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Diode not behaving as short circuit I recently started to play with online circuit simulator and in a very simple circuit I can't understand the behavior I have a voltage source connected through a current limiting resistor to a diode and an inductor in parallel. As far as I know, a diode should behave like a short circuit when its anode is connected to the positive terminal of the voltage source. In this simulator something very strange happens: as I close the switch, a lot of current passes through the diode (and very small amount through the inductor) and after couple of seconds no there is significant drop of the current through the diode until it stops completely. Why is that? link for my circuit in the simulator: Link (click on the swtich to close it and watch the simulation) <Q> Initially, the inductor resists the change of current, making the diode the path of least resistance and causing it to carry most of the current. <S> When the magnetic field in the inductor builds up, the voltage across it decreases as it allows more current to pass. <S> The diode has a forward voltage drop (typically 0.6V) to account for, so it won't conduct any current after the voltage across the inductor falls below the diode's forward voltage. <A> Yes the previous posters are right. <S> So it behaves like this whenever the voltage is lower than 0.6 V <S> no current will flow and when the voltage is over this threshold current flows. <S> The inductor responds to sudden changes in current in a different way, it exhibits something called impedance, that is a way to say that while it has a resistance R it also has an inductance L, a component that is directly dependent on frequency. <S> So an inductor when suddenly connected or disconnected from a voltage supply reacts by spiking up voltage for a brief while and the current is initially almost zero, only to settle a brief moment later with smaller currents and voltages approaching zero. <S> The diode in the circuit sees this increase in voltage (while the current is still almost zero in the coil) and it closes, letting the spike flow through it, reducing also the excessive voltage on the coil and thus the large current in the diode that flows for a very short time. <S> A very common arrangement usually called a SNUBBER is what you will find in some switching relays or even solid state devices. <S> Its function is to stop the excessive voltage spike from breaking coil insulation by conducting temporarily the large voltage spike and then to close as the voltage on the coil returns close to zero. <S> I merely translated the above equations and observations in layman terms, hope it helps. <A> When enough time has passed, the inductor behaves as a short circuit and bypasses the diode. <S> That means \$V_d=0\$ and no current flowing through the diode. <S> OTOH, at the precise time of switching, the inductor will try to keep its current (that happens to be zero because it is not energized). <S> Because of that, it briefly behaves like an open circuit and all current flows through the diode, which will be forward biased until the voltage in the inductor falls below \$V_f\$. <A> For an inductor, \$V = L\dfrac{di}{dt}\$ <S> In any steady state condition, there is no change of current against time hence, the voltage across the inductor MUST be zero. <A> As others have pointed out, a diode is not a "perfect" short (or open) circuit. <S> However, if you understand its "limitations," then you can use the idealized behavior every were, except the limitations area. <S> For your particular circuit, you have to know that an inductor initially appears as an open circuit , and then as a short circuit after reaching steady state. <S> What this means is that initially, your circuit behaves as if only the resistor and diode (in series) are connected to the supply. <S> So the diode is forward biased and acts like a short. <S> As the inductor reaches steady state, the voltage across it goes to zero, and therefore the voltage across the diode goes to zero. <S> Since the diode needs at least .6V to be forward biased, it stops conducting when the voltage on the inductor goes below .6V. <S> At this point, the circuit behaves as if only the resistor and inductor (in series) are connected to the supply. <S> I hope you can now see that your simulator is showing the correct circuit behavior.
| To further clarify, a diode is not a short circuit but a threshold device, it starts conducting whenever voltage across it (when oriented properly to conduct) is greater than some value, typically 0.6V (but may differ for special types) .
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Design to replace several jumper on a dense PCB I'm working on the design of a board which include a lot of Jumpers (around 40). They are used in order to provide a possibility to connect or disconnect parts of the schematics. Signals are all analogic. As my board is small and I already have quite a lot of components, I'm thinking of a way to reduce the space used by jumpers (header with 2 pins , 2,54mm). It is possible for me to cut a group at a time (lets say, unplug 8 jumpers at time). I am thinking of using of CMOS switches, such as CD4066, with output control pins themselves drived by jumpers, but I am not sure if it is the best way to go. I am also paying attention to the noise induced. So my question is, do you have better suggestions to improve space reduction and / or noise induced ? Additionnal information on signals used : +/-12 Vpp range, 15Hz-100kHz frequency range. <Q> There are a number of factors that need to be considered here. <S> Analogue circuits can be sensitive to routing things around just to make certain other features work. <S> So the proposal of how to reduce jumpers has a lot to do with just how those jumpers are used. <S> A. <S> For the cases where an isolation is needed you simply remove the component with a tweezer type soldering iron. <S> B. <S> If the jumpers are used for configuring the circuit so that it can be calibrated then it could be a great idea to use the analogue switches as long as the added component count and circuit complexity can be justified. <S> There is a temptation with this approach to use multi-switch packages and the extra routing added to analogue signals could be a problem. <S> Switches also add some capacitance and switch impedance that could become factors to consider in certain types of analogue RF or low signal level circuits. <S> C. <S> If the jumpers are also used as test points to allow connecting scope probes and as signal injection points to apply signals from external testing gear then you really have no choice but to leave the jumpers in the design. <S> You could consider using the smaller type of jumpers that have 2mm spacing or 50mil spacing. <S> It is possible that after you analyze your circuit that you will find that the current implementation is a mix of A, B, and C above and you can set to optimize each accordingly. <A> One trick that I've used is called a 'Cuttable Jumper'. <S> Basically it's an SMT resistor footprint (0603, 0805, or whatever) with a trace connecting both pads. <S> If you later need to disconnect the signals then just slice through that trace with an X-acto knife. <S> If you want to connect them again then populate a zero-ohm resistor. <S> If I know that non-expert solderers will be doing the work then I try not to go any smaller than 0805; it's easier. <A> +/-12V signals mean rails of more like <S> +/-15V so the 4066 is already not suitable due to absolute maximum supply voltage. <S> There are (relatively expensive) <S> analog switch ICs that may do the job. <S> Keep in mind that a jumper is almost perfect at low frequencies- low 'on' resistance, very high off resistance, potentially very little leakage. <S> Just to mention one possible issue, the on resistance of a switch might be of the order of 100 ohms and it will vary with the signal (and temperature) <S> so it could introduce distortion in the signal. <S> The 4066 does not have very tight specs, but it still can be useful. <S> One thing to keep in mind that it does not have level shifting built in so the control signal on, say, +/-5V will have to go below ground <S> (maybe -4 to +4 to be reliable). <S> The higher end switches usually allow 3V or 5V control signals to control a switch that can handle +/-15V or more.
| If jumpers are used to simply isolate the circuit sections in the case a board is under debug trying to root cause a fault of some sort then it can be a good idea to simply use SMT components commonly called zero-ohm resistors to link the circuit sections.
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Electric Shock at 5V and 40 A I am using an array of WS2812 leds for an application of mine. The length used is around 10 m with 60 leds/m. The current for one led is 40 mA, this makes the total current to be around 25 A. I wanted to know if this is a dangerous current at 5 volts. I want to know this as I have to put my circuit in an insulation class where a single fault would result in electrical shock to the user. <Q> This low voltage is insufficient to push a harmful current through dry skin. <S> So if that's their only point of contact, you're OK. <S> However, it depends where you're planning to deploy this as to how far you have to insulate it and protect from contact. <S> 5 <S> V @ <S> 25 <S> A is more than enough to hurt the user if they put the wires in their mouth. <S> Consider this if, for instance, the wires on your lights are accessible to a baby crawling on the floor, such as with decorative lights. <A> A 5V 40A power supply can output enough power to make fire hazard a possibility if a short circuit happens at the end of thin wires which are not able to carry the full current without overheating. <S> Fuses rated to protect the wires solve this problem. <S> This is important if you have connectors which will be handled by the user and other places where shorts can happen. <A> Addressing the question straight: no, under normal circumstances. <S> Having a PS that can deliver 25A with output set at 5V does not mean that it will deliver the 25A. <S> If this is going to be a commercial product you should give it more thought, since users can do surprising things. <S> Anyways, from the WS2812 datasheet ( http://cdn.sparkfun.com/datasheets/Components/LED/WS2812.pdf ) it is seen that the current needed for the LED is 20mA, not 40. <S> Also, you must add a resistor to each led to set the proper current.
| Your 5 V is far underneath the standard hazardous voltage levels (50 Vac, 120 VDC) and far below anything that can cause harm.
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Difference between load resistance and Thevenin/Norton resistance? In the Norton/Thevenin theorem, what is the definite difference between the load resistance and Thevenin/Norton resistance? I cannot find any sources defining the difference, all of the sources use them as if I already were to know the difference. Take this example. Here I know the load resistance is the 40 ohm resistor. But not really why. Is it because of the direct connection to ground? If the circuit was more complex, I might not have been able to see it as clear as in this picture, because I don't really know what qualifies 'the load resistor' to differ from the rest of the circuit. <Q> In the example you've posted, we see two Thévenin circuits that share a 40-ohm load resistor. <S> But the question is what differentiates between the "load" resistor and the "Thévenin" resistor. <S> I've redrawn the circuit below: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The Thévenin equivalent circuits are enclosed in boxes as shown. <S> Alone <S> both Thévenin and Norton circuits have open-ended terminals. <S> Whatever impedance is found across those terminals is deemed the "load" that the equivalent circuit is driving. <S> For a Norton equivalent, we would see the following circuit: simulate this circuit <A> Difference between load resistance and Thevenin/Norton resistance? <S> Generally, the load is the something that is connected to the output of an electrical circuit. <S> It can be anything like a resistor, capacitor, inductor, LED or anything. <S> Now coming to your circuit, <S> Here I know the load resistance is the 40 ohm resistor. <S> But not really why. <S> Is it because of the direct connection to ground? <S> It's not because of the direct connection to ground but the given terminals \$a\$ and \$b\$. <S> Basically, in Thevenin/Norton theorem analysis, we assume the resistor connected in between the terminals say \$a\$ and \$b\$ is the load resistor. <A> Thevenin & Norton Resistances are the same. <S> The difference is whether you model the circuit with a Thevenin-voltage-source or a Norton-current-source. <S> This page should help you <A> These theorems are usually used to find the current through (or voltage across) <S> "LOAD" resistor, <S> obviously Load Resistor is one through which you wish to find <S> current (or across which you wish to find voltage drop). <S> Obviously, Choice is yours!!! <S> Ultimately you will have Thevenin's and Norton's equivalent circuits which are nothing but either a voltage source or current source with their internal resistance, connected across the load resistor. <S> With Ref. <S> To question fig. <S> you posted, anyone of 3 resistors can be considered as load resistors while obtaining equivalent Thevenin's or Norton's network, you have to decide through which resistor you want to find current. <S> I can also find Thevenin's and Norton's equivalent circuits across other two resistors and I will still be right. <S> The original question which you have to tackle is not "Difference between load resistance and Thevenin's resistance" <S> but "Which shall be my load resistor? <S> " <S> and the answer to that lies in "your requirement. <S> " <S> Suppose if you were told to find current through resistor connected to +ve terminal of either battery and node A, you'd have selected that resistor itself as load resistor. <S> Actually the reason of calling that resistance as load resistance is the Thevenin's and Norton's equivalent circuits. <S> As I said earlier, these are the simplified forms of complex network; containing either single voltage source or single current source (with internal resistance) supplying the load resistor. <S> So the word "Load" has originated from these equivalent networks and not from non-reduced ones and thus you don't have to worry about predicting "Load Resistor" in non-reduced network, <S> you just have to know which current or voltage <S> you want to find.
| A Thevenin/Norton resistance is the equivalent resistance of the network when looking from the output terminals by shorting voltage sources and opening the current sources. A load resistance is simply a resistor which is being used as a load for an electrical network.
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What is this component and where can I find more details? I would say it is some kind of transformer but it has the same wire girth on both primary and secondary windings and the same loops count. It had the following label: CLTE 250V, 0.5 A, 45 U, 2 × 500 uH VDE I need to know what it does. It is located between a capacitor and contacts that close the circuit Additional info It is part of a Pfaff 92 sewing machine pedal. Right between what looks like a capacitor / a group of them and contacts that gradually touch themselves and (with the help of some resistors, I guess) send current of different values to the motor (or maybe they do something else). This little thing (the size of an AAA battery) had a short and I would like to replace or wrap new wire. That is what I need to know what it is before <Q> The current and voltage is much too high for the apparent size of this thing for it to be a transformer. <S> It looks way too small to transfer 125 VA from primary to secondary. <A> It is a crude old fashioned large but effective CM choke or balun. <S> They are made much better these days for reducting conducted and radiated spike noise in both directions. <S> They are used everywhere for AC line filters and switched motor or SMPS supplies. <S> http://www.mdipower.com/content/ApplicationNotes/html/dcpower06.htm <S> The purpose is to raise CM impedance with L and shunt differential noise with C which has low impedance depending on f and C/L ratio. <S> But the C shunts current to ground so because the AC leakage spec limit of 0.25 to 0.5mA at line frequency. <S> using line rated X or Y caps for different purposes with plastic film caps only. <A> 500 uH is rated at 500 micro Henrys, micro Henry's is a measument for inductance.
| Just the picture alone could also suggest some kind of transformer for RF, but 250 V and 500 mA on the label means it's not. This is a guess, but based on the label showing voltage and current, this is probably some kind of common mode choke or balun . It is definitely an Inductuctor, the purpose of an Inductor in a circuit is similar to a capacitor, Inductors contain a magnetic field it can store current where as the capacitor stores voltage.
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What does the Amp rating on a Diode Mean? I have two rectifier diodes, a 1N4007 and a 1N5391. The first one is rated at 1A while the second is rated at 1.5A. What does the current rating mean? Will I damage/overheat either if I connect a device (such as a wall wart/barrel connector) that has an output of 2A or more? <Q> The key is to read the datasheet carefully. <S> The 1A or whatever rating is under some particular set of conditions that may be optimistic for your situation. <S> In the case of the 1N4007 (what I assume you mean, since a 1N1007 is an old germanium diode) a "1A" rectifier, the rating is specified as follows: <S> So we are talking average rectified output current with the lead temperature <S> maintained at 75°C 9.5mm from the case. <S> Note that they effectively define this as the ambient temperature which is.. rather dubious. <S> There is an obvious danger if you have long leads and are running close to the maximum current. <S> On the plus side, the actual RMS current may be significantly higher than the average rectified output current. <S> A graph on the datasheet shows linear derating to zero at 150°C (again that's the lead temperature 9.5mm from the case). <S> If you are running at higher current (or even close to 1A) <S> I suggest you use a higher rated part such as 1N5407 or a Schottky 3A diode which would run cooler (because less voltage drop) <S> eg. <S> 1N5821 . <S> The 1N5821 datasheet I linked is a bit more honest in that it refers to Tl (lead temperature) rather than ambient temperature, however it is specifically the cathode lead and only 1/32" (0.8mm) from the case! <S> There is detailed thermal design material in that datasheet, and it's pretty straightforward, just simple arithmetic. <S> Of course if your adapter can supply 2A but you are sure your load can never draw more than (say) <S> 0.5A then a 1A diode is fine. <A> Read the rest of the datasheet. <S> Vishay's datasheet for 1N5391, for example, has these two charts: <S> So a continuous current of 2 A is not allowed. <S> Even 1.5 A is not allowed if your operating environment does not keep the lead temperature below 75 C. Up to 50 A current surges may be allowed if the surge duration is short enough and the surges are infrequent enough. <A> If you use these with a 2 A or 6 A power supply makes no difference .... <S> it's the amount of load current that flows that is the limitation. <S> For example if you have a 2 A power supply but are only using a load that draws 1 A, you may be fine with the 1N1007 (though it will get warm/hot). <S> If you used a 1N1007 in an automotive project (the battery is capable of 100's of Amps) <S> the same applies .... <S> as long as the load current for your project (flowing through the Diode) is less than the continuous current rating for the Diode, everything will be fine. <S> So the current rating of the supply is irrelevant, it's the current rating of your load that must be considered, and must not exceed the Diode rating.
| The rating is the continuous current limit of the Diode.
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What is the best practice for driving IC enable pins? When using ICs that use enable pins, what is the best practice for driving them in a microcontroller system (particularly when there's no especially pressing need to disable the device at any time)? For example, I'm using a chain of parallel-to-serial shift registers to aggregate a large number of physical device inputs into a single serial bitstream to my uC, and these registers only operate when a clk_en clock enable pin is driven low. The inputs will be polled for the lifetime of the uC, so at no point do I NEED to explicitly disable the clk_en signal. Is it considered bad practice to have this pin permanently pulled low (i.e. with a pull-down resistor)? <Q> I don't think that would be bad practice, just choose an adequate pull-down resistor. <S> In order to make the design flexible though, you could also add a zero ohm jumper to that enable pin (a do-not-stuff component) along with the pull down resistor, just in case you need to do some debugging in the future, never know. <S> That zero ohm jumper could connect to one of the GPIOs, Vcc, or some header pin you may want to add for debugging or whatnot. <A> If the device is to be permanently enabled, I would tie the Enable pin to Vcc or Ground, whichever will enable the device - should be no need for pull-up or pull-down resistors. <A> The purpose of an enable pin is to allow disabling the part (or some function of it) when you do not need it. <S> Typically, you want the output(s) of a chip to be disabled while it is powering up, so you tie the enable pin through a resistor to "disabled", and later force it on with a GPIO output from your microcontroller, after the initialization is done.
| If you do not care anout the state while powering up, and if you do not want to disable it later, you can connect the enable pin directly to V CC or ground.(But for debugging, it might be useful to add a jumper or pull-down resistor to allow to change it.)
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change cell protection value for 18650 from DW01? From what I read, the tp4056 with protected circuit can still damage the cell. According to data sheet, VOCP at 4.3v, VODP at 2.4v. Is there an "easier" way to lower VOCP, 4.1v, AND increase VODP, 3.0v? TP4056 data sheet DW01 8205A Thx edit For your info, i'm not a expert, just a DIY guy. I think i dont explain correctly. I need to protect my cell, 18650, from over use, (Overcharge or Overdischarge) I bought several TP4056+DW01 from Aliexpress .Frankly, no schematic is needed, data sheet is enough? I read several forum about TP4056 and DW01. Those chip are not perfect. TP4056, is a OK a charger. My concern is DW01 and restrict of space for my usage. DW01 should protect the cell not to Overcharge and Overdischarge. All I want is to change the cutoff circuit from DW01. Overcharge detection voltage [VOCP] (V) : 4.250±0.050 Overdischarge detection voltage [ VODP ] (V) : 2.40±0.100 Here the perfect chip, FS312F-G Overcharge detection voltage: 4.25±0.025 V Over discharge detection voltage[ VODP ]: 2.90±0.08 V I found a perfect chip for my need, but why should I buy if i could trick the DW01. Could I ? I check the CN305, it seem good, but not easy to find here :( Till now, i could not find someone that could change the cutoff, probably because it it not possible with DW01 ??? :) What do you think of this option, 1. Use TP4056 + DW01 + voltage divider for EACH cell a. TP4056 charge to max 4.263V, good for not to Overcharge b. DW01 + Voltage divider, i can adjust to cutoff around 3v, good for not to Overdischarge, but my cell stuck with TP4056 permanently. 2. Use TP4056 + FS312F-G (replace DW01) a. TP4056, as a charger b. FS312F-G, Cell protection *sorry for my bad english, *thx for good comment or suggestion :) merci! <Q> Update: The DW01 can have it's low voltage turnoff voltage increased by use of a divider that presents Vbat <S> x K <S> (K <=1) to the IC. <S> The divider will increase the overvoltage protection voltage by the same ratio but, as long as the charger is competent the overvoltage protector is less needed. <S> Operation of the OV protector could be obtained by disabling the divider by shunting the input resistor with a MOSFET which is turned on for Vbat <S> > <S> V_lowvoltage <S> and well prior to Vbat = Vbatmax. <S> Is there? <S> The TP4056 is intended to NOT damage batteries. <S> Where did you find suggestions that it does? <S> - links useful. <S> TP4056 datasheet here V_chg_max is said to be 4.2V +/- <S> 1.5% or 4.137 / 4.2 / 4.263 V min/typ/max. <S> That disagrees with your 4.3V figure. <S> I assume that your Vodp = Vtrikl in the above cited datasheet. <S> Vtrikl = <S> 2.8/2.9.3.0V <S> min <S> /typ/nax which about matches your desired 3V figiures. <S> This is well above your cited 2.4V. <S> It sounds like your datasheet is suspect as the ones in the sheet I've referenced sound reasonably standard. <S> Can you provide a link to the 4056 <S> ds you are quoting? <A> I don't think there is an easy way. <S> The IC is pretty complicated (3 feedbacks) and adding a resistor divider probably will fail. <S> If OCP (over current protection) is not critical, you can look at CN305 - you can easily setup the OVP and UVP thresholds. <A> Schematic created using CircuitLab <S> This way, when the battery is being charged, TP4056 will see the voltage equal to \$V_{BAT}+V_{D1}\$ and stop charging when that sum reaches 4.3V <S> So you want a diode which has a forward voltage drop of 0.2V at 100 mA <S> (look for a Shottky or Ge diode) <S> In a similar way, when the battery is being discharged, TP4056 will see a voltage of \$V_{BAT}-V_{D2}\$, so the over-discharge protection will kick in when that value drops to 2.4V. <S> This calls for a regular Si diode with about 0.6V of voltage drop (at your load's typical current). <S> Of course this is a hacky approach that may work for one-off indoors-only prototypes where you can actually measure the voltage drop of the diodes you're going to use, and keep those diodes at the same temperature all the time. <S> Also note that your load will see the voltage of \$V_{BAT}-V_{D2}\$ as well. <S> If you need something efficient and production-friendly, get a charger IC with programmable thresholds.
| If I wanted flexibility I'd try to use another IC unless there was a compelling reason to use the TP4056. You could try adding diodes in series in both directions, like this: simulate this circuit –
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Does a NYY or NYM cable affects a CAT6 cables transmission if they are crossed? I get an unfinished job and really hard trying to complete it. Ex-engineer of my company had designed the cable tray system as a cable tray should used for both light current cables and heavy current cables. I am in doubt about using light and heavy current cables in a tray. I will use a separator to keep them separate but somehow they have to pass over eachother in some points. I am transmissing voice and image with CAT6 cables (entryphone system) and heavy current will be in 5x6 NYY cables. Is the NYY cable affect the CAT6 cables image and voice when they are crossed? <Q> It depends on the quality of the CAT6 cable. <S> Industrial cables have larger cross section and very thick shield, also two layer shield. <S> If you put a separator is yet better, the best would be separate trays. <S> If the cables go underground, you should have also two separate hoses. <S> It is almost inevitable to cross the power cables at certain section. <S> Try to lay the CAT6 as distant as possible from power, even when you cross it. <A> Is the NYY cable affect the CAT6 cables image and voice when they are crossed? <S> Crossing cables can affect each other but the general practice is to use balanced drivers and balanced differential receivers on the circuits that might be susceptible to noise. <S> It's still important to use the "right" cable of course but, just using the "right" cable and not paying attention to drivers and receivers is asking for trouble: - The top picture shows an unbalanced driver - noise can more easily get onto one wire than the other and, as a result the receiver output is affected by that noise. <S> The bottom picture shows a balanced driver and, as expected, noise will affect both signals about the same but, because signals are inverted, the effect of the noise is cancelled at the receiver. <S> Balancing not only means making sure impedances are balanced (thus noise affects both lines the same) but can also mean (and be enhanced by) using differential drivers. <A> Generally, the maximum coupling occurs when wires are in parallel, and minimum if they cross at right angles.
| It is a common practice to place the signal cables on one side and power cables to the other side of the tray.
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Is electricity wasted when you've connected, say a mobile charger but are not using it? I know that the answer is yes. It is consumed, but in very little amount. My question is how it consumes electricity. When the phone isn't connected the circuit is open at the phone end of cable, then how does electrons move around the circuit, because electricity is consumed when a circuit is completed and electrons can move around (I think so). <Q> Yes, some electricity is consumed. <S> A charger is a relatively complex device that has multiple "circuits" internally. <S> Some of these circuits dissipate power regardless of whether anything is connected to the charger's output. <S> The amount of power is miniscule, however — on the order of tens to hundreds of milliwatts. <S> You'd have to leave it plugged in for a year before it consumed a single kilowatt-hour of energy. <S> In other words, each charger you have plugged in is adding about a penny to your monthly electric bill. <A> Because there are other components that use up energy. <S> Timer circuits, any IC, the led, transformers, reference resistor dividers, etc. <S> These are in parallel to the device to be charged. <S> Multiple circuits are completed thus energy flows and is used. <S> Look up quiescent current. <A> In any power supply, some current is consumed by the inefficiencies of the circuit, as leakage currents and as operating currents by the internal regulator circuitry. <S> But a mains charger is a switch-mode power supply (SMPS). <S> Its regulator works by rectifying the mains then using a PWM-controlled switch to send that to a transformer going to the output. <S> It only sends through the transformer the power the load requires, plus a little extra for what's lost by its own inefficiency. <S> You can find plenty about SMPS their internal design and operation on the internet (search for 'smps basic operation'). <A> No component is ideal, capacitors leak, transformers require magnetising currents. <S> Control/regulation circuits require power to operate. <S> So there are many ways current can flow other than into the load. <S> On a modern charger based on a rectifier followed by a flyback converter no-load losses are generally very low. <S> Under light or no load the flyback converter will switched to a discontinuous mode with very short bursts of operation to make up for the small amount of energy lost to leakages and powering the controller. <S> OTOH on an old-fashioned charger that has a mains-frequency transformer there are significant no-load losses. <S> A noticeable amount of magnetising current is needed by the transformer core and this current leads to resistive losses in the transformer windings. <A> If the phone was connected directly to mains then no, it would not use power while disconnected. <S> However, mobile phones (for example) are NEVER connected directly to mains. <S> Instead, they are connected to a step-down circuit that steps down mains voltage (120VAC or 240VAC depending on what country you live in) to usually around 5VDC. <S> This can be done in various ways, but the two most common methods for chargers are step-down transformers and buck converters. <S> Transformers consist of two windings around an iron core. <S> One winding (with more turns) is connected across mains, and the other winding (with fewer turns) is connected to a rectifier, filter capacitor, and then to the phone. <S> Even when the phone is not connected, current is still flowing through the primary winding (the one with more turns connected across the mains plug) and thus power is dissipated. <S> Buck converters are a type of switching supply that turns on for a short period of time and turns off for a longer period of time. <S> This happens tens or hundreds of thousands of time per second. <S> The ratio of the on-time to the off-time determines the average voltage, which is passed to the output. <S> However, this circuitry runs and draws current even when the output is not connected to the phone, so it too dissipates power. <S> In the case of transformers and switching supplies, there are closed loops in both cases, so the charger will use power even if the phone is not connected to it.
| If there is a closed loop across the mains, there will be current draw and power dissipation. To stay in regulation with no load, the circuit contains a minimum load within itself and this uses a small amount of power.
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Buck/Boost circuit not working I'm a co-op student working on a hardware design for an industry client, and I'm trying to use a TPS63000 to get a Vout of 3.45V. Vin is 5V. When I power this circuit up, I read 5V on Vin but only 0.7V from Vout. Here's the PCB design: I'm not sure what approach to take in troubleshooting this. Any help would be appreciated. EDIT: Unfortunately I don't have easy access to an oscilloscope, but when I do, I will definitely try to confirm that there's a pulse across the inductor.Yesterday I assembled another PCB, and I'm getting a very similar result. Vout actually stabilizes at 322mV, and does so with a high degree to accuracy. During startup, I do see Vout shift drastically, but always settles around 322mV. Sorry that that's not very much to go on, an oscilloscope will definitely help give a better picture. So it seems to me possible causes are: A. lack of R3/C3 on the Vin pin, B. narrow traces on the PCB, or C. bad chip. Thanks for all the responses so far. <Q> You have no delay on the VINA, EN and PS/SYNC pins. <S> It is unclear what mode the thing will go into when connected like that. <S> The device specs shows a suitable delay circuit. <A> Regardless of the following, I fail to see why your circuit does not work. <S> \$R_1\$ and <S> \$R_2\$ are correct for 3.45V. <S> \$C_1\$ , <S> \$C_2\$ and \$L_1\$ are within parameters. <S> Why your circuit does not show some life is strange. <S> It is scope time! <S> Calcs have been confirmed. <S> Bad chip (is always lame)? <S> Try connecting up a load during power up. <S> Try connecting up \$W_1\$ , when your source has stabilized. <S> I do not think the addition of \$R_3\$ <S> and \$C_3\$ is the answer. <S> The app-note does not even mention them. <S> But it is certainly worth a try. <S> From <S> TPS6300 datasheet : <S> As for all switching power supplies, the layout is an important step in the design, especially at high peak currents and high switching frequencies. <S> If the layout is not carefully done, the regulator could show stability problems as well as EMI problems. <S> Therefore, use wide and short traces for the main current path and for the power ground tracks. <S> The input capacitor, output capacitor, and the inductor should be placed as close as possible to the IC. <S> Use a common ground node for power ground and a different one for control ground to minimize the effects of ground noise. <S> Connect these ground nodes at any place close to one of the ground pins of the IC. <S> To lay out the control ground, TI recommends to use short traces as well, separated from the power ground traces. <S> This avoids ground shift problems, which can occur due to superimposition of power ground <S> current and control ground current. <S> Irregardless, you have skinny and long, as opposed to short and really, really fat, so you will definitely have production problems. <A> A few guesses/pointers: - Try fitting link W1 Check R1 isn't 1.8 kohm (or a wrong value) <S> Is exposed thermal pad properly earthed? <S> Have you got too much output load? <S> Check <S> that oscillations are running <S> about 1350 kHz Is your meter working correctly? <S> Good luck. <A> My guess is that you don't have a proper power supply bypassing. <S> You might be able to succeed without R3 (but you shouldn't), but you definitely need C3, and the controller side power supply should be separated enough (with at least a thin wire) so that the highly fluctuating current between C1 & the power side of the IC (pin 5) does not affect the controller side power supply of the IC (pin 8). <S> You should also look at page 16 in the data sheet and read the layout recommendations (note that the layout recommendation picture shows a differing schematics, with a different C3).
| The feedback divider should be placed as close as possible to the control ground pin of the IC. You have omitted the separate bypass cap C3 from the typical application schematics, where R3 & C3 form a low-pass filter to separate the power supply of the power part from the controller part of the IC.
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connecting a LED to a PC case fan header On my motherboard (ASRock B150M Pro4S/D3) i have an unused 3-pin case fan header. Can i use it to drive a single LED with only a resistor in series? The fan can be configured in the BIOS to change speed according to the CPU temperature, so would like to use the LED as a visual indicator for that. (I expect the LED to blink faster as the CPU temperature increases.) <Q> Pin 1 is GND. <S> Pin 2 is 12V power to the fan. <S> If the motherboard supports fan speed control via a three pin header, this 12V pin is PWM controlled by the motherboard . <S> Pin 3 is tachometer feedback from the fan. <S> The fan pulls this pin to GND, usually twice per revolution. <S> This pin has a pull-up resistor on the motherboard. <S> So <S> With three pin headers, there is no standard PWM frequency, but I expect the frequency to be much too high for you to be able to percieve any blinking. <S> What you will see is the LED getting dimmer as you "slow down the fan" in software. <S> And for completeness: <S> With four pin headers, the 12V is constant. <S> Pin 4 has a 5V 25 kHz PWM signal that tells the fan how fast to spin. <S> 4 pin headers can often be configured to operate in 3 pin mode. <A> I= <S> (12V-Vf)/R <S> I don't expect it to blink <S> but you can run it in parallel with your CPU fan. <S> They use PWM to create a DC voltage for the fan. <S> So if RED 20mA, Vf=2.1V then R=9.9/20m =495 ohm or nearest <S> Also, Pd(R)=20mA*10V = 200mW so <S> 1/4W will get hot. <S> but then the resistor can be near the fan ( lol ) <S> R=(12V-3 <S> *2.1V)/20mA <S> = 5.7V/20m = 285 ohms or nearest R then Fan @ <S> 6V LEDs dim or off and Fan @ <S> 12V full RPM, = full brightness. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> From: ASrock B150M User Manual <S> So CHA_FAN_SPEED is an output from the motor as a tachometer to tell the CPU what speed it is going. <S> So using it as a blinky light for temperature feedback is not possible. <S> You may have better luck with the 4-pin connector. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This should work off of the 4-pin header, using PWM.
| With a 3 pin PC fan header, it should be possible. If you want better UX, use 3 RED LEDS in series then when the fan is running very slow at say 6V, the LEDs are very dim. yes, if you connect a LED with a current limiting resistor between GND and the 12V pin, it should vary in brightness according to the fan speed set in the software for that header.
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Fruit juicer motor generates very low voltage I'm planning on connecting an elliptical trainer to an AC motor in order to charge some batteries and use the energy later. So I found an old 300 W juicer which I disassembled. Then I connected the bars to a multimeter and a drilling machine to the shaft. So when I measure the voltage with the drilling machine on (at maximum power), I get maximum 1.8 V (see picture). This is a very disappointing value for me, as I expected more...How much more? I don't know, but definitely not such a low, useless value. The voltage of the drilling machine battery is 14.4V and the energy is 21.6Wh. Not sure if this has any relevance. Can anyone help me to understand why do I get such a low voltage from a 300W motor? And if it is normal, why is it so low and what to do to increase it, considering that I want to connect it to the elliptical trainer. Edit: Please see the picture with the specifications of the juicer. <Q> This depends on the type of motor. <S> For example, if this is a AC induction motor, then it is only due to accidental residual magnetism that you are getting anything at all. <A> The rotor of AC asynchronous motors is simply a lump of aluminium and iron. <S> As there isn't any magnetically hard material involved, there is little to no residual magnetism stored and so, the motor, as it comes from the shelf, has no field. <S> When connected to AC power, the AC in the stator first has to induce a field inside the rotor. <S> Then the rotor starts to run slowly. <S> Because its field has smaller frequency than the stator field due to the overlay of rotation speed and outer field, the rotor has a driving torque in result and goes up to a speed slighty smaller than the sychronous speed (for a two-pole motor: ≈2950rpm on 50Hz, ≈3500rpm on 60Hz) <S> M: momentum (torque) <S> ; n: revolution speed <S> The only way to make an AC asynchronous motor into a generator is by connecting it to AC and connect a drive to its rotor which runs that one at a slightly higher than synchronous speed (for a two-pole motor: <S> > <S> 3000rpm on 50Hz, <S> >3600rpm on 60Hz). <S> Then, the current direction (in relation to voltage) on the AC input reverses and the motor actually delivers electrical power. <A> Such appliances are mostly made with asynchronous AC (induction) motors with cast rotor and higher gap between rotor and stator. <S> There are 3 reasons - the first is that it requires fewer (zero) external components to run, the second is that this motor is relatively cheap to produce, and the third (and most important) is that if the motor stalls mechanically, it will not draw much more current than its nominal (depending on design, but not more than twice, which is acceptable for a certain period until the operator turns it off) and will not require any additional protection. <S> After removing the reason for which it got stuck, it will work again. <S> These motors cannot be used as generators. <S> To charge batteries you will need a DC motor. <S> If you take a DC motor with a permanent magnet from some toy it will do. <S> Not perfect, but will do.
| Not all motors work like generators without any electrical power applied.
