If you desire 1 heatsink per 1 component then a gluetype heatsink will do. <S> If you require 1 heatsink for multiple transistors the transistors might not be mounted exactly equal in height or planarity.	Then go for a heatsink with screw/bolt mount into the PCB with a heatpad to get proper thermal connection.
Differential amplifier and different ground I am designing a circuit to measure the voltage across the output of buck converter. With different ground for the buck converter and the micro controller, with respect to which ground will the ADC measure? For the circuit shown above, the ADC will measure the output with respect to which ground? Isn't the input to the ADC floating? If the above circuit is not a good design practice or if it doesn't work, will the below circuit work provided the output voltage of the first op amp is within the voltage range of the second op amp? <Q> The grounds are floating. <S> There are two problems with the designs: <S> The design will be susceptible to any kind of noise, because the GND2 is floating with respect to GND1. <S> EMI could cause GND2 to be a higher voltage (and the second board will function like an antenna) <S> There is no return path for the current flowing from the board with GND1 to the board with GND2 (in the diagrams in the OP). <S> This means the only return path for current will be through the air and the current will be very very small, not enough to overcome the input bias current needed almost all amplifiers. <S> You'll need some kind of connection between grounds. <S> These amplifiers use chopping to pass the analog signal through an isolation transformer, then reconstruct it. <S> (you still need to provide a ground from one board to the next, but the grounds are not tied together on the same board). <S> If your trying to save on wiring, there is simply no way to pass an analog signal between boards without a return path for the current (it's like trying to get current from a battery without connecting a wire to the other terminal). <S> Source: <S> https://www.analog.com/en/products/adum3190.html <S> will the below circuit work provided the output voltage of the firstop amp <S> is within the voltage range of the second op amp? <S> For the circuit shown above, the ADC will measure the output withrespect to which ground? <S> Isn't the input to the ADC floating? <S> will the below circuit work provided the output voltage of the firstop amp <S> is within the voltage range of the second op amp? <S> Probably not, it will depend on the voltage of GND2 and if GND2 is only connected by air, then no current will flow between boards and there will be no way for a voltage to be determined by the buffer amplifier. <A> In these topologies you would ground R3 to your microcontroller ground GND2, not GND1 if you want the amplifier output to be the voltage across R2 referenced to GND2. <S> The TLO81 doesn't have a common mode input range that includes ground. <S> Select another amplifier or use split supplies. <S> 100 ohms is kind of low, 1K is probably the minimum you would want to use. <S> You probably want to use much more if you want your two grounds isolated. <S> How far apart can your grounds drift? <S> If there's no limit you can't guarantee being within the common mode or <S> input range specs on your op-amp. <S> The common mode rejection will be dependent on the matching of your resistors around the op-amp. <S> If I remember right 1% resistors get you about 48dB worst case (or somewhere around that). <S> So that should be factored in as an error term along with offset, etc. <S> I don't see a need for an additional buffer after the diff amp. <S> For better accuracy and isolation between the two circuits you could consider an instrumentation amp, or even an isolation amplifier. <A> But that's not the case here is it?. <S> The Opamp is referenced to GND1, so its output will also be referred to GND1. <S> The ADC won't be able to read the value. <S> I do see one solution though. <S> Please correct me if I'm wrong. <S> I see that you are measuring the voltage across R2, and the gain is 1. <S> The voltage across R2 is 3.3V. <S> So the output of the opamp would be 3.3V. <S> If you can increase V2 to 5V, and power the opamp with V2, then the ADC will read 3.3V. This way you would be able to sense up to 4V without saturation, that's sensing rage of 0-36V. If you need to increase the range, increase V2 accordingly. <S> That circuit will work. <S> You won't have to connect the grounds together.	There will be no reference for the ADC to base the voltage because GND1 is floating, it will pick up a reading but it would be similar to connecting a wire to the ADC and air. The ADC is always referenced to it's ground GND2 (many ADC's have built in voltage references that are referenced to ground). If you want to isolate the grounds between boards use an isolation amplifier such as the adum3190 .
Cause a voltage drop of about 1V I am currently working on a battery powered project (4 x AA rechargeable batteries) with an Attiny (max vin = 5.5V), which works perfectly so far. But if I would insert normal batteries into the holder instead of the 4 AA rechargeable batteries (the voltage of a normal battery is much higher than that of a full rechargeable battery), the maximum input voltage of the Attiny would be exceeded, which wouldn't be so great. For this reason I am looking for the most efficient, cheap and above all power-saving method to lower the total voltage of the 4 AA batteries or rechargeable batteries by about 1V, so that it doesn't matter anymore if I use rechargeable or non-rechargeable batteries for the project. Should I simply use a diode (voltage drop around 0.7V) or resistors? What are your suggestions? <Q> You are in need of a voltage regulator . <S> A diode drop is not ideal (no pun intended). <S> It's only 0.7V for a certain current for a certain diode make for a specific diode. <S> i.e. Given several diodes, they will all measure a little different. <S> Buck regulators are very efficient nowadays and can commonly be found in the upper 80%. <S> If you don't need 5V, then you can skip the extra battery. <A> There are three ways to create a voltage reference: <S> Resistor and diode (least stable and accurate, but least expensive) <S> Voltage regulator (more expensive stable in the ~1mV range depending on the IC) Voltage reference (built to provide a voltage reference that is stable and accurate below 1mV and sometimes into the uV's) <S> If you need something that is more stable than 1mV, I'd use a voltage reference. <S> This article goes into more detail: https://www.nutsvolts.com/magazine/article/build_a_01_accurate_voltage_reference <A> A diode is the easiest place to start. <S> A 1N4004 drops about 0.8 V, and a 1N4148 drops about 0.7 V (both dependent on the actual current through the diodes). <S> Those calculate out to above 85% efficient. <S> The downside comes when the batteries are partially discharged. <S> Now you don't need the voltage drop, and it is subtracting from the useful battery life. <S> An LDO (low drop-out) regulator circuit can get you the regulation at higher voltages, but only about 0.1 V drop as the batteries discharge. <S> However, it is a more complex circuit.	It really depends on the accuracy needed, and how stable (temperature affects most electronic devices and can cause them to vary) I would suggest adding another battery in series, then use a buck regulator to turn it into the low voltage (5V or 3.3V) that you need.
Is most of non-isolated power supply "safe"? I need a large power supply (a little over 1kW) for a consumer appliance that only has a single on/off switch and some non-conductive handles. An off-the-shelf isolated supplies run at least $60 for a sketchy Chinese supply that only supports either US or EU mains, whereas I could probably produce something similar to the following schematic for <$20 that would support both US and EU mains: A lot of the cost savings would just be because I can afford to run the supply "dirtier" (higher ripple current) on input bulk capacitor(s). Even the output can be somewhat dirty, but I would plan to clean it up by using "smart" switching on the input to the buck converter. However, safety is a concern. While from what I could tell from regulations there is no requirement that the power be isolated for my application, I would feel pretty bad (not to mention the financial ramifications) if someone got hurt somehow using a non-isolated design (e.g. got water into it and got shocked, if that's even possible). However, it seems like only a small portion of a non-isolated circuit is actually dangerous. As annotated in the schematic, is it true that mains is only dangerous to the extent that high voltages are involved, and that stepped-down but non-isolated voltages later in the circuit are "safe"? Also note: I would plan to earth ground the metal case of the device (such that a short to live, and potentially neutral, would blow a fuse). And also, I would definitely have whatever final design I came up with reviewed by some experts; I'm just wondering if this is a plausible direction, given that using an isolated supply would probably make the product economically infeasible. Edit: I'm going for 48V and actually closer to 1.2kW. Sorry for not clarifying, but that ground symbol was not meant to mean that the circuit was connected to Earth (which would short the bridge), but rather just something I placed as required by the simulation software. The output side would be disconnected from any of L/N/G prongs, which I guess could cause some common mode voltage swing. However, the metal case would be earth grounded (and isolated from the rest of the circuit). Then if the live or the ground reference of the output side connected to the case the fuse would blow, but I guess if the pre-bridge neutral connected to the case, that may not be detected if there isn't an RCD (or I could add a low-rated fuse on the Earth ground path). <Q> No, even if that was only a 5V supply for digital logic, it is hazardous and unsafe. <S> So at least your rectangle of unsafe voltages is too small. <S> It will extend to all components and wires going into the box. <S> That is because what you drew as the 0V or ground symbol, will actually have peaks of minus 325V in EU and half of that in the US during the negative sine wave cycle. <A> It is not only unsafe as stated by Justme, it will also blow imediately one of the rectifier diodes (one of the two diodes connected to GND on the left side of your rectifier symbol) because it shorts the AC input. <S> The reason is that one of the AC terminals is also grounded which is not shown in your diagram (the other one of those two diodes will never be conducting). <A> Look at how appliances are designed to be safe. <S> Something like a stove may have isolated control circuits but the power circuits will not be isolated from the mains. <S> Similarly with an air conditioner, a washing machine and its VFD motor controller, even a lowly microwave oven or toaster oven. <S> That means that the low power parts may need an isolated supply and some control circuit isolation and anything connected to the mains needs to be protected against accidental contact, for example by the ceramic heater insulation in a stove element, surrounded by a grounded sheath. <S> If the insulation fails, a circuit breaker or GFCI trips and removes the power.	So your supply voltage is not safe and it must not come contact with humans and it must not have connections to any other devices external to the box. There are no appliances I can think of off the top of my head that use 1kW-ish power and have that power isolated before it is consumed.
