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ultrafast catch diode for relay Is there any reason to not use a UF2008 as catch diode on a DC relay, solenoid, or electromagnet? IN4001 costs $0.07 and UF2008 costs $0.03 and seems to be better suited in all ways simulate this circuit – Schematic created using CircuitLab <Q> Either diode is fine in this application as long as the relay coil current is less than 2 amps. <A> 1N400x <S> (and the SMT equivalents) are very cheap in quantity and can be found in all kinds of consumer goods, now and in the future. <A> Do we really need nanoseconds diodes for this application? <S> Cheap high performance components might not be always available. <S> Don't count on them unnecessarily. <A> The title of the question is "ultrafast..."; the release speed of the circuit you show has virtually nothing to do the the speed of the diode. <S> When the transistor turns off, the current continues to flow in the coil, and the relay continues to hold in the ON position until that energy is dissipated. <S> It takes a long time because the effective diode resistance is quite low, so the L/R time constant is very long. <S> Solution is to insert a resistor in SERIES with the diode, start with a resistance about the same as the relay coil DC resistance. <S> This will allow the collector voltage to fly up to about double the supply voltage (as opposed to supply voltage plus a diode drop), but will allow MUCH faster operation.
| Transistors switching the relay solenoids not always driven in such a way where it turn on or off in nanoseconds, as such, the di/dt in the coil could be slow enough for low speed diodes. If you're doing small quantity or are flexible you may be fine with them, but keep in mind that the real manufacturer is apparently not known and it may be a surplus lot that will be unavailable the next time you go to buy, according to this source.
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What's the advantage of lead-free solder containing silver? I'm looking to buy some lead-free solder. There appears to be two different kinds available, tin-copper alloys and tin-copper-silver alloys. What advantage does the latter have, as it is more expensive? <Q> There are far more than two types available. <S> RoHS and lead-free solders including silver have the following advantages and disadvantages: Pros: <S> Higher melting point, higher working temperature. <S> Stronger bond, less susceptible to mechanical fatigue, more reliable joint. <S> Addition of an impurity to tin (silver, copper) reduces chance and/or rate of tin-whisker formation. <S> (Note that silver itself can whisker in humid, hydrogen-sulfide environments.) <S> Cons: <S> Too much silver can form inter-metallics that cause grittiness and formation of pimples on the solder surface. <S> Higher melting point = <S> higher fabrication process temperatures. <S> Higher temperatures mean rework can be more difficult. <S> Stronger bond = more reliable, but also more brittle, having a lower ductility and higher Young's Modulus . <S> More expensive due to silver content. <S> What usually matters more is the mechanical properties and assembly/rework-ability. <A> In general silver makes the solder stronger and has a higher melting point. <S> We use it in high temperature applications such as downhole. <S> Copper in the alloy lowers the melting point and makes it somewhat easier to work, and has some chemical advantages when soldering to copper conductors. <S> Neither alloy has sufficient resistivity to matter much as far as the actual solder joint is concerned. <A> Contrary to what some other answers suggest, the Sn-Ag-Cu ternary eutectic melting point (217 °C) is below the 22/78 Sn-Ag (wt.%) eutectic of 221 °C and the 59/41 Sn-Cu eutectic of 227 °C. <S> Those 10 degrees may be important.
| While it is true that silver is a better conductor than most other metals, the resistivity of a typical solder joint is so low that any small gain in conductivity would matter only for very high-current applications. Improved resistance to fatigue from thermal cycles.
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Why is the voltage across the capacitor constant? I've connected a 6V DC voltage source via a 1Ω resistor to a 1F capacitor. I was expecting the simulation to reveal a less constant t-vs-v graph: it would start at 0 and after a while stabilize at 6 volts. But it seems that the voltage across the capacitor begins at the source level and remains so. Two questions: Why is the voltage across the capacitor constant for the 10 seconds? (Secondary) Why isn't the current through the resistor I = V/R = 6/1 = 6 A (at least while the capacitor is being charged)? I'm making my first steps in electronics. Any feedback would be very helpful to me. Solution. Thanks to everyone for the feedback. Lack of reputation doesn't allow me to upvote them, but all were very helpful to me. Turned out that I didn't check "use initial voltage", so while the initial voltage of the cap was set to 0, it wasn't being used: Circuit designed and simulated using SystemVision. <Q> 1) <S> The simulator starts by calculating a DC solution of the circuit. <S> In the DC solution capacitors are charged to the voltage that would be on a node after a very long time. <S> The transient (time) simulation you performed starts calculating at time = 0 <S> with this DC solution . <S> So that is the solution where the capacitor is already charged. <S> If you want to see the charging behaviour of the capacitor, you must convince the simulator to calculate a DC solution where the voltage across the capacitor is zero. <S> In some simulators this can be done with an "initial condition". <S> However it is much simpler to replace your DC voltage source with a pulse-source. <S> Let it make a pulse that starts at 1 second and ends at 100 seconds. <S> Then you will see the charging curve. <S> 2) <S> You're not charging the capacitor so that is why. <S> If you were you would see that predicted current briefly when the voltage across the capacitor is zero. <A> However, depending on the value of R and C, the capacitor will eventually charge, and when it is "full <S> " it will not allow any current to pass. <S> It acts as an open circuit, meaning no current will flow (and thus, no current will flow through the resistor). <S> Your simulation is showing the steady-state (after the capacitor has charged), so it shows the full 6V across it and no current flowing through the circuit. <S> It seems that your simulation thinks your capacitor is fully charged when it begins. <S> This is definitely not what I would have expected to see. <S> I would have expected a voltage curve on the capacitor increasing up to around 6 volts in around 4.5 seconds, and then level off (and the current with an inverted version of the graph, eventually dropping to 0). <S> Check the initial conditions of your simulation to make sure it doesn't treat your capacitor as charged on startup. <A> The simulator tries to find a reasonable initial condition for your circuit. <S> In this case it assumes that the capacitor is already charged. <S> Either use a pulse (square wave) as an input source or specify an initial condition directly.
| When you first apply a voltage across a capacitor, assuming the capacitor is discharged, it acts as a short, and thus will show 0 volts across it.
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Why should we set input source equal zero for calculating output resistance of common source amplifier? This lecture (page 10) gave derivation of common source output resistance. What I don't quite get here is why we should set the input source equal 0 for calculating output resistance. Could anyone explain that? <Q> You are considering the small signal model <S> so we have to short all large signals. <S> But in your case you also have a small signal source at the gate(input) . <S> As many have already answered, its just a matter of perspective. <S> You want to measure the output resistance. <S> Even when you consider any normal circuit in real life and when you have to find the resistance of it with a multimeter. <S> You de-energize <S> all the sources (Short independent voltage sources and open circuit independent current sources) and then measure the resistance. <S> So in your case also you do the same. <S> You could of course have the small signal source <S> Vs (large signal source <S> Vbias should be shorted by any means because you are doing a small signal analysis and all large signals should not be considered) but now if you trry to calculate the rresistance from the output it would be incorrect because you have other sources influencing the resistance, one case where we need to have it is when we are findinng the transfer function where you need to find the Vout/Vin. <S> So to find the output resistance we short the source, since Vg=Vs=0 <S> and Vsource=0(gnd) <S> Vg-Vsourcce=Vgs=0. <S> So gm*Vgs=0, this is what is done in the equations you have mentioned. <A> <A> The reason is that when you try to get a Thevenin equivalent circuit, using the 'test source' method, you set all the independent voltage and current sources to zero. <S> However, you keep the dependent sources intact. <S> \$V_t\$ is your test source and to find the equivalent resistance looking into that terminal you need to find \$I_t\$. <S> Then \$R_{equiv}=\frac{V_t}{I_t}\$. <A> Simply, the measured test current would be the sum of the circuit response to both the test voltage and the source voltage, and an incorrect output resistance calculation would result. <S> The formula you included shows it quite nicely: if vgs ≠ 0, then vt = it(r0||RD) + gm·vgs, and Rout= r0||RD + gm·vgs <S> /it, which gives the wrong result. <A> Measuring the output resistance of a circuit is essentially the same thing as measuring its Thevenin resistance. <S> When measuring the Thevenin resistance, all independent sources are turned off. <S> The (small signal) <S> input source is such an independent source so it is turned off -- and turning off a voltage source means setting the voltage across its terminals to 0. <S> The large signal input source (the bias voltage) sets the dependent current source \$g_{\text{m}} v_{\text{gs}}\$ of the MOSFET because it determines \$g_{\text{m}}\$ and other small signal MOSFET parameters, so it can't be entirely ignored. <S> Once it is used to determine the dependent sources it too is turned off. <S> Because of the topology of the circuit, 0 input means \$v_{\text{gs}} = 0\$: the MOSFET has ideally infinite impedance looking into the gate, so no current can flow through \$R_{\text{S}}\$ <S> and thus there is no voltage across it. <S> It turns out that the dependent current source \$g_{\text{m}} v_{\text{gs}}\$ is 0 as well. <A> Let me cover it all, DC signals are neglected in small signal analysis( when amplitude of signals is quite small ) because they cause constant current and provide a bias level, so pretty much neglecting them will be like neglecting a constant in a phase or frequency dependent equation which makes it simpler to understand. <S> Ok there goes the DC sources MOSFET have <S> pretty high input impedance means nothing or no current flows through the gate terminal, that is shown by a open( red ) at Hybrid-pi model of small signal analysis, <S> Now we know that the conductance of the channel in the MOSFET is effected by the voltage at the gate terminal, but the input circuit is isolated from the output circuit, this relation is established by concept of transconductance , which is the ratio of change in the drain current to the change in the input gate voltage vgs ( where small v and gs represent the small signal nature of the source ). <S> Now comes the superpositioning theorem Which states that the current from two sources in any branch of the circuit( linear circuits ) is independent of other sources, and the resultant current is the algebraic sum of current calculated when individual sources are acting alone. <S> Hence here by grounding we are removing the effect of the input voltage, which also makes the gm.vgs ( where Is=gm.vgs is a dependent current source with dependency upon vgs ) acting as a current source zero, hence now we can calculate the sole resistance of the channel by connecting a small signal source vt at the output when the small signal input is zero. <S> Though the vgs doesn't play a direct role but works as a Transconductance amplifier, hence the effect of other power supply is removed by grounding vgs. <A> My answer is written as if dealing with a real circuit amplifier but the same principle is involved when calculating values. <S> You don't have to set the input source to zero if you want to do it the harder way. <S> That harder way would test the impedance at a specific frequency other than what the input source is producing then you would use a tight band-pass filter (or fourier/spectral analysis) to analyse JUST the signal at the frequency you want to measure and calculate output impedance. <S> Yes, you can do it that hard way or, just set the source voltage to zero and do it the easy way. <S> Alternatively you can use the input signal to generate a voltage and then place a load at the drain/collector and when that AC voltage seen drops to 50% you have the equivalent of the output impedance.
| My answer in one sentence: When you try to determine the current through the unknown output resistance (caused by a voltage source that is connected at the output node) there must be not another current caused by the input signal.
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What can I replace a BSS138 N-Channel MOSFET with? I'm replicating this level shifter by Sparkfun but they use a transistor that I don't have. Frankly, I know next to nothing about transistors so reading the specs is way over my head at this point. I do have a "kit" that I bought that has a bunch of them that might be able to replace the BSS138. Here's the list of the transistors that I have. Can anyone suggest a replacement for the BSS138? 2N4401, 2N3904, PN2222, 8050, PN4393, J113, 2N3055, TIP41C, TIP31A, 2N5458, 2N3904, 2N3906, 2N5088, 2N4401, BC547. EDIT: Commenters saying that none of these parts will work as they are not the right kind of component. Having said that, could you suggest a through hole part that would satisfy these requirements? <Q> See here -> <S> Digikey has the BSS138 <S> and they can be obtained from them relatively quickly. <S> They do not seem to have through hole versions. <S> Once you play a little you will find that the SOT23 smd pkg is not too hard to hand solder - and you can add wire leads if needed. <S> That level shifter circuit is trickier and more critical of part characteristics than most simple circuits are. <S> The BSS138 is an "N Channel MOSFFET". <S> It has a much lower turn on voltage (Vgson) than most similar MOSFETs and this is important in this circuit. <S> Alternative parts in through hole are available but they are less liable to be available to you than a BSS138. <S> eg see Digikey for TN0702 <S> or maybe the LP0701 - but <S> BSS138 is available from the same supplier. <S> If you can learn to handle the SOT23 pkg a lot more parts will be usable. <A> I know <S> the question was answered years ago, but I had a similar problem, a supplier (LCSC) didn't had the BSS138 as a "basic" component <S> and I had to find an equivalent one that was "basic" to reduce assembly costs. <S> So, to make things easier for someone else who stumbles into this question... <A> Just to add to the excellent accepted answer.. <S> Have a look into IRLML2402 , they have a low Vgs/th of 0.7V and are very fast. <S> For level shifting you want Vgs/ <S> th (threshold voltage) and on/off delay timings to be as small as possible, and Vgs and Vds max rating should be sufficiently high for your application.
| For low (a few kHz to a few hundred kHz, "serial port" speeds) speeds, the 2N7000/2N7002 worked fine, and they're readily available both in SOT-23 packages with the same pinout as the BSS138 or in PTH TO-92 packages.
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Best way to get components for a new electronics lab I recently joined a lab that is still in the process of getting set up. Our approach to acquiring electrical components (e.g. resistors, caps, opamps, etc.) so far has been to just wait until we decide we need something, and then when we do need it we buy a bunch of that component, along with maybe some related components. Unfortunately, I'm finding that this work flow leads to a lot of time spent waiting for parts to arrive. So my (soft) question is: What's the best way to stock a new lab with (standard) electronics components? Are there e.g. kits one can buy, or should I just go systematically through a vendor website and buy 20 of everything I could possibly need? (Or perhaps the approach we are taking is best, and we should just suffer with it for a month or two until we are eventually all stocked up...) <Q> Search on ebay or aliexpress and very often the price for 2 pieces is the same as for 100 pieces. <S> SO search what you want and enter something like: "50 pcs" or "100 pcs". <S> And mostly they are not much more expensive and you get 100 pieces instead of 2 for the same price. <A> It completely depends on what type of components your lab needs. <S> If we are talking microprocessors, motors, servos, transformers, high voltage relays etc. <S> It will be very expensive to keep all you need in stock. <S> However apart from this cheap components I would not recommend keeping anything else in stock unless you are planning to go big. <S> Delivery times from Farnell and Mouser etc. <S> within Europe and US are 1-2 days. <A> For passive components you could easily find a kit with wide range assortmets of a common value resistor, capacitor (plastic film, electrolytic, ...) etc. <S> If you talk about professional lab (I think it's the case), I prefer to buy from secure channel, like RS Components, Farnell, Digikey , etc, Also the delivery time is short. <S> For a home lab you could buy from ebay or similar without problems. <S> With less than 20 euro I bought a lot of assortments from AliExpress (resistors, capacitors, button, AVR programmers, etc). <S> I done a little lab. <S> But the delivery time is really long, one month or more.
| However if we are talking passive components, transistors, wires, diodes etc, then I would recommend going onto Mouser or Farnell or similar store and buy a 100-1000 of each of these inexpensive components, that will not cost you much. My suggestion overall is: buy basic components for your lab and keep them in stock, and do some planning before each project, this way you should be able to order things and get them to your place in reasonable time.
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Why are there no easy-to-solder SMD diodes in 0603 packages? What seem to be the most popular SMD diodes currently come in SC-79 packages, but they super tiny and really hard to hand solder... ...and they also can be a hassle even for reflow. So I thought I'd replace them with easy to solder 0603 equivalent parts, but all the parts I can find on DigiKey have footprints with bottom exposed pins that are even harder to hand solder... There are also some SOT-23 diodes that would be easy to solder, but are not as easy a swap out because they are a different shape.... Why don't diodes come in normal, easy to hand solder 0603 packages? Is there some other 2-pin package that they do come in that is easy to solder and is easy to swap out for the SC-79 footprints? NB: Of course Light Emitting Diodes do come in all manner of nice 0603 packages... ...and for prototypes I have been using them in place of the normal diodes (sometimes the feedback is actually nice!), but it is not always practical to use LEDs in place of a normal diode (if you need a low-drop shottky, for example). <Q> 0603 is 1.5 x 0.85 mm. <S> Digikey lists ~250 part numbers, from over a dozen manufacturers, for single rectifier diodes in this package size. <A> While Mr Photon manages to perfectly answer the spirit of your question (and by the way, there are many, many diodes that are well doable, but whether "they fit your application", who knows other than you) <S> , I felt like adding the literal answer to your question as well: Because parts in 0603 and similar sized packages are generally not designed for hand-placement or "hobby-shop" grade reflow, but for professional, high volume manufacture. <S> A small diode with bottom pads may be made, to facilitate the growing need from industry to have reasonable performance in a package that can be jammed right onto the next package without creating shorts. <S> Imagine your run of the mill 0603 resistor or LED. <S> If you had to place them as close together as possible... how big a gap do you need, even in professional reflow, to guarantee no shorts or solder bridges between the exposed ends? <S> That's pure loss in mobile phones and/or wearable tech. <A> It's not quite the same as 0603, but I often see MELF packages for SMD diodes: <S> These are pretty easy to hand-solder. <S> The main downside is that the cylindrical packages have a tendency to roll away if you're not careful about handling them! <A> In answer to why there aren't any "easy to solder" diodes, as others have shown there are some . <S> The machines have no problems with the existing packages, so why would large companies want to spend money on packaging in additional packages just to cater to a very small market. <A> I use a simple trick to overcome any soldering issues. <S> Increase your component's recommended land pattern by 0.5mm or so and that should ensure that you get enough space to place the component exactly where it should be. <S> It also makes soldering the component fairly easy. <S> If you are unable to find the "easy to solder" parts, this trick can make sure that you are not stuck in a situation like this again. <A> CDSU4148 is about as close as you will find in a 0603 package. <S> This is functionally equivalent to a standard 1N4148 diode. <S> CDSU4148 on Digikey <S> The downside is that they are relatively expensive compared to similar diodes. <S> FWIW - I'm in the process of changing one design from CDSU4148 to BAW56 in the SOT23-3 package BAW56 on Digikey <A> I'm assuming its simply because of the different voltage and wattage capacities of the diodes. <S> At 70V and 50mA, compared to a 0603 led at 3.3V and 20mA, the diodes have more heat to dissipate that the smaller gull wing type package can't handle.
| There are diodes available in "SOD-323" (aka SC-76) package, which is 1.7 x 1.25 mm, pretty close to an 0603. However the important thing to remember is that the "hobby" market is tiny compared to the industrial market, and for the industrial market, they use Pick and Place machines - you won't see many if any mass produced products that are hand soldered.
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How to switch separate circuit from doorbell? My doorbell rings as long as somebody pushes the button. I measured 10V AC at the bell when it is ringing. Now I would like to use this AC voltage to switch another circuit. How can I achieve this? Here is what I want to happen when the doorbell button is pressed: Let the doorbell ring. That is currently the case. Close another circuit, which is independent from the first one. Edit: As Olin says in a command, the proposed duplicate is about how to get the doorbell signal into a arduino and not about closing another circuit. <Q> What you want is called a "relay" . <S> Ideally you want one that runs from 10 VAC. <S> That's not a common coil voltage, but 12 V and 6 V are. <S> You could use a 6 VAC relay with a resistor in series, or full wave rectify the 10 VAC to make close enough to 12 VDC on the capacitor after the diodes. <S> Either way, the relay is basically a electrically operated mechanical switch. <S> That gives you the option of using it as a switch that is normally open except when the button is pressed, or normally closed except when the button is pressed. <S> Here is a schematic of what I'm talking about: <S> Note the capacitor after the full wave bridge. <S> This is necessary to keep the average rectified voltage high enough to reliably activate the relay. <S> The particular relay I'm showing in this example draws 27 mA at 12 V. <S> The peaks of the 10 VAC waveform will be 14.1 V. <S> The full wave bridge subtracts two diode drops for about 1.4 V, leaving 12.7 V peaks. <S> That will be what the cap gets charged up to twice per power line cycle. <S> For a 12 V average, the voltage on the cap can drop to 11.3 V before the next peak, for a total drop of 1.4 V. <S> You didn't say what your power line frequency is, so I assumed 50 Hz. <S> That means the cap will get recharged every 10 ms. <S> This gives us enough to compute the target cap value: C = <S> (27 mA)(10 ms)/(1.4 V <S> ) = 193 µF <S> So the common value of 200 µF will work fine. <A> simulate this circuit – Schematic created using CircuitLab Figure 1 (a) original circuit and (b) with monitoring relay. <S> How it works <S> The 10 V AC is applied to the bell. <S> Diodes 1 to 4 (or a small bridge rectifier) rectifies the AC voltage to give DC. <S> C1 smooths out the DC and holds the voltage up during AC voltage drops. <A> If I understand your question correctly, you want to connect two wires when the switch is pressed and supply both of them with the AC power. <S> To do this, replace your switch with a SPDT Switch.
| The 12 V relay will energise when the bell-push is pressed. The contacts should be rated for greater than or equal to the voltage and current you are switching. At minimum, you will probably find a SPDT (single pole double throw) relay.
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Is there such a thing as rubber band resistor? I'm doing a project with some tight requirements, in which I need to measure the distance between two points, which might not be in direct view of each other, and on parts that are too small to mount anything on them. An elegant solution would be to tie some kind of conductive rubber band between them, which would change resistance as it stretches, and then measuring the resistance, but all I've got from Google is stuff about physical resistance of exercise bands. <Q> There ABSOLUTELY are such devices, made of conductive rubber of some sort. <S> For example, HERE , from RobotShop.com <S> I suspect there is a bunch of hysteresis, and that the resistive properties will change over time and use. <S> Perhaps one of the other suggestions would work better <A> I'm not aware of any "rubber resistors", but there is a thing called a string potentiometer that implements pretty much the same function. <A> A spring converts a displacement into a force. <S> You can then use a "load cell" to measure that force. <S> Include words like "miniature", "micro" or "SMD" in your search. <S> There are several manufacturers of devices that fit your space requirements. <S> You could also use an actual rubber band, but a metallic spring will give more consistent and linear results. <A> Yes there is. <S> It normally changes its resistance on compression rather than tension, though there may be tension operated versions, and it uses quantum tunnelling. <S> The first available version was the "QTC pill". <S> Availability was difficult a couple of years ago but seems to be improving again... <S> Here's one source Wiki article <S> Some background from the Nuffield Foundation <S> These are manufactured by Peratech <S> who now offer a range of form factors for force sensing, as well as custom products. <A> While far from a rubber band, you might consider a rotary encoder with a spool of steel wire. <S> Reset at a certain distance, maybe 0 cm, and then count the revolutions of the spool and calculate the distance. <S> Although you would have to hold the device close to one object if too small to mount the device on. <A> There are three options that I know of for stretchy conductive strings, none very good. <S> A much better alternative would be something like acoustics, ultrasonics. <S> It can even go other than line-of-sight, if that's a problem to be solved. <S> You don't say what stability is required, either short or long term. <S> Although the speed of sound in the atmosphere is dependent slightly on pressure, temperature and humidity, it is relatively simple to install an additional calibration recevier at a fixed reference distance from the transmitter, and make what is effectively a comparison between the reference and the measured distance. <S> So, a review of the stretchy strings ... <S> a) <S> Any metal wire, obviously steel has larger elastic range than copper. <S> The change in resistance will be tiny, and easily swamped by the temperature coefficient. <S> A wire like constantin will have a much lower tempco, and may be more useable. <S> b) <S> A rubber tube filled with mercury has been used in the past for exactly this purpose, however its toxicity rules out its use in this sort of application these days. <S> There are other less toxic metal liquids available, though they are pricey and difficult to obtain. <S> An aqueous solution of some suitable salt, measured with AC, say copper sulphate between copper electrodes may be possible, but temperature coefficient would still be a problem. <S> c) <S> This has quite a marked tension/resistance response, but it is weak, and prone to creep. <S> A broad equivalent is the carbon-filled plastic anti-static bag material. <S> The same problems of weakness and creep apply. <A> Why not forego the mechanical solution and go with a purely electrical solution? <S> How about using laser distance measurement? <S> It won't have reliability issues and you won't have to worry about tolerances on the piece that measures tolerance. <S> Also with the mechanical solution it's accuracy <S> will most likely decrease over time. <S> Here's an example of a laser tape measure hack: https://www.sparkfun.com/tutorials/323
| You can get silver-filled rubber cord, for making conductive gaskets.
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Unsual sizes of TFT, LCD or OLED panels? On Stackoverflow there was a question about a round TFT display, but that answer was difficult and pricey. What is the possibility of finding rectangular LCD/TFT/OLED displays that are small in height, or long in width? Like a LED strip banner display, but then in either color or monochrome. Like one third the height of a normal 16:9 monitor. So 16:3. <Q> For larger displays, there is no demand and as such the cost of the NRE of first runs are so high that there is ZERO reason to do this. <S> If you do need such displays and have the $$ <S> then there are cool companies that can take also any display and cut it down to size. <S> Of course they can only cut to modulo driver chip size (i.e. they have to cut between driver chip lines) but they can do it. <S> One such company is http://pixelscientific.com/ <A> A little something I want to add: some display technology allows for borderless displays. <S> You can combine several of those borderless displays to make what is effectively one big display. <S> Although this does mean that your controller chip can churn out that many pixels fast enough (so something with a proper GPU may be called for, like a nVidia Tegra which comes with an onboard GeForce GPU) <A> When designing in a TFT displays, go with a standard. <S> Like $200K or more just for the tooling. <S> Try to go with a standard size TFT or switch to a monochrome.
| In short No, there are some companies that provide smaller displays for wearables and small electronics displays, but those are standard sizes , like electronic shelf labels etc. The cost to tool up a custom TFT to meet your needs of one that is short and long is reallllly high.
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Connecting 74HC165 to 74HC595 This is a rather noob and potentially silly question. I am planning to allow the user to randomly connect the 8 output pins of the 74HC595 to any of the 8 input pins of the 74HC165 shift out and shift in registers. I am trying recreate a digital Enigma machine. I am wondering if #1 and #2 from http://www.rugged-circuits.com/10-ways-to-destroy-an-arduino/ will allow me to add some extra protection to the input and output pins to the shift registers. <Q> If your purpose is simply to be able to detect which output is connected to which input, you should be able to add a resistor up to 100K or perhaps more in series with each input or output without affecting operation. <S> Adding capacitance will significantly increase the time required for signals to propagate; the approximate time in seconds will be twice the resistance value in ohms, times <S> the capacitor value in farads (or if you prefer, resistance value in megohms times the capacitor value in microfarads). <S> Using 100K resistors and <S> 0.1uF <S> capacitors would mean that you'd have to wait about 0.02 seconds for signals to propagate, but your board would be extremely robust. <S> Reducing the caps to 0.001uF cap would cut the time by a factor of 100 but your board would still be pretty robust (connecting to AC120 would probably not faze it). <S> If you want to be robust against AC240, that could be achieved if you use half-watt resistors but otherwise leave the circuit as-is. <A> In my opinion you have to use few resistors at the output of 74HC595 and at the inputs of 74HC165 like follows; +5V \ <S> / <S> \ <S> 10K ohm / | | 110 ohm <S> | 110 <S> ohm74HC165 input -----/\/\/\/\----|---o-------o----/\/\/\/\-------- <S> 74HC595 output <S> * <S> ** are connectors on PCB for connections to users <A> Masterleous' suggestion is a good one to protect the HC outputs when several are shorted together. <S> Wrt the HC inputs, I would be most concerned about ESD damage with lot of plugging/unplugging. <S> In order to protect from that, I would consider a small zener diode at each input (connected from input to ground after the 110Ω resister suggested by Masterleous), with Vz~4.5 V.
| If you want somewhat more protection, you could add some small capacitors on the inputs to the 74HC595.
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What are all the reasons that transformers are considered safer? I am making a circuit for measuring AC mains voltage and current and I want to avoid using a transformer because it introduces distortions. But everyone says that the circuit will be considered unsafe (without explaining why). I know one of the reasons for this is that using a transformer separates the neutral and it is harder to complete a short circuit. But this isn't a big downside for safety in a case enclosed circuit (or at least that's what I think). What are the other reasons? How is using a transformer different than using a voltage divider (apart from the neutral separation)? Is it because of possible malfunction of a resistor causing it to short? <Q> Transformers are safer because they are isolating your voltage from the mains (even if your transformer is 1:1 and there is no voltage difference). <S> The reason is the ground connection (please see this article for difference of ground and neutral). <S> A really good answer with pictures <S> you can find here <S> To sum it up. <S> If your ciruit is not isolated and you touch a single wire of it the current can flow through your body to ground. <S> If your circuit is isolated by a transformer, the floor you are standing on has no connection to the secondary circuit of your transformer. <S> The voltage potential will be only between the secondary wires of your transformer. <S> So touching a single wire will not let any current flow through your body. <S> Only when you touch 2 positions of your circuit with different potential, current will flow through your body. <S> Its considered safer as the chance to accidentally touch 2 positions in your circuit is less then only one! <S> A voltage divider does not have this effect, as it will not isolate your circuit from ground. <A> I am making a circuit for measuring AC mains voltage and current 1) <S> If your looking to accurately track the AC mains voltage (L-N) over a wide range an isolation transformer will give you poor non-linear results. <S> As the AC voltage drops lower the transformer core transfers energy less efficiently, so your low voltage readings are actually higher than your meter indicates. <S> 2) <S> To get accurate voltage measurements use probes with differential inputs, rated for 500vac or twice the voltage your testing. <S> They have a common ground which can be tied to earth ground if not done by design. <S> I used that type of setup to measure the voltage and current on a 0-30KVDC wire used to charge a capacitor bank for surge testing. <S> 3) <S> For voltages below 1,500vac you can use analog-isolators such as those that Analog Devices make, if you do not mind the $30 USD cost. <S> How is using a transformer different than using a voltage divider? <S> 4) Avoid voltage dividers because you do not have galvanic isolation from the AC mains, where as the methods I mentioned do, including use of a isolating transformer. <S> 5) Think about what your needs are, and ignore the 'he said/ <S> she said' statements, or you would be to scarred to build anything. <S> Be careful no matter what method you use. <S> If you use a transformer use a fuse on the primary in case of things going wrong downstream. <A> As others have said, galvanic isolation is recommended when interconnecting a low voltage circuit with a high power circuit. <S> There are various commercial off-the-shelf components that provide galvanic isolation—e.g., <S> isolation amplifiers and opto-couplers . <S> Also research how to properly route air gaps on/in circuit boards to prevent arcing from the high voltage section to the low voltage section. <S> (There are various questions/answers on the subject of routing air gaps on PCBs this forum.)
| If your tracking an AC voltage over a narrow range then a transformer is good enough.
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Undeletable EEPROM or other storage for micocontrollers? Is there such thing as permanent EEPROM storage? The data written should be un-deletable. Once written it should be there as long as the chip is not destroyed or damaged. Any other alternatives to EEPROMs are welcome. I use DVD drives for PC. But I need a similar chip or system for microcontrollers. Important: I don't want this to be implemented in my microcontroller Software by custom code. The Chip or IC or System it self should have that by default. Link on Security.SE: https://security.stackexchange.com/questions/128160/undeletable-logging-or-file-writing/128161#128161 <Q> Yes -- what you are after exists, at least for small sizes. <S> While most OTP EPROMs are parallel devices designed to be "burned" in dedicated EPROM programming hardware, Maxim makes exactly the part you're after in the form of the DS2505 -- 16kbits of in-system programmable OTP EPROM, with a unique serial number lasered into the chip as well that can be used to thwart chip replacement attacks. <S> You'll need a wee bit of drive circuitry as well as some bit-banging code to talk 1-Wire as the DS2505 requires its Vpp pulse to be muxed onto the 1-Wire interface along with normal power and data. <A> One of your comments suggests it has to be resistant to an 'inside attack'. <S> Which will be difficult. <S> What do you actually want to achieve? <S> a) 100% reliable access to your data? <S> If so, a denial of service attack, damage of the data or the chip would be a lose for you. <S> b) 100% reliable trust in the veracity of your data, so no alteration without detection? <S> As you mention data valid until damaged in your OP, this could be sufficient, depending on whether 'destroyed' means the physical chip (or system it's embedded in) or the data on it. <S> If a) then you have an insurmountable problem if the attacker has physical access to the system. <S> If he can zap the chip, or the system, or hit it with a hammer, really all bets are off if he has access. <S> If b) then you don't need any physical controls at all. <S> A sufficiently strong cryptographic checksum stored with the data will tell you whether it has been altered in any way since writing. <S> So OTP and write bits are neither necessary (for b) nor sufficient (for a). <S> Make sure you pick a solution that meets your real needs, and understand what you really need. <A> Yes, it's called a ROM, read only memory. <S> A One Time Programmable (OTP) ROM is what you want. <S> Alternatively an EEPROM with a Write Protection or Write enable pin that you can tie to VCC or GND to prevent rewriting. <S> It's a less than 100% solution, if you have physical access, though you could just pour epoxy or potting compound over the EEPROM afterwards to make it really hard to write enable. <A> Other people have asked you to cover your ground rules already, so we know that whatever you put on can be damaged or removed if the aggressor has access to the components. <S> I don't know how much ROM data you need. <S> But looking at what can be programmable by you and then stay 'tamper-proof', I'd suggest using a Lattice iCE40 FPGA such as an iCE40LP1K. <S> It contains OTP configuration Flash, so you alone can programme it but no-one can read it or reprogramme it. <S> As per p6 of Lattice TN1248 'iCE40 Programming And Configuration': <S> "The NVCM contents are entirely contained within the iCE40 device and are not readable once protected by the one-time programmable Security bits. <S> Furthermore, there is no observable difference between a programmed or un-programmed memory cell using optical or electron microscopy. <S> The NVCM memory has a programming interface similar to a 25-series SPI serial Flash PROM. <S> Consequently, it can be programmed using Diamond Programmer (version 2.2 or later) before or after circuit board assembly or programmed in-system from a microprocessor or other intelligent controller. <S> " <S> You then design an FPGA circuit to implement a ROM of your data and your communications port. <S> That circuit is then loaded into the FPGA configuration from the internal OTP memory. <S> Your ROM communications port protocol is of your choice. <S> It can be to a standard (e.g. SPI, I2C) or a made-up one that's difficult to understand by an intruder with a 'scope. <S> I appreciate that obfuscation of data will not save you in your application but it will slow those intruders down a bit and its a cheap option. <S> Which device you need and whether an iCE40 can hold all your data depends how much you've got <S> and I don't know that. <S> Oh, and I think the iCE40 also has an internal oscillator <S> but I'd need to check that. <S> Putting a conformal coating on to make it very difficult to observe pin activity seems a cheap option afterwards, if its of use. <S> You can either have it done professionally or get medieval with a packet of Araldite. <A> Some chips have OTP areas that are for serialization information etc. <S> It depends on the manufacturer and product line as to whether that is available on-chip though.
| But EEPROM can be 'permanent' if you write your program so that it can't be overwritten!