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Power rating on Zero Ohm resistor I was ordering some resistors online, and I saw that 0 Ω resistors have a power rating. Why is that? Power through a resistor is calculated with the equation \$P = UI\$ or \$P = RI^2\$. Since \$R = 0\ Ω\$, \$P=0\ W\$. According to this post ( How to calculate Power Rating for Zero Ohm Resistors? ), a 0 Ω resistor has no power rating... But Farnell tells me the opposite: <Q> While it may be true that distributors don't want to check every single part individually, in this case it is not down to laziness that the 0Ω resistor has a specified rated power of 125mW. <S> As pointed out by @BumsikKim's answer, the datasheet for the series does in fact specify this rating - the distributor product page is correctly representing the manufacturers specifications. <S> From Page 5, we have the following table entry: <S> Notice how for the entire RC0805 size series, there is a specified rating of 0.125W (1/8W). <S> This includes the 0Ω resistors in that series. <S> There is also however crucially another specification - Jumper Criteria . <S> This column specifies the rated current for an 0805 jumper (i.e. 0Ω resistor). <S> We can see from the table your jumper is rated for 2A, with an absolute maximum of 5A (presumably short pulse). <S> So why might a "zero ohm" resistor have such ratings? <S> Simple, it's not a 0Ω resistor. <S> Unless the manufacturer of the resistor you are using have secretly made a room temperature superconductor, the jumper is actually still a resistor, just a very small one. <S> According to the datasheet it is specified to be ~50mΩ or less. <S> Because the resistance is non-zero, some power will be dissipated. <S> If we plug in the provided numbers, we actually find that the power rating is real and sensible: $$P = <S> I^2R = <S> 2 <S> ^2\times0.05=0.2W$$ <S> So in the worst case resistance of 50mΩ, and at the rated current of 2A, it will be dissipating more than the 125mW rating. <S> Still think the rating is silly? <S> In a power supply design I had the pleasure of surge testing, the designer had added an 0805 0Ω resistor in series with a 24V DC input, just prior to a TVS diode. <S> During the test, we charged a 10mF capacitor up to 200V and then connected the capacitor to the input of the power supply. <S> Naturally the TVS started conducting, and the 0Ω resistor turned quite literally into a firework... <A> It is not really 0Ω. <S> According to the datasheet , page 5, the resistance of the jumper (0Ω resistor) is less then 50mΩ, not the perfect 0Ω. <A> I mean, if you're Farnell, you aint gonna pay someone to manually create each product entry for the E96 series into your database. <S> You would have a software tool which would create the product records according to a template. <S> Like, enter the common data from the datasheet only once (brand, series, power, package, photo, etc), and then automatically create all values in the resistor series using these common datasheet values. <S> Since I once saw a mistake in a resistor manufacturer's part#, I guess the part# would be entered manually for each value too. <S> Now, 0R resistors aren't exactly 0 ohms, more like a couple tens milliohms, so yes, they do have a max current and max dissipation power. <A> It's really stating the power rating of the resistor family it belongs to. <S> Some 0R resistors are in place of a different value in future. <S> If you place this 0R part on a board, that position will be able to accept any resistor in that family. <A> 0ohm resistor are not perfect. <S> You can take 1mohm as value for you calculation. <S> This will lead you to a very low power. <S> You shouldn't bother much about it. <A> As wikipedia says: The resistance is only approximately zero; only a maximum (typically 10–50 mΩ) is specified. <S> [ <S> *] <S> A percentage tolerance would not make sense, as it would be specified as a percentage of the ideal value of zero ohms (which would always be zero), so it is not specified. <S> In the ideal world the 0ohm is the ideal wire. <S> In this case the power is calculated as: $$P=RI^2$$ for current-driven applications and no power is consumed by the ideal wire. <S> $$P=\frac{U^2}R$$ for voltage-driven applications and infinite power is consumed by ideal wire. <S> In the real world neither the ideal wire neither the actuall 0ohm resistor exists. <S> That means some (little) power is consumed in current-driven applications. <S> That's why there are different 0ohm resistors with different power ratings; they do dissipate heat so they can be overloaded and burnt. <A> A physicist's perspective on a resistor \$R = 0 \ \Omega\$: <S> Applying a constant current source with finite current \$I\$, there is zero power dissipated. <S> In this case, \$P = I^2 R = 0\$, because \$I\$ is finite. <S> Note that \$V = <S> I R\$ is zero, and <S> so \$V^2 <S> / R\$ isn't infinite, even though \$R\$ is zero. <S> Applying a constant voltage source with finite voltage \$V\$, there is infinite power dissipated. <S> In this case <S> , \$P = V^2 / R = \infty\$, because \$V\$ is finite. <S> Note that \$I = <S> V/R\$ is infinite, and so \$I^2 R\$ isn't zero, even though \$R\$ is zero. <S> More practically, if \$R\$ is small but nonzero, then by similar arguments: Applying a constant current source, the power dissipated is small <S> Applying a constant voltage source, the power dissipated is large <S> The point isn't whether or not the resistance is exactly zero, but that applying a constant voltage source to a small [zero] resistance results in large [infinite] current in such a way that the final power dissipated is large [infinite and definitely nonzero]. <A> Value of R should be rounded off and close to zero because that looks nothing like a superconductor. <S> Safe to say that all electronical components have a non zero R value, even wires.
| The most likely explanation is that the resistor is part of a product series, and all product pages on Farnell have the same information for all values in the series.
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Powering ir led with a 9V battery I'm using this break beam https://www.adafruit.com/product/2168 with an arduino. Currently both sides are are powered by the 5V from the arduino. I want to move the led half of the break beam to the far side of a doorway so it can be used to count foot traffic. Is there a way to hook up the ir led to a 9V battery without using a breadboard? Can I use electrical tape to affix a resistor between the battery and the led? Here is my project page https://devpost.com/software/door-counter thank you in advance for any advice on how to power the led separately from the the receiver. <Q> Yes, you can <S> but you have a few problems. <S> And making electrical connections to a 9 V battery requires a snap-on battery connector. <S> But if you insist on proceeding, here's the circuit: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Note that the 180 ohm resistor is contained inside the transmitter module. <S> I'm guessing the internal circuit, since ADAfruit don't provide details. <S> Since the light beam expands as it travels further, you may find that detecting the beam across a doorway may be difficult, and will require larger current from the battery (too much for the LED to dissipate). <S> "Electrical tape" is used because of its protective insulating properties, not because it promotes electrical contact. <S> It should not be used to secure <S> a wrapped connection - reliability is not assured. <S> Soldering or crimping is more reliable. <A> IR LED starts from 1.2~1.6V depending if pulsed up to 100mA with an ESR near 5 ohms, so 9V is very inefficient. <S> That's why LiPo 3.6V is preferred or some other low ESR , low voltage source. <S> 9V is higher V ,higher ESR source, thus lower mAh capacity in terms of density or size. <S> You can trickle charge with 3.7~3.9V without fancy charger, if you can use adequate super low dropout , 3.3V LDO on arduino. <A> I might use a battery for a short-term test, but for any permanent or long-term installation, I would want the whole system, including the LED, powered from AC. <S> Otherwise you will have to replace the battery frequently, and won't know the battery needs replacement until you realize the counter is not working.
| A 9 V battery will discharge quickly when continuously driving the infra-red LED. Soldering is the only reliable way to connect the LEDtransmitter + resistor to the battery snap-on connector. An alternative circuit is shown, using two batteries in series (1.5 V each) that will last somewhat longer, yet still provide similar light intensity.
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LED in a circuit with alternating polarities I am building a project that simplifies a dolly moving on a track. When the dolly reaches the end of the track, it triggers a microswitch that cuts the power to the motor, and redirects the power to a LED to light up to indicate that the switch has been triggered. My problem is that since the motor controller I am using changes the polarity of the motor to make it go left or right, I am getting positive and negative voltage every other time. (depending on if the dolly is moving right or left).I have tried to use a 12V relay to trigger and light the LED, but the problem is relay needs at least 6 volts to latch, and when the motor is running slow, the motor controller only provides 3-4 volts. Is there anyone that has a solution on how I can make this work? I am just starting to fiddle with electronics, and I am on a budget with this build, so therefore: Is there a relatively easy fix for this problem? Are there relays out there that has a latch range from 3V to 12V? Schematics of the circuit with the LED (and resistor) connected directly to the microswitch. The problem here is that every other time the positive and negative switches. Schematics of the circuit with a relay attached to the 2nd output of the microswitch. The problem here is that the relay need 6 volts to latch, and the motor controller doesnt provide that at slow speeds. <Q> Take your first schematic. <S> Place another led and resistor in parallel across the first pair, but in opposite direction. <S> You could make it a different color to indicate which direction it was going. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The second version has some better reverse protection. <A> You can use a bridge rectifier to do this. <S> You can either make one out of four diodes or buy a single component containing the four diodes. <S> simulate this circuit – <S> Schematic created using CircuitLab D1 is your LED, as before, but R1 needs to be of a lower value to what you had. <S> This is because you have an extra 1.4-ish V drop caused by the bridge rectifier so have a lower voltage across the resistor to push the LED current through it. <S> You can calculate your new resistor value using Ohm's Law or go for trial and error if you'd personally find it quicker. <A> The simplest solution is to replace the LED with a BI-colour LED. <S> No other changes required to the circuit. <S> Added benefit <S> you will have the colour of the indicator change as polarity of the signal changes. <S> https://www.digikey.com/products/en?mpart=LTL-293SJW&v=160
| I understand then that you need a circuit that lights an LED from a supply and that supply can be of either polarity.
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Overflow results into wrong sign-flag bit set in 8085 microprocessor . In case of 8085 microprocessor when the MSB bit of accumulator is 1 then sign flag becomes 1 ( simply copy the result's msb) , my question is that : When we add two positive numbers say 44H(01000100) and 43H(01000011) , the result would be 10000111 , and thus the sign flag becomes 1 , but the both numbers are positive the how sign becomes 1 ? How microprocessor is able to handle this , please Help me out ? <Q> You must study binary arithmetic. <S> Sign flag is not enough to identify the result of the signed addition. <S> There should be another flag called overflow which will identify if resulting value is correct or not. <S> 8085 does not have overflow bit, unlike Z80's P/V bit, and you will not be able to identify correctness of the result using S flag only. <S> It will be your task of programmer to ensure the overflow condition either by: examining signs of input data. <S> In your example, both are positive, but output is negative meaning there's an overflow. <S> In general, values of differing signs will never trigger overflow; if signs of operands are the same, but result's sign is different, the result has overflown. <S> expanding size of the values to hold overflow bit. <S> For this, in case of 8-bit registers, you may only have 6 significant bits, with two MSBs being sign [7] and overflow [6] respectively. <S> If, after operation, these bits differ, it means overflow condition. <S> Your example extended to 9 bits 0.0.1000100 + <S> 0.0.1000011 = <S> 0.1.0000111 44h is positive 7-bit value, 43h is positive 7-bit value; result is having its sign and overflow bits set differently, 0 and 1, and it is overflow condition. <S> Another example - 1.1.1100000 <S> + <S> 0.0.1110010 = 1.0.0.1010010 = <S> > <S> 0.0.1010010 with carry <S> discarded <S> we see that sign and overflow bits are the same, thus result is correct, and in decimal it is -32 <S> + 114 = 82. <A> Adding signed integers which yield a sum outside of the range of representation is formally invalid. <S> To make it practically work, you would need software logic to figure out what happened and create a correct result in some viable representation, for example you could test that both operands were positive and re-interpret the result as a positive value in a wider (or simply unsigned) format. <S> However the most common higher-level language scheme (for example in C) is to require that you first change the representation of the arguments to one wide enough to hold the result, so for example you would store your 8-bit values in 16-bit registers and add them to produce a 16-bit result. <S> While in one view inefficient (if this is wider than your ALU and thus requiring multiple additions), from another perspective this actually is efficient, since the operations are always the same irrespective of the data, and in (later) pipelined and otherwise accelerated designs, making program-flow decisions can cost more time than performing arithmetic. <A> Short answer:
| Sign flag is only dependent on the bit D7 of the result you get in the accumulator and nothing else.
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Low frequency transformer at high frequency I'm getting interested in high voltage circuits, and I already have drawn out a couple of circuits (I have not yet built anything). However, I'm just wondering as to what would happen if a pole transformer rated at 120V and 50-60 Hz were run at 120V and 50-60 kilo hertz? Would the transformer overheat? Explode? <Q> A transformer that's designed for 50/60Hz will not work as a transformer at frequencies much above that. <S> The core will be designed with laminations 'just thin enough' for reasonable eddy current losses at 60Hz. <S> As the frequency rises, losses in the laminations increase. <S> 400Hz aircraft transformers have much thinner laminations than 60Hz transformers. <S> Fortunately, as you raise the frequency, the peak flux in the core drops for any given terminal voltage, so the hysteresis losses per cycle in the core drop. <S> However, as there are more cycles per second, the core losses tend to rise anyway as the frequency rises. <S> By the time you reach kHz, the transformer behaves more as a heater, with the windings coupling the input power into core heating. <S> Pump in enough power and it will get hot, with all the normal bad things that that entails. <S> If you want to operate a transformer at kHz or 10s of kHz, then you need to switch to a ferrite core material, with higher resistivity (to defeat eddy current losses) and lower hysteresis losses. <A> Yes, it would likely overheat or explode. <S> It would not make a very good high frequency transformer anyway, because (aside from the horrendous losses) the leakage inductances which are manageable at 50/60 Hz would become a problem at 50-60 kHz. <A> Other answers mention inductance, but transformers also have interwinding capacitance. <S> On such a large transformer it won't be negligible. <S> Depending on the wiring and construction, the HF impedance of the primary might be capacitive enough to draw more current than expected, or the primary waveform might couple to the secondary through the interwinding capacitance. <S> This is hand-waving as it depends a lot on how the transformer is built, but please keep capacitance it in mind. <A> The inductive reactance of 50Hz / 60Hz transformer will be too high at KHz. <S> That alone will drastically reduce the amount of power you can get out of the transformer.
| The core material in a low frequency transformer is not designed for kHz operation, and the core losses would cause overheating to destruction.
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capacitor power supply for 1 Amp output current I'm trying to make a fairly small AC(220V) to DC convertor to power microcontrollers and some actuators. It needs to output a current of 1 amp to the load. I tried looking at using transformers, but all the ones I could find that are rated for 1 Amp output current are too big for my design. So I changed my design to be a capacitor power supply. I'm thinking of using the circuit provided in this website: http://www.circuitsgallery.com/2012/07/transformer-less-ac-to-dc-capacitor-power-supply-circuit2.html But this one only supports around 150mA, It seems that if I change the X rated capacitor to 11uF from 2.2, I can get an output current of 1 Amp. Does this seem feasible. I am fairly new to electrical design, any help would be very appreciated! <Q> A capacitor input power supply is only appropriate for specific, low current, non-isolated applications, like tickling a SMPS into life, or driving an isolated low power LED bulb, where safety is not an issue, and a constant output current is OK. <S> As you want to drive microcontrollers and actuators, safety and isolation is paramount, and this is the wrong design to use. <S> In addition, 1A is quite high, and the load will vary, so this type of supply is triply inappropriate. <S> If they are too big, then you need to rethink your project dimensions. <S> When you do your own designs, it's often quite easy to get the behaviour you intend, say 1A output. <S> It's more difficult to avoid the behaviour you don't intend, like electrocuting yourself. <S> That's the reason that people buy isolated power supplies rather than build their own, anticipating all the bad things that can happen, then avoiding them, takes a lot of experience. <S> You're unlikely to get it right and safe first time. <A> Without knowing what output voltage you actually need, or the exact dimensional constraints of your system, many of these off the shelf power supplies could suit your purpose - I've found it more and more rare that rolling my own power supply is necessary or cost- or time-effective. <A> If you felt adventurous, you could use something such as TNYxxx series IC's. <S> They do not need much, but winding a transformer may be hard. <A> Use of capacitor's to replace a transformer is not new. <S> One can obtain any output of desirability just by changing capacitor values. <S> You are really dealing with Time-Constants of the capacitor/load resistor. <S> I recall one television maker using a capacitor input to replace a transformer, (about a 200 watt load) and it worked well. <S> Effiency is very high, higher than a switching power supply, so long as the dielectric of the capacitor does not heat up! <S> (due to wrong dielectric and/or input frequency involved.) <S> There is no regulation here. <S> This system is also used today in small multi LED table lamps that are very bright by overloading the LED into a short life! <S> It too uses a 2.2 ufd capacitor. <S> The value is too high! <S> Reducing this cap. <S> from 2.2 ufd to 0.47 ufd reduces the table lamp brightness to a more realistic value without pushing the LED to destruction! <S> I also got rid of the smoothing 20 ufd capacitor on the bridge rectifier outputs as this only increases the output voltage to a maximum of 1.41rms volts, which is not needed. <S> This is driven by the full ripple output of the rectifier but is not seen as a human eye <S> cannot see this, again due to the time constant of the eye-ball! <S> This is a good system to drive LED's of any array, usually 36 LED's as one series string. <S> But can be adapted to parallel strings, but not really necessary! <S> No regulation is needed on an LED, its only a resistor! <S> Small voltage variations cannot be seen. <S> And this power supply uses no electrolytics or regulation which spells extremely long life for the lighting system. <S> Cap's do fail due to very high current demands which drives up internal temperatures. <S> (as in switching power supplies) <S> If this is a problem, use multi cap's in parallel to reduce current load on each cap.
| If an iron transformer based supply is too large for your application, see if you can find a complete SMPS solution, like a USB charger.
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RPi GPIO relay interference with input I'm trying to use a RPi3 to output/control a lamp using this relay module. I also want to be able to read from a push button and a switch. The problem I'm having is that it looks like switching the relay is introducing noise in to the GPIO input and creating false positives. The RPi and relay module are in a outdoor enclosure 190mm x 145mm x 140mm. They are right to each other. I tried using a solid state relay and that solved the problem of the interference, but I would prefer to use a conventional relay because of size and pricing. I have tried to see the noise picked up by the GPIO using a cheap Hantek digital USB oscilloscope, but unfortunately wasn't able to see anything. Still waiting to check with my analog oscilloscope. In addition, I'm thinking about trying shielded cable from the pushbutton and switch to the GPIO and also wrapping the relay module in grounded metal mesh to create a Faraday cage. Till now I've tried putting a low-pass filter for each inputs. The first circuit I tried was: C1 - 10n and 100n simulate this circuit – Schematic created using CircuitLab Then I tried a low-pass filter: R1 - 100 and 1k C1 - 1u, 10n, 100n simulate this circuit Also I tried with GPIO pull-up on and off In any case none of them improve the result. On the software side of things I'm using Python, RPi.GPIO event_detect; I tried increasing the bounce time and inserting a delay after event-detect to "filter" human generated events, but this would only work for the push button case, not switch. Any advice would be appreciated. UPDATE : This Relay Module insert a lot of noise, I tried with the RPI HAT slice of realy and tha same code worked fine, could detect any false positives on a quick test run. this is the code: #!/usr/bin/pythonimport RPi.GPIO as GPIOimport timeGPIO.setwarnings(False)GPIO.setmode(GPIO.BCM)GPIO.setup(24, GPIO.OUT)GPIO.setup(26, GPIO.IN, pull_up_down=GPIO.PUD_UP)GPIO.setup(20, GPIO.IN, pull_up_down=GPIO.PUD_UP)GPIO.setup(23, GPIO.IN, pull_up_down=GPIO.PUD_UP)def read2(channel): print "GPIO 2"def read20(channel): print "Button"def read23(channel): print "Switch"if __name__ == '__main__': GPIO.add_event_detect(26, GPIO.BOTH, callback=read2, bouncetime=300) GPIO.add_event_detect(20, GPIO.BOTH, callback=read20, bouncetime=300) GPIO.add_event_detect(23, GPIO.BOTH, callback=read23, bouncetime=300) while True: pass Regardless I'm still looking for how to be able to use that board. <Q> I've faced exactly the same problem and just before going insane, I found the root cause and the solution. <S> To overcome this problem, just put a 10ms sleep delay in your callback definitions and read the pin status again; if still same than perform action, otherwise it was just a spike. <S> Capacitors can help, but have to be designed in a specific way so the timing is perfect, also they have a different loading and unloading behavior, so this is rather difficult! <S> Please have a look at: https://www.raspberrypi.org/forums/viewtopic.php?t=134394 for some real proper information! <S> Hope you will succeed! <A> You should be able to debounce your switch in software. <S> It is always a good practice to "pull" the input in the opposite direction of your switch action using a 10k ohm or so resistor. <S> So I would try placing a 100 uF capacitor or larger directly across the power supply terminals of the relay board. <S> This will help "hold up" the 5 volt supply to the board when the relay is energized. <A> I am not sure whether it applies, but there are multiple possible problems with these relays. <S> They have 5 V coil, thus requiring level shifting from RPi's 3.3 V GPIO. <S> The coil resistance is only 70 Ohm, which would cause to draw large current from the GPIO pins, possibly destroying it. <S> This could be solved together with p. 1 <S> A Relay needs a flyback diode to protect GPIO pins from potentially destructive EMF. <S> Unless the relay module provide for them (couldn't find the module data sheet), you must do it yourself. <S> Now, given that the module is designed for Arduino, which, AFAIK, has 5 V GPIO, there is high probability that at least p. 1 is on spot. <S> That is, the module expect 5 V as control voltage and trying to control it with 3.3 V RPi GPIO leads to unstable operation. <A> At last a 100uf cap on the relay board between VCC and GND solved the issues of false positives. <S> No level shifting needed. <S> Thanks to all. <S> UPDATE : <S> False alarm! <S> was testing whitout a load (LAMP) <S> it dosnt work with the relay module or RPi HAT. <S> I'm starting to believe that I'm traying to do something imposible. <S> think I only have the options of SSRs and/or software "debouncing" left...
| Regarding the relay switching transients, it appears that there are snubbing diodes across the relay coils. Your problem is that the edge detect is very, very, very, very sensitive for voltage spikes. You probably will see false detections even when you switch on or off other equipement in your house as well.
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Simulate bad cell phone reception in the lab I'm currently developing an internet of things device that utilizes a uBlox SARA-U260 gsm/3G modem. In field testing, we have had significant software/firmware issues due to poor cell phone reception in our deployments in sub-saharan Africa. I'm having a hard time duplicating the problems in the lab because the cell reception here is too good. Even if there is no antenna plugged into the modem, it is still able to connect to the cell network internet! So, I'm looking for the best way to thwart my modem into thinking it has bad reception. One thought I had was to cut the coax cable of the antenna and solder a resistor between the gnd shroud and the center conductor. Would this work? What size resistor would make sense? Probably a fairly low-value resistor (say 50 ohm?) I'm confused about why the modem can still connect even without an antenna. I thought that during normal operation the modem must short the ground and center conductor when it is transmitting (hence the high current associated with a transmission burst). Under normal circumstances, is the short locate inside the antenna? If so, wouldn't the center conductor of the coax always be at ground potential (i.e. during reception)? EDIT: Thanks for the replies. I have a big metal box sitting on my desk ready to mount the DUT (thanks for that acronym @Ali Chen ) tomorrow! <Q> Normally, the WWLAN testing and debug is done in electromagnetically shielded cages in labs when external tower signals (or local 3G-4G repeaters/re-translators) are too strong. <S> The tests are usually done with special instruments like Agilent LTE tester , which has all abilities to change signal levels and simulate weak reception. <S> In normal operation the LTE tester is connected in place of antenna on the Device Under Test (DUT), bypassing the DUT's own antennas. <S> Alternatively, you can connect a simple onmidirectional antenna right to the LTE Tester output, and use the DUT natve antenna, but then you will need to control the distance and do re-calibration of the RF channel. <S> Keep in mind that the issue might be not in your software/firmware, but in HARDWARE. <S> Wireless devices are normally using Automated Gain Control not only in receiver channel, but also in transmitter channel. <S> When the reception is weak, the DUT rightfully will assume that it is far away from cell tower, and therefore MUST use higher transmit power. <S> The power of transmission bursts can be up to 1-5 W, which will interfere with board layout, and if the PCB design is done poorly, it can disrupt all other DUT functionality, all other sensors, etc. <A> They are composed of multiples resistances building a network that will attenuate RF power very predictably, both in RX and TX, while maintaining the proposer RF impedance as seen from the modem and antenna. <S> They have a limited power rating. <S> Do not run your modem with its RF output open or short-circuited. <S> The RF amplifier won't like that, at all. <S> Small attenuators with SMA plugs typically cost around 10-20€ ( https://www.minicircuits.com/WebStore/dashboard.html?model=VAT-3%2B ). <S> The SARA-U260 can radiate a maximum of 33dBm (2W) peak, less on average. <S> Start to dissipate the bulk of the power with a 2 or 3dB attenuator (rated for 1W) connected on the modem side <S> and then you can add more attenuators to decrease power even more and finally plug you antenna. <S> Systems without antennas can behave erratically, and attenuators are not a great substitute for antennas. <S> For large attenuations (> <S> 40 dB of attenuation, I would say) <S> stacking attenuators is not a solution as some RF power always leak in and out of a modem by ways other than the antenna port (small sections of RF trace on your PCB, power supply, etc). <S> Then, you need an RF shielded box (such as http://www.jretest.com/jre-0709-P.htm ) with some attenuators inside the box, and some attenuators outside the box. <S> Your modem is put inside the box, your antenna stays outside, and the box is equipped with RF connectors to bring signals in and out of the box in a controlled manner. <S> If you have a budget for it, half a dozen of fixed attenuators, a shielded box, and a variable attenuator (eg. <S> https://www.aliexpress.com/item/2Watt-0-90dB-Coaxial-Adjustable-Key-Press-N-K-K-RF-step-Attenuator-Stepping-DC-to/32779942411.html ) can be very convenient to simulate dynamic RF conditions in a controlled manner. <S> If the poor reception is due to nearby sources of interference, you will need a RF signal generator (plus an RF "tee" and maybe an isolation box) to emulate that issue. <S> This is not cheap. <A> RF is tricky. <S> I wouldn't mess with the antenna, I think the product should be tested in its final configuration. <S> The reason for this is simple: when reception is really bad, your transmitter will increase its power to its maximum value, and you WILL be interested in knowing if the (rather enormous) amounts of RF it will crank out will crash your micro, corrupt your analog sensors, or whatever other black magic RF can do on innocent bystander circuits. <S> From the point of view of the chips on your PCB, your GSM module cranking out a few watts of RF is like a mini nuke... <S> way more than enough to any opamp go bonkers. <S> You really, really want to test for that, and modding the antenna for lower efficiency <S> will NOT achieve it! <S> I suggest putting the whole product inside a Faraday cage, like a microwave oven, a pressure cooker, or something like that: <S> Also available in see-through... <S> Now, if you run wires through your Faraday cage, then the RF will ride on them, and you'll need to use proper feedthrough filters, and basically Do It Right, which I guess isn't the point. <S> As an alternate solution: do the test in your building's basement. <A> Bad reception can not only stem from low signal, but also from the distance to the cell tower. <S> Especially GSM is only designed to go over about 35km. <S> Otherwise the "timing advance" between sender and receiver of the RF signal would be too large for the protocol in use. <S> See <S> https://en.wikipedia.org/wiki/GSM and other resource for more information about that. <S> This is hard to simulate without proper equipment. <S> You should probably hire a lab for this. <S> Large telecoms companies have such testing capabilities. <A> Some possibilities: Connect the antenna to a dummy load <S> Use a step attenuator between the antenna and the radio using double shielded coax <S> Note that a proper attenuator is not a simple resistor. <S> It consists of a resistive pad that has impedance matched inputs and outputs. <S> Look up T pad resistive attenuator. <S> Most are specified as dBs of attenuation. <S> Stepped versions with > <S> 100 dB of attenuation are readily available as lab accessories. <S> Use them with double or quad shielded coax to reduce leakage paths. <A> In field testing, your device is affected by environmental (technically geographical ?) <S> issues, not issues with the device itself. <S> So why not just isolate the device with some physical blockage such as walls or metal boxes( from the replies above). <S> Some buildings have built in telephone antennas (whatever you call them.) <S> so maybe try out your device somewhere else?
| To "simulate" a poorer cell reception you can use RF attenuators between your board and the antenna. Conduct your experiments in a shielded room / enclosure
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What set the speed in a cassette tape recorder? I was wondering what set the rotational speed in a cassette recorder? I assume that the speed must be constant, but that means that the tape data would have different effective densities depending on where you were on the tape. Was it a stepper motor and a timer to create the rotations or was something more interesting used? <Q> The hubs do not rotate at constant speed in normal tape recorders - there is what is called a capstan that is kept in contact with the tape by a rubber pressure wheel. <S> The capstan rotates at constant speed and so drives the tape at constant speed. <S> The take-up reel is usually driven through a friction wheel so that it keeps reasonably constant tension on the tape but the hub speed can vary as needed. <S> The variations in quality for voice are usually acceptable. <S> The standard for cassettes is where the magnitude of the magnetic flux on the tape represents the instantaneous value of the signal being recorded. <S> There are a couple of techniques used to improve the quality of the signal played back: 1) <S> A high-frequency AC bias is added to the signal being recorded to avoid the non-linearity inherent in the flux recorded vs the current in the recording head, without this there would be a non-linear response around zero. <S> A cheaper alternative where low quality can be tolerated is to use DC bias, that will give a lower signal to noise ratio on playback with more hiss. <S> 2) <S> Another technique is to use equalization where high frequencies are boosted in recording and attenuated on playback to improve the overall signal to noise ratio (i.e. reduce hiss). <S> There are standards set by the RIAA so that tapes are interchangeable between different machines. <S> Some also use noise reduction techniques such as Dolby Noise Reduction. <S> The motors used for driving the tape were usually brushed DC motors with either a centrifugal governor (a small weight on a spring that open contacts at a defined speed to slow the motor) or by using the back-emf of the motor to sense and control the speed. <S> (See another question I answered DC Motor speed control .) <S> Stepper motors were never used in conventional cassette recorders. <S> They tend to be inefficient (i.e. consume more battery power), have unsteady speed (e.g. cogging) and are not so easy to drive as DC brushed motors. <A> In almost all tape recorders (cassette, open reel audio, various video cartridge types, 2" Broadcast videotape, etc...) <S> the tape is driven by a capstan and pinch roller, not by the tension from the take-up reel. <S> In audio cassette recorders, the takeup reel is probably driven by a friction drive from the capstan motor. <S> In professional open-reel machines, there are often a separate motors to drive the takeup reel and to maintain tension on the supply reel. <S> These motors and their drive systems are designed to maintain a suitable tension on the tape during record/play operation but can produce higher torque for fast forward or rewind operations. <A> I seem to recall someone (Colecovision?) had a cassette recorder without a capstan, where they dragged tape across heads using fairly precise control of the drive reels. <S> The drive mechanism was surprisingly naked, with much of the typical guts of a cassette transport just conspicuous in its absence . <S> It was obvious they had reduced production cost considerably. <A> The repair shop I worked in used a special calibration tape to set the speed precisely. <S> It had a number of prerecorded tones on it used for various adjustments and measurements. <S> The tone was monitored with a frequency counter which allowed precise speed adjustment using the technique described by Harper above. <S> Other sections of the tape were used to measure frequency response and adjust the head azimuth (tilt) ensuring good high frequency response. <S> Wow (slow speed variations) and flutter (fast speed variations) could also be measured if the right equipment was available. <A> The speed for a conventional audio cassette is set at 1 7/8 inches per second. <S> As others have said, this is by controlling the rotation rate of the capstan. <S> The take-up spool is on a friction drive, and runs at whatever speed it needs to in order to take up the tape.
| Some tape recorders such as ones for voice recording often do drive the tape by rotating the take-up reel at a constant speed - in that case the recording density will change depending upon how much tape is on the reel. The speed was set using a section on the tape recorded with a 1 kHz tone.
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How to convert a range of 0.5 to 4.5V for a range of 0~3.3V? ACS712 20A current sensor The exact part number is ACS712ELCTR-20A. Considering it is supplied at exactly 5V, it's output will present 2.5V when there's no current flow on its main pins. Output offset = VCC/2. As its sensitivity is 100mV/A: When instantaneous current is 20A on a given direction flow (-), its output will present 2.5V (offset) - 20x100mV => 0.5V. When the instantaneous current is 20A on the oposite direction flow (+), its output will present 2.5V (offset) + 20x100mV => 4.5V. Is there any way to convert this range (0.5~4.5V) to a range of 0~3.3V just by using a single (or a max of 2) OpAmp ports? When input = 0.5V, output = 0V. When input = 4.5V, output = 3.3V. When Vin < 0.5V, Vout = 0, when Vin > 4.5V, Vout = 3.3V I already simulated the circuit described below, but it would require a good PCB area. I'm looking for a simplier circuit which would require less area. It would be great if that 'simplier circuit' doesn't require the use of a negative power supply. ACS712 output (0.5~4.5) -> OPAMP1 [2.5V voltage subtractor] -> OPAMP2 [amplifier] -> OPAMP3 [precision full wave rectifier] -> OPAMP4 [voltage buffer] -> 10 bit ADC 1Msample max, GND and 3.3V references, of a MCU. I can't change the ADC references to 5V/GND. Max Vref+ is 3.3V. OPAMP1 and OPAMP2 is an IC supplied with +5V and -5V. OPAMP3 and OPAMP4 is an IC supplied with +5V and GND. This is the only signal read by the ADC in the whole circuit. I can add a second ADC channel to read ACS712 supply voltage by using 2 resistors of 1% also. Regards. EDIT: I'll try to use ACSxxx sensors for 2 applications. Some months ago I produced some PCB prototypes and included some circuits for tests on the same panel. I have a circuit (schematic below) for possible tests related to the question of this topic. I didn't had enough time to test it yet. The circuit below is for application 1. This is a future application... Here I just need to estimate the AC current consumption of the load (AD/DC switched supplies or smartphone chargers for example), without precision. Using a 8-bit MCU which can do 16MIPS, if possible using little processing in firmware. For this project I'll use a 5A sensor (ACS712). Sensor and MCU are supplied at 5V. ADC ref = 5V/GND. The sensor should withstand an AC supply of up to 250VAC and have an electrically isolated output. Application 2: The current application. Here the MCU needs to be able to know if there's or not current flowing through the load (inductive) and maybe (not required) detect some possible over-currents on it. The project uses a 32-bit MCU which can do 40MIPS. Currently, MCU is supplied with 3.3V and sensor ACS-712-20A supplied with 5V. At sensor output I'm using a 5.6K/10K divisor, which is read by the MCU and I do the job in firmware.After some answers on the topic, I see that I can try other ACSxxx sensors instead of ACS-712-20A, one which could be supplied with 3.3V. The sensor should withstand an AC supply of up to 250VAC and have an electrically isolated output also.The circuit is minimal in this application. I just wanted to check what could be done in terms of hardware. <Q> A basic schematic is the following, <S> The transfer function maps 0.5 to 4.5 to 0 to 3.3. <S> Though I would suggested limiting the output swing a few 100 mV above/below the rails. <S> Edit: Showing thevenin of ADCREF/2, Scale resistors as needed. <A> Do you want to connect a ACS712 current sensor to a 3.3V processor (for example an Arduino board) ? <S> In the datasheet there is a "Application 4" that shows how to scale it into 3.3V range. <S> You don't need the rectifier with the diode, so only two resistors are needed to scale the output voltage. <S> I suggest 4k7 (or 5k6) and 10k. <S> Have you tried a ASC712 ? <S> It is noisy and sensitive for magnetic influences from wires and transformers. <S> For low voltages, the high side current sensors are more accurate. <S> For example a INA169.For high voltages, for example the mains voltage of 110V or 240V, most ACS712 modules don't even meet the specifications for that. <S> Some work with 3.3V and 5V. <S> For example this one: Pololu current sensor <S> Have you heard about the XY-problem ? <S> xyproblem.info <S> You are trying to make the ASC712 work with a lot of effort, while only two resistors are needed, and other current sensors are more suitable. <A> Use acs711 (3.3V range) . <S> It's true that all of those suffer from noise, but for low cost applications you will not find anything better. <A> Do you really need to do this? <S> These current sensors are not precision devices. <S> Error and non-linearity are specified at 1.5%. <S> That means you don't really have to worry about extracting the most dynamic range from your ADC. <S> Just use a pair of resistors as a 3:2 voltage divider and (if needed) <S> add a unity gain opamp buffer. <S> Let the zero offset and scale be what they will be, and adjust for it in software. <S> simulate this circuit – <S> Schematic created using CircuitLab
| If your outputs are low-impedance or buffered, than you can use a differential amplifier. The ACS712 is for 5V (4.5 to 5.5V). Other ACS current sensors are available for 3.3V. All your calculations and simulations do not show what the ASC712 is doing in a real project. The sensor is always going to be the limiting factor.