Do LEDs (of the same type) in parallel consume different current? In a part of my circuit, there will be three LEDs (of the same type) in parallel. Is it true that they will not automatically get the same current and have the same brightness? Somewhere I read about that. The LEDs are: UV Voltage Typ. 3,4V - Max. 4,0V Current: 20mA Typ., 30mA Max Up to now they do not have series resistors individually, but all three have one. The whole thing will be an ultraviolet light show, regulated by a Raspberry Pi via 16 GPIOs, each of which will activate a mini panel of three LEDS. <Q> Yes, as a load such as your UV LEDs, when connected in parallel will consume the same amount of driving voltage by the virtue of parallel circuit combination but will consume varying levels of current depending upon the internal & external resistances associated with the UV LEDs & the circuit itself. <S> Hope <S> this answers your question. <S> If you have any other queries hit me up :) <A> The V vs I curve will be slightly different for each LED even if they are the same type. <S> This is simply due to manufacturing tolerances. <S> This means if one LED starts to consume more current, its voltage will rise a bit, in turn reducing the current. <S> Current sharing based on the internal resistance of LEDs is not the most accurate though. <S> If you want accurate current sharing, use a separate resistor in series with each LED. <A> Given the exponential nature of the relationship between current and voltage, even the smallest variation in forward voltage results in significant current change. <S> Even among "matched" LEDs, there is some variability due to the intrinsic non-ideal characteristics. <S> A simple way to look at this is by examining the well known exponential curve for a diode: $$I_D = <S> I_S <S> e^{V_D/ <S> V_T}$$ <S> Where <S> \$I_S\$ <S> is the scale current, \$V_D\$ is the forward voltage across the diode, and \$V_T\$ <S> is the thermal voltage at room temperature ( \$V_T=26\text{mV}\$ ). <S> Say now that you have two diodes, "matched" ( <S> \$I_S\$ the same for both) diodes with currents \$I_{D1}\$ and \$I_{D2}\$ , then the ratio between those two comes back to be: $$ \dfrac{I_{D1}}{I_{D2}}=e^{\frac{V_{D1}-V_{D2}}{26\text{mV}}}$$ <S> Which means that if \$V_{D1}-V_{D2}=26\text{mV}\$ , that the ratio of \$I_{D1}\$ to \$I_{D2}\$ is 2.7. <S> A difference of 60 \$\text{mV}\$ , represents a ratio of 10, you get the idea. <S> It shows that one of the diodes will be taking up a lot more current than the other pretty quickly. <S> Now imagine that \$I_S\$ <S> isn't exactly the same for the diodes, which is the real case, then you could see that the ratio of the \$I_S\$ will scale the current ratio further. <S> You may ask why should there be a difference in the voltage if the diodes are place in parallel and that goes back to the non-ideal characteristics ( <S> like \$I_S\$ <S> being different). <S> Think of each of the diodes having some internal resistance in series, in that case the perfect parallel connection situation falls apart. <S> All this analysis could have been done without using the exponential model, and instead using a linear approximation of the exponential model around the nominal \$V_F\$ of the diode--that would more clearly show the "internal" resistance since the model would have the form <S> \$V_D = V_{F0} + I_D R_{Di}\$ , where \$R_{Di}\$ is the slope of the linear approximation of the exponential I-V plot around the nominal voltage \$V_{F0}\$ . <S> Hope it helps.	If the LEDs are from the same batch, you should be able to get away with a single current limiting resistor by virtue of the fact that as that each LED has an internal resistance (albiet small).
How does charge redistribute in a capacitor? Suppose, you have to capacitor plates as shown: And, according to the solution key to a problem which I was doing, in the steady state behaviour of the circuit, the charges would redistribute such that each plate gets charge of equal magnitude. Like so: Now, this was the method taught to us for solving problems where extra charge is introduced into capacitors but I'm not sure why the charges redistributed such that each plate has average of both plates? Why does at end state the charge redistribute so that each plate has the same amount of charge <Q> If you provide a path, the electrons may move from/to a plate. <S> If there is no path, the electrons have to remain on a plate. <S> You have not drawn any paths. <A> I think you may have created an anomaly in the way you have drawn your picture. <S> Capacitor plates that are parallel and of the same size will have equal and opposite charge. <S> But for there to be unequal charge on two capacitor plates there needs to be a difference in the plate areas and, this creates "fringing" to a third party (usually ground) like this: - The overlap area between left plate and upper right plate will have +500 and -500 values for charge. <S> The lower right plate (representing the rest of the universe) will have +200 and -200 charge values. <S> You could also redraw it like this: - <S> But, by definition of a capacitor, it is a device that HAS <S> equal and opposite charges on its plates meaning that the +200 charge surplus on the +700 plate has to produce leakage flux to other stuff. <S> This means that if the "other stuff" is a much greater distance away than the two larger plate's gap, then the net average voltage on the two larger plates can be quite large with respect to the "other stuff". <A> This is actually a simple result of Gauss Law. <S> Consider a cylindrical closed surface whose cross section is shown below (in dotted lines): <S> Due to the symmetry of the plates the electric field is pointing in direction perpendicular to the surface of the plates as shown in the figure. <S> So how much flux crosses the cylindrical surface? <S> The answer is zero because the electric field is pointing along the surface in air/dielectric and electric field is zero inside the conductor (of-course assuming electrostatic conditions). <S> From Gauss's Law, $$\Phi_E = \frac{q_{enc}}{\epsilon_o}$$ <S> Since \$\Phi_E\$ is zero <S> the net charge enclosed inside the cylinder is zero. <S> The final charge configuration is thus, as shown below: Note that inner surface of the plates have equal and opposite charges and outer plates carry the residual charges.	This is only possible if the charges on the two plates are equal and opposite.
Why does attaching the case to the PCB ground plane cause the audio to cut out on my cassette player? I bought a NIB (new in box) cassette recorder on eBay recently but have had problems with the audio from the start. The issue was that touching the dials could cause the audio to crackle, go quiet, or cut out entirely. It took some time and a lot of triage to narrow the issue down to something unexpected. The problem is with the screw that connects the PCB ground to the case. The reason it appeared to be the dials was because the PCB was being nudged slightly against this screw (it was very confusing to locate this problem!). Here is an overview of the PCB: This is the point at which the PCB ground plane connects to the metal inside (the housing is all plastic): The issue is when I add the screw here and begin to tighten it. Screwing it in enough to achieve continuity between the case and the ground plane does not create any noticeable distortion. But if I continuing to tighten it then the audio will begin to crackle, become very quiet, and then cut out entirely. This only affects the audio signal; the tape continues to play as the motor spins. I tried adding a steel washer to ensure a good contact with the ground plane but it made no difference with the previous description: I made sure that this was not a fault of the PCB due to pressure or flex. I flexed the board manually and applied pressure to the location where the screw would be with my finger as well as the back of my plastic screwdriver. The audio is perfect without the screw being present but this also remove all continuity with the case: What I don't understand is why this only happens if the screw is tightened. Simple continuity with the case does not cause the audio to cut out. For example, the tape is playing just fine as I bright the ground plane with the case here: This issue occurs whether the cassette player is powered by batteries or the DC barrel jack. The player also has an option to be powered by AC with what I think is an AC/DC converter inside (not visible in the photos). I could use a non-conductive screw or a plastic washer to "fix" the problem but then the case is no longer grounded. I am interested in a solution that solves the problem safely but I also want to u the audio cuts out only when the screw is tightened and not just from achieving continuity with the case. I am new to electronics. Shouldn't the case always be connected to ground for safely in the event that it somehow becomes live? UPDATE I added a small piece of heat shrink between the screw and the ground plane contact to test for sure whether or not this was a mechanical issue. I was able to tighten the screw fully without any of the previously described issues. This does not appear to be a problem with mechanical stress around the area. Continuity between the screw and the case was confirmed as well as no continuity between the screw or case with the ground plane of the PCB when the heat shrink was used. UPDATE The answer provided by JRE indicated the cause of the problem. Sure enough, there was a large filter capacitor on the reverse side of the board that was just barely touching another metal standoff. The degree of which it touched probably changed as I handled the board but it would always short itself when the ground screw was fully tightened. I bent the capacitor back and the problem went away. It was probably some kind of filter capacitor that connected up with the speaker; grounding it caused the audio to cut out. I'm glad that I tested it only with batteries or my fused power supply connected. I guess that means that the chassis could have been live if I'd used the AC adapter? <Q> Your PCB is attached to a metal plate. <S> There are components on the PCB between the PCB and the metal plate. <S> If you tighten the screw really tight, the component touches the plate and causes a short to ground. <S> Take out the PCB, turn it over, and look for parts that stand up higher than the others. <S> See if those "too tall" parts can touch the metal plate. <A> Obviously, it isn't the electrical connection to ground that's causing the problem. <S> Instead, the physical flexing of the PCB is either causing an open or a short somewhere nearby. <S> In older equipment, it would not be surprising to find a hairline crack in the PCB itself or one of the nearby solder joints. <S> Or, it could be that the tip of the screw itself is touching something that it shouldn't — possibly a wire whose insulation has been abraded. <A> Sounds like an issue with the potentiometers <S> the dials are manipulating. <S> The sound issues you are having are common symptoms of a dirty or failing potentiometer.	As long as the screw isn't completely tight, there's a tiny gap between the component and the plate. Very likely there is a component that is just a little too tall.