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Auto power off for 220 V AC cable breaks Can anyone suggest a circuit to automatically cut power when 2 core 220 V AC cable tied on poles powering 12 V SMPS at different locations breaks between any poles? There is lot of monkey menace. __________________________________ EXAMPLE ONLY of use case: There is lot of monkey menace Sumatra, 2009.:-) Genuine photo(s) but badly edited. <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Using a "remote-control over mains" transmitter and receiver to prove line integrity. <S> There are a range of devices using the X10 protocol for home automation. <S> I can't recommend any particular product (and we don't do that on this site anyway) <S> but there should be something suitable for the scheme shown in Figure 1. <S> On power up the RESET button is pressed. <S> This energises RLY1 and provides power to the 220 V line. <S> An X10 transmitter at the far end of the line is wired for constant 'ON' transmit. <S> An X10 receiver at the feed end monitors this signal and once received it latches RLY1 on. <S> The RESET button may now be released. <S> If the transmitter signal is lost the receiver should drop out and remove power from the relay coil. <S> The line power will be disconnected. <S> X10 devices may be expensive - I don't know - but there are sure to be many similar systems available in various price ranges. <A> The problem appears to be connectivity to the end of the cable. <S> I assume that current monitoring is not sufficient, as the wanted load can vary, so lacks discrimination with respect to a break at the penultimate pole. <S> A suitable host, say a tablet or a Beaglebone, continuously pings it. <S> If the pings time out, it cuts the supply at this end. <A> If I understand the question properly (?), you want to essentially "turn off" the power when it is interrupted. <S> I have used used a simple relay exactly like the one shown at the left end of the circuit shown by @transistor. <S> You must press the "Reset" button to turn on the power to the load (the SMPS), but if the power is interrupted, the relay drops out and prevents powering the load until it is "Reset" again. <S> This is necessary in cases where you don't want the load to re-start by itself (without supervision) for whatever reason. <S> Or in cases where the power is intermittent or unreliable and when the power is interrupted, it is prudent to wait until the problem is located and repaired before attempting to re-power the load. <S> You don't describe exactly what is the motivation for needing this function, so we can't directly address your specific issue. <A> The simple and obvious answer is to use a residual current breaker. <S> Any current not flowing through both wires simultaneously is a strong indication of cable tampering or failure. <S> RCDs use a ferrite core that both AC wires pass to measure a difference current and when this difference current excedds about 30 mA <S> the device trips: <S> - Here's the problem and solution more pictorially: - <S> And here's an RCD: <S> - In the US and Canada they are called GFCI (ground fault circuit interrupter) and please note that this works without measuring or needing an earth wire. <S> Will 30mA kill a monkey is a different question that I can't answer but <S> 5 amps will!!
| An 'out of the box' solution would put a 'LAN over mains wiring' LAN adapter at the far end of the cable.
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How does the ELPF and XTAL inputs on the ADV7180 recommended circuit work? I am looking at the recommended circuit for the ADV7180 by Analog Devices and trying to understand some of the reasoning behind the design choices. I realize some of these questions may be a lack of understanding of some fundamental concepts and would really appreciate if someone could explain them to me. I know that the XTAL inputs are used for the 28.63636Mhz crystal but I don't understand how they came up with the circuit shown. What is the purpose of the additional resistors and the capacitors in this case? For the ELPF input, from reading the datasheet all I could get was that it is the "External Loop Filter" pin and must be placed as close as possible to the IC. What is the purpose of this pin and again, how does the recommended design work? The data sheet can be found here: http://www.analog.com/media/en/technical-documentation/data-sheets/ADV7180.pdf <Q> The capacitors for the crystal are crystal load capacitors . <S> The actual value will vary across crystal manufacturers, but the suggested values are a good start. <S> The datasheet is light on details of the required load capacitance for the internal oscillator. <S> The ELPF components are part of the loop filter for the internal phase locked loop . <S> I have extensive experience with this specific range of decoders, and if your application has any real temperature variation, the use of C0G / NPO capacitors in the ELPF is not merely recommended, but just about mandatory for stable circuit performance. <S> Note that ADI does not mention this <S> ; I found it out the hard way. <S> The component values in the ELPF are set by Analog Devices, and I was never told just why those specific values were used despite my best efforts to find out (my application was a bit unusual, so understanding all these things was rather necessary). <A> The ELPF pin is the output of the chargepump and at the same time the input of the VCO which are both on-chip. <S> the components have to be close to the pin as not to introduce extra inductance. <S> This is needed so that the filter suppresses the pulses coming out of the charge pump properly. <S> If Analog Devices could have integrated the loop filter on chip they would have but some components, in this case the 10 and 82 nF capacitors, are far too large to implement on-chip <S> so they have to be external. <S> The PLL locks the VCO frequency to the accurate frequency from the crystal so that the VCO can also be at an accurate frequency. <S> Without the PLL the VCO's frequency would be inaccurate and varying with temperature, supply voltage and aging. <S> A PLL eliminates all this. <A> This design choice is not ADV7180 exclusive. <S> It's how a quartz crystal oscillator works. <S> The circuit is called a Pierce Oscillator .A quit good explanation can be found in this MX COM application note. <S> The capacitors C1 and C2 form the load capacitance for the crystal. <S> The optimum load capacitance (CL) for a given crystal is specified by the crystal manufacturer. <S> The equation to calculate the values of C1 and C2 is Where CS is the stray capacitance on the printed circuit board, typically a value of 5pf can be used for calculation purposes. <S> Now C1 and C2 can be selected to satisfy the above equation. <S> Usually C1 and C2 are selected such that they are approximately equal. <S> Large values of C1 and/or C2 increases frequency stability but decreases loop gain and may cause start-up problems. <S> The resistor Rf around the inverter provides negative feedback and sets the bias point of the inverter near mid-supply operating the inverter in the high gain linear region. <S> The value of this resistor is high, usually in the range of a 500KΩ ~ 2MΩ. <S> Some of MXCOM’s ICs have this resistor internal, refer to the external component specifications in the data sheet of a particular chip. <S> R1 is the drive limiting resistor, the primary function of this resistor is to limit the output of the inverter so that the crystal is not over driven.
| The components on the ELPF pin is the Loop filter for the chip's PLL, read this to learn about PLLs .
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What are the uses of Lugs/Ferrules? I've seen various projects using copper lugs and compressing them with bench vices but what is the actually use? Can't you just use wires without the lugs? Image Source <Q> 1) It's not easy to make a safe and reliable connection under a screw head to stranded wire, especially wire with hundreds of strands ( <S> like very flexible very large gauge wire). <S> The barrel of a lug is designed to keep all the wire strands safely contained and under control. <S> This is especially true where solder is an unreliable mechanical bond for the wire to the equipment (e.g. in power distribution systems) <S> 2) <S> If the wire needs to be removed, replaced, moved, or added to the device, then a lug and lug socket (or screw/bolt) is a more field-configurable type of connection, and can be mated/unmated many more times than bare (especially stranded) copper wire. <A> When's the last time you've had to make wiring repairs out in the field? <S> Or heck, even re-doing the battery connection on your car. <S> It's much easier to take a set of crimpers and lugs with you than having a propane solder torch (not to mention you don't have the chance to set adjacent things on fire). <S> Add to that <S> when you've got massive wires, soldering is error-prone (strand oxidation if your timing/temp is off w/) and ugly. <S> Lugs are stupidly simple. <S> And that's a good thing. <A> The problem with using the bare wire is that copper is very ductile. <S> Given enough time it may work itself loose from even very tight compression.
| Steel is far less ductile so using a lug that holds not only onto the copper wire but also onto any sheathing as well will result in a much more robust connection.
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Aluminum sheet as an underlay for electonic equipment I am planning to build a NAS into a (sufficiently ventilated) closet. For flame protection and heat spreading, I mean to put a 2x700x550mm aluminum sheet under the equipment. Is this a good idea? Specifically: Should I ground the plate? If I ground the plate using the earthing conductor of the same power outlet I use for the NAS, will this influence the signal quality of the Power LAN that I mean to use as a network connection for the NAS? Will an aluminum basis wreak havoc on wifi reception, which I mean to use as a backup connection? 3 sides of the closet are brick wall/plaster, the front is wood. The router is maybe 5m away in the next room. For wifi connectivity, does it make a difference if I ground the aluminum sheet or not? <Q> I don't think a plate is going to help very much for any of the things you are trying to do. <S> If you actually need to prevent spread of flames, you would probably want to put metal under, and up the walls all around the NAS. <S> But that would adversely affect wifi. <S> If you do it anyway, I would not bother grounding the aluminum. <S> I don't see how putting aluminum plate or sheet under and appliance will help keep it cool. <S> Especially in an adequately ventilated closet. <S> There are other materials besides metal for fire resistance such as gypsum board (which provides some limited resistance) and some rockwool materials with excellent fire resistance and insulation properties. <S> Roxul is one brand. <S> If you use aluminum anyway, I suggest you put the aluminum between your NAS and the insulation. <S> And I suggest you make sure there is a continuous insulation barrier between your NAS and any combustible material such as wood. <S> Note <S> : I am not saying fire protection is required around an NAS. <S> That is your premise. <S> I am assuming it is true for the purposes of this answer. <S> But if you do, you might as well do it right. <A> It's unlikely to prevent a connection unless it's directly between the communicating nodes, or within one wavelength of either one, although you might see some signal degradation (or improvement -- <S> interference patterns can be become very complex, depending on frequency and what other conductors and absorbers are already in the area). <S> Worst case, you can use a coax cable to place the antenna farther from the NAS. <S> If you're going to ground the plate, do so before tuning antenna placement. <A> I'm wondering if the material that would work best for you is a ceramic tray. <S> I tend to repurpose things, so I'd be looking through a medical supply catalog for this. <S> I'm with you on the gypsum. <S> Material, glue and finish materials-who knows what substances you're working with.
| If you really need fire protection, I recommend using fire rated insulation. In reality, I am not sure you really need to put fire protection around an NAS. If your WiFi interface has an typical jointed external antenna, you'll almost certainly be able to position it for a good connection. A plate of metal (or anything conductive) will affect your antenna pattern, and it will do it whether grounded or not (although in subtly different ways).
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How accurate is a single GPS at detecting distance traveled? I read that consumer grade GPS is accurate 5-10 meters at resolving global positioning coordinates. However, my question is how accurate is consumer grade GPS at detecting distance traveled. For example: Assume I record a position with GPS and consider it my origin of (0,0,0). One second later I recheck my position & calculate the difference which result is (10mm, 20mm, 30mm). For this sort of example how much error should I expect? Thanks!! <Q> Relative positions are more accurate than absolute positions for a L1 C/A user ("consumer grade GPS"). <S> Differential GPS and RTK positioning take advantage of this effect. <S> Measurement of the carrier phase would yield centimeter level accuracy, as the L1-signal has 20cm wavelength. <S> Users without access to the military code can however not relate this carrier phase to the timing information exactly , there is always an ambiguity. <S> You can therefore only use code phase, which gives the 10m accuracy at best. <S> The situation is better <S> if you have access to information from two receivers or from one receiver, that had continuous lock on (some) satellites while moved. <S> You do then know the difference of carrier cycles. <S> For two receivers, you are able to resolve the ambiguity after some observation time, for one with continuous lock, the ambiguity will cancel off when subtracting. <S> For a commercial receiver, where you do not have access to the raw observables but only to the position output, this still means, that the traveled distance is more accurate than the absolute position. <S> This has to do with that fact, that receivers use "carrier phase aiding" and the last known position fix to some extent to improve their solution. <S> The fix would be much more noisy without such tricks. <S> I would expect around 20cm distance accuracy in optimal conditions (no multipath, short time (some seconds) between observations, no loss/acquisition of signal between observations). <A> The answer is "it depends" and you need to test in your application to find out if it will be suitable. <A> You can expect several meters of inaccuracy. <S> Unless you are moving several 10s of meters between "samples" then you should expect no useful delta position information.
| It can be quite accurate, but depending on the conditions of the environment (open field vs city canyon) and the position of the GPS (in your pocket vs sitting open with a good sky view) it can have a fair amount of random walk.
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Ciruit Board with traces on top and bottom? I was looking at re-making a Circuit board. The Circuit board I have has the same exact traces on top and bottom. I cant figure out why this would be done. Any one have any idea why you would have the same layout for traces on top and bottom side? ThanksGM EDITPretty much its a back plane that takes a 50 pin ribbon cable from the back side of the board to a 65 pin quick connect to front side of board. with the other 15 pins, they go to other connectors that are connected to DMM , Spec Anlizer , o scope ect.. <Q> For lower volume PC boards manufactured in the U.S. or Canada, it is usually much, much cheaper to just use a board house' cheapest and most standard setup: 2 layers, 0.0625" thick FR4 board, green soldermask, standard drill sizes, etc. <S> This way, they don't have to change the work flow. <S> If that is the case, then you get the bottom side for "free" <S> so you might as well use it to decrease resistance and improve reliability. <S> Duplicating the top layer is also a no-brainer. <S> Don't bother laying out the bottom layer. <S> Just copy the top gerber file and rename it as the bottom gerber file. <S> Any text will read backwards though :) <S> You can use a text merge program (used by SW dudes) to see if there is any difference between the top and bottom gerber files. <S> They are simple text files. <S> If exactly the same, this is probably what the previous engineer did. <A> PCBs come with different weight copper (that is, how thick the copper is ). <S> Hardware designers are sensitive to BOM cost. <S> Especially on high run PCBs. <S> Another possibility, if the traces are offset by one from the top to the bottom, it may be a way of adding shielding. <S> It is common to interleave ribbon cable signal with ground. <S> If this was done here for this back plane, that would mean a ground trace would be above (or below) and to the left and right of each signal trace. <A> Ideas that spring to mind are: (a) to increase the current capability with the thin tracks you need between these connectors; (b) to let you put the board in backwards and solder the through-hole connectors to the other side. <A> When a PCB has significantly more copper on one side than the other, the board can actually warp or bend. <S> Copper foil and fiberglass have different temperature coefficient of expansion, and with an asymmetrical layer stack the board will warp. <S> It's more of a problem with large boards or very thin boards (like 0.031" thick instead of the usual 0.62"). <S> The warping problem can be addressed by mounting board stiffeners, but these add cost. <S> Duplicating layer 1 to layer 2 prevents the board from warping, by keeping the internal stresses symmetrical.
| It may have been that the cost of a light weight double sided copper board was lower than a heavy weight single sided copper board.
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Bode plot of transfer function? I am having trouble where to start an attempt at sketching the Bode plot for the following transfer function: I know I need to rewrite it into its proper form, making both the lowest-order term in the numerator and denominator unity... but the denominator is what's throwing me off. If I keep all of the terms separate in the denominator, how can I find the poles of the transfer function? <Q> For the numerator you have \$(s^2+s+1)=0 \Leftrightarrow s_{n_{1,2}}=-\frac{1}{2 <S> } \pm j\sqrt{\frac{3}{4}}\$ <S> For the denominator you have \$(s+1) <S> =0 <S> \Leftrightarrow s_{d_{1}}=-1\$ as well as $$(2 s^2+1) <S> =0 \\\Leftrightarrow s^2=- {\frac{1}{2}}\\\Leftrightarrow s_{d_{2,3}}=\pm j\sqrt{\frac{1}{2}}$$because a product is zero if one or more of its factors are zero. <S> Therefore your transfer function is$$W(s)=\frac{(s-s_{n_{1}}) <S> \cdot <S> (s-s_{n_{2}})}{(s-s_{d_{1}})\cdot(s-s_{d_{2}})(s-s_{d_{3}})}$$ Here is a picture of what the bode plot should look like (sorry for just using Matlab instead of drawing by hand): <S> Please note that the transfer function drops with 20dB/decade after the resonance. <S> As for the resonance: @Mario has explained very well that you see a resonance peak because of the complex conjugated pole pair on the imaginary axis. <S> Because it is in the denominator, the peak points upwards. <S> As there is no dampening (the peak is on the imaginary axis), you get an arbitrary high value. <A> The poles can be found by checking where denominator becomes zero. <S> In this case you have a product of two terms and if one of them is zero <S> the whole expression is zero. <S> (1+s) == 0 results in s = <S> -1 <S> (1 + 2 s^2) = <S> = 0 results in s = \$\pm <S> j/\sqrt 2\$ <S> The complex conjugate pair indicates that you have resonance. <S> It will appear as a peak in your bode plot <A> Solve for numerator = 0 (for zeros) and denominator = 0 (for poles). <S> For a pole-zero diagram, plot the imaginary values to the y-axis and the real values to the x-axis. <S> I use Mathcad to do so, but you could do it by hand of course.
| You need to calculate the roots of numerator and denominator.
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How to Choose a Lead Acid Battery Charger? It has been suggested that you can charge sealed lead acid batteries with a normal power supply . On the other hand, chargers are available for purchase , but cost almost $100. Is there a reason to go with the store bought battery charger? If so, I have six 6V lead acid batteries that I need to charge. The maximum allowed charge current is 1.5A. Many chargers (unlike the one I linked above) only put 2A or more. Is there a way to make that sort of charger easily work? And do I really need to spend almost $600 on chargers if I want to be able to charge all of my batteries at once? Or can I hook them up in parallel. I have been reading pretty extensively on this over the last couple of days, but am finding conflicting information everywhere. The Battery University source seems good, but it also isn't immediately clear to me what to purchase. I don't want to screw up and end up buying an expensive charger that doesn't work, but I also don't want to spend days figuring out how these work, when this is just a small component of an experiment that I need to have working ASAP. (Therefore, I will go with whatever is easiest!) <Q> The Amp hour (Ah) rating of your batteries will make a difference on what size charger you should use to charge your cells. <S> Any charger providing a current with a value of more than (Ah/4) is considered a fast charging unit for that battery. <S> If you need your batteries to last, you should not use this method. <S> The method of parallel charging is acceptable, but will slow down the charge rate, essentially splitting your current by the number of batteries in parallel. <S> This method, combined with the knowledge above could bring down the current from a 'fast charging' unit to an acceptable charging range. <S> The biggest problem with using a regular power supply to charge your batteries is that the charging will have to be very closely monitored. <S> If the batteries become over-charged, electrolysis of the water in the cells will start to degrade the batteries. <A> I've always thought that this model looked interesting: https://www.amazon.com/NOCO-G750-UltraSafe-Battery-Charger/dp/B004LX3AS6/ref=sr_1_4?ie=UTF8&qid=1467226159&sr=8-4&keywords=lead+acid+battery+charger <S> It has a big-brother version that can charge 4 at a time. <S> I'd be strongly disinclined to charge them either in series or in parallel. <S> A better approach would have separate regulators for each connected battery, as does the unit I linked to. <S> The lead-acid charging is pretty simple, as these things go, and is readily accomplished with a lab power supply having both constant current and constant voltage (as they generally do), with just a little attention to when to switch from 14.4V accumulation charging to 13.7V float charging. <S> I expect you've seen that already in the question you linked to. <S> I don't see how you'd accomplish that in series or in parallel. <S> In series they all get the same current, which could overcharge one or more if they are out of balance. <S> In parallel they all get the same voltage, which I suppose could possibly work OK if you used a conservative current limit, but that would lengthen charging time. <S> Anyway, I'm thinking one charging circuit per battery is most reliable. <S> Whether or not these regulation circuits all feed off the same "bulk" supply internally would be a question of pricing and convenience (size and sprawl I expect :). <S> I like this page on lead-acid charging: http://www.evdl.org/pages/hartcharge.html <S> It's a little rambling in style but seems to hit all the important points well above the fold :) <A> A maximum charge rate of 1.5A suggests a small capacity battery (7.5Ah at 5 hour rate) which is probably an SLA (Sealed Lead Acid) type. <S> SLA batteries should be charged slowly (10 hour rate or longer) and never allowed to reach gassing voltage. <S> Unlike flooded lead acid batteries (which can be balanced simply by overcharging and then topping up with distilled water) <S> SLA batteries must be balanced by voltage alone. <S> So if you want to charge a bank of several batteries in series then you should charge each one separately beforehand to make sure they all have the same voltage. <S> You can then use a charger rated for the total voltage, but you should check the balance regularly and charge them separately again if needed. <S> If you intend to use the batteries separately then just get sufficient 6V SLA chargers to do as many as you need to at once. <S> Low current SLA chargers are cheap enough that you should be able to afford one per battery. <S> You could also use them to charge and balance the individual batteries in a series connected bank.
| A battery charger will have the necessary circuitry to maintain a constant voltage, limiting the possibility of over-charging.
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Help for identifying pinout of unknown mp3 chip I'm trying to use a cheap mp3 player to play song using Arduino, I couldn't find the datasheet of the main chip but I made some progress identifying some pins. Pin 1 16 Busy Play/Stop IO1 2 15 GNDPrev/Vol- IO2 3 14 Next/Vol+ IO3 4 13 5 12 6 11 SPK - 7 10 SPK Right 8 9 SPK Left This looks very similar to the picaxe by8001 and the WT5001 I will need some ideas to find the datasheet or at least find RX/TX/ADC KEY pins. More pics here Note: Why not use mp3 module for arduino? I try the WTV020 with no luck, I could buy a dfplayer or the WT5001, but this mp3 player cost 1/5 of those modules. <Q> It is a Jerry mp3 AC1082, a cheap encoder Chinese IC <A> And the labels seem to be pure confusion or some kind of serial number. <S> There’s a version with and without LCD display, the chips are probably very simlar. <S> The LCD version displays "ic type: <S> ht2836 firmware ver v1.00". <S> The chips seem to be sold by chipkingdom (the site also has a few "datasheets", look out for the GPDxxxx MP3 stuff).GP <S> is for General Plus who designed these chips to begin with, but no further information is available on their website. <S> Here are some further more or less interesting links that deal with the chip: <S> http://psmay.com/2014/09/18/teardown-1-45-chinese-mp3-player/ <S> http://tarkus-notes.com/en/chinese-mp3-player-ipod-shuffle-clone/ <S> http://overskill.alexshu.com/cheaphack-a-2-5-chinese-mp3-player-for-its-128x64-mini-lcd/ http://homeant.000space.com/articles/4/4/?i=1 <S> https://forum.arduino.cc/index.php?topic=361444.0 <S> http://ameblo.jp/akinas/entry-11978758574.html <S> http://web.archive.org/web/20160302021815/http://www.ieyong.net/en/chanpinzhongxin-259928-0-item-459063.html <A> After what felt like ages of googling, I found that it's made by Jie Li Electronics, a company based in Taiwan that mainly makes transistors, resistors, and capacitors. <S> They also make these ICs <S> but I couldn't find any info on them. <S> I'm looking for a similar IC from this same company, but I can't find any info on these. <S> So that logo isn't a Pi <S> , it's actually a JL like most of us assumed it was.
| There seems to be a common design for these cheap mp3 players.
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What switch do I use so as to be open when pressed, and closed otherwise? I need a switch that turns the circuit off when pressed on, but turns on when lifted off. this is to go under a lid I am making to turn on the lights when lifted. What is this switch called? it is the the only missing piece of the puzzle... Update: I need a switch for one circuit. low voltage @4.5v. What is the switch called, or is the more than one switch needed. <Q> You want a Normally Closed push button - the switch contacts will be connected unless the button is pressed. <S> A single pole, double throw button may be easier to find - that would give you both <S> Normally Closed and Normally Open contacts. <S> A common name for a switch that is operated by things rather than directly by people is MicroSwitch (TM Honeywell, I think). <S> Microswitches are available with a wide variety of actuating levers. <A> Search push switch on wikipedia. <S> https://en.wikipedia.org/wiki/Push_switch <S> There is mention of 2 basic type of push-button switch 1. <S> Normally Open / Push to make (the well-known one) <S> 2. <S> (that you're seeking) <S> Here is the circuit-symbol of 2 kinds of push-button switch. <S> The symbols carry a logical-hint of their actions. <A> You require a Normally closed SPST switch. <S> A very good example is an Emergency switch, which breaks off the circuit when pressed and also illuminates RED(usually) to give an indication that it has been activated. <S> Something similar to this http://www.digikey.com/product-detail/en/omron-automation-and-safety/A165E-LS-24D-01/Z1425-ND/456231
| There are switches available, which illuminates when the switch is pressed. Normally Closed / Push to break.
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Detect AC current in series with load I have a challenge at work to build a device to monitor the health of ceramic cartridge heaters, similar to this . It operates with an AC transformer and there are 5 heaters I'd like to monitor independently. The transformer operates on our 120V mains and is estimated to provide 1A to the heater. First I picked up a non-invasive current sensor but in the limited testing I've done so far, I think the < 1A current and the gauge of the wires won't provide the necessary field needed for useful monitoring. Additionally I'd have to scale the device's 0-1V output to 0-5V logic of my microcontroller. Then I thought about sticking an LED indicator inline with the heaters, like this one . This brings me to my main question: if I put this LED in series with the heater, as in Transformer > LED > heater, will the current drawn by the heater just burn out the LED? Would a resistor in the circuit allow the indicator to work properly without also limiting the heater and/or just melting? <Q> If you are just interested in a visual hint (rather than an integration into an automated system), just grab a bag of cheap current indicators. <S> Save yourself fiddling with a micro controller board. <S> Like these . <S> Not trying to advertise those specifically, just to get my idea across. <S> They were the first to show up on Google. <A> Current transformers with indicator LEDs are available. <S> (Search for "remote current indicator LED".) <S> The wire to be monitored is passed through the centre of the core. <S> These give the advantage of electrical isolation from the mains wiring. <S> Pay attention to see if there is a reverse protection diode in with the LED and keep polarity the same. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1a: <S> Likely internal circuit. <S> 1b: <S> Possible alternative with back-to-back LEDs in one package. <S> 1c: <S> Replacing the LED with an opto-isolator for connection to your monitor / alarm circuit. <S> Note that a certain number of ampere-turns (amps in wire \$ <S> \times \$ turns through the core) will be required to turn on the LED. <S> This will be listed in the datasheet. <S> If the device turns out to be not sensitive enough you can fix this by threading more turns through the core. <S> e.g., Load current is <S> 3 A. Current sensor rating is <S> 10 A. Then thread the conductor through four times as \$ 4~turns \times <S> 3~A = 12~At \$. <A> A lot of these are used in temperature control systems, but can be applied to others. <S> They come in a variety of current and voltage ranges and can sense down below 1 V, even in the mA or mV range if necessary. <S> Here are some examples, not playing favorites with manufacturers: OMEGA: <S> DMD1080 <S> Moore Instruments: Limit Alarm Trips and Switches <S> I have more examples including Acromag and Action I <S> /O and more from Omega, but since I don't have much rep, I can't post anymore links, you can search for them if you'd like.
| If required, you could cut the LED off and connect to an opto-isolator to pass the signal through to your controller. I don't know your budget, but you can look for a current (or voltage) limit sensor or limit alarm.
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What can I use to replace an SV-04F bias diode? I had a nice Wega V3840-2 audio amplifier, but sadly, it broke . During my attempts to test what was wrong, a diode, attached to the heat sink, failed with a nice whiff of magic smoke. According to the Kundendienstanleituing (user service manual), this particular diode is a SV04F. I cannot check the markings on diode itself, since it's embedded in epoxy in a little clamp that attaches it to the heat sink. I cannot find this particular diode at my local stores, so my question is, what can I use to replace it ? For your convencience, I also added part of the schematic. The top rail is +32V (also the leftmost connection coming from above), the bottom rail -32V. The connection center left is the actual audio signal, the output on the right goes to the speakers. The failed diode is D201, that's attached to the heat sink of T210 (which is in turn attached to the main heat sink of all four final stage transistors). Note about the circumstances it broke under for those interested: I replaced T210 and T211 with a BD2430 and BD244C in an earlier attempt to fix the amplifier (see first link of this post). I now realise that the trafo is set for 220V even though we have 230V these days, perhaps I'll set it for 240V. Other than that, I still don't know how or why my amp is no longer working... <Q> This is from an old Sony service bulletin. <S> It mentions SV04S (not F) <S> but part numbers show SV04, SV04S and SV04F as equivalent, so it should work. <S> It's a Vbe multiplier, also known as an adjustable diode or rubber diode. <S> It's made from an NPN silicon transistor (in this case 2SD1585) and two resistors. <S> You could replace both the 1k and 3k3 resistors for one 5k linear trim pot, so you can precisely adjust the DC bias for the output transistors, like so (fig 2.5b): <S> I built a similar device to fix an old Sony receiver before I bumped into this solution, so I know it works fine. <S> I'm not sure if the resistor values are ok for your particular application, so maybe the 5k trimpot should be the way to go, at least until you find the exact resistance values to use. <S> As it was said before, it's important to thermally couple this assembly to the output transistors (mounting it to the heatsink close or at the same place as the output transistors should do) to get thermal feedback. <S> I just replaced my previous fix (BC550 NPN transistor + similar resistors) with this very device on my Sony TA1630, and can confirm it works fine <S> so it should work in your Wega amp as well. <S> I built the device right on top of the D1585 transistor (you may prefer using a PCB, but I think it's overkill. <S> YMMV). <S> With some patience and 1W metal-film resistors, it can be built lifting the base pin, and putting both resistors in front of the transistor. <S> Then, I tapped two holes on the heatsink, on top of the PNP power transistors and fixed the device there. <A> The purpose of D201 is to bias the push-pull darlington output stage. <S> Since that output stage has 4 B-E drops between the bottom and top of D201, this is probably 4 diodes in series. <S> It's not clear how much current the diode needs to handle since it's hard to tell how much of the total 64 V drop T206 and T207 eat up. <S> The remainder of the 64 V that isn't across the diode and transistors is across the sum of R275 and R276, which is 200 Ω. <S> The absolute worse case is 300 mA, but the actual maximum could be substantially less. <S> You can replace D201 with 4 diodes in series. <S> These should be rated for 1/2 A, which rules out the 1N4148. <S> These diodes are always forward biased and are only used as voltage sources. <S> If D201 fried, then probably at least one of T206 and T207 has failed shorted. <A> Not sure if this is the same device <S> but I came across this while reverse-engineering an old calculator: <S> I took some measurements and it looks to be consistent with 4 diodes in series: Current (mA) <S> Forward Voltage <S> 0.01266 <S> 1.871 <S> 0.0522 <S> 2.057 <S> 0.1022 2.143 <S> 0.360 <S> 2.31 <S> 0.402 <S> 2.325 <S> 0.719 <S> 2.407 <S> 1.002 <S> 2.454 <S> 1.226 2.485
| Just about any ordinary silicon diode that can handle the current should work here.
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Can you measure voltage along a wire a multiple points simultaneously? Basic EE question here. Suppose I have a piece of wire a foot long. Suppose there is some signal (for example a clock signal) going through the wire. Is it possible to simultaneously measure the voltage on this wire at (for example) 4, 5 and 10 inches along this wire? Is it correct to assume these readings would not all be the same? <Q> Your question assumes 2 things.1) <S> The signal is a constant current DC, typical for measuring ohms. <S> 2) <S> A low frequency ac signal, such that the wavelength is at least 20 times the length of the wire. <S> I mention these first because your question also having simultaneous measurements at different points along the wire. <S> With DC current and LF AC readings will be stable and predictable. <S> Above a few hundred MHZ you will see dips and peaks in signal amplitude along the length of the wire, due to standing waves. <S> If you keep the same sample points and increase the frequency these peaks and dips will move along the wire. <S> At some point, when the wire is equal to 1/8th to 1/4 wavelength, grounding one end is a moot point because so much energy is radiant, and the very connections you make for testing begin to affect the signal. <S> At some point your equipment becomes very expensive and you would use a spectrum analyser (about $30,000 USD) to probe the wires RF performance. <S> Readings along the length of the wire will always be different as long as there is current flowing , whether its DC, or LF AC. <S> HF AC or VHF/UHF frequencies will have peaks and dips in the signal, until your outside (too high or too low) of the resonant frequency of the wire. <A> If the frequency of the current is high enough to the extend <S> where the foot length wire exhibits significant impedance, and the magnitude of the current itself is significantly high, <S> then yes. <S> There will be a ranging voltage drop along the length. <S> I will say this can happen only at radio frequency of VHF band and above. <S> This actually happens in the telescopic antenna <S> mounted on ordinary FM radio but the voltage induced on those antennas are in microvolts, not measurable with ordinary multimeters. <A> Is it possible to simultaneously measure the voltage on this wire at (for example) 4, 5 and 10 inches along this wire? <S> Certainly. <S> But note that when you start getting up into very high frequencies, this becomes very difficult to do accurately. <S> Is it correct to assume these readings would not all be the same? <S> That depends on the frequency of your signal. <S> But at high frequencies, your foot long piece of simple wire becomes an improperly-terminated transmission line and <S> what you measure at different points becomes very messy. <S> So it comes down to what were you imagining in your hypothetical scenario. <S> If you were talking about frequencies under a few hundred MHz, then you have the simple case. <S> But if you were imagining higher frequencies (100s of MHz, or GHz, etc.) <S> then it is a much different game.
| At low frequencies, all the readings will be identical for all practical purposes.
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Dual power with lm1117 5V I am working on RGB light strips. I have created the circuit that includes at Atmega 328P processor. See attached image. I have question about applying dual power supply. If using only 5V supply, I could feed the Atmega chip directly and also the led strip (addressable that uses 5V supply). Using only 12V supply, it will feed different led strip (non addressable) that runs on 12V source. Also it will feed power to +5V regulator. At its output, it will feed 5V to the processor. Now the question is... If connecting two power source ( to drive two different strips at the same time), as in the diagram, is it safe to run as in this circuit? I haven't tested with dual power yet.Note: +5V source is also connected to the output of regulator. <Q> You mentioned that your LEDs and 5V and on 12V both need around 10A of current. <S> And you have two power supply with 5V and 12V. <S> Thus just connect the Grounds and use the 12V supply for the 12V stripes and the 5V supply for the Arduino and the 5V stripes. <S> That's it. <A> No, because if the two voltage supplies have even a little bit of variance in the voltage each supplies then you could burn out one or both of them. <S> If you add a ballast resistor to the external 5V input then the voltage difference would result in a dissipation of power in the resistor rather than either supply. <S> A 0.5ohm resistor with a 0.5V difference would consume 500mW, which a 1W or 2W resistor could handle (although using a 5W device would not in any way be a bad idea). <S> At the same time, it would only have a 75mV drop if the MCU were to draw 150mA from the external supply if it was the only one connected, resulting in a voltage well above what the ATmega328P calls for. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Hmm I am thinking of the alternative... by adding the "power source" switch to the MCU circuit. <S> Note: The switch would be "ON-OFF-ON" switch ("12V-OFF-5V")
| You also want to reverse-bypass the LDO with a schottky diode in case someone shorts the 12V input to ground with the 5V supply connected so that the current doesn't flow through the LDO's internal parasitic diode, thereby burning it out.