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Purpose of a capacitor between op amp inputs in buffer circuit I have stumbled upon the below circuit. At first I thought it is a voltage follower, but I can't understand why there is a capacitor between the positive and negative inputs. Can anyone please explain to me the purpose of the capacitor in this circuit? <Q> At DC, and low frequencies where the gain and speed of the amplifier is sufficient to keep its inputs at more or less the same voltage, it does nothing, as it's 'bootstrapped out'. <S> Any amplifier has a finite bandwidth. <S> If a high speed step is applied to the input, then for a moment, there will be the full step voltage across the inputs. <S> Some op-amps misbehave under these conditions. <S> A small capacitor across the inputs like this reduces that effect. <S> Most applications of op-amps do not have very high speed inputs, so do not need this protection from them. <S> It's usually better to explicitly filter any inputs, rather than use this crude method. <S> However, in the case of an emergency retro-fit, putting a small C directly across the inputs is a convenient improvement. <A> Such a capacitor between both inputs (very often, in series with a small resistor) is a kind of external lag compensation . <S> This is a simple method to stabilize an opamp (with strong feedback) that is not unity-gain compensated. <S> The explanation is simple: For rising frequencies the feedback factor is reduced and, hence, the tendency to oscillate is reduced. <S> As a consequence of this compensation method, the usable closed-loop bandwidth is reduced considerably. <S> This kind of compensation can an also be used for unity-gain compensated opamps with strong feedback to improve the phase margin (step response with smaller overshoot) <A> Both Op Amp output and source have a current limit and bandwidth. <S> Thus If you have a voltage differential at some frequency <S> can you see how voltage is limited with rising f by passive C due to source impedance which is a function of frequency in negative feedback GBW limited unity gain <S> Op Amp (Zo rises with f) and some unknown source with stray noise.
| In circumstances where there is RF pickup on the input leads, which can also make op-amps misbehave, a small capacitor here can also improve things.
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Resistance of a fully conected graph? How would one approach to solving for the resistance between any two nodes of a fully connected graph with an arbitrary number of nodes n and arbitrary resistance between them ? (resistors omitted from image for clarity) <Q> The equal-resistance case <S> In the case where all resistances are equal (call it \$R\$), the solution is straightforward. <S> Without loss of generality we can label the two nodes we are measuring between as node 1 and node 2. <S> We consider a voltage applied between the two nodes. <S> We can re-draw the graph with the entry and exit nodes separate, and the others in a line: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Here I have omitted many resistors which joins nodes 3 to n with each other. <S> However, by symmetry, it should be obvious that nodes 3 to n are all at the same potential, so no current flows between them and the resistors joining them are irrelevant. <S> It should now be obvious that the resistance between Node 0 and Node 1 is the parallel combination of R and N-2 instances of two resistors in series. <S> Thus the total resistance is: \$R_{1,2} = \left(\frac{1}{R} + <S> \frac{(n-2)}{2R}\right)^{-1} = \frac{2R}{n}\$ <S> The arbitrary resistance case <S> There is (as far as I know) <S> no simple solution. <S> For a graph with \$n\$ nodes, you will get \$n\$ current conservation equations, and \$n^2\$ resistance equations. <S> You now have \$n^2+n\$ equations, and the \$n^2+n-1\$ unknowns, so you can solve the system of linear equations. <S> This may be quite time consuming. <A> Node voltage analysis? <S> Make an equation for each node such that all the currents flowing in total 0. <S> Solve the set of simultaneous linear equations. <A> I don't think that there can be single general analytical solution. <S> With n nodes, you'll have either n/2 or (n-1)/2 possible resistances depending on symmetry which forms a discontinuous solution. <S> So you'd have at the very least two sets of equations for odd and even numbers of nodes. <S> More as you navigate the periphery. <S> And that's for all resistances being equal. <S> If the resistances are arbitrary, the number of discontinuities increases for each different resistance. <S> This makes the whole thing totally unwieldy. <S> The best approach is to consider all resistance equal and use a numeric simulation. <S> It's fairly easy then. <S> Simulate 4 nodes with your favourite simulator. <S> Ground node 0. <S> Apply voltage to each node in turn going around the periphery till hitting the symmetric point, dependant on evenness. <S> So for n = 4, measure current n01 and n02. <S> Calculate equivalent resistance. <S> Plot total nx along x <S> axis. <S> Plot resistance along y axis. <S> Numerically fit a curve to the results for n = 4. <S> Loop and repeat for all 4 < n < 16. <S> I think for n > 15 results will become asymptotic. <S> Colour graph to taste. <S> So you'll have 12 curves. <S> You might also be able to convert this graph to 3D, but obviously you wouldn't be able to form a continuous surface. <S> This is your best way forward. <S> You can see that this is all cumbersome. <S> Now imagine this extended for multiple arbitrary resistances. <S> You wouldn't be able to stop at n = 15 either. <S> Your original exercise becomes infeasible. <S> PS. <S> I've excluded n = 2 as trivial, and n = 3 as atypical as there are no resistors passing though the interior of the hull.
| You must assume some voltage between the nodes of interest, and then write down ohms law for each resistor in the graph, and current conservation equations for each node.
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Serial Monitoring of DB25 Parallel Port Data I would like to examine the string of data being outputted from an industrial equipment's DB25 connector that is being sent to a printer. I was thinking I need to convert this 25 pin parallel connection to USB type A, which would then be plugged into my PC and then examined using a serial monitor. My question is what kind of hardware could I use to convert this parallel data to USB? Do I only need the data 0 - 7 pins and ground pin on the DB25 connector to analyze the data on my PC? Is there any arduino library out there that could convert this data? I appreciate any help. This is my first time dealing interfacing a DB25 connector to USB. I've done some research on the parallel communication and seem to have hit a wall. <Q> Tap it like the old serial break out box days. <S> Either make yourself a Y cable to repeat the pins to another 9 pin connector, or buy one of these:- <S> Then you just read the 8 data pins with a Arduino as a port read instruction (say PINC). <S> You may have to play with the wire order to get sensible codes. <S> Then send the byte over the Arduino's USB cable (using Serial.print(byteValue) ) to the PC running terminal emulation software like SCREEN or PuTTY. <S> Try to run the printer at a low speed, say 10 kBaud. <S> The Arduino software is surprisingly simple. <S> I estimate 10 - 15 lines plus a quick interrupt service routine. <S> You'll find that in order to discriminate one byte from another passing along the cable, you'll also have to access the /STROBE line on pin 1. <S> If you connect this pin to the Arduino as an interrupt, you should be able to read the 8 data bits on every strobe signal. <S> The old Centronics ports ran fairly slowly so the additional impedance of the Arduino's input ports and Y cable shouldn't load the industrial equipment's printer port. <S> The interface was 5V so get a 5V Arduino to avoid level issues or catastrophic explosion of a 3V one. <S> Mixed signal oscilloscopes are good for this too. <A> You don't just "convert" a DB-25 to USB. <S> A DB-25 is just a connector. <S> USB is a set of connectors and a whole protocol. <S> USB is much more complicated. <S> You can't just connect the right pins on the DB-25 to a USB connector and plug it in. <S> Then there is the question of what protocol is being transferred over this DB25 connector. <S> This might be a old "parallel port" printer interface. <S> In that case, it's 8 data lines with a few control lines. <S> Look up the parallel port spec, then probe around on your 25 lines with a scope and see if it seems to match. <S> Once you know the protocol, it would probably be easier to sniff the data by programming a microcontroller to do the low level reading, then send the data on via a UART or something. <S> The low level electrical protocol likely has some timing constraints, which will hard, if not impossible, to meet with a PC running user code on a modern operating system. <S> There are micros with 8 bit parallel interfaces built in. <S> It shouldn't be hard to harness that to read whatever 8 bit parallel interface you have, assuming that is what you have. <S> You may need a little glue logic to get the right polarity strobe and the like, but it shouldn't be too hard. <S> However, again, the first step is determining what each of the wires do, and what the low level protocol is. <S> Only then can you find or make something to capture the data as it is going by. <A> I've done it a year or two back. <S> I used a Pic32MX. <S> Basically there is a strobe signal that you use to get the 8 bit data from the port <S> - I used this to trigger an interrupt. <S> The data went into a buffer which was eventually written as a file to a USB stick. <S> It worked fine - but be aware that you will need to do some tricky conversion to turn the data into an image. <S> It worked fine for me, but I used a piece of software called PrintCapture which it seems is no longer available. <S> I am actually now investigating doing the conversion to bitmap on teh PIC, but it means parsing the printer language and writing to an image file, which may take a bit of work. <S> As far as I know there are no FOSS libraries for this <S> (if I am wrong I'd be very interested to know!) <S> It could also be a nice job for a Raspberry Pi or similar, which has enough IO pins and a lot more memory. <S> There are quite a few bits of old gear out there that have parallel printer ports so this might be quite an interesting project.
| Just parallel plug it into the Arduino. Essentially you'd build a logic analyser, which you could also just use if there's one lying around.
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Can a 30V USB power a 12V light strip safely I'm a beginner in this field. I'm making a desktop LED Sign, for that I'm using a 12V LED Light Strip. I have a standard USB Cable from an old phone charger, it says 30V on the cable, so if I connect that cable to the LED, and power it from a USB port on my PC, will the LED break or get ruined? Also another question, on the right side, it says +12V so does that make a difference if I power it through a 30V cable? <Q> USB ports (usually) output 5V. <S> The "30V" on the USB cable just means that the wire is rated for 30V, not that it will actually supply that voltage. <S> So, in short: no. <S> If your LED strip needs 12V, you need to connect it to a 12V power supply, not a USB port. <A> You need 12V,0V and Logic Data in DIN to address the serial data chip. <S> There is a separate 0V for DIN. <A> Keep in mind that current goes down as voltage goes up (so if your supply is rated 5V @ 1.2A, you'll get less than 0.5A at 12V from a converter). <S> Keep in mind that this particular LED strip has a microcontroller (the little black chip) and a DIN line, as Tony mentioned in his answer, so <S> how it behaves without data on the DIN line is anyone's guess. <S> The strip may not light at all, or (and I'm assuming this is your goal) <S> it may just come on full brightness with no control, which isn't a terrible thing in and of itself.
| As @dandavis suggested in the comments, if you really want to power it from USB, you can get a 5V->12V step-up converter to get 12V from your USB supply. For that small strip of LEDs you show, this should be sufficient, but do not try to run a six-foot length of that stuff off your computer's USB port -- your computer is gonna have a bad day (in all probability the strip just won't light, but you could pop an internal fuse on your USB port, which may or may not be self-resetting; if it's not, hopefully it's a discrete component that can easily be replaced by someone with the proper electronics/soldering skills).
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Isn't my ground different from the factory's ground? My understanding is that ESD safety things (mats, wrist straps, specially marked soldering irons) are designed to bring everything that can touch a component to the same electrical potential energy – ground. But it seems unreasonable to expect that there's no voltage between my desk and the factory where my components were produced. After all, the factory is likely halfway across the world, and the resistance between here and there is significant. So, say a component is carefully packaged and shipped to me in one of those little ESD-safe bags. Before opening the bag, I carefully ground myself and my workstation. Despite this, the component is destroyed as soon as I touch it because the ground that I tied myself to is much different from the ground that the component was tied to when it was produced. What precautions are taken against this? Is it just something that can happen in theory but that isn't an issue in practice? <Q> Components are damaged by two or more of their pins being at a large enough potential difference. <S> If the component has a conductive case, or pad, then that counts as a 'pin' too. <S> It's possible to break them by trying to charge them up to a new potential through one sensitive pin, while the voltage of the other pins is held more or less constant through capacitance to ground. <S> That can be the situation where you, perhaps charged to 15kV with respect to ground, pick up a component that's at ground potential by (say) the gate lead. <S> Conductive packaging shorts all the pins together. <S> What you do is to bring the conductive bag to your potential first. <S> Any charging current that has to flow into the component does so through all pins, so does not damage the component. <S> Let's say an insulated carton of components in conductive bags charged to 100kV arrives at your workstation. <S> You and the workstation are grounded. <S> You open the carton, and as soon as you touch a component bag, a current flows between you and the bag to discharge it down to ground potential. <S> Meanwhile, the bag has maintained all the component pins at the same potential, so no damaging voltage is applied across the component. <S> Now you and component are at the same potential, you can open and touch. <S> Why did the component arrive at 100kV? <S> Surely the other factory ground is not that different to yours? <S> No, but the last bit of the trip might have been carried by a guy with nylon shoes. <S> When stuff is properly packed, it doesn't matter if intermediate stages of the journey take it to potential way different from ground. <A> Hopefully your parts are packaged in an ESD-dissipative tray or bag. <S> Then when you set them down on your ESD mat in your lab, any charge that's built up on them can drain away through the packaging and the mat. <S> They won't discharge quickly enough to damage the components because both the bag and the mat have substantial resistance (1 megohm to ground is common for ESD mats and wrist straps). <A> My understanding is that ESD safety things are designed to bring everything that can touch a component to the same electrical potential energy <S> That's when you're right. <S> ground. <S> And here's when you're wrong. <S> There is no such thing as "universal ground". <S> Not even Earth is. <S> You just pick a point of a circuit and say "Hereby, by the power vested in me by Science of Electrical Engineering <S> , I declare you as The Ground and all other grounds as null and void." <S> and instead of a sword, you touch it with an ESD safety thing. <S> That's it. <S> As you've said, the ESD safe equipment is designed to bring the part safely (read: slowly) to your ground. <A> A logical extension of this would be sending a replacement PCB to the ISS or a satellite where "ground" in the sense of what you stand on is a long way away and separated by vacuum. <S> Large charges can build up in space , so there will be a significant potential difference. <A> Basic electrostatics: <S> If a bag or metal box or some other conductive container completely surrounds some object, the potential of that object is effectively zero, regardless of whether it's connected to earth ground or a 10KV power source. <S> (I suspect there's an exception for high frequency AC, but we can ignore that.) <S> When the object is bagged in the factory, the bag is grounded to the factory ground for the transfer. <S> It may go through several thousand volts fluctuation between there and where you are, but the potential between any two points inside the bag is still zero. <S> When you receive the bag you first ground it to your desk ground, then open it. <S> As your hand reaches in it is a zero relative to desk ground because your wrist strap assures that, and the component is at desk ground because the bag (and everything in it) is also grounded to your desk.
| In an electrical sense there's no problem so long as you bring the two grounds together properly (slowly through a significant resistance, and without causing a large potential difference across the circuit -- see Neil_UK's answer) It doesn't matter if your ground is same or 1000V different than ground they've used at the factory.
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How to electrically shift the phase of a signal? Let's say I have a sine wave and I would like to shift it by varying degrees, like 90, 180, and 270. How would I do this? <Q> An op amp integrator will give 90 degree phase shift. <S> Integrate again for 180 etc. <S> If you want, say, unity gain and you know the frequency, <S> \$\omega\$, then choose \$\small RC = <S> \large \frac{1}{\omega}\$ <A> It depends - on the frequency, the bandwidth you need to operate over, whether you want a phase shift or a time shift (different when a wider band is involved), whether you are generating the signal yourself, and quite a few other things. <S> A classic way is to build a low pass filter including varactor diodes, and vary the voltage on them. <S> Another is to use a quadrature hybrid, with a pair of varactor diodes as the loads. <S> Another is to digitise and regenerate the signal with an ADC and DAC pair, with delay or more complex DSP processing in between. <S> And quite a few more, tapped delay lines <S> , all-pass networks ... <S> These will have different domains of applicability, and different strong and weak points. <S> Pick your operating situation, and pick your method accordingly. <A> if it's a pure sine wave, then an easy digital way is to multiply the signal by: $$ e^{i\phi}$$ to achieve the desired phase shift. <S> This can be done without the need of huge banks of memory. <S> But as a consequence you need a fairly fast processor to do this in real time (frequency dependant of course).
| Another is to put the signal into a pair of SSB mixers, and change the local oscillator phases.
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increasing impedance of headphones I have headphones with differential input and impedance of 8 ohms. I need to drive this headphone with some DAC with built in headphone driver (Currently I have chosen ADAU1772). DAC's datasheet mentions that the minimum impedance of headphones it supports is 16 ohms.Can I increase impedance of my headphones so that I can drive them through this chip? What will happen if I drive this 8 ohm headphone through a chip supporting minimum impedance of 16 Ohms? <Q> You have several choices: Get a chip that can drive 8 Ω. Put a buffer between the D/A and the headphones. <S> This buffer must then be able to drive a 8 Ω load. <S> The D <S> /A only sees the input impedance of the buffer. <S> Use headphones with 16 Ω or more impdance. <S> Put a 1.4:1 audio transformer between the D/A and the headphones. <S> The D <S> /A output will need to be 1.4x larger than before, but will see a 16 Ω load when 8 Ω headphones are connect to the output of the transformer. <A> Nominal 8 Ohms for headphones is not much, and will make the DAC deliver the double current (compared to nominal 16 ohms). <S> You can not do much about it, as the impedance of the headphones are from the coil. <A> I connected the speaker with 8 ohms resistors in series between headphones differential wires and dac output. <S> I was getting a hear able sound. <S> I removed resistors and connected the headphones directly, the voice got louder. <S> However my application was outputting only male voice and no music. <S> Dac was also not heating up while headphones were directly connected.
| If the impedance of the headphones were only a resistance, you could put a resistor in series (with power loss), but as the impedance is very frequency dependent, you will ruin the frequency responce.
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Switching a 220VAC/20A Load with a 5VDC Coil Voltage? I have a Rasberry Pi running inside a device, which I want to connect to a relay to switch a 220VAC/20A load. The load 'spikes' to 20A when the motor is starting up, but then runs nominally at 7.7A when the motor reaches it's max. I currently have an electro-mechanical relay rated for 10A contact current and am afraid that the 20A peaks from the motor startup (~7-8 spikes each for ~1.0 second) will cause the contact to 'arc and stick,' if not cause a worse failure. I do not want to use a SSR, bc of the heat dissipation requirements, if I can help it. I was thinking I might use the 10A-rated relay to drive a contactor switching the higher-power load, but Im not confident in that design. Thanks! <Q> Rather than guess at how much to oversize the relay, it is best to use a relay that is rated based on motor power. <S> The IEC motor standards include a rating system that rates relays on more specific application information. <S> In the USA, relays that are suitable for use with AC motors have a horsepower rating. <S> Look at my more specific answer to this more specific question: <S> Max allowed inductive load (pump) for a given relay <A> Unless the data sheet specifies the relay can tolerate such spikes, don't use it. <S> Using a under-rated relay is a fire and safety risk. <A> There's no reason why you can't use a small relay to drive the coil of a larger relay. <S> It may be simplest to use a second relay with a 220V AC coil, if you can't find a suitable 5V one. <A> Those relays, often called contactors or motor starters, have a nominal rating and an overload rating. <S> You won't find a 5V coil motor starter relay, so you will need an small 5V coil general purpose relay connected to the Rasberry Pi to switch the motor relay. <S> Do not forget the buy the large contactor with snubber (attachment), otherwise your next question will be why the Raspberry Pi keeps restarting and why your control relay isn't working anymore. <A> Your application targets directly a Solid State relay there is no question. <S> The SSR's need very little current to switch the load since they work with light to trigger (inside the puck). <S> You must always have a load for them to work properly and that is what you have, a motor. <S> Note that there will always be a very small current flowing throught the SSR/motor, typicaly 150mA.
| Use a motor starter relay. Also the relay may appear to work, but die after some time.
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Long range communication for a wearable device We are building a wearable device for low resource settings and are trying to figure out a way to be able to send the data collected over a long range say 5-6 miles at a base station/hospital. What would be the best protocol to use in this case? Also, since this is a wearable device we will have to take into account any possible effects on the body on using a particular kind of communication protocol. Apologies if this is vague. We are still fleshing out the final details and data transmission is a very critical piece which we are grappling with. <Q> You have a collection of options: Mobile phone networks: any of the options, GPRS upto 4G depending on cost your willing to pay and data <S> you need to transfer. <S> Probably the easiest and most reliable, assuming that the mobile networks are available in the operational area. <S> Some analogue radio communication. <S> I'm pretty sure that all of the frequencies that will give you the range are license controlled (in Europe, Asia and N America), but it might be worth investigating depending on how much you can work with the countries you plan to operate in. <S> High power consumption, license fees and large antenna are all issues with that. <S> Satellite phone: <S> lots of low level data communication takes place over satellite phones already, so I'm sure you could add something to that. <S> Advantage is they cover all the world (depending on you provider) as long as you're outside (some work inside, sometimes). <S> Power consumption is high, as is the required equipment cost. <S> Mesh network: <S> if you have enough of these sensors spread from where you need the data to the furthest node, you could use a mesh network of nodes, one every 20 metres or so, and link one to another to get to the data centre. <S> Clearly this is lots of nodes, lots of cost, and each node needs it's own power supply. <A> Standard solutions are just few:the Nordic packet-data transmitter, it cost about 1 USD and is capable UNDER 1 km; the LORA(tm) product, packet data also, capable for 3 km yet costs 10x more; the ordinary most prost transmitter, yet for 5-6 km You will need at least 5...15 Watts; The WiFi, yet such is the maximum maximorum ever gained between two very high cost machines with large sharp-directed antennas; and cellular phone networks, where Arduino phone sub-module costs few USD. <S> I would vote for the last solution, and if dislike, then for 15W prost radio. <A> A mobile phone network seems the most practical way.
| No doubt there are other methods, but looking at miles, you are best to use existing networks; mobile phones or satellite phones.
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How do I figure out the windings of a transformer I have a transformer. It is a custom wound transformer for a old linear powersupply. As is common, it has two primary windings for supporting 115V and 230V operations (or, 120V and 240V as is now more common. the powersupply is old). So, the options are the following: The owner of this powersupply bought the device in the USA and is now in Europe, so he wants to change the wireing such that this is possible. They desolered the transformer windings, but then got cold feet and stopped, worried they would damage the device (or themselves) if they rewired the transformer incorrectly. It was then handed to me and I now have to make sure it's done right. However, the transformer doesn't have the wires labled. I wondered, what is the correct way to verify what pins are what coil, and more importantly, what orientation the coil has. It's obvoius in this case what the primary windings are since they were labled on the PCB, and those are the only ones unsolderd. I also know what two pins belong to the same windings with a simple continuity check. What I'm now worried about is the "orientation". What I don't want to do is this: I have only limited tools to my disposal for now since I cannot use univerity instruments. All I have is a simple multimeter and an old analog oscilloscope. I was thinking about applying, with a sound card, a low-frequency (say, 1kHz) signal. Since it is so low voltage, and higher frequency as the transformer is rated for, I expect that the voltages on the secondary would be on the order of milivolts, not able to turn on any of the circuitry. I could then apply this signal to one of the windings. By then using two channels of the oscilloscope, one for each winding, I could see what way the windings are in respect to each other (if I have one connected the "wrong" way I expect a near 180 degree phase shift, otherwise almost perfectly in phase). Is this a good method? how is this done "propperly"? I would rather not fully dissasemble the powersupply to unsolder the secondary windings as well since this requires me to pretty much take the entire thing appart. <Q> A mains filament light bulb would be suitable (make sure it's filament, not LED or CCFL). <S> Or a mid power resistive load like curling tongs. <S> Whatever load you choose, it should not draw more current than your transformer input is fused for. <S> This does mean you'd have to connect things mains side. <S> If you're not happy doing that safely, then you shouldn't be inside with the transformer mains connections anyway! <S> Wired correctly, the transformer will produce a significant output, and the bulb will be dim or off. <S> Wired <S> incorrectly, the transformer will present a short circuit to the input. <S> It will produce no output, and the bulb will light at more or less full brightness. <A> Drive your sound card signal onto the secondary and look at the relative phases of each primary winding is my advice. <S> However if you did exactly this it would work: - <S> If you don't know how it was originally wired <S> then you can't do that. <S> If there is any doubt about the transformer's state or condition get a new one. <S> Mains AC voltages can kill and burn. <A> Apply 120 V to one of the primary windings, and connect ONE wire of the second winding to one wire of the first. <S> If wrong, you will get 240 V. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> I love when the quiet voices give the solution. <S> @LeonHeller nailed it with his comment to OP. <S> Backfeed the transformer secondary with lower voltage than it's rated for, then check to make sure sensible voltages are coming out of the primary.
| Perhaps the simplest way is to wire the transformer up and test the whole unit, with a current limiting load in series with a primary . Measure the voltage from the free end of the second winding to the other end of the first - if you have the polarity correct, you should measure virtually no voltage.
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How is the remote control working I am for the first year in electrical engineering so sorry for this question but i couldn't test it by my own. One of my lecturers said that the remote control of a TV saves the channel we choose, and that when we press the "last channel" button, the remote sends the channel and not the "return" signal. Is this true? Thank you all for your answers, they were all good ways to test it.I will inform my instructor about them! <Q> It might be true, but it could just as well work the other way around too. <S> Probably real systems do this different ways. <S> I know that my TV has non-volatile memory in it, since it powers up to the same channel it was shut off with, without the remote being active. <S> Here is a simple way to test this: <S> Use the remote to set the TV to channel A. <S> Then use the remote to go <S> to channel B. Bring the remote to a different closed room so that it's <S> IR can't get to the TV. <S> Select channel C on the remote. <S> Select channel D on the remote. <S> Bring the remote back to near the TV. <S> Selet last channel. <S> If the TV goes to channel C, then the last channel is stored in the remote. <S> If the TV goes to channel A, then the memory is in the TV. <A> One of my lecturers said that the remote control of a TV saves the channel we choose, and that when we press the "last channel" button, the remote sends the channel and not the "return" signal. <S> Is this true? <S> Almost certainly not. <S> Outside of complicated "smart" remotes, most remote controls are very simple devices. <S> They have one signal associated with each button -- the TV is responsible for deciding what to do with that signal. <S> They do not keep track of the current channel in any way. <S> For a remote control to function the way your lecturer described, it would probably need to be in two-way communication with the TV, so that it can be aware of things like what channel the TV was in when powered on, when channels are changed using buttons on the TV, what channels are available in your area… it'd be a much more complex design, with absolutely no benefit to the end user. <A> When a button is pressed a configured binary value is generated which is then combined onto a carrier wave which will then be sent to the TV unit through a IR LED. <A> Simple way to test this is to select a channel, change it, then hit "Last Channel" to confirm that it works. <S> Then remove the batteries from the remote for about 1 minute (just in case), put them back in and hit "Last Channel" again. <S> If the remote stores that info, it will do nothing, if the info is stored in the TV and the remote simply sends a command to the TV to revert, it will go to the last channel. <A> An additional practical reason why remembering the last channel is an operation that would be carried out by the TV is there are many situations where people may use multiple remotes and having a "Last Channel" per remote would be pretty annoying to the end user. <A> No. <S> Absolutely not. <S> The "Last/Return/Previous Channel" button is a unique function code that the TV interprets! <S> For the most common IR control protocols, Philips' RC5, Sony's SIRC or NEC, the remote is normally an application-specific integrated circuit (ASIC) with hardcoded TV codes for fixed remotes or a microcontroller with a reprogrammable eeprom for learning remotes (programmed via special codes or through a programming interface like JP1 or even IR learning). <S> The remotes have a single code programmed into each button, and that does not change without intentional reprogramming. <S> These codes are arbitrary for the remote and depend on the TV to interpret. <S> You can see an example for Insignia TVs here: http://www.getzweb.net/jp1/data_returns/DeviceEFCs.php?type=TV&devid=1204&webpid=++396 <S> Each of these codes are single function, unless you get into more advance macros <S> which are a group of codes assigned to the same button that get sent in order. <S> Unlike some of the dubious test methods listed above, you can actually decode the actual code sent by your remote . <S> All you need is a common IR receiver module and a microcontroller (MSP430, Arduino) or computer like the Raspberry PI. <S> Once you do, you will find that the "Previous" button will always send the same exact code, regardless of what channel you have switched to and when. <S> The remote does not actively track what channel or ANYTHING you do. <S> A different way would be to use a IR Learning Remote like the Harmony remotes. <S> If the previous button was an active copy of the last channels selected, then you could not clone the feature. <S> Further note, the On/Off button is typically a single code that the TV interprets as toggling the current state, but discrete On and discrete Off codes often exist for any given tv, as well as dedicated "Input Source X" button instead of "Next Input" type codes.
| From my understanding I don't believe that a TV remote saves any information as it has no memory.
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Non-mechanical load suggestions for a 12VDC / 6A PSU? I need to provide a load for the 12VDC/6A power supply. I used a small motor before, but something less mechanical (more reliable) like a light or (anything that's not mechanical) would be ideal. <Q> The typical test load is a light bulb, or a power resistor, like a high wattage low resistance ceramic resistor. <S> Just ensure proper heat sinking. <A> 12 V car light bulbs should give you a selection of power ratings you can combine to give your 6 A load. <A> How about using electronic load?If you change often the test load condition. <S> I recommend the electronic load. <S> Below is related products of electronic load in google link. <S> http://www.tek.com/dc-electronic-load/series-2380 <S> https://www.google.co.in/?gfe_rd=cr&ei=ifwxWcaXF4_z8wfdjJWADw&gws_rd=ssl#q=electronic+load&spf=1496448127996 <A> However, there is a problem with testing power supplies with them, their cold (switch-on) resistance is usually more than an order of magntitude lower than their hot (running) resistance. <S> This means that (for instance) a 50W 12v halogen bulb that takes ~4A running could easily draw 50A when cold. <S> Different power supplies react to overload in different ways. <S> If it has a constant current mode, then the bulb will eventually heat up. <S> If it trips, then it's a race between the bulb heating up and the over-current trip timer running out, it may be impossible to put the lamp directly onto the supply without tripping it. <S> The same goes for starting a power supply with the lamp already connected, some supplies are smart enough to tell when they're being asked to start into a 'short circuit', and will refuse. <S> As bulbs are so convenient (high dissipation, self indicating, and while the world is switching over to LED, cheap, stock up on your filament lamp offers before they disappear!) <S> it's worth finding some workarounds <S> so they can be used. <S> a) Use multiple low power bulbs, and connect them one at a time, so the inrush for each is manageable. <S> b) Use a string of several high power bulbs in series (you did stock up on these things when I told you, didn't you?), and short them out one at a time so the inrush for each is manageable.
| My normal go-to load is a filament light bulb, as many of the other answers have suggested.
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is my schematic correct? (Optocoupler with triac) I designed a pcb that contains an optocoupler and a triac to work as a dimmer for AC But when I give a high value to the GPIO corresponding the triac doesnt fire So I am looking for the mistake I did ... Is my schematic wrong ?? if its correct , what could be the problem on the pcb leading to this problem ?? How can I check it ?? I want to search for the problem but I dont know where to look <Q> It appears that you do not have sufficient current through the opto LED. <S> 15 to 50 mA is recommended to meet the transfer function. <S> First check that your uP can drive at least 15 mA from its output. <S> If so, then a series resistor of ~ 220 ohms should be used. <S> If the output cannot drive that much current, add a simple NPN or FET driver to control the higher drive current. <A> You can find more info in this application note from Fairchild Semiconductors about the MOC30XX family that I share with you: https://www.dropbox.com/s/hwek0h4thuyi5go/MOC30XX_AN-3003.pdf?dl=0 <S> About the firing problem I think Glenn has answered it in part. <S> Can you supply the minimum current needed? <S> I would activate the moc with a signal in low connected to the pin number 2, so you can connect the pin 1 to Vcc (5V) through a pull-up resistor of 220 ohms. <A> Yes, the optocoupler resistor should drop to approx 220 ohms, since it needs to turn on harder than what the 470 ohm resistor will do. <S> Also, you should have a snubber circuit on the Triac for any inductive load, such as a motor. <S> or compressor. <S> You might not need it with a resistive load, like a heating coil, but It can't hurt.
| About the schematic, at the first look, you should (at least) add a snubber circuit in parallel to the TRIAC, to prevent a quickly slew-rate to turn it on (turn the TRIAC or the MOC!).