Using a 9V battery instead of a DC power supply I have a DC power adapter that has the following specs: Input Voltage: 100-240V AC, 50-60Hz, 0.5A Output Voltage: 9V DC, 1.5A I am interested in taking a 9V battery and a snap adapter so I can use my device away from my outlet. However I am not sure how to ensure the output will be 9V DC and 1.5A (specifically the 1.5A). How can I take a 9V battery plus a snap adapter and create a power supply that would be equivalent to the DC power adapter? Do I need to add a 6 Ohm resistor in the loop to ensure that I = V/R or is that too simple? <Q> 9V/6 ohms= 1.5A which will kill the very cheap low power batteries extremely quickly. <S> Why kill the battery with a parallel resistor? <S> Rayovac does not show performance graphs <S> so I show one of an Energizer cheap old 9V Zinv battery like what you asked about. <A> 9 volt batteries are designed for low current long life applications. <S> Even if you could draw 1.5 Amps at 9V in an ideal setup, with a typical 500 mAh 9V alkaline battery, you would get less than a 3rd of an hour life on it. <S> In reality the battery chemistry will reduce the voltage and current capacity as the current draw increases. <S> From https://www.powerstream.com/9V-Alkaline-tests.htm <S> you can see that a 1 Amp draw for a good battery may only give 90 mAh of capacity and the voltage drops below 7V real quick. <S> That means you will get 7 minutes of battery life. <S> Expensive. <S> Ideally what you need is modern technology. <S> Aka a nice lithium battery pack with 9V output. <S> You can find these offthe shelf with a usb Power delivery enabled power bank with a USB PD or Quick charge trigger module to get 9V out. <S> These will have safety measures in them and should easily give a few hours at what you need. <S> Alternatively hack your device. <S> Some devices will have internal regulators or components because they run at a lower voltage, and you may be able to bypass them. <A> The exact current that will be output depends on the circuit. <S> You can think of it like this: the adapter outputs a specific voltage and then the circuit draws <S> whatever current it wants till the adapters reaches its maximum. <S> Kinda like with water faucet in a bathroom, water pressure will offer to push water, and you controlling the faucet will determine the amount of water flowing. <S> Voltage is the "Pressure pushing on electrons to move", while Current is the rate at which they move, note the quotes as this is a simplified view. <S> Back to you case, a common 9V battery, will not be able to output 1.5A, they are usually rated for 0.25A max, with high quality ones rated to 0.3A. <S> You can either select a different type of battery or add 6 of them in parallel. <S> Adding batteries in parallel increases the max current and capacity while keeping the voltage the same.	Power adapter are rated by their maximum current and voltage, a 9V, 1.5A rating means it will output 9V max and 1.5A max.
Charging lead acid batteries in parallel? I recently got my hands on eight 12V 9AH lead acid batteries. Could I simply place them in parallel and charge them with a battery tender without any special measures? Are there any risks involved? Could it damage the batteries? <Q> Batteries don't often fail low voltage or short circuit, but if they do, then a 'very parallel' arrangement could be bad as all the good batteries gang up on the bad one to force a high current through it. <S> Protect each battery with a fuse in series. <S> Before connecting them in parallel, make sure one or more of them aren't duds, check each individually into a load. <S> You may have to charge them individually to do this. <S> Before connecting them directly in parallel, roughly equalise the voltages. <S> This is more important with lithium than with lead or nickel chemistries, but it's still worth making sure that you don't have large imbalances in the voltage, even with lead batteries. <A> Most lead-acid batteries charge at a constant 144 volts, so charging several in parallel is really just a charge-current issue. <S> If the charger cannot supply enough current it will likely lower the charge voltage to protect itself. <S> As the batteries charge up the voltage will rise, but should NOT go over 14.4 volts, or you could "cook" the batteries, releasing deadly vapors and ruin the batteries ability to hold a charge. <S> In a nutshell, you can do this but stay away from "wimpy" chargers. <S> Ideally it should have charge current equal to the sum of the AH values of all connected batteries. <S> This means their is a limit to how many you can charge in parallel, as the charge time gets to be untenable. <S> A good 20 amp charger should be fine. <S> Note that many lead-acid batteries may have a 13.8 volt idle charge rating <S> , so be sure before you invest in a charger. <A> Parallel connection of lead-acid batteries is done routinely in a lot of cases - including almost all UPS devices, small boats, offroad cars, etc... <S> The more identical batteries are, the better. <S> They ABSOLUTELY must be the same voltage. <S> They MUST be of the same type <S> (flooded/gel/AGM, starter/traction/standby) <S> , it is good if they are the same brand and even better if they are all brand new. <S> It is also good for them to be in the same state of charge (i.e. fully charged) when you connect them first. <S> You get a battery pack having the same voltage as each of them, the sum of capacities and if the connections are symmetric - the sum of their CCA. <S> In most cases, if one of the batteries fails and you don't notice, the rest of them follow quickly.	Once you've connected them in parallel, you can treat them like a high capacity battery.
Why have extra resistor in switch debounce circuit I am using a Laird BL654 module in my project, and using the dev board schematic to help me. On the reset switch they have an external pull-up of 10k (R112) (in addition to the internal one on the module of 13k), which I understand is to give robustness against ESD. The capacitor C63 I understand is for debounce. What is the 150 ohm resistor for (R34)? <Q> For many reasons. <S> First of all, the external 10k pull-up is not connected by default, and the 10nF capacitor is not populated either. <S> So forget that those exist. <S> If there is capacitance directly over a pushbutton, pushing the button would short the capacitor and it would discharge with a large current spike. <S> These pushbuttons and capacitors are not rated for such large currents and repeated button pushing could damage the pushbutton contacts or the capacitor. <S> Also the pushbuttons are not perfect, there might be several rapid transitions shorting and releasing the button contacts when it is pushed. <S> This causes rapid current changes, or high di/dt. <S> The wiring has some inductance, so high di/dt causes voltage spikes due to the inductance, and the voltage spikes may exceed what the module reset pin tolerates without damage. <S> Also there can be high rate of change in voltage dv/dt which can couple capacitively to neighbouring wires. <S> This will result in lower changes in current, which will lower the voltage spikes from inductances. <S> Also dv/dt is lower <S> so there is less coupling. <S> So in short, the resistor is there to limit and slow down current and voltage changes so they don't cause trouble or damage. <A> R34 is to limit discharge current noise the switch. <S> As the switch contacts bounce a few times they produce a tiny amount of local RF noise which the 10 nF capacitor suppresses. <S> R34 takes this further and keeps the noise from the IC RESET pin. <S> You do not even want the trace to the switch to have noise on it as it can migrate to adjacent traces. <S> If the cap is not installed R34 still reduces slew rate to less RF is created. <S> Overall this is not much of a filter circuit, so possibly digital filters are used to be-bounce the switch. <A> yes I agree that limiting the large current spikes with a 150 Ohm resistor will in the long term save the switche's contacts.	So when the button has resistance in series, it limits discharge current from any capacitances to safe level that the button and capacitors can handle. Without knowing any details what the reset connection looks like in the module, it is quite typical for microcontrollers to have a capacitor on their reset pin, so inside the module the MCU could have a capacitor on the reset pin too.
In a circuit schema, why the voltage at the battery anode is conventionally 0V while the voltage at cathode is usually the voltage of the battery? I will try to explain where my question is coming from. Knowing that voltage is the electric potential energy per unit of charge at a given point.And knowing that electrons are pushed from - to go to + To me that means that the electrons won't have potential energy when arrived at +And they have the most when starting from -They should have more potential energy where they are leaving from and less where they are arriving. That would mean that electrons have more potential energy at - than at +So in an AA battery (1.5V) I would expect that the voltage at the anode (-) is 1.5V and the voltage at the cathode (+) is 0V But I am learning the opposite from my course and I do not understand why it is so. <Q> Conventional current is "upside down" with regard to electrons, due to some arbitrary choices of labelling made centuries ago. <S> That's why I normally suggest ignoring them and focusing on conventional current and voltage analysis. <S> Electrons are negatively charged. <S> Each electron has a charge of minus 1.602176634×10−19 coulombs. <S> You could label the "negative terminal" -1.5V and the "positive terminal" <S> 0V. <S> Then your negatively charged electrons flow away from the negative side of the field (because similar charges repel) towards the positive side of the field. <S> (Electrons themselves do not really "carry" energy, the energy is embodied in the field, and the ability of the field to do work on electrons is the energy.) <A> 0v point of a circuit is completely a matter of choice. <S> In the above drawing, yo can as well put 0v at point A <S> and then you'll have -1.5v at point B. <S> It is only the difference that matters. <A> The battery positive terminal is called the cathode: - Picture from Battery university <A> As I learned, the arrow of the voltage is commonly drawn from + to - like this: - + <S> \|\| \| <S> \| 1.5V\| <S> \| \|\|_\| - ˅ <S> Edit: Image of a complete schematic: <S> Note that all of U_0, U_R, I are positive and <S> of course U_0 = <S> U_R.	As the technical direction of current is defined as from + to - it totally makes sense to label the Cathode the higher potential because the energy of a positive charge is higher there.