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Can I use two DC-DC Converter modules to generate a +5V and -5V Rail? These converters from Pololu look great ( https://www.pololu.com/product/2831 ), but I need both a positive and negative rail. Can I use two of these to make a +5V and -5V rail? For example. if I tie the + terminal on one to the - terminal on the other and tie that node to ground and then use the respective output terminals as +/-5V rails, would that work? <Q> The short answer is not the way you have described: these are non-isolated converters. <S> As you can see in the picture (from the web site), there is a single ground pin (for both input and output). <S> To achieve what you want to do you can either use isolated converters or you can use two of them in this manner: use one to get 5V; power the second one from this 5V to get a negative output provided your total load is within the current allowed from the first regulator. <S> As these are rated for 36V you should be fine - the second regulator will see an effective 10V in to out. <S> I made a picture of two regulators showing how I would hook them up for +5V and -5V <S> : <S> Hopefully that is clear. <A> Look at the schematic of this converter. <S> If you do what you suggest, it means you tie the 5V output of the second converter to ground, thus shorting it. <S> You need isolated converters to do this. <S> Or another option would be to use a non-isolated inverter DC-DC converter. <S> They are easy to find and would be cheaper and smaller than isolated ones, I think (There are none on polulu, it seems, though. <S> Try mouser/digikey). <A> No you can't because that would short circuit <S> the - 5 V converter as the input ground and the output ground are the same node. <S> It actually has only one connection pin ! <S> What you need is a regulator with an isolated output for the -5 V. <S> Then the output is floating and you can use it to make - 5V. <S> Some examples of these are here .
| Isolated converters have a transformer that allows this. So, no, since they are not isolated, you can't wire these converters as you suggest.
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How can I calculate the inductance of brushless DC motor? I want to solve the differential equation of brushless DC motor $$ L\frac{di(t)}{dt}+Ri(t)=V-k_e\omega$$ How I can calculate the inductance (L) of brushless DC motor? <Q> Take a 1 k\$\Omega\$ resistor, put it in series with one phase, then apply a sine waveform from a signal generator. <S> Adjust the frequency to get same amplitude on both resistor and the motor phase. <S> Then use \$ Z= <S> j2 \Pi <S> fL <S> \$. <S> The slight rotor movements should not be a problem since it will not achieve any significant speed. <A> I "think" inorder to calculate the inductance , you have to know the number of turns and wires Gauge and other manufacturing properties of the motor , which most likely are not available. <S> What's the easy way to measure a DC hobby motor's inductance? <A> 'Calculate' may prove difficult, in the absence of further information. <S> To measure \$\small L\$, lock the rotor, apply a small input voltage step, and measure the time constant, \$\tau\$, of the resultant exponential stall current using an oscilloscope. <S> This will give the value of \$\small L\$ from \$\tau =\frac{L}{R}\$ <S> (\$\small R\$ can be found from steady state voltage divided by steady state current). <S> You may need to add a small series resistance to measure the current - connect this in the ground supply line to the motor so that you can measure the voltage across it without shorting anything to ground. <S> And remember to include this resistance value in the time constant calculation.
| BUT you can measure the inductance using an LCR meter , or by some experiments using an oscillscope and signal generator:
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What is this style of button / pad called? How can I us them in my own projects? I really want to make some custom MIDI controllers. I've gone through the exercise of making a microcontroller talk USB to my PC and breadboarded some button circuits on perfboard. I'd like to step it up to the next level though and use "real" buttons and pads. Problem is, I don't know how to learn what to do next. I've attached a picture below of the style of pad I'd like to use. I thought this would be a "membrane button" but that isn't really what I want. Can you provide some guidance on what that style of button is and how I can use them in my own project? <Q> Wiki calls them Dome-Switch or Chiclet keyboards: <S> Dome-switch keyboards are a hybrid of flat-panel membrane and mechanical-switch keyboards. <S> They bring two circuit board traces together under a rubber or silicone keypad using either metal "dome" switches or polyurethane formed domes. <S> The metal dome switches are formed pieces of stainless steel that, when compressed, give the user a crisp, positive tactile feedback. <S> These metal types of dome switches are very common, are usually reliable to over 5 million cycles, and can be plated in either nickel, silver or gold. <S> The rubber dome switches, most commonly referred to as polydomes, are formed polyurethane domes where the inside bubble is coated in graphite. <S> While polydomes are typically cheaper than metal domes, they lack the crisp snap of the metal domes, and usually have a lower life specification. <S> Polydomes are considered very quiet, but purists tend to find them "mushy" because the collapsing dome does not provide as much positive response as metal domes. <S> For either metal or polydomes, when a key is pressed, it collapses the dome, which connects the two circuit traces and completes the connection to enter the character. <S> The pattern on the PC board is often gold-plated. <S> Both are common switch technologies used in mass market keyboards today. <S> This type of switch technology happens to be most commonly used in handheld controllers, mobile phones, automotive, consumer electronics and medical devices. <S> Dome-switch keyboards are also called direct-switch keyboards. <S> Keep in mind that instead of custom poly domes or metal domes, you could use the rubber part over mechanical switches. <A> That is a silicone rubber keypad . <S> They are usually custom-made, but SparkFun sells 2×2 arrays similar to those in that picture along with the other parts required to form an entire assembly. <A> Adafruit calls them: "elastomer keypad". <S> They were also available from Livid Instruments , but most of their products seem to have disappeared.
| Sparkfun calls them "silicon rubber button pad".
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Choosing decoupling cap for M74HC595B1R shift register I want to use a bunch of cascaded M74HC595B1R shift registers in my project. VCC will be 5V, clock frequency 8-10 MHz. Outputs of these registers will be driving gates of 2N7000 N-MOSFETS if that matters.My questions are Should I install decoupling caps for my shift registers and what value if yes? VCC and GND pins are rather far apart on the opposite corners of package. Does it matter if I physically place the cap as shown with red lines (unequal distances to VCC and GND pins) or as shown with blue lines (equal distances from pins on the opposite side of the board) <Q> You definitely should have decoupling capacitance on your board. <S> I would recommend a 100 nF ceramic or similar per IC. <S> You also should have bulk decoupling capacitance distributed across the circuit, as befits the load on the gate outputs and the distance to the source power supply. <S> For myself, I would put a 10 uF capacitor for every 3..5 ICs. <S> I would also have a 47 uF capacitor by the power supply, be it local regulator or input connector. <S> As an aside, will you have a series resistor between each IC output and its loading FET gate? <S> Otherwise, on an IC output rising edge, it's trying to charge the gate capacitance which first appears as an instantaneous short and dips your rail. <S> The resistor limits the gate charging current and gives your IC a better life. <A> ST has helpfully included that information in the datasheet: <S> Note that these are not propagation delays (which are listed beneath). <S> The name of the game in decoupling is to ensure that the decoupling path is short enough to not cause power pin droop due to transition effects; taking 0.1 of the output transition time determines a safe distance for the decoupling path. <S> Using the shortest time of 6ns is informative. <S> 6nS on FR4 is about 3 feet , so getting to 10% of that (a useful rule of thumb) means the path should take less than 3.6 inches. <S> The manhattan distance between the Vcc and ground pins on the device is just over 1 inch. <S> Even using the option on the left will not bring the decoupling path anywhere close to having any issues, so from a PCB decoupling perspective, either solution works. <S> [Update] Although this device lists transition times in the datasheet, many do not; newer devices will usually have an IBIS model which has (amongst other things) the same information in most cases. <A> Yes, you should install VCC decoupling caps. <S> 10nF will be adequate as you are only driving a light load. <S> Given the size of the chip dominates the loop inductance of capacitor <S> , it does not make a huge difference which connection path you choose. <S> If you do have problems with your scheme, it will more likely come from trying to distribute the 10MHz clock and latch signals to 'a bunch' of 595s. <S> There are several ways you can get this wrong and get poor clocking, depending on how many and how physically distributed 'a bunch' is. <S> Perhaps that should be the subject of a different question. <S> Parenthetically, the 2N7000 is not a 'logic level' FET. <S> It will work, and it is specified at 4.5 and 5v, but it is better specified at 10v. <S> There is little difference between the maximum 5ish ohms RDSon, but the typicals are fairly different.
| In addition to the other answers (yes you should use decouplers), the path length (so you can use the optimal path) for decoupling needs to be short relative to the output transition time.
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RTC - is it worth it? I need to design a circuit which will count the number of incoming impulses from an external source (the signal`s frequency goes from about 0.05 Hz to maximum 100 Hz) in a preset time interval (that can be be 20, 30, 60 or 120 seconds) and then display this number (on a LCD) I am planning to use an Atmega mcu with the code written in arduino. My question is: is it worth it to use a external RTC or for this kind of resolution (1 sec) the internal clock is safe to use ( with milis() )? My goal is to get something as low-cost as possible and with as few components as possible, this is why I`m considering to avoid using a RTC. <Q> Firstly, a real time clock is not what you want. <S> A real time clock is designed to give you the current time, as in year, month, day, hour, minute, second etc. <S> It usually has a battery, so it can keep track of time when the device is switched off, and will be much more complicated than a basic timing device. <S> You want to measure relative time, not absolute time, so you need an oscillator. <S> So you have a few options: <S> Use the internal oscillator in the AVR, as is. <S> Most (all?) <S> ATmega series chips come from the factory with an internal oscillator which is accurate to about ±10%. <S> Use the internal oscillator, but calibrate it first. <S> You can measure the frequency of the internal oscillator, and adjust it by writing to the special variable OSCCAL . <S> You will end up with a oscillator which is stable to about ±0.3% but depends slightly on the temperature and supply voltage. <S> Use an external crystal oscillator. <S> Actually, if you are using an Arduino, there is one built into the board. <S> The accuracy will depend on the oscillator and how much you pay for it, but typical is 20ppm (ppm = parts per million, 0.1%=1000ppm). <S> Expensive crystals can be a thousand times better than that. <S> Any of the above will give the MCU a time reference, and you then need to write some code to measure time using that reference. <S> millis() is a good start but you'll get slightly better accuracy using the AVR's built in timers. <A> In your application an RTC would add the complexity of communication and software required to read it. <S> With a crystal-controlled device you should be able to get good stability but may need to perform some calibration to check each device you build. <S> A low voltage signal derived from the mains (50 / 60 Hz) may suffice for your application. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> A mains-derived reference signal. <A> Both are only as good as their oscillators are accurate. <S> The internal R/C oscillator of an MCU is not very good <S> but you dont need to go so far as an RTC to make that better you can just use an external oscillator. <S> If you are interested in higher accuracy time get a GPS receiver. <S> The GPS system is designed for very accurate time, that is their problem how they solve it. <S> I have found an off the shelf RTC to be not very good compared to a GPS, was that strictly its oscillator? <S> Who knows. <S> You are the only one though that can define what accuracy you need, then you need to find a clock source with that accuracy. <S> If you dont care about time of day then an RTC is just an extra expense, use an oscillator of equal accuracy that you would have used for the RTC and just feed the MCU with it. <S> If you need something more accurate then a GPS based solution or there are some solutions based on cell tower signals (CDMA for example) that give the same output as a GPS. <S> Servers that care about time use this equipment, so it is available but often rack mounted and pricy. <S> Cheap GPS receivers can be had for around $10 to $15.
| Real-time clocks (RTCs) are generally used where calendar and time of day are required.
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Which term is right? "No load" or "infinite load"? If I have a generator that rotates in open circuit, do I call it "the generator has no load" or " the generator has infinite load"? <Q> From an electrical perspective loading refers to what is impeding powerflow. <S> "weak pullup", refers to a high ohmage resistor. <S> Likewise with generation and other such powersources a low load is one that requires low power flow & this <S> implies higher impedance. <S> Loading and impedance are inverse terms. <S> Low load -- <S> > High resistance High load -- <S> > Low resistance <S> Infinite load --> short circuit <S> No load -- <S> > infinite resistance <A> Load represent how much current (power) is drawn from source. <S> For example, if there's no load, no current (power) is drawn, voltage is still there, it's just an open circuit which has infinite resistance. <S> If you short it out, you effectively have zero resistance (well, something near that because wires still have some). <S> If you turn your supply on, your drawn current (power) will be set according to Ohm's law and it's going to be very high. <S> So, no load is the right term <A> Infinite load is not a common term, but might be understood to mean short circuit. <S> It makes sense when you think about the current or power being taken from the source: <S> Big resistor = small current flowing = <S> small power = small load Small resistor = <S> big current flowing = <S> big power = <S> big load <S> No resistor = <S> no power flowing = <S> no power = <S> no load <S> Also, in the last case there is literally no load attached to the generator. <S> There is nothing plugged in. <A> If a truck is loaded up then it struggles compared to its unloaded situation. <S> Infinite load on a truck would cripple it just as infinite load on a battery would effectively short circuit it. <A> No load. <S> You could also call it open circuit. <S> Infinite load is not a phrase that is used by professionals, and as you've noticed, it causes confusion, <S> so there's every reason not to use it.
| You would usually use the term no load.
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Narrow TVS margin between operating and overvoltage Devices like the PIC16F18323 have an extremely narrow margin on I/O pins. The standard logic-high voltage level is 5V, but the absolute maximum is Vcc+0.3V, i.e. 5.3V. Reading about TVS devices, in particular those offered on DigiKey, for a diode whose Vwm is 5V, the lowest clamping voltage I can find is 7.5V for the 1N6373. I can assume that by that point my I/O port will be destroyed. So what gives? Should I be using a TVS diode with a lower Vwm and have it already in avalanche when the device is under normal conditions? Or just cross my fingers with a diode whose avalanche and clamping voltages are significantly above what the pin will tolerate? <Q> The reason why the absolute maximum Vcc on the pins is Vcc+0.3V is because above that voltage, the internal clamp diode to Vcc will turn on -- as long as there's a series resistor between the I/O pin and outside world to limit that current to what the clamp diode can handle, it will not permanently harm the IC. <S> (Modern datasheets give an Iik, Iok, or Ik specification for this maximum current.) <S> So, use the higher clamping voltage TVS to absorb the bulk of the energy and a series resistor to keep the port pin's clamp diode from being burned out. <A> The usual way of protecting I/ <S> O pins with narrow margins (and pin injection currents in the order of 5mA usually) is with a resistor and perhaps an external clamp schottky: simulate this circuit – <S> Schematic created using CircuitLab <S> The reason for the 0.3V is usually that there is a schottky internally, but it cannot handle much current; adding one (or two for bidirectional protection) that can handle the current due to a voltage excursion event is a simple and effective means of protecting the pins. <S> The resistor limits the current in the schottky and therefore maintains \$V_f\$ at a nice low voltage. <S> TVS devices have wide variations of turn-on voltage, full clamping voltage and those parameters are also thermally dependent, making them unsuitable for protecting I/ <S> O pins in this particular case. <S> Typical arrays for mult-line protection can be found at <S> some manufacturers Note that if you expect a high voltage (too fast even for a front end TVS as JonRB alludes to - some have slow reaction times) you may want to use pulse-withstanding resistors. <S> These maintain their resistance even after a large transient event; standard resistors may not. <A> You need to choose a TVS that has a working voltage greater or equal to the maximum normal voltage you will see. <S> This means performing some tolerance analysis with regards to the supply & expected IO (5V could potentially be 5.25). <S> Take a littelfuse part ( 1.5SMC series) <S> a 5.8V working has a min & max breakdown of 6.45 -- <S> > <S> 7.14V & this could RISE to 10.5V <S> (depending on the avalanche current). <S> All this would imply that you are unable to use these devices right on the inputs. <S> Except you do not need to just use a TVS. <S> Additional steering diodes right at the IC in question will clamp the I/O to Vcc+0.3 (schottky used). <S> Additional input L-C will again reduce what the actual pin will see simulate this circuit – <S> Schematic created using CircuitLab <S> A strike (indirect lighting or ESD) is a short duration waveform and thus "waveform shaping" is a viable option to ensure the IC IO does not exceed its maximum rating & other components do not exceed their current or power rating
| In the case of I/O pins with schottky devices and a rated pin injetion current (usually the maximum sustained current the internal schottky can handle) a simple resistor can suffice if the maximum voltage can be defined.
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Thickening wire to prevent voltage drop Context: I want to extend a wire that hooks up to the power momentary switch of a PC ( which has a built in LED). This extension will be about 10 meters long. While this distance might be that long to matter, I want to make sure that there is no voltage drop so I have read that thickening the wire will help. A cat5 cable has 6 wires. I'd like to use this cable as it looks nicer and is easier to route. Question: Does soldering three of the cat5 cables to each terminal on the momentary switch effectively thicken the wire (thus preventing a drop in voltage)? Or am I misled in thinking this would work? More information on the switch:-The switch is rated up to 3A/250VAC, but the LED is rated for 1.8-2.8V, if you want the switch and LED to work under 12V, you need to add the resistor <Q> The 'power momentary switch of a PC' should take very little current at all, at least with respect to cat5 cores. <S> It is a logic input to the mobo, so a few mA at most, probably < 1mA. <S> If OTOH there is significant current running through it, then I would not use cat 5, even multiple cores of it which would 'thicken the wire'. <A> If I've read you right, you're trying to put the POWER push-on/off switch from your PC on the end of about 10 m of cable. <S> This circuit is powered from the standby rail coming out of the PSU, so it's always powered while the PC has mains coming in. <S> There's a good chance that the power switch signal will be 5 V or 3.3 V and may pick up noise or interference on its 20 m trip up your cable and back. <S> The thing is, you can't readily add noise filtering without knowing/experimenting what the effect will be on the PC circuit. <S> One solution would be something like the circuit shown below. <S> This replaces the power switch with the transistor of an opto-isolator. <S> That leaves you to drive the opto-isolator's IRLED and you can put plenty of noise filtering on that. <S> Here, R1 and R2 provide the IRLED current and the ~1.6 <S> mA drawn by R3. <S> D2 protects the IRLED from reverse voltages, simply because IRLEDs aren't good at it rather than because they're likely to come along but <S> you'll wish you had it there if the unforeseen happens. <S> C1 and R3 stop high-frequency transients briefly powering the IRLED, if they ever had the current to. <S> R3 also discharges C1 so transients have a long hill to climb. <S> R2 stops cabling faults accidentally shorting out the PSU's standby rail. <S> R1 protects the diodes from unexpected power supplies. <S> It's all a lot of protection against a lot of possible hazards by a few cheap components. <S> The supply is shown to be taken from the PC's standby power supply but you might use other methods. <S> But in place of that, a decent load with a charging time for an RC filter is just as good. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This just sketches out an option. <S> With this, using a single RJ45 wire would be fine but its up to you if you want to do something different. <A> The answer is: it depends... <S> If it is a logic level (5 volts or less for on state-such as turning on 2 computers with 1 switch), you should be OK. <S> If, however, you are trying to turn on something that draws serious current, you would have to calculate the drop according to Ohm's law E=IR, where E is the applied voltage, I is the current in amperes to be drawn by the remote circuit, and R is the characteristic resistance of the wire in ohms. <S> One other consideration is interaction between the two circuits. <S> So you need to know EXACTLY what you are connecting. <S> The best idea would always be to isolate the two circuits using a wired-or with 2 diodes, a gate or even somehow replace the switch with one that is double-pole/single throw to completely isolate the two circuits. <S> Otherwise you could risk damaging either your motherboard or the remote circuit or both. <S> I hope this helps.
| If one circuit runs at or near 5 volts and the other uses a different logic level like 2 or 3 volts, you could potentially cause a short circuit between the 2 power levels and possibly damage to the remote circuit. Inside the PC, the switch connects to the motherboard and this goes into the allsorts power circuit that controls the PSU's power on/off input. There's other ways of handling the noise filtering, such as by introducing a time delay. In some ways, a higher supply rail going up your wires and a higher threshold at the detecting end would be nice. It depends on how much current will be drawn by the circuit at the other end.
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Do power MOSFETs have significant disadvantages over normal MOSFETs? I have noticed that many power MOSFETs are cheaper, have a much greater \$I_{D}\$ and \$V_{DS}\$, and have comparable \$R_{DS(on)}\$ and \$V_{GS(th)}\$ values to normal MOSFETs amongst other advantages (shorter \$t_{d(on)}\$ times, etc.) without any apparent disadvantages. For example, the IPS105N03L G is cheaper and (apart from having a lower \$V_{DS}\$) appears to be better than the 2N7000 in every possible way. IPS105N03L G \$I_{D}\$ = 35A \$V_{DS}\$ = 30V \$V_{GS}\$ = \$\pm\$20V \$R_{DS(on)}\$ = 15.5mΩ (max) \$t_{d(on)}\$ = 3.7ns \$t_{d(off)}\$ = 14ns \$V_{GS(th)}\$ = 1 - 2.2V 2N7000 \$I_{D}\$ = 0.2A \$V_{DS}\$ = 60V \$V_{GS}\$ = \$\pm\$20V \$R_{DS(on)}\$ = 5Ω (max) \$t_{d(on)}\$ = 10ns \$t_{d(off)}\$ = 10ns \$V_{GS(th)}\$ = 0.8 - 3V What am I missing here? Do power MOSFETs have any significant disadvantages as such, or are they suitable for general purpose use? <Q> Of course, they have some disadvantages, and in some applications, they are totally unsuitable. <S> Both types of mosfets offer some compromises on different aspects. <S> First: regarding the price, I don't know where you got the estimations, but they're inaccurate. <S> Mouser offers 2N7002 starting from about 0.12€, whereas IPS105N03L is about 0.68€. <S> So it's not exactly cheaper. <S> Maybe you were looking for a specific packaging of 2N7002 (TO-220, perhaps), which made the price much higher, but 2N7002 is actually one of the cheapest mosfet available. <S> Now, even without looking at the price, there are parameters other than I D , V DS and R DS(on) <S> that characterizes a mosfet. <S> For example: V GS(th) , the gate threshold voltage. <S> For power mosfets, it is usually higher than small mosfets. <S> Small mosfets can be driven directly with a 3.3V voltage, which is convenient. <S> This is not always the case for power mosfets. <S> However, for the two mosfets you compare, there is no significant difference. <S> The gate charge. <S> Here, for IPS105N03L, we have 14nC of total gate charge, whereas for 2N7002, it is less than 1nC. <S> Basically, this means that, when the mosfet changes state, the power mosfet will require a lot of current through its gate to overcome the miller effect. <S> So you need big drivers to make it switch fast. <S> However, the 2N7002 can be very easily driven, and even a simple MCU output pin can drive it as fast as you want. <S> So it makes your design much simpler. <S> This is where small mosfets shine. <A> Those two are hardly comparable. <S> The IPS part is about 10mm x 6mm and can dissipate 38 watts, input capacitance 1100pF typical, the 2N7 part is SOT-23, dissipating a fraction of a watt, with 20pF typical input cap. <S> Lower power parts always seem to be over-priced in hobby quantities, as the package, carrying stock, shipping etc tend to go more like per item rather than per current carrying capacity. <S> If you want to drive an array of 200mA loads, then the 2N7000s would be adequate, smaller and cheaper than the IPS parts. <S> If you only wanted to keep one part in your personal stock, then the larger one would be more flexible. <S> The IPS part is a much more recent design, using a smaller, cheaper process than the older 2N7000, so you are getting greatly increased performance, current, RDSon etc for 'free'. <A> Did you happen to compare cost? <S> You can get BSS138 or 2N7002 for around $US 0.015 (in high volume). <S> The BSS138 can be used in 3.3V CMOS logic circuits. <S> There are versions of them available with very low gate leakage current, so they can be very useful for certain types of circuits in battery powered devices. <S> Usually power mosfets <S> (ans some versions of above mentioned devices, too) <S> have Zener diode protection built-in on the gate. <S> This causes them to have very high leakage currents at the gate. <S> Sometimes that is OK, and sometimes it isn't. <A> And less stability when biasing the devices up in analog mode. <S> Most of the time there will not be a performance penalty.
| If you use a power MOSFET where a small signal device would suffice, there is a price to pay in dollars, size, capacitance which could muck up high speed circuitry.
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MOSFET as a switch to control 12 solenoid valves using arduino I am working on a project that requires me to control 12 solenoid valves using arduino. I find out that I can use MOSFET as a switch to turn on or off my solenoid valves. I have minimal experiance in EE field and wanted to ask if the diagram I made is a correct one for my application.Just for completeness, here is the link to the solenoid valve that I am using. Here is the diagram So this diagram shows only 1 solenoid valve, but in my application there will be 12 of them. Is this a viable method for controlling my valves ? Also if I can use MOSFET for my application, which type of MOSFET should I use ? Thank you. <Q> This should work. <S> You may want another resistor in series with the gate (in between the MOSFET gate and processor). <S> Maybe 1k in this application. <S> When you pick a MOSFET make sure it is rated for Vds of more than 12V. <S> I think 20 is enough, but I see the other answer suggested 30. <S> I don't know what is the IO voltage of arduino. <S> Is it 5V? <S> If so, well, most MOSFET's will probably be ON when gate is 5V. <S> But if it is 3.3V, double check that the MOSFET can be fully turned on at 3.3V. <S> Your solenoid says the power is 4.8W. <S> That implies a current of 400 mA at 12V. <S> So you want a MOSFET that can handle Id of 400mA without getting hot. <S> I would suggest that you look for a MOSFET that lists Rds at of 100mOhm or less at Vgs = 3.3V or lower (or 5V or lower if you have a 5V IO voltage). <S> The other important thing is you need to add a diode in parallel with the solenoid. <S> This diode should be pointed so it is reverse biased when the solenoid is on. <S> Otherwise you will blow-up your MOSFET when you turn off the solenoid. <A> Your MOSFET needs to have: <S> Low enough Vgs, gate threshold voltage so that your 3.3 V from the arduino will switch it. <S> High enough drain source breakdown voltage to withstand 12 V plus margin. <S> 20 V is stretching it. <S> 30 V should be fine. <S> Low enough RdsON, static on resistance to not heat up excessivly from the rated solenoid coil current. <A> I recommend the MOSFET IRLZ44N, it can use at 5 Vgs and can drain 55Amps, with an Rds ON 22mOhm. <S> As "mkeith" tells is a good idea to add a 1k resistor between mosfet an the arduino pin, but I will change resistor between G an S form 10K to 100K. <S> Finally I will recomend to put a flyback diode between 12v and G, ie, in antiparallel with the selenoid. <S> I hope it helps,
| There are lots of MOSFET's out there that can do this, and they shouldn't be terribly expensive.
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Making LM317 output voltage adjustable down to 0 V I want to design a variable voltage regulator that will get from it`s maximum output (lets say 10V) down to 0V (or very close) considering the input as regulated 12V (from a PC power supply). All the designs that I found use the 317 IC and can not go under 1.25V and I am pretty sure that there must be a way to do so. I could not find any tutorials that explain in an easy way the way that the 317 behaves (beside the classical configuration) so any additional explanation is welcomed. <Q> This dual LM317 circuit consists of a current limiter based around LM317(1) and a voltage regulator based around LM317(2). <S> The voltage regulator section is relevant to this post as it is adjustable down to zero volts. <S> Source: ON-Semi datasheet . <S> The control pin of LM317(2) is pulled low by Q2 which is wired as a simple constant current sink pulling several milliamps from the adjust pin. <S> D3 and D4 clamp the top of Q2 at two diode drops (2 x 0.7 V) below zero (-1.4 V). <S> The 240 \$\Omega\$ resistor and 5k pot can then adjust from zero up to the supply limit. <S> The problem with this circuit is that you need to generate a negative supply capable of sinking the few milliamps. <S> My answer to Smartest way to use current limit using LM317? <S> (where I explained the current limiting section) may help in this regard. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> If a centre-tapped transformer is used the negative rail can be generated quite easily. <S> In this example a half-wave rectified signal is smoothed by C2 which doesn't have to be very large. <A> A somewhat nasty solution is to place two silicon diodes of suitable current rating in series with the LM317 output (after the voltage-setting resistor divider). <S> This would reduce the output voltage by about 1.4 volts. <S> This will degrade the voltage regulation slightly, as the diode voltage drop will vary with load current. <S> A better solution is to provide a -1.25 volt (or so) <S> low current supply to the bottom of the voltage setting resistors, rather than connecting that point to ground. <A> A regulator is an amplifier which stabilizes the output at a target voltage. <S> The LM317 does this by comparing the output to a 1.25V reference, and while you can get higher voltages than 1.25 (by voltage dividing the output) you cannotget lower (voltage dividers top out at 1:1). <S> Instead of an LM317, you can use an operational amplifier that sensesnear ground, and for a 20V range, divide output by a fixed ratio ( <S> 20:2.5)and amplify difference of that divided output to a second divider on a2.5V reference voltage ( <S> TL431 being a suitable reference source). <S> When the reference divider is at 1, output stabilizes at 20V; when itis at 0, the output stabilizes at 0V. <S> See figure 13 here: http://www.ti.com/lit/an/snoa589c/snoa589c.pdf <A> if there is sufficient loading, try one LED to drop a fairly constant voltage. <S> You an always bury it where it can't be seen.
| To get the LM317 down to zero volts you need to bring the control pin down to -1.25 V. Figure 1.
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Add ground earth to a Chinese music player I have this Chinese Portable Rechargeable Mp3 player. It has a two pin plug connector behind that is used to recharge it and some generic rechargeable non removable battery inside it. It has the following connecting points on the top: Speaker out Line in Usb Sd Card Whenever I play it with the plug connected to wall outlet, it has mains current flowing through the line out and speaker out. When I insert one end of the 3.5mm pin in the speaker out, I get full 220 volts shock in the other 3.5mm male pin end. I tried doing this with the other line in port too, even that is releasing 220 volt current. This happens only when it is plugged in for charging and used. It does not happen when unplugged and played. I am uploading few pics. The red wire is for the antenna I want to correct this design flaw. I want to add ground to this device. (am I right with my decision?). where should I solder a third wire on the circuit board for ground? ______________________________________ RM: IT IS ALMOST CERTAIN THAT THE RADIO IS "DESIGNED" TO BE CHARGED FROM A LOW VOLTAGE AC SUPPLY AND NOT FROM MAINS. While it is not 100% certain what all the circuitry present is it seems that the diode bridge that converts AC input to DC is connected directly to the battery. Even a low voltage power source would damage most batteries if used this way. Connecting AC mains is liable to damage the device, the user and the battery. Red is (probably) positive. Green is (probably ) negative. Purple is AC. <Q> There is no way to fix this. <S> Look at the picture: The mains input goes onto the PCB without any means of isolation. <S> There is no transformer, and there doesn't even appear to be any cheapo alternative that might, probably, be acceptable - like <S> two safety rated capacitors that might allow charging the batteries while maintaining isolation. <S> Those capacitors would have to be very small in value because otherwise, they would allow a significant (read: dangerous) current to earth. <S> At worst, the earth current would set stuff on fire. <S> Having no earth wire, it would have to be a class II device. <S> This means that the entire circuit must be isolated to the user in a proper (double/reinforced) way. <S> This is not happening here <S> (antenna, battery compartment may be accessed, pushbuttons, headphone outlet, speakers, ... pretty much anything on this f-iretr-ucking piece of lethal junk. <S> Also, there is no way the two little black wires coming from the input connector <S> are rated for mains voltage levels. <S> They may cause a short between the two input contacts, causing a fire. <S> This device is a death trap, and it has taken the idea of making stuff cheap way, way, way too far. <S> Return it to the dealers. <S> Tell them you're serious about putting them out of business if they don't stop selling such warez, immediately. <A> This thing cannot be fixed. <S> Here's why <S> :There seems to be no isolation between mains side and most part of the circuitry. <S> If you try to earth something in the circuitry you probably will produce a short circuit. <S> Most likely the device will let out much of the blue smoke then. <S> If you are unlucky it will wait until you leave and incinerate your house. <S> If you are lucky your RCD and/or circuit breaker / fuse will blow. <S> A missing isolation cannot be fixed by trying to pull the potential of the mispowered parts of circuit to earth, because the connection to the AC mains has to be severed first. <S> The analogy to the broken roof is not bad <S> but I think it fits better if you try to put a garden hose into a hole in a hydro dam to redirect the water shooting out. <S> In fact: despite your tries to fix this thing it will keep violating any electricity code of any country and regulatory body of the world. <S> It is in fact by its very design a criminal act. <S> Mains voltage must be separated by isolation from touchable conductive parts. <S> This cannot be implementated later on. <S> You should take measures that this thing leaves the surface of the earth without being plugged into any outlet again. <S> I recommend removing the two prong mains receptable before throwing it away. <A> From the picture you uploaded it's clear that there's no insulation whatsoever between mains voltage and the rest of the player. <S> You shouldn't use this thing when connected to mains, especially with 3.5 jack earbuds <S> (220V applied to your head is insta-death). <S> Running it on battery may be OK, but every time you charge it you risk getting shocked, which may also be fatal (getting a shock while touching some grounded appliances in your house is all it takes to make the shock fatal). <S> There is no way to fix this thing with grounding, as any contact with ground will create a short circuit and burn the player down. <S> The only practical way to go about it would be to return it to the seller and get your money back.
| If you were to introduce an earth connection, it would, at best, cause your fuse to blow.
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DC rating for specific AC rated relay I am interested in buying a high amperage relay which is AC rated ( https://www.panasonic-electric-works.com/cps/rde/xbcr/pew_eu_en/ds_61A04_en_dqm.pdf ). In particular it is rated for 60A @ 240VAC and I would like to use it at a max 12.4VDC circuit. Could you please advice me what is the equivalent amperage rating for DC @ 12.4VDC?Thanks in advance and I am looking forward to hearing from you.Best regardsGeorge <Q> Usually, the contact current rating is independent of the contact rated voltage. <S> The application is limited by the contact resistance, and for all types is listed as: Contact resistance (Initial) Max. <S> 30 m\$\Omega\$ <S> The maximum current in the contacts is a heating limited function, and therefore the DC capacity will be the same as the RMS capacity. <S> I would not use this part anywhere close to the rated capacity <S> (calculate \$I^2 R\$ to see why). <S> [Update with comment from JimmyB] <S> Who quite rightly states that the break current can be much lower for DC than AC, which is another reason to not use this anywhere near the stated rated current. <S> The test for contact resistance is a DC test: <S> (By voltage drop 6 V DC 1A) <S> so the only current guaranteed to be safe is that test current. <A> Panasonic are a competent and intelligent manufacturer. <S> Both their components and their documentation are usually both high quality and relatively comprehensive. <S> So ... <S> If a data sheet for a relay series makes NO mention of DC contact current rating and conditions for a specific relay or relay family <S> then it is extremely likely that they are not intended for use to switch DC. <S> While it would be possible to do so, if Panasonic says "don't" then it's probably a bad idea. <S> Or worse. <S> Others have suggested that DC current rating of a relay is usually below the AC current rating. <S> While this may be true to some extent, the main limitation for switching DC is usually maximum voltage. <S> Relays that are specified for switching eg 230 VAC often have maximum DC contact voltage of 30 V or less. <S> Higher than this is unusual and much higher <S> is very unusual. <S> The main limitation in switching DC is the contact set's ability to break the arc that forms at the contact surface. <S> When AC is switched the voltage and current fall to zero and reverse polarity at twice the AC frequency (ie 2 zero crossings per cycle). <S> This drop to 0 Volts and polarity reversal leads to extinguishing of any arc that forms (within the rated operating conditions.) <S> With DC there are no zero crossing points and arcs can form and sustain at far greater distances for a given voltage / current. <S> It is possible to increase the DC rating of relays using magnetic blowout. <S> Magnetic arrangements are not complex but very few non integrated solutions of this sort are seen. <S> This causes dissipation in the semiconductor switching device only at open/close transitions and allows modestly rated semiconductors to be used. <A> AC goes through zero many times per second, this assists in breaking the arc that forms when you interrupt a large current. <S> With DC, the supply is steady, so other methods are needed to break the arc. <S> This can be by having big heavy (cold) contacts, in combination with a low enough current, which will quench the arc. <S> A high capacity AC relay will have big heavy contacts, which is why derating the current capacity may well be sufficient for DC. <S> Some relays specifically designed for DC (automobile types for heavy loads) have interesting contact metallurgies, some use magnetic fields in the contact region so that the arc drives itself away from the contacts. <S> An AC relay will not usually have this sort of arc suppressor. <S> Relays tend to have two ratings, which can be very different, depending on whether the load being broken is pure resistive, or has a significant inductance. <S> An inductive load (motor for instance) will generate a high voltage when the relay attempts to break the current, which will tend to sustain the arc. <S> The inductive breaking capacity will always be lower than the resistive breaking capacity.
| Another way to increase maximum DC operating voltage is to place a semiconductor switch (MOSFET or other) across the contacts and to activate it when the contacts open or close.