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Comprehensive way to indicate error code using single LED I am implementing Altera remote update mechanism, and it does not work, configuration falls back to the factory one. I need to get the cause of it, and the only output I have is single LED.Remote update state machine outputs 5 bits ( page 35 , 111 in the table). My thoughts: Blinking number of times binary represents is bad idea :), thus LED should somehow display bit states; Circuit is having input clock, thus I can time the blinking for easy value readout; Separating start, 1s, 0s and end: display "message" in loop, starting with 1 second long turn on (start), then one short blink for 0 or two short blinks for 1 within the second, then pause of a second, then display next bit, and at the end of 5-bit sequence the start goes again (1 second long turn on). Is there any better way, or even standard and proven way in semaphore signalling , to achieve value readout effectively by the human? P.S. I do not know which tags for my question are the right ones! Please edit if needed. <Q> I have used a lock-in amplifier which would blink out its IP address using a single LED. <S> It would flash 1-10 times to encode each decimal digit, with gaps between digits and longer gaps for the groups. <S> It didn't work very well, watching and writing it down was quite error prone. <S> Fortunately you have 5 bits not 32 bits, <S> so it should be easier. <S> Suggestions: <S> You are trying to communicate 5 bits, not a 5-bit number, so sending it in binary (as you have suggested) is a good idea. <S> If you were sending a number from 0 to 32, with 32 different meanings, you could consider decimal or octal or similar. <S> Representing bits with long and short flashes might be easier than one/two flashes. <S> Five bits is not too many, so it should be possible for the user to remember all of them, without stopping half way to write them down. <S> This means they can go faster than you've suggested. <S> My initial solution would be to have five flashes, at half second intervals, then a one to two second wait, then repeat. <S> This means that a 0 would be represented by 100ms on, 400ms off, and a 1 would be represented by 400ms on, 100ms off. <S> If I was then using such an interface as a customer, I'd probably watch the light for a few repeats whilst muttering to myself "long, long, short long, short", until I was sure I had the right pattern, then write it down and look it up. <A> The pulse sequence is followed by a longer off time and then repeating the sequence of long and short pulses until the user comprehends the code being reported and resets the unit. <S> In my implementations, either in MCU software or in FPGA logic, I have found the following pulse timing parameters to be optimal. <S> LONG PULSE <S> ON TIME = <S> 700 msec SPACE BETWEEN LONG PULSES = <S> 500 msec DELAY FROM LONG TO SHORT PULSES = <S> 200 msec (after last long pulse space) <S> SHORT PULSE <S> ON TIME = <S> 100 msec SPACE BETWEEN SHORT PULSES = <S> 500 msec DWELL TIME TO REPEAT = <S> 1900 msec (after last short pulse space) <S> User instructions should indicate that long pulse counting comes before short pulses. <S> Experience with this system shown that the optimum number of long or short pulses should not exceed five pulses for the most common error conditions. <S> That gives a total of 25 possible common error reports. <S> The most likely error conditions should be assigned to combinations with the lowest number of pulses. <S> I often group the error reports into classes based upon the number of long pulses and <S> the short pulse count is the error within the class. <S> Error codes are nicely documented as 11, 12, 13....21, 22....31, 32, 33 etc. <S> I have never found a reason in any given product to have more than about 15 error codes. <S> In rare instances where more than 25 are needed you can resort up to 6 long and short pulses to get up to 36 codes. <A> If all you really have is one LED, the best way of doing this (from an end-user perspective) will probably be to use a set of counted blinks. <S> Consider the following scheme, for example <S> : Break the 5-bit code into groups of 2, 2, and 1 bits. <S> (3 bits would lead to a group of up to 8 flashes, which is too close to the limit of what a user can reliably count.) <S> Add 1 to the value of each group, resulting in numbers from 1-4, 1-4, and 1-2 in each respective group. <S> (This is important -- having any of the groups display as zero will make the resulting sequence difficult or even impossible to read!) <S> Repeat the following sequence <S> : Blink the LED on and off at 200 ms intervals to count off the first group. <S> Wait for 500 ms -- just a little bit longer than it'd take to blink one more time, to make it clear that this is a gap. <S> Blink at 200 ms intervals again for the second group. <S> Wait for 500 ms again. <S> Blink at 200 ms intervals again for the third group. <S> Wait for 2000 ms <S> before starting the loop again, to make a long, clear gap between each time the code appears. <S> Here's an example of what this would look like for 4/2/2 blinks (binary 11/01/1). <S> Users should be able to describe this pretty reliably as "four flashes, then two flashes, then two more flashes".
| A far better approach for signalling error conditions on a single LED is to using a concept of a series of long pulses followed by a series of short pulses.
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What does Computer-in-the-loop or CIL mean related to embedded systems Im trying to learn more about embedded systems. During some of my research I have come across references to CIL or HWIL (apparently Computer-In-Loop or Hardware-In-Loop). I have not found much of an explanation in terms of CIL. Could someone elaborate a bit more on what this typically references? Is it possibly related to testing scenarios in a simulation environment? <Q> These are terms I have never used before because I was never in the 'merican military which use 3 and 4 letter abbreviations like crazy. <S> But basically you start with precise parameters and values for every source of error on every interface and then validate the design on those parameters. <S> e.g. <S> Monte Carlo runsets using known CIL/HWIL interface errors, such as scale factor, bias and noise, are used to create minimum-maximum boundary value plots. <S> Although I have worked for and with many American Corporations, the common terms we used are: Design Specs; DVT plan, PVT plan, HALT, HASS and Tolerance Stackup. <S> DVT = <S> Design Verification ( or Validation) <S> Test plan (& results) that verfies EVERY spec and in my case, test to failure or margin to failure. <S> Uses every Design Spec. <S> PVT= <S> Production or Process Verification Test <S> Uses DVT results to fine tune for production tolerance verification and ALL <S> design Specs. <S> HALT= Highly Accelerated Life Testing ( like shock and vibe with Temp stress while running voltage margins on supply with crystal tolerance errors , then increase until failure occurs) <S> HASS = <S> Highly Accelerated Stress Screening ( like above but not to stressed to failure mode) and used for quality screening and burnin. <S> For complex process with many e.g. 10 variables such as pre-flux and wave-soldering, Taguchi Method is used. <S> For embedded code work, this means doing worst case input and output test vectors to ensure all faults are detected, reported or corrected. <S> and every Function is validated. <S> Design Specs are a MUST for HIPO design specs. <S> Heirarchical Input Process Output. <S> ( an IBM acronym) <S> For complex IO processes, TAguchi Method can be used to reduce number of experiments to determine process sensitivity of each variable. <A> When I have hear of CIL used in the context of embedded programming, it usually means automated testing done on a mock system, rather than simulation. <S> One way to test an embedded microcontroller would be run in a simulator, wither by mocking the IO routines or using logs generated by the simulator. <S> So your test is "under stimulus X <S> , data Y needs to be sent to peripheral Z within t clock cycles." <S> This is nice if you can do it because it can be run completely in software. <S> This makes it easy to automate with a build system and do lots and lots of runs. <S> It also may be easier to inject faults or do internal inspection like monitor state machine transitions. <S> This approach has some limitations as well. <S> First, it is simply difficult to do. <S> Embedded systems are designed to "talk to the world" constantly, and generating a simulation environment that captures that sufficiently well can be difficult. <S> Second, the tests don't necessarily reflect the true constraints -- lets say that peripheral Z above is some SPI or CAN bus chip. <S> What you actually care about is that the peripheral does what it is supposed to. <S> If your test doesn't capture the timing requirements of that peripheral, then your test won't detect that type of error. <S> So a HWIL test would mean that when you build the program, you flash it to a "test" platform, then perform the tests on real hardware, and e.g., log the results with a logic analyzer or oscilloscope. <S> If your test is that "if the accelerometer goes above some value, fire the airbag", then you are actually testing the pulse that would go to the airbag inflater, rather than simply that the correct IO routine gets called on the microprocessor. <A> While there are probably official definitions for the terms, the line between CIL and HWIL often vary from group to group. <S> Ask your customer for clarification as to what they generally mean. <S> You then exercise the software using simulated inputs and outputs. <S> As I am lead to understand it, HWIL means you have not only the actual computer, but you also have other hardware devices such as sensors or actuators involved. <S> (because we all know that every hardware product behaves exactly as simulated) <S> As an example, if you had an embedded solution which was a motion detector using an IR camera, a CIL setup might generate simulated imagery and pipe it into the CPU as a stream of data. <S> A HWIL setup testing the same product might generate simulated imagery, project it onto a screen in front of a real IR camera, and attach the IR camera to the CPU.
| HWIL testing lets you shake down errors which may arise from the sensors and actuators not behaving exactly as they did in simulation. As I am lead to understand it, CIL means you have the actual computer in the loop, meaning the actual CPU and other data processing chips.
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How did my rabbit not get electrocuted? So, I'm sitting here looking at a chewed up extension cord that I just unplugged from the wall out of fear. Both the neutral and hot sides (no ground) have been chewed to (and part way through) the bare copper. My rabbit is nuts, I know, but did she really continue to chew while receiving 120 volts AC for her trouble? This is throwing everything I thought I knew about electricity into a tailspin. Maybe she wasn't grounded well, but I've been shocked myself when I was probably less grounded. Any help greatly appreciated! <Q> Why did the rabbit leave the cable alone after a while? <S> Perhaps she did receive a tingle. <S> Teeth have a high resistance. <S> It's unlikely a fatal current could flow through them. <S> A fatal scenario would need the rabbit to be well grounded at a remote body point, and to be contacting the bare live with lips or tongue. <S> Contacting both live and neutral with lips might leave a burn (and be very educational for the rabbit) but be unlikely to kill. <A> The short circuit would have been, well, short, and not through a critical, electrically sensitive organ like the heart. <S> That it did not die is simply one outcome of chewing on electrical cords. <S> Keep in mind, that while the conductors may have been exposed, unless they were shorted out with something like spit or flesh, mains voltage is not high enough to conduct across a nominal air gap. <S> Aside: My old electronics teacher in high school used to tease us with touching mains voltage between two fingers on the same hand. <S> Due to skin resistance and the non-harmful path, this is technically safe, but I still didn't want to risk it, even knowing the science behind the trick. <A> Common misconception. <S> Touching mains isn't 'automatic death' . <S> It's Russian Roulette. <S> One is skin conductivity (conductance is 1/resistance). <S> Another is how much current flows and each of the different routes through the body. <S> (And now the paralleling resistors formula may make more sense, you are simply adding the conductivities). <S> Obviously, good practice and use of PPE (such as GFCI/RCBO) can control those factors nearly 100%, that is their job. <S> But a lot of electrocutions are wildcard situations like finding (or gnawing) <S> a point of failed insulation, using a landscaper's bucket truck (which isn't insulated) too close to distribution lines, etc. <S> In these wildcard situations you just can't control enough factors, and that's what makes it Russian Roulette.
| The rabbit may have felt a shock, tingle, burned fur or flesh or other effect but that does not necessarily mean it was safe. Several factors have to converge in order for it to kill you (or not kill you). Current follows all parallel paths in proportion to their conductivity.
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How to use Zener Diode in Parallel RC circuit What I'm trying to do is allowing the current to move through only one component. In the circuit shown, DC voltage is running two components (Z & C), also charging the capacitor When voltage cut, then capacitor give current to Z & C. What I want to do is allowing the current to flow through only C but not Z.I tried to put zener diode on the line indicated by light blue but the current still flowing through Z (zener diode is not working). How can I make this work? What can be the problem? <Q> No Zener diodes need to be abused: <S> When V1 is present, it will supply current for R1, R2, and charge cap C1. <S> When V1 is disconnected, the cap will supply current to R2 but not R1. <S> Of course the energy in the cap is limited, and the voltage across R2 will decay exponentially towards zero over time. <S> There will also be a voltage drop across the diode, even when it is forward biased. <S> For example, if V1 is 10 V, then there might only be 9.6 V or so across R2, depending on the current. <S> If this is not acceptable, then a more complicated topology that switches a FET is needed. <A> A Schottky diode as D1. <S> Be careful of the voltage and current specifications of the Schottky you choose. <S> In this case: at least 40V and 2A to be safely above maximum ratings. <A> And the current in Z is because the Zener is operating in breakdown region. <S> So either you need to select a Zener which has cut off voltage greater than 28V or just use one of the standard diodes as mentioned above. <S> A standard diode should work perfectly for your circuit.
| Using a standard diode (eg in4007) should work for your application.
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Book Suggestion - How to calculate the power dissipation of a BJT I did a exam that asked for "power dissipation" of a given BJT circuit. The question was simple but my professor considered my calculation of dissipated power wrong, because my answer was \$P = V_{CE} \cdot I_C + V_{BE} \cdot I_B\$ . But he told me that the correct answer is \$P = V_{CE} \cdot I_C\$ . Can someone indicate a textbook or a scientific article confirming my answer so I could talk to him again ? I found some links on the internet confirming my answer, but he would not consider a website... <Q> From a theoretical POV your prof is WRONG. <S> You correctly calculated the total power dissipation of your BJT. <S> Your professor might be right only in an approximate way: if both the BJT quiescent point is in the active region AND its \$h_{FE}\$ is at least 10-20, then he is (approximately) right, because then \$I_B\$ would be much less than \$I_C\$ <S> and \$V_{BE}\$ would be about 0.6V, so that the \$V_{BE} \cdot I_B\$ contribution would be negligible. <S> In this case you cannot be deemed to be wrong, but only to be overzealous. <S> Note, however, that if the BJT is saturated, i.e. both its junctions are forward biased, \$h_{FE}\$ <S> no longer relates \$I_B\$ to \$I_C\$, i.e. \$I_B\$ cannot be assumed to be much lower than \$I_C\$. <S> In this case the base-emitter junction dissipation can be as relevant as the collector-emitter junction dissipation, and your professor would be completely wrong. <S> This latter case is particularly important for power BJTs, since their \$h_{FE}\$ is quite small (~20, or less sometimes), so \$I_B\$ is not so much lower than \$I_C\$ even in the active region. <S> Thus, to bring them into full saturation you sometimes have to provide an \$I_B\$ that is comparable to \$I_C\$. <S> Considering that \$V_{CE(sat)}\$ could be less than \$V_{BE(sat)}\$, this means that in some particular (but not theoretical) case base dissipation could be slightly higher than collector dissipation. <S> Neglecting it would be disastrous, then (especially when choosing the right size for the heat sink!). <S> BTW, If your professor doesn't think that the \$I_B\$ term is right, simply redraw the transistor circuit with the base powered by the Thévenin equivalent of the bias network (and whatever is connected to the base), so that it can be modeled by a simple two-terminal network. <S> Then calculate the power that that equivalent circuit provides to the base of the BJT and ask where does that power go, if not into the BJT? <S> Conservation of energy is a basic law of nature. <S> I hope your professor doesn't need a book reference to believe that! <A> Long story short <S> : You got it right, he is wrong. <S> I would challenge him (caveat: that may fire back): <S> Get a bit NPN power transistor like the 2N3055. <S> These can take up to 7A of base current. <S> Short C and E so that Vce is zero. <S> Then apply 6A of current between base and emitter. <S> Ask him to put his finger on it. <S> If he is right, the power dissipation should be zero, so the transistor will not heat up at all. <S> If you're right <S> Ib <S> * <S> Vbe <S> does dissipate power and converts into heat. <S> I'm pretty sure he'll change his mind about BJT power dissipation within half a minute or so. <A> Unfortunately you may be stuck in a position where your professor isn't going to admit he was wrong (or at least less correct than you) in order to save face for himself. <S> I don't have a book suggestion, but you could download a free SPICE program like LTSpice, build the circuit in it, and measure the various voltages, current's, and power dissipation. <S> http://www.linear.com/designtools/software/#LTspice <S> That could be both useful for you to learn how to use a SPICE tool like that <S> (it isn't overly difficult), and you could have the fun of listening to your professor explain to you how somehow SPICE must be wrong. <A> Consider the following circuit: simulate this circuit – Schematic created using CircuitLab <S> Total power is (Vi x Ib) <S> + <S> (VCE x Ic). <S> If your professor is correct, then the power dissipated in the resistor is Vi x Ib, but this implies that Ib = <S> Vi/R1, and <S> this is only true if Vbe is zero, which is not true. <S> Instead, VR = <S> Vi - Vbe. <S> This leaves a total power <S> Vbe x Ib unaccounted for. <S> Good luck on getting your professor to admit his error. <S> He has so obviously mishandled the problem that admitting his mistake will be extremely difficult for him, and he may lash out at you. <S> At the very least, make sure you approach him in private, as a public discussion is unlikely to go well. <S> Even in private, there is a distinct possibility that he will react with hostility, so be prepared.
| Your professor is, sad to say, wrong.
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The rate of change of magnetic flux linkage with a rotating coil When does the rate of change of magnetic flux linkage with a coil rotating in a uniform magnetic field equal zero? I think the answer would be: when the coil completes one revolution. as the the the change in the magnetic flux in the first half of the revolution is in opposite direction to the change in the magnetic flux in the second half, or I am wrong? Because someone told me that it will be when the coil is parallel to the magnetic field lines, so what is correct? <Q> Write down an equation for the flux through the coil as a function of angle. <S> $$ <S> \Phi =BNA\sin <S> \theta$$ <S> where A is the area of the coil, N is <S> the number of turns in the coil and theta is the angle between the B field and the plane that the coil lies in. <S> Differentiate the equation once with respect to angle. <S> The result is the rate of change of flux through the coil as it rotates. <S> $$\frac{d\Phi}{d\theta <S> } =BNA\cos \theta $$ <S> Set the result equal to zero and solve for the angle. <S> $$0 <S> =BNA\cos \theta $$ <S> $$\theta = 90, 270$$ <A> The above picture holds the answer. <S> So, why should maximum voltage occur when the coil is in-line with the lines of magnetic field (as shown). <S> Remember, the formula for induced voltage is proportional to rate of change of flux "cut" by the coil. <S> This is an instantaneous quantity and not something that is "accumulated" over one rotation. <S> When the coil is in the position shown, there are no flux lines "cut" but one instant <S> afterwards it is indeed "cutting" lines of flux so. <S> the rate of change of flux lines cut is significant in this transitory area. <S> Its rate rises from zero to some value dependant on the new angle of the coil and the speed of rotation. <S> Now consider what the rate of change of flux lines cut is when the coil is repositioned by 90 degrees (vertical to the picture shown). <S> The maximum number of flux lines are passing through the coil but there are the same number of flux lines being cut slightly before and slightly after and therefore, the rate of change is actually zero. <S> A few further thoughts. <S> I can see how this might be confusing because if the magnetic field (from N to S) were produced from an open iron core and coil fed from an AC voltage, the maximum induced voltage to the rotatable coil would be when that coil is in the vertical position. <S> This possibly seems contrary to the spinning coil and magnet shown in the picture. <A> 'when the coil completes one revolution' is when the total change in flux from start to finish adds up to zero. <S> This is not what the question was asking. <S> This occurs when the plane of the coil is normal to the field, the axis of the coil is parallel to the field, when the coil encloses the maximum possible flux.
| The rate of change of flux linkage is zero when there is no change in flux linkage with a small change in the angle of the coil.
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PIC I/O pin status on power up I have an LED connected to pin RA1 of PIC18F25K20 (powered by 3.3V supply) in the following way: 5V rail -> Resistor (200 Ohm) -> LED Anode -> LED Cathode -> RA1 RA1 : Input -> LED is off RA1 : Output Low -> LED is ON. Is this the correct way to connect the LED, as the PIC is powered by 3.3V and LED is connected to 5V? I initialized RA1 to be an input as shown in the DEVICE_Init routine. Code is written in C language - Microchip C18 compiler. void main(){ DEVICE_Init(); //wait one second while (1) { TRISA = 0; //led is on //wait 2 seconds TRISA = 0x06; //led is off //wait 4 seconds }}void DEVICE_Init (void){ OSCCON = 0x73; //PRIMARY INTERNAL OSCILLATOR PORTA = 0; LATA = 0; TRISA = 0x06; //RA1, RA2 INPUT ....} When I first power up the chip, the LED turns ON for a fraction of a second and then turn off. My understanding is that the IO pins are input on reset. So why would the LED turn on for a brief moment at power up? <Q> PIC <S> I/ <S> O pins wake up in high impedance state on power up. <S> The reason you see the LED glitch is probably because the 5 V supply is coming up faster than the 3.3 V supply, and the LED cathode is temporarily held low thru the protection diode on that pin in the PIC. <S> This is not good, and can lead to latchup on power up. <S> However, it is better to use a transistor, or to run the LED from the 3.3 V supply. <S> The latter will also waste less power, assuming your 3.3 V supply is reasonably efficient. <A> Your LED is turning on for a split second because the outputs are reset as Hi-Z (high impedance) or floating. <S> To fix this issue, put something like a 10k resistor from the output to ground. <S> I asked a similar question here: https://stackoverflow.com/questions/41557868/atmega128-output-flicker-on-startup <A> Do not initialize them as inputs only as outputs. <S> Outputs can source & sink. <S> Any pull down resistors will not do as in this configuration diode is sourcing current. <S> Do not pull-up!!. <S> Just initialise ports asap. <S> Your diode has to have Uv <S> > 1.7V.
| If you can arrange to be sure the 3.3 V supply comes up before and goes down after the 5 V supply, then your scheme can work. This is known as a pull-down resistor and will make the output low on startup.
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ADC power supplies monitoring - How to reject short overvoltage pulses I have an MCU application where the MCU switches on/off and monitors the various power supplies of the board (3.3V, 1.8V, 1.0V etc.). When an under- or over-voltage is detected, an alarm to the system should be raised. Until now the implementation was a bit simplistic: I only took one sample every 5ms and if it was above/below the relevant threshold, for overvoltage/undervoltage respectively, then the alarm was raised. That led to the problem that on one board, where the dc/dc converter design was not optimal and short over-voltage pulses were present, the alarms were thrown. Such a behavior is of course not desired. Now I am trying to optimize the c code of the MCU to cope with the problem. I know that ideally this should be solved at its origin (correct the HW design or implement an analog LP filter) but let's assume that this is not possible for the time being. My first idea was to require more than one consecutive samples to be outside the limits before an alarm is triggered. But somehow I came to this thought: what if I happen to sample exactly when the over-voltage samples occur? I am not sure if they are periodic at all, but it may also be the case. I cannot exclude it. In such a case such an implementation would also fail; it would trigger a false alarm. What do you think? Does this thought make sense at all? And then my next idea is to implement a digital low pass filter. Probably an IIR recursive filter. But here the questions are even more. Would this really solve my concerns? I mean, if the samples are taken exactly when the periodic pulses come, then the filter solution fails too. Do you have something else to propose? Do you think the digital LP filter is a good idea? If yes, I suppose I should somehow set the 3dB frequency of the filter close to DC. Is that anyhow possible? <Q> Edit: My original answer did not meet the design goal. <S> This is a different approach. <S> Count each under and over voltage event. <S> When either count exceeds a threshold of say 3, trigger the alarm. <S> Reduce the over and under counts each time <S> an in-band reading is taken in order to provide basic digital filtering. <S> Ensure you do not under flow the counters. <A> The issue with ANY sampling system is it is prone to frequency matching the source signal and if that frequency is undefined you are always in a gamble / probability situation. <S> I recall the " Viking " Mars explorer which had a camera which took panoramic pictures by scanning and transmitting vertical lines of image data using a rotating camera head. <S> You can see them as those two vertically slotted pots sticking up in the image above. <S> It took a wonderful picture for it's time.. <S> Unfortunately, as one of the astronomers on the team pointed out, you could march a brass band past the camera and if the camera took its slices in the gaps in the band, you would never see them. <S> The same applies to your requirements. <S> You can keep running average values in code, basically creating a low pass filter, and or use an event counter to require multiple out of ranges before reporting, but if things happen to line up with the deviations you will still have the issue. <S> Then when you detect a good value, reset the procedure. <A> As you pointed out, filtering should be done right at the source. <S> Maybe you can place an LC LPF with a cutoff frequency of 10kHz. <S> This should chop off most of the HF content. <S> About spikes: If you get 10 samples in a short period and 2 or 3 non-consecutive samples show irrelevant values (over- or under-voltage) then just ignore them and average the rest.
| In addition to those methods I'd suggest, on detection of an errant value, change the sample rate on that particular channel to various time values to try to detect an in-range value. But if you have no chance to make a hardware revision then averaging a limited number of consecutive samples will be the simplest solution.
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Measuring current with a multimeter - What did I do wrong? I wanted to measure the current of a generic lithium ion battery charger. I set I plug my red lead in the A 10A max fused and black into COM. The charger is rated at 600 mA and my other lead plug is fused to 250 mA so obviously that wouldn't work. So I put my leads on either side of the battery and there's a small spark. Did I break something? <Q> You placed an ammeter in parallel to a voltage supply, which created a short circuit through the meter and most likely blew the 10A fuse in the ammeter socket. <S> What you want to do is put the meter between the charger and the battery, and then connect the other end of the battery to the charger. <S> Diagram will help: simulate this circuit – <S> Schematic created using CircuitLab Edit: If everything still works fine, it's likely you tripped some kind of over-current protection in the charger before the fuse in the meter blew. <A> You used an ammeter in parallel to a voltage supply, and not in series. <S> This resulted in more current than the ammeter could handle. <S> You have thus blown some fuse. <A> The principles behind your idea were OK. <S> You wondered what the maximum current that the PSU could deliver. <S> So you connected a very low resistance across it (the DMM) to draw the maximum load and to observe the measurement. <S> It's the execution of it <S> that's wrong, I'm afraid. <S> A simple linear supply, commonplace decades ago, could be shorted like you tried to. <S> They consisted of a bridge rectifier and a smoothing capacitor. <S> Very shortly after though, the high current would overheat the diodes and/or draw a relatively very high current from the supply and probably take out the supply fuse. <S> But a modern switch-mode PSU is more sophisticated than that and contains protection, such as output current limit and 'hiccup mode'. <S> The latter will switch the PSU off if its output is low voltage while the current is high. <S> It repeats this cycle for as long as the output overload persists. <S> The practical way to determine your PSU's maximum output current is to connect your PSU output as (V+)->ammeter->Radj->(V-) where Radj is an adjustable load resistance i.e. a high-power rheostat. <S> You also connect a voltmeter across (V+) and (V-). <S> You set the rheostat to draw (say) 75 % max current, so 450 mA here, and switch the PSU on. <S> You then decrease the rheostat resistance to draw increasing current and observe the output current as the output voltage starts to fall. <S> You will be able to see when the voltage goes outside of its specified voltage range and therefore when the PSU can no longer provide the increase in load current. <A> If the meter still works you didn't break anything. <S> A lithium ion charger has both a constant voltage mode and a constant current mode. <S> So it acts sometimes as a voltage supply and sometimes as a current supply. <S> It decides which mode to use at a given time by monitoring the battery voltage, current, or, in some cases, the temperature (battery has a sensor built in, which the Samsung phone that caught fire apparently did not). <S> If you put the meter in series with the battery, as Chris M. said, you see how much charging current is being supplied to the battery, which may tell you about the state of the battery <S> (low current means it's mostly charged and being sustained, high current means it's mostly discharged) but not really anything about the charger's maximum current.
| In either mode, connecting the ammeter in parallel with the operating charger+battery circuit will pull current from both the battery and the charger (depending on, as one poster said, its output current limit). It then waits a short while and switches on to try again. So if you get a reading, it doesn't tell you anything specific about either the battery or the charger.
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what is the name for this DC connector? Question 1: what is the name for this DC connector? Question 2: Why are there 3 connections for a DC connector, and what are they for. <Q> Rev A <S> but in the category of barrel-power-connectors . <S> which normally have a sleeve switch "sleeve=outer conductor" with output on the side pin. <S> Normally today 0V return path is on outside <S> and +V is on centre pin. <S> So we might call this application a low side SPST or sleeve switch with 3mm center pin and 5.5mm diameter sleeve male receptacle. <S> Although we do not use the sexist term male and female for (floating) plugs and (fixed) jacks anymore, we prefer to call them Plugs and receptacles for mechanical reasons. <S> But for limiting exposed power, connectors are usually "pins" for the electrical load receptor and "Sockets" for the electrical source while connectors termed "Jacks" <S> are simply Fixed mounting connectors and Receptacles refers to the electrical connection. <S> But unlike AC outlets, it is both an electrical and mechanical receptacle to receive a plug which takes the electrical load. <S> Some DC barrel connectors are explicitly defined in sort columns such as at Digikey under Internal Switch as "Does not contain switch" but these are the exception. <S> However, Digikey and others carry no stock. <S> credits to @Passerby for the nudge on "all" Other Mfg's describe it as JACK, 2.5MM PIN, SHORT BUSHING <S> Connector names should be abbreviated in BOM's with description in reverse order of significance, thus a common abbreviation is simply. <S> CONN <S> PWR JACK 2X5.5MM <S> SWITCHED <S> SOLDER <S> However many major connector factories find define any connector with (M) or (F) which I support as being easier to understand especially when users have the acronym as effective series inductance. <S> (ESL) <S> UPDATE <S> Not all pins are the same diameter, and this is critical for good contact. <S> ALthough I said 3mm and 2.5 <S> this was a generic size <S> THere are many others. <S> prompted by comments @DanDavis and @Passerby (keep up the good work) <S> I looked up the website in the photoNow that you know how, dont ask again. <S> The part number and description is DC-005 5.5-2.1MM 2.0 Power supply DC <S> but generically it is a CONN, Barrel, DC 5.5mm-2.0mm <S> ( I suspect the difference is the pin and socket diameter of the center conductor http://www.xinlgo.com/product_detail-1850.htm <A> DC barrel jack of unknown size . <S> A coaxial power connector is an electrical power connector used for attaching extra-low voltage devices such as consumer electronics to external electricity. <S> Also known as barrel connectors, concentric barrel connectors or tip connectors, these small cylindrical connectors come in an enormous variety of sizes. <S> It is a switched jack. <A> The exact name requires a part number or stamp. <S> In general, that kind of connector is called a "Power Jack" by manufacturers. <S> There are typically three connections instead of two in order that a "plug in detect" is possible. <S> Two of the wires are shorted together when there is no plug in the socket, and they separate when a plug is inserted. <S> The jack you show a picture of, is very similar to a manufacturer called CUI. <S> Perhaps you have a foreign made knock-off in the photo. <S> An example data sheet is here for a CUI part: http://www.mouser.com/ds/2/670/pj-002ah-516007.pdf
| It's called a DC POWER JACK The third pin will connect either to the center pin or outside pin while no plug is inserted, and will break open when a plug is inserted.
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Combine outputs from 1kW and 2 kW Inverters I have a 3 year old 1kW PV System and recently bought a 2kW system. Both are off grid. And since the batteries on 1kW system are old, and since I availed subsidies for both systems I cannot change the configurations. Both inverters are 230V ~50Hz How can I combine both the inverter outputs to get a 3kW output?I came across some combiner boxes but most are for grid-tie systems or way too expensive. Is there any other way? How can I build my own combiner box if it's the best option? I have a 3-phase electricity connection. Is there a way to make use if it for this? <Q> There's an elaborate procedure for putting a generator in sync with another. <S> Traditional generators stay in sync because of load dynamics and backfeed forces (a generator a few degrees lagging will have its load eased, causing it to race faster and catch up). <S> The problem is semiconductor inverters are absolutely oblivious to these forces, and will simply ignore them and bang their drums ever more out of sync until sparky smoky things happen. <S> I'm not a huge fan of using inverters generally in solar systems intended to run off-grid, as the overhead of the inverter is a total loss you must pay for dearly. <S> I would be moving as many loads as possible to the DC side and see if you can put those on the smaller system and deprecate the inverter entirely. <S> Ideally, only run the inverter when you have a non-DC-able load. <S> If you really must tie the two systems together into a common AC waveform, that's a hard problem. <S> You could use DC as the shared language: tie them together on the DC side, with diodes, and both feed a single inverter. <S> This of course depends on their battery voltage being equal and their batteries being of same chemistry. <S> use inverter B's battery as the shared language: the smaller one's AC output feeds a battery charger for the larger one. <S> use shaft torque as the shared language: using a magnetic rotary machine (motor-generator set) to sync them, either each turning a motor which turns a common shaft to a generator, or the smaller one having an M-G set, which you then manually sync to the larger one, the protective circuitry here would be an adventure. <S> use water head as the shared language: install small hydro, and have both solar supplies backpump. <S> It helps if you have a hill and a creek. <S> I know some of these solutions are pretty appalling, but these consumer tier inverters are just not made to do this. <A> To do this your both inverters must be capable of paralelling using a third wire for frequency synchronization and to be of same manufacturer and model/series. <S> This feature is not common for cheap inverters. <S> If yours dont have it you cannot tie them. <S> You could either buy a single 3kW inverter or split your mains to 2 different circuits - one for each inverter. <A> What you have to check is what happens when the grid fails. <S> Unfortunately there is not an easy answer on this. <S> It is very unlikely any of the both converters is going to work without the grid when they are paralleled - even if they work without as singles. <S> If they are good quality, they should detect a failure condition and switch off automatically. <S> If not, one of them may blow.
| If both converters are grid-bound, you can simply wire them in parallel to the grid, same live wire.
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What is the reason for a fuse next to a thermostat? I'm repairing a heater that someone threw in the trash (this model): It has an internal thermostat next to the heating wires, plus a thermal fuse. What is the reason for a fuse in addition to the thermostat? It seems to me that the thermostat alone is sufficient protection against overheating, since the fan does not produce heat. <Q> The fan doesn't produce heat, but if the fan never blows, the heating element might overheat and start a fire. <S> Thermostats fail. <S> Safety regulations generally work on a "single fault" principle. <S> Meaning no single fault in a product should lead to a safety hazard. <S> In this case, the thermal fuse provides a backup to prevent a fire in case the thermostat fails (or, as @winny points out, in case the fan is mechanically blocked). <A> In addition to the fused closed thermostat issue already mentioned, the other issue that can happen with heaters is the thermostat can be arranged such that it measures the ambient air not the temperature inside the heater itself. <S> This can mean the thing will call for heat <S> but if the fan is not turning or is blocked by that carelessly discarded garment..... <S> (imaginations run wild here)... <S> the coil and internals will get really hot, really quickly. <S> As such a second internal protection device that is more sensitive to the heater coil temperature is required. <S> Having said that, the proximity of the thermostat and the fuse in that particular heater does not look like it lends itself to that characteristic. <A> Note that as drawn, if the thermostat opens the fan keeps running to cool the element/enclosure. <S> This might happen if you partially block the airflow or return the hot air into the heater, even if the temperature control realises the room is cold. <S> Then the thermostat will reset and everything is fine (your room just heats up more slowly than you expect). <S> It also protects from failures/complete blockages of airflow. <S> For the thermal fuse to blow, on the other hand, something must be really wrong. <S> This should be investigated rather than returning to an operating condition, so the fuse will open permanently. <S> In practice most users would throw out the fan and buy a new one rather than investigating (this isn't something that will trouble the manufacturer). <A> The thermostat regulates the temperature within the normal working range. <S> The fuse protects against exceeding that range. <S> You have also already noticed that while the thermostat controls only the heating coil, the fuse powers down the entire appliance. <S> Their jobs are complementary, not mutually exclusive. <S> They're like a regular brake and an emergency brake. <A> A fan motor that is stuck (maybe from a worn bearing) or suffers from a winding short - maybe by plain material fatigue, production defect, or foreign object damage - can become very good at producing heat. <S> And smoke for good measure. <S> After a fan failure, the heating element designed for forced convection heating will act as a radiation heater instead - likely heating (and trapping enough hot air) in the plastic casing to melt (or even light) it within minutes. <A> This is common in anything with a heater, and also AC motors. <S> It is there as a failsafe in case the thermostat gets stuck on, otherwise a stuck thermostat could cause catastrophic damage, perhaps even burn your house down. <S> I'm not sure, but I would assume it would have to be there to get CSA approval. <S> The fan motor most likely has one internal to it as well, in case the fan seizes. <S> source: <S> apprentice electrician that likes to take things apart
| If the fan was not working it would cycle on and off still, based on the location of the thermostat.