Can someone explain the mysterious non-behavior of the DAC0808 and DAC0832, is there a difference between them on the output side? So I have been very happy successful with my R-2R network ladder DAC powered by 74LS245 or similar bus transceivers, with 1% resistors, for nearly perfect output voltage going from 0 to +5 rail, with pretty good precision . Enough to produce voltage ramps to drive the X and Y inputs of my oscilloscope to produce a pretty decent raster display. But I wanted to learn how to use a proper DAC chip and compare if perhaps the precision I can get with those might be better, and the space it takes up on the board. I have here the DAC0808 and the DAC0832. I know the '32 has some magic with dual latches on the digital side where one can stage the new input while the previous value is presented on the analog side, and then instantaneously activate the new stabilized (latched) value, etc. That digital stuff is easy for me to comprehend and control. I'm pulling my hair out over the apparent complexity of the analog side of these two DAC chips. I cannot make this work for the life of me and I am flabbergasted about why this is so complicated. First, all datasheets are telling me to use an external OP-amp. I understand why I would need that to drive a load, but I don't care about load right now, I have a high impedance load on my scope, and just reasonable voltage is all I want to see right now. Besides, if I have to add an op-amp I am already using up nearly as much space as in my home-brewed ladder-DAC. Besides, I have already established that just any op-amp might not have the frequency characteristics that I would want aside from the hassle of procuring one that doesn't have voltage loss , if it can deal with a single GND-Vcc supply without needing negative power rail. So much hassle! But there is intricate magic going on with these 4 inputs and outputs that I have been searching all over to find someone who explains it to me real slow, and found nothing. +/-Vref Rfb Iout1 Iout2 it appears that the principle is somehow not voltage but current based, but frankly, I just don't understand it. And why is it so complicated? All I want is to have the DAC produce output between the 0 and 5 V digital rail. Isn't there some minimal circuit that does just that? I have seen some, and only in passing on video, and when I build it I get nowhere, because I don't understand why it is like that to begin with? Here is an example of such an in-passing shot of a simpler circuit, but then the guy shows me how he programs his stupid Arduino, and I don't care about the digital side. If you don't want to explain it yourself, but you know a good tutorial introduction page I can read, I would much appreciate it. <Q> Strangely you've chosen a current (as in amps) <S> output DAC ( DAC0808 ) <S> and, to get a voltage out you need an op-amp added to the output: - Same story for the DAC0832: <S> - Both these devices are as old as the hills so you might want to choose something more modern and accurate. <S> Maybe a voltage output DAC like this might suit: - <S> Try this link to Farnell <S> - it's set up for 8 bit DACs with an 8 bit parallel port. <A> That is a current mode DAC. <S> Switching currents is faster than switching voltages, plus it allows for multiplying an input signal. <S> The magic is that there are 8 weighted current mirrors inside the chip, and the data bit inputs control whether current is taken from GND and fed to negative supply, or taken from OUTPUT and fed to negative supply. <S> I suggest at least opening the DAC0808 datasheet from TI to have a quick look inside how it works. <A> Consider using a common-base bipolar circuit. <S> Its very fast, but not extremely accurate. <S> For extreme speed, use a common-base bipolar device, a single NPN transistor. <S> Connect either of your 2 DACs to the emitter. <S> Have 1,000 ohm in the collector to +5v. <S> Have 51 ohms in the emitter to ground. <S> Now to bias the base --- use a similar transistor, with base_collector shorted. <S> have 510 ohms in the emitter to ground. <S> have 20,000 Ohms from collector/base to +5v. <S> tie the collector/base to your first transistor's base.	It means that you must externally set the output pin voltage to GND with op-amp, and the current that is drawn from it is converted with feedback resistor to voltage output on op-amp.
How to safely power a large LED strip project I'm making an LED matrix with a board and 4X WS2811 strips (5 m length, 150 LEDs). Each strip is 45 W, takes 12 V and is 3.75 A. I have connected them in series and injected a 12 V, 60 W/5 A adapter for every strip. It's powered by an Arduino Uno and all grounds are connected to the ground in the Arduino. My question is, can this safely be power on from a single household outlet? I believe here in Canada each outlet can handle 15 A. So if this is true, is it unsafe to power all of these from a single outlet? The reason I want it to come from a single outlet is because I can turn on all adapters at the same time from the switch on my power bar. If this is not possible, can the adapters be spread out from different outlets and simply turned on by connecting or disconnecting the Arduino? I've attached a primitive schematic I made to the best of my abilities. <Q> Canada uses 120V outlets. <S> The amperage of the circuit is typically 15 or 20 amps. <S> But that is at <S> 120V. 12V 5A is provided after the power supply regulates the input down. <S> Without taking into account efficiency, 12V 5A is the same power as 120V 0.5 Amps, or 60 Watts. <S> Add in efficiency costs as no power supply is perfect, it's likely 0.6A input. <S> Look at the label on your supply and it will tell you what the input amperage is. <S> That's the amperage you need to worry about on your power strip/outlet/circuit. <S> If this is on a small localized display, you should consider using a 200 to 240 Watt 12V supply, like maybe a PC power supply. <S> Less issues with multiple supplies. <A> The power supplies can each deliver 5 Amps at 12 volts, but they will only draw about 0.5 amps from 120 V, so there is no problem running them all (and more) from a normal 15 Amp 120 Volt outlet. <S> A power supply or transformer passes power ( <S> volts times amps) <S> not just voltage or just current. <A> In short, yes you can power these from the same outet. <S> You're way under the limit. <S> You need to do some additional math: <S> Each of your supplies is 5A @ <S> 12V, which is 60W. <S> Assuming 85% efficiency, that means that it draws ~71W from the wall. <S> That's just about 588mA at 120V. <S> All four of your supplies should draw less than 2.5A all together. <S> To put things in perspective, hairdryers and heaters commonly weigh in at about 1200W (~10A from the outlet). <S> You would need like 17 of your power supplies to start approaching that limit. <A> Other answers are good, but your curcuit can be improved: <S> Don't connect your PSUs in parallel. <S> They are never made exactly equal and you have to do additional things in order to distribute the load evenly. <S> You can easily overload some of them. <S> Power a stripe from a single PSU (or few stripes from a single, bigger one). <S> Don't power one stripe from another stripe's end. <S> Don't allow the current to travel more than 5m stripe. <S> Their power rails have non-negligible resistance. <A> Fuses! <S> If you read the manufacturers spec carefully, DC PSUs usually have an overcurrent cutout, but this is often 1.5 x <S> the rated current. <S> Make sure that the current in the event of a fault is not too large for the cables you're using. <S> This is so often the case at 5v, because it's tricky to solder thicker cables to LED strip. <S> ATO blade fuses are perfect for this kind of thing, and if you use just one PSU to power the whole project, which is definitely best practice for something this small, you can use a fuse block to hold the fuses, and also do the power distribution. <S> Otherwise use some inline fuse holders and wagos. <S> For the last project I did, we used these, which were useful because they also have a busbar for the 0v terminals too. <S> Other models are good too, but unusual to find one with 0v busbar. <S> https://www.amazon.co.uk/dp/B07SGQ3QNR/ref=cm_sw_r_cp_apa_i_DP8iFbJWQB5QB Just make sure the wire that goes from the PSU to the fuse block is rated for the full max current the PSU can supply, not just what you expect to pull under normal demand. <S> To see what can happen if you don't use fuses (which is why you have been advised not to use a single chunky PSU, check out this video on YouTube: https://youtu.be/nWC0PkHB8O4 Have fun but be safe!	Whether you use one power supply (better) or several, even when the PSU is rated for a max output just above what the max demand current is for what is connected to it, fuses are really important to reduce the risk of fire in the event of a fault.
What's the effect of shorting the wiper terminal with one of the ends of a trim pot? EDIT: I realized I didn't draw the schematic properly. Previously, the 100 Ohm resistor wasn't there. I thought I'd done the math correctly for this, but clearly something conceptually isn't straight for me. I need a 5K potentiometer, but the sources I've looked through didn't have it in the package I'd like. I did see a 10K pot in the package though. So I thought by shorting the wiper terminal with one of the ends of the trim pot I'd be able to half the maximum possible resistance offerable by the pot. $$\frac{1}{R_{total}} = \frac{1}{R_{wiper}} + \frac{1}{10000}$$ $$\frac{1}{R_{total}} = \frac{R_{wiper} + 10000}{10000 \cdot R_{wiper}}$$ $$R_{total}=\frac{10000 \cdot R_{wiper}}{R_{wiper} + 10000}$$ Since the highest resistance Rwiper can reach is 10K, the resulting highest value would be $$R_{total}=\frac{10K \cdot 10K}{10K + 10K} = \frac{10K \cdot 10K}{2 \cdot 10K} = 5K$$ I could also make a graph by plotting the function of Rtotal. By this point I was convinced. I tried to measure the resistance on a spare 10K pot lying around, but I saw that maximum value remained 10K. Why is this? simulate this circuit – Schematic created using CircuitLab <Q> If you connect to the top and wiper, you can use the potentiometer as a simple variable resistor. <S> That can: reduce crackling noises in an audio circuit when the track is old or dirty prevent loss of signal or things blowing up due to wiper failure <S> prevent gain (and volume) <S> attempting to reach infinity in circuitswhere gain is proportional to resistance <S> If you have to design circuits to tolerate component failure, this is worth remembering. <A> If the wiper is at the bottom of the resistance element in your schematic, the whole 10K resistance element is in circuit. <S> At any point in the wiper's travel, you will only have one part of the resistance element in the circuit. <S> There is no way to get two parts of the resistance element in parallel, as your calculation would require. <A> The formulas do not match the picture. <S> In the picture, the resistance between ground and 5V can be set between 0R where the wiper bypasses the resistance and 10k where the wiper selects full resistance. <S> To whatever position you set the wiper, it will short out the unused portion of the resistance when it is connected like in the picture, so it will have no effect.	If you connect to the top and wiper, AND short wiper to bottom, you are still using it as a variable resistor, but with one important difference : if the wiper goes open circuit, the resistance value is 10K rather than infinity.