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Easy 9-12Vdc to 6.3Vdc/300mA circuit? i'm in need to reduce the 9 to 12V from my power supply to 6.3V, for a current consumption of 300mA. Here's the requirements: Cant be a switching solution (so Buck converters are not an option) Doesn't have to be a very clean 6.3v, as long as it stays within 5% tolerance Can't evacuate a lot of heat (small closed enclosure). (my first idea was the usual LM317, but it does dissipate a LOT of heat with this requirements) I know it's not an easy solution (considering i eliminated the 2 most logical options) and eventually i'll have to compromise (i'll probably end up using the LM317 and hope i don't cook everything).Any ideas would be apreciated(oh, i already thought of 7 nasty diodes in series, i was really hoping it didn't came to that) (for those who didn't already guess what could use 6.3V/300mA, it's a vacuum tube heater) any ideas would be welcome <Q> I'd really like to post this as a comment, but I don't have the rep. <S> I'll try to make it a reasonable "answer". <S> You didn't specify your tube complement, or what the circuit does. <S> Would you care to share this information? <S> If you are flush with tubes, you can always add a second tube and put the two filaments in series across twelve volts. <S> 6 V vs. 6.3 V won't make a discernible difference. <S> Maybe the second tube could be used in the circuit? <A> There are a few specs that are mutually exclusive, and you're basically ruling everything out. <S> Let's see: Doesn't have to be a very clean 6.3v, as long as it stays within 5% tolerance <S> How accurate must the rail be over load? <S> Is that still 5%? <S> This is why voltage regulators were invented; they provide a constant output voltage given a sweep of voltages and load currents. <S> A diode, resistor or any passive dropper "solution" have rather horrific line/load regulation specs, that will likely surpass 5%. <S> A diode can drop anywhere from 0.3V up to 1.0V depending on the type, the load current, etc. <S> These rules of thumb can work, but you have to remember these will only work for 1 load current and also for the particular diode you have chosen. <S> Same really for resistance; 20 ohms may work at 12V, but maybe not at 9V.Is there a large difference in load current between a cold and a warm tube? <S> But then you also require: Cant be a switching solution <S> (so Buck converters are not an option) <S> Can't evacuate a lot of heat (small closed enclosure) <S> 2W is plenty of heat, especially with no heatsink on a TO220 package. <S> A switching regulator is a better solution, but I understand for logistics reason you can't use it. <S> Any "dropper" circuit using earlier said components will still dissipate a similar amount of power, however it may be spread across multiple components. <S> This can be an advantage and disadvantage; i.e. more components to cool (if each of them are running excessively hot) or multiple components to spread the heat. <S> If that will not last, you could bodge something together but only expect it to work under 1 set of conditions. <A> Using linear regulators you are looking at a maximum of 2W heat output. <S> If you are okay with the regulator dissipating that much heat you can safely go ahead and use LM317 . <S> If not you are pretty much stuck with a switching regulator like LM2596 .
| Any linear regulator/solution will dissipate (excess) voltage x current as a heat power output. My advice: get a switching regulator OR a linear one with proper thermals (heatsink/ventilation).
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How do you design a CMOS buffer with exact same delay of a CMOS inverter? Everyone knows that a CMOS inverter is simply a PMOS connected to an NMOS. There are situations in asynchronous design that we need to compensate for the inverter propagation delay in a parallel signal running along side of the other signal. In that case I can see that designers add a buffer in gate level schematics. But I need to implement that buffer by myself using CMOS cell libraries and it seems to me the most rational approach is just to put two inverters in series which inverts the signal twice hence acts like a buffer. But it seems to me that the propagation delay also will be doubled. How one can have a buffer with EXACT same propagation delay of an inverter? <Q> I would suggest an exclusive-or gate. <S> If you tie one input high, you have an inverter. <S> If you tie one input low, you have a buffer. <S> The propagation time should be the same. <A> A common solution is to have to two path as shown in the figure below. <S> The first path is a inverter chain to buffer the signal, or it could be a single inverter if the load is low. <S> The second path is almost identical to first, but one inverter is replaced by a transmission gate like structure. <S> (Please note that the bulk should be tied to ground and vdd for the NMOS and PMOS, respectively. <S> It's not drawn correctly.) <S> simulate this circuit – <S> Schematic created using CircuitLab <A> One approach is to generate both buffered and inverted signals via the same stage, and there are two classic analog approaches to the problem. <S> I'll not translate them into CMOS, as I have no expertise at or below the cell library level. <S> They may require matched stages for voltage level conversion before or after the stage itself. <S> The first is the classic phase splitter: <S> based on a common source amplifier, its gain is approximately -R1/R2, or -1 with equal value resistors. <S> (Component values are otherwise nonsensical). <S> In this form it clearly isn't CMOS, though matched active loads would substitute for the resistors. <S> One drawback is that its maximum voltage swing is only half the supply rail, and the DC level on each output is different. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The second is the long tailed pair, which steers current from one leg to the other. <S> Same remarks apply to replacing resistors with active loads, and nonsensical component values. <S> I am not convinced it will achieve such good balance of propagation delays, but at least the outputs can be at the same voltage levels and are not limited to half the supply. <S> simulate this circuit
| Using a simulator is should be possible to equalize the delays.
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Why bridge rectifiers are used in case of DC power supply.? I was searching for 12v to 3.3v/5v DC voltage conversion. I found many schematics and the voltage regulators. But in most of the schematic, they have included a bridge rectifier at the starting. As I am a beginner, I thought bridge rectifiers are only used in case of AC to convert AC to DC . I found this similar question but they are discussing about AC volts. So why bridge rectifiers are used in case of DC power supplies. Here is the schematic which I found: <Q> As Vladimir Cravero mentioned in the comments, using a bridge rectifer in a DC application would be make the input polarity insensitive. <S> The image below is a single diode used for reverse polarity protection. <S> If V1 < 0.7, R1 sees no voltage/current, and the load is protected. <S> simulate this circuit – <S> Schematic created using CircuitLab simulate this circuit <S> In the image below, if the V1 is wired correctly and V1 > 1.4 , then D1 and D2 conduct. <S> If V1 is wired incorrectly, meaning that <S> + and - are accidentally reversed and V1 < -1.4, then D5 and D3 conduct. <S> When V1 is between <S> +/- <S> 1.4, there is not enough voltage to overcome the diode forward voltages, so R1 sees no voltage/current. <S> So while this does protect against polarity reversal, I like to think of it as polarity insensitive at the cost of a reduced output voltage of 2*Vd. <A> The examples you have found certainly included it to be on the safe side, in case <S> the wall-wart has the wrong polarity, or if a simple AC transformer is used instead of a DC output wall-wart. <S> In your case, you can omit it if you don't require the device to work whatever the input jack polarity. <S> Last advice: the schematic you have here is a linear regulator, not a DC converter. <S> It works well, but will waste a lot of power if you need to draw significant current. <S> If you need to draw more than about 100-150 mA, I'd use a DC-DC buck converter instead (there are modules readily available for this). <A> You have just found schematics that describe AC to DC power supplies. <S> Bridge rectifier is the most obvious way to do rectification. <S> It does not hurt if you provide DC at the input of the power supply you have shown (beside small voltage drop and slight heating of the bridge). <S> If you use a bridge at the input of a DC to DC power supply you gain reverse polarity protection and it will also work no matter where you connect input + and - . <S> If you want just to make 5V DC from 12V DC look at the datasheet and reference design of 7805 or LD1117. <A> A bridge rectifier is required to convert AC to DC. <S> In you had 12V AC which has to be converter to 5.5V DC, you needed a bridge rectifier. <S> The rectifier part of the circuit actually converts 12V AC to 12V DC and the rest converts 12V DC to 5V DC. <S> For your case, since you already have 12V DC supply, you can simply eliminate the bridge rectifier part of your circuit and start from C3 on the left. <S> 2 very simple 12V-5V and 12V-3.3V circuit schematics are given below. <S> simulate this circuit – <S> Schematic created using CircuitLab
| You are right, bridge rectifiers are useful for AC to DC conversion, and can be omitted when the input voltage is DC already.
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Using voltage divider in a circuit I am designing a circuit for SIM900 and PIC MCU . I am facing few problem on power supply for them. I will be giving 12v to the circuit. Now PIC works on 5volts and SIM900 works on 3.2v - 4.8v. At first I thought of using two different voltage regulators. But Now i am using 7805 to convert 12v to 5v and then using a voltage divider to convert 5v to 4v. Below is the schematic: Here input voltage is 12v and using 7805 to convert it to 5v which will be given to PIC MCU. This 5v is also used in voltage divider circuit to convert it to 4v for SIM900. I have used R1=100 and R2=400 for voltage divider circuit. Is this circuit safe to use. Does voltage divider circuits are good to use. Please help. Thanks.! <Q> almost fixed current consumption circuits <S> If you have a digital circuit with different current consumption it means you have a variable load <S> then your output voltage will vary in a wide range. <S> This will affect performance of your circuit. <S> For better efficiency I suggest use a two stage regulator, first stage is a switching regulator with efficiency up to 85%. <S> This stage will deliver you a 5 volt regulated voltage. <S> In second stage use a linear regulator to produce 3.3 voltage. <S> In this scenario you will have a good efficiency behind a stable circuit for your digital circuits. <S> Always remember : power supply is the most important thing in a circuit for working stable and desirable. <S> - <A> No. <S> If you connect something at your voltage divider that have input resistance less than infinite (ideally) <S> you load the divider and change the voltage ratio. <S> Here you can see recommended circuit to power the Sim900. <A> A voltage divider is not generally used for this purpose. <S> The reason for this is the output voltage of a voltage divider is calculated by the resistors used. <S> The load (in this case your SIM900) will present a resistance/impedance in parallel with R2, hence changing the output voltage. <S> An example of where you would use a voltage divider is when the load is high impedance/resistance, and therefore does not affect the output voltage. <S> An alternative for your application is using something like the LM317, which is an adjustable voltage regulator. <S> Just use the calculation in the datasheet to achieve 4V output based on your 5V input voltage. <A> Using a voltage divider to provide a power voltage is a bad idea. <S> The first thing to do is to step back and look at the overall problem. <S> Many PICs can run from somewhere in the 3.2 to 4.8 volt range. <S> Use one of them instead of some relic that requires 5 V. <S> Now you can use a single buck converter that makes 4 V or so from the 12 V input. <S> If you really really needed separate supplies, then you have to start with the current requirements of the two supplies. <S> If one of them is much higher than the other, then consider a buck switcher to make the high current supply and a linear regulator to make the low current supply.
| Don't use voltage divider unless in these situations: high impedance load inaccurate voltage If supplies need to source similar current, then use a buck switcher to make the 5 V supply, and then linearly regulate that down to make the 4 V or 3.3 V supply.
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How to measure mains AC voltage with an ADC from a microcontroller? I want to measure the Mains voltage (230V/50Hz) using an ADC channel from a microcontroller. This is my current approach ... Below I have two schematics that add two voltages (one direct, one alternating). Added to that, the both the AC and DC signals are going through voltage divider, but that isn't important. I'm guessing that the output voltage on the second schematic is lower because of the impedance of the capacitor? <Q> If you only want to monitor the MAINS VOLTAGE, then you don't need to convert the entire sinusoidal waveform, add offset, etc. <S> etc. <S> You seem to be asking the wrong question. <S> The common way of monitoring mains voltage is to FIRST ISOLATE the voltage with a transformer. <S> This could be a very small transformer as found in a discarded wall-wart power supply, etc. <S> Then rectify and integrate (filter) <S> the voltage so that you have a DC voltage that is faithfully proportional to the mains voltage. <S> This DC voltage can be simply scaled with a voltage divider, perhaps a potentiometer, and fed directly into the ADC input. <S> Here is a typical circuit which is very good for monitoring mains voltage... <S> Ref: https://mlabsbd.wordpress.com/2013/11/16/how-to-measure-ac-voltage-with-micro-controller/ <A> The usual way to do this depends on the frequency of the AC signal. <S> This may limit the current that can be delivered at DC. <S> At RF frequencies you can use a bias tee. <S> This is similar to your right side circuit, but removes unnecessary resistors and adds an inductor in the dc input path: simulate this circuit – Schematic created using CircuitLab <A> What is the correct way to add a direct voltage and an alternating voltage? <S> Added to that, the both the AC and DC signals are going through voltage divider, but that isn't important. <S> It turns out, from your additional information in the comments, that it is important: <S> Basically, I want to measure the Mains voltage (230V/50Hz) using an ADC channel from a microcontroller. <S> I think your question is really, "How can I (safely) add a half \$ <S> V_{REF} \$ <S> offset to a signal potentially divided from the mains voltage? <S> " <S> What you seem to be proposing is simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> OP's proposal. <S> Note that there is no mains isolation. <S> If you are proposing the circuit of Figure 1 then I suggest that you redesign. <S> There is no isolation between the mains and micro. <S> At best you will have neutral voltage appear on your micro ground. <S> Do not assume that this will be zero as any current flowing from other circuits through the neutral wiring back to the mains source will cause a voltage drop along the resistance of the cable.) <S> A far more serious situation will arise if the live and neutral connections were swapped. <S> In this case the micro GND will become live. <S> Also, since you seem to be in 50 Hz land it is likely that your mains voltage will be 220 to 240 V. <S> In this case a single resistor may not have adequate voltage rating for the high voltage supply. <S> Two series resistors would be required for the high-side resistor. <S> simulate this circuit Figure 2. <S> A small mains transformer of a few VA will provide isolation, a signal to work with and an easy means of adding a DC bias to the AC signal. <S> A 6 or 9 V transformer would be adequate for this circuit and provide a safe working voltage with no isolation or safety issues. <S> Size R1 and R2 to pull a few tens of mA from the transformer and reduce the peak voltage to the operating limits of the micro.
| At low frequencies you might be able to use a summing op-amp circuit.
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Using a large resistor instead of a flyback diode In a video lecture a professor is explaining how to prevent an electric arc caused by the switching operation in an RL circuit. Here is the section (at around 18:26) where he explains it: https://youtu.be/FYDLG_A2P1I?t=1106 I know that a flyback diode is used for this purpose: ... yet in his example he uses a very large resistor to create a path for the reverse current. My questions are: If a very large resistor is used, wouldn't it be a problem if a continuous switching is very fast comparing to LR time constant? If a diode is used, would the inductor burn if the back emf current is too high? In almost all examples they just use a diode without mentioning any possibility about this. <Q> You need to be aware that the professor in the video is skipping over a few things. <S> Note that at about 22 minutes, he writes the equation for the current through the resistor $$i = <S> I_0 e^{\frac{R}{L}t}$$ but conveniently fails to evaluate $$ I_0= iR <S> $$ <S> In other words, for his proposed 10,000 ohm resistor, a 1 amp current will provide a 10 kV voltage spike. <S> Using a resistor instead of a diode is one of those entertaining tradeoffs. <S> A diode will limit the switch voltage to only a bit over the source voltage. <S> This is extremely useful when the switch is a solid-state device such as a MOSFET which absolutely cannot handle kV spikes. <S> On the other hand, the voltage limit restricts dI/dt, so it takes a long time for the diode to bleed off the current. <S> A resistor will allow much higher pulse voltages, but they will typically be much shorter. <S> So: <S> 1-) <S> If a very large resistor is used, wouldn't it be a problem if a continuous switching is very fast comparing to <S> LR time constant? <S> Yup. <S> You don't do it with switching regulators. <S> The professor was referring to very low switch rates, such as applying and removing power from a circuit. <S> Very old school. <S> 2-) <S> If a diode is used, would the inductor burn if the back emf <S> current is too high? <S> In almost all examples they just use a diode without mentioning any possibility about this. <S> Nope. <S> The "back emf current" is simply the value of the current which existed when the switch was opened. <S> The diode needs to adequately sized, though - it must be able to carry whatever current the inductor was carrying. <A> This is an addition to what others have said - not a complete answer by itself: A "feature" of using the resistor is that it can dissipate the stored energy more rapidly than the diode. <S> Diode loss initially = <S> Vf_diode x Iinitial <S> whereas Resistor initial loss = I^2_initial x R. <S> For more than about 1 Volt drop the resistor energy dissipation rate starts higher and above a few volts its higher or much higher overall. <S> In some case this matters - eg relay release times are "slugged" by a diode and may be adversely extended. <S> A compromise is a resistor in series with a diode <S> so dissipation is higher/ <S> faster, Vmax can be designed and back current is blocked in the on state. <S> _____________________________________ <S> Added: <S> I was thinking using the resistor instead of diode would bring a longer exponential decay so it would be problem if the switching speed is much higher than LR time constant. <S> Don't you think a large R would increase the time constant? <S> No. <S> This is not a "normal" situation. <S> In an inductor, at turn off I_existing WILL continue to flow. <S> If there is NO series path then 1/2Li^2 energy will convert to 1/2CV^2 where C is available capacitance - added or stray. <S> If only stray capacitance is present it is often small . <S> For small C V must be very large - you can get kiloVolts of spike from a low voltage supply worst case. <S> The time constant for an LR combination is tc = <S> (L/R). <S> ie as R goes up tc goes DOWN. <S> This is because Iexisting MUST flow so V = <S> Iexisting <S> x R. <S> For large R you get large V and larger power as V^2/R or I^2R. <S> ie <S> increasing <S> the value of a shunting resistor will dissipate more power per time period (and more peak power) and take less time for Vspike to decay but voltage increases with R and energy dissipation rate with R squared. <A> V=IR. <S> This is something you cannot work around. <S> With the switch closed such that the inductor is charging or has charged up, a parallel resistor will also provide a "leakage" path. <S> If the R is large enough this might not be a concern. <S> But when you open the switch the inductor is going to want to keep the current flowing and now the path is via the resistor: <S> V=IR and thus the voltage across the inductor will rise to satisfy this. <S> You might be able to tolerate this in which case ... <S> Sure and its a balance between the losses due to the leak path and the resultant voltage to maintain current flow. <S> Or you could just use a diode <A> A more practical non-diode-based snubber network uses a capacitor in series with the resistor -- this allows a small resistor (say on the order of 30-100 ohms instead of 10k ohms) to be used without dissipating any power in the resistor under steady-state conditions. <A> That professor is talking out of his a... <S> No current through the resistor when the switch is closed, and the resistor will take up the current when open. <S> Well, the first statement is just dumb, and the second is correct - but he fails to mention the size of the voltage across that resistor. <S> Let's say the current through the coil is 100 mA, the resistor is of 1 kOhm - then the voltage across the resistor will peak at 100 V when the switch is opened, which again means that the resistor has to dissipate 10 W. <S> That's one expensive component compared to a diode. <S> You might want to use a bleed resistor in series with a flyback diode, but only to expedite the dissipation of the power stored in the coil - ie. <S> to make a relay switch faster. <S> But that's another story.
| In general, the use of a resistor alone for flyback suppression is frowned upon due to the conflict between making the resistor large (to avoid excessive steady-state dissipation when the coil is being driven) and making the resistor small (to limit the back-EMF voltage in a meaningful way as opposed to the uselessness of the 10kOhm resistor given in the question).
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Electronic device that is open for 6V and passes current once it reaches 12v? Please excuse the way I am this question as I do not have much of an electronics background... I am trying to upgrade the headlights in my car to HID projectors. One of the issues I'm running into is that the daytime running lights circuit runs the low beam bulb at 6V when the system is active, and then 12V when the full power low beams are needed. I am trying to find a device that I can put between the power supply and HID ballast/ bulb so that it only turns on when 12V are available on the circuit. Does such a thing exist? Thanks in advance! <Q> A simple solution is to use a 5V relay, and put a zener with about 7V breakdown voltage in series with the coil. <S> Something like 1N5342 or 1N5921 (6.8V) would be appropriate. <S> This way, the relay will see 5.2V when there is 12V, and nothing when the voltage is only 6V (or anything under the breakdown voltage). <S> This is a very simple circuit, which doesn't rely on some unspecified behavior. <S> Just put the zener the right way <S> (cathode towards the positive supply). <S> And size the relay appropriately <S> (coil rated for 5VDC, contact specified for at least 20VDC or more, and check max amp capabilities against your projector's requirements). <S> Schematic: simulate this circuit – Schematic created using CircuitLab <A> You could use a relay which requires 12V to turn on. <S> You will then connect your light to the relay so it is only connected to power when the relay is connected to 12V <A> http://www.digikey.com/product-detail/en/intersil/ISL55141IVZ/ISL55141IVZ-ND/1976806 <S> Then the circuit is super simple : On the negative input you connect the 12V, on the positive input you connect the system voltage and the output goes to your HID.
| Hi you can use a voltage comparator like this one : Better use some automotive-grade relay, by the way.
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Television flyback transformer: does it do 2 tasks ie. horizontal flyback and E.H.T.? Television EHT transformer is also called Flyback-transformer? E.H.T. (Extra High Tension) generates the cathode ray; ie. acts in same direction of the main pathway of running electrons , i.e. "normal" projection to centre of screen from cathode. Does-it do something with horizontal movement of electron beam on the screen??? Also, in the terms, "Flyback Transformer", "Horizontal Flyback", "Flyback-pulse" etc; there is a common-word "Flyback". Are these all flybacks the same thing? <Q> The electrons are "boiled off" a heated cathode and modulated and accelerated by a series of grid and anode electrodes in the neck of the CRT, conventionally called an "electron gun". <S> In a colour CRT, each of the red, green, and blue colours might have a dedicated electron gun, which "fires" the electrons at the phosphors on the front glass of the CRT. <S> After impact, a "cloud" of free electrons begins to gather near the front glass, and would repel new incoming electrons, until sufficient numbers had gathered to prevent any new electrons from actually reaching the phosphor. <S> The EHT is actually connected to a conductive layer inside the "bell" of the CRT/picture tube, and acts as a positive electrode to attract and "collect" these free electrons after they have impacted the phosphor. <S> The flyback circuitry is so-called because the horizontal deflection waveform that controls the horizontal sweep of the electron beams resembles a sawtooth with a long linear ramp up, and short steep ramp down. <S> The ramp down is the period in which the electron beam is returned from the extreme right to the extreme left of the CRT/TV screen, and called the "horizontal flyback" because the electron beam "flies back" to the left in preparation to beginning another horizontal sweep of the screen. <S> Because the flyback is a very steep ramp, it has a high di/dt. <S> This steep ramp is the "flyback pulse" and is used to generate the EHT through a step-up "flyback transformer" and rectifier. <A> This is "horizontal flyback" when talking about CRT screens: - <S> This is unrelated to the term "flyback transformer" other than the fact that the "method" and the "device" are both "used" inside an old TV or oscilloscope. <S> The term "flyback pulse" probably refers to the well-known problem that flyback SMPS's have when switching - not all the energy taken-in by the primary can be released to the secondary and this creates a back-emf that needs to be snubbed out else the output transistor may fail. <S> It usually looks like a pulse and is controlled (limited) on the primary side of the transformer by Rs, Cs and Ds shown below. <A> TV tubes use magnetic deflection to move the beam across and down the screen. <S> The magnetic fields are produced by a coil assembly (called a 'yoke') which surrounds the neck of the tube. <S> These coils need a lot of power to get full deflection. <S> To get the required horizontal beam movement (scanning linearly from left to right to produce a horizontal line, then 'flying back' to the left during the horizontal sync pulse) requires a sawtooth current waveform. <S> However the deflection coil has inductance so it generates an opposing voltage proportional to the rate of current change. <S> This is fortunate because it means the drive circuit only needs to switch a fixed voltage across the coil to generate the ramp-up current, then switch the current off to complete the sawtooth and return the beam to the left side of the screen. <S> During the flyback period the rate of current change is much higher than during the forward scan, so a very high voltage is produced. <S> To get the current down to zero the energy stored in the coil's magnetic field has to be removed. <S> Rather than waste this energy, it is used to help generate the EHT voltage which accelerates the electron beam in the tube. <S> This is achieved by coupling the horizontal coil to the primary side of the EHT transformer, and then both are driven with the same rectangular voltage pulse. <S> So the beam 'flyback' at the end of the sawtooth deflection is the same as the 'flyback' pulse in the EHT transformer that generates the EHT voltage for the tube. <S> In a TV set these two 'flybacks' refer to the same thing. <S> However the same principle is also used in other switching power supplies, in which case 'flyback' refers only to using the stored energy in the transformer to produce output voltage. <A> Television EHT transformer is also called Flyback-transformer? <S> Yes <S> E.H.T. <S> (Extra High Tension) generates the cathode ray; <S> No. <S> The E.H.T. ACCELLERATES the electron beam generate at the cathode. <S> Does-it do something with horizontal movement of electron beam on the screen??? <S> That signal goes to the "yoke" around the neck of the CRT. <S> The "yoke" is a pair of coils to deflect the electron beam in the vertical direction, and another pair of coils to deflect the electron beam in the vertical direction. <S> The flyback transformer connects directly to the horizontal coils in the yoke to move the beam back and forth. <S> Also, in the terms, "Flyback Transformer", "Horizontal Flyback", "Flyback-pulse" etc; there is a common-word "Flyback". <S> Are these all flybacks the same thing? <S> Yes. <S> The word "flyback" refers to the rapid return of the electron beam to start the next line sweep.
| The flyback transformer generates the horizontal sweep signal. In a strict sense, the EHT does not generate the electron stream ("cathode ray") in a CRT.
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Does a dashed line mean "when energized"? I'm having a hard time understanding this circuit diagram: I'm interested in the following component sates, and how they affect power distribution: the breakers: the buses: the AVIO MSTR switch: I'm not well-versed in schematics, but I'm under the impression, that the dotted vertical line means "when energized"? Does that mean the Avionic Buses cannot be powered unless the MASTER CB is popped? This is my best guess on deciding whether the avionic buses receive power: <Q> I'm not well-versed in schematics, but I'm under the impression, that the dotted vertical line means "when energized"? <S> Dashed line means that there is a mechanical connection. <S> Used in switches and relays in your diagrams. <S> In effect it means, that when the essential bus 1 relay is energized (current flows from X1 to X2) relay pulls the switches A and B into the down-position (marked with triangle). <S> So, when the relay is not energized, you have connection between A2 and A3, and B2 and B3. <S> When relay is energized, there is connection between A2 and A1, and B2 and B1. <S> It is important to note that unless otherwise told in the drawing, they are drawn in not energized / off state . <S> And for the switch S70003 <S> (Avionic Master) dashed line <S> just indicates that it is a switch. <S> Does that mean the Avionic Buses cannot be powered unless the MASTER CB is popped? <S> According to the diagram you provided, avionic bus is powered whenever essential bus 1 relay and shed bus 1 relay are not energized. <S> So if you pull the master cb, you are powering avionic bus and render avionic master switch useless. <A> Figure 1. <S> Cropped version of OP's schematic to improve legibility. <S> Standard practice is to draw switches and relays in their un-operated state. <S> Push-buttons, if normally open type, will show the contacts open. <S> Relays will show their contacts in the de-energised state. <S> In Figure 1 we can see that B2 is connected to B3 when de-energised and connected to B1 when energised. <S> The dotted line shows that all the contacts driven by the associated coil. <S> When is the coil energized? <S> Without understanding anything else about the application we can deduce the polarity of the supply on account of the diodes incorporated in the relays (1). <S> These are "snubber" diodes and give a path for current to flow when the relay coil is switched off. <S> When the relay coil is energised the diodes must be reverse biased - otherwise they would present a short-circuit to ground - so that means thatthe right-hand side is + and the left -. <S> That in turn means that positive supply or battery +, or whatever it is, is connected to (2) and negative to (3). <S> When both X2 and X1 receive current? <S> You haven't got your terminology right here. <S> The coil will be energised when current flows through the coil. <S> This will happen when X1 is connected to + and X2 is connected to -. <S> For this to happen to K70013 CB70009 must be closed and AVIO MSTR <S> I switched OFF. <S> Does that mean the Avionic Buses cannot be powered unless the MASTER CB is popped? <S> The master CB must be closed (un-popped?) <S> for the relays to be energised. <A> In the drawing of the single relay, both moving contacts will move down (connecting B1 to B2, and A1 to A2) when the relay coil is energised.
| The dashed line in a multi-pole switch or relay means that the contacts the dashed line join all move together when the relay is operated or the switch is moved.
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Why use FREERTOS instead of the interrupt mechanism on a microcontroller? This might be a very stupid question. I have experience with embedded software on bare-metal and just started with FREERTOS. However I don't really understand why one would use FREERTOS instead of the built in interrupt mechanism. The goal of multitasking is to give the impression some sort of parallellization is going on on a single core (similar to multithreading). If you have a main function that is running specific functions continuously, next to this you can have different interrupt routines with different importance levels that can interrupt that main function and then return to the main function. With an embedded OS you have some task running and it then gets interrupted to run a task with a higher priority. So far in my eyes the final result seems pretty similar, could someone correct me and/or give some more explanation? <Q> You said, "the goal of multitasking is to give the impression some sort of parallellization is going on on a single core". <S> That is way off the mark. <S> You might be able to achieve the first part of this goal using interrupts alone. <S> But if you put an entire task's worth of functionality into an interrupt handler then you're going to worsen the system's responsiveness. <S> Imagine a monitoring device that: continuously processes samples from an ADC connected to a sensor <S> andperforms some signal processing on the samples, has a simple user interface with a display and a few buttons, and has an Ethernet interface to communicate data on a network. <S> To do all of this from a main super-loop would make for a pretty complex super-loop. <S> You could simplify the super-loop somewhat by moving functionality into interrupt handlers. <S> But imagine if the ADC interrupt performs all the signal processing, the button interrupt performs all the display updates, and the Ethernet interrupt performs all the network message parsing. <S> This system would not be very responsive because it's spending relatively long periods of time in interrupt handlers. <S> With an RTOS, you divide requirements into tasks making each task relatively simple to implement. <S> Interrupt processing is kept to a minimum and the scheduler determines which task to execute. <S> This can make for a highly responsive system. <S> If you've designed the tasks smartly then it's also more robust and extensible than the complex super-loop. <A> Perhaps this comes down to a misunderstanding of what an RTOS is. <S> There's no magic happening that you can not create yourself. <S> It does make a lot of things a lot easier though. <S> Let's make up an example of a simple state machine that has to do different things depending on what has happened in the machine so far: while(1) { switch( state ) { case 1: if( has_sample ) fft_segment( sample_buffer, output ); state++; break; case 2: if( has_packet ) reply_to_packet( network_state ); state++; break; <S> case ... }} <S> You don't want either of these running in an interrupt handler. <S> fft_segment and reply_to_packet can potentially take some time. <S> Let's say that fft_segment takes a long time to run, and there's an incoming packet to reply to. <S> It has to wait until your FFT is completed. <S> With an RTOS in place, you will run these in tasks with a different priority, and the OS can preempt <S> fft_segment if necessary, to reply to the network request. <S> This without blocking more important hardware interrupts. <S> Additional benefits are separation of code, which of course helps when your projects grow. <S> Nothing stops you from hacking this together "by hand", for example by splitting your heavy processing up in smaller parts that can run after each other, but with an RTOS you don't have to - it allows you to code in a way that's more natural. <S> I don't really understand why one would use FREERTOS instead of the built in interrupt mechanism. <S> You don't use it instead of the interrupts, you use it in addition to the interrupts. <S> An RTOS gives you a more flexible way to manage priorities and different hardware. <A> This is from the FreeRTOS site itself: http://www.freertos.org/FAQWhat.html#WhyUseRTOS albeit a bit out of date. <S> Whether you can benefit from an RTOS or not depends on the size of your application and the requirements of your application - but that is stating the obvious. <S> Look beyond "trying to make things look like they are running in parallel" though - that is important for a multi-user system (like Unix) <S> but almost irrelevant for a small dedicated system. <S> What is more important is things like <S> , how simple can I make my design <S> , how maintainable can I make my design, what is the least amount of code I have to write to implement my requirements, how can I get as much functionality onto the smallest processor possible (correcting a misconception that appears time and time again, making your system completely event driven and never executing code [checking a state, polling an input, etc.] <S> unless some processing actually needs to be done, means you can fit more functionality onto a processor when using an RTOS than when not), how can I have maximum responsiveness simultaneously with minimum power consumption ...... <S> and so on. <A> Yes an RTOS carries a lot of baggage, it is a lot of code that is not yours but <S> will get you fired if it fails <S> and you own the project. <S> When you are able to manage the tasks without it, great, no problem. <S> If you get to where your software is starting to resemble an rtos with the features you are needing to add, etc. <S> There is a tipping point where you should consider trying the rtos to see what it does or doesnt give you. <S> It is likely overkill <S> but it may have many of the wheels you were re-inventing. <S> Obviously there is nothing in there you couldnt do on your own. <S> Likewise all the chip resources and features are equally available to you as they are to the RTOS, priority interrupts if present, <S> etc. <S> At the end of the day it comes down to personal preference.
| I would say that the goal of multitasking is to divide a set of complex requirements into more manageable chunks while increasing the overall responsiveness of the system.
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Can an AND gate be used to toggle an HDMI connection? I am looking to turn on and off the throughput of an HDMI signal via a microcontroller. Would four ICs such as the SN74LS08N AND gate be enough? Or do HDMI signals need something more complex? <Q> Absolutely not. <S> First: HDMI uses TMDS signaling. <S> This is not compatible with TTL logic gates; it's a current-mode differential protocol, very different from the single-ended outputs used in TTL logic. <S> Second: <S> The TMDS signals used by HDMI run at over 1 GHz. <S> This is far faster than the 74LS parts you're looking at can handle; indeed, I don't think it's compatible with through-hole parts at all. <S> (The pins alone will have too much inductance for this fast of a signal.) <S> Consider using a dedicated HDMI switching part, such as the TI TMDS261B ("2 to 1 HDMI switch") or the Fairchild <S> FDHDMI08 ("Wide-Bandwidth Differential Signaling HDMI Switch"). <A> No, the signal frequency is way too high for typical AND gates. <S> The signalling is also differential TMDS . <S> You may be able to apply your AND gate to the HPD (Hot Plug Detect) signal. <S> You can then emulate unpluggnig the HDMI connector. <S> Most sources will then stop transmitting, which is what you want. <A> As you would need dedicated HDMI switching chips to do the switching, which might be difficult to source and a PCB with the right footprint, etc. <S> For example, this 5-way HDMI selector on eBay has an infrared remote control. <S> You could use the microcontroller to generate the infrared signals, or use a contact closure to press the button(s) on the provided remote control. <S> Just a thought. <S> Maybe it helps.
| Allow me to make an alternative suggestion which might be easier for a one-off project: Get a cheap, ready-made HDMI switch.