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Why is Rds(off) never listed on MOSFET datasheets? I can always find the Rds(on) values for the MOSFETs I use, but the Rds(off) value is hardly ever listed. Is there a special reason for this? In the datasheet listed, the Rds(on) equals 0.54 ohms but Rds(off) can't be found. http://www.vishay.com/docs/91015/sihf510.pdf <Q> The concept of "resistance" assumes that voltage and current are at least roughly proportional. <S> Diodes and by extension transistors tend to have a roughly constant leakage current over their normal operating voltage region (though there can be significant variation with temperature), so it doesn't make sense to characterise that leakage in terms of a "resistance". <A> There a drain and source dependence, as well as a temperature dependence. <S> For a nFET and starting from the EKV model, I came up with $$R_{off}=\frac{L \kappa e^{\left({\kappa V_{T0}-V_{g}}\right)/{U_T}} \left(V_d - V_s\right)}{W U_T^2\mu C_{cox} \left(e^{{-V_{s}}/{U_T}}- e^{{- V_{d}}/{U_T}}\right) } <S> $$ <S> where \$\kappa\$ <S> is the channel divider, and \$U_T\$ is the thermal voltage. <S> Most people do not really care that much about actual performance when a device is ``off"; however, due to the drain dependence, it can be significant. <S> (I've ignored \$\sigma\$ <S> on the drain term) <A> Rds(on) is normally used to determine the on-time power loss. <S> The power loss is always assumed to be zero when the FET is fully turned off i.e. Rds(off) is infinite.
| Rds(off) is so high that it is not relevant for the vast majority of MOSFET applications (mainly power switching applications).
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What will be the output voltage when 9 volt given to LM7812 I'm thinking of using 9 - 25 volt, 1750mA constant current DC input. In the circuit, there is LM7812 after A component in order to decrease voltage to 12V . My tests shown that when Voltage Is above 12V up to 28V, things look not bad. Output Voltage Is ~12V But I was wondering that what will be the output voltage if input voltage is less than 12, Ie, 9.9 Volt. My another question: What will happe if I give voltage from output to input to LM7812? Will it work? will there be flowing current? or will It behave like A diode? <Q> It's not quite as simple as a fixed dropout voltage. <S> Here is a plot of the dropout characteristics of the LM340 which is sold as the same part as the LM78xx series. <S> This is a 5V part so imagine the graph extended up to +12 rather than 5. <S> Nothing very exciting there, though you can see the drop decreases from 1V at 3V in to about 0.5V at 4.5V in (with zero current draw), and below about 2.75V the part has virtually no output as the internal biasing fails to keep the 'lights on'. <S> A fixed dropout voltage would have a slope of 1.00 and extend all the way to the 'X' axis. <S> The quiescent current is well-behaved: <S> There was at least one older bipolar regulator (Low dropout = LDO type) that had Iq spiking near dropout as the internal circuit tried to maintain regulation by sucking lots of current through the (very low gain) <S> lateral PNP pass transistor base. <S> Keep in mind that about 3V headroom is required to keep the regulator happy with 1A flowing under all conditions- <S> so the valleys in any input ripple on a 7812 should not go lower than about 15V. Applying a negative voltage to the regulator input relative to the GND pin, if more than a diode drop will cause a lot of current to flow and can easily destroy the regulator. <A> Linear regulators like the LM7812 have a minimum "dropout voltage" or "headroom"- for the LM78xx family, this is about 2 volts. <S> This means that the minimum input voltage required for the regulator to work is about 2 volts above the output voltage, so you'll need at least 14 volts into the LM7812 to get 12 volts out. <S> With 9 volts into an LM7812, I'd expect about7 volts out. <S> The actual voltage drop across the regulator will vary somewhat with the current drawn from the regulator. <S> I would expect the regulator to be destroyed by a negative input supply - but consult the apropriate datasheet for full details of the operation of this, or any other IC. <A> About 2 diode drops at very light loads, and up to 2 V at 1 amp. <S> Reverse voltage will likely destroy the part. <S> So will Vout <S> > Vin, so a good practice is to use a clamp diode from output to input. <S> (See Fig. 15 in the datasheet.)
| According to the 7812 datasheet , the drop-out voltage (the difference between Vin and Vout) depends on the load and temperature.
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How can I rotate a small 240VAC induction motor with 12VDC? I have an induction(?) motor that might be 30 years old. It's principal dimension is ~50mm so fairly small. It's actually the fan motor from a fan heater. It has one winding and runs off 240VAC with a DC resistance of 375Ω. There is just one simple stator winding. It's just like:- I also have a 12VDC source of sufficient capacity. How can I get the motor to rotate using the 12V direct source? All I want is rotation. It doesn't have to do any useful work just go round. And the speed is irrelevant, any constant rate is fine. Beyond any rhyme or reason I'm assessing the feasibility of using it as an on /off indicator in a Steampunk project. Steampunk gives us licence to do stupid stuff. Initial thoughts: some form of siney wavey generator + 12V audio amplifier chip like a LM384 (5 Watts)? Horrible impedance mismatch but might it rotate? Or is this just too stupid? Can feasibility be assessed without building it? I though that running AC synchronous motor with DC might help but that motor is properly voltage matched and seems to be performing work. <Q> (I once built something like this at work to power a three-phase 240 V 300 Hz motor) <A> If you want to operate the motor without an electronic circuit, you can build an electromechanical inverter like this: It requires three parts, an auto radio vibrator , a six-volt battery and a transformer. <S> I don't know the operating frequency the vibrator, but you can probably find that if you search. <S> It is probably not very high. <A> An AC motor is wound to rotate at a specific speed/Hz with a specific minimum current to sufficiently magnetize the rotor. <S> If you double the frequency, you must double the voltage to get the same current. <S> If you halve the frequency, then you halve the voltage to get the same current. <S> This beautiful simplicity is ruined by one thing... resistance. <S> At a certain lower frequency, the resistance will prevent enough current to flow for the core to magnetize properly. <S> The relationship goes like this:Close to DC frequency, we have a maximum of 12V/375 Ohm = <S> ~32mA available. <S> So, basically, if the motor will spin with significantly less than 32mA current at 240VAC there's a very good chance you can get it to spin from 12V. <S> To make it work, however, you would need to lower the frequency of the power going to the motor. <S> At 12V, you would need to have around 3 to 6 Hz to make it work. <S> The 12VAC (chopped) would spin the motor very slowly ... like 20x or 10x slower than when plugged into the wall. <S> In order to figure out if your motor even has a chance... all you need is an AC amp-meter, and a way to plug your motor into 240VAC (use a clothes dryer outlet, or you can try 120V, it might still work.) <S> If it spins at less than 32mA... <S> you've got an excellent chance. <S> You might want to add teflon oil to the bearings to improve your chances. <S> ;) <S> Otherwise, the transformer (doorbell transformer) idea is necessary to step up your 12V to something that can overcome the DC resistance of the AC motor. <S> It might also be possible, if the current is small enough, to use a capacitor transistor circuit to generate pulsed DC at higher than 12V from your 12V supply. <S> These kinds of circuits are actually pretty easy and cheap to build.
| You could use a 60 Hz oscillator feeding an audio power amp, followed by a 240 Volt->12 Volt transformer used "backwards" as a step-up transformer. The probability of success increases the smaller the required current draw is in comparison to 32mA. In general, there is a linear relationship between volts and Hz that will cause the motor to spin with the same current.
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Effect of temperature on Radio Components Just starting to gain understanding on how various Radio components work (PA, LNA, LO, Mixer, etc.). Every component has ratings from the component manufacturer. I am curious to know, what effect we may see when these radio components are exercised to extreme temperatures. How does temperature affect the Radio performance? <Q> A receiver is rated for a certain sensitivity, and if thermal noise exceeds it, you have a reduction in signal-noise ratio as thermal noise contributes to the overall noise floor. <S> Thermal noise power is calculated in dBm/Hz by the formula 10Log_10(kTB*10 <S> ^3) where k is Boltzmann's constant (1.3803*10^-23) <S> , T is temperature in Kelvin, and B is bandwidth in Hz. <S> At room temperature with a receive bandwidth of 1 Hz, thermal noise is usually about -174 <S> dBm/Hz. <S> To see the effect on a receiver, you need to subtract thermal noise from your receive signal level (RSL). <S> This gives you your SNR. <S> All of these will change at any given time due to bandwidth, frequency, temperature, and a myriad of other variables, so it is beneficial to learn the calculations and put them in a spreadsheet to make quick use of them when you need them. <A> The ratings might indicate temperatures at which the component completely fails - either temporarily or permanently. <S> Less dramatic effects are changes in phase/group delay, voltage fluctuations, etc. <S> Additionally, boards can flex causing weak connections or causing problems at any cold solder joints. <A> With FETs in radio circuits, at higher temperatures the amps/volt drops off and the FETS provide less charge/ <S> picosecond thus the frequency response drops off. <S> When you say "extreme" does that mean outside the semiconductor process models?
| As far as radio performance goes, temperature has a measurable effect on receiver performance.
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What is the shortest range radar system? I've been reading up a bunch on radar and most applications exploit the long range capabilities (rightfully so), but I'm wondering what is the smallest range a radar system can operate at? I'm interested in synthetic aperture radar on small scales. Specifically, I would like to measure the ocean waves over a 5m radius (yes 5m, not 5km) and build a "depth map" showing the "instantaneous" wave heights and surface currents (maximum frequency is 10Hz). I know this is done from space satellites, so why can't it be done from a pier or drone? I'm not a radar person, so excuse my ignorance, but doesn't the concept of radar scale? I would imagine you can build a miniature antenna array that scales to the range of interest? So if my range of interest is 0-100m (yes meters), couldn't the antenna be as simple as a wifi antenna? <Q> Yes, radar scales down; due to diffraction, you need shorter-wavelength radio to image smaller objects. <S> Milimeter-wave radar can be used at airport security. <S> Antenna complexity is another question. <S> You need either enough antennae to do synthetic aperture, or a physical moving or scanning directional antenna. <A> , it gets harder to tell the two apart. <S> This is why radar uses RF chirps as it allows you to convert distance (well, "time of flight" really) into a low frequency which is easy to measure. <S> Many radar antennas do share a lot with directional wifi <S> patch antennas (usually buried inside phones and tablets etc.), but not so much with monopole/whip antennas which you'll see on routers and the like. <S> That's not to say you can't use a bunch of whip antennas <S> , it's just harder to do so as they radiate all around their axis normally. <S> Having said that, Google has a radar system that can track <S> hand gestures <S> so the minimum range of radar can be quite small. <S> What you are probably looking for is an automotive collision avoidance radar as they commonly operate both on the scales you want and on small, slow watery things (i.e. pedestrians). <S> There are also a near infinite number of microwave motion sensors on eBay, most operate at 2.45GHz but some sit at around 10GHz. <S> With a bit of probing, you might be able to find the demodulated output signal and thus net yourself a functional short-range radar for a couple o' bucks. <S> Good luck. <S> Shahriar at The Signal Path recently posted a teardown and in-depth explanation of an automotive radar unit if you want to learn more about how these systems work. <A> Yes the concept of radar scales. <S> It all depends on the type of radar and the RF frequency. <S> I work on 60 GHz and 80 GHz radars and their range resolution can be less than 1 cm. <S> Also the shorted distance we can measure is a few cm. <S> For 0 - 100 m I'd recommend to look at radars operating at 10 - 30 GHz. <S> There are products (car radars) that operate at 24 GHz. <S> No the antenna will usually not be similar to a simple WiFi antenna as a WiFi antenna is mostly designed to transmit in all directions while radar antennas are far more directional. <S> They can be simple designs though but not the same as a WiFi antenna.
| I don't think radar has a minimum distance per se , but I'd imagine it's harder to use lower frequencies for close range applications because as your received signal gets closer and closer to your transmitted signal
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Schrödinger and electrical measurement issue During the 19's, Schrödinger state that it is impossible to make a measurement without disturbing a system. To illustrate his statement, he made the famous experiment of the cat and the box. Poor cat btw.. This is also true in electronic nowadays. A multimeter is use in parallel of a load for a voltage measurement. This "add on" will change the equivalent resistor value and then change the "true" voltage value. This is also true for current or other kind of measurement. How is it possible to compensate this lack of precision? I don't need to be that accurate, but I ask that question by pure curiosity <Q> Schrödinger made an statement about a quantum mechanical system . <S> So I disagree your conclusion <S> "This is also true in electronic nowadays." <S> because in common electronic systems QM effects are not observed directly; i.e. other effects are more dominant (by several orders of magnitude) than the QM effects you are referring to. <S> So normally there is no need "to compensate this lack of precision" because it doesn't matter at all. <S> E.g. normally no QM effects prevent you from measuring voltage arbitrarily accurately without affecting current (many other effects you will need to overcome; but they have nothing to do with Schrödinger's statement). <A> If circuit has a impedance of 10 ohm, and the multimeter is 10Mohm, then the change is 0.0001%. <S> With the impedance of the circuit I mean the load, the power source, and everything combined. <S> Together they determine the impedance of the circuit. <S> Sometimes the difference can be calculated, but not every circuit has a linear transfer. <S> Sometimes is makes a big difference, for example with high voltage and low current applications and with current measurement. <S> So yes, you have to be aware of it. <S> For example a Geiger counter with a Geiger tube that needs 400V with a very low current. <S> Suppose the voltage measured with a multimeter is 400V, but after the multimeter is removed, the voltage may raise to 450V without you knowing it. <S> One solution is to have a voltage divider with two resistors always connected to the high voltage. <S> Once calibrated, the low voltage of the voltage divider can be used to calculate the actual voltage. <S> To do a current measurement <S> the right way is a difficult subject on its own. <S> For example, the wires themself could have influence. <S> Do you know about 4-wire shunts ? <S> Measuring high frequency current is very hard. <S> And so on. <A> As you said, multimeter has an internal resistor to make different kind of measurement. <S> That value can be found in the documentation. <S> If you know that value, you can then make the calculation and find the impact of the multimeter on the circuit. <A> No measurement can be made without disturbing the source. <S> That is not only true for electrical measurements but for all kind of measurements. <S> Nowadays we use instruments that consume far less energy then in the past <S> therefore the influence on the source can be neglected in most of the occasions.
| In general it can be stated that no measurement is possible without transfer of energy and therefore the source always becomes disturbed.
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Push Button with Voltage Rating (AC) : 125V, Only having power supply of 3.5 or 5V I have 3 buttons (R13-81 PUSH SWITCH) for a school project. In the specifications, it is written that they take Voltage Rating (AC) : 125V ( Source ) Problem is I only have 3.3V or 5V as power supplies. Will it still work ? I don't understand if it takes between 0 and 125V or if it has to take 125V every time. Also, the output needs to be 3.3V maximum... What should I do ? <Q> Your question is wisely asked. <S> AC ratings are usually higher than the DC rating, if available, for the exact same switch. <S> There's a reason for that. <S> AC has zeroes , twice per AC cycle. <S> A manually opened (or closed) switch happens "randomly," relative to the AC cycle. <S> Any "arcing" drawn out as the metal contacts open up will soon be extinguished because the AC cycle will hit one of those zeros. <S> DC doesn't have any zeros. <S> So there is no inherent self-extinguishing process and therefore they may have to cope with more prolonged arcing when operating at the same voltage. <S> In your case, the DC voltage is so low and is also so much lower than the AC switch rating, that there is almost certainly no problem using it. <S> I wouldn't worry at all. <S> But it was still wise to ask. <S> (There is also the problem of current compliance, but you've not mentioned it. <S> I'm assuming from the context of your question that we aren't talking about large currents.) <S> P.S. <S> A manual switch is just a bit of mechanical components arranged to make it easy for a human to operate, and includes metal conductive contacts that are placed into direct contact with each other when the switch is engaged. <S> They will make contact, regardless, because the mechanical design for operation is independent of the voltage it is switching. <A> The voltage rating on switches is the maximum voltage that the switch can safely handle. <S> Likewise, the current rating of a switch is the maximum current the switch can safely handle. <S> Your switches are fine for use with 3 or 5 volts. <A> (In this case the rating is basically for the insulation within the switch, and the distances between bare metal components at different voltages. <S> Staying within the rating assures that the insulation will not "break down" due to voltage stress and that arcing between parts of the switch will not occur.) <S> There are two separate concerns, however, unrelated to the voltage rating. <S> First off, one must keep the current through the switch below its max current rating. <S> Your switch has a half-amp rating, so you need to keep current below that level. <S> Most uses for signal and control applications don't exceed a few milliamps, but if you are, say, powering a motor you could see higher currents. <S> Secondly, arcing. <S> When the switch "breaks" (turns off) and the contacts in it pull apart, it's possible for there to be an arc formed between the contacts, causing them to briefly heat up and be damaged, and, in rare cases, causing the current to continue even when the switch is supposedly off. <S> This is mainly a problem with "inductive" loads -- relays, solenoids, motors, etc. <S> If you are switching such loads you need to study up a bit on how to deal with this situation. <A> SPST off-(on) .5A <S> @ <S> 125VAC <S> Plating undefined ( possibly silver micro-plated over nickel over beryllium copper) <S> Usually these are derated for DC to about 1/4 of the AC rating due to <S> inductive load arc needs to be extinguished rather than <S> zero crossing current of AC. <S> All relays are however gold plated upto 2A to permit low current circuits to conduct without oxidation risk (high) <S> However when switch supplies, there is often a low ESR filter cap and thus a surge current. <S> THis can be a bad thing or a good thing. <S> If the cap is too big, and very low ESR, it can pit the silver oxide and raise the resistance from 50 mOhms to much higher value. <S> But a little "wetting " current usually 10% of rating is all you need to burn thru oxidation insulation on silver plating. <S> So depending on your surge current on contact, 0.5A should be OK but contact will wearout some material with a low ESR cap of 1 Ohm giving V/1Ohm= 5A. Ceramic and plastic caps have even lower ESR and gives higher peak currents. <S> So beware Low ESR caps are good for noise reduction but bad for switch load surge currents and contact wear. <S> In well designed system, it could last for 50k cycles. <S> If really well designed, even more and much less if poorly design surge current is applied on contact. <S> So designer beware.
| In general, the voltage rating of "passive" components like switches and connectors is their max rating, and lower voltages may be used with no difficulty. The metal contacts are just physical bits of metal and do not have to have a specific voltage in order to work. In general, an AC-rated switch should be derated for DC use.
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LiFePO4 in cold: charging I have used LiFePO4 batteries, only in discharge mode, by down to -15 deg C (charging was done 'back to the warmth'), and they worked fine. But from what I read it looks like Li-* batteries do not like to be charged by minus temperatures. I would like to use large LiFePO4 cells with a solar panel by negative temperatures (typically -15 deg C, may be down to -30 deg C). I will have quite low charging currents, max 1 to 1.5Amps for a 40Ah cell. Do you think it may be OK to charge at such low rates (this guy seems to say no: Solar powered single cell LiFePo4 charger circuit ; but here it sounds like very low charge may be ok http://batteryuniversity.com/learn/article/charging_at_high_and_low_temperatures )? Has anybody more specific experience with this? Or has anybody knowledge of a place where I may get military / space grade cells that accept cold if usual LiFePO4 is a no go, or alternative battery chemistry solution? FYI, here are the cells I use as by now: http://www.batteryspace.com/lifepo4-prismatic-module-3-2v-40-ah-10c-rate-128-wh-----un38-3-passed-dgr.aspx <Q> When working on the fringes of specifications, it is always best to speak with an applications engineer at the manufacturer. <S> They often have access to data that is not part of their published specifications. <S> With that being said, you should give careful consideration to thermal management within your design. <S> For example, a large insulated thermal mass in contact with the cells will retain heat generated during the exothermic charging cycle. <S> This will help prop up the ambient temperature of the cells during non-charging periods - such as at night. <S> The choice of enclosure color can also play a role in raising the average ambient temperature of the cells. <S> You may also go so far as to use your solar power to warm the cells until they reach an acceptable charging temperature. <S> While this is not the most energy efficient approach, it is a pragmatic one that may enable the application. <A> There is newer LiFePO4 vendor which allow you to charge battery at negative temperatures. <S> I am sure there is more vendors like this <S> but you have to do some research on them. <S> If vendor doesn't list negative temperature charging in specs - don't charge because you will damage the battery irreversibly. <A> Basically to prevent damage to negative electrode that has high ESR at cold temps, you need to heat up the electrolyte evenly before you can push 1C charge or more. <S> This means it is slow. <S> Otherwise if you heat up the layer around the electrode too fast then permanent damage may occur. <S> REF: <S> https://relionbattery.com/blog/lithium-battery-cold-weather With the low-temperature RELiON RB100-LT 100Ah battery, it takes about an hour to warm from -20 <S> °C to <S> +5°C before charging begins. <S> A thermal analysis with ESR needs to be done with your pack to before can say for sure how to lower the ESR safely and start charging at rated currents. <S> This can be done by feedback of testing ESR or by design with insulation and computing time to raise temperature safely. <S> That's my advice only FWIW. <A> I agree with both answers. <S> This is not advise, but my opinion that you should NOT charge your LiFePO4 battery below its rating. <S> Instead, you can discharge it to some extent to help warm it if you have the capacity or use an external already warm smaller LiFePO4 battery with a temperature sensor to warm your main battery via a heat mat or a current regulated Ni-chrome wire. <S> It will increase it's life, is much safer, and will increase performance. <S> For example, say for an ATV using a LiFePO4 battery, turn on the headlights to discharge the battery for a few minutes, then try to start it, <S> if it does not start, wait a few more minutes then try again. <S> As the battery discharges, it will warm the battery. <S> I would not recommend charging on a cold LiFePO4 battery, call the manufacturer as stated above <S> but they sell heat mats for batteries just for your case. <S> And a "GOOD" BMS is always recommended. <S> I have not seen these, but it makes good common sense. <S> As the battery charges, it gets warmer, but lower charge rates at lower temps are good if the temperature is above the manufacturers rating:) <S> Again, just imho. <S> Food for thought: SSD's (solid state drives) hold data longer when the drive is hot, and have lower long term storage if the drive is cold. <S> Batteries perform best when charged and discharged at specific temperatures. <A> Alternate thought, why is the temperature so low? <S> If possible bury the battery in the ground as much as possible, and use the higher temperature of the earth itself. <S> The battery does not out-gas like lead-acid, so no explosive mixture. <S> Trying to put the battery high up on a pole will be counterproductive from a charging point of view.
| Ideally, the charger will have a temp sensor that disallows charging below a certain temp and has a built in temperature curve that increases the charge rate as the temperature increases to optimize charging.
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Resistors between Arduino pin and transistor I'm trying to activate a relay using an Arduino Uno. Most schematics I have seen in the web include two resistors between the Arduino pin and the 2N3904 transistor. What is the reason for that? <Q> R9 <S> In your application, you want the transistor to switch on when the arduino pin is HIGH. <S> To calculate that resistor value, you have to how much current will flow across the transistor when saturate (\$I_c\$). <S> You can then determine the needed current coming from the arduino to toggle the transistor : \$H_{fe}=\frac{I_c}{I_b}\$ with \$I_b\$ <S> the current delivered by the arduino. <S> Knowing <S> \$I_b\$ <S> , you can determine the resistor value. <S> R10 <S> This resistor is here to set a default value a the transistor's base, commonly called "Pull Down resistor". <S> When the arduino is just power up, pin's voltage level is uncertain, that's why you need such a resistor. <A> The series base resistor (R9) is to limit the base current drawn from the Arduino output pin. <S> The resistor between base and GND (R10) is to conduct away the I/ <S> O pin's leakage current when it is tri-state. <S> This will be the case after your circuit powers up (when reset configures <S> then I/O pin as an input), until the Arduino CPU is out of reset and its software configures the I/O pin as an output. <S> The leakage current may be enough for your transistor to conduct. <S> The resistor value can be much higher than what you have, with 10 K to 47 K commonplace. <S> Calculate it from R= <S> Vilg <S> /Ilk where Vilg is the input voltage from a good logic low (recommend 0.1 V) and Ilk is the I/O pin input leakage current (always see datasheet). <A> R9 is there to protect the logic driver. <S> Consider the schematic below. <S> Without a series resistor, when you drive the pin high, you are basically shorting the Vcc rail to ground through a diode. <S> This will exceed the maximum current the pin can deliver and ultimately result in damage to the device. <S> The resistor needs to be large enough to perform that function while small enough to deliver sufficient base current to drive on the transistor hard enough to power the load. <S> This is one of the reasons MOSFETs, which are voltage driven, and often preferred for this role. <S> simulate this circuit – <S> Schematic created using CircuitLab R10 is required to ensure the base of the transistor is held low while the micro initializes and has it's pin in a high impedance state.
| The resistor is here for current limitation.
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How fast can I cycle an AC SSR? I have a PID/heater setup I am driving with an arduino. I'm using the Opto 22 MP240D4 solid-state relay to switch about 1A 120V 60 Hz (resistive heater). It's not clear from the product sheet or the datasheet whether it is zero-cross turn-on, or not. Also, just a bit of context, the application is a very low mass heating element, so it is a bit unusual from the perspective of typical PID plants. I want to minimize the granularity of the pulse width/density modulation so as to get the best performance out of the system. Obviously, my absolute upper bound is 120 Hz since that's the number of zero crosses. But is there any harm (in terms of thermal fatigue, stresses to the chip, etc) to running it at high pulse speed? Also, is there any advantage to either pulse-width or pulse-density modulation? <Q> There is no harm in turning on or off a zero-crossing SSR at every zero crossing. <S> I did exactly that once in a system that had to control 24 heaters simultaneously. <S> The control algorithm produced a 0-255 value proportional to how much each heater was supposed to be driven. <S> The low levels used a Bresenham algorithm to decide each half-cycle whether each heater should be on for the next hal-cycle. <S> It worked very well. <A> No worries switching on and off at 120 Hz, but as you mentioned you are hard limited to a pulse width of 1/120 Hz. <S> Pulse width modulation is a natural here since your pulse rate is always going to be a multiple of 120. <S> In a heating application, a 8-bit PWM with feedback turning "on" once per 2.13 (256/120) seconds and "off" on a 120- or 60-second boundary should be good if the load resistance is appropriate. <A> I agree with @JackCreasey <S> there is nothing in the datasheet that says zero crossing. <S> You could get aliasing, depending on your frequency. <S> The SCRs will stay on for the remainder of the half-cycle regardless of when you trigger them. <S> I suggest going with about a 2 second cycle minimum, especially if you don't want to synchronize to the mains zero crossings, a bit more would be better. <S> Unless you have a really, really good reason that's plenty fast enough for thermal setups. <S> About a 2-3 second cycle is worst-case for causing thermal fatigue on the power semiconductors. <S> We saw a multi-million dollar screwup some years back caused by the die bonds fatiguing causing overheating and failure (in the on-state typically) of the power semiconductors. <S> Modern ones are better, and it helps to keep well away from the maximum rating.
| If you are running directly from a wall switch, you are ok using 1/2-cycle as your minimum pulse width, but if running through a transformer, use an even number of half cycles to prevent having a DC bias. Often 10-20 seconds is more than good enough (except for things like IR heaters which change temperature significantly in seconds).
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What would be the lightest batteries to run a 12v/12.5A skillet for 2 hours? I like to go camping and use various gas stoves to do my cooking. I was curious with the current (no pun intended) state of battery technology, how heavy, bulky, and what sort of wiring it would take to cook via battery. I found an electic skillet designed for RVs online, and its specs are 12v, 12.5amps, 150 watts. What would be the lightest combination of batteries (motorcycle, car, AAs (NiMH, Lithim, Alkaline)) that would be required to satisfy this, and how should they be wired. I had read that AA's although they may be rated to ~2000+ mAh, can't consistently put out more than .5 A. So my thoughts were 25 in parallel to get 13 amps, then those 25 in series with another 7 sets of 25 to get up to 12v.In theory that would provide the necessary amps and volts, for as much as 5 hours, but I'm sure there's more to it than that, and I'm not sure if a 12v motorcycle battery would be far smaller and lighter and still satisfy the requirement. And all of my calculations might be way off. Thanks <Q> A butane-propane canister contains 12000 <S> Wh per kg of gas. <S> A Li-Ion battery contains about 100-200 <S> Wh per kg. <S> As you can see, energy density is merciless. <S> I'm tempted to say "end of story" here, although you should also consider cost, which leans heavily on the gas burner's side. <S> Also if you want to cook a steak, you need a 2kW burner, which is readily available... bit more complicated on battery power... <A> 12V x 12.5A is a power of 150 W (Watts). <S> 150W x 2h is an energy of 300 Wh (Watt-hours). <S> These are the key factors for estimating <S> the size of battery needed. <S> Lithium-ion batteries are about the lightest batteries currently available. <S> A quick Google/Wikipedia search shows that they have a specific energy in the range of 100–265 Wh/kg, and specific power of 250-340 W/kg. <S> Let's use 200Wh/kg and 300 W/kg because they are round numbers. <S> 150W means you'll need at least 0.5kg of 300W/kg batteries to be able to deliver enough power. <S> 300Wh means you'll need at least 1.5kg of 200Wh/kg batteries (to hold enough energy to run for two hours). <S> We can see that energy is the limiting factor here <S> and you'll need at least 1.5kg of batteries. <S> You can do similar calculations for other battery types, but I guarantee that this is the lightest you'll get. <S> An electric stove of this type would be suitable for an RV, because: You don't need to lug the battery around - the RV's engine does that. <S> Because of this, the RV has a bigger, heavier battery. <S> If the battery gets low, you can turn on the engine to convert petrol into electricity. <A> If you are that concerned about weight, on a budget, LiPo is still best. <S> Match the voltage and get extra Ah for safety margin more than 25Ah for > <S> > <S> 300 <S> Wh or about 3.6V x 5S8P or 40 cells running PWM at 85%=12/14.2V. <S> That will cost about 100x more than a propane gas used in a tank. <S> You can get burners to sit right on the tank <S> and it cooks a lot faster. <S> Conclusion: bad idea.
| Petrol has a much higher specific energy (over 10000Wh/kg) than any kind of battery.
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Two phase clock on a breadboard How can I make a two phase clock on a breadboard (i.e. no surface mount components) without using a microcontroller? Phase 1: _ ___| |___| |__ Phase 2: ___ ____| |_| |_ The exact duty cycle is not important. Phase 1 must go low at least 50ns before phase 2 and go high at least 50ns after phase 2. Ideally, the clock needs to be adjustable (via resistors, capacitors, etc.) between 1Hz and 100KHz. I have been thinking about ANDing and ORing two 90 degree out of phase 50% duty cycle clocks - but I still need to generate those. <Q> Use a 2-bit binary counter (two D flip-flops), then use some simple combinational logic to generate the phase1/phase2 outputs. <S> This could be individual AND / OR / NOT gates, or it could be a decoder such as 74__139. <S> There are four ways to map the counter state to the phase1/phase2 waveforms, you can choose whichever seems the easiest to implement. <S> Remember to consider the initial state of the counter as well. <S> At 100kHz there shouldn't be any major issues with using solderless breadboard. <S> Up around 10MHz solderless breadboard starts to reach its limits. <S> ------|-------|------a1 <S> a0 <S> |phase1 <S> |phase2- <S> -----|-------|------0 <S> 0 <S> |0 <S> |00 <S> 1 <S> |0 <S> |11 <S> 0 <S> |1 <S> |11 <S> 1 |0 <S> |10 <S> 0 <S> |0 <S> |00 <S> 1 <S> |0 <S> |11 <S> 0 <S> |1 <S> |11 <S> 1 |0 <S> |1 <S> phase1 <S> = <S> a1 AND (NOT a0) <S> phase2 = <S> a1 OR a0 <S> ------|-------|------a1 a0 <S> |phase1 <S> |phase2 <S> ------|-------|------0 <S> 1 <S> |0 <S> |01 <S> 0 <S> |0 <S> |11 <S> 1 |1 <S> |10 <S> 0 <S> |0 <S> |10 1 <S> |0 <S> |01 <S> 0 <S> |0 <S> |11 <S> 1 |1 <S> |10 <S> 0 <S> |0 <S> |1 <S> phase1 = <S> a1 AND a0 <S> phase2 = <S> a1 <S> OR (NOT a0) <S> ------|-------|------a1 a0 <S> |phase1 |phase2 <S> ------|-------|------1 <S> 0 <S> |0 <S> |01 <S> 1 |0 <S> |10 <S> 0 <S> |1 <S> |10 <S> 1 <S> |0 <S> |11 <S> 0 <S> |0 <S> |01 <S> 1 |0 <S> |10 <S> 0 <S> |1 <S> |10 <S> 1 |0 <S> |1 <S> phase1 = <S> a1 NOR a0 <S> phase2 = <S> a1 AND (NOT a0) <S> ------|-------|------a1 a0 <S> |phase1 |phase2 <S> ------|-------|------1 <S> 1 <S> |0 <S> |00 <S> 0 <S> |0 <S> |10 <S> 1 |1 <S> |11 <S> 0 <S> |0 <S> |11 <S> 1 |0 <S> |00 <S> 0 <S> |0 <S> |10 <S> 1 |1 <S> |11 <S> 0 <S> |0 <S> |1 <S> phase1 = <S> (NOT a1) <S> AND a0 phase2 = <S> a1 NAND a0 <A> Divide your input signal using a flip-flop to half the frequency. <S> The run the original frequency and the divided frequency into the S1 and S0 inputs of a CD74HC253 multiplexer Datasheet . <S> This will output the voltage present at the inputs 1 to 4 at the output in serial order. <S> You can use these four inputs as a generic table and 'program' any four step sequences that you need. <S> For your signals: Apply these at the first multiplexer inputs: 1I0: <S> low1I1: <S> high1I2: low1I3: low <S> And for your second sequence: 2I0: <S> high2I1: <S> high2I2: <S> high2I3: low <S> You'll get your two phase signal at the outputs 1Y and 2Y at half the clock frequency. <S> To compensate for the lower frequency, either run the circuit at twice your wanted frequency or use a PLL to double the frequency of an existing input clock. <A> A CD4001BE dual NOR gate with a dual CD4011BE NAND gate would do the job. <S> They have a 60ns delay each. <S> Wired as: simulate this circuit – Schematic created using CircuitLab <S> This will drive 74hcXXX logic correctly: Can original CD4xxx logic drive 74HCxxx logic?
| You can do this using two 4:1 multiplexers and a frequency divider (a simple flip-flop will do).