Laser scanning: Why are things like deflection and mirrors used instead of just a mount and servo? In the Wikipedia article for laser scanning , it is said that laser scanning is the controlled deflection of laser beams, visible or invisible. It then lists some of the technologies used, such as mirrors. But why is deflection necessary? Why do we need to use these intermediary components, such as mirrors, to direct the laser beam? In particular, couldn't we just attach the laser to a servo, as done here and here , and do laser scanning in this way? What benefit does deflection have over using a servo? It seems to me that using the intermediary components for deflection is just superfluous, outside of, perhaps, specialized use-cases. I would greatly appreciate it if people would please take the time to explain this to me. <Q> A few thoughts: <S> Weight: moving a lightweight mirror will be easier than moving the laser assembly. <S> Speed: related to weight, the system response can be much faster. <S> Reliability: <S> no flexing of wires which may fracture or break. <S> Infinite rotation: Mirrors can be made in a prism fashion to do a raster scan by continuous rotation. <S> (See the checkout laser scanners for example.) <S> With constant rotation speed there is no non-linearity caused by acceleration and deceleration. <A> Moving mass affecting bandwidth is the main problem. <S> It also gives you twice the deflection for any given angle of turn. <S> A secondary problem is the conduction of power to the laser. <S> A moving mirror needs no wires. <S> These problems become magnified if we talk powerful lasers such as used in concert displays, or guide star projectors. <S> A good front surface mirror will not get hot, and can be mounted well away from the heat dissipating laser. <A> In particular, couldn't we just attach the laser to a servo, as done here and here, and do laser scanning in this way? <S> What benefit does deflection have over using a servo? <S> You can in principle move the laser, it just isn't very practical in most cases. <S> From your examples I suspect you're thinking of laser points or other small lasers, moving at very low speeds <S> but it's pretty common that lasers are large and very fragile. <S> For example, on recent system I designed used a 100 lb, liquid cooled laser that needed to be mechanically decoupled from the room vibration on a floated air table, while the scanner spec was 16,000 Hz. <S> Moving something that heavy at that rate is physically impossible. <S> Actually, even moving a tiny laser diode at that speed would also be hard. <S> This is an extreme case, but you'll run into similar problems even with small diode lasers except at very low speeds, and usually people want to scan as fast as possible. <S> Even then, diode lasers still have collimating optics, and shaking them back and forth on a motor can be complex to tolerance. <S> All of the deflection technologies are implemented using mechanical components (such as servos); otherwise, how would we expect them to move? <S> There are akinetic beam scanners as well, which can reach higher speeds since there is no moving mass. <S> The most common type is the acousto-optical deflector, which uses a sound wave in a lead glass to diffract (rather than reflect) the beam at an electrically controlled angle. <S> There are also electro-optical scanners that modulate refractive index using an external electric field, but these are not very practical in most cases. <A> A scanning rangefinder isn't just a point rangefinder made to be retargetable so you can slowly move from point-to-point to measure the distance to different targets. <S> They scan to form an image which means they scan fast. <S> No gears and no reciprocating motion allow for higher reliability and scanning speed. <S> These scan at least 90 degree arcs at least 10Hz. <S> Try moving your back and forth hand like that and just imagine what is happening to whatever you are holding, as well as feeling all the stress. <S> And no slip rings for 360 degrees scanning, because slip rings suck.	A tiny mirror can have orders of magnitude less moment of inertia than a laser, so can be scanned correspondingly faster.
Why are the parts more expensive than a Pro Micro? I have just started studying electronics and I am having a good time. In the past my experience would have been to use cheap boards, such as the Pro Micro. What I find confusing is that parts, such as an oscillator, is often more expensive than a cheap $3 Micro Pro, which could have done the same thing with a few lines of code! What are the benefits of using electronics parts, when a cheaper development board could do the same thing? <Q> Manufacturers don't pay $3 for an oscillator. <S> a hundred of these is under $25 https://lcsc.com/product-detail/SMD-Oscillators-XO_HD-830260003W1_C655083.html <S> and that's still retail pricing. <A> If you start at mouser or digikey or equivalent and look at the multiples of thousands of part prices. <S> so something might be $1.00 in single quantity but it might be $0.20 in quantity of 10,000 and also understand IF and when you buy in those quantities you can do better than the mouser/etc price going through a distributor. <S> They make a profit, early raspberry <S> pi's perhaps not, but in general there is a profit, in quantity, with quantity prices, on an automated production line, (and other cost reductions like quality), the whole board with profit can cost less than the most expensive single part on the board. <S> If your goal is to make a robot on your garage then just buying already made boards is often the way to go. <S> If you are instead evaluating a part for some mass produced product you want to make, or if you want to learn more about the part like how to field upgrade it <S> or you want a similar part from that vendor that does not have a hobby/eval board <S> then you would either sample a part for free or buy a few at full price, make a pcb at full price, etc. <S> So you may end up paying $10-$20 for something that only has a few components and is less complicated than the $3 board (and you solder yourself), but it is specific to what you want rather than something generic. <S> Then if you go into production then you shop around, you decide if you are buying the parts or having the contract manufacturer buy the parts (they buy in very high quantities and get the best pricing, but they add profit when they essentially sell you the parts), how to test it, etc, the whole process. <S> Another way to look at this. <S> For whatever parts are available start shopping for car parts, doors, hood, quarter, panel, battery, wheels, driveshaft, transmission, engine, etc. <S> Same deal you cant build that car for remotely the price that you can just buy an assembled one. <S> Economy of scale. <A> A Pro Micro, or at least the knockoff/clones you can get for 3 dollars, are not using high quality parts with warranties for lifetime. <S> Plus you are not paying for local reseller profit margins or regulatory compliance costs. <S> You pay for a 3 dollar board from china <S> is not comparable to a genuine US manufactured product and name.	Pretty much everything on the boards are cheap parts and things like the microcontroller often straight counterfeit.
Can a voltmeter itself disturb the exact measuring of voltage? I'm trying to understand the effect on a system when the voltage of the system is measured by connecting a voltmeter. For example let's say we are trying to measure the voltage of a battery. To measure the voltage of the battery, we must connect the voltmeter. Each probe will be connected to each side of the battery (positive and negative lead.) Assume this voltmeter is a high-end one, there is almost zero current flowing through the voltmeter (infinite input impedance.) Although it is, I'm concerned that this behavior itself could disturb the voltage of the battery because the Fermi level of each side of the battery will change. When a probe of voltmeter is connected to one side of battery, their Fermi level would be in alignment for thermodynamic equilibrium because both the probe of voltmeter and the lead of battery are metal. If their material is different, the Fermi level of probe-lead connected metal would be somewhere between that of pristine probe and lead. Anyhow, by connecting the probe to the lead for measurement, this changes the original Fermi level of the battery. The same thing may happen when we connect the other probe of voltmeter to the other side of the battery. So the voltage displayed in the voltmeter would be somewhat distorted because this is the result after some electrons from the voltmeter flow into or out of the lead of the battery. Since the voltmeter measures Fermi level difference between two probes, the metal of the probe will probably have an effect on the result. I've never seen or heard this kind of effect discussed. Is it because the contribution of the effect mentioned above is negligible or am I just wrong? <Q> The characteristic of an analogue voltmeter, which would decide how much it, by itself, could load the source and change the voltage being measured, is known as its 'sensitivity'. <S> The sensitivity of an analogue voltmeter is expressed in ohms/volt. <S> The best analogue DC voltmeter would have a sensitivity of 20,000 ohms/volt. <S> Such a voltmeter, with a full scale reading of say 1.0 V, would present a load resistance of 20,000 ohms to the source. <S> Should the actual voltage be 0.2 V, the current drawn by the meter from the source would be 0.2/20000 or 10 μA only. <A> The disturbance should cancel out as far as the voltage reading goes because both terminals on the battery (the exposed terminals where you connect the meter) are the same metal, and likewise both of your probes are the same metal. <S> On a car battery the terminals are lead or a lead alloy. <S> Your meter probes are, let's say, nickel plated brass. <S> As you follow the circuit around the voltage goes up and down across all these metal/metal junctions. <S> But the net effect is zero because all the junction drops or rises cancel out. <S> If you introduced an asymetric metal/metal junction on, let's say, the positive wire, by adding an extension wire made of aluminum, then you would have an uncompensated drop that would introduce an error. <S> Practically speaking, with a good voltmeter, you can trust the battery voltage on the display to 0.01V. <S> With a better meter you can trust it to 0.001V maybe. <S> After that you may have to worry about measurement setup things that most engineers never worry about. <A> No it can't. <S> Hmm the voltmeter is connected in parallel to the branch we want to measure its voltage. <S> A perfect voltmeter has infinite resistance, and if you connect an infinite resistance in parallel to the load then the total resistance doesnt change at all <S> and we have the same voltage drop on the load. <A> OK I'll consider an infinite input resistance voltmeter but it must have some input capacitance due to the insulation between + & <S> - conductors must be non-zero capacitance. <S> So what do voltmeters measure? <S> Volts, of course with the potential to do work and conduct current but not until it is buffered by this zero-input current amplifier. <S> So there must be a charge transfer to the input of x pF. <S> This equalization depends on the source having enough stored energy to transfer. <S> It doesn't have to be a voltage source of 0 Ohm impedance. <S> It just must not be affected by the energy dump to charge up this voltmeter <S> $$ <S> E=½CV²$$ <S> let C be really small value in the voltmeter = <S> \$1pF=10^{-12}F\$ <S> All batteries are like UltraCapacitors with an intrinsic chemical cell voltage as the minimum. <S> The Li-Ion cell is about \$10kF=10 <S> ^ <S> 4F\$ <S> A drop of water between 1mm electrodes might be 100pF, as the dielectric constant =80 relative to air. <S> Yet the battery will not notice the capacitance charge transfer to the voltmeter, but a charged drop of water might notice that load of 1pf. <S> However, keeping the voltmeter attached might be sensible if the capacitance ratio was known and corrected in any resulting acquired voltage. <S> Solid State Electrometers have an input impedance of <S> \$10^{14}\Omega\$ and can be modified with a reference to store a minimal energy of a few percent of a picocoulomb. <S> [pC]	Yes, an analogue voltmeter, by itself, could disturb the precise measurement of a voltage by loading the source and changing its voltage.