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How to check for short circuit in coil I have 70 turns of enameled copper coil, 1.25 mm diameter and 25.4 m in total. I wound it on top of eacher and soaked the whole thing in a thermally conductive compound. What I fear is that by pulling the coils while winding them, I scraped some of the enamel away. I am not sure how resistant it is. That would cause some of the coils to come in contact with each other thus short circuiting the whole thing. How can I check for this ? Should I see a fluctuating resistance on a multimeter? Is there any obvious tell? <Q> Your copper wire, based on the measurements given, would have a theoretical resistance of: $$R=\frac{25.4\mathrm{m}}{\pi \times (0.5\times1.25\times10^{-3}\mathrm{m})^2}\times1.7\times10^{-8} \Omega\cdot\mathrm{m} \approx 0.352\Omega$$ <S> Measuring this resistance accurately may be difficult with the multimeter you have on hand and would have required measuring the resistance of the fresh wire before winding. <S> If your multimeter is not accurately calibrated and/or is not accurate at this resistance, you can't use the theoretical calculation to 'see' a short inside your coil. <S> If you can repeatably measure a value from the coil, you could unwind the whole coil and compare the loose wire resistance to the coil resistance, but that sounds difficult with your thermal compound. <S> If your are using this coil for fairly low voltage applications (<60VDC) and don't need a lot of reliability out of it, you probably didn't scrape the enamel in two places such that a short inside the coil has formed. <S> If you scraped the wire a lot or need to use it for a long time, you would probably want to make a new coil. <S> To answer your other questions: you probably won't see a fluctuation in resistance on the multimeter; that would require an intermittent short circuit, which is fairly unlikely and would be difficult to accurately detect. <S> Unfortunately, there is not an obvious 'tell' that there is a short circuit in a coil like this. <A> A shorted turn in an inductor will present as either very low inductance or as having very low "Q" factor - or both. <S> Measuring the inductance is fairly simple and easy. <S> You can then calculate the expected inductance of your big coil based on the actual measurement of the test coil with few turns. <S> Ask if you need help with any of these. <A> According to what other answers state, you could measure inductance since I suppose you have compute the number of turns in the coil to achieve a concrete inductance value. <S> A simple way to do it is attaching in series (or parallel) a capacitor and look for the resonant frequency. <S> Given the nominal value of the capacitor, you could figure out the value of the inductor as: $$L = <S> \frac{1}{4\pi^2 f^2_0 C} <S> $$ <S> Where \$f_0\$ is the resonant frequency and C the nominal value of the capacitor. <S> If you detect that there has been a noticeable drop in inductance, a short circuit in the turns may have occurred.
| You can either calculate the expected inductance and see if the measured value is close to that or wind a similar size coil but with fewer turns.
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Switching 1 BT Module between 4 Arduino devices I have a design for an electronics workbench that has 2 Arduino's (a Mega and an Uno), and several component testers that have Atmega328 MCU's at their core, and I was wondering if there is a way to have one BT HC-05 module that I can switch between the devices to program them individually one at a time (rather than having 4 BT modules ie. one for each). They can share a common 3.3V rail, ground, and LED indicator pins as all the devices share the same power and ground. AFAICT the only 3 bus lines that need to be switched between the devices are the RST (using the HC-05 pin 32 & 100nF Capacitor), TX & RX lines. What components & circuit can I use to select and switch these three bus lines to the devices I wish to program? The basic circuit for one device with one BT module is here: -> Arduino BT programming circuit Though I will be using the bare module, not the version on a 5V adapter board. Could I use 3 quad optocouplers (one each for the 3 lines that need to be swiched between the 4 devices) and some mechanism to select when to turn the one I want in each OC package on? I understand how to turn on the OC (probably with the ATMEGA328 or a latching logic gate triggered by a push button, but would I need any components around the OC on the data line side (the EC side of the OC) or would it be pretty much a simple placement of the quad OC transistor side into the circuit? I'm stretching my knowledge here a bit, but you have to learn somehow I guess. <Q> Have all of the subsystems talk to it over UART, SPI, whathaveyou, and let this router handle the BT communications. <S> This way sure does have some benefits - you can keep the connection to the BT module at all times, which may even result in Better UX™; you can make it present a simpler/better-suited protocol to the other subsystems, without those needing to care about which particular BT module you happen to use, if any! <S> Abstraction! <S> You seem to have no shortage of microcontrollers in the project already, <S> so tossing another one in shouldn't be that much of a problem, eh? :D <A> One thing first, you do not want to connect PIN32 in your configuration. <S> PIN32 is intended to signal when the HC05 is paired ( see page 4 ). <S> If I'm judging the intent of that circuit correctly, when the HC05 transitions from paired to unpaired, RST (which is active low) will be pulled low by PIN32 momentarily, causing the Arduino to reset, which I am guessing is not what you want. <S> For switching between devices, is it important that you can do it remotely? <S> If not, UART is very simple and the default data rate of the HC05 is low. <S> A double pole triple throw switch would suffice. <S> Just make sure you have pull up/down resistors on any lines that would be left floating when its not connected or in the process of switching. <A> You can use a bilateral switch like CD4066. <S> Here's a simple tutorial on how to use it. <S> This semiconductor switch is bi-directional, so while I'm going to use terms "input" and "output" they can be swapped around. <S> So, place one switch at each of your programming targets lines (RX, TX, RST), connecting switch output to the target. <S> Then connect all RX switch inputs together and to the Bluetooth module. <S> Repeat for TX and RST switches. <S> Connect all control inputs for each programming target together, and add a pull down resistor at each target switch group's control inputs. <S> Then use a good old mechanical switch to select one target control input and connect it to +5V. This will allow you to use a simple single pole mechanical switch (slide or rotary) to control three lines of signals.
| Another possible approach is using yet another micro as a router of sorts. Alternatively, you can get a three-pole switch and use it without any semiconductors, but three-pole switches tend to be bulky and expensive.
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Improve Rise Time on 1Hz Signal I am using a MCP79410 real time clock to generate a 1Hz square wave signal (50% duty cycle) for a low power display ( Sharp Memory LCD ). Without this signal charge builds up and the pixels show burn-in. The problem is that the RTC output is open-drain, so a pull-up resistor is needed.Because I need to save power I cannot use a resistor smaller than ~1MOhm.Unfortunately with this resistor the display does not recognize the signal anymore, 10kOhm works perfectly fine.Now my guess would be, that this is due to the slow rise time of the signal.The frequency and duty cycle don't have to be very accurate. What would be the most simple way to solve this? Maybe a (schmitt trigger) inverter could improve the edges of the signal. EDIT: I soldered a Schmitt trigger inverter ( 74HC1G14 ) onto the board. It works!! Thanks for your help! Overall power consumption in standby is under 10uA now. I can't say much about the dynamic consumption but it should be OK. For the final design I might switch to the SN74AUP1T14 as user2943160 suggested. <Q> You don't say what signal it is you are driving, but most of the inputs in the datasheet you linked to specify a maximum rise time of 50ns. <S> The larger resistor and any input capacitance will certainly exceed this rise time requirement. <S> A simple CMOS buffer will help by squaring up the edges as it has a much higher drive strength than the pull-up resistor. <S> The key is to find one with a low input capacitance which will allow you to use a higher resistance pull-up which while keeping the rise time at the input of the buffer reasonable. <S> As an example the SN74LVC1G34 has an input capacitance of around 3pF. Even with that low of a capacitance, you are still looking at a 2-3μs rise time at the buffer input. <S> This may well be acceptable, but as you can see, it is still very slow. <S> That will mean that if there is a very long rise time on the input signal you will have better noise immunity to stop the output toggling at high frequency during the transition. <S> But again, you need to make sure the input capacitance is minimised. <S> Another alternative would be to get an ultra-low power comparator. <S> Microchip do some nano-amp comparators. <S> I've used some of these in the past coupled with 10MΩ bias resistors to create a low power draw Schmitt trigger for a coin cell application. <S> The whole circuit including resistors draws less than a 1μA (it was too low for the ammeter I was using to measure). <S> You can tune the resistors to match the required hysteresis, but this was the circuit I used in that application: <A> Changing to a logic gate will not do you much good if the leakage current of the transistors in your buffer/inverter draws significantly more current than the 1M pull-up resistor. <S> Use of a Schmitt trigger input is probably advised due to the very slow rise time of the input signal. <S> This will also help avoid additional power consumption due to the input of a 'normal' logic gate spending a long time at an invalid logic level. <S> Since your signal polarity doesn't matter to the display, using an inverter is a completely valid option. <S> One such Schmitt trigger, 3.3V compatible, low quiescent current option is the SN74AUP1T14 inverter. <S> There will probably be a fair number of footprint-compatible options you can research. <A> Low power battery operated designs can be very difficult. <S> A good knowledge of hardware is often necessary to identify potential problems and how to fix them. <S> And always step back and consider alternative solutions. <S> For instance, perhaps the Microchip RTC can be programmed to have a short low duty cycle. <S> Thereby mitigating the problems associated with a small valued pull up resistor. <S> Perhaps the signal only needs to be sharp on the rising edge of the signal. <S> Using a transistor to the positive rail and a large pull down resistor may provide such a wave form.
| You can also use a Schmitt trigger. You want to find a CMOS buffer with extremely low quiescent current .
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Short life of LED parallel to Solenoid as indicator I have a relatively simple circuit: 12 V to a switch; on the other side of the switch is a solenoid and an LED in parallel. Nothing special about either component; the LED is a 20 mA typical LED rated for 12 V and the solenoid is a solenoid on an LPG gas line. The idea is when the switch is flicked, the solenoid does its thing and the LED lights up to indicate the switch is on. The LED is just from the switch to ground. The problem is that the LED doesn't last more than a month. I can put it a brand new LED in and flip the switch and it'll light up, but after a couple of weeks the LED will die. After the first time it dies, I can turn the switch off again and then slowly ease the switch back on and the LED will gradually come back on (weird behaviour that I can't explain?). However this only works a couple of times and after that the LED is dead. Any ideas on how to rectify this? I'm thinking the LED is getting ruined by spiking voltage but am not sure and don't know how to fix it if this is the case. <Q> The LED may already have a series resistor, the OP states that it is "rated" for 12V. <S> I suspect so because without current limiting an LED wouldn't last a second with 12V across it. <S> In that case it may be the inductive kick from the field collapsing during turn off that is avalanching the LED and causing the short life. <S> It should be rated for more than 12V (around 30V would be good) and for the same current as the solenoid. <A> Your LED is being destroyed by the substantial voltage spike generated from the collapsing magnetic field in your solenoid when you remove power from the circuit. <S> The solution is simple: Add a diode across the coil "backwards" which will absorb the brief spike. <S> This is called a "flyback" or "snubber" diode. <S> Ref: https://en.wikipedia.org/wiki/Flyback_diode <A> There is no such thing as a 12 V LED. <S> I will therefore assume the LED is not being abused when lit. <S> The key clue is that you said this LED assembly is directly across a solenoid. <S> The problem is happening when the solenoid is being switched off. <S> The current thru a inductor (which is what the solenoid looks like electrically) can not change instantaneously. <S> This isn't how the physics works, but picture the current thru a inductor as having momentum. <S> You can't suddenly stop the flow. <S> When you try, you end up with a very large and short-lived spike in voltage. <S> When your inductor is switched off, it makes a high voltage for a short time that is reverse from the voltage that was applied to it. <S> When the solenoid is on, there is 12 V across it and the LED. <S> When the rest of the circuit attempts to shut off the inductor, it could be making 10s to 100s of volts in reverse. <S> This large reverse voltage is applied to the LED, which damages it each time. <S> The solution is simple. <S> Add a diode with reverse polarity across the inductor. <S> This provides a path for the inductor current to go when switched off. <S> The reverse voltage will only be one diode drop, which is well within the LED's ability to withstand. <S> L1 is the solenoid, R1 and D1 are your LED assembly, and D2 what you need to add. <S> D2 needs to be able to withstand the power voltage in reverse, and conduct the solenoid current in forward operation. <S> A ordinary 1 A 50 V diode should work.
| To "rectify" this put a regular diode across the solenoid coil so that it's reverse biased in normal operation and carries the inductive current when the solenoid turns off. However since you say your "LED" is rated for 12 V, I'll assume it's a complete assembly that includes the appropriate resistor so that the LED is run normally when 12 V is applied to the whole assembly. Eventually it is damaged to the point of not working at all anymore during normal operation.
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Inverting op amp used with low frequency oscillator I am designing a low frequency oscillator. An image of the schematic is below. The oscillator part works fine. My intention is to scale the triangle wave at the output of U1 to +/-2.5 V by using U4. In the simulation this works correctly; the gain of U4 is of course -(68/100) = -0.68, or 0.68. In the simulation the triangle wave at the output of U4 is approx. +/- 2.5 V. However, on the breadboard, Vtri at the output of U4 appears as a triangle wave equal in magnitude to its original wave. If I increase or decrease R9 it has no effect on the gain. I am not sure why. Perhaps the way I am trying to couple the output of U1 to U4 is not working the way I want? I also tried AC coupling the output of U1 to a noninverting amp, but this didn't work either. I thought maybe somehow my 074 was damaged; I popped a 324 into its place and I got the same result. I am not sure why. Can anyone suggest a way for me to amplify the output of U1 by 0.68, preferably with my 074 op amp? Thank you. <Q> The circuit should work as diagrammed. <S> I think you have a breadboard wiring error. <S> There's 4 opamps in a TL074, and you appear to only be using 3. <S> Try using U3 instead of U4. <S> This will force you to rewire the breadboard, perhaps correcting the wiring error in the process. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> (a) What you want. <S> (b) What might have happened? <S> Just suggesting you check for silly stuff: if the output from your U4 is in phase with the input it would suggest that you have somehow wired U4 as a unity gain buffer as shown in Figure 1b. <A> What you show should work. <S> Check the datasheet and note how two of the opamps in the quad package have mirrored pinouts of each other. <S> That answers your question, but a few other things need to said about your circuit. <S> That's one convoluted way to make a triangle wave oscillator! <S> The slope of the triangle wave segments are dependent on the voltage that U2 clips at, which is based on its output headroom from each power rail. <S> The headroom of a TL0xx is quite large and poorly specified, so the slope, and therefore the frequency, will not be well determined up front. <S> I can't even guess what you think R6 and R8 are doing for you. <S> All they do is slightly load U2, which makes no sense. <S> What's with the mess of R7, R15, RRate1, and RRate2? <S> All you need is a single resistor there. <S> The slope of the triangle segments will be inversely proportional to that resistance. <S> You don't need some fancy attenuator. <S> R7 and RRate1 would be better replaced by two back to back zeners. <S> Those would clip the large signal from U2 to a smaller and more predictable range. <S> RRate2 would then be smaller, like maybe 10 kΩ. <S> That puts more current thru the zeners, which creates a more accurate clipping threshold. <S> You would then use the output of this zener clipper as the hysteresis feedback into U2 thru R5. <S> That then also makes the hysteresis thresholds predicatable, which set the peak voltages of the triangle wave. <S> R15 would stay there, and then would be the slope controlling part. <S> You could make it a variable resistor to get varying frequency at a fixed amplitude. <S> Make sure C1 is of the right type. <S> Many ceramics exhibit decreased capacitance at higher voltages. <S> That would make the triangle segments non-linear. <S> A polypropylene or polystyrene cap might be appropriate here. <S> The right ceramic could work too, especially if it is rated for significantly higher voltage than you are using it at.
| Most likely you have a wiring error arond U4.
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If there is a branch and the two paths lead back to the same ground, will they both recieve power? I'm using an online circuit simulator found here . When I have a single path through both LEDs, they each illuminate: However, when I add a wire to create a cycle, one of the LEDs loses power: Why does this happen? If I were to write my own simulator, I would check to see if a particular path leads to ground. If it leads to ground, it gets power, if doesn't lead to ground, it doesn't complete the circuit and doesn't get power. In this case, I would say both paths lead to ground, and thus they should both get power. I thought the simulation might be correct if "electricity takes the path of least resistance". But then I read that this is not true. Is the simulation correct? Would only one LED light up in real life? <Q> One property of circuit theory is that all things connected to the same node are at the same potential (voltage). <S> If there is no voltage across your LED, no current will flow, and no light will come out. <S> To do what you are wanting to do, you should put the second LED on a different branch entirely. <A> The resistance through the "off" LED is so much higher than the short that not enough current flows to light it up. <S> So even though some current does flow, it is in the picoamperes or less. <A> By ohm's law, V=I/RThe resistance at LED is higher than the wire, therefore, lesser current will flow through the LED and more will flow through the wire.
| When a wire and LED is in parallel, voltage going in is the same for both.
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Does running LED at higher voltage damages the LED? I want to run 12v 6 watt LED at 19v by keeping maximum watts constant ( current controlled ). By doing so will damage the LED? I.e 12v x 0.5000 A = 6W 19v x 0.3157 A = 6W update1: I am running these LED with buck based PT4115 home made led driver <Q> The idea sounds good but won't work. <S> On the understanding that it doesn't, if you increase the voltage a little bit across a partially conducting LED, it will draw a lot more current and quite possibly enough to near-instantly fry it. <S> Take the example of a 24 ohm resistor. <S> With 12 V applied it takes 0.5 A but with 19 V applied it takes 0.792 A. <S> There is no power regulation implied in ohms law. <A> If you were able to simultaneously control voltage and current across a device, all the Current-Voltage Characteristics (including the Ohm law) would be nonsense. <S> If you regulate in current the voltage will step down, if you regulate in voltage the current will step up, but you will always be on this curve: <A> If you are using a constant current supply, it will attempt to adjust it's output voltage until the load draws the expected current, within its voltage range. <S> So if your led setup has a forward current of 500 mA at a forward voltage of 12V, the CC supply will try to lower it's Voltage output to 12V. <S> If it can't go that low, it may result in a higher voltage and current than you want, blowing your led. <S> The other option is using a simple series ballast resistor to soak up the extra voltage, as @EM has answered. <S> But a 5W power resistor is not exactly an ideal solution. <S> Update: As Op has listed the driver they intend to use, from the data sheet introduction: <S> The PT4115 is a continuous conduction mode inductive step-down converter, designed for driving single or multiple series connected LED efficiently from a voltage source higher than the total LED chain voltage. <S> The device operates from an input supply between 6V and 30V and provides an externally adjustable output current of up to 1.2A. <S> Depending upon the supply voltage and external components, the PT4115can provide more than 30 watts of output power. <S> The IC is designed for this very function, stepping down a higher input voltage, to a current controlled LED. <S> Just set it to 500mA, and it will adjust the voltage down until the led load pulls 500mA. <S> If that means 12V, it will adjust down to that. <A> How can you expect that using more voltage will allow you lesser current flow? <S> (uploaded from <S> http://www.sengpielaudio.com/ohms-law-illustrated.gif ). <S> If I agree, a diode does not obey the linear (proportional) relationship of Ohms Law, still, increase of voltage (Volts) increase current (flow) (Ampere) in a non-linear manner as shown in USER @DarioP 's answer. <S> If you're thinking about a current-limiting resistor in series with that LED-light (as done indeed) , that is, at-the-same-time also decreasing voltage (or in simple-words, electrical-pressure), to that individual LED. <S> Though to the whole series-combined circuit's 2 terminus, the total applied voltage remain (almost) same. <S> However the question asked, is conceptual and basic.
| LEDs cannot magically adjust their incoming current to suit the prevailing applied voltage unless it has some type of switching regulator built into it.
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When should one use an external crystal for this MCU given that the internal oscillator is much faster? I am looking at this MCU and was wondering if it makes sense to use an external crystal. Extracted from the datasheet pg1, *Clock management – 4 to 32 MHz crystal oscillator – 32 kHz oscillator for RTC with calibration – Internal 8 MHz RC with x6 PLL option – Internal 40 kHz RC oscillator – Internal 48 MHz oscillator with automatic trimming based on ext. synchronization* The internal oscillator can be up to 48Mhz. The external crystal is between 4 - 32 Mhz. Why would one use an external crystal when the internal one is faster than 48Mhz given that external crystal costs money and occupies space? When should one use an external crystal? <Q> The internal oscillator is much less stable than an external crystal oscillator. <S> If I'm reading the datasheet correctly, the internal 48 MHz oscillator is only factory calibrated to within 2.9% of the specified frequency - not even good enough for RS-232. <S> There are ways to synchronize it to an external clock <S> , I think it's designed to be used in a USB device situation where you can lock the PLL to the USB bitstream. <S> An external crystal is typically accurate to around 20 ppm , parts-per-million. <S> That's 0.002% from the specified frequency. <S> If you need even better, there are even temperature compensated, ovenized crystal oscillators. <S> Additionally, you may want an exact clock speed at a different frequency, typically for communication with a device or master over an asynchronous communications channel. <S> For this you might need an oscillator at for example 29491200 <S> Hz (115200*256). <A> The internal is an oscillator, usually an RC oscillator. <S> These oscillators are far less accurate than crystals. <S> Also these oscillators tend to drift with temperature changes. <S> Crystals on the other hand can be as accurate as the money you want to spend. <S> The accuracy is needed when, for example, high speed communication is used such as USB. <S> USB has very tight tolerances. <A> See this diagram on page 16: <S> It looks like you can also use the crystal as an input to the PLL, allowing you to achieve internal clock speeds of 48MHz from a slower crystal. <A> The crystal oscillator feeds into an on-chip Phased Locked Loop (PLL) which can multiply and divide the crystal frequency by values between 1 and 16. <S> So with a 8MHz crystal you can generate a 48MHz main clock for the processor.
| As others have said, an external crystal is more accurate and stable than the on-chip high speed oscillator.
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What value PTC Thermistor should I choose for over-current protection for a child's electronics kit? What is an appropriate value for a PTC thermistor used in a 3-cell AA battery pack to provide over-current protection if the terminalsare shorted? Current draw during normal operation will be only a few hundred milliamps) I'm looking to make something similar to this: SnapCircuits 3 battery holder so my son can experiment with connecting switches, leds, motors, etc. together without a risk of melting something. <Q> Perhaps that is what you intended? <S> Similar to a glass fuse, a resettable fuse will trip at a specified current and go high resistance to limit the current. <S> It will remain high resistance until the current is removed. <S> Then, after a short time to cool down, it will self-reset and can be used again. <S> Have a look at these: <S> Bourns <S> If the highest current from your battery pack in normal operation is 300mA then you need to use a MF-R030. <S> This has a Ihold current of 300mA, so it is specified not to trip with 300mA going through it at an ambient temperature of 23 degree C. <S> It's Itrip value is 600mA, so it will definitely trip with 600mA going through it, although it will take a nearly 2 minutes to do so. <S> With a current of 1.5A, such as a short circuit, it should trip in 3 seconds. <A> I would use something like a 0.5A leaded PTC. <S> For example, a Littelfuse RXEF050 , which is a radial leaded type. <S> It has a resistance of less than 1\$\Omega\$ when 'on' and a hold current of only 50mA. It will conduct at least 0.5A indefinitely without opening, and will trip in about four seconds at 2.5A (at reasonable temperatures). <S> With PTC resettable fuses (and fuses in general) there are a couple of other things to consider- maximum voltage (when open) and maximum current that the fuse can interrupt. <S> Some battery technologies can supply enormous current, but I'll assume you have alkaline cells which can produce <S> less than 10A. Checking the ratings on the RXEF050, we have: <S> So it can interrupt a fault current of up to 40A at 72V, which should be more than safe for a few AA Alkaline batteries (perhaps not for NiCd types though). <S> If he leaves the short in place it will drain the battery relatively slowly if it trips, however if the batteries are getting weak and cannot produce enough current to trip the fuse, they will be drained much, much faster. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Current limiter and short-circuit indicator. <S> Something similar to Figure 1 might work and could be installed in the battery holder. <S> R2 limits the current to 450 mA in the case of a short circuit. <S> As the voltage drop across R2 increases towards 1.5 V D1 should begin to glow and will be very bright at about 10 mA on a short circuit. <S> The downside of this scheme is that even at 100 mA a volt will be lost across R2. <S> Maximum power in R2 under short-circuit conditions will be \$ P = \frac {V^2}{R} = \frac <S> {4.5^2}{10} = 2~W \$ so choose a suitably big resistor and protect it from little fingers (while allowing ventilation). <S> If you can find a suitable PTC thermistor you could use it in place of R2 and still have the indication.
| Rather than use a PTC "thermistor" you should use a PTC Resettable fuse.
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Help with electronic components Hello im completely new at this subject, so this might be a "newbie" question, but i need some help with finding some components, if they exist. I would just like to know the names of the components and know if im using them right. Component 1: As you might see, i want to reverse the motor spin direction with the component " S2 ". Questions: Name? How to use it? (Using the switch) Can it be done with a remote controled circut? Component 2: This might be a less obvious drawing, but my goal here is that i have 1 energy source, and the remote reciever has to recieve a lower voltage (I forgot to draw the resistor) than the component on the other half of the circut (LED).Which also is the why the component LED is not connected in a serial connection. In the real project the LED could be anything, but i drew it to clear it up.So I want to be able to turn switch S1 ON / OFF with a remote control. The voltage running through S2 is lower than the one in S1, and i therfor want them to not get in contact. Questions: Name? How to use it? (Using the switch) Currently i do not know, which battery, motor or reciever im going to use. I hope i made myself clear, Thanks in advance. :) <Q> You have drawn it correctly. <S> When the switch is thrown the motor will reverse. <S> The second circuit is as clear as mud. <S> I think you just want to switch a load with the remote receiver. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> A typical method of solving this. <S> Use a voltage regulator to step down the voltage to your receiver. <S> When the receiver output turns high it switches on the transistor which switches on the load from your unregulated supply. <S> Any small signal NPN transistor will do. <S> ... <S> but my goal here is that i have 1 energy source, <S> That's the 9 V battery. <S> ... <S> and the remote receiver has to receive a lower voltage <S> (I forgot to draw the resistor) than the component <S> That's the 7805 voltage regulator. <S> A resistor isn't suitable as the voltage at the end of it varies with current. <S> The regulator maintains a constant voltage output; +5 V for the 7805. <S> That may or may not be what you want for your receiver. <S> C1 and C2 prevent the regulator from oscillating and are recommended in the datasheet. <S> The receiver has to have an output to be any use. <S> ... <S> receiver has to receive a lower voltage ... <S> than the component on the other half of the circut (LED). <S> If you want to use a low-voltage receiver to switch a higher voltage load the standard approach is to use a transistor as shown. <A> Component 1 Switch S2 is a "double pole, double throw" type. <S> See also this article titled " Switch Basics " on the SparkFun.com website. <S> A circuit that can be automated (e.g., controlled via software running on a microcontroller), that can reverse the voltage polarity across a motor or other load, is called an "H-bridge" circuit . <A> For component 2, I think you want a relay. <S> A relay is essentially an electrically operated switch. <S> Conventional mechanical relays have a coil (electromagnet) that moves the switch contacts when energised.
| Component 1 is a 2-pole, 2-way or double-pole, double-throw (DPDT) switch.
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Does a helicopter battery drain when it is "off" but attached to a battery hot bus? I'm reading about the DC Power System for a civilian helicopter. It shows the following diagram for when the battery is off: However, it shows that the battery hot bus will still be energized. I assume the battery hot bus is grounded completing the circuit. Doesn't this mean the battery would be draining with the helicopter off? What is the purpose of a battery hot bus, and how does it affect the battery? <Q> This is the same setup as a typical automotive/car/truck electrical system. <S> You have a Switched bus <S> , that's only on when the key is on Acc <S> /On. <S> The Key Ignition switch enables a power relay to power most of the car, like radio, windows, AC, and further for the Starter Motor. <S> High drain devices that only need to be on if the user is already in the car. <S> Then you have an <S> Always On bus. <S> These are directly connected to the battery, without any relay. <S> Typically, this will be alarm systems, keyless entry, door locks, some headlights, some cigarette/power outlets, clocks, the ODB-2 or CAN-Bus system, some lights. <S> Most of these are very low current draw systems, or not constantly active without user interaction (roof light, headlights, motorized windows, electronic locks). <S> This is why you are typically expected to use the car on the regular, to prevent this from even being notice. <S> Letting a car sit for a week isn't normally an issue, but Long term sitting/non-use , like weeks (or a month for new/good batteries), will drain it enough to need a jump. <S> Long term storage recommendations are to disconnect the battery completely, or use a battery tender. <S> This is also why leaving the headlights or map lights on will drain the car overnight. <S> These are Always On/Hot systems. <S> Same for any <S> Always On cigarette outlets with a cell phone charger or similar. <A> It depends on the particular model, but typically a hot bus is used for things like maintaining clocks, and keeping medium-term memory for things like GPS almanacs (but not database updates, which would be stored more permanently). <S> In most applications, the drain is low enough that the battery should be left connected all the time, but I know at least one instance of a business jet that the users typically disconnect because of high drain on the hot bus. <S> It's possible the only way to know would be to ask someone who uses it regularly, or use a multimeter to measure the load when the master is off. <A> I am no helicopter expert, but I know something about battery systems. <S> When you have a battery powered system, even when it is fully off, there usually is something on. <S> Think of your smartphone: when you turn it on you just briefly press a button. <S> This button is connected to an integrated circuit, which is then energized, so that when it senses that the button is pressed it can power the CPU and wake up the system. <S> For a nice smartphone I would expect the battery to last forever. <S> The standby power consumed by this IC is probably way less than the auto discharge current of the battery. <S> Now to your helicopter. <S> The battery hot bus does not seem to be connected to anything and no, it is not grounded. <S> Circuits are completed to ground through a load . <S> But I suspect that's not the full story. <S> On a system complicated as an helicopter I would be very surprised if there is no system permanently connected to the battery hot bus, there probably are some low power blocks that are kept on in order to speed up the boot process of the machine. <S> This blocks are probably listed somewhere in the manual. <S> I bet that disconnecting the battery for long term storage requires a precise procedure and checklist, you might want to check that section out and see if there is any warning about a long power on time because of some systems that need time to boot, and are usually kept always on via bat hot bus. <S> A thing that comes to mind is GPS, possibly helicopters use something like AGPS to get a quicker fix. <S> If this is not stored in permanent memory, after a battery disconnect the data needs to be downloaded from scratch.
| Yes, the battery is slowly draining with the Always On bus, it's just designed to be very low draw, with active use/recharging. Looking at the schematic it seems that when BR is disengaged the battery is not powering anything.
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How does the Electrical Grid React to Large Changes in Supply or Demand? Background I'm performing some research for a software project that monitors power generation and consumption for a large fictional spaceship. My rationale is I'd like to first understand how real-world power generation works before starting to make the app feel as authentic as possible when using it. Basis My question stems from my current understanding of how the electrical grid works. Energy is generated through whatever mechanism (steam/mechanical, photo-voltaic) and is then stepped up via a HV substation for long distance transmission. Large factories (Steel Mills, Microchip Plants, etc.) may have a substation dedicated to them for their operations. Otherwise, a substation steps HV down to MV for smaller scale distribution (factories, large office buildings, etc.). This is repeated again for LV, for delivery to homes and small businesses. It's a simple model, albeit misleading because it presents a linear chain of flow from source to load, with a single source generator of power. In the real world, there are multiple stations that are running to meet the demand, and they adapt as the demand changes over time. Question Suppose a large event such as a generator station unexpectedly shut down. What equipment would be involved in "rerouting" power to minimize the possibility of blackouts? Or, if a rolling blackout was temporarily implemented because of high demand, what equipment or process would be involved? Related Questions For coordinating multiple generator stations: How are multiple power sources synchronized in a grid that uses a distribution ring? For transient heavy loads: What is the effect of heavy loads on the electrical grid? Modelling of Electric Power Grid <Q> What equipment would be involved in "rerouting" power to minimize the possibility of blackouts? <S> Transmission switching stations re-organize things when there are problems with generators or section of grid: - As you can see, every generator connects to the (national) grid via a TS. <S> This wiki page should help. <S> Pictures taken from here or here if not a Quora member <A> Suppose a large event such as a generator station unexpectedly shut down. <S> What equipment would be involved in "rerouting" power to minimize the possibility of blackouts? <S> For generator failures: There must be enough spare generation running to cover the loss of the largest generator. <S> This is called "spinning reserve". <S> Note that having spare generators sitting still doesn't help you. <S> A cold generator takes minutes to hours to start, but the spare generation must take over from the failed generator in a period of seconds. <S> Note: This means that generation companies get paid for keeping generators turning over, but producing no power. <S> People wonder why they are getting paid for producing nothing! <S> The generators aren't producing any power, but they are providing an essential service. <S> It's also important that the spinning reserve generators be able to react quickly to changing load. <S> " <S> Transient response" to "load swings" are the key words to look for. <S> Big coal / nuclear power plants are no good for this, as they respond slowly to control inputs. <S> Gas turbines are much better at taking swings. <S> For transmission infrastructure failures i.e. transmission lines, substations - I think Andy's answer has that covered. <S> We do engineer transmission systems to have at least N-1 redundancy, so we can lose any one component and still keep most of the system running. <S> I don't have any experience with HV transmission systems, so I will refrain from saying anything more. <S> Or, if a rolling blackout was temporarily implemented because of high demand, what equipment or process would be involved? <S> There are two ways to manage excessive load. <S> The traditional system reacts to excessive load by turning off supply to entire load centres, i.e. an entire substation, or an entire feeder. <S> This is a "load shedding system". <S> Loads would be turned off in priority order. <S> You would turn off a factory before you turned off a hospital. <S> You keep turning things off until the load re-balances with the available generation. <S> The other way is to pro-actively manage the load, i.e. using ripple control signals. <S> This turns off specific customer equipment, such as air-conditioners, at times of high demand. <S> This won't help in severe power shortages, so a load-shedding system is still required. <A> The actual piece of equipment in a transmisson substation that "switches" electricity around is the Quadrature Booster, otherwise known as the Phase Angle Regulating Transformer . <S> Essentialy, by changing the phase of power with respect to the source, you change the resistance to energy flow, thereby regulating the flow of power through a particular segment of the grid.
| Power grids are engineered to withstand the failure of any one component.
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GND signals not connecting to Polygon in Eagle I am designing a board layout in Eagle. Before manual/auto routing, I made a polygon around the corners and named it as GND. Then I used the autorouter for routing. After the routing is complete, there are few yellow wires left and these wires are of GND connections. As you can see in the image, there is a yellow wire attached to the GND of the LED1 and the 3rd pin of the JP2 . According to my understanding, this should not happen because I have made the Polygon of GND. What could be the problem. Am I doing something wrong. Please help. Thanks.! <Q> Well something has to give - either the ground plane has to give up space for the auto routed tracks or <S> all the tracks cannot be autorouted. <S> Personally I would put ground on the bottom layer and use a double sided board - much more robust in terms of EMC. <S> It also looks like your copper clearance rules are too tight. <A> The remaining yellow wires are where the autorouter was unable to make the connections. <S> The polygon connecting to the cathode of the LED is not connected to the rest of the ground plane -- it has its own island that is electrically isolated. <S> This is a problem, because there is no return path (hence the yellow airwire). <S> I am not going to take the time to review your board, but I expect you have a similar problem with the other airwire. <S> You'll need to rearrange your components and re-route (by hand would be more reliable -- Machines are very stupid) to make sure you have a solid ground plane and that all connections can be made. <S> You are severely limited by the single-layer requirement. <S> You need to think very carefully and try multiple times until you get something that works. <A> Why limit yourself to single layer? <S> The PCB houses all do double layer as a standad product. <S> I'd doubt you can even find one that will do single layer. <S> Go double layer, with GND plane on both sides, add some vias and Name <S> then GND also, place as needed to connect all the GND areas together.
| Your traces are too fine and closely spaced for home production to be reliably made.