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increasing torque of a sewing machine motor I have a very small table saw (Harbor Freight "Mighty Mite") that is rated at 0.9 amps and 1/12 hp with max RPM of 14000 (a sewing machine motor). I use a jewelers saw blade and cut a very small channel approx. 1/8 of an inch deep for guitar frets. It works fine on soft woods but will sometimes bog down on hard woods. If I replace it with a 1.5 amp sewing machine motor with a 9000 RPM would it increase the torque to more likely make the cuts on hard woods? <Q> This is a tricky question. <S> On the one hand, it's clear that the larger motor will give you better torque. <S> The combination of higher current and lower rpm should give about twice the torque. <S> On the other hand, there is no guarantee that this will be enough. <S> A basic question arises - if your saw is bogging down, why don't you just reduce your feed pressure and cut more slowly? <S> Assuming that you realize that you could just cut more slowly, but don't, then there must be something else at play, such as the cut quality with slower feed and harder wood. <S> So it's possible that using a beefier, slower motor will not give you an acceptable cut. <S> And I don't know of a way to tell in advance. <A> The published ratings for motorized consumer products are very difficult to interpret. <S> They usually describe motor test conditions that can not be achieved during normal use of the product. <S> The motors used must be either universal motors or permanent-magnet DC motors. <S> In either case, the speed will decrease considerably under load. <S> The current given may or may not be a safe operating current for an extended period of use. <S> It is likely that the 1.5 amp motor can safely provide more torque for an extended use time, but there is no way to know for sure without trying it. <S> Even if the motor doesn't overheat, mechanical wear may shorten the life of the tool. <A> There are many of different types of sewing motors. <S> The most used is universal motor with variable series resistor aka pedal, or some SCR controlled AC speed controller. <S> If you hear the 50/60Hz noise then it's an old fashioned sewing motor. <S> Newer and costly sewing machines have permanent magnet DC motor with encoder feedback. <S> If you want a cheap and good solution, then a permanent magnet DC motor with speed controller is the way to go. <S> More costly and yet better is the BLDC motor + controller. <A> A quick but rough analysis: 1.5A is a 67% increase over 0.9A, a 67% increase in electrical power into the motor. <S> Assuming the two motors have the same efficiency and are similarly rated, there should be a corresponding 67% increase in mechanical output power. <S> Power is proportional to speed times torque, so a decrease from 14 to 9 krpm corresponds to an increase in torque of 56% if output power is the same. <S> So you might expect to see 1.56 <S> * 1.67 = 2.6 times as much torque from the larger motor, though at a lower speed.
| To begin with, a motor will only bog down if the load is too great, and for a saw the load will be pretty much proportional to the feed rate.
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Is a cigarette lighter battery voltage meter accurate? Recently, I bought a USB car charger that shows my batter voltage at about 13.7v when the car is running. Is it possible that this is accurate? Since you can get a lot of gimmicks on eBay, I want to be sure if this is even possible. <Q> A car cigarette lighter or 12V power outlet in a car is, like most everything else, connected directly to the battery positive/alternator output, through a fuse (typically 10 Amp). <S> Sometimes it is connected through the main switched power relay. <S> In any case, aside from the negligible voltage drop across the fuse and wiring, is an accurate measuring point for the voltage in the car. <S> As mentioned, 13.7 is a typical voltage for a car under load while the alternator is running. <S> It will have a resistor divider to bring the voltage down to the microcontroller voltage level. <S> But the lower the quality of the item, the more likely that the resistor is not a very precise, and the Analog to Digital converter in the microcontroller they used will not be very accurate. <S> For the most part, this may mean it's off by a fraction of a volt. <S> Not precision laboratory grade. <S> But good enough for your average user. <S> Sidenote, I use one in my car due to alternator issues and it's as accurate as my free harbor freight multimeter. <A> 13.7V is in the normal range for the 12V system in a car. <S> It will vary over 12.5 to 14.5v when the car is running and down to about 11.5 when it isn't. <S> Depending upon the car the lighter connection may be active when the ignition is switched off or not. <S> (Japanese cars tend to have them switched by the ignition, American cars tend to have them always active). <S> They are usually fused for 10-20A. <A> First of all the voltage at cigarette lighter jack is always somewhat less than actual voltage on battery terminals. <S> There are ground losses thorough car's chassis, losses of "positive" as it goes through couple of relays, fuses and wiring. <S> Any load (like stereo, GPS, dashboard illumination, etc.) drops a bit the voltage you will read on the cigarette lighter jack. <S> Second - you can't be sure of this meter's accuracy. <S> I don't know the exact product, but most chinese meters may have difference of about +/-0.3V. <S> Third - you don't have to worry much. <S> The main purpose of such a meter is to know <S> does you alternator charge or not. <S> If you see anything over 12V-12.5V it charges and in most cases it charges OK. <S> If your alternator fails and stops charging you will read values of about 11-11.5V or less. <S> Of course there are cases where alternator can charge poorly and the actual battery voltage is 13.2-13.6V or can overcharge (more than 15V) <S> but it happens quite rare. <S> Although rare, I recently had both of these problems on my Ford Mondeo MK3 equipped with Visteon alternator. <S> This kind of fault can be seen by measuring battery voltage directly on its terminals. <S> This means you should make additional wiring just for the voltmeter. <S> Another solution for a modern car (over 2000's) is to buy an OBD display. <S> It shows you some stuff read by digital bus from the Engine Control Unit (ECU). <S> ECU knows battery voltage with much better accuracy compared to what you measure on the cig.lighter. <S> On some cars ECU has digital comunication with alternator's charging regulator.
| The cars cigarette lighter is connected directly to the cars 12v system but as others have mentioned may vary slightly from the battery voltage due to the resistance of the wiring. As to the accuracy though, keep in mind that the trinket you got is a cheaply made mass produced item.
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What would cause a 120 VAC universal motor to run slower in reverse? I got a bargain on a few small universal motor that run on 120VAC and I'm trying to reverse the direction. I don't have a lot of experience with these types of motors but I do know they run on AC or DC and the direction can usually be reversed by simply swapping the wires on the brushes. On this particular motor swapping the wires to the brushes does reverse the direction but the RPM is slowed to about 1/3 of that in the forward direction. I have read that manufacturers sometimes tune the placement of the brushes to reduce inductive reactance but I'm not sure how to confirm if this is the case or what to do if it is. The schematic below shows my current test setup. I added a DPDT switch to be able to reverse the direction easily but my goal is to run it in reverse only. There are two small inductors on the outputs of the field windings just before the brushes but I assume these are just for tuning and would not be the cause of the issue but I could be wrong ( they appear to be identical on both sides but I have not measured them). Is there anything I'm missing that would cause the motor to run slower in reverse or anything I can try to resolve the issue? Motor Schematic Motor view 1 Motor view 2 Motor view 3 Motor view 4 (tuning inductors) <Q> Your motor is 'timed' to operate in one direction by advancing the brushes. <S> This is clearly visible in picture #2, where the brushes look to be rotated anticlockwise by about 12º relative to the stator. <S> Advanced timing increases rpm in one direction, but reduces it in the other direction. <S> A motor that is 'neutral' timed will go the same speed in both directions. <S> The timing on your motor does not appear to be adjustable. <A> The motor is not constructed for bi-directional rotation. <S> You can see, that brushes are not exactly 90 degrees aligned with respect to the stator. <S> This is due to the compensation, it reduces sparking and EMI. <S> Further, if you switch the direction of rotation then the brushes might also zap because they are trapezoidal shaped at the end. <A> Those are likely just the Tuning Inductors, used because it is intended to be fed with AC and operate at higher speeds. <S> You're right, that would not affect the direction.
| Most likely it's because this is a used motor (from the looks of the commutator) and it was never reversed in its previous life, so the brushes have worn into a shape that, when you reverse the rotation, makes poor contact. You could try removing the brush holders and springs from the board and drilling new holes for them, but it may not be worth the effort.
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Why is 5.1V the standard Zener voltage and not 5V? Looking on Digikey there are ~1300 results for 5.1V Zener diodes compared to 6 for 5V ones. Is there any physics-related reason for this or is it something else? <Q> Zener diode voltages largely follow the E24 resistor value intervals, a range of a intervals specified with +/-5 % accuracy. <S> Zener diodes have not particularly developed as a technology, with accurate voltage references using bandgap techniques instead. <S> The intervals have stayed the same for many years. <S> As an aside, Zener diodes are less commonplace in circuits than may be expected to electronics newcomers, who can sometimes regard them as appealing 'magic voltage drops'. <S> Their combination of basic inaccuracy, notable variance with temperature and current and a fairly large working current for stability lessens their usefulness compared to other design approaches and components. <A> So, quite often you'll find 4.3 volt, 4.7 volt, 5.1 volt, 5.6 volt zener diodes etc.. <S> As an aside, the 5.1 volt zener diode in several supplier ranges has the lowest temperature coefficient of voltage change with temperature: <S> - So, if you had asked why a lot of designs use 5.1 volt zeners I would say it is down to physics. <A> As Andy said, the available zeners are forced into the E24 series, like other components. <S> This is done by fiddling with the physics- lower voltage (less than about 5V) are actually Zener diodes, and higher voltage diodes are avalanche diodes. <S> The avalanche diodes work a lot better (in terms of having a sharp voltage-current 'knee'). <S> Here are some characteristic curves from Onsemi <S> that show the variation of characteristics with zener voltage (which they show as a continuous variable): <S> Note that zener impedance is plotted on a lot-log graph- <S> the impedance of a ~3V zener at 5mA is about 80 ohms, whereas that of a 10V zener is more like 8 ohms. <S> So if we use a 400 ohm resistor on the former (from a 5V supply). <S> If we similarly use a 16.7V supply and a 10V zener with 1.33K in series, the percentage regulation is more than 25x better with the latter. <S> So lower voltage zeners are pretty useless. <S> There is one niche where the zener voltage that is picked is based on physics, and on the above characteristics - and that is for voltage references. <S> For many applications band-gap references have replaced zeners- <S> they have many advantages, especially for non-critical applications- low power and low voltage operation etc. <S> They are noisy and have other disadvantages, however those have gradually been addressed. <S> Some of the very best component references were a zener diode with an integrated series diode to effectively zero out the temperature coefficient of voltage. <S> This combination only works at a single voltage/current combination- <S> about 6.2V or 6.55V, so <S> the underlying +2mV/degree C zener is about 600mV less than that. <S> An example of these parts is the 1N829 . <S> Those devices, while very stable, are less popular these days, partly because they cannot be trimmed so the voltage tolerance cannot be made extremely tight. <S> Modern bandgap references with trimmed resistors can be made to very tight tolerances. <S> Stability may not be as good as the zener though. <S> Some of the very best references available are still integrated Zener diodes (subsurface "buried" zeners with added temperature compensation and trimming, often in ovenized form). <S> An example of that is the LTZ1000, probably the best commonly available reference in terms of tempco and stability (albeit expensive, power hungry and a bit needy in other areas). <A> Zener Diodes tend to follow E24 series like Andy aka stated. <S> The stated device to device tolerance of garden variety Zeners is 5% .When <S> variations in temp and Zener current are factored in things <S> can get a lot worse than 5% .The <S> difference between 5V and 5v1 is only 2% .If <S> you tested two bags of say 10 zeners that were bag A 5V and bag B 5V1 you might not be able to tell which bag was which .The <S> same reasoning applies with 5% resistors.
| Well, like resistors and capacitors, zener diode voltage values tend to follow a range of preferred numbers such as: -
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Is the average of instanteous power equal to real power? simulate this circuit – Schematic created using CircuitLab I'm trying to understand if the average of instantaneous power is equal to real power. My reasoning is that the reactive components average out to zero and from there the average of that signal will give you just the power dissipated by the resistor. Is this true or do we always have to use P=I V pf. Thanks <Q> Instantaneous power is v x i from the perspective of the power source. <S> So, you average that to produce average power. <S> Is the average of instanteous power equal to real power? <S> It sure is - <S> that is what generates the heat in your resistor and that ultimately is what you get billed for. <A> I'm trying to understand if the average of instantaneous power is equal to real power. <S> Yes. <S> My reasoning is that the reactive components average out to zero ... <S> This is true only if their impedances match and cancel. <S> In your example $$ Z_C = <S> \frac { <S> 1}{2 <S> \pi fC} <S> = \frac { <S> 1}{2 <S> \pi \cdot <S> 60 \ 1 <S> \mu} = 2653 \; <S> \Omega $$ and $$ Z_L = 2 <S> \pi <S> f L = 2 <S> \pi \cdot <S> 60 <S> \cdot 1 <S> = 377 \; <S> \Omega $$ <S> so they don't balance. <S> There will be a reactive component in your supply load. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Since the reactive and inductive vectors don't cancel out the result is the kVA vector and it is inductive. <S> ... <S> and from there the average of that signal will give you just the power dissipated by the resistor. <S> Is this true or do we always have to use P=IVpf. <S> If you are measuring V and I at the supply then you need to use the power factor in your calculations. <S> (If you measure V and I at the resistor then the power factor there will be unity.) <A> The reactive components average to zero. <S> Internally to the load circuit, some or all of the reactive components may cancel out. <S> Any reactive power that does not average to zero inside the load must average to zero between the load and the supply. <S> The real and reactive power can be completely calculated from instantaneous measurements anywhere in the circuit. <S> As long as you are careful about the +/- <S> signs of the instantaneous values, you can calculate the real and reactive values for each circuit component and see that the real power furnished by the supply is equal to the real power dissipated by the resistor and that the reactive VA of the supply, capacitor and inductor sum to zero. <S> The same results would be found using RMS values and power factors.
| If you measure the instantaneous voltage and current at the power supply, the real power is the average of the instantaneous V X I.
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Wiring for 10000W 220V SCR Voltage Regulator Speed Controller I live in the US and have the Twist Lock Electrical Plug 4 Wire, 30 Amps, 125/250V, NEMA L14-30P. It has 2 hots, 1 ground, and 1 neutral wire. I saw the following scr controller on ebay and am confused how one might do the wiring. There's a wiring diagram on the side of the picture. Am I supposed to combine 2 hots IN, shared neutral for both IN and OUT (COM slot), and 2 hot wires from OUT? I'm used to the controllers that have 4 slots, 2 hots IN and 2 hots OUT. <Q> You are correct - <S> this is European style, where they run 220 instead of 120, so they only need one phase - a hot and a neutral. <S> Here we use two phases (2 hots) for 220, so you can't get there safely. <S> See if the seller makes a US version. <A> From your "2 hots" description it sounds as though you have a two-phase + neutral supply. <S> You need to either get two of these - one for each phase - or get a two-phase controller. <S> If getting two the control side inputs can probably be wired in parallel but you haven't provided enough info to assess that fully. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> I have the device in use as a controller for a 120 volt quartz tower heater. <S> The unit is designed to be in series with the load and is in reality a two terminal device the common terminal being unused in the circuit (thus the broken line on the diagram). <S> The common terminal is just a tie point <S> so you don't need a wire nut. <S> For 220v use just consider one line as a return.
| If you combine two hots in you will short circuit the phases together. That is a single-phase controller as clearly indicated by the diagram on the side.
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Making Hot and Cold room at the same time with Peltier stack I am thinking about 2 small boxes with peltier stack middle of them to get a cold room and hot room at the same time. Is it possible to get -40 degree celsius cold room and 100+ degree celsius hot room? Don't care about power dissipated. I will create DC power supply for that if it is possible. <Q> ( My experience is limited to building a demonstration board for our own in-house training, using our own (Maxim Integrated) <S> MAX1968 <S> /MAX1969 TEC regulator IC. <S> I used a single TEC element with a simple heatsink + fan IC cooler arrangement, and did not address regulating a temperature chamber. ) <S> Study the datasheets carefully <S> , for example CUI Inc CP85 , the CP85438 is rated 15.4V <S> 8.5A best case deltaT max 75C <S> (measured in a vacuum). <S> But the real story is found in the typical performance chart: <S> This chart shows a family of curves representing the typical achievable temperature difference, as a function of drive current. <S> At 1.7A, the temperature difference is about delta- <S> T=35C when no heat is being pumped, but drops to delta-T=0C ( no heat flow ) when 20W is being pumped. <S> At the maximum rated 8.5A drive current, a temperature difference of over <S> delta-T=70C is possible as long as there is little or no heat transfer. <S> At 8.5A this part is capable of pumping a maximum of just over 80W of heat. <S> This is what limits the performance of this component. <S> Also note how close the 6.8A and 8.5A lines are, there is diminishing returns from increasing the drive current. <S> It is also important to realize that as soon as drive current is removed, heat begins flowing back across the junction, to equalize the hot and cold side. <S> So when the device is powered off, the cold side may feel a sudden temperature increase. <A> Read the Peltier datasheet. <S> You are asking for a 140 °C difference between the hot and cold sides. <S> Most Peltiers can't do that. <S> Even if so, they will be horribly inefficient at cooling. <S> A better answer is to use a Peltier between the cold chamber and ambient. <S> Even then, -40 °C is a stretch. <S> You can then use plain resistive heating for the hot box, or possibly a Peltier between it and ambient. <S> Two separate elements are also the required if you want to regulate the temperature of both boxes. <A> I would agree with Mark, that you are wanting to achieve a bit much of a swing in temperature from ambient. <S> Assuming ambient is around 23C - then for any kind of efficiency your probably restricted to around 80-85C as seem most commonly in a number of cheap and more expensive cool box / peltier fridge devices. <S> How big are the chambers? <S> The main issue will be temperature control as each chamber will directly effect the other, on different days and with different ambient it will be impossible to control the temperature of both accurately with such a setup. <S> So if the application is critical I would say it will not work. <S> I have tried a similar setup with two columns of water, not as a fixed setup but as an experiment for testing some peltier devices, and controlling the hot side temperature we had significant (in our application) variation on the cold side temperatures day-to-day based on small fluctuation in ambient temps. <S> -40C is not going to be easy to achieve with such a setup either.
| It might be possible with careful component selection and design attention to insulating the hot and cold chambers. In my limited experience, 140C is far too much temperature difference for typical Peltier stack.
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Thickness of Depletion region, transistor I came across a doubt regarding thickness of Depletion region in different regions of transistor while reading an online article. It says that the thickness of Depletion region in collector is more than base, but i think thickness in base should be more because it is lightly doped, am I correct? <Q> No, you're incorrect. <S> The Doping doping levels are: E: highest doping level B: less than E C: <S> weakest doping, less than B <S> It needs to be like that because the ratio of B/E doping sets the beta <S> The collector has the weakest doping level so that the depletion layer will be large so that it can handle a large reverse voltage. <S> Remember that in the active region the BC junction is reverse biased ! <S> The thickness of the base must be small, the base must be thin because the carriers coming from the emitter should be have already passed through the base and enter the collector region before they realize: Oh, oops ! <S> Too late to recombine in the base . <S> And, Oh, blast now we're pulled to the collector contact (because that will have a large positive voltage, for an NPN that is). <S> That is what makes the base current small and forces the carriers to go to the collector instead of recombining in the base. <S> Your assumption that the base has a lighter doping than the collector is not true. <S> It is the collector that has the lightest doping. <A> You are mistaken. <S> It is so because the doping level is emitter > base > collector. <S> In the active mode of transistor, the depletion layer is weakened slightly by the additional holes in the middle layer. <S> The barrier is now low so that electrons can diffuse into the depletion layer to reach the point at which they are pulled to the collector by the electric fields inside of the transistor. <S> This makes the width at collector side more and at base side less. <S> Hence, the electrons can now move from the emitter to the collector. <S> The transistor became conductive across the emitter-collector line. <A> You need to stand firmly on two grounds if you want to remove your doubt. <S> They are:- 1) <S> the doping levels of various sections of transistor.2) the biasing of different sections. <S> Lets cover them one by one. <S> The doping levels of various sections are :Emitter>base >collector And the EB junction is usually forward biased and $CB$ junction is reversed biased. <S> Since the EB junction is forward biased so charge carriers moves towards the junction. <S> Now since the base is lightly doped recombination does not takes place appreciably in the base region . <S> And a large amount charge carriers are attracted towards the collector due to its high biasing applied to its junction. <S> Since the CB region is reverse biased heavily ( by heavily <S> i mean the potential across CB region is very high ) while the EB region is forward biased lightly <S> so the depletion region should be more on the CB region as compared to EB region . <S> And The depletion region is more diffused to the collector side because it is less doped than the base. <S> Hope this helps!
| The depletion region in the base will be smaller than the depletion region in the collector because the base has a higher doping level !
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1V to 5V transistor amplifier for arduino digital inputs I'm trying to use a soundcard to send a trigger signal to an Arduino. However, the software I'm using limits the output of the soundcard to between -1 and 1V, which is insufficient to trigger the digital pin on the Arduino. Basically, I want to try and amplify this 1V signal to a 3.3-5V signal in the simplest way possible, maybe by using a transistor to route the Arduinos 5V supply to the digital pin. This is somewhat time sensitive, and the only components I have available at the moment are some NPN transistors (2N2222) and a broad selection of resistors. Can I build a transistor amplifier to get the 5V I need? I'm new to this, so I am also not sure how/where to put resistors in this circuit. Any help would be greatly appreciated! edit: Just to clarify, I'm using the soundcard to present a sound on one channel, and this trigger signal, in the form of a short pulse, to accurately estimate the onset of the sound from the arduino/for other experimental synchronization purposes <Q> This will turn on the transistor when the audio voltage exceeds approx. <S> 0.7V, this will pull the IO pin low. <S> It will only trigger on the +ve going part of the audio signal. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> I assume your sound card output is capacitively coupled ...if not then you could put a capacitor (1 uF or more) in series with the input. <S> The Arduino input pin should toggle for signals above about 0.7-0.8 V positive peak. <S> If you use low frequency sine/square waves you may be able to get both presence and a frequency indicator. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> You can use the below circuit. <S> As you haven't any diodes to hand, Q1's base-emitter diode is used to protect Q2's base from large negative voltages, such as static or transients. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> This is complicated by a general CMOS requirement to drive an input very close to logic low, or very close to logic high - lingering near half-way stresses the logic gate with excess shoot-through current, or may cause oscillations. <S> So a fairly high-gain amplifier is required, or a comparator. <S> Another complication involves AC coupling that a sound-card invariably employs: its average output will always be zero volts, even if you program full amplitude +1V or full <S> -1V. <S> So your threshold circuit will only be able to detect changes of amplitude. <S> simulate this circuit – <S> Schematic created using <S> CircuitLab This inverting amplifier will boost small changes at its input to logic-level changes at its output ( Logic out ) for a microcontroller. <S> With no input, or too-small input, base biasing is arranged that "Logic out" will be high, pulled up via R3. <S> Positive-going pulses at the input will turn on the transistor, and produce a logic low at the output, for a short time only (probably milliseconds). <S> Audio output must have its ground connected in common with the microcontroller's ground. <S> Since this is a non-linear amplifier, output is very distorted, and would sound unintelligible. <S> A continuous, full-amplitude tone at the input would produce logic-level pulses of the same frequency at the logic output.
| It would seem that you require a non-linear amplifier, because the digital input will only recognize a logic high (near Arduino + supply) or a logic low (near Arduino's ground).
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LED anvil and post are reversed! Typically, through-hole LEDs seem to have the cathode (-) side connected to the "anvil" and the annode (+) connected to the "post" (see diagram below). ...But I now have in front of me an LED that does not follow that convention. MPN: C4SMF-RJF-CT0W0BB1 The light is coming from the annode (+) side. Note that this LED does not have a "flat spot" as seen in the diagram. The lens is actually ovular instead of round. The cathode is still the shorter leg, so I assume that convention that is more universal. As for the anvil/post, why would this one LED be different, and is there any purpose to either convention? <Q> The LED chip can be made to bond in a cathode up or anode up configuration. <S> The pole that is down is typically bonded with a thermal epoxy that transports the heat from the die to the lead frame. <S> The top pole is typically gold wire ball or wedge bonded to the other part of the frame although aluminum wedge bonding is also used. <A> Was your LED essentially free from China? <S> Are they really Crees? <S> I have a bag of similar LEDs from eBay in pretty blue. <S> They came without any information, just a ziploc bag that was made of the thinnest plastic I've ever seen. <S> Cathode is still the short lead. <S> Your statement regarding that standard might be true. <S> But the flat is on the anode(+) side. <S> They work satisfactorily. <S> This is the first experience I've ever had with atypically formed LEDs, so perhaps it's not all that common. <S> We could do a poll if SE were into that kind of thing... <A> When making 5mm LEDs you can choose; beam angle , round or oval, lambertian or batwing, diffused or not, colored or clear; ESD protected or not. <S> THe leads may be have stoppers or not and the case may be smooth or D coded for cathode. <S> ( the flat edge being the bar on the diode symbol)e <S> THe flange is useful as a stopper when paired with the lead stoppers when a soft edge is avalable. <S> But a hard edge causes interference, so flangeless in a matching hole allows more height variation without interference. <S> So no D and no flange. <S> By convention, Anode (+) has the longer lead. <S> THese are conventional flangeless parts with lead stoppers. <S> (not wrong) <S> Only once have I seen D cases on the wrong side and that was the 1st and last time I ordered 10k of those parts by mistake 12 years ago.
| Regardless of the bonding orientation, manufacturer's should follow the convention that the shorter lead is the cathode, the longer lead is the anode, and the flat side of the package indicates the cathode.
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How to Generate Heat with Joule Effect? I'm going to build a coffee heater for a school project. I know I have to take advantage of the Joule effect (Joule heating). I am restricted to using a resistance between 330 Ω and 1 MΩ. The voltage supply is 5 V, via USB connection. If I use resistors, then the circuit won't get hot. If I use diodes, their temperature will increase but not enough. How do I generate enough heat in the circuit to heat coffee? <Q> If you study physics you should learn that $$ t = \frac <S> {\Delta T \times m \times SHC}{P} <S> $$ where P is power in watts (W), <S> t is time in seconds (s), SHC is specific heat capacity (kJ/kgK) and m is mass (kg). <S> SHC for water is 4.2 kJ/kgK. Testing for a 2.1 kW kettle with one litre of water at room temperature and rearranging the formula <S> we get $$ t = \frac <S> {\Delta T \times m \times SHC}{P} <S> = \frac {(100 - 20) \times <S> 1 \times 4.2}{2.1} = <S> 80 <S> \times 2 = 160 \; s$$So a bit under three minutes to boil a litre of water from room temperature. <S> This sounds about right. <S> Since you want to know what power to use you should rearrange the formula: <S> $$ P = \frac <S> {\Delta T \times m \times <S> SHC}{t} $$ Plug in your \$ \Delta T, <S> \; <S> m, SHC \$ and \$ t \$ and calculate the power you think you need. <S> If you know your supply voltage <S> you can now work out the resistor value from $$ R = <S> \frac <S> {V^2}{P} $$ <S> Finally you can work out the current drawn from your battery from $$ <S> I = <S> \frac { <S> P}{V} $$ <S> All of the above assumes a well insulated container. <S> An open top or conducting sides will cause heat loss. <S> The power supply is with a USB cable. <S> You can probably get 1 A at 5 V from a USB power supply. <S> From \$ P = VI \$ you can see that you have 5 W for your heater. <S> Plug that into the formula for calculating \$ t \$ and see if you think heating coffee from a USB port is feasible. <A> Sorry, but there it is. <S> A USB power supply will put out a maximum (typically) of 1.5 amps. <S> Total power is volts times amps, so this is a maximum of 7.5 watts. <S> However, with a 330 ohm resistor, current is 5 volts divided by 330, or .015 amps. <S> If you were to connect a bunch of resistor is parallel, and all were 330 ohms, each would produce .075 watts separately, so 100 of them would produce the maximum power. <S> So your demand that 330 ohm to 1 megohm resistors must be used condemns you to failure. <A> It sounds like you're using the wrong resistors. <S> The power, P is given by P = <S> V²/R <S> or P = <S> I²R <S> Use whichever is most useful to you. <S> Swap the equations around using basic algebra to work out whichever term you don't know <S> (probably the resistance you need). <S> Depending on how much coffee you're wanting to heat, you might need anything from a few tens of watts up to a few kilowatts. <S> You don't say what your power source is. <A> Consider the Max current on the power supply and use maybe 100 watts with thermal pasted and clamped ring heater insulated nichrome wire <S> Maybe 5A at 12V will keep it warm, but not boil cold water. <S> Try 50 Watts from 5V 50 Watt supply or more (ATX PSU is good) using V^2/W <S> =R = <S> 25/50 <S> = 0.5 <S> Ohms net resistance = <S> 330 ohms <S> / 660 resistors in parallel = 0.5 Ohm . <S> This means 1/4W resistors with < 1/10W per R <S> but since touching side to side over the whole plate it should get to 85'C . <S> like this and use 24 to 30 AWG magnet wire also embedded in the epoxy. <S> What I'd do is figure how to solder 100 resistors together like firecrackers so that the ends do not short on the plate. <S> using the cut resistor wires to make a bar across as many as it will reach in a neat flat bundle like firecrackers. <S> In bulk there are only about a penny each and 10% of that in high volume. <S> ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| For safety add a thermistor or wire 100 resistors at a time and try to fit them on a baseplate in parallel strings so you can always switch off the outer strings sepearately. If you're using off-the-shelf resistors, make sure they are rated for the power you intend to produce, with a bit of safety margin on top. Power is then 5 volts time .015 amps, or .075 watts, and this is just not enough to warm anything. What you are doing is not going to work. Wiring a home-made coffee heater directly to the mains is probably a bit dangerous, so a bench power supply might be more sensible.
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MOSFET overheating in LED driver I'm trying to design an isolated LED driver using this IC. The problem is that when I turn on the driver the MOSFET heats up too quickly and it fails if it stays on for a few seconds. I have attached a moderately sized heat sink suitable for To-220 package to the MOSFET. I have tried changing the gate resistors and even shorted them because I thought the gate was not turned on fully but the result is still the same. I don't have a temperature sensor but the heat sink heats up to extremely hot to touch in just a couple of seconds (~5 seconds). Obviously the MOSFET shouldn't dissipate this much energy when I'm running a 15W LED. What could be the reason for this? <Q> Designing switched power supplies is not easy. <S> (Not for beginners) <S> Well, a power transister becomes hot out of two possible reasons: The current that flows when it switches through multiplied with the voltage that remains between Drain and Source. <S> You can connect an oscilloscope and measure the remaining voltage between drain and source when it is switched through and calulate the current by measuring the the voltage at the resistors RS1 - RS5. <S> Form that you calulate the power that the transistor is consuming. <S> If this power is 1 <S> Watt you already feel the transistor is warm. <S> If it exceeds 3 Watt it will become so hot that you cannot touch it anymore with your finger if there is no cooler. <S> The other reason for extreme heat (and that is more probable) is that the signal at the gate is not sufficient square. <S> If the voltage rises and falls too slowly you will have a lot of loss in the transistor. <S> So to answer your question you must provide the oscilloscope signals measured between gate and source and between drain and source and at RS1 - RS5. <S> If the signal at the gate is OK I suppose that your transformator is not well designed. <S> Generally the information you give is very basic. <S> What voltage is DCPOS ? <S> What are you connecting at the output? <S> Does the transistor also become so hot when the output is without load? <A> Gate resistors are very high. <S> And the 100k across G-S is very high, too. <S> That resistors ( R5 || [R3+R4] ) and the input capacitance of the MOSFET form a low-pass filter having a cutoff frequency of f=1/(2 pi 340R 1.2nF) <S> = 390kHz. <S> This can seem high enough, but it can be low enough to smooth the sharp edges of the gate drive signal which can cause the MOSFET to overheat due to insufficient driving. <S> Remove R3(10R) gate resistor, put a 0R. R4 can be between 1R and 4R7. <S> R5 can be in 1k-10k range. <S> Another possible reason is the ringing across D-S caused by switched primary inductor. <S> There is a snubber formed by D3-R6-R7-C5 but they may not be sufficient. <S> Note that the snubbers in a Flyback should be designed carefully. <A> If i may weigh in, there are some circuits controlling massive amounts of power and hardly getting warm at all. <S> The trick is either on or off, and not too much time in the transistion period.
| Some devices benifit from dampers across the device (diodes, caps,) but often its a case of adequate drive: on or off, nothing between.
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Substituting a matrix keyboard push button with transistor I have a remote like this I removed the plastic sticker and below, we have a series of push buttons. Two pieces of copper in SSS form. When you push and the plastic bends, a little metal piece touches the SSS and you close the circuit. The device seams to be performing some kind of scan over the rows and the columns since the buttons clearly share "pins" of the "chip". I tried short-cutting with a piece of wire the two sides of one button and I can easily fake a finger press. I tried substituting my finger by a transistor powered from the RasPi and .. there is no way to get it to work :( I removed the battery of the remote and powered straight from the 3V3 of the Pi. The device works (short cutting still works). I connected emitter and collector to the proper places. (To discover the polarity I short cut it with a diode. It only works in one direction as expected.) I then connected a 200 omh resistor to the base and connected to the 3V3 and ... nothing happens ... What can be happening? I tested the transistor in a protoboard with an LED and it works as expected. <Q> This means there is no common potential you could refer to, but you need that to place transistors across more than one of the keys, and use a raspberry to control them. <S> I recommend YOU to use reed relays. <S> There are ways to get around that limitation but they are not easy to implement and require to re-engineer the schematic first. <A> I understood that you got a proper fake pressing generated by connecting a diode. <S> They do not care the potential difference between your Pi and the matrix keyboard reader. <S> I have once generated fake pressings by connecting 2 analog multiplexers between the X and Y rows of an existing matrix keyboard. <S> It was year 1982. <S> I wanted to get high quality printed output from a computer that had Centronix type printer interface. <S> I built a box that made an electronic typewriter to Centronix compatible printer. <S> The needed X and Y numbers for each character were stored in EPROM. <S> If you haven't it, then try optocouplers or even relays that are even more robust. <A> This is similar to <S> How do I read a Board? <S> in that the author was trying to perform a similar modification. <S> Figure 1. <S> A typical keyboard matrix. <S> Scanning the keyboard is done by pulling each row low in sequence and reading which, if any, column is pulled low by a switch. <S> This arrangement allows a matrix of R x C buttons with only R + C pins on the chip. <S> Check the battery supply on the remote. <S> If it is 3 V and the Raspberry Pi is 3.3 V <S> then you can probably remove the remote battery and power it from the Pi. <S> This will make control much simpler. <S> See the linked question for one means of performing the switching action using a CD4016 CMOS analog switch. <S> Another idea would be to try an FET opto-isolator. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> Pulling the GPIO pin low will turn on D1 in the FET opto-isolator shorting out the switch contacts. <S> A regular transistor opto-isolator might work too <S> but you'd have to get the transistor connected right way around on the matrix. <S> The FET should work either way. <S> I've never tried this <S> but it might work!
| This most likely is a matrix keyboard. Then you shoul succeed also by inserting optocouplers. They are small and they don't give you a headache. It's well possible that you can do the same, except no EPROM is needed, only 2 analog multiplexers and six outputs to it from PI. This unfortunately needs exact knowledge of voltages and polarities. Note that with the arrangement shown in Figure 1 that neither side of any switch is connected to ground so your simple transistor idea won't work.