Is it possible to wire a DPDT switch so that each position creates an output? (3 outputs) I am controlling three LEDs, and need to be able to toggle between them individually with a switch. Is this achievable with a DPDT switch? I'm not sure where I can find a switch that has one input and three possible outputs. <Q> Here's one way to do it using an ON-OFF-ON switch like this one: one half of the switch in either on position shunts D3 preventing it from illuminating, this wastes about 20mA in the off position the LED lights <S> the other half selects between the other two LEDs when on simulate this circuit – Schematic created using CircuitLab <A> It may be identified by a higher pitch between one terminal and common. <S> Should SP3T not be available DP3T may be used. <A> @Jasen has a very simple way of doing it that is probably what you should use. <S> Just for interest, here is a circuit that (at the cost of a few more parts) uses less power because it does not depend on shorting out an LED in two of the positions, <S> so it's better suited for battery power. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Here's a way you can do it with a single pole ON-OFF-ON switch. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> It relies on the center 'OFF' position LED D1 needing more voltage to turn it on than the others. <S> When either of the 'ON' LEDs is switched in its lower voltage drop robs current from the 'OFF' LED. <S> D4 increases the voltage required to turn on the 'OFF' LED, and R2 ensures enough current passes through D4 to get a good voltage drop (without it D1 may glow weakly when one of the others is switched on). <S> Depending on the LEDs used, D4 and R2 may not be required. <S> If D1 is Green or Yellow and one of the others is Blue or White then 2 diodes in series may be required to get sufficient voltage drop. <S> Each LED is lit through R1, so the circuit is efficient. <S> However the LED currents are not individually adjustable, which may be a problem for different colored LEDs with greatly different brightnesses.	A SP3T slide switch (single pole triple throw) may be used. If D1 is Blue or White and the others are Red, Green, or Yellow then their voltage differences alone may be enough.
Do beam-forming devices expect a particular antenna configuration? Do beam-forming devices expect a particular antenna configuration? For instance, do they expect the antennas to be a certain distance from one another? Or parallel to one another? Etc. Background Info: I have a beam-forming capable WiFi router located in a utility room. I would like to detach the antennas and add extender cables, so I can run them to various parts of the attic. I am concerned that whatever calculations are necessary to perform beam-forming may include hard-coded values based on the current configuration, e.g., each antenna is placed exactly 2 inches apart, and parallel to one another. <Q> There are two classes of beamformer. <S> One type has the antenna configuration hard-coded into its mathematics, and attempts to construct geometrically beams towards wanted sources, and/or nulls towards unwanted sources. <S> The other type uses general complex matrices to maximise SNRs of expected signals, and to the extent that it 'beamforms', the positions of the antennae appear implicitly in the operation of the optimaistaion during channel measurement. <S> The 'beamformer' in a MIMO WiFi base station is of the second type. <S> There may be second order effects if the algorithm used expects the antennae to be within a wavelength of each other, as would be the case in a physically small router. <S> Widely separated antennae, although better than close antennae in some respects, have a larger dPhase/dAngle_of_arrival, which may potentially cause some training algorithms to arrive at sub-optimal solutions. <A> Yes, it will be important to keep all antenna extensions equivalent to ensure the propagation delays are equal. <S> You will want to keep any changes equivalent to each antenna to minimise efficiency loss. <S> I have done this before and found that in the technical doumentation the manufacturer recommended how to reposition the antennas and defined a maximum cable length, so it might be worth looking if your model has the same. <A> Do beam-forming devices expect a particular antenna configuration? <S> Well, they need to be designed to work with the particular antenna configuration you have. <S> But that constellation doesn't inherently have to be a particular one. <S> Sure, some shapes might be advantegous for specific directions, but you'll find beamformer arrays in all kinds of arrangements; from simple antennas in equal distances on a line to honeycomb constellations to satellite constellations in space. <S> Note that for beamforming behaviour to emerge , i.e. happen "coincidentally", a MIMO precoding (on transmit side) or maximum ratio combining (on receive side) <S> algorithm doesn't need to know anything about the geometry of the antenna array at all - it happens automatically if the channel coefficient are such that there is a strong dominant path from a single direction, and there's enough antennas to deliver the degrees of freedom.	It’s probably best to keep the spacing the same, if you do increase the spacing, it's probably best to increase it between all the antennas.
electric guitar grounding for safety As I understand it there are two reasons to ensure your electric guitar strings are connected to the common ground of the instrument. To dissipate the EMF/RF interference created by the human holding the instrument, which reduces the noise level on the circuit Safety I understand the first. But the second reason I struggle with. The oft sighted example... it's possible for the common or neutral side of the line to be out of phase with another power outlet or piece of gear. So, instead of the chassis and all common parts sitting at ground potential, they're at line potential relative to another piece of gear. So far so good. if you're holding onto a guitar with "grounded" strings that are correctly grounded and walk up to the microphone with a "grounded" housing that really is at line potential it'll knock you flat on your ass. Done properly, it can even kill you. So surely, in this situation the, the fact the the guitar strings are grounded makes touching the strings more dangerous. If the strings were not attached to the circuit you'd be safe? So I'm thinking, if the strings are not grounded at all the current might pass through my body and kill me that way, but at this point isn't the guitar irrelevant? So grounding the strings is pointless? EDIT:So looking at that link in the comments. The 3rd example in Kyle's link sees the path to ground from a faulty mic either going through the guitarist's feet OR through the earthed strings, but the latter is said to be more likely lethal. So the question remains, why ground the strings of that makes the legal path more likely? <Q> I think what's missing is an appreciation of why an electric guitar is the way it is. <S> The wiring of an electric guitar is basically UNCHANGED since the introduction of the iconic Les Paul and Tele/Stratocasters in the 50's. <S> So what you have is an instrument that's designed to work with the amplifiers of the day. <S> Virtually every electric guitar from every manufacturer today is wired basically the same way. <S> In 1955, the vast majority of power outlets did not have a 3-rd wire ground or even have one blade bigger than the other, so it was not only easy to get hot/neutral reversed <S> , it was EXPECTED to happen 50% of the time. <S> To accommodate, a guitar amplifier chassis had a 'ground capacitor' switch such that the user could tie his instrument 'ground' (i.e. the plug/jack) to either of hot or neutral through a high-voltage capacitor. <S> The user would pick one of three positions (connect to either AC wire, or no connect at all) until the point of lowest hum was found. <S> This capacitor has since become known as the 'death cap', because a failure of it places live AC voltage on the strings. <S> ( https://robrobinette.com/Death_Cap_and_Ground_Switch.htm ) <S> It's just a crappy electrolytic, not a safety device, and such a scheme would never fly today. <S> Guitar amplifiers have had to keep up with changing UL standards etc.. <S> so yes today you would find proper 3-prong plugs. <S> But people don't throw away old guitar amplifiers - Quite the opposite, they are coveted. <S> It would be super easy to design a 'perfect' guitar plug that was noise-free, high reliability, and safe. <S> But that would be incompatible with the vast array of already existing gear. <S> Nobody would buy it and it'd be a market flop. <S> The thing is, guitar players are almost religious about keeping designs "vintage". <S> That's why the Les Paul/ <S> Strat/etc have been in continuous production nearly unchanged for 70 years. <S> I challenge anybody to find any example of such a long-lived design in any other field. <S> Guitarists are strange creatures --- <S> Tweak the design, they won't want it. <S> So it's the market that DEMANDS that guitars remain wired the way they are <S> even though EE's know it's "wrong". <A> it's possible for the common or neutral side of the line to be out of phase with another power outlet or piece of gear. <S> So, instead of the chassis and all common parts sitting at ground potential, they're at line potential relative to another piece of gear. <S> No professionally designed equipment will have that possibility unless multiple faults occur. <S> The chassis and circuit ground is always isolated from both live and neutral through a good old-fashioned heavy-metal 50/60 Hz transformer or through the switching transformer of a more modern SMPS (switched-mode power-supply). <S> We never assume that the live and neutral will be presented in a particular orientation. <A> The purpose of grounding electric guitar strings would be only to ground the guitarist in contact with them and thereby prevent ambient electrical noise, picked up by him/her, being passed on to the amplifier. <S> Safety to personnel would be ensured only by proper chassis grounding of all audio equipment through the 'third pin', using fuses of the right rating and ensuring 'line' and 'neutral' are connected to the right terminals in the plugs as well as receptacles. <S> Of course, the ultimate protection would be provided only by a Residual Current Device (RCD). <A> A properly grounded system will improve safety, not make it more dangerous. <S> But that may not be easy in your situation. <S> ... <S> reasons to ensure your electric guitar strings are connected to the common ground of the instrument. <S> A lot comes down to what is meant by "common ground." <S> A safety ground provides a path to earth with no current running along it; any significant current running along your ground can and will negate its usefulness for safety. <S> In particular, an AC neutral line should never be used for grounding. <S> That neutral is the return path for the same current delivered by the hot line; if you connect yourself to neutral, you could become an alternate return path. <S> So in an ungrounded system, where only hot and neutral wires are available, then what is the answer for safety? <S> Either completely isolate from the power supply, or add a ground. <S> I nearly rewired a house to run a ground to ungrounded circuits; rewiring an amp is also possible. <S> It may not be cheap or easy, but the question is asking about safety. <S> So surely, in this situation the, the fact the the guitar strings are grounded makes touching the strings more dangerous. <S> No! <S> Any number of things can ground you; you can argue about how current would enter/exit your body, but by that time all bets are off. <S> With proper grounding, in the guitar and in the microphone, your scenario would be prevented altogether. <S> You may decide that proper grounding is impractical for your situation, but that is your choice. <S> Grounding is still good practice.	Such a grounding would make for a quieter guitar and would neither complement nor compromise the safety of the guitarist, who would anyway be exposed to the danger of electric shock from the pickup/connector, microphone etc.