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Low side and High side switches Why do we go for a Low side / High Side Switch ? Why do I find more of Low side switches than High Side in Automobile electronics ? <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> (a) NPN open collector switch. <S> (b) <S> A failed attempt to make a PNP version. <S> The low-voltage logic just has to feed sufficient current into the NPN transistor to turn it fully on. <S> There is a temptation to think that we could do the same trick with an PNP transistor as shown in Figure 1b. <S> The problem is that the emitter-base junction is always forward biased. <S> This will apply the 12 V to the chip output and destroy it or, if there are protection diodes on the output, the current will flow through the protection diodes into the micro-controller supply (shown as 5 V in this case). <S> The effect of this current flow is to turn on Q3 and the load can not be switched off. <S> There are ways around the problem of Figure 1b <S> but they are all more complex. <A> what the guys said is totally correct . <S> but your initial statment is not accurate <S> "Why do I find more of Low side switches than High Side in Automobile electronics ?" <S> Proof in this document: HIGH SIDE <S> SWITCH Stmicroelectronics <S> I will QUOTE : <S> Since the body of a car is metal and 95% of <S> the total car is ground, the short to ground <S> is much more <S> common than short to VCC b) - <S> High Side <S> Drivers <S> cause less problems with electro-chemical corrosion. <S> It is of primary importance in automotive systems because the electrical components are in an adverse environment, specifically adverse temperatures and humidity and the presence of salt. <S> For this reason <S> the series <S> switch <S> is connected between the load and the positive power source. <S> Therefore when the electrical component is not powered (that is for the greatest part of <S> the lifetime of the car) <S> it is at <S> the lowest potential and electrochemical corrosion does not take place. <A> As Jim pointed out , cost is a driving factor, because n-channel MOSFETs are easier to produce with low on-resistance and high current capabilities. <S> Other than that, ease of construction <S> : With a single MOSFET low-side, you can switch a (basically) unlimited voltage, because the voltage that decides about whether the device is on or off is the Gate-Drain voltage; Let drain sit on Ground, then you see that you can use a typical logical level 0 or 1 to switch the device on or off. <S> If you put the switch on the high side, you inevitably need to have some circuitry that translates between the logic levels, which are typically much closer to ground than to the voltages of the high side.
| The low side switch is very simple as shown in Figure 1a. Almost every electronic switch used in a modern automobile application is a high side switch. This configuration is preferred for automotive use because: a) - This configuration protects the load from continuous operation and resulting failure, if there is a short circuit to the ground.
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Nomenclature: How to describe this switch I'm looking for a particular type of switch, however I don't know what it is called. The switch I'm looking for is actually a bank of switches. When one of the switches is pressed, the others are depressed, such that the circuit is always closed, but each circuit changes the input source. It is the type in this product . What is this switch called? <Q> They used to be called radio buttons but a search for those will return all software version for GUI applications. <S> Figure 1. <S> Pressing one button cams the holding bar sideways to release the other buttons. <S> You will find better results searching for "interlocking push-button switches". <S> Note <S> : The English word "depress" (in the context of buttons or levers) means the same as "press" so you may cause search confusion. <S> Use "un-press" or "release" instead. <A> It is called a "gang switch" or "ganged switch" Where the switches are mechanically interlocked to release the activated switch when the next one is pressed. <S> These were once very common in products, but in this modern electronic age, they are antiques. <S> You can still find them as surplus parts, or you could probably have them custom made if you need 100s of thousands of them. <S> But we do those things electronically in the modern age. <S> A whole microcontroller costs less than a single section of those old switches. <A> A Multiplexer <S> I'm not sure I completely follow your question, but it sounds like you have many inputs and one output, and you want to pick one input to feed that output. <S> If so, a multiplexer is the tool for the job. <A> If we are talking about the switches themselves, or the switch assembly, then it could be called a 'one of N', or a 'one by N' switch.
| Interlocking push-button switches.
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Can a board have two power supply circuits? I have a PCB holding a Video decoder, FPGA and a USB controller, powered by USB3.0. Can I have two different power circuits generate power from same input and deliver to the same output? <Q> No, each power rail must be supplied by only one power supply. <S> (Having multiple power rail voltages is common, but each power rail is supplied by a single power supply.) <S> Consider two power supplies that are both "nominal" 3.3V output. <S> Suppose one supply has its setpoint regulating at 3.301V and the other has its setpoint regulating at 3.299V. <S> If these two outputs are wired together, then the (slightly) higher voltage supply ends up doing all of the work. <S> Even if both supplies happened to be at exactly 3.300V under room temperautre condition, the setpoint will change with temperature variation. <S> And once one supply begins carrying more of the load, that supply will get hotter, causing more mismatch. <S> The right way to approach system power supply design, is to determine the total load current needed and then design or purchase a power supply module capable of delivering the required load current. <A> Multiple power supplies can be paralled for additional current, but as MarkU says, you cannot simply parallel two circuits and expect it to work. <S> You need to implement some kind of current/load sharing topology to equalize the load placed in each supply. <S> The simplest method is a series resistor: if one supply is taking more load, then there's more voltage drop and the other supply will naturally take more load. <S> This is inefficient however, so more complicated solutions can be used: generally the feedback to the regulator is modified somehow, based on sensed current. <S> This TI whitepaper goes into detail about a few possible implantations. <A> Often there will be two sources of power for the same supply rail, e.g. one for DC coming from a wall-wart, and the second from a battery. <S> But these will be isolated by diodes, so they don't interfere with each other. <S> This simple schematic assumes <S> the wall-wart voltage and the battery are the same. <S> The more likely case, a voltage regulator would be used to generate the same voltage from either source. <S> Schottky diodes are usually used in these circuits to minimize the voltage drop across the diodes. <S> Sometimes there is a switch built into the jack that disconnects the battery when the wall-wart is plugged in. <S> Then no diodes are needed. <S> However this scheme fails if the user unplugs the wall-wart and expects the device to run on battery power. <S> This same scheme is used when a device is powered by either a USB cable or a wall-wart.
| The problem with attempting to drive a load with two different power supplies, is that the supply voltages will not match exactly.
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What is the power specification for SATA? We are designing a product which needs to provide data and power to an SSD connected using SATA. However, we cannot find a specification of the power requirements for a SATA connector. How much power should a power supply behind a SATA power connector provide at a minimum? I have acccess to the SATA specification. However, we did not find a specification of a minimum power supply. It only mentions the 1.5 A rating of the individual pins. It seems to define how to deliver power, but not how much. I found an answer on power connector pins and the wikipedia entry on SATA power connectors. However, it seems to me that the 1.5 A specified there is the minimum the connector should support. That is, the connector should not melt if I put 1.5 A per pin through it. I don't think it means the SATA device connected to it can expect 1.5 A per pin to be supplied. <Q> As @bitshift says, you'll likely need to get the actual spec to be sure. <S> 1.5A may well be a real supply requirement, though. <S> The cables used are actually rated to 4.5A per supply rail . <S> The connectors contain 3.3V, 5V, and 12V rails. <S> A regular (magnetic 3.5") <S> hard drive can dissipate almost 10W . <S> This energy is probably split across the three available power rails. <S> Since I have no idea what the split may be, the worst case numbers look something like this : All 10W @ <S> 3.3V - 3.33AAll 10W <S> @ <S> 5V <S> - 2AAll <S> 10W @ <S> 12V <S> - 0.83A <S> The actual spec could help you understand how the load is allowed to be spread over the three supplies and what the effective ratings needed are. <S> If you don't care about it being SATA compliant, you could assume the load is lower by a factor of 2 or so since you're only using SSDs. <A> 1.5A per pin is the design spec, though in practice you'd de-rate it. <S> The MOLEX specification for the connector is just that, and is agnostic to what you should run where. <S> For power requirements look at the ATX specification. <S> The SSD is expecting the pin-out, voltage, and ripple to be within the ATX specification. <S> So long as you meet it when the SSD is going full out. <A> It looks like to get a definitive answer you may need to purchase the spec from The Serial ATA International Organization . <S> If not, getting in touch with them may be valuable anyway.
| I don't think there's a standard minimum per connector, as there are a lot of embedded low Wattage DC power supplies that are ATX rated.
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How many amps can a car battery supply? Googling this hasn't given me much information. I've heard numbers ranging from 30A to 300A. My question isn't how many amps a car battery does supply in normal operation, it's how many amps I would measure if a car battery is shorted across a multimeter (assuming the multimeter doesn't explode). <Q> First, it highly depends on the battery. <S> Some cars have much beefier batteries, measured in Amp Hours. <S> We arn't even talking about Electric Vehicle battery banks which are massive . <S> Then it depends on the type of battery. <S> Some chemistries are different. <S> Some are 24V instead of 12V. <S> Some cars have more than one. <S> Etc. <S> That said, the normal peak current is the Cold Cranking Amps. <S> This is the amount of current the battery should provide for starting a cold engine at 0°F. <S> 300 to 1000 Amps is not unusual. <S> This white paper describes a dead short test :  <S> Finally, each battery was “dead shorted”, connected to a “shorting circuit” consisting of a shunt (5000A+ 0.25%), Hall effect transducer [model LEM LT 4000T (4000A+ 0.5%)], 26 feet of MCM-550 cable and a knife switch. <S> A 2 channel Fluke 190 Scopemeter with automatic triggering was attached to the Hall effect transducer and to the battery terminals. <S> Current and voltage readings were recorded at 0.2 millisecond time intervals from 0 to 0.2 seconds. <S> An Agilent 34970 data-logger was used to monitor the shunt current and battery terminal voltage at 40 millisecond time intervals from 0 to 30 seconds. <S> The “shorting circuit” had a resistance of 1.80 milli-ohms, as measured with a Biddle DLRO micro-ohmmeter. <S> The inductance of the circuit was not measured. <S> To determine the effect of temperature, sets of UPS12-140 (12V-33AH) batteries were float charged at 13.65V (2.275 volts/cell) for 48 hours at 2, 11, 24, 33 and 40oC in a temperature-controlled environment. <S> OCV, impedance and conductance readings were measured and each battery was “dead short” tested using the test method described above. <S> In theory, with a perfect conductor you are looking at over 2000 Amps. <S> With their test, they saw 1700 Amps. <S> And these are just 33 Amp Hour batteries, small compared to most cars. <S> These are UPS batteries! <S> My car has a 150 AH battery with 750 CCA, and it's not even a premium battery. <S> But any multimeter worth a damn will have high resistance for voltage measurements, and a fuse for current measurements, that should blow before you see much current flowing. <S> But it will blow if you short it. <S> No if and or buts about it. <A> Hundreds of amperes. <S> For example, my truck has a battery rated at 625 amps. <S> Each battery should have a rating. <S> Many auto parts stores have the ability to test the battery for you to make sure it is putting out the correct current. <A> Assuming the battery and the ammeter both have internal resistances of 1 milliohm, then from Ohm's law, $$ I <S> = \frac{E}{R} = \frac{12V}{0.002\Omega} = \text {6000 amperes} $$ <A> The reason you're seeing such a large range is because a battery is better thought of as a fixed voltage source, not a current source. <S> If you have a 12V battery and <S> you're asking how much amperage can it kick out <S> , the answer is however much or little <S> it has to to satisfy Ohm's law <S> , V = IR. <S> The less resistance you have in a circuit, the more current will flow and vice versa. <S> The absolute extreme of this would be if you had zero resistance (an ideal short circuit), then the poor battery would try to crank out infinite current to maintain the relationship. <S> That means kaboom. <S> Of course, there will always be some resistance in the real world so your battery will probably only have to try to crank out thousands of amps - still kaboom. <S> It is not a fixed value for any one battery or class of batteries. <S> Even the resistance of the circuit is not necessarily a fixed value, it would depend on factors like the level of corrosion in the terminals and the temperature of the conducting wires. <S> If you want a ballpark of how much current your battery sometimes supplies, check the cold crank amperage rating.
| To answer your question: How many amps a battery supplies depends entirely on the voltage of the battery and the resistance in the circuit. In short, we are talking about literal Thousands of Amps for a dead short.
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What is the name of this connector component? First of all, I recognise how hopelessly difficult it will be for other users to find this question based on a StackExchange search. Such is the nature of image-based questions. In any event, I'm looking to identify this connector, in order that I can buy another one. For reference, in this application at least, it's part of a 3.6V battery charging mechanism. <Q> I believe that is a 2-pin Molex " <S> Mini-Fit Jr." nylon line plug <S> There are other, generic, manufacturers - but I'm providing a specific name brand as I don't see that any of the generic versions have a specific name. <S> Searching for ATX style or EPS style may help. <A> I think there are many brands, so I cannot give you a specific name <S> , most of the time I see such connectors on RC car batteries. <S> Such as <S> this might be suitable. <S> Or this . <S> Or this from another shop. <A> Looks like one of these: - Available from here . <S> You'll probably need a crimping tool or maybe you can solder the wires up to the crimps successfully. <S> Also from here
| If your are looking for purchasing one I recommend you to check RC Hobby or model sport shops.
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How to connect PSU to these PCB pads I have the Maxim Integrated evaluation kit for one of their chips ( MAX14830EVKIT ), and would like to understand what the intended way is to connect the two PSUs (24V and 3.3V, both 100mA) to this PCB. The PCB has two pairs of small pads, each with two small holes: I could solder wires onto the PCB, but that does not seem appropriate for a "fully assembled and tested" board. I got some test probe grabbers, but they don't go into the holes. The pads and holes seem too small to screw some binding post into. What is the right tool to use here? <Q> You are indeed meant to solder your connector pigtail wires to those pads. <S> If you prefer to use clip leads, standard 0.025" square pin headers will fit. <S> And in this context fully assembled and tested refers to the Evaluation Kit board itself, we can't know what connector you will use, if any. <S> We use a custom test fixture with spring-loaded "pogo pins" to contact those pads from below. <S> This approach lets you eliminate contact resistance or use whichever connector you want. <S> (I am one of the EV kit designers at Maxim Integrated... <S> although I'm not the designer of the MAX14830EVKIT, I have designed many similar EV kits.) <A> It was probably designed for test points like this example . <S> These are popular and available from many different manufacturers. <A> Solder wires directly to the oblong pads. <S> Solder male "header pins" into the holes (assuming 0.1 inch / 2.54mm spacing) <S> Then you could use 2-pin female mating connectors for connecting and disconnecting the power wires. <S> Clearly each of these options (and perhaps others) have their advantages and disadvantages. <S> Since only you know all the conditions you are working with, only you can decide which method makes the most sense.
| Or you can make a half loop with bare wire and solder each end to a hole.
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Problem with capacitor after bridge rectifier I wanted to make a simple power supply so I made a bridge rectifier with 4 diodes rated 6 amps, then I connected the bridge to a transformer (220 volts to 9 volts (measured 9.4 with voltmeter) 50 Hz) and when I measured the voltage after the bridge the output was 9.1 volts. Then I added a 4700uF capacitor (50 volts) in parallel and when I measured the output voltage it was 13.8 volts but when I disconnect it, voltage is 9 volts again. I first thought that maybe the voltmeter is broken but when tested with an small motor it was actually running faster with the cap. Is it pulling more pressure on the transformer to draw 13 volts or something? Because I'm really confused right now. By the way will it help if I use an smaller cap? Because that would increase the ripple voltage? <Q> Your capacitor will charge to the peak bridge rectifier's output voltage, minus the drop through the diodes. <S> For a transformer with an output voltage of Voac, your capacitor should charge to somewhere around ( <S> Voac*sqrt(2))-1.4 where the 1.4 is the voltage lost across the two conducting diodes in your bridge rectifier. <A> The answers already here are correct. <S> I just wanted to add that since you are using a Full-Wave Rectifier (4 diodes), your ripple voltage is determined by: $$ V_{ripple}=\frac{V_{peak}}{2fCR}$$ or equivalently,$$ V_{ripple}=\frac{I_{load}}{2fC}$$ <S> Now, the thing here is that you want to keep the ripple voltage 'small.' <S> A good number is within 10% of \$V_{peak}\$. From your post, I see \$V_{peak}=13.8{\mathrm V}\$ If you do the math, in order to keep the ripple voltage, say, 10% of \$V_{peak}\$, the maximum current you can draw from the rectifier is :$$ I_{load}=2(1.38{\mathrm V})(50{\mathrm{Hz}})(4700\mu{\mathrm <S> F})$$$$ I_{load}=0.65{\mathrm A}$$ <S> That is the maximum current you want to draw if you want keep the ripple voltage at a maximum of \$V_{ripple}=1.38{\mathrm V}\$ or 10% of the peak voltage. <S> I saw in one of your comments that you were concerned about damaging components. <S> The capacitor should be fine (since it is 50V), but they may be a lot of current running through the circuit at startup (when the capacitor is fully discharged). <S> So you want to make sure your diodes are rated to handle that much current, and also check the reverse voltage ratings for those diodes. <S> The reverse voltage across those diodes are theoretically the same as the peak voltage for a full wave rectifier with four diodes (in your case the diodes should handle more than the 13.8 volts you are getting at the output). <A> That's correct. <S> You are storing the top voltage value in your capacitors so you will see an 1.41 times increase. <S> Google top wave rectification. <S> Regarding their size, you need to make a trade off between output voltage ripple and input current/power factor/THD. <S> How big is your load? <A> The capacitor, as others have told you, basically stores peak voltage with little ripple. <S> Assuming that your load is mostly constant and that the transformer is a voltage source, the ripple consists of a long discharge time and a rather short charge time where the voltage from the transformer replenishes the ripple. <S> This short charge time will draw a lot of current and you want to avoid voltage loss, so your connections should be solid enough. <S> Now the whole period time is 10ms and the recharge period will be a fraction of that: <S> for discrete circuits, that's too short for heat death, so you only need to worry about the sustained power losses. <S> For diodes (with their constant voltage drop) they are proportional to the current through them, for resistive material they are proportional to the square of the current. <S> This includes the transformer coils. <S> So yes: using that capacitor will likely cause your transformer to get warmer than otherwise. <S> But their specs for continued operation will support the kind of peak load that standard rectifying circuits with capacitors will cause. <S> Killing a mains transformer with overload usually requires rather drastically exceeding the specs.
| You won't damage them as long as they are rated to handle the voltage/current they are being provided with.
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Amplified microphone signal output to ADC input. How can I eliminate DC offset? I'm making simple sound level meter. I try to make it precise. I'm using ADC ref voltage (2,048) so rms calculation can be simpler. Can I use this ref voltage to set offset voltage to (Vref/2) and measure this signal by diffrential measurement so it eliminate DC offset form readings? I must use battery supply and 3V3 is from voltage regulator. Entire module is supply by 3,3 volts including ATxmega microcontroller. I'm using electret condenser microphone. I'dont want to use dual polarity supply if this is not necessary. <Q> Assuming you cannot measure negative voltages, your schematic is good. <S> You have to compensate in your software for the DC offset. <S> The latter has the advantage of being fully automatic and compensates for drift. <A> Can I use this ref voltage to set offset voltage to (Vref/2) and measure this signal by diffrential measurement <S> so it eliminate DC offset form readings? <S> Assuming that your signal into the ADC MUST NOT drop below the ADC_NEG input then this won't work. <S> Assuming that this is a fairly conventional pseudo differential input ADC, then connect ADC_NEG to ground. <S> This will force ADC_POS to be mid-scale DC on the ADC input. <S> If in doubt, download LTSpice and simulate it. <A> Use a non-inverting amplifier. <S> R2 and R3 feed in the Vref offset. <S> If you want to amplify the signal, this topology has the advantage that DC gain is 1, because C2 is blocking DC. <S> Therefore the Vos (input offset voltage) of the opamp will not get amplified. <S> This topology only works, if your gain is greater or equal 1. <S> To realize smaller gains, you can use R4 to attenuate the input signal, because R4 basically forms a voltage divider with R2 || R3. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Why not omit the inline capacitor at all? <S> You can use the DC_bias of the JFET and trim it via differential action in the opa to your target value? <S> Attention: <S> A RRIO Opamp is likely to be needed, if u want to dc-couple. <S> You can also put the whole JFET inside the NFB-Loop of the OPA, if u have access to the gate of the JFET - you can also trim DC via standard methods, to your target value. <S> If you need amplification, math can get a little complex - if buffering is sufficient, the calculation becomes quite simple.
| Compensating is as simple as subtracting the offset; you could implement a calibration procedure, or use a long-term average of the input values to correct for this offset.
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What switch do I need to enable/disable fan with wind speed measurement by a DIY anemometer? I'm going to be making a device for burning man that will effectively evaporate grey water from our camp. This will be mostly wind powered, however when there isn't any wind, I plan on using a high torque, low rpm 12V dc motor to power the device. The power to the motor will be provided by a set of Rynology 100W solar panels which will power up a battery store. Here is what I would like to do. I want to be able to determine if there isn't any wind in order to engage the power to the motor to take over operations until the wind returns. I plan on achieving this with a DIY Anemometer that is attached to an old CD player motor. When the little CD motor is spun by the wind, the anemometer will produce a little voltage. I would like a switch of some sort that can detect that little voltage, and if its detected, to close the 'gate' that controls the power from the battery store. It seems I would have 1 lead coming into the switch that will carry the small voltage from the motor, and then two leads one end attached to the battery bank, and the other end attached to the 12V dc motor -- which can be interrupted if there is voltage detected on the aforementioned lead. Is this a thing? Can I pick this voltage detection gizmo up at a small electronics store? What is this thing called? EDIT: here are more details of the parts I'm looking at in terms of the motor. This is the motor I will be using to drive the device when there is no wind https://goo.gl/2TWwNS . I haven't decided yet but I'm pretty sure something like this will be used in the aeronmeter https://goo.gl/l8cFUd which seems to be able to produce a little voltage when spun.. here it is in this youtube video... I dont have enough reputation to put more than 2 links in my post. So I left the youtube link in the comment to Transistors suggestion <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Anemometer wind detection circuit. <S> Assuming your anemometer can put out a few volts the circuit of Figure 1 should work. <S> 12 V relay, RLY1, will provide power from the battery to the motor. <S> When the anemometer puts out some voltage it will turn on Q1 which will turn on Q2 which will energise the relay. <S> When the relay is energised the contact will change over and the motor will be disconnected from the battery. <S> You need a relay to match the battery voltage. <S> The contacts should be rated >= to your load current. <S> Your local auto shop should have something suitable. <S> Check the anemometer polarity in the wind before connecting up. <S> Tip: you could test the device and cut-in speed by driving around with the anemometer on the roof of your car. <S> Edit <S> after motor specified: Figure 2. <S> Motor (generator) specification from Mabuchi Motor . <S> I didn't watch the video but at 3 V it will spin at about 2500 RPM. <S> This means that to generate 3 V you will have to spin it at 2900 RPM. <S> This is close to 50 revs/second and seems most unlikely for an anemometer. <S> If it ever happened then Burning Man is over. <S> You need to come up with a different wind detector. <A> Maybe a relay like this: http://www.mouser.com/ProductDetail/Crydom/DC60S3/?qs=mNyg5qXQ%2FsdpD8JEee%252brpQ%3D%3D&gclid=CMn1xMnX-M0CFQ-oaQodFeUP-g <S> You can see in the specs it is good for up to 3A, from ~3-48V. <S> Also the control input will draw <S> ~2.2 mA <S> to 25 mA. <S> There are lots of different relays for whatever specific applications you need. <A> , the flap opens and the switch gets pushed (or released). <S> Mind you, I'd be tempted not to solve this at all and just run the system all the time - because why not? <S> If it's solar powered it's not costing you anything, if the panels are there ready and waiting then why not just leave it running 24/7?
| I'd be tempted to solve this mechanically: Rather than an anemometer, just use a flap pointed into the wind (easily done with a weather vane style mount) with a microswitch - when the wind blows
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New soldering iron smokes from inside the handle (not from tip) - is this a problem? I've started using a brand new soldering iron for the first time and it's not working as expected: After about 10 seconds of power, smoke starts coming out from inside the handle, next to the metal shaft. After 30 seconds on, there's a lot of smoke pouring out, like a candle that's just been extinguished. I've turned it off and on and this still happens There's no smoke from the tip, and a very small amount from where the tip attaches to the shaft The tip doesn't get hot enough to melt solder after being on for a little over a minute. I haven't tried leaving it on for longer than that because the amount of smoke from inside the handle seems to just keep increasing. The handle doesn't seem to get noticeably hot in this time I bought it in a 230V country (UK) and am using it in a 230V country. It's a 40W Maplin soldering kit . I expected the tip to smoke, but not the inside of the handle. Looking this up online, I see a lot of similar questions being mis-read as being harmless tip smoking, and a lot of comments that they're often coated in oils that burn off. It seems to me that these wouldn't be inside the handle, though? Or would they? Some other comments seem to think this is a sign of a defective iron, but don't offer anything to back up this opinion beyond "I've never seen that". Just found in the paperwork: The first time you use the soldering iron, it may smoke slightly and there may be a smell of burning as the heating element dries out. This is normal and should only last a short while ...which sounds unusual and is a little vague. This seems like it'd explain the small amount of smoke where the tip joins the metal shaft, but I'm not sure if this would explain large amounts of smoke from within the handle. Update: I tried leaving it plugged in outdoors under supervision (fire extinguisher at the ready). After about 7 minutes it seemed to stop smoking, so I took it back inside (keeping the fire extinguisher handy), let it warm up again for a few minutes, started soldering, then noticed it was smoking again. It stopped smoking pretty quickly after killing the power, and was still hot enough that I could finish my soldering, but I think I'll be looking for a replacement. <Q> Maybe I bought one of those from Maplin a year or so ago. <S> It produced a lot of acrid smoke the first time I turned it on, and I was also concerned about it. <S> It's a very cheap and nasty product. <S> I dismantled it insofar as was practical, but couldn't see anything wrong with the wiring, but there was some residue on the hot parts which was burning off. <S> I put it back together, plugged it into a RCBO protected extension lead, and left it turned on sitting on a fireproof mat in the garden for a while. <S> The smoke died away in a few minutes, and the smell was gone after an hour. <S> The iron has been perfectly serviceable ever since. <S> Now, I haven't seen your iron, so I can't say for sure whether it's the same. <S> Can you safely leave it somewhere for the smoke to die away? <S> Somewhere fireproof, just in case. <A> Yes this is a problem. <S> Don't inhale the smoke, and return it. <A> It's common for manufactured items to have some trace amount of surface contamination; in particular, metal items may have residues of oil or similar substances. <S> So, a device intended to get hot may smoke a little bit when new until the contamination is burned off. <S> That said, nothing inside a handle should be getting hot, and even at the tip, there should be no more than a few wisps of smoke. <S> It sounds like your iron has a poor electrical connection inside the handle, which could be getting hot, even if it isn't noticeable; the smoke would then be coming from burning insulation. <S> It seems like your iron is defective, should not be used, and should be returned.
| The tip should not really smoke either, except for when you are putting flux on it (a lot of solder will have flux-cores, so solder will make it smoke some too).
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How to measure water heater temperature? I'm building a summer cottage and plan to add some sensors for monitoring system health, with a Raspberry Pi and/or Arduino. Specifically, since electricity is not always available there, I'd like to measure the water temperature in a stock water heater (50..100 litres, 13..26 gallons) that doesn't have any extra connectors to use, eg. a Jäspi 30 or 60 litre model ( pdf ). What kind of sensor should I get? How do I plug/plumb it securely? Some notes: I'm aiming for a "ballpark" accuracy, maybe something like ±5°C..±10°C (±9..±18°F) I'm not planning to control the water temperature, just for logging I'd rather not damage the heater or its insulation I'd prefer a solution that would not void my house insurance (in case of an unrelated water damage) It doesn't have to look pretty :) <Q> I need a similar setup for recirculating the hot water in the boiler through an adjacent solar water heater tank. <S> This would increase the storage capacity of the system as it moves heat to the boiler when the temperature of the solar water tank exceeds the temperature of the boiler. <S> In my case I'm more interested in the temperature at the bottom of the tank. <S> A possible solution would be to replace the drain valve with a "T" such that one "path" goes to the tank straight and the other one forms a 90 degree angle. <S> On the straight through path insert an immersion well for the thermocouple on the other path the drain valve. <S> All standard parts and no holes to drill. <S> The immersion well will constrict the flow to the drain valve a bit which is not really an issue on the drain port. <S> For the hot water port it might still work with a 1/2 inch immersion well on a 3/4 pipe. <A> Electric water heater thermostats get their reading by being mounted to the outside surface of the tank behind the insulation. <S> In my opinion this is the ideal location to also mount a thermocouple to measure temperature since it is identical to the control point. <S> Any temperature differential (actual vs. measured water temp) should be identical to that sensed by the adjacent thermostat (upper or lower). <S> Just take the access panel off to expose heating elements and thermostats. <A> Why do you need to put the temperature sensing device inside the water heater, which as the comments indicate, will either be difficult, or void the warranty, or both. <S> I would think the temperature of the water in the hot water line immediately adjacent to the hot water heater would be a close enough approximation. <A> Consider tapping into the hot water outlet pipe with a 'T' and having a sensor that protrudes down into the tank. <S> It should be a straight run down into the hottest water at the top of the tank. <S> The only way I would consider this is with a compression fitting on the T and a thick-walled welded-end stainless steel (eg. 316 SS) tube containing the sensor going down into the tank. <S> The T-fitting and sensor and seals will be subject to your water system pressure (and likely some water-hammer impulses) and failure <S> will cause water to spill out continuously, possibly causing a lot of (tens of thousands of dollars' worth or more) damage. <A> The only way I'd be willing to do this in a house I own would be to tap off of the existing thermostat, or to replace it by a smart thermostat that's web enabled, like this . <S> Any drilled hole that you put in on your own is likely to eventually fail. <S> Indeed, you should assume it will, and plan accordingly. <S> If you want something robust, you should talk to a plumber. <S> I suppose you should ask yourself if you want to measure the tank temperature, in which case you should go to the consumer market, or if you'd like to build something that will measure the temperature. <S> If the latter, I'd advise finding some other project, as some projects are just not DIY friendly.
| Put a T-coupling between the hot water heater and the hot water line, and at the "bottom" end of the T, you can screw in a cap in which you have drilled a hole for the temperature measuring device, sealed with epoxy or something similar.
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can a simple diode and resistor in series qualify as an inverter? I'm working on a single-sided PCB and I find transistor pins are too close together for tracks to run through, so I'm wondering if I could roll off my own inverter by using the cathode of a diode as input and anode connected to a pull-up resistor as output. Would I be able to apply this circuit to logic gates without problems if my resistor is 10k and the diode is the 1N4007 kind? I want to avoid using an inverter IC just for one inverter. EDIT: this is the proposed solution <Q> When IN is driven low, it should be fairly obvious that OUT will then be driven low thru the diode. <S> OUT will be one diode drop above IN, or Vcc, whichever is lower. <A> I'm working on a single-sided PCB <S> and I find transistor pins are too close together for tracks to run through, ... <S> Just spread the legs of the transistor. <S> Figure 1. <S> Spreading the legs of a transistor is quite acceptable if machine insertion is not being used. <A> Thanks for pointing out my mistake, Photon. <S> This does not create an inverting function. <S> For a more solid, proper inverter, I would recommend CMOS instead, requiring one PMOS and one NMOS only. <S> This would still require 2 components (6 vs. 4 connections, though) and would give full rail logical signals, with lower output resistance. <A> This is the circuit I think you're describing: simulate this circuit – Schematic created using CircuitLab <S> This circuit does not invert. <S> The output is low(ish) when the input is low, and high when the input is high. <S> This circuit might be useful as a non-inverting level translator between a high voltage logic family and a lower-voltage logic family.
| No, your circuit does not invert. Diode logic, without any transistors, by nature creates non-inverting functions (and, or, buffer).
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Power supply circuit for powering SG90 servo. How to handle high current requirements? I need to power a small SG90 servo motor. I could easily use a 5V battery but why bother? :) Aside from this not funny joke, I'd like to put to good use my little knowledge of electronics gained at University. The datasheet of the servo does not mention any current draw (max, min, average, etc., ...). However I know that my USB port can barely handle Arduino with some sensor and a servo moving around so I guess it could easily be 100 mA. This figure is nothing I know, but it's a start at least. Furthermore the servo does not do any "heavy lifting" it just rotates a small aluminium bar which is super light. There should not be any "stuck" positions. Since my main PS is a 12 V battery I thought I could step down the voltage using some resistors and then use an op amp (LM741) and a NPN transistor (2N3904) to set voltage and current, respectively. See the schematics below. By using this circuit I should be able to output 100 mA while keeping the heat dissipation of the 2N3904 NPN within the admissible range of max 0.5 W. I guess I could even go up to 200 mA if I work around the circuit a bit more, however, before going on I ask myself and you, is this circuit able to handle such high current loads such a servo motor (i.e. do you see any flaw in this)? <Q> Don't reinvent the wheel, just get a voltage regulator. <S> You can probably get one for < $10, all parts included. <S> For your circuit, I would be concerned about R3. <S> If you want to output 0.1A, the voltage drop across a 100 ohm resistor is going to be 10V. 12V - <S> 10V = 2V. <S> Also, there's probably going to be other problems that I haven't foreseen yet <S> (I'm not an expert on voltage regulator design). <A> Although I haven't used the SG90 servo, most RC servos want to see about 6 Vdc with significant current capability. <S> Even though the servo might be lightly loaded, you still have to get the motor spinning everytime the servo begins to move. <S> The motor stall current can be significant - anywhere from several hundred mA all the way to to several Amps. <S> Because the servo is lightly loaded, you can probably get away with a really simple linear regulator. <S> The simplest such regulator is a simple Zener diode followed by a series-pass current-boost transistor. <S> Something that I have used in the past is as follows: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This is good for up to 1 Amp or more for brief surges and is perfectly adequate for powering a RC servo. <S> Note that the regulation is fairly lousy but the servo won't mind. <S> [Edit] <S> Someone commented that this servo doesn't want to see more than 5V. <S> This is an easy fix: simply change the Zener diode to a 6.2V part instead of 7.5V. <S> That would be 1N4735 or 1N5234. <S> This will provide a nominal 5V output. <A> For completeness, here are the specifications of the SG90 (from FEETECH (Fitec) <S> FS90 9G Mini Servo with Accessories ), showing the current draw, at idle, no load and stall: <S> The FG90 is similar in specification to the SG90. <S> From the website: The FS90 is a 9 g analog servo from FEETECH (formerly known as Fitec) which is similar in form and function to the Towerpro SG90. <S> The specifications state that this servo has <S> a 120° operating angle for standard servo pulses between 900 µs and 2100 µs. <S> As with most servos, the pulse range can be expanded to achieve an expanded operating angle, but the limits of which are not specified by FEETECH. <A> The rated current for a SG90 is 220 ±50mA, stall 650 ±80mA. <S> So budget on 270mA worst case. <S> A simply 12V/5V BUCK should handle this and be relatively small.(long link)Check www.aliexpress.com for Mini360 or similar.
| Digikey sells efficient switching stepdown regulators that will output 5V @several amps (like this one ).