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Voltage Regulated by Resistor in Generator Circuit So I am tinkering with making a circuit that will allow a motor to generate power to charge a battery. I have this, what I presumed to be an, AC induction motor I salvaged from an old Christmas decoration. After removing the motor from its casing, I have learned it is a synchronous motor. This one with the ratings of 120VAC 3.8W 4.2/5 RMP It can output upwards of 200 Volts AC in short circuit (or DC using a rectifier bridge) at ~ 6.7mA. I could only get the amperage reading from a short circuit through the rectifier. Could be my $7 multimeter not doing well with AC or my ignorance on reading AC amperage. Strangely enough (at least to me), no matter how I worked it: the amperage would stay at a consistent cap of ~ 6.7mA. In my tinkerings I figured out there is, at least almost, a straight line which shows that the resistance given to the circuit will output a maximum voltage I can get from the motor itself. The diagram posted is a test circuit to gather data points on this. I'm wondering if there is anyone with an idea as to what is causing this phenomena? Here is a chart and graph of the voltages across the entire circuit (from either end of the rectifier bridge), given the different values of R1. Definitely some good answers. Not sure which is the best answer.I appreciate all of the input, and will select the best answer once I get back from work and have time to do some more testing, and take apart the motor to see what's really going on inside. To clarify: the end game is to maximize on incoming voltage so I can reduce the voltage later in the circuit and bump up the amperage to charge a battery somewhat efficiently. Also to understand why there seems to be this constant 6.5mA coming from this motor. I may just go back to my research and choose a best answer for now. If I run into anything interesting down the road, I'll make sure to post again. <Q> Great job on the experiment! <S> The motor that you are using as a generator has a high internal impedance largely due to the windings and the internal magnetics structure. <S> You can think of this as a resistor inside of the motor that is in series with its output. <S> Of course this is not a real resistor, just a way of modeling it. <S> You didn't mention it, but when you are experimenting consider that the load placed on your motor acting as the generator may cause the driving motor to speed up or slow down. <S> This will of course affect the results of your experiment. <S> In electronics, we know that maximum power will be transferred to a load when the load impedance matches the source impedance. <S> You may find it interesting to add a column to your chart that shows power in the load (= V 2 /R) to see if you can find the point of maximum power transfer. <S> You will need to extend your experiment with higher values of resistance most likely. <S> Once you have determined the maximum power you can obtain from your generator, you can then see if it is suitable to power your target device. <S> If it does have enough power, then the most likely the solution will require a buck regulator to efficiently step down the higher voltage. <S> Keep up the good work. <A> Over the range you measured, the generator is largely acting like a current source. <S> To first approximation, the generator can be modeled as a voltage source in series with a resistance. <S> The voltage is directly proportional to rotation speed, and the resistance is reasonably fixed. <S> You say you are getting 200 V open circuit voltage, and about 6.6 mA short circuit current. <S> Assuming that the generator is still spinning at the same speed when shorted as when open, the internal resistance of the generator is (200 V)/(6.6 mA <S> ) = 30 kΩ. <S> This value will be skewed if the generator actually slowed down when the output was shorted. <S> Here is the simplified model of your generator and rectifier diodes: <S> If the above is correct, then you will get largely constant current for loads significantly less than 30 kΩ. <S> At 30 kΩ, you should get half the short circuit current at half the open circuit voltage. <S> That is the point where the generator produces maximum power. <S> At significantly above 30 kΩ load, the generator will look largely like a voltage source of 200 V. <A> An induction motor would not generate very much voltage into a rectifier circuit. <S> The rotor might have a small amount of residual magnetisim to allow it to generate a little voltage acting as a permanent-magnet synchronous generator. <S> If the motor is generating more than 50 volts, it may be a permanent-magnet synchronous motor like a clock or timer motor. <S> When using a salvaged motor, it is very helpful to find all of the information marked on the motor and on the product from which it was removed. <S> If the product contains other electrical components, that are connected to the motor, it is important to have those components and be able to re-connect them after they and the motor are removed. <S> It is also helpful to know about any other use of electrical power in the product. <S> Before attempting to use a motor as a generator, the motor should be tested as a motor. <S> It is best to test it as it was originally used. <S> Determine the voltage, current, power and speed with the original load and with no load. <S> Measure the DC resistance. <S> Detailed pictures and dimensions can be helpful. <A> We learned in the motor class at UCLA that every motor is also a generator. <S> Syncronous motors will generate and buck the power draw according to the phase relationship between the applied voltage and the motor position when it is under load. <S> When the load is negative (somebody is turning the crank and trying to make the motor go faster), the power draw becomes negative. <S> That is how a synchronous motor becomes a generator. <S> You regulate the output by regulating the mechanical power applied to the shaft. <S> The whole thing is an exercise similar to financial book keeping: <S> account for all of the energy. <S> I don't think Tesla had DC in mind when he invented AC motors.
| It could also be a permanent-magnet DC motor with a commutator, but that would be unusual for a small motor at that voltage level.
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Feasibility of reverse polarity feature If I were to design a circuit for a small appliance that works in DC and I would want to allow it to work with the batteries in reverse polarity could I feasibly use a diode bridge? Would it make sense? Would the energy efficiency drop significantly? <Q> It would work. <S> The loss in efficiency would be given by the diode drop divided by the operating voltage. <S> To give a couple of examples: Using a standard 1n4148 diode, with a forward voltage drop of 0.65V, and a 1.5V battery, efficiency is only 55%. <S> Using a 0.4V diode with a 48V lead acid would give 99% efficiency. <S> These numbers are to provide protection only - plugging it in backwards wont work but won't damage anything either. <S> For it to work both ways, you need two diode drops, and efficiency is 10% and 98% respectively. <A> Have a look at the LT3240 Ideal Diode Bridge Controller. <S> The LT®4320/ <S> LT4320-1 are ideal diode bridge controllers that drive four N-channel MOSFETs, supporting voltage rectification from DC to 600Hz typical. <S> By maximizing available voltage and reducing power dissipation (see thermograph comparison below), the ideal diode bridge simplifies power supply design and reduces power supply cost, especially in low voltage applications. <S> Source . <S> Using MOSFETs rather than diodes reduces voltage drop. <A> Old School simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Old school relay solution. <S> If V1 is positive RLY1 picks. <S> If V2 is positive RLY2 picks. <S> Voltage drop is zero but the penalty is a little current for the relays. <S> You can think of this as a very slow full wave mechanical rectifier. <A> You could use a diode bridge. <S> A silicon bridge would lose you nominally 1.4v, a schottky bridge 0.7v. <S> A better solution would be to use a MOSFET bridge, which being resistive when on would result in near zero drop. <S> For a range of input voltages > Vgs(th) but < Vgs(max), the following should work ... <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Four complementary FETs are shown, with their intrinsic body diodes. <S> Neglecting any FET operation, the body diodes form a normal bridge rectifier. <S> Now consider Va high, Vb low. <S> Va high turns on M1. <S> Vb low turns on M4. <S> M1 shorts D1, which is conducting anyway to connect Vb to the output. <S> M4 shorts D4, which is conducting anyway to connect Va to the output. <S> Result, happiness, as long as Vin exceeds Vgs(th) for both types. <S> Obviously the complementary thing happens for Vb high. <S> Another way to look at it is to notice we have two complementary FET inverters.
| Obviously, efficiency would be maximised for a low-drop diode, and a high operating voltage.
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How can I connect more than device on the same Isolation transformer? Why isolating transformer should be used to protect only one item of equipment at a time? I tried to search and I found this answer , but I could not understand it. This is why, ideally, an isolating transformer should be used to protect only one item of equipment at a time. With one item a fault in the equipment will probably not produce a dangerous situation. The transformer has done its job. BUT with N items of equipment - if one has a fault from neutral to case or is wired wrongly this may defeat the transformer such that a second faulty device may then present a hazard to the user. <Q> However, in some cases this makes the setup less safe. <S> Whatever is powered from the isolation transformer requires two connections to some other object (like you) for current to flow thru that object. <S> Equipment that is powered from a unisolated source can cause shocks with only one other connection. <S> That is because one connection has already been made to ground. <S> The more equipment you power from the same isolation transformer, the more chances of one accidental connection to the rest of the world, thereby making everything else less safe. <S> This is what the paragraph you quoted seems to be referring to. <S> However, there are also safety advantages to powering more equipment from the same isolation transformer. <S> You are still more protected by two connections being required before shock. <S> If the multiple pieces of equipment need to be directly connected together anyway, then you must power them from the same isolation transformer if you want there to be any isolation. <A> An isolation transformer is a transformer used to transfer electrical power from a source of alternating current (AC) power to some equipment or device while isolating the powered device from the power source, usually for safety reasons. <S> Now independent device may have different safety ratings. <S> (safe current, safe power etc.) <S> So that's why it is obvious to use separate transformers. <S> But if the case is the safety requirements for all the devices are same then you can connect multiple devices to same transformer. <A> To the best of my knowledge, isolation transformers are used for a couple reasons; Safety, and Troubleshooting/Testing. <S> Both of which benefit from the isolation which decouples/isolates the Earth ground connection of the primary and the equipment/circuit ground of the secondary. <S> The secondary becomes its own isolated loop. <S> General practice has always been to use a single isolation transformer for each system that needs isolation because real isolation transformers come in many different configurations, usually specific to a particular piece of equipment or circuit. <S> However, I do not see why you could not hook up more than one if you really wanted to. <S> Just be sure both circuits meet the specifications of the ISO XFMR. <S> Here is a link to an article about isolation transformers. <S> https://www.allaboutcircuits.com/technical-articles/transformer-isolation/ <A> The link in your question answers why to use isolation transformers but doesn't state that only one piece of equipment should be connected to it. <S> As you seem to think, it should be possible to connect multiple devices to the one transformer. <S> One reason for not doing this is that, for some reason, the device needs to be floating (isoltated) relative to the other devices to prevent electrical interaction or strange current return paths between them. <S> In Europe where 230 V phase to neutral is the norm portable tools are powered on 110 V, centre-grounded transformers. <S> This gives a 55-0-55 secondary (with 110 V between the outer terminals) effectively limiting the shock hazard to 55 V with respect to ground. <S> If either phase shorts to ground its fuse will blow. <S> In the event that the ground connection is lost (which might not be unlikely on a building site) some reduction in risk is provided by the reduced voltage of 110 V. simulate this circuit – Schematic created using CircuitLab Figure 1. <S> European tool transformer wiring. <S> Note that the grounded centre-tap limits the phase voltage to 55 V AC. <S> In this case it is also permissible to connect multiple tools to the one transformer. <S> The first one to fault will blow the fuse and stop them all.
| You can power multiple devices from the same isolation transformer.
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Can a RF 433.92 Mhz transceiver communicate to a RF 433.42 Mhz Receiver? I am working on automating my Somfy Blinds. The idea is to avoid using the remote control which works on 433.42 Mhz and use a simple RF transmitter and an Arduino.The thing is that I can not find any transceiver at such frequency. I only can get RF 433.92 Mhz transceiver like RFM69 @ 433 MHz among others that only mention 433 Mhz without precision. I read from other projects that this will work just fine for short distances, let's say 5-10m.So before buying a module that won't work I just wanted to drop the question here for the expert <Q> A 500 KHz offset frequency would be substantially weakened by the filter of a decent OOK receiver at these frequencies, however, some lower cost appliances may use fairly wide receivers. <S> But that is really beside the point here, as the RFM69 is not a fixed frequency radio, rather it is configurable to a range of frequencies and modulation types, including the ~433.4 MHz or therabouts OOK used by a Somfy shade. <S> So the premise of the question comes only from misunderstanding, and is moot. <S> Still, your task is not that simple either, as Somfy shades use a rolling code scheme. <S> You will not be able to control one without transmitting a code which changes at each transmission in accordance with that scheme, and pairing your code sequence to the shade receiver. <A> Somfy use 443,42Mhz for carrier and RTS rolling code. <S> Independantly of rolling code, 443 generic transmitter and receiver (as low cost chinese modules) use 443,92Mhz Frequency. <S> Using this frequency on receiver as Somfy motor tuned on 443,42 works but only at short distance <S> (around 5m depended of course of walls in sight of views)About RTS and rolling code you can read lot of info on https://www.domoticz.com/forum/ <A> The README of the following github project contains some valuable information regarding this topic: https://github.com/Nickduino/Pi-Somfy <S> However those often use the frequency 433.92 MHz and require soldering skills to replace a 3-Pin 433.92 MHz chip with a 433.42 MHz chip using the same pin out. <S> I was yet unable to find a source which sells 433.42 MHz transmitter modules ready to use which don't require such a soldering patchwork. <A> Somfy in the US have a product called URTSI II which can be intefaced to using <S> RS-232, RS-485 or IR <S> so that might work for you. <S> Not sure where you are located <S> but I believe Somfy US use 433.42 and here in Australia and possibly Europe use 433.92.
| In summary: Some cheap 433MHz transmitter modules are availabe.
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Reference for color sensor I am trying to make a colorimeter using TCS3200 sensor and white LED light source. After weeks of research and understanding on how different white light sources work and what are the algorithms involved in calibrating the sensor in CIE1931 space. Now I am facing problem in getting a reference for getting a calibration curve. For example, if I want to map R values, then I also need some reference for which I know R values. I initially planned to use RAL cards to calibrate my sensor, but the RGB values are not shared on their official website and other websites provide sRGB values and not the actual RGB values. I even thought of printing varying shade of Red on paper, but I believe there will be a lot of unknown factors in this (printer calibration, RGB space used by the software, etc). So is there any standard source of known colour against which i can calibrate my sensor? Edit:My application is more to sense the change in shade rather than getting the actual color of the subject and that's the reason I was thinking if I can get away with calibrating it against some standard color swatch of known RGB values. <Q> You need a light source with a spectrum of known power vs. color (i.e. frequency or wavelength) distribution. <S> (As much as I understand your problem you don't need absolute intensities (e.g. Watt/Steradian) but just relative intensities with respect to color). <S> There come two relatively simple sources to mind: <S> black body radiation (e.g. of a incandescent light bulb) whose spectral distribution is directly known by a physical law, Planck's law (you need to know the temperature of the filament) sunlight (after passing through the atmoshphere till e.g. sea level) whose spectral distribution is empirically known (well measured) and you have a high chance of finding published spectral distribution data. <A> long story short <S> In case of colorimeters, that is something of known color illuminated with light of a known color temperature (and usually, also a known intensity). <S> That's a bit of an expensive test setup you're aiming for, if you need to build it yourself (these lamps don't come cheap). <S> Why not simply buy a sufficiently calibrated colorimeter, or a colorimeter calibration toolkit, and calibrate your device against the readings / known values of the commercial one? <S> Seems both the cheapest and easiest approach. <A> If you're only interested in matching colors you don't need to calibrate. <S> A stable colorimeter will give find differences in color which might, in themselves, be inaccurate, but will show color differences. <S> Your reference should be the sample you're trying to match. <S> The only problem is that you can't match using two different colorimeters.
| : for calibration, you're right, you'll need known quantities.
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Can a CPU function with nothing more than a power supply and a ROM, using only the internal cache as RAM? Can a CPU (such as the Intel i3/i5/i7/Xeon) with on-chip cache RAM use that as its only functional RAM, without any external memory banks attached? Or must there be external RAM, and the cache cannot be accessed or used alone? Modern desktop/server CPUs often have more internal cache RAM than many 1990's computers had in entire system memory, so there should be plenty enough there to run simple code. CPUs from before cache existed such as the 6502 would be unable to do anything, as the internal CPU RAM only amounted to a few bytes for the address counter and accumulators. This is not a question of running any sort of modern operating systems, but running simple code programmed into a custom ROM, or hand-entered with a hex input keypad. <Q> See this extremely detailed account of the PC boot sequence: http://www.drdobbs.com/parallel/booting-an-intel-architecture-system-par/232300699?pgno=2 <S> Since no DRAM is available at this point, code initially operates in a stackless environment. <S> Developers must write extremely tight code when using this cache-as-RAM feature because an eviction would be unacceptable to the system at this point in the boot sequence; there is no memory to maintain coherency. <S> That's why processors operate in "No Evict Mode" (NEM) at this point in the boot process, when they are operating on a cache-as-RAM basis. <S> In NEM, a cache-line miss in the processor will not cause an eviction. <S> Developing code with an available software stack is much easier, and initialization code often performs the minimal setup to use a stack even prior to DRAM initialization. <S> You can observe this by running a PC without RAM: it will play a series of beeps. <S> The program that plays those is run from the BIOS Flash ROM. <S> I've also seen this behaviour on some ARM processors. <S> There will be configuration registers inside the SoC that allow you to use the cache as RAM early on in the boot sequence, in order to run a program that finds, enumerates and configures the DRAM. <A> Generally, the cache memory is not addressable. <S> A program cannot store or retrieve data intentionally from it. <A> While this does not directly address the processor families specified in the question, the scheme below would work on the earlier x86 processors so, yes it is possible to operate without either RAM or cache, although this approach requires some creative programming skills. <S> Back in the 1980's I came across a design for a radio receiver that decoded <S> the MSF time signals broadcast in the UK. <S> This design used a Z80 processor and only had a ROM for the program storage. <S> All of the processing and data storage was performed using the internal registers within the processor. <S> This obviously meant that there could be no subroutine calls as there was memory available to hold the stack. <S> Back then the cost of RAM was high and as this was a hobby project, keeping costs down was important, quite apart from it being an interesting academic exercise. <S> This was also before the days of widely available microcontrollers (an 8751 with eprom cost over £100 IIRC).
| Most modern processors have an internal cache that can be configured as RAM to provide a software stack.
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Can I identify an LED's characteristics from the transistor used as its current limiter? I'm a total NOOB, but I am teaching myself electronics while in retirement. I want to change the color of the "Power ON" LED in my mod of an ATX-PSU, but I don't know the characteristics of the LED. Current to the LED is controlled by a Fairchild C1815 NPN transistor. Can I use the C1815 data sheet to ferret out the LED's numbers? <Q> Usually NOT. <S> That said: there is usually significant latitude when swapping one LED for another that has vaguely similar characteristics. <S> Changing the LED to a different color or package will most likely work just fine because a small change in the forward voltage won't affect the LED current noticeably. <S> Usually, that is. <S> The easiest way to tackle this project is to just do it and observe the results. <A> The transistor will usually be used as a switch. <S> There should also be a resistor in series swith the LED which will limit the current. <S> You can determine the current used by the LED by measuring the voltage across that resistor, and its value - then calculate the current using Ohm's Law. <S> For common small LEDs, you can look at any distributor's website to find the "usual" maximum current and forward voltage rating for each colour of LED - but LEDs used as indicator lights are usually operated a much less than the maximum current - perhaps 5 - 10 mA <S> where the Absolute Maximum current in a datasheet will be 25 - 30 mA. <A> Can I use the C1815 data sheet to ferret out the LED's numbers? <S> No you can't <S> and you also need to (more than likely) <S> know the value of the current limiting resistor used <S> so the steps are: - Measure the open circuit voltage (when supposedly on) and this means disconnecting the LED. <S> Then measure the current flowing into the LED. <S> And to be absolutely sure, you need to establish if the LED is being driven from a full-wave rectified supply that is current limited by the resistor because this makes a difference to how you choose the LED. <S> Of course you can just swap it out and hope for the best <S> but where's the learning in that?
| Most LEDs are supplied with a voltage source that is significantly higher than the LED forward voltage.
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Butterworth filter with non arbitrary component How do I design a filter with certain cutoff frequency and as flat pass-band as possible, given that: one of the filter components is non arbitrary, and that filter component does not match the corresponding Butterworth prototype (i.e., the impedance conditions and cutoff frequency)? For this question, I understand that performance will be degraded compared to real Butterworth filter. Still, is there known analytical or numerical method for designing this to match the Butterworth filter as closely as possible? As you see in the schematic below, I can arbitrarily choose values for \$L_{1}\$ and \$C_{2}\$. However, I cannot change the source or load impedances (since they do not change with frequency), nor can I change \$C_{fixed}\$. simulate this circuit – Schematic created using CircuitLab <Q> The quickest method is plug the circuit into a simulator like LTSpice to find values for the unknown components that produce what you feel to be the best response. <S> Sims like this are free but have a steep learning curve but with something as simple as this is probably worth the effort <S> and you'll never look back once you have got the hang of it because you'll be able to use the sim for all manner of electronic circuits. <S> Algebraically, this could be an awkward problem to solve <S> but numerically it'll be a breeze to a sim. <A> It saves me a lot of back and forth calculations in situations like yours. <A> You probably can find proper source and load impedances that allow the fixed capacitor (= previously selected) and still give to you the wanted frequency response. <S> By adding 2 resistors to both input and output you can fool the LC-circuit to see the right source and load impedances, but you must accept the caused losses. <S> You can prevent the losses if you can insert buffer amps just after the source and just before the final load. <S> Probably only one amp is needed because there are available prototype filters with different impedance ratios. <S> Finally there's allways possible to use circuit analysis to find if the result still is acceptable regardless some wrong values. <S> Nearly any AC analysis capable program can be used, But high end circuit analysis programs surely have tools for tuning the circuit without compromising too much the frequency response. <S> ADDENDUM: <S> User Glenn W9IQ just showed us one such tool in his answer. <S> Still better: A free version for simple circuits is available. <S> Upvote to him is well used! <A> Like the Butterworth itself, this is an optimization problem. <S> Pick a way to parametrize "flatness in the Pass Band" as an error term (call it E ) -- <S> maybe rms deviation from flat, or some such -- figure out the closed form solution <S> E in terms of your fixed and floating parameters, and minimize it.
| There is a free filter design tool called Elsie that allows you to fix components or find closest commercial values while showing you the effect on the passband, stop band, etc.
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Need more current than 4 mA in 4/20mA loop current We know that in a 4/20 mA loop current, 4 mA represent 0% and 20 mA represent 100%. For loop current we use an IC like XTR115, and my circuit is fed from this IC. If my circuit needs higher current than 4 mA to run, how can I obtain it? <Q> For example, if you need 3V at 20mA (60mW) you could drop 17V or so and if your DC-DC is efficient enough you will stay under <S> the 4mA. Chances are you will have to design this, most chips and modules use too much quiescent current to make this work. <S> Optionally, and this is common practice if you are not even close on the current, is to use an external supply and DC-DC isolation to provide the necessary current. <S> There are many kinds of devices that cannot operate from the small amount of power in a loop-powered instrument. <S> Of course it increases the wiring cost and may make intrinsic safety more complex. <S> Another option (probably less popular these days) would be to use a different analog current loop standard such as 10-50mA. <A> When designing a transmitter with the XTR115/6 <S> your options are limited by the chip specifications. <S> In particular the 5 volt regulator output current is limited to < 12 mA. <S> There is a knee in the voltage output at 1 mA of current where it drops to about 4.5 volts where it remains up to I max . <S> You will find that if you are designing industrial 4-20mA devices, this is about the maximum power budget you have available since the loop specifications are made to limit power for intrinsically safe environments. <S> Note that if instead of using the internal regulator, you add a buck converter to your circuit, for example, you are adding "energy storage" devices to the design which will force you to go through certification if you desire an IS rating. <S> If you already have other "non-simple" devices in your design, this may not be a deciding issue. <S> Edit: <S> You may find this app note from TI regarding harvesting energy from a 4-20 mA loop applicable to your situation. <S> They walk through a design using a TPS62125 to provide a 10 mA supply with some higher peak current capabilities. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> (a) 2-wire 4 - 20 mA sensor and (b) 3-wire sensor configurations. <S> You now have a greatly increased power budget. <S> Figure 2. <S> Different arrangements for 2-wire and 3-wire transmitters. <S> Source: e2e.ti.com . <S> 2-wire 4-20mA <S> Transmitter: <S> Accepts loop-supply from 4-20mA receiver Sinks a return current between 4-20mA 3-wire <S> 4-20mA Transmitter: <S> Accepts power and GND from a local supply Sources an output current between 4-20mA <S> For the 3-wire transmitter you may be better using the XTR111 . <S> Figure 3: XTR111 application note extract. <A> You have a few options but it depends on how much current you require and how much burden voltage you can tolerate. <S> The obvious solution is to use a small DC-DC converter to step the incoming voltage down to a lower value. <S> This will give you more current at the expense of increasing the minimum burden voltage. <S> Please modify your question stating exactly what your requirements are.
| You can increase the voltage drop and use a (very efficient) DC-DC converter to convert the higher voltage drop to a higher current. One way out of the problem - but perhaps not easily achievable with the XTR115 chip - is to add a third wire.
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74181 ALU not working properly cascaded to add 8 bit numbers I'm trying to make a homebrew CPU but having problems with the 74LS181 ALU, which I'm simulating in Logisim software before diving into real ICs and breadboards. I've cascaded two 74181 following instructions found in many sites such as this . This works fine with most of addings I tried, but some don't work. For example: 0010 0111 (39 in decimal)+ 0000 0001 (1) leads to a 0010 0000 (32) result. I know it should have something to do with the fact that 74181 uses signed numbers represented in two's complement. The least significant nibble in the example (0111) represents the max positive number with 4 bits, so when incremented by one goes to zero... Is there any way to make 74181 work with only positive numbers? Or solve this issue in any other way. I'm frustrated because with this problem the ALU is completelly usesless. link for 74181 datasheet link for project (has image of error on simulated ALU) - link on comment bellow <Q> Please show how you are interconnecting the two devices, particularly the carry from the LSB to the MSB device. <S> It is up to you to interpret the values in the form you need. <S> The two devices should be interconnected similar to this: <A> First, get rid of the 2nd 181. <S> Concentrate on getting a 4-bit result and understanding it. <S> Second, combining 0111 and 0001 to get 0000 is very strange, but for instance, if you are applying 0001 to the A inputs, and 0111 to the B, and a Mode select of 1111, you'll get A - 1, or 0000. <S> So I'd recheck your select lines and your inputs. <S> I'd especially check things like a swap between pins 2 and 3, which would confuse A0 and S0. <A> Based on your example of the error, you don't have Cn and Cn+4 connected correctly. <S> Your Cn input is floating and inputting a phantom carry-in and adding an unintended +1 to the low stage. <S> And the upper stage which should receive a carry-in looks like it is tied low instead. <S> Your problem is not related to signed vs. unsigned binary numbers. <S> The adder works the same on each bit stage regardless of your interpretation of what the binary represents. <S> That's the beauty of 2's complements. <S> The logic circuits for each bit stage is the same (generally). <S> You just interpret the values differently. <A> All the answers helped me in finding the bug. <S> Actually the error was in the implementation of 74181 <S> ALU in Logisim, as I supposed. <S> There are 3 file versions of the chip, but the right one is (strangelly) <S> the v2 not the v3. <S> I will put a comment on the author's page to warn others in the future. <S> Besides this the Carry output/input connection was inverted, should be from the ALU of Low nibble (4 least significant bits) to the High nibble. <S> I made the opposite. <S> My mistake. <S> Thanks everyone.
| It looks like that may not be wired correctly. The 74LS181 doesn't deal with signed numbers, just unsigned integers.
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Rectifier for two phases (220 V 50 Hz) I have a switch circuit which works with two phases. I need a flyback SMPS power supply for my logic circuit. I want my control circuit to be alive if any or both phases are receiving power. This can be achieved if my bulk DC capacitor receives power from both phases simultaneously. Currently, my power circuit receives power from a single phase like this: To add the second phase, I am planning to do this: Will it work? I think it should because the capacitor's voltage will block any voltage lower than it's voltage right behind the diode itself, at any point of time. Or, am I wrong? At the bulk capacitor, will there be two voltages fighting each other? In case it works, can I do something like this: I have removed the redundant neutral supply in the bottom rectifier, thereby saving me two diodes. Besides this, is there anything else that I am missing out? Will the voltage on capacitor remain same whether single or both phases are supplying power or will it differ? EDIT: After looking at the answers, I realize that if I use full bridge rectifiers as I shown in previous images, peak output on bulk capacitor will increase significantly depending upon the phase difference of the voltages. This increase will create difficulty due to these two reasons: 1) Higher voltage means a higher voltage rated capacitor 2) It might get difficult to find a flyback IC which operates at this high voltage. 3) Even if capacitor and IC are available, it might be costlier as compared to what I can use with a lower voltage which I get in 1 phase design. As such, I can think of another option. Using half bridge rectifiers to rectify each phase and use Neutral line as my reference: Since my output power requirement is less than 5 watts (1-2 watts typical), I feel that half bridge will work even when only a single phase is available. This also ensures that my bulk capacitor will not see a voltage higher than sqrt(2)*220 V = 311 V peak. In worst case where Vrms = 270V, it will give 382 V peak. Thus I don't need to modify my existing single phase power supply circuit after the bulk capacitor. Am I right in thinking so? Is there any better way to do this? <Q> You're almost there, but not quite thinking about the problem enough in a generalized way. <S> If you have N leads with arbitrary AC voltage between them, then you need 2N diodes to make a full wave rectifier. <S> Each lead is connected to one diode going to the positive DC output, and one diode coming from the negative DC output. <S> Note that your bottom circuit is like that, only that you've drawn it to make it not immediately obvious. <S> Here is a generalized full wave rectifier for N AC inputs: <S> You have 3 AC inputs. <S> Your mental block seems to be thinking about the neutral line as special in that regard. <S> It's no worse than just another AC input line, so you can make a full wave rectifier with 3 pairs of diodes. <S> Rectifying two of three AC phases and neutral is a odd thing to do, but is no special case for the general circuit above. <S> Here is a plot of the voltages: <S> This example uses 120 V AC at 60 Hz and assumes ideal diodes. <S> The rectifier output is the greatest difference between any two AC inputs at any one instance. <S> To make this easier to see, the max of all the AC inputs is shown in red, and the min in blue. <S> The resulting difference is the green trace, which is the instantaneous output of the rectifier. <S> This is the voltage the cap smoothes out, and is what you'd get if there was no cap and you put a small resistive load on the output. <S> Note that the green trace is NOT relative to the same reference as the AC inputs. <S> It is showing the DC output when that is taken in isolation. <S> The individual DC outputs with respect to the same reference as the AC voltages are the red and blue traces. <S> For contrast, here is what full 3 phase rectification looks like: <S> Note that this has much less ripple, and that ripple is at 6x of the AC frequency. <S> Your two out of three phase rectification results in ripple at only 2x the AC frequency. <A> Figure 1. <S> The voltage available to charge the capacitor will be the greatest difference between inputs. <S> This will be much higher when two phases are present. <S> In the 2-phase mode the neutral will carry current in the shaded zones - but even then only if the capacitor voltage has dropped below the L-N level. <S> With either L1 or L2 present <S> + N you will get the corresponding voltages shown in the L to N section of the curve. <S> This will peak at \$ V_{RMS}\sqrt 2 \$. <S> When two phases are present it is clear from the L1 to L2 voltages that a much higher potential difference exists and your capacitor voltage will be much higher. <S> I will leave it to the reader to calculate the peak voltage difference between two phases. <S> Will the voltage on capacitor remain same whether single or both phases are supplying power or will it differ? <S> It will be higher with two phases. <A> "I think", the last schematic should work as long as the neutral is the common line. <S> Alternatively, you can use half wave rectification as your power demand seems low. <S> Connect neutral to -ve terminal of the capacitor and connect each phase via diode to +ve terminal of the capacitor. <S> Cathodes of diodes connect to +ve terminal of the capacitor. <A> The basic idea will work but there are a couple of things to be aware of. <S> The voltage will be significantly higher when both phases are present than when only one phase is present. <S> If the two phases are 120 degrees out of phase then the voltage will be \$\sqrt3\$ times the voltage with only one phase present. <S> If the two phases are 180 degrees out of phase then the voltage will be twice the voltage with only one phase present. <S> Your capacitor will have to be selected (the one you have selected has far too low a voltage rating) and your fly back <S> converter designed with this voltage range in mind. <S> If the neutral becomes disconnected or one of the diodes fails then there is a potential for back-feeding. <S> This may or may not be a safety hazard depending on the larger context of the system. <S> Overall <S> I suspect you are probably better off using two separate isolated power supplies and putting your combiner-diodes on the output side.
| With your 2-phase + N rectifier scheme there will be a difference in rectifier output when running one-phase or two phases.
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Why are there two diodes in series in this test circuit? Look at Figure 2 in the following datasheet: http://www.onsemi.com/pub/Collateral/MMBT3904LT1-D.PDF Why are there two diodes and not just one? Thanks. <Q> The circuit includes two diodes to set the negative bias at the base during the turn-off phase of the measurement. <S> A single diode would only give about 0.7v reverse, two give twice that. <S> If they didn't have any diodes at all the voltage would exceed the reverse breakdown of the base-emitter junction that is in the region of 7v with silicon planar transistors. <S> I agree that it is somewhat unusual way to do the test rather than defining the voltage from the pulse generator. <A> I would say the schematic is incorrect. <S> If they were pointing down, they would give a combined voltage drop (taking the average) of 0.63 *2 = 1.26, putting the voltage in the middle of the active region of the transistor. <A> The diodes in such configuration are used for temperature compensation when such circuit gets to work in hot/cold environments so that the target output remains temperature invariant.
| They should be pointing "down".