What are some quick, cheap hacks for determining the source of EMI in my house? TL;DR - Without dumping a bunch of money, is there a quick MacGuyver-y way to help diagnose what's causing the EMI in my house? Longer - I have the curse that I'm a little too technical to be helped here by dummies.com, but not sophisticated enough to have or use the tools I imagine most of you would recommend. Several times per day - sometimes hours without issue, sometimes firing once every 5 seconds - the wifi blips for a few seconds. I've caught command line ping dropping packets for about 2-5 seconds, on computers throughout the house. Notably, we realized the baby monitor also cuts out for just a few seconds at the exact same time ping goes down. So it would appear my house is regularly awash in brief bursts of EMI, that's causing connectivity issues (especially a problem with video conferencing in the current covid-19 lockdown world). But I have no idea what to buy or rent to help me triangulate where this is coming from or what it could be. I have some dim memory that magic things can be managed with an oscilloscope and simple length of metal plugged into it operating as an antenna. How can I find EMI? (Meta - if there's a better place besides electronics.stackexchange.com for this please leave a comment) UPDATE (for more information, as requested in comments) The Baby monitor and its receiver is an Infant Optics DXR-8, audio and video over 2.4ghz wireless (NOT wifi / tcp-ip, but some point-to-point proprietary). In fact, shielded twisted pair cat5 appears undisrupted (but not a fix for everything getting disrupted in our wifi world). The wifi is 5ghz AC + 2.4 ghz N (both ranges/protocols impacted). To my knowledge, I'm not particularly close to any unusual EM sources - e.g. airport, military base, transformers - though I guess you never truly know. It is relatively dense population (single family residences every 35 feet) - I'm wondering if a neighbor has one of these "weaponized" routers that disables other people's wifi, but need some kind of proof before I go pounding on doors. All of this aside, to keep the question within format for stackexchange, assume generic American residential suburbia, and let's see if we can dream up a gimmick that works generally. Anything further peculiar to my immediate situation I can answer in comments. <Q> A yagi antenna and a cheap SDR might be good to play around with. <S> Make sure the antenna and SDR work in the frequency range of interest <S> (so if you want to look for wifi signals, then the antenna and SDR need to work in the 2.4GHz range.) <S> MSI SDR is an SDR I like because it has a range of 10kHz to 2Ghz <S> (but you need 2.4GHz). <S> RTL-SDR shows an SDR solution for 2.4Ghz . <S> The antenna could also have a lobe in the back so read up on antenna patterns (or look at the one you buy, antennas are not perfect by any means, they have different lobes and nulls which changes the gain with direction) <A> You can make a UNTUNED energy detector, with a diode, 2 resistors, 2 filter caps (to set the upper frequency limit) and a switch to change that frequency limit. <S> You'll use a 9 volt battery, powering a series connection of 1Megohm resistor, the forward-biased (schottky) diode, and 10,000 Ohm resistor to ground. <S> Across the 10,000 ohm is 100pF capacitor. <S> At 1GigaHertz, this has about 2 ohms reactance, thus the WiFi energy (if the antenna picks up any) ripple across the capacitor will be quite small, and your readout device ----- a DVM ---- should be accurate. <S> To explore Low Frequencies --- Air Conditioner spikes --- place 1uF across the 100pF, but SWITCH the 1uF in and out. <S> Keyfobs may still be detected, and 60Hz or black-brick trash will still be detected. <S> Build this over a FR-4 board with standard copper foil. <S> At the top of the diode (the shared node of diode's anode (the base of the arrow, not the bar) and the 1MegOhm resistor), install 1" solid_copper wire, approximately 90 degrees from the copper ground plane. <S> To look for other interference, simply clip some test leads to the 1" wire. <S> Now your interference may be VERY QUICK. <S> May be tough to catch the miscreant in action. <A> Phones already have the correct antenna for measuring signals in the wifi frequency range, may as well use it. <S> You can see all the nearby networks being broadcast, the bands they are on, and the signal strength. <S> Some apps even offer signal strength as a time graph, or even allow you to make maps. <S> This may not identify other sources of EMI, but it will tell you more about your own router and nearby routers to help narrow down the problem. <A> Also, check for an occasionally used electric welder in the neighborhood - especially one of very high power. <A> See if you can locate your antenna in an open central area. <S> The ceiling might be ideal with vertical direction for signal strength to reduce the dispersion and Ricean Fading effects from reflecting surfaces. <S> If you can get a connection close to the router, that would help correlate to me your signal level. <S> Specify your path distance and room separations to give us a better idea. <S> 11 Mbps rates are the most robust data rates for a lower signal level than 54Mbps and also higher for wall reflection Ricean Fading loss. <S> Setting this in the mobile will help for remote mobiles or change router to 802.11b only using narrow channels on 2.4GHz. <S> some Router antenna wires can break if tightened by rotating too much. <S> Back off try again. <S> I have used Wifi Signal plotting software (forget name) on a Windows Laptop <S> It can be used like directional Radar by plotting the RSSI Signal level in -dBm in realtime and aiming the laptop towards and away from the signal or interference. <S> I once used this to bounce a signal off a neighbour 2 doors down when routers were open by default 10 yrs ago. <S> I found I could gain a few dB above -80dBm by bouncing off leaves in the back yard tree and adjusting the laptop on a bay window by 1 degree. <S> That worked well until cable service was installed after moving in.	SDR's are software defined radios, you can download spectrum analyzers for free and see which bands are carrying radio information. Keep in mind that you may not be able to find exactly what the source is because of reflections and attenuations of walls and other objects. There are some apps available for analyzing wi-fi, e.g. "wifi analyzer", you may be able to spot what's going on with other networks or your own network.
Arduino analog input value oscillating, why does the DMM work and arduino not so much I am measuring voltage output from a device with 10 kohm output impedance with an arduino analog input pin. I see what I would expect with a multimeter, but with the arduino I get either garbage, or at best, oscillating measurements that average out to about what I would expect. I know the impedance of the analog pins to be 100 Mohms, while on the DMM, it would be around 10 Mohms, so I wasn't expecting the Arduino to be useless while the DMM works fine. I finally found that shorting the outputs with a capacitor gives me what I'm expecting, but don't really understand why. Resistors helped but I still saw oscillation, and I assumed, being an issue seemingly with impedance, that resistance mismatch was the culprit. <Q> The MCU in Arduino has a successive approximation ADC. <S> It works by briefly taking a voltage sample via a multiplexer into a small storage capacitor to handle multiple input channels with one ADC. <S> So the impedance of an analog input pin is not 100 Mohms, as it is momenrarily charging a 14pF capacitor with 1kohm series resistance, and there can be current in or out of the pin. <S> Therefore the source impedance must be low enough to charge the sample/hold capacitor to within 0.5 ADC counts during the sample time. <S> Assuming the MCU on your Arduino is an AVR, the ADC specifications say it works best when the source impedance is 10k or less. <S> It seems that your sensor is having a high output impedance. <S> Also if the sensor output cannot handle the gulps of periodic sampling capacitor charging, it might become unstable and exhibit ringing when the empty capacitor is suddenly connected for charging. <S> The same is basically when you do have a filter capacitor at the AVR input to have low enough short-term AC impedance, so you can take one measurement without much affecting the value. <S> But if the filter cap is charged by a 1 Mohm resistor, it will still have a high long-term DC impedance, so it takes long to charge the capacitor back to original value, so taking measurements too often will slowly discharge the filter capacitor. <S> So there are many reasons why analog measurements made with Arduino won't work when compared to a multimeter. <S> The Arduino does not have a built in signal conditioning and buffering like a multimeter has. <S> The required analog input stage should be built between sensor and MCU. <A> The meter will have either a real capacitor or a software-simulation of one internally. <S> If it didn't, you'd never get a stable number to be displayed and it'd be impossible to read. <S> You can be sure this has nothing to do with any 'resistance mismatch'. <S> A little story.... <S> When I was in college, I was taking a CMOS class - <S> We laid out integrated circuits and had them fabricated. <S> My team decided to do a 'voltmeter on a chip' --- It had the analog inputs, and would output directly to a typical 7-segment LED digital display. <S> When we powered it up, it read "888". <S> Of course, we were super bummed, until somebody suggested adding a small cap to the input. <S> Suddenly it worked. <S> So what was happening is, the numbers were bouncing so rapidly, all the LED segments were lighting up so fast they appeared to be steady-lit. <S> But such an environment doesn't exist. <S> There will ALWAYS be something perturbing the wires to a degree. <S> So the Arduino isn't lying to you. <S> It's reporting what it sees at the pins. <S> Just so happens the value at the pins is moving around. <S> Placing the cap there introduces a smoothing action to the voltage, so it becomes stable enough to make a reading. <S> That said you do need to give the Arduino a hand by making the voltage stable as possible. <S> For example, using tightly twisted wire instead of two separate leads nearly always will yield improvement. <S> Noise-reduction and proper grounding techniques aren't something I could describe in a short post, you could fill textbooks with the stuff. <S> But if you simply can't get the stability you want, posting some pics of your setup may get you useful feedback here. <A> Following up on Kyle B answer, if you're not using a GROUND PLANE, then do so. <S> I assist a guy in sensing/digitizing musical instrument transients. <S> Over the last year, he/we have learned how to reduce the error floor from 90 milliVolts to about 0.3 milliVolts. <S> Have <S> a 0.01uF cap on input to the ADC Have a unity_gain buffer before that 0.1uF cap. <S> Don't use switching_regulator power supplies. <S> If you must use SwitchRegs, then pay for the better grade. <S> Use a Ground Plane. <S> You may need common_mode chokes, on VDD/GROUND wiring to the SwitchRegs. <S> In the VDD to any opamp buffer, have 10 Ohms in series, and 100uF shunting.	With a high source impedance, the sample/hold capacitor may not have time to fully charge, and thus the sample of the voltage does not resemble the actual voltage. If you could somehow operate your Arduino in an environment with zero electromagnetic noise, you'd get perfectly stable readings without the cap.