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Determining drill size for panel mount components Many components can be panel-mounted by drilling a hole through the panel and then fixing in place with a washer+nut. However, sometimes the datasheet does not specify what is the hole size to drill. For example, the datasheet for this panel-mounted push button includes diameters at different points in the component, but no hole size: Should the hole size just be 7mm (major diameter of an M7 thread)? Or is it recommended to make the hole slightly wider? <Q> It is ASME B18.2.8-1999: "Clearance holes for bolts, screws and studs". <S> There is a similar ISO standard as well (for metric only): <S> ISO 273-1979. <S> For each thread size, they give three hole diameters (for close, normal, and loose-fit categories). <S> Here is a PDF containing the chart, but you can easily find them using the "hole clearance chart metric" search terms on the internet. <S> In your specific case, for M7, it is 7.4mm for close fit, 7.6mm for medium, and 8mm for loose fit. <S> You choose. <A> A correctly produced M7 thread will be slightly less than 7mm across, so should fit in a dead-size 7mm hole. <S> But it doesn't hurt to make the hole 7.5mm. <A> What you show seems quite clear. <S> The threaded body is M7 size. <S> Surely there are references out there for the maximum diameter of a M7 thread. <S> It's the datasheet's job to tell you the characteristics of the product, not how to use it in your particular case. <S> It shouldn't tell you the hole size, although it should tell you the outer diameter of the thread without you having to look it up elsewhere. <S> Once you know that, you can come up with your own hole size, knowing the slop in your process, how much clearance you want, etc. <A> The answer is actually right there on your data sheet. <S> The circles on the right are an end view of the device. <S> The smaller of those 2 circles represents the diameter that would be going through the drilled hole you are asking about. <S> You can see that they have indicated it as 7.4 (and i'm assuming this datasheet is in millimeters?) <S> So, if you make a hole the same diameter as that, you are going to get a pretty snug fit. <S> There is a down side to that. <S> First, its actually tricky to put it in, cause the body and the hole have to be perfect with regards to tolerance on the measurement, and if they are, you still have the issue that if the hole is not perfectly perpendicular to the surface, the top bezel may not sit flush with the surface, rather one side will have a small gap. <S> You can avoid both of these by going with a loose fit, and make that hole just a tiny bit bigger. <S> The device will go into the hole easier, and will have a little wiggle room for the front to go flush when you put the nut on. <S> On the other extreme, if you make the hole too big (in this case, 9.5) your device will fall right through the hole. <S> All right there on the data sheet. <S> You could also look it up, but they are nice enough to give us that data sheet <S> so we don't need to, cause they already did... <S> so if everybody goes looking these things up on their own, the guy they pay to look it up looses his job. :)
| Make the hole diameter whatever you would normally make it to accommodate a M7 thread without interference. Amazingly, there is actually a standard that defines what should be the hole sizes to accomodate standard threaded screws.
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How can I connect BeagleBone to 2 slaves with 1 "Chip Select" pin? I have a BeagleBone white and I want to use it as a master which is connected to 2 slaves via an SPI connection, besides SCLK, DIN, and DOUT there is one pin of chip select "CS" "just one" and I need 2 because I have 2 slaves, what I already knew is that I can use any GPIO pin as chip select, but when I saw the "CS" pin I get confused, can I use any GPIO pin as a chip select ? if yes, why there is a pin with the name of chip select ? <Q> Yes, you can use any digital out as a CS. <S> Bit banging will never get any easier than that. <S> Alternatively, you might consider just NOT-ing the one chip select, and seeing if you can get away with controlling two slaves that way. <A> I assume you use SPI0. <S> Another option is to use SPI1 instead which has both CS0 and CS1 available on the expansion headers. <S> (At least in theory. <S> I'm currently trying to achieve the same thing, but I've failed to enable both CS0 and CS1 for the pins connected to the expansion header using TI's pinmux tool. <S> There seems to be no available IO-set that includes all the required pins. <S> Any hints appreciated...) <A> With micros you have a dedicated spi "block" if it includes a CS, it may be able to assert the CS line automatically and there is usually a designated pin. <S> Most SPI hardware blocks do this is <S> so when preforming writes you don't have to spend precious CPU cycles to assert a gpio pin if you need fast timing. <S> You don't have to use the SPI block's CS lines, you can use any GPIO line, however you have to write code to turn it on and off. <S> As long as fanout works out for you GPIO <S> (which shouldn't be a problem for two devices) and the devices have the same logic level you should be fine. <S> But why would you want to? <S> You have plenty of pins so you could have two different GPIO's and control them seperately or at nearly the same time. <S> The only application I can think of is simultaneous sampling with an ADC to make sure you start at the same instant. <S> If you bit bang the SPI port you can even have two seperate SDO's so you trigger the CS line to start the conversion at the same time, and write the same value into both ADC's and then read back both values simultaneously on two different GPIO's connected to two different SDO's of the slave devices.
| If you roll your own spi code you have more options. Use some other pin for a second chip select, and bit bang it.
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What is the need for precharging in SRAM/ DRAM memory cell? Why is precharging so necessary while preforming read operation on SRAM/ DRAM cells? For example in the SRAM 6T cell shown below, Reading 0 requires bit line to discharge to 0; Reading 1 requires that bit line voltage is equivalent to logic '1' value. Right? So why can't this charge (in case of Q=1) occur via M4 and M6? i.e. why dont we consider M4-M6 as path strong enough to pull up the bit line? Why do we rely on precharge value to give logic '1' on bit line? Is it because NMOS M6 is a weak highpass?(due to the fact that NMOS is weak highpass) Also, In case of precharging, is the precharge voltage = Vdd or Vdd/2 ? <Q> The answers here are good regarding how in normal practice the bitlines will be charged to VDD/2. <S> However that doesn't really answer the question, because it: does not apply all the time (depends on the cache requirements and process technologies. <S> I have seen plenty of caches that precharge to VDD because at low voltage operation VDD/2 can be too risky) <S> The 'canonical case' everyone learns first doesn't precharge to VDD/2, and this is the situation he is asking. <S> There is still a good reason they go to VDD, though. <S> The main reason they charge the bitlines HIGH (in the circuit he is showing) and let them discharge is because the pass transistors are NMOS. <S> This means they pass a very solid ' <S> 0' <S> but they pass a degraded '1'. <S> So rather than start the bitlines low and let them pull up through the NMOS (slower and weaker, can only pull to VSUPPLY-VTH), they will start the bitlines high and let them pull down through the NMOS (which can pull down more strongly, to a solid '0'). <S> Another very good reason are the constraints on transistor sizing which must be met for proper writability/readability. <S> Read operation: <S> M1 must be stronger than M5, so that the voltage divider formed between M5/M1 does not flip the bitnode. <S> Write operation: <S> M2 must be weaker than M5, so that M5 can overcome the feedback loop when writing a '1'. <S> So, M1 > M5 > M2 (and M3 > M6 > M4). <S> The PMOS are the weakest transistors in the whole cell, why use that to pull up? <S> On top of that, traditionally NMOS have been faster than PMOS. <S> This is less true today in the lower process technologies (22nm, 14nm, 10nm, etc) but is still normally assumed. <A> Precharging is not intended to deal with the ability to source drive current into a bit line. <S> If there is no precharge, the maximum voltage swing in a readout is from a "0) to "1" (or vice versa), which happens in T01. <S> This results in about one half the transition time (1/2*T01), and results in a faster memory. <A> For DRAM, purpose of precharging is to close the current row and to allow activation of another row. <S> In the case when the same row cells are to be read, precharging can be done later when using Fast Page Mode DRAM. <A> You can't drive a high-value through an NMOS switch. <S> The output will stop rising fast once the output gets close to Vdd-Vt. <S> Pre-charge allows the circuit to only have to drive a 0 value, which it can do well. <S> I have attached a simple schematic to demonstrate the idea. <S> simulate this circuit – <S> Schematic created using CircuitLab
| Precharging ensures that the bit line is driven to voltage midway between "0" and "1", so that when the actual cell is read out, the line need only be driven from the midway voltage to either "0" or "1". Precharge is intended to minimize propagation delay time.
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Junction between curved and straight edge in KiCAD I have a round PCB with a tab that extends outwards for a connector. I am unable to get the outside lines to properly resolve and connect together at the ends. Here is an image of the complete board and a zoom on the problem area. How can I properly draw the outline so that the company will accept the file? <Q> It seems that it is not possible to snap to the circle. <S> Eagle has the same difficulty. <S> However the problem can be solved using some relatively trivial maths. <S> Its basically a geometry problem of trying to find the coordinate of where the line intersects the circle, and then setting the XY coordinate of the end of the line equal to that point. <S> A diagram will help: <S> You know both the origin coordinates of the circle \$(X_0,Y_0)\$ and also where the line is on the x-axis - the \$X_1\$ part of the coordinate. <S> Finally you also know the radius \$r\$ of the circle. <S> That is all the information required to find \$Y1\$. From the diagram, we can say that: $$\mathrm{d}X = <S> X_0 - X_1 \tag <S> 1$$ <S> From there we can simply use Pythagoras's Theorem: $$\mathrm{d}Y = \sqrt{r^2 - \mathrm{d}X^2 <S> } \tag 2$$ <S> Then: $$Y_1 = <S> Y_0 + <S> \mathrm{d}Y \tag 3$$ <S> Putting (1) and (2) into (3) <S> it becomes: <S> $$Y_1 = <S> Y_0 + \sqrt{r^2 <S> - (X_0 - X_1)^2} \tag 4$$ <S> If you solve (4) for each of your two lines, that will give you the y-axis position for the end of each of the lines such that the line ends on the circle. <S> In Eagle I would then draw an arc which ends on both of those two points. <S> You can work out the total angle of the arc by using trigonometry to get \$\theta\$ for each of the two lines and find that \$\theta_{arc} = <S> 360^\circ - (\theta_1 - \theta_2)\$. That draws an arc that is the correct radius and intersects with the two lines. <S> I assume you can do the same thing in KiCAD. <A> That is pretty much nontrivial, because snapping to graphics primitives does not work. <S> I'm going to file a bug about this. <S> For the time being, I think your best bet is drawing the outline in inkscape or some other vector tool, export it as DXF, and import in KiCad into the Edge. <S> Cuts layer. <A> I was trying to do something similar, connecting the end of a curve to a straight line for edge-cuts. <S> The way I got it to work was:Draw a reasonably close approximation. <S> Then use 3d-viewer and it will tell you there is an error at the end of the curve (you might have to rotate your drawing so that the error occurs at the end of curve instead of the end of the line). <S> This will give you the coordinates of the end of the curve to 6 decimal places. <S> Change the end point of the line to these coordinates. <S> Now the other end of the line should be 'on the grid'. <S> TLDR:
| KiCad's grid system only goes to 6 decimal places and kicad will work out the curve's end point for you.
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Absolute cheapest, (easy) way of starting programming on ARM microcontrollers I want to start using ARM processors, migrating from PICs which I've been using for too long. 8 bit models were available for under $1, programmers for under $10, and I have been spoilt by the ease and low cost of getting started. But seeing that their power is dwarfed by some ARM chips at the same price, and being fed up with closed-source software, I want to make the switch. I would rather avoid getting any development boards, and instead get straight into it with a cheap generic break-out pcb ($1), and on a breadboard using whatever external components are essential to get it going. What are my options for programming ARM chips in this way? Are there programmers such as the PICKit2 which use USB to connect to a PC, programming via a simple serial connection (like ICSP) with the chip? How much of a difference in setups is required for different ARM manufacturs, ARM versions and individual chips? (ST, Atmel...) E.g. does each manufacturer need their own compilers, programmers, IDE etc? Or are there common tools for all? EDIT: Alright so after more research I believe I have come up with a relatively cheap solution, an stm32 dev board can be flashed with DAPLINK firmware, but I dont believe that the official github firmware will work natively(all of this is speculation until I get my stm32 dev board in the mail). But I found that the daplink_usb board included with the readbear mk20 is running an stm32 chip, they have released the firmware, which needs to have a line changed to make it compatible with the 8mhz crystal(Detailed in forum post linked below). Otherwise change out the crystal with a 16 mhz one. Ill update once I've confirmed this when my dev kit arrives. GITHUB REPO Good Forum Resource here . Redbear Github fork <Q> The stm32F0 and stm32L0 line have discovery boards running about $10, and Keil will provide a full function IDE for this line at zero cost. <S> The Keil free pro MDK install <S> instructions are HERE <S> Also, ARM has a white sheet on migrating to Cortex M3 from PIC that you might find helpful <A> Easiest way- shell out > <S> $10K USD for a full-functioned Keil Pro compiler, buy their JLINK debugger (another $1K maybe- <S> there are cheaper ones with some limitations). <S> IAR is another expensive possibility (examples are provided for STM32F7 Cortex M7 processor that work on the 30-day IAR demo) <S> Get a JLINK clone for $20 or so, which I think will work okay- not tested yet, for debugging. <S> There are detailed instructions for the latter on the net, however they make certain assumptions. <S> Expect to spend a day or more getting it going, especially under Windows. <S> Don't expect to be able to use many of the examples provided for other IDEs without some work. <S> Impressively, the free toolchain can use 'packs' ('experimental' right now). <S> There are other systems such as Rowley Crossworks (which uses gcc, I believe) which are less painful financially. <S> Atmel Studio is another, but I've had bitter complaints from my very experienced firmware developer about it (have only briefly played with it myself). <S> If your code needs are less than 32K you can use the same Keil system for free (code limited version), but be aware the upgrade path is easy but rather expensive. <S> For example, it won't compile the simple Ethernet examples for the SAME70. <S> Fine if you're replacing PICs or AVRs with low-end ARMs, but not so great if you are going ARM because you actually need to talk to LCD displays and run complex communication protocols (possibly pre-compiled modules can be included without affecting the 32K limit, I've not investigated that particular angle). <A> Here's what I use: STM32F103 "minimum system board" (see <S> e.g. here , Cortex-M3 core ), runs on 3.3V or USB power without any external components, clones come for about US$3 each. <S> This fits nicely your request to "get straight into it with a cheap generic break-out pcb, and on a breadboard". <S> ST-Link V2 USB programmer clone (looks like this one ), starts at about US$2 and supports on-chip debugging aswell. <S> EmBitz (formerly Em::Blocks) as IDE with gcc toolchain, US$0 STM's CubeMX to help getting started with new projects, US$0 <A> The easiest start probably is one of the third-party clone boards. <S> Random example from ST . <S> That requires a programmer using the 'SWD' protocol. <S> ST make 'ST-LINK' branded ones, I'm not sure if you have to use ST-LINK ones with ST devices or if it's really generic. <S> Some combination of SWD and JTAG plays the role of ICSP on ARM systems, giving you programming and debug capabilities. <S> Software-wise, it's usually possible to work with GCC and OpenOCD on most chips. <S> The details are slightly different for each device. <S> Professionals often use the Keil toolchain, which is quite expensive. <S> Some devices (e.g. of the Kinetis series) have USB bootloaders: the device appears as a mass storage device, you download a BIN file onto it and press a button. <S> Easiest possible solution, no programmer required. <S> Atmel AT91 have a USB bootloader that works with a proprietary protocol called SAM-BA. <A> On the IDE front, Silicon Labs provides Simplicity Studio , which is based on Eclipse . <S> It comes as standard with GCC. <S> There is built-in support for all the starter kits they sell, making getting started relatively painless. <S> Programming uses a Segger J-Link driver for the starter kits (free). <S> Just connect the kit to USB and go. <S> Atmel have their Studio which is based on the Visual Studio IDE and can connect to any of the Atmel debuggers. <S> This also ships with GCC. <S> Both vendors have numerous (very numerous) examples to drive their devices. <S> I have used both and although the documentation is not perfect (it never is), it was certainly sufficient to get me going relatively painlessly. <S> Many of the ST kits (and others) are mbed enabled. <A> Cypress makes PSOC-4200-based (ARM Cortex M0) breakout boards with a DIP-40 form factor that include a USB-based programming adapter in a break-off portion. <S> Price for the breakout board and attached (detachable) programming adapter is a whopping US$3.99 from Digi-Key. <A> In my believes TI,NXP(consist of NXP+Freescale) & ST are major role player in cortex M world,especially ST and NXP offers a really competitive tools for newcomers,one another important parameter is popularity in open source community which cause amount of tutorials,libraries,device drivers,tools & etc. <S> then : mbed online compiler + a mbed board <S> A STM32 discovery <S> board(like STM32F407discovery) + built in st-linkdebugger +SPL or HAL (STM CUBE) framework + free license keil IDE. <S> A LPC discovery board + LPC-link 2 debugger + LPCOPEN framework+LPCXPRESSO free license IDE. <S> Number 2 & 3 are better choice for upper level than hobby. <S> Also remember with using free license of professional tools like IAR,Keil or even LPCXPRESSO you have less pain while moving to the professional area. <A> I would recommend <S> CooCox <S> - it is the same GCC + Eclipse combination, but no need to configure toolchain manually, just install it and start coding.
| Cheapest way- download and install a (free) GCC-ARM + Eclipse toolchain with JLINK debugger plugins.
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How can I safely use an aluminium enclosure without an earthed socket For a mains-supplied circuit I want to use an aluminium case, so the case can act as a heat sink.However, I want it to be safe, also when there's no earth connection in the socket where the circuit is plugged into (this is still quite common in Europe). What is the common way to make the circuit safe to use in an aluminium case? I know that normally you would connect the earth-wire to the case, so in case one of the wires internally touches the case, the circuit breaker will turn off the power, protecting people from touching the enclosure while there is 230V on it (see also this question ). But of course this does not work when the socket has no ground connection... <Q> Under IEC rules a mains applicance that does not have a ground (earth) connection is a class II appliance , a.k.a. ' <S> double insulated'. <S> This means that any mains conductors must be protected by either two layers of suitable insulation, known as 'basic' and 'supplementary' insulation, or by 'reinforced' insulation. <S> If you can construct your appliance using a power supply unit that qualifies by itself as class II equipment - i.e. it would be safe and legal to use outside the metal enclosure - then I believe (though IANAL) you can also install that PSU inside a metal enclosure to make a piece of equipment that also counts as class II. <S> Of course you can't interfere with the mains wiring to that PSU in any way, so you would have to either route the intact mains lead through a cutout in the case, or have a cutout allowing a mains lead to be plugged in to the PSU's mains inlet. <S> Also of course, you need to follow good practices for strain relief of cables, protecting them from sharp edges etc, and be aware that heat dissipation inside a case will not be as good as in free air. <S> If you are OK with having a class II power supply outside your equipment, this is also a safe option <S> and I think means your equipment then classifies as class III , although I haven't seen this symbol on common devices that do use class II external power supplies so it's possible it only applies when the extra-low-voltage power supply is a fixed installation. <A> EVERY part of the circuit MUST be abundantly ("double") insulated from the outside conductive enclosure. <S> Of course, you have revealed nothing about what this circuit IS, so we don't know whether this is practical. <S> If the circuit has input and/or output connections then this is impossible. <S> You have also not disclosed WHY you are unable to use a proper grounded mains connection? <S> Or whether you can use a GFCI/RCD as alternative safety measure. <A> I don't think you can, legally or safely. <S> The normal way round this is to move the mains-to-DC conversion into a power brick. <S> The other approach is "double insulation", which would require an insulated sub-enclosure inside the aluminium case.
| The common way to make the circuit safe to use in a conductive case is to make the gadget DOUBLE INSULATED . There must be NOTHING that the user can touch that is connected to the circuit in any way.
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Help needed with soldering SPI flash (MX25L8006E) Ok so I want to first remove a SPI flash ( MX25L8006E ) from the pcb and then after modifying its content using Arduino put it back on. How to do this safe? First on I only have a soldiering iron (30W) (of-course I also have tinol and rosin flux). I watched this video but I'm not sure if it's safe judging by the comments. What I'm worrying about is that the temperature range specified on the site for the chip is -40℃ to +85℃ but I've looked up the soldering temperature is way higher (315C). And I don't want to brake the chip. Any advice? Foot note : I'm a total noob at soldering. My only experience with it was making my old PS2 unusable by trying to install some hack-chip on it. So sorry if the question is stupid for you. <Q> All soldering profiles are high temperatures. <S> A typical reflow profile will peak at about 210C (leaded) or 250C <S> (lead free). <S> There is no guidance on hand soldering (the usual guidance is 300C, 10 to 30 seconds). <S> The maximum rating in the datasheet is not the soldering profile; it is the maximum temperature where normal operation is guaranteed . <S> Note that the storage temperature (not in use) is -65C to +150C. <S> There is no guidance in the datasheet for the specifics of the reflow profile either. <S> I searched the manufacturers site for soldering guidance and there was no results. <S> You do not state which package you have <S> (there are 5) as some packages are more amenable to removal with a good chance of not destroying the device than others. <S> So you do not need to worry about the 85C; what you do need to worry about <S> is whether hand soldering is appropriate. <A> I know this isn't the answer you want to hear, but my best advise is to MAKE OTHER PLANS. <S> In any case, count on tossing the IC after desoldering. <A> The answers made by Peter and Scott are valid. <S> Adding to this that you consider yourself a Noob, I strongly advise against desoldering the component as it does imply risks that you will not be able to resolder it. <S> An alternative is that you leave the component on the board, but in order to read/change its contents, you simply solder wires on the pins which you connect to your Arduino. <S> You "just" have to ensure that nothing on your original board is driving the pins. <S> To do so you may need to power just your SPI flash and not the board. <S> You may need to separate the power supply to your SPI flash from the rest of the board. <S> That may be possible by cutting the power trace on the PCB and fix that later or by lifting the power pin of the memory or whatever is possible with the orignal board and your competence. <S> I personnally find it easier to solder a few wires to a component than solder it off and on again.
| ICs are designed to survive at least one SMD soldering cycle, but I simply don't believe that you will be able to unsolder an SMD IC with a 30W hand iron and then expect it to work when you're done. Either rework your system so you can alter the memory in-circuit, flash your brand-spanking new memory off the board before you mount it, or build some sort of adapter board that is easy to remove, maybe with a connector, and program the memory on that.
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Does LiPo4 battery amperage output increase as voltage decreases from 4.2V to 3.57? I'm trying to size up a diode I'm trying to determine the max amp output of a battery pack that has 28 cells in it. To be more specific towards what I'm doing: I'm trying to size up a diode for the charge port; the battery charges at 29.4V max. On the website where I bought the e-bike it says it's a 466Wh battery. So, doing conventional math 466 / 29.4 = 15.85A. But, theoretically if a cell says it's rated at 7.56Wh, and the voltage drops from 4.2v to 3.7v, the battery would have to increase amps to maintain 7.56Wh. In my case my battery may be putting out 15.85A at 29.4V, but when the battery is low at 25.9V does it put out 17.99A? One website recommended using a diode that's twice the rating as the output to reduce the need for a heat sink. In this case I would be looking at a diode rated at 58.8v and 35.98A. Update 10:44AM: I need a diode between the battery harness and battery charger; the charger gives me an error that says it's receiving current from the battery, so it's unwilling to charge the battery pack. It makes sense as it would be better to not have current flowing from the charge port. <Q> Dividing energy [Watt-hours] by voltage [V] does not give amps [I], you have to divide power by voltage to get amps. <S> The power output is V*I with voltage determined by the battery and current output determined by the load. <S> The current capacity of the battery pack is not specified by the energy capacity but you can calculate it if you have the C rating. <S> A 100Whr battery can deliver 100 watts for 1 hour or 10 watts for 10 hours, it depends on the load. <S> As far as needing a diode, it sounds like your charger is not correctly configured for the battery pack. <S> The only way for the battery to source current into the charger is if the charger voltage is lower than the battery voltage. <S> No charging will ever happen if that is true. <A> The maximum current that a battery is designed to safely discharge is known as the C rating. <S> I believe usually a laptop or cell phone battery is considered 1C where rc batteries (more ideal for ebikes) are usually 10C+ with 40 and 50C not being uncommon. <S> Now in your example you divided 466Wh/29.4V = <S> 15.85 <S> A h <S> not A <S> so since the capacity with respect to the C rating is based on Ah the maximum current you should be able to pull from (or push into) <S> the battery safely is a multiple of that. <S> The current that actually gets pulled is up to the way you try to discharge the battery so with great power comes great responsibility and that's up to you. <A> What you're getting confused here is that WattHours is not Volts x Amps. <S> WattHours are synonomous with Joules, with the exception of the scale factor. <S> what's you're looking at is the wattage rating, or the C rating. <S> wattage Watts are simple enoug to calculate, it's simply the V eff X A. Voltage, being measured in Joules/culomb, and amperage, being measured in culombs/second, multiply out to give the units Joules/second, or Watts. <S> C rating the C rating of a battery is usually only used for LI-Po batteries, because of their composition. <S> The basic unit of a 'C' is Joules/(Culomb hours).This seems simmilar to voltage, with the exception of hours being on the bottom. <S> Following good ol' PEMDAS, we find that we multiply this by (Culomb hours/second), which is also known as AmpHours. <S> For example, if a battery is rated for 20C, aand can hold 5Ah, then the maximum discharge rate is 100A. To answer the first question, No, the amperage does not increas as the voltage drops <S> , it decreases, as shown here:
| The voltage will decrease as the battery discharges and for a fixed resistive load the current will also decrease according to Ohm's law.
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How do I compute the average and RMS of a series of triangular pulses? I'm trying to calculate the Average and RMS values of the current through a diode with the waveform image attached. I'm struggling to work out which equations to use. Any feedback would be appreciated. <Q> Average current is the sum of the current over a cycle, divided by the length of the cycle <S> Your current is a triangle with \$I_{peak}=1.8\$ amps, and a base of about \$t_{on}=.9\$ divisions, so you get an area of \$\frac{1}{2}BH=.81\$ (some unit of charge, since it's current x time). <S> You then spread that out over \$T=3.7\$ divisions (charge/time = <S> current again), which gives an average of .22 amps. <S> Alternately, if you work the calculus to get a general form for the average current of a series of triangular pulses: \$I(t)=I_{peak}(1-\frac{t}{t_{on}})\$ between \$t=0\$ and \$t=t_{on}\$, and zero elsewhere. <S> \$I_{AVG}=\frac{\int_{0}^{T}{I(t) <S> dt}}{T}\$ by definition of mean. <S> Substituting in I(t) from above: \$I_{AVG}=\frac{\int_{0}^{T}{I_{peak}(1-\frac{t}{t_{on}}) dt}}{T}\$. Rearranging and executing integral: <S> \$I_{AVG}=\frac{I_{peak}}{T}(t-\frac{t^2}{2t_{on}})]^{t_{on}}_{0}\$ \$I_{AVG} = <S> \frac{I_{peak}D}{2}\$ <S> Where \$D=\frac{t_{on}}{T}\$. <S> This gives us the same answer as above, for your duty cycle of about .25. <S> RMS current is a little more complicated. <S> \$I_{RMS}=\sqrt{\frac{\int_{0}^{T}{I(t)^2 <S> dt}}{T}}\$ by definition of RMS. <S> Substituting in I(t) from above: \$I_{RMS}=\sqrt{\frac{\int_{0}^{t_{on}}{I_{peak}^2(1-\frac{t}{t_{on}})^2 <S> dt}}{T}}\$ <S> Rearranging: \$I_{RMS}=\frac{I_{peak}}{\sqrt{T}}\sqrt{\int_{0}^{t_{on}}{(1-\frac{t}{t_{on}})^2 dt}}\$ \$I_{RMS}=\frac{I_{peak}}{\sqrt{T}}\sqrt{\int_{0}^{t_{on}}{(1-\frac{2t}{t_{on}} + \frac{t^2}{t_{on}^2}) dt}}\$ \$I_{RMS}=\frac{I_{peak}}{\sqrt{T}}\sqrt{(t-\frac{t^2}{t_{on}} <S> + \frac{t^3}{3t_{on}^2})]^{t_{on}}_{0}}\$ <S> \$I_{RMS}=\frac{I_{peak}}{\sqrt{T}}\sqrt{\frac{t_{on}}{3}}\$ <S> \$I_{RMS} = <S> \frac{I_{peak}\sqrt{D}}{\sqrt{3}}\$ <S> So your RMS is about .52 amps. <A> The basic principle of taking an average (in this case the artihmetic mean) is: <S> Take every point of a single period of your waveform <S> Add together all those values Divide by the number of values <S> \$ average = <S> \frac{1}{n}(x_1 <S> + x_2 + x_3 <S> + ... <S> x_n) <S> \$ <S> The basic principle of RMS is: <S> Take every point of a single period of your waveform Square that value <S> Add together all those values Divide by the number of values <S> Take the square root <S> \$ rms = <S> \sqrt{\frac{1}{n}(x_1^2 <S> + x_2^ <S> 2 <S> + <S> x_3^2 <S> + <S> ... x_n^2)} \$ <A> Your triangle wave current is about 1.7Amp peak at a 25% duty cycle. <S> The RMS current of a continuous triangle wave is:Irms(continuous) <S> = 1/sqrt3 <S> x Ipeak. <S> This is multiplied by the sqrt of the duty: <S> Irms = <S> Irms(cont) <S> x sqrt(duty <S> ) = 1/sqrt3 <S> * sqrt(duty <S> ) * Ipeak.so, in your case Irms = <S> 0.577 * sqrt(0.25) * 1.7Amp = 491mArms. <S> I won't help you do simple algebra. <S> If you're an engineer you should know this.
| The average current you can easily calculate from the waveform if you estimate it to be a triangle wave.
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What do you call a cable that has string along with its conductors? I'm trying to replace the cable that connects the magic wand in a VTech Magic Wand toy laptop. It's an interesting beast. Inside its outer insulator is a bunch of strands of string plus the two conductors. Each of the two conductors is a braided bundle of enameled wires around a string core. It's clear to me that this was designed this way so that it wouldn't break under tension or repeated flexing as a kid uses the magic wand to point-to-select things on the toy's LCD screen. But what do you call such a bundle? I'd like to replace it with something similar to the original cable, but I'm having trouble naming the product that I'm searching for. <Q> You frequently see microscopic conductive "ribbons" (vs. round wires) wound around non-conductive fiber cores. <S> Very common in headphones, etc. <S> Tinsel wire is produced by wrapping several strands of thin metal foil around a flexible nylon or textile core. <S> Because the foil is very thin, the bend radius imposed on the foil is much greater than the thickness of the foil, leading to a low probability of metal fatigue. <S> Meanwhile, the core provides high tensile strength without impairing flexibility. <S> https://en.wikipedia.org/wiki/Tinsel_wire <A> According to this presentation http://slideplayer.com/slide/5851896/ <S> The cord can have one of a few related purposes. <S> Structural Filler: <S> Inert extra materials added to help the cable maintain the proper shape. <S> Strength member: <S> An element mainly used to maintain cable rigidity. <S> Aramid Yarn or Kevlar ® : <S> Used for strength <S> (longitudinal or “pull”). <S> Dupont Kevlar ® 39 Definitions: <S> Structural Gel Tape: A tissue-like material that expands into a gel-like substance when exposed to water. <S> Rip Cord or string: <S> A string placed just inside a cable jacket, used to cut the jacket material away. <S> Usually make of Dupont Kevlar ® 40 Filler material being to bulk up a cable to a standard size due to missing inner conductors (cord is cheaper than copper). <S> Not all jackets are shrunk to any arbitrary size, and manufacturing process may be easier to use one standard size with filler than changing to a smaller jacket. <S> Strength to make the cable last longer when constantly flexed. <S> Rip cord to make the outer jacket easier to strip or pull back. <S> It also allows a long cable to be pulled, like for large wiring runs. <A> Sounds like fiber core Litz wire . <S> Multiple insulated strands are used to reduce the skin effect at high frequencies. <S> Types 4, 5, and 6 at <S> this link use fiber cores (where a standard 7-wire bundle has the central wire replaced with a fiber core, and in the case of the "type 6" the 6 outer wire bundles also have their central member replaced with a fiber core.
| That is commonly called "tinsel wire" and the technique is used to make super-flexible cables that will survive orders of magnitude more flexing than ordinary wire or cable. Common in bulk networking cable.
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What is *actually* a *standard part*? This is something that has bothered me for ages. I wasn't sure it could be made up as a question here, until Olin Lathrop suggested it to me in a comment to another thread . We all know what we are talking about when someone mentions, for example, an 1N4007 or 1N4148 diode. The first is a general purpose rectifier with 1A max forward current and 1000V max reverse voltage, just to be explicit. And this information is completely independent from the manufacturer of the specimen we have at hand (I'll bet anyone with a bit of experience in EE would go crazy if he saw, say, a DIAC labeled as 1N4007 produced by a reputable manufacturer). So we know there are "standard" part numbers which corresponds to devices with well-known characteristics. As Olin pointed out in that answer to the question in the thread I mentioned above, although we all know what a standard part is, sometimes different manufacturers give slightly different specs for the same part. Still, an 1N4007 is an 1N4007, this little differences notwithstanding! So my question is, what defines what a standard part actually is? Historical reasons (the first manufacturer got the part number; second sourcing the part to others made that part a de-facto standard)? Industrial agreements (some manufacturers decided it would be good not to do cutthroat competition on low-tech parts, so they decided to standardize)? Official standardization organizations (Olin mentioned JEDEC in that comment; may it be some other institution as well)? To really gild the lily the answer should also provide references to possible official documentation about common parts or procedures (if they exist) to standardize a part. <Q> After some hunting, this organization claims to be able to supply a print or PDF copy: https://global.ihs.com/doc_detail.cfm?&input_search_filter=JEDEC&item_s_key=00207654&item_key_date=290303&input_doc_number=1N4001&input_doc_title=&org_code=JEDEC <S> Most manufacturers parts are defined better by the manufacturer (finish, lead length, marking, packaging) so it is rare that a designer will refer to the original standard rather than the manufacturer's current data sheet. <S> A requirement for lead-free finish on the wires was NOT part of the old 1N4007 spec, but can be very important nowadays. <S> So, too, can feed belt packaging versus box-of-loose-parts. <A> Historically, standards were used by the US military to set the requirements of various parts but vendors are free to implement a part in any way they choose provide the required performance is achieved. <S> Note that the overall performance can vary widely; the key is "does it do what this test requires"? <S> There is a specification finder for 'standard' microcircuits, but beware: these specifications have holes large enough to drive a very large truck through. <S> I have been bitten by this when trying to source an ADC to replace a very obsolete part and it did not operate the same way in my application as the original part <S> (an X-ray showed the die had different dimensions). <S> Most microcircuits are defined within MIL-STD-1562 <S> (warning - paid for at this link). <S> All the specifications are here . <S> To put these numbers on a part, vendors need to apply the specific tests required. <S> Capacitors can be to many specifications such as CECC30801-x and MIL-PRF-55365 <S> (it is a long list). <S> Likewise, resistors can be to a number of standards such as CECC40401-x or perhaps to MIL-PRF-55342-x <S> Industry standard parts are more correctly known as popular parts that happen to be in very wide use; a few decades ago, the 'standard' general purpose op-amp was the venerable 741, but it would an unlikely candidate for a new design. <S> Fortunately, many op-amp footprints are now compatible across many parts (particularly in SOIC for single, duals and quads for ordinary parts - no special features such as shutdown and so on). <S> Likewise, we have standard footprints as answered at this question <S> and although there is little enforcement, it is in the interest of vendors to follow this guidance. <A> There are at least two types of standard, with one or two slightly grey levels in between as a part slowly gets promoted from the less to the more official. <S> There are many official industry standards, documents produced by organisations like JEDEC, IEEE, 3GPP etc, that define the parts. <S> Then there are de facto standards. <S> These are neither officially documented, nor created for the purpose, but just happen, because I don't need to use an extra parts bin for half-amp diodes if 1N4007s are cheap enough, and enough other people think the same way. <S> What is a 'standard' changes as time marches on. <S> 30 years ago, leaded E12 resistors, and the BC109 were 'standard', whereas today surface mount E24 resistors and perhaps the ULN2803 are. <S> It also varies strongly by engineering sector. <S> What's the most 'standard' component? <S> I'd vote the 10k 1% resistor ;-)
| The "1N4007" and "1N4148" part numbers are JEDEC standards, and presumably there is a standard document available from http://www.jedec.org .