Explanations regarding the blocking of diodes in single phase rectifier I have started an analog circuits course and have trouble understanding the behavior of such opAmps and rectifiers. This is the circuit given as example, however there's no extra explanations provided regarding the functioning of diodes and how the amp distributes the current. LE: Does the opening/closing of diodes depends on the direction of current? Am i misunderstanding that the direction of current changes when Vi changes sign? Thanks <Q> This is not a correct schematic for ideal diode rectifier. <S> The feedback must go to the negative input of opamp. <S> The negative feedback will eliminate the voltage drop of the diode D2. <S> This is an "ideal diode" single phase rectifier. <S> I've added a test point at the opamp out and smaller input voltage for better understanding. <A> As has been mentioned, the schematic should be changed so that the feedback is to the negative terminal of the op amp rather than the positive terminal. <S> Here is my updated circuit: SPICE simulation of the circuit gives data for the following plot: I used the following model for a mostly ideal op amp: <S> * Mostly ideal op amp <S> * in+ in- <S> vcc vdd out.subckt opamp 1 <S> 2 3 4 5Rin 1 <S> 2 10GCin <S> 1 2 100pFRout 6 0 0.1 <S> * Ignores power rails (more ideal) <S> * Bout 3 6 v={100kv(1,2)} <S> Has less amplification for low voltages and approaches supply voltagesBout <S> 3 6 v={(v(4)+v(5))/2+v(4,5)atan(100kv(1,2))/pi}.ends <S> As to a simple, intuitive explanation as to why this circuit works, remember that: A diode may be loosely approximated by a small voltage source (usually about 0.7V) and a small resistor which allows current flow in only one direction. <S> When Vin is positive, current will want to flow to the right through R1. <S> Nearly all of this current will flow through D1 (low resistance), causing Vx to be about -0.7V (since the input is brought to ground and the diode drops about 0.7V). <S> Since nearly all of the current will flow through D1, no current flows through R2 and the output voltage is nearly the same as the input voltage (0V). <S> When Vin is negative, current will want to flow to the left through R1. <S> Since almost no current will flow backwards through D1, this current has to travel through R2. <S> By Ohm's law, the current will be Vin/R1 giving an output voltage of -(Vin/R1)R2=-(R2/R1)Vin. <S> This voltage will be positive since Vin is negative. <A> The schematic is incorrect. <S> The inputs have been swapped. <S> See imgur.com/a/RAKtLwR for further slides from the tutorial for example. <S> It pretty straight-forward. <S> Remember that: The op-amp has very high input impedance so usually we can ignore any current in or out of that pin. <S> With negative feedback the op-amp will settle down with V IN- = V IN+ . <S> Since V IN+ = 0 <S> V both inputs can be considered to be at 0 V. <S> (V IN- is a "virtual ground". <S> Now consider the two cases for V i : <S> When V i > 0 current through R2 can only go through D2 as D1 is reverse biased. <S> The op-amp output will settle at about -0.7 V or so (depending on V F (the forward voltage) of D2. <S> When i < 0 current can only come back through R2 via D1 as D2 is reverse biased. <S> Since R1 = R2 = <S> R <S> the gain of the circuit is -1 so <S> V O <S> = - V i .	An ideal op amp with its output connected back to its negative input will try to output whatever voltage causes its inputs to be the same (i.e. it will make the negative input voltage match the positive input voltage).
How to connect unused MOSFET from Dual N-channel MOSFET? This question may be stupid but how should I connect an unused FET in a dual-channel MOSFET ? I using this one multiple times on my board but in one case I'm using only one MOSFET. I'd like to know how I should connect the unused one ? I guess I should connect the gate to the source (through a resistor) but what about the source pin ? <Q> Instead of 0 volts, if it's inconvenient connect source to the same potential as the other source and gate the same or via the 100 kohm resistor. <A> Assuming you can't switch to a single MOSFET package which would probably be cheaper, you can simply terminate the pins. <S> Three options: <S> If the source of the MOSFET you are using is grounded, then connect all three pins (Gate/Source/Drain) to ground (or even just gate and source). <S> No need for any resistors. <S> Again, no resistor required. <S> Leave the pins floating, but with gate and source tied together. <S> There isn't much danger of anything conducting, as the circuit is not like digital inputs whereby you might get shoot through currents from floating inputs. <S> Though keeping the gate and source tied together is no bad thing. <A> There should be no issue with just leaving all pins floating. <S> It's not like an op-amp that still gets power even when unused. <S> You could also just short all terminals to GND or whatever nearby rail you want. <S> Or, if your application allows, connect it in parallel with the other MOSFET so both are used.	If the source of the other MOSFET is not grounded, connect the gate and source of the unused MOSFET both directly to the source of the used MOSFET. Connect unused gate and source to 0 volts or Connect source to 0 volts and gate to 0 volts via a 100 kohm resistor and Leave drain unconnected
Reason of using diode between common mode choke legs I'm investigating SMPS circuits. And I couldn't understand why there is a diode beetween common mode choke legs (I don't know exactly but it seems like signal diode) What is the main reason of using diode on there ? Circuit guess Bottom side Full photo <Q> It is a gas discharge tube (GDT) in a glass package. <S> In some power supplies they are shunting one of the common mode choke coils, so this is not an unique design. <A> My speculative answer: <S> You wouldn't put a unidirectional device across an input CM winding for clamp suppression. <S> It would conduct every half-cycle, after all. <S> A bidirectional TVS or a gas-tube would work for clamping - in both cases, should sufficient voltage be present to cause breakdown, the TVS or gas tube would clamp the voltage. <S> (This sort of event is usually only induced by lightning strikes or related input disturbances.) <S> I would argue that: The device was originally designed with a TVS for surge clamping. <S> For whatever reason they used a polarized diode PCB footprint for the TVS. <S> Later testing found that a passive damping element (like a resistor) was sufficient to protect the unit from surge damage, so they stuffed a resistor into the diode's location. <S> EDIT: It may actually be a gas-tube upon closer inspection. <S> (I still don't think it's a diode.) <A> I think its a Flyback Diode . <S> When the circuit is turned off after reaching a steady-state, The coil has back EMF which is reversed in polarity. <S> This will result in high voltage, potentially creating an electric arc. <S> The flyback diode is reverse biased when the circuit has power, but when turned off it provides a path to dissipate power, avoiding an electric arc.	It makes somewhat sense that it seems to shunt high voltage transients so that they bypass the common mode choke neutral coil. Clearly the part that is stuffed in that position isn't a diode - it looks more like a resistor to me.
Input voltage control by resistor, is this method is correct or not? In my application I have a requirement to switch 40V to 5V with the help of a linear voltage regulator. My load circuit consumes approx 7 to 8 mA. Due to the unavailability of any specific model of a linear voltage regulator, I am using 7805 for this purpose. But the input voltage limit of 7805 is 35V. I already tried a voltage divider method put before 7805 to control the input voltage level, but that method has other issues because of the internal resistance of the 7805. Here I applied one input resistor to control input voltage flow. Circuit: Now due to 7mA of load, and (2.2Kohm, 1-watt) resistor, the voltage drop of the resistor is (7/1000)*2200 = 15.4V. The input voltage of 7805 is ( 40 - 15.4 ) = 24.6 V in this scenario which is less than the safe limit of 7805. Is this solution is suitable for this type of application, or there is another efficient way to solve this kind of thing? <Q> Here I applied one Input resistor to control Input voltage flow. <S> If you know the load will always be there and always draw at least 3 mA, then this is viable. <S> If you want to be safer about it, you could add a 30 or 35-V zener in parallel with C1. <S> or there is another efficient way to solve this kind of thing. <S> You could choose another regulator that allows more than 40 V on its input. <A> Use a voltage divider in front of the 7805 for safety. <S> simulate this circuit – <S> Schematic created using CircuitLab or a zener as Photon says (preferable method)I put 10V zener, as 7805 recommended operating voltage is 25V. simulate this circuit <A> If you really care about efficiency, use a Buck converter like MC34063 - it can handle input voltage up to 40V. Using a regulator for converting 40v to 5v is not an efficient way. <S> But since your load is 8mA max, you can find a regulator with higher input voltage or use 7805 with a voltage divider or with a Zener diode as told by @ThePhoton. <A> Just use a single shunt zener or shunt regulator like this: - Maximum current into zener is 35 volts / 2200 <S> = 16 mA and power dissipation is 80 mW.	This solution won't work if there's a possibility for the load to be disconnected or for its current to drop significantly because then the 7805 input will be subjected to ~40 V.

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