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How to filter noise from fan I try to make controller for 12V fan, where I can vary voltage from 3V to 12V on this fan. It works OK, but it makes electrical noise on 12V power line.How could I get rid of it? How to filter it, or how to modify my circuit? this is the noise: UPDATE: Fan draws 220mA @ 12V, 60mA @ 3V.This is the modified circuit, according to answers. Noise is lowered down to ~20mV peak-peak:Added 100nF across pins 1 and 2 of opamp, 100nF cap across power pins of opamp, 2200uF cap from 12V to ground. UPDATE2: I put 3k resistor from Q1 emitter to ground and removed capacitor that was directly on fan connector (there is just 2200uF on 12V line). This lowered noise once again. <Q> Nothing dramatic going on here. <S> You just need to install some basic bypass capacitors on the power supply feed and the op-amp feedback path. <S> Insert a .1uF capacitor from pin 1 to pin 2 of the op-amp. <S> This will stabilize the op-amp <S> so it is not prone to oscillation burst. <S> Replace C1 with a .1uF capacitor and move the 100uF capacitor (+) lead to the 12 volt power supply, as close to the mosfet as possible. <S> Take readings again and the servo-control loop should be stable with little or no noise on the 12 volt line. <S> EDIT 1: <S> OP made final(?) changes which included a .1uF across the op-amp pins 1 and 2, 100uF across the diode/motor, and adding a 2,200uF capacitor across the 12 volt to ground power feed. <S> The result is a noise drop from 3V p-p on the 12 volt line to 18mV p-p. <S> This is an improvement of 166 times lower, or 44.436 dB lower, as a voltage ratio. <S> EDIT 2: <S> OP removed C1 and installed a 3K resistor in the emitter of Q1, leaving other parts the same. <S> Reports lower noise level but did not give values. <A> I have tried all advices and hints, and experimented. <S> This is my final circuit: And noise at 12V line looks like this <S> : it is ~12mV peak-peak <S> I managed to reduce this noise further more only by increasing size of C3. <A> Couple of things here: <S> 1) <S> The output stage following the op-amp has significant voltage gain. <S> This may be causing instability. <S> Try adding a 1k or so resistor in the emitter of Q1. <S> If this improves things but doesn't completely get rid of the noise, increase the value of the emitter resistor. <S> The value of the emitter resistor has to be low enough to ensure that Q2 will turn on fully - you will need to check the datasheet for the MOSFET to find out what the maximum G-S voltage is to ensure full enhancement. <S> Assuming that Q1 saturates, you can figure out the maximum allowable value of the emitter resistor to guarantee that Q2 can be fully enhanced. <S> 2) <S> This may or may not be related - <S> but I often have to add a LC filter network in the power leads of a brushless DC fan if I am having noise issues elsewhere in the circuit. <S> The noise shows up in audio circuits as a whine, where the pitch of the whine is directly proportional to the speed of the fan. <S> The quick check for this is to simply use your finger on the center hub of the fan blade to slow it down while listening to the whine. <S> If the pitch changes, the whine is being caused by the fan. <S> A typical LC filter is two simple components: an inductor with a high-enough current rating so as to not saturate and a capacitor. <S> Inductor values that I use range from 330uH all the way up to about 1200uH (1.2mH). <S> Capacitor is a Sanyo AX or GX at 100uF, <S> 35V. The Sanyo AX and GX capacitors have extremely low ESR and work cold <S> and we just keep lots of them around. <A> You will need to add bypass caps across connectors pins 1,2 for higher freqs. <S> Perhaps a 100nF, and a 1nF. <S> They should be as close as possible to the connector's pins 1,2. <S> That deals with the conducted noise. <S> You will also undoubtedly have radiated noise. <S> The wires of pins 1 and 2 need to be twisted together to get rid of the radiated noise. <S> If you don't eliminate the radiated noise it will show up everywhere.
| The high freq noise will be on the cable from the connector to the fan. To prevent odd behaviour install a .1uF capacitor directly across the power supply pins of the op-amp.
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How to identify pinout of 7 segment display when data for it is not available online? I found some 1-character 7-segment LED display in my toolbox with unusual pinouts. There is one however I can't seem to figure out and I even googled the part number with no results. The part number printed on the side of it is "MAN4840A350G82", and my camera was good enough for me to show you pictures of how the pins are arranged on each side of the display. There are no other labels. If I can't find the pinout on google, should I just assume some generic pinout? I want to be able to use this display in a schematic created with eagle (yes I'm using version 4.17) but I don't want to pick the wrong LED display. How do I identify what each pin represents on this display? <Q> A diode tester or a coin cell battery or a bigger battery with an appropriate resistor (1k or so). <S> Map out every pin combination by hand. <S> Since it's not a multiplexed display, it should be one cathode or anode to multiple of the other. <A> Apply power through pairs of pins until a segment lights up. <S> Now just move one probe at a time until another segment lights. <S> The pin in common between the first segment and the second is the common anode or common cathode. <S> Then simply test all the pins (with the appropriate probe on the common connection) until you know what segments are connected to what pins. <A> They tend to be rather standard. <S> Find something with the same number of pins in the same places on the same-size package and it's likely to be the same pinout, even if it's a different brand. <S> An appropriately limited low voltage test supply (5V, 300 Ohms,for example) and poking at pins. <S> Depending where it came from (ie, if it's a takeout) it could be partly dead or all-dead. <A> I found this from the MANXXX part number, which is obsolete and a current cross-reference from Liton. <S> Though if it was not a standard part, probing with a 1K resistor in series with a 5V supply is fast and easy and will verify functionality. <S> Your camera is impressively bad- what the kids call a 'potato'. <A> I have a similar display as your picture. <S> I use a digital volt meter to get the pinout. <S> On the side with the 2 pins at each end, the middle pin is the GND.Hold <S> the negative probe of the digital volt meter on that pin, use the red probe + to touch the other pins and you will see that each segment light up when you go from pin to pin. <S> There is another GND pin on the side with 3 pins and 2 pins at the other end. <S> The side that has the 3 pins the third pin counting from the end towards the middle is the other GND.
| You can find the pinout here . You should be able to tell if it's the common anode or common cathode. Also not uncommon to have a pattern which includes features (decimal point or sign) that don't light up on some models in the line.
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Questions regarding using an HDMI cable to transmit SPI, I2C, and UART serial data? I'm currently working on a design in which I need to transmit serial data over 3 different serial channels. I won't be transmitting data far (1" - 1') but I didn't want to use plain wires. Instead, I thought it may be useful to implement the standard HDMI connector and cable to transmit this data. However, I have a few questions. 1) Are HDMI cables designed for only transmitting data on specific conductors? It appears that some of conductors are 'paired'. I assume I would just need to be cautious on what conductors I place the signals? My serial data will be transmitted much slower than most audio/video data streaming that HDMI is typically used for. 2) I'm going for a compact design, I've thus considered using micro HDMI for the small form factor. Besides the form factor, are there any other differences between micro and standard HDMI connectors/cables? 3) Where are we at with regard to the life time of HDMI cables? I'm trying to create a device that will last a long time. Is HDMI/Mini HDMI/Micro HDMI on there way out any time soon? I know they've been around for close to a decade now, are there any other rising standards that I should be considering? Thanks! <Q> Yes, they are. <S> Data goes over the data pairs. <S> HDMI uses high speed differential pairs for data. <S> No difference. <S> Not really. <S> No replacement for HDMI right now. <S> But if you are using this commercially, note that using a standard connector for non standard use is fairly bad idea. <A> It is totally possible, if inadvisable (as mentioned in the question comments, you could damage an HDMI device if you have the two connected) to use an HDMI cable, including a microHDMI cable, to transmit your serial data. <S> However, some considerations: <S> The HDMI specification already has an I2C bus (pins 15 and 16) that you can use. <S> If you use each wire of a differential pair separately as a data line, you'll get a lot of cross-talk because they have very high capacitance. <S> You could get around this by sending, for example, MOSI on both Data1 <S> + and Data1-. <S> However, this reduces the number of conductors you have available. <S> However, there are also the CEC, Pin 14 (reserved/HEAC+/Utility), and Hot Plug Detect pins, which you could use if you're not worried about accidentally plugging it into a real HDMI port. <S> Shields are not meant for conducting signal and should be dealt with as shields. <S> Given that, you do still have enough lines. <S> If you use GND and +5v properly, you still have 7 misused data lines available plus SCL/SDA. <S> To answer your earlier questions: <S> Yes <S> and yes. <S> See above, the main issue is the capacitative cross-talk if you put different signals on the + and - lines. <S> The pinout looks identical, and the cables should be exactly the same. <S> They're meant to be different form factors for the same interface. <S> You might want to consider the rise of video over USB-C. <S> This will likely make microHDMI not catch on as well, but I don't think that it will limit the lifetime of your device. <A> I happen to be debugging a HDMI interface and have noticed that the signal differential pairs are on 3V common mode. <S> If your serial channels run at 3.3V or lower voltage, I don't see why it would damage the sink (Television) or the source (PC). <S> When you place one signal to both the + and -, there would be no differential voltage on the pair. <S> Given the above constrains, I'd say using HDMI cable seems to be a great idea. <S> Though just my opinion. <A> What i gather: You want to transmit serial data at low speeds <S> Commercial ready made cables should be long-term available (to get replacement parts?) <S> Connector should be small <S> Why not use USB-C (over HDMI for reasons mentioned in other answers here) <S> - if i read http://www.ti.com/lit/wp/slly021/slly021.pdf correctly, then it should be possible to build your device in a way that is does not damage connected USB-C devices. <S> You could use it close to the intended use (since it has enough lines which are TX lines to support your purpose).
| Although the cable is designed with most of its conductors as twisted pairs (Data1 - Data3 and clock in the HDMI pinout ) for differential signaling , you could still use them for transmission of data.
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Measuring Current in Combination Circuit with a Shunt I've done a fair bit of looking/asking around for help with this but I'm struggling to come up with an answer .... I'd like to measure and record current in a circuit. There is a 12 VDC / 2 A power supply which provides power to two instruments. They draw 75 mA, and 640 mA at 12 V respectively. They're in parallel with the supply, like this: Instrument 2 is a water pump which operates at a nominal rate of 3 L/min. However, if it gets clogged for some reason, it draws more current in order to try maintain that 3 L/min output. SO, I would like to be able to measure and log the current supply to these devices. From my research, it appears as though a shunt is a possible option. I would place it just before the negative lead on the power supply, and record the voltage drop across it using an external DAQ device (which measures from -10 to 10 V), much like this: I guess my questions are: 1) Is this the best way to solve this problem? If not, what would be a better alternative?2) If it is a good option (or even if it isn't, maybe humour me a bit ....) what "type" of shunt should I get? Is this the right way to wire the shunt. Unfortunately, the instruments themselves cannot be re-configured. I would be happy to provide more detail as required. <Q> There are very convenient modern IC solutions for measuring high-side current. <S> For example consider the TI INA169 60-V, High-Side, High-Speed, Current Output Current Shunt Monitor . <S> You can even get a breakout-board ready to use from vendors like SparkFun, et.al. <S> Unless the entire circuit is within your control (no external connections). <A> You are on the right track but generally we try not to disturb or "load" the circuit too much when taking measurements. <S> That usually means that when measuring voltages we use a high input resistance meter and when measuring currents (as in this case) we use low resistance shunts. <S> As you've indicated on your drawing, the shunt could cause a drop of 10 V leaving only 2 V for the motor! <S> In practice the shunt would cause so much voltage drop that the motor current would drop too and the 2 V situation would not be reached. <S> To meet our criteria of not disturbing the circuit too much we might decide that we could tolerate a 0.2 V drop at 1 A. <S> From that we could calculate the shunt resistance as \$ R = \frac {V}{I} = \frac {0.2}{1} = 0.2~\Omega <S> \$. <S> That's the shunt sorted out. <S> Now we have the problem that 0.2 V isn't going to give us good resolution on our 0 to 10 V DAC <S> so an amplifier is required. <S> We can calculate the gain, \$ G = \frac {V_{OUT}}{V_{IN}} = \frac {10}{0.2} = <S> 50 \$. <S> This would usually be done using an opamp in non-inverting amplifier mode. <S> You could put a shunt into the leg of each load to measure both independently. <S> As an alternative you could investigate Hall effect current sensors. <S> Figure 1. <S> The Hall effect generates a voltage across a current carrying conductor exposed to a magnetic field. <S> These sensors run a small, known current through a portion of the chip which is placed in the magnetic field around the current carrying conductor. <S> The small voltage induced across the conductor is amplified to give a signal suitable for direct interface with a micro ADC circuit. <S> Figure 2. <S> To enhance the magnetic field a toroidal core is usually employed and the Hall effect sensor inserted in a small gap. <S> One or more turns of the current-carrying conductor are passed through the core to obtain the required ampere-turns ratio to optimise the resolution of the complete circuit. <A> Your drawing is plenty good enough, and you do not need a shunt for such low currents. <S> A precision resistor of .1 ohm .1 <S> % 5 watts maybe cheaper than a shunt. <S> Precision shunts can be expensive, while .1% tolerance resistors are becoming common. <S> 1 amp flowing through a .1 ohm resistor = <S> 100mV. <S> Also, you do not need a bipolar DAC if your not reading a negative current flow. <S> A low cost 12 bit DAC will give you 3-1/2 digits of resolution.
| You may consider a digital panel meter by Modutec for a direct reading and use the DAC with a Raspberry Pi board as a data logger, though I do believe some versions of Raspberry Pi have analog input options. It is generally not desirable to put a shunt in the ground-return side because of possible interactions with external ground connections.
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Get -50V to 50V range from battery output I have the following problem: on the system which is battery-powered(4S1P Li-ion battery, 14.8V) I need to create outputs from -50V to 50V. The system has microcontroller on-board. I am baffled on how to do it. Initially, I planned on using bipolar DAC which would give output of +/- 5V or so and then amplify it with gain of 10 to get total +/- 50V. The problem is, how to get power supply with range from -50V to +50V from the 4S Li-ion battery? I am familiar with concept of having two 50V sources, then connecting the + of one source to the - of another and using that point as common ground. Total current needed at +/-50V is 300mA. Loads are resistive. Am I allowed to connect in this topology using two boost converters(switchers)? Would it work? Thanks;D. Petric <Q> If you want to do this with a single SMPS circuit, you need some kind of transformer or capacitive coupling. <S> You can do this with SEPIC topology <S> http://www.linear.com/solutions/7345 <S> TI has app note with transformer and capacitive coupling options for boost topology <S> http://www.ti.com.cn/cn/lit/an/slua288/slua288.pdf <S> Buck can also be used for this <S> but obviously it does not apply here. <S> If you want a straightforward solution, use boost for +50V and buck-boost for -50V. <S> You can often use buck converter as buck-boost, texas has app notes for a few different solutions. <S> As a bonus this way both outputs are well regulated, dual output can have problems with cross-regulation. <S> TI appnote shows you the efficiency and regulation suffers when you start draining more than 100mA. <A> <A> Am I allowed to connect in this topology using two boost converters(switchers)? <S> Would it work? <S> No, not if you are talking about the classic boost converter in this picture: <S> The problem is that the negative terminal of the boosted voltage is the same as the negative terminal of the power supply (battery pack). <S> If connect a second one of these to the same battery and then connect its positive terminal to the negative terminal of the first one, you short-circuit the second one. <S> It's not going to like that. <S> You can use a normal boost converter for the +50V, but for the -50V you will need what is called an inverting switching regulator.
| A switch-mode power supply (SMPS) is typically the most efficient and straightforward way of converting a battery source into bipolar supply rails.
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Can I measure voltage and current at the same time by using two multimeters? I think the title is revealing. I'm planning to use 2 multi-meters to get a better perspective of my circuits. Is there any drawback in this situation i should be aware? <Q> Yes, that works. <S> However, keep in mind that no meter is perfect. <S> Ideally, voltmeters have infinite impedance and current meters 0 <S> impedance. <S> Real ones don't, of course. <S> To the extent the meters aren't perfect, particulary the current meter, it will affect the system while trying to measure it. <S> The current meter will cause some voltage drop proportional to the current. <S> You have to be aware of this and possibly account for it. <S> For example, if the voltage drop across the current meter is too big to just ignore <A> Yes, you can. <S> No real draw back unless you get to very small fractions of a volt or micro amp ranges and your precision requirements are tight. <S> Or very high voltages and currents that your multimeter wasn't designed to handle well. <A> Bear in mind we're assuming digital voltmeters here, an analog meter will have an ohms per volt <S> characteristic (google is your friend if you don't understand this). <S> Digital meters usually have a fixed impedance but this value will dependly largely on the quality (and price) of the meter. <S> I agree about probe fuse resistances being high, stick the meter onto ohms and short the probes, <S> on mine it's about 8 ohms. <A> But calibration matters before you taking the reading. <S> Do it couple of times. <S> So you can stick to the precise value
| , you have to consider carefully on what side of the current meter you measure the voltage, and whether your system can tolerate the voltage drop in the first place. I don't think it is a big deal, you can use.
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Is each individual CPU calibrated before being shipped? The quartz movement in battery-powered watches is plated with gold and trimmed individually using laser to achieve a vibration frequency of exactly 32,768 Hz to keep time accurately. The crankshaft in pistol engine is adjusted by hand or computer to make sure the center of gravity is in the middle of the shaft for smooth rotation. The 50 ohms resistors on an impedance standard substrate for high frequency measurements are laser trimmed as well for ultimate accuracy. Therefore, does each individual CPU get calibrated by any means before being shipped? I would imagine the answer to be yes as CPUs are intricate products that should require some tweaking. But how is CPU calibration done on a hardware level? <Q> In a CPU per se, no part requires calibration. <S> Pure digital logic does not need it. <S> But this depends on the microcontroller capabilities. <S> If calibration is required, it is done directly on the wafer during production. <S> Once the chips are made, some probes measure the elements needing calibration (e.g specific resistors, RC network frequency, reference voltage, ...) and either adjust their value using laser trimming, or, in some cases, just store the value read in some part of EEPROM, for later use by the firmware. <A> MCUs that have internal RC oscillators are typically calibrated by storing a number in non-volatile memory that trims the oscillator frequency. <S> That number might be used directly, but more often it is loaded into a register to trim the hardware oscillator. <S> As a rough example, the total range (hardware limited) might be something like 25% and the center point (and range) chosen so that virtually all processors can be tuned to close to the desired frequency. <S> Some processors might be toward one or the other extreme depending on manufacturing tolerances. <S> Similar methods exist for calibration voltage sources etc. <S> , not always with visible (to the programmer) methods. <A> There is typically nothing inside a CPU chip that requires any kind of "calibration". <S> CPU chips are certainly tested to confirm they operate properly. <S> And many are "graded" for speed and/or other operational parameters.
| In a microcontroller (combination of CPU+periperals in a single chip), the only parts that would need calibration would be the internal oscillators, and eventually a few specific elements for particular peripherals (internal references for ADC/DAC, maybe some internal termination resistors for USB, ...). The simple and basic answer is no. The calibration is done in testing.
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Can a NOT gate be used to achieve 180 degree phase shift? I have seen from various sources which say that a NOT gate cannot be used to achieve an 180-degree phase shift. Is this claim true? Edit: The question is definitely sounding unclear because that is how it was worded, but one thing I missed was it was asked in the context related to FPGA's. So I assume we are dealing with digital signals here. Anyways giving a sine wave as an input to the NOT gate would change the shape of the output itself so I guess there would be no question of comparing/calculating the phase difference. The solution given was to use a DCM in an FPGA and it was explicitly mentioned not to use a NOT gate, but I don't understand why it won't work because inverting a wave is definitely an 180-degree phase change right? Please correct me if I'm wrong! <Q> What is 180 degrees phase shift? <S> When the signal is a sine wave , a 180 degrees phase shift delays the signal for half the period of that sine wave, the sine wave then looks inverted: Can an inverter do this? <S> No , because it has signal gain, the output would be a square wave, not a sine. <S> When the signal is a square wave with a 50% duty cycle , then something similar happens as with the sine wave: <S> Can an inverter do this? <S> Yes <S> But now let's look at a square wave with a 25% duty cycle ,and see what happens when I would <S> NOT that signal: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> So you see that it is indeed possible to use a NOT gate (inverter) to 180 degree phase shift a signal, but that only works on a square wave with a 50% duty cycle . <S> I would not use the term "180 degrees phase shift" in the context of digital signals and NOT gates. <S> I would call that inverting a (digital) signal. <S> Calling this 180 degrees phase shift is confusing and wrong in my opinion because it is not a phase shift, but an inversion. <A> The claim is misleading. <S> An inverter deals with digital signals and digital signals contain frequency harmonics that meander off to infinity (theoretically). <S> Take a square wave for example: - The one above is very approximate showing only the fundamental, third, fifth and seventh harmonics. <S> If you inverted this square wave you could certainly say the fundamental is phase shifted 180 degrees. <S> Then, if you (say) took the 3rd harmonic in isolation, and looked how that phase shifted it would also be phase shifted by 180 degrees but, in relation to the fundamental that phase shift is only 60 degrees. <S> The claim is correct in some respects but it's misleading in others. <A> <A> Since no answer so far (and one has a score above 20!) <S> actually mentions FPGAs, I'll add the answer <S> I think you are looking for. <S> You can't invert a clock on an FPGA with logic for various reasons: <S> Skew . <S> The inversion is not instantaneous, so there would be a transitional time with both clocks matching. <S> You probably don't want this. <S> Architecture . <S> FPGA's logic elements have dedicated clock input lines that are independent of the data lines. <S> To route from data to clock nets involves connections that are suboptimal. <S> If you are trying to get an inverted clock inside the IC, you'll have to use one of the architecture's native clock modules. <S> If you are trying to drive an output with an inverted clock, the major vendors all have DDR logic that you can use to send the "data" of a continuous 010101 at "double data rate" which is actually the clock inverted. <S> The dedicated circuits will handle any corrections needed. <S> This lets you avoid using global clock resources for the inverted clock and also gives you a way to invert the clock in the field with a control register. <A> A practical NOT-gate will invert the input, but after some finite delay. <S> Your digital signal from one positive edge to another is 360 degrees, the phase shift provided by the NOT-gate will be more than 180 degrees.
| A not gate has too much gain to provide a clean 180 degree phase shift but with the right amount of negative feedback some not gates can do it.
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HF transformer fed fed diode rectifier not working As part of my pure sine wave inverter, I have a DC - DC subsystem in which asignal stepped up through a HF transformer is put through an uncontrolled dioderectifier to get a DC output which I am o feed into my PWM inverter (not shownin circuit). I am using a 12V to -12V square wave as input to the transformerand getting the required 475 to -475 V wave output, but the moment I connect adiode rectifier to it, the output of my transformer sinks to a range of 4 to -4V on the secondary. I am using a threshold voltage of 1.7 V and a dioderesistance of 0.57 ohms. Rpri = 2.48 mOhms Rsec = 0.411 Ohms Lpri = 1.909 uH Lsec = 2.98 mH Lm = 27.51 uH Npri = 2 Nsec = 79 SWG 19 wire (1mm diameter) used on primary coils SWG 24 (0.5 mm diameter) for secondary MOSFET ON STATE resistance = 0.015 Ohms <Q> Well, obviously, the HV supply you've built can't source enough current! <S> Hence, the voltage drops as soon as you connect a load. <S> I don't know the specifications (winding thickness, winding count, core material, transformer topology) of your transformer, but that's where I'd start looking. <S> Also, I don't know the effective resistance of your MOSFETs in ON state; you might also consider the fact that these device are frequency dependent, too! <A> So, you're putting +/- <S> 475 volts into a bridge rectifier, followed by a 596 ohm resistor. <S> In a perfect world, you'll get about 950 volts out of the rectifer, and then your trouble starts. <S> 596 ohms? <S> Really? <S> (And where in the world did that extra .07 <S> ohms come from, anyways? <S> Are you truly measuring resistance to 5 significant figures?) <S> Apply Ohm's Law. <S> 950 / 596 is 1.6 amps! <S> 1.6 amps times 950 volts is 1500 watts! <S> Have you really got a 1.5 kW transformer, and a 1.5 kW resistor? <S> You need to rethink your design. <S> From your trace, it's apparent that you're driving this at about 60 kHz, and you should be aware the designing a transformer for these currents and frequencies is not simple. <S> And it's pretty clear that you've done something wrong. <S> Oh yes, <S> and you have not specified the diodes in your rectifier. <S> You're not using 1N400x types, I hope. <A> So it's just a simulation isn't it? <S> All your secondary circuit, bridge, filter, load is floating, i.e. it has no reference to 0V node. <S> Most simulators are not happy with this. <S> So I didn't read problem's data carefully enough <S> and I apologize for this. <S> Switches: <S> a 500W system run from 12V would take over 40A. Just rds(on) and primary windings resistance score up to 18mohm: this alone drops nearly 1V from 12V and dissipates over 30W statically only. <S> Then add switching lossesMoving up to 24V or 48V would be a very good choice unless really impossible. <S> In any case such an inverter will take several MOS for each switch and for sure transformer primary winding will not be made of 1mm copper wire. <S> Wound copper stabs or several parallel windings are mandatory. <S> Talking of transformer: so far it is no clear to me wether <S> we are talking of measured or simulated data but inductances tell that coupling between primary and secondary is very loose. <S> With Lpri=3mH, Lsec=1.9uH and perfect k=1 coupling Lm should be around 75uH. <S> If you have 27uH it means k=0.35 or so and hence around 2.6mH of series inductance. <S> Around 1kohm reactance at 60kHz. <S> First you need a much better transformer, but IMHO inverter design is also a far cry from a good working circuit. <S> I swapped Lpri and Lsec <S> but it's just the same <S> , they are very loosely coupled , they cannot transfer much energy each other. <S> How have you designed it? <S> What is the core size and shape? <S> Where have you got those inductances? <S> Simulators are toghether the most usefull and the most dangerous to any field of engineering. <S> I am afraid that used with no background experience, sensibilty to results and experimental work will not give any good results. <S> By the way designing SMPSU transformer is one of the most demanding tasks. <S> It is made of educated guesses, trials and errors back and forth electric, magnetic and thermal design at the same time. <A> So I didn't read problem's data carefully enough and I apologize for this. <S> The matter is much simpler: it cannot work. <S> Switches: <S> a 500W system run from 12V would take over 40A. Just rds(on) and primary windings resistance score up to 18mohm: this alone drops nearly 1V from 12V and dissipates over 30W statically only. <S> Then add switching lossesMoving up to 24V or 48V would be a very good choice unless really impossible. <S> In any case such an inverter will take several MOS for each switch and for sure transformer primary winding will not be made of 1mm copper wire. <S> Wound copper stabs or several parallel windings are mandatory. <S> Talking of transformer: so far it is no clear to me wether <S> we are talking of measured or simulated data but inductances tell that coupling between primary and secondary is very loose. <S> With Lpri=3mH, Lsec=1.9uH and perfect k=1 coupling Lm should be around 75uH. <S> If you have 27uH it means k=0.35 or so and hence around 2.6mH of series inductance. <S> Around 1kohm reactance at 60kHz.
| The matter is much simpler: it cannot work.
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Why would a time signal receiving watch have an airplane mode? I've been looking at radio controlled watches online. I noticed that some radio controlled Casio watches have an 'airplane mode', which disables GPS and terrestrial time signal reception: Now, why would signal reception be restricted? The watch in question does not have a Bluetooth module or similar. <Q> Most useful radio receiver designs utilize some form of the superheterodyne architecture, where one or more radio-frequency local oscillator signals are used to shift the frequency of the received signal to an intermediate frequency where it is more readily processed. <S> Typically, the local oscillator signal will radiate from a receiver to some degree - how much depends on the specific design, but it is not uncommonly detectable in near proximity. <S> Since a local oscillator theoretically radiates, the argument was that you should not be operating one, which effectively meant that you should not be operating a radio receiver. <S> Of course digital processing circuitry typically radiates clocks and their harmonics, too... <S> But then, the shift in the last several of years towards services like in-flight WiFi also indicates a shift in thinking away from vague theoretical concerns. <A> I think that there is a very simple answer to this question that has nothing to do with electronics. <S> A number of airlines prohibit the use of GPS receivers in flight. <S> Their rationale does not actually matter. <S> In order to comply with their rules, the watch must be able to turn off the GPS receiver somehow or it would be prohibited from being on the plane entirely. <S> Here is a list of airlines which do not allow GPS receivers: <S> http://gpsinformation.net/airgps/airgps.htm <S> Does GPS actually cause interference? <S> Doubtful. <S> http://gpsinformation.net/airgps/gpsrfi.htm <A> Mobile phones, GPS receivers, laptops or indeed smart watches have no effect whatsoever on avionic equipment. <S> Most kit is designed and tested for restricted emissions and EMI tolerance in reverse. <S> The reason this equipment is banned has nothing to do with safety, but financial expedience. <S> There's a lot of electronic equipment in the world. <S> To rigorously prove my assertion in a way that would absolve an airline in a damages court case would take £100Ms. <S> No one wants to pay that, especially if it would have to be passed onto customers. <S> Better just ban the stuff. <S> People will still keep flying as they're a captive market. <S> You'll fly naked sat in a transparent plane if necessary. <S> Try leaving your phone fully on the next time you fly a long way across Europe. <S> If you survive to land at the other end, you'll find perhaps 10 -15 messages on it welcoming you to various mobile networks that you over flew. <S> No problem. <S> Edit following comments: There are approximately 1 million people flying in the air at any one time across the world. <S> If 1 in 1000 people randomly forget to switch off their e devices in flight (plausible), this theory is tested 1000 times/day across all randomly selected airliners in the world. <S> That's 365,000 times a year on every conceivable aircraft type. <S> That's a million hours /year of testing based on a 3hr mean flight time (plausible). <S> That's a pretty extensive systematic experiment. <S> Nothing dangerous happens. <A> All radio receivers have circuitry that could emit RFI if not properly shielded. <S> Better to be safe then try and determine the possible harm
| Restrictions on radio frequency sources on passenger airlines have historically tended to err on the side of caution, restricting anything that could theoretically emit moreso than being specifically tailored to demonstrated concerns. If you wish to keep GPS enabled during flight, check with the pilot.
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Why do we want higher power factor in ac motors? Here is a simplified 3P electric motor below: This motor is simply combined of three coils facing toward to the rotator (in the center) 120 degree apart to each other. The user wants his rotator to rotate (real work expected from the motor). Rotation force comes from the magnetic flux. And magnetic flux comes from the current flowing through the coils. So magnetic flux has nothing to do with voltage, only current is its concern. Am I right so far? If so, look at the graph below: This is an illustration of the power factor. Current being out of phase to the voltage. Think that I am applying 3 phase AC voltage (120 degrees apart from each other) to ports 1-4, 3-6 and 2-5 in order. In that case, current flowing from the AC voltage source, directly flow through the coil as expected. But, I am going to pay for only real power? Isn't it right? So why do I want power factor to be high? Secondly, take a look at the equivalent circuit of an inductor(or coil whatever) below. Even if I pay for the negative power, still though, why do I want power factor to be high? To make power factor high, I have to increase the series resistance on the inductor(or coil whatever), that means, since I can not increase or decrease the voltage, I will get less current to flow on coils. Therefore, I will get less amount of magnetic flux. Therefore, I will get less performance from my motor. This is maybe a really dumb question. I am aware of that, but please, can someone help me to understand it? I am trying to understand this for almost a month! <Q> But, I am going to pay for only real power? <S> Isn't it right? <S> So why do I want power factor to be high? <S> If you are a small user you will only pay for real power. <S> Your poor power factor will make little difference to the electrical grid. <S> It is not worthwhile for the power company to meter you for the reactive power (kVArh). <S> If you are a large user and your power factor is poor you will cause problems for the grid: <S> Your reactive load will cause higher currents to flow than would flow for the same (real) power in a resistive load. <S> This will cause increased voltage drop along the supply line. <S> If you are only metered for real power (kWh) then the power company has to bear the loss. <S> To solve this a dual meter is installed for large users. <S> This measures both kWh and kVArh. <S> Typically there is no surcharge if the power factor is kept > <S> = 0.95 <S> but penalties are applied if too much reactive power is used. <S> To make power factor high, I have to increase the series resistance on the inductor ... <S> No, we can correct for inductive power factor by adding capacitance across the supply. <S> Figure 1. <S> Adding power-factor correction capacitors can partially (as in this example), fully or over-correct inductive power-factor. <S> Source: PowerFactor.us . <S> In industrial applications a bank of power-factor capacitors is typically installed at the main distribution panel. <S> A controller monitors the incoming voltage and current, calculates the power-factor and switches in contactors to connect the capacitors across the supply. <S> Since most industrial premises are three-phase the capacitors are supplied in delta configuration inside a three-terminal metal case. <S> Figure 2. <S> An industrial power-factor correction bank with controller (mounted on door), fuses, contactors (relays) and capacitors. <S> Source: Direct Industry . <A> To answer the title "Why do we want higher power factor in ac motors?":if <S> you are a domestic user you pay for W. <S> If you are an industrial consumer you pay for VA. <S> For example, if you are a water supply company, the largest cost of your operation is pumping the water. <S> If you can increase the PF of your electrically-powered pumps then you can save a lot of money. <S> Related: <S> Why do power companies never bother residential customers about power factors? <A> Low power factor means that the load is drawing a higher current than is necessary to deliver a given amount of power. <S> Since the excess current is not delivering power, the only cost connected with it is the cost of losses due to the current flowing in the transmission lines and other distribution equipment. <S> There is also an indirect cost due to equipment being used to deliver current without power rather than delivering current with power. <S> Most of the disadvantages are felt only by the power supplier. <S> For small users, the suppliers just lump those costs into the total cost of doing business and don't have a metered charge for it. <S> However, we all pay for power company inefficiencies indirectly in the price of the actual power that we purchase. <S> Users are expected to deal with the resulting power factor in whatever way is appropriate under their particular operating conditions. <S> For small scale users, that generally means that no action is required. <S> For large scale users, that may mean installing power factor compensation capacitors at each motor or in banks in their facility.
| Induction motors are designed to operate at as high a power factor as possible without making the motor too large and expensive or inefficient.
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