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Measuring tension force on rope Suppose I want wo measure how strong someone (or somthing) pulls on a rope, i.e. I want to measure the tension (=dragging) force on the rope.The pulling would come from only one direction.I assume a maximum force of let´s say about 1000N (could be more, but let´s forget about the exact range for now). What would be a reasonably priced method?I searched a bit on the web and found lots of force sensors which should do the job, just one example out of many would be , e.g. https://www.burster.de/de/produkte/p/detail/8417/ Now, as far as I can see, this part has "just" some resistance strain gauge(s?) with a wheatsone bridge incorporated. Nevertheless, It is priced at about 800 EUR per piece, something I cannot afford. I thought about building an own setup, i.e. sticking discrete strain wire gauges to some kind of cylinder. Of course the question arises which material to use for the cyclinder. Can someone share experiences in building "custom" strain gauges for such a scenario?Maybe resistance strain gauges are even a bad idea, maybe an LVDT (Linear Variable Differential Transformer) in combination with a spring would be better suited? EDIT: The solution should be small enough to be integrated in some kind of handheld device (let´s assume a box of 20 x 5 cm) Environmental protection such as IP5X is not needed since ideally, the solution can get some kind of enclosure later on. Thanks! <Q> You can get your preferred style of sensor from China for less than 10% of the price of the German unit. <S> In other words, maybe you should try harder. <S> There are also (probably less convenient) sensors in digital luggage scales that are even cheaper, even if you buy one and throw away the electronics. <A> You don't need something that goes in series with your rope. <S> https://www.youtube.com/watch?v=aDb0SnNdSh8 <S> I don't see any great difficulty in home brewing this kind of tension meter. <S> A spring loaded plunger and a way to measure how far the plunger moves. <S> Calibrate it with weights on the rope hanging vertically, and a look-up table in an Arduino Nano or similar simple micro. <A> If you don't wish to homebrew your own transducer, look into "strain type load cells". <S> These are typically found in S and Z type configurations. <S> An example of an S type cell that is available for $130 USD can be found here . <S> You will have to add your own means of mechanically fastening the rope ends to the cell but most come with tapped holes for easy attachment of eye bolts or similar. <A> Get a few of these from eBay, £50 for 10:- <S> Glue it on with some epoxy and solder a couple of wires on. <S> This is the standard technique for ad-hoc force measurements. <S> Wheatsone bridge + opamp /instrumentation <S> amp <S> and you should be there. <S> There's a good guide here as to gluing them on. <S> Glue them onto anything metal <S> that's rectilinear. <S> I mean it's long and thin so that you get even strain across the cross section:- <S> You could even bend a steel strip <S> say 500mm long to form hooks either end. <S> Stick a gauge each side of the strip. <S> Some simple formulae will allow you to determine the overall strain and allow for any bending moment. <S> The type of metal doesn't matter as long as you know what it is to get the Young's modulus. <S> Just don't make it too thin as to neck it at the gauge location when you pull. <S> 1000N is nothing. <S> Get a 10 x 2mm strip from somewhere that'll allow you to identify the metal. <S> This is really one for engineering. <S> SE
| Google "cable tension meter" and you will see a lot of configurations that work on the same principle as this
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PIC on 3 volts power I made a personal count down timer with PIC16F628A and two multiplexed seven segment LEDs. The system was tested on breadboard at 5 volts USB . However, I intend to work them on batteries at low power consumption. Is it advisable to power the whole system by 3 volts by AA batteries ? <Q> No. <S> The PIC16F628A only works down to 3V, according to the datasheet. <S> The LF version works down to 2V. Even with brand new AA batteries, the voltage will quickly drop below 3V, and can cause issues. <S> Solutions. <S> Use 3x AA for 4.5V nominal, or use a boost converter to bring up 1x AA or 2xAA to a stable 3~5V. <A> The PIC can operate with the lower voltage. <S> The only thing I am unsure of is if your seven segment LEDs will have enough forward bias voltage to operate at the lower voltage. <S> For some segment displays 3V isn't enough. <S> My suggestion is (if you can't find the required forward bias voltage) try it and if it doesn't work add another AA battery in series to get 4.5V. <A> As said, it's not a good idea to do it. <S> In fact, all your component will work at 3V but this voltage level isn't guaranteed overtime due to battery discharge. <S> This solution will consume a bit more power (due to regulator power loss) but it is safer for the whole circuit. <S> By the way, if the input voltage change, you may also need to change some resistor value for LED driving. <S> It's not that important since the voltage difference is pretty low but still !
| What you should do is to increase a bit the input voltage and then use a regulator to have a fix voltage output.
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Using Neodymium Magnets as wire connectors? A new project I'm working has to be highly modular. All the electronics will be hidden in a case, so I was wondering if there's any way to use magnets and pass current through them for the project? It's well known that neodymium magnets could lose their magnetic capabilities in extreme heat, so simply soldering wires to the magnets is out of the question. Is there anything I could do to solve this problem? Also the magnets might need to run close to 4 Amps of current, which shouldn't be a problem if the metal coating on the magnet is good. <Q> I would use a method that does not rely on the magnet as a conductor. <S> Use the magnet to hold a conductor against another. <S> This brings a number of advantages that result from isolating the requirements of conductive connector components from the properties of the magnet. <S> This eliminates the need for coating a magnet with gold, much easier to coat the contact itself. <S> Magnets tend to be fragile, and the conductor can be very slightly sprung; enough to soften impacts (which also cause de-magnetisation if I'm correct) and may help with connection reliability. <S> Otherwise, I would still use a simple mechanical method for connecting to a magnet. <S> For example, wedge a wire between the magnet and enclosure. <S> Very low cost to implement. <S> It may even be stronger than any soldering/welding technique since some coatings on magnets have quite a weak bond. <A> You could use a pressure contact or a spring clip of some kind- or spring-loaded pins such as "pogo pins" used in test fixtures or the type used for battery packs. <S> You might be able to use a low temperature solder- if you can find one that is usable below the maximum working temperature of the magnet, which is actually quite low for the neodymium types, less than 100 degrees C, much less than the >300 C Curie point of the magnet- <S> so I am guessing that won't work without degrading the magnets. <S> Or try to spot weld the wire as @Passerby suggests, the heat should be localized and will not demagnetize the magnet. <A> Neodynium magnets fall apart after snapping together a few too many times. <S> I have some which have never had much use, but they have crumbled from being snapped together, pulled apart, snapped together, pulled apart... <S> The Nickel casing flakes away first, followed by chunks of magnet. <S> I offer you a related, but alternate approach. <S> clamp and bond two magnets to the inside of the case, either side of contacts, one with north facing away from your assembly and one with north facing into your assembly attach metal connectors on the outside of the chassis, which will make the electrical circuit. <S> I'll leave it to your choice as to the connector/ terminal. <S> On your complimentary module (which connects to this one), create a similar assembly, but reverse the polarity of the magnets. <S> When you push the two modules together, The like poles will repel preventing misalignment, but dissimilar poles will attract causing alignment. <S> The magnetism forces should hold the equipment together. <S> The Magnets will not suffer compression & shock forces of snapping together and should enjoy a longer lifespan. <S> You will not encounter heating issues through solder, welding, or passing high current through the magnets they will be isolated from the rest of the circuit. <S> Rudimentary Diagram (drawn in gimp) <S> Here the individual magnets are represented by a north (red) <S> and south(blue) poles <S> Thin yellow strips with dark grey on them represent isolated terminals, i.e. not able to conduct into a metal chassis. <A> Look at the the design of the macbook power connector - called magsafe contacts and magnetic and comes apart with sufficient applied force. <A> I have had this on my CNC router Z-probe for a few years now, used to temporarily connect the measurement wire to the cutting tool shaft: <S> That is simply a wire soldered to a 5 mm diameter neodymium magnet. <S> It also has a coat of epoxy, which originally also covered the wire insulation and was supposed to protect the "fragile" solder joint. <S> But it turns out that the epoxy has wore out while the solder joint is holding up just fine. <S> The magnet might have been slightly damaged due to soldering, but it is still pretty strong. <S> But one difference from your situation is that the current I'm using is only a few milliamperes. <A> The neodymium magnet materials are crumbly, usually there's a nickel plating over them. <S> While you COULD make contact with the nickel, it'd be easier to justuse the magnet to attract a slug of iron. <S> Then a set of electricalcontacts that is pinched between the magnet and the iron will be pressedtogether. <S> You could get elaborate, and hide the contacts inside a slot in the case, knowing that they will be extended only in response to the (magnet) connector or cradle. <S> Imaginative mechanical connections ought not to preclude unsafe conditions ifhazardous voltage and current levels are involved, of course.
| Another option would to use a nickel plated ferromagnetic chunk of metal with a wire soldered to it or held with a cross hole and set screw, and let the other end of the magnet glomp onto it.
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The luminous intensity and beam angle of LEDs Do the luminous intensities of LEDs mean the peak intensities of the LEDs, or the intensities averaged over different beam angles? Typically the data sheet of a LED specifies its luminous intensity and beam angle, like the following. https://www.sparkfun.com/datasheets/Components/LED/COM-09590-YSL-R531R3D-D2.pdf The data sheet says the intensity of the LED is between 150mcd and 200mcd. Would it mean that the intensity is 200mcd at the beam angle of 0 degree? Or is it the intensity averaged over entire beam angle? <Q> Luminous Intensity, Iv for visible LED's is always peak maximum and <S> then roughly 50% at 1/2 the BW angle to either side. <S> Your LED spec is 50° ±10° as the total beamwidth \$2θ^{1/2}\$ at half intensity <S> IR LED's often with very narrow θ were once all defined as \$θ^{1/2}\$ meaning <S> the peak was half angle and not always dead centre. <S> Recently to avoid newbie confusion, some IR specs show the full angle. <S> sage advice <S> A rule of thumb on diversity gain of the lens is that when you reduce the \$2θ^{1/2}\$ by 50% the Iv intensity doubles but due to lens loss -10% each time to magnify 2x or reduce the angle from no lens which is called the "Lambertian" response curve of 160° like most SMD LEDs. <S> Thus to compare your 50 ° LED should be about 50% of the Iv of an equivalent chip with 28~30 ° or other words a 30 ° Iv could be 2x {150~200mcd}[50deg] = <S> 300~400 mcd <S> Now I only use 30° 5mm LEDs for most applications and only Iv > 10,000 mcd with tighter tolerances and get in bulk 10k MOQ but usually have lots left over in many colours and white > 16,000 mcd and unlike most LEDs these are Zener protected. <A> The Candela measures the luminous intensity, or luminous flux per unit angle, per steradian, of which there are \$4\pi\$ in a sphere. <S> This means that in two light sources with identical total power, but different beam angles, the device with the tighter beam will have a higher luminous intensity measured in cd, or mcd for smaller sources. <S> The Lumen measures the total luminous flux. <S> One candela is one lumen per steradian. <S> If a source emits one candela in all directions equally, it emits \$4\pi\$ lumens. <S> Needless to say, when manufacturers advertise their products, they quote whichever measure makes their specs look better. <A> Luminous intensity is the power (adjusted for human eye perception called the luminosity function) of light in a unit solid angle. <S> The SI standard for candela, the unit of measure for luminous intensity is 1/683 watts in a steradian at a wavelength of 555 nm. <S> When an LED manufacturer specifies candela for their LED, they are picking the solid angle with the maximum luminous intensity. <S> So if their LED outputs one lumen at 555 nm, and they focus all of that power evenly into one steradian, the LED would be specified as one candela. <S> But with that same LED chip, if they narrow the beam to 1/2 steradian, they would show 2 candela in their spec sheet.
| Luminous Intensity is a human perception weighted sum of the emitted light power, not the power that would be measured by a thermal method.
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How do I repair the solder of this component? The chip is a chip that is surface mounted to the main PCB.This solder gave out and separated from the PCB as shown below. I'm not very well skilled in soldering, but it seems a bit too fine for a normal soldering iron. Should I use solder paste and a heat gun? Or is there a better way? <Q> Apply flux liberally, then put a little solder on the iron tip and rake it down the pins on an angle to heat up several pins at once (or use a 3mm spade tip like I do). <S> The flux should prevent solder from bridging the pins. <S> If a solder bridge does occur then use desoldering braid to remove it. <A> If you do not want to use solder iron, then here is an alternate method: <S> Apply liquid flux on the pads of the IC. <S> Apply some solder on the pads and then place the IC carefully in place, and Use the hot air blower/heat gun to secure it firmly in place. <S> PS: <S> Where your IC pads are, I can see some capacitors and other components. <S> So, you'll have to use your heat gun very carefully. <S> I'd personally prefer to solder this IC than relying on heat gun to do the job. <S> If the heat gun displaces the caps while placing your IC correctly, re-placing those caps would be much tougher later. <A> I would go at it like this. <S> Use solder wick to clean the pads of both boards. <S> You want the pads of both to be clean and smooth. <S> Use a soldering iron with a fine tip, and 0.5 mm solder. <S> It can be done with thicker solder, but I use 0.5mm for just about everything. <S> Heat one pad one <S> the main board with the iron, and melt a bit of solder onto the pad. <S> Put the smaller module in place and line it up properly. <S> Melt the solder on the one tinned pad, and push the smaller board down on the larger one so that they are flat on one another. <S> Remove the iron and let the pad cool. <S> Check alignmeny. <S> If crooked, reheat and adjust. <S> Solder all of the pads, including the first one. <S> It won't be perfect because of the reheating, so it must be redone.
| Just use a normal soldering iron (preferably temperature controlled) and liquid solder flux .
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Torque-Slip Characteristic Stability Analysis for Induction Motor I am having confusion for this stability question of induction motor.The real question is here I searched and searched online, keep reading and reading, but still not enough to answer my doubt. My understanding (that I learnt from here http://electricalbaba.com/understanding-induction-motor-stability/#comment-2645 ) on why point A is stable is like this: "If I increase the load torque, as the load torque increases, the new operating point has higher slip compared with that of point A which in turn means that speed of Induction Motor has decreased, and hence operating point A is stable". Is this reasoning correct? But how about the condition that should be satisfied for the stability of the operating point A? Hope some people can kindly help me.Thank you! <Q> At the operating speed indicated by the point marked, the torque capability of the motor is equal to the torque required to turn the load. <S> Thus the motor has no tendency to provide the additional torque that would be necessary to accelerate the load. <S> Also there is no tendency for the load torque to overcome the torque provided by the motor and decelerate. <S> Any difference between the torque required to turn the motor at a given speed and the torque capability of the motor at that speed is applied to the inertia of the load as accelerating or decelerating torque. <S> Any change to in the load or motor characteristic curve will move the operating point to the new intersection of the two curves. <S> Here is an illustration: <S> Note also that the motor curve can be expected to change. <S> Normal (or abnormal) increases and decreases in the supply voltage will increase cause the available torque to increase and decrease. <S> Motor warmup upon starting and ambient temperature <S> changes also have a small effect on the motor's torque curve. <S> That means that operation at the peak of the motor curve is not a stable operating point because the slightest decrease in motor torque or increase in load torque can cause the motor to stall. <S> Of course operating continuously at that point is normally a severe overload condition. <S> Summary <S> There are conditions that are more complicated than the diagram shows, but they are beyond the scope of the question. <S> As the situation is presented, when the motor is switched on, it simply accelerates to the stable operating point. <S> At all speeds lower than that point, the motor curve is above the load curve and accelerating torque is applied. <S> At all speeds above that point, the motor curve is below the load curve and the deficit in motor torque allows the load torque to decelerate the load. <A> (rev C) <S> We know from linear and rotational acceleration that F=ma. <S> Thus a constant speed is achieved (stable) when the forces are equal at a point on the load line. <S> Where it is unstable is below the Tmax. <S> This is a boundary conditions where Tload < Tmax and where equal determines RPM with no acceleration force and constant velocity. <S> The other is starting up if the load is between Tmin and Tmax, it wont start up, so startup caps and coils are used to boost the motor current until a speed where the transfer switch has about the same torque as Tmax <S> Otherwise the motor would never start up and the load torque must be less than the starting torque to start up. <S> Now the slip effects cause this low torque at 0 RPM and this loss of torque declines with rising speed ( ie torque rises) such that when it generates BEMF which ends up limiting the voltage drop internally and thus current drops with available torque. <S> Past this peak, Tmax Torque is now dominated by BEMF as the slip effects are now minimal but depends on the magnetic design. <S> analogy <S> Just like a voltage regulator with resistance. <S> Voltage drops with more load current V= <S> Voc-IR or like a spring in the linear stable part of the curve that deflects down x with force in the same direction, F for x=F/k for some spring constant, k. <S> Thus we see there <S> two factors that reduce torque in AC induction motors. <S> 1) Slip frequency which demands a boost circuit for startup and 2) <S> back EMF , which generates a voltage opposite to the applied voltage and for Induction Motors <S> it becomes 0 torque at some integer ratio of the line frequency and number of poles. <S> This optimal peak ratio of RPM / max RPM depends on the motor design and there are standard torque curves that differ from your example. <S> The maximum power is at a higher RPM than Tmax since Power = speed <S> * Torque when using MKS units. <A> The reasoning doesn't require the notion of slip. <S> Consider two cases: Motor torque higher than load countertorque <S> → the speed increases, which leads to a lower motor torque. <S> Motor torque lower than load countertorque <S> → the speed decreases, which leads to a higher motor torque. <S> What you see at a stable operating point is not only the equilibrium of motor torque and load countertorque, but also the conclusion a motor torque which is too high for a given load countertorque will lower itself by the M/n characteristic of the motor. <S> And vice-versa. <S> This is true for <S> all motor-load combinations, not only those which operate on slip as an AC asynchronous motor does. <A> If the constant load torque is below the zero-speed torque (that the motor can produce) then point <S> A is achieved when the motor is switched on. <S> If the load torque is higher than the zero-speed torque then there is a problem and the motor would never run at the required speed. <S> However, if the higher load torque were applied once the motor was up and running then the system would be stable because it could back-down (a bit) on the graph without going past the point of no return (maximum torque).
| The essential condition that defines a stable operating point is that it is the point where the torque requirement curve crosses the motor capability curve in a region where the motor torque capability is declining as speed increases.
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12 potentiometers with multiplexer I'm going to connect 12 potentiometers (100k) to a couple of cd4051 analogue (8->1) multiplexers and to 2 of Arduino (Uno) analogue pins. Now I think each pot draws 0.5mA (I'm going to measure it to be sure) so around 6mA for all 12 pots. I think it is ok the Arduino permits 200mA as I know. Also I read that it is recommended to use 10k potentiometers (because of the ADC) but I have also read that the 100k are going to work. So is there anything I miss here? (before soldering all the power and ground terminals). (the multiplexer in the schematic is a single module) simulate this circuit – Schematic created using CircuitLab <Q> This has mainly to do with how long it takes to charge up the internal circuitry to get a stable sample. <S> Since you don't really want to go into the guts of the Arduino libraries to wait extra time between initiating the sample and reading it, you're best off following the 10K recommendation, unless your application is not a precision application (and you haven't told us). <S> When you're multiplexing 12 channels though, things generally run pretty fast. <S> If you're really trying to go low current, you can keep using the 100K's, and just put a voltage follower on the output of the multiplexer. <S> Yes, 100K pots will cause more noise than 10K pots. <S> How important this is to you depends on what you're actually trying to do, how precise your measurements need to be, etc., <S> and we have none of this information. <S> I suspect that the noise difference will not be an issue for you. <A> stray inductive noise L currents and C voltages may be induced on cables connected to pots or on board into R of pots. <S> So 100k is worse for I(f)R <S> than 10k. <S> But shielding and ferrite CM chokes improve the results and may be mandatory anyways with twisted pairs, since the inputs are unbalanced. <A> Now I think each pot draws 0.5mA <S> Don't guess. <S> Do the math. <S> (5 V)/(100 kΩ) <S> = <S> 50 <S> µA. <S> 50 µA times 12 instances = <S> 600 <S> µA. <S> That's a negligible amount of current considering you are running a arduino off the same power supply. <S> I think it is ok <S> the Arduino permits 200mA as I know <S> That is irrelevant. <S> This is either the worst case power current spec or the maximum all I/ <S> O pins can source or sink together. <S> Either way, it has nothing to do with how much current the ... <S> Oh, I just noticed that you accepted another answer as I was typing this, only 29 minutes after you asked the question. <S> No point wasting more time here now. <A> A CD4051 inherently generates a leakage current from its analogue pins. <S> If you read the data sheet this can be as high as 100 nA at ambient temperatures. <S> This means that for 11 of the 12 switches there could be a total of 1.1 uA flowing into the only closed analogue switch and feeding this current back to the pot wiper. <S> At mid range position that pot will present an equivalent resistance of 50 kohm so, this 1.1 uA will produce a DC offset error of 55 mV. <S> If you are happy about this then no problem. <S> I would also suggest you use a 100 nF capacitor on the common line to remove noise and allow a more accurate ADC result. <S> This may also require you to limit the speed at which you toggle through the multiplexer to ensure that the pots can be read accurately but if you are happy with a potential error of +/- <S> 55 mV (due to leakage current) then this won't be an issue.
| The input circuitry on an Arduino will work "best" when you use pots that are 10K or less. Your guess was off by a factor of 10.
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What happens when a Thermoelectrical generator exceeds its maximum temperature ratings? I have a commercial Thermoelectrical Generator used to generate 12V of power from the heat of stove. The data sheet says the maximum Temperature rating is 452 C°. What happens when this temperature gets exceed? Will the energy production become inefficient again or will certain parts of the peltier-element melt? Will it cause irreversible damage? <Q> Since there are electronic and plastic components in the unit, irreversible damage will likely occur. <A> The TEG element will die with too much heat .The internal crystals melt .The <S> lower the temp the longer the life <S> .Power output is a function of delta T so keeping the cold side cool means reasonable power output at modest hot side temperatures. <A> When you exceed the maximum the manufacturer recommends, damage may occur. <S> It's unlikely to happen very fast at 453°C compared to the maximum of 452°C <S> but it is not good for the unit. <S> Examples include, but are not limited to, melting of the solder used to assemble the Peltier module, damage to the semiconductors themselves and damage to wire insulation and damage to the fans. <S> The module with its semiconductors itself is probably not rated for more than 200°C <S> so the product depends on the fans running. <S> The fans are probably not rated for more than 70°C <S> so the fans will be destroyed unless they cool themselves. <S> Hence all the warnings about how the fans must be spinning . <S> Clearly, if you put that device into an oven and bake at 452°C it will be destroyed. <S> Engineers typically don't like operating devices near the 'maximum' as usually there is some grey area where lifetime will be compromised. <S> Brief excursions to close to the maximum and sometimes even beyond may be acceptable (for example, during soldering we typically exceed the maximum storage temperature of a device), but usually we like to stay well away from limits.
| At higher temperatures damage will occur.
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Is three-phase power in a residential building 'good'? I live in a three story building. Last night, the top floor lost power for about 30 minutes while the rest of the building was not affected. When I asked a local electrician, he said this was due to the three-phase power coming from the local power company. I am a software engineer, not well-versed in electrical topics, so I would appreciate insight into what's going on. Is this type of configuration of wiring by the power company something that is good, or should I ask them to change it? Where I live, the mains voltage is 220 v. <Q> What you have is 380V three-phase "wye". <S> Think of the three phases as a triangle. <S> "Wye" means there is a "neutral" wire in the center of the triangle. <S> Any phase to neutral is 220V as you are accustomed. <S> (Distance is voltage). <S> That is how power is wired on five continents and New York City. <S> Three-phase power is brought to the pole in back of your house, or inside your apartment building. <S> Some houses get only 1 phase (that's all they need for household loads), some get two phases (doubles available power, only 1 more wire), and some get all three either because they need a lot of power, or to run fairly large loads (A/C system, heat pump emergency heat, etc.) <S> Three-phase power is ideal for motors. <S> So for any group of houses or apartments, they will put 1/3 of them on each phase, with the idea that they will tend to average out. <S> If you have more than one phase in your house, they will spread your loads around as equal as they can. <S> What happened is one of the "hot" phases came loose or had a breaker trip. <S> This shut off one phase, and killed 220V power on the 1/3 of your outlets that used that phase. <S> Heaters would be 2/3 out. <S> Motors would not run. <S> If you were skilled with electrical, you could have opened up your service panel/consumer unit and moved the blacked out loads onto one of the phases that was still working. <S> The bigger worry is losing a neutral . <S> Remember the triangle I mentioned? <S> If your neutral wire had failed instead, neutral is no longer held in the center. <S> It will float around anywhere inside that triangle, depending on the load on each phase. <S> That means any phase to neutral voltage could be as low as 0 and as high as 380. <S> If the problem was inside your house, check your neutral wire also. <A> Three phase is used to supply large loads and large buildings have their loads equally divided between the phases to even out the load for the generator. <S> This happens with houses the first house on one phase the second house on phase 2 etc. <S> So, for your building first floor one phase, second floor second phase etc. <S> The third phase may have gone down due to someone else causing a problem. <A> Is this type of configuration of wiring by the power company something that is good, or should I ask them to change it? <S> Three-phase wiring is common, if not standard, for AC-current. <S> From Wikipedia : "Three-phase electric power is a common method of alternating-current electric power generation, transmission, and distribution. <S> It is a type of polyphase system and is the most common method used by electrical grids worldwide to transfer power." <S> I'm not an expert in this field though, so it's possible I've missed an important detail. <A> As already stated it is normal to spread distribution over three phases to multiple domestic users. <S> if all the apartments were on the same phase, as well as the load on that phase being much heavier requiring larger cables, all the user would have lost power instead of just your floor.next time a phase goes down for a short time it may well be the phase you are asking about having your supply changed to. <S> In short, stay as you are, it's normal. <S> John
| Three-phase power is best loaded evenly. So, to answer your question, no, you shouldn't need to ask them to change the wiring.
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EEPROM Values Corruption - Only Some times I am using 25LC640A EEPROM with a 32 bit Micro controller. This EEPROM can store 8kb of data, with SPI serial Communication. In my case during every power down sequence MCU will write some block of data into EEPROM and then it will shut down. And also its working fine. I have tested as many time as possible. No issues with EEPROM Data Issue : But, Some times the EEPROM values got corrupted. My code throwing checksum error after turn on the MCU. i have checked code thoroughly, i am not able to find the bug since its working fine for so long time. So any one here please suggest me what is the cause of this corruption of data. I have attached schematics also. <Q> This can be caused by a faulty (or missing) brownout-reset circuit. <S> It should stretch any glitch to long enough that the MCU is guaranteed to reset properly. <S> The word 'guaranteed' is not used lightly- <S> there are some internal MCU BOR circuits that are not guaranteed to work when you work out the tolerances and so on. <A> What you are trying to do sounds reasonable, but this hints at something to look more carefully at: <S> power down sequence MCU will write some block of data into EEPROM and then it will shut down. <S> You have to be careful that the EEPROM is done doing the last write before you shut down power. <S> EEPROMs can take ms to do writes. <S> That's a very long time for even a modest microcontroller. <S> During normal operation, it's actually good to initiate the write, then let the chip go off and take its time while the software does something else. <S> The software then only blocks waiting for the EEPROM to finish the write if another EEPROM operation is requested. <S> In many cases, the write and other software operations overlap nicely. <S> Even if another operation is requested immediately, that's never any worse than waiting explicitly after each write. <S> If you are using layered routines you didn't write, this may be how they work under the hood. <S> In that case, the final shutdown needs to be handled differenctly. <S> You make the call to do the last write, <S> but then you must call something else that explicitly waits until the write is finished. <S> Powering down before then will cause corruption. <S> EEPROM libraries designed as described above will have a call that explictly waits for the EEPROM to be idle. <A> You mention that it has been working for a long time. <S> I have had engineers not appreciate the number of lifetime writes of the device specification. <S> With one million write cycles, if the uP does a write every second, the specification will be exceeded in less than 12 days. <S> If it does a write every minute, the spec will be exceeded in less than two years. <S> Once you exceed the specification, you should anticipate anomalies. <S> If you are convinced that it is not a lifetime issue, then you should add some diagnostic routines to dump the data, including the CS, when powering down and up. <S> Run this in a test environment to see if you can spot the problem.
| I highly recommend using a circuit (internal or not) that is guaranteed to inhibit the MCU's operation at invalid levels and is guaranteed to work down to a voltage at which the EEPROM cannot possibly perform a write (less than 1 volt will usually do it). When the MCU is operating at invalid Vdd voltage levels it can do just about anything, including going berserk and running random bits of code to write to an external EEPROM.
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How does a benchtop power supply measure and control current? This may be a simple question, but it is something that I can't seem to wrap my head around. I have a +12v 5A power supply at work which has two knobs, one for voltage and one for current. There are various devices that upon plugging them in I have to increase the current otherwise they will not work, but when I do this, the voltage level stays the same. My question is, what exactly is happening when I turn the amperage knob on a benchtop power supply? My guess would be that we are changing the resistance being supplied to the device but I am unsure and would love a real explanation! Thanks! <Q> For most uses of a bench power supply you're interested in providing a steady voltage. <S> That's what the voltage setting is for. <S> The idea is to keep your electronics from frying if you've made a mistake. <A> You most likely have a switching power supply that measures the current through a shunt resistor before the output voltage is measured - because the two are in series the power supply knows the output current is the same as the shunt current <S> The logic then follows something like if the voltage is less than the knob set, and the current is less than the knob set, then turn on the switch, and if it's higher on either, turn off the switch - then after the switch there is a smoothing capacitor or similar to reduce the ripple to where you won't notice it <A> Here is one current-limiting circuit simulate this circuit – <S> Schematic created using CircuitLab
| The current setting is a current limiter -- if your load draws more current than you've set the limiter for, it lowers the voltage until the load draws only the amount that you set the limiter for.
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How should I properlly draw on a schematic a shunt going through the center of a toroid coil? So I'm trying to reverse engineer the schematic for an Avair AV-200CN SWR bridge. Two toroid coils, used for sensing, are mounted on a shunt, so that the shunt goes through the center of the coils. I'm not 100% sure how to draw such an arrangement on a schematic. Was initially thinking of drawing it as a transformer with two secondaries, since the shunt acts as a one-turn primary, but I'm not sure if there's a better way to draw it, since I'd like to minimize the amount of thinking needed to understand the schematic. Here's the picture of the circuit: <Q> Draw it as a transformer where one of the windings has got a single turn. <S> T301 and T302 are the cores with sensing windings. <S> W303 is the shunt. <S> This schematic should translate well into the PCB layout. <S> The toroids aren't rigidly connected attached to each-other, or to the shunt. <S> So, in this schematic they are separate components. <S> If you need even more flexibility during layout, the shunt could be represented by two separate pads, instead of one component with two pads. <A> Here's one example where a strict adherence to schematic symbols can cause confusion - that single-turn "winding" can easily be done wrong. <S> Perhaps not clear is the phasing of the multi-turn winding of the transformer <S> - the "dot" convention is easy to get wrong. <S> It might be easier to leave the labels of input connector and output <S> connector off until it is built. <S> Then discover which direction gives a null with a dummy load in place. <A> It is not a shunt. <S> It is considered a one turn coil on the toroids. <S> This is very typical for SWR meters.
| Many SWR schematics do a pictorial representation of a toroid that helps make clear the single-turn nature (one wire through the toroid's centre) of the current transformer:
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Purpose of components in op-amp radio circuit I am trying to understand the purpose of a couple of components in a schematic I am looking in at: My questions are: 1) What exactly is the 100k resistor attached to the diode doing? 2) Why is the output of the rightmost op-amp fed into a transistor? Could the audio signal not be pulled directly off the output of the op-amp? Thank you for your responses. <Q> The \$100\:\textrm{k}\Omega\$ resistor is part of an RC filter (as well as a DC return path.) <S> Together with the \$300\:\textrm{pF}\$ capacitor, its filtering purpose removes the RF that is still present in what passes through the diode detector (smoothing it out) while also leaving the relatively low frequency audio signal envelope undamaged. <S> Take a look at the RC time constant. <S> Also imagine that node without the resistor <S> present -- all you see then are a couple of capacitors and a diode feeding it. <S> It really needs a DC path added, as well. <S> That last bit uses an emitter follower as a driver. <S> It can source fine, but is dependent on the \$470\:\Omega\$ resistor to ground for pulling down and sinking current. <S> I'd definitely arrange things differently. <A> The resistor discharges the capacitor charged by the diode hence (in conjunction with the 300pF cap) it sets a high frequency audio cutoff of \$f_m = \frac{1}{2\pi R C }\$ or about 5.5kHz. <S> It filters out virtually all of the carrier frequency since the carrier frequency of AM radio is > <S> > 5.5kHz. <S> The output transistor is supposed to boost the op-amp output a bit- <S> but with that 470 ohm resistor(!) <S> I don't think it does much useful. <S> The op-amp alone is capable of driving a modern sensitive headphone well enough. <S> Eg. <S> a typical in-ear headphone has a sensitivity of 100-120dB SPL/mW <S> (that's lots of volume) with an 18 ohm impedance. <S> That means ~7mA is plenty, and the op-amp alone is fine since it can typically drive 20mA or so. <A> The diode is the demodulator stage. <S> It rectifies the RF signal created across the tuned circuit, the rest of the op amplifiers are simply audio amplification. <S> The op amp does not have a high power output capability. <S> Typically you would use a 32 Ohm headset for a simple circuit like this, though the transistor may allow a small 8-16 Ohm speaker to work.
| The diode, resistor and capacitor form the demodulator. The output transistor increases the current capability into the speaker/headphones.
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Will multiple chips outputing onto a bus for a few nanoseconds cause damage? I'm working on a home-brew CPU design, with the usual mix of parallel EEPROMs, static RAMs and registers, tri-stated onto a single 8 bit bus. My /output-enable logic for three tristate-able chips on the databus is: /Ken = /a/Ren = /a nor /b/Men = /a nor (/b nor /b) Do I need to worry about the few nano-seconds during which more than one chip's /oe pin will be low, due to the differing number of gates? Will current flowing out of a high output of one chip, into a low output of another chip, for less than 10ns cause damage? If it would cause damage, how was this situation avoided in the 1970's-80's? Update: Chips are: http://www.farnell.com/datasheets/32783.pdf http://www.farnell.com/datasheets/1911297.pdf http://www.farnell.com/datasheets/2047758.pdf <Q> Typically the situation was avoided by the outputs being open-collector instead of tristate, with a pullup resistor to +5V completing the circuit. <S> If one device drove the shared line low and the other went open-collector, the shared line would stay low. <S> For a practical example of this, the CAN bus is widely used in industry (particularly in cars) and operates in exactly this way. <S> Every device's output is open-collector with a current limit, and one device (and one only) contains the pullup resistor. <S> In an industrial/automotive context where devices can and do go wrong, and wires can also be shorted high or low, this ensures that no device's output can damage another device, however it goes wrong. <S> In addition, the CAN bus driver for each device monitors the bus to check that it goes to the expected state, and reports bus errors to the application if it finds a conflict where someone else is stomping over its data. <S> In practise this is unlikely to damage the chips. <S> However it will give instantaneous current spikes on outputs where high and low are shorted, which does unpleasant things to EMC emissions. <S> Designers in the 70s and 80s were much less concerned about EMC, so it's likely that many circuits went out of the door with exactly this problem. <A> There are a number of ways to create nonoverlapping enables for bus devices. <S> Perhaps the simplest is to add the clock signal itself to your equations. <S> Then, only one device at a time is enabled while the clock is high, and no devices at all are enabled while it is low. <S> (Or vice-versa if you're using the rising edge of the clock to capture data.) <S> Normally, the output-enable function of most devices is fast enough that "wasting" half of each cycle in this way does not cause a timing problem. <S> But if it does, one workaround is to modify the duty cycle of the clock as needed. <A> If it would cause damage, how was this situation avoided in the 1970's-80's? <S> manual design that compensated the delay. <S> Whether this will cause damage depends on your technology, so no general advice can be given; however, for a transient of 10ns to have effect, your system needs to have a bandwidth <S> > 100MHz <S> , so that's something you can actively avoid. <A> If a device is designed to output data as quickly and strongly as possible when /OE is asserted, bus contention could result in substantial peak currents which may cause unwanted noise on the supply rails even if they don't cause physical damage. <S> On the other hand, such a device may be able to drive the bus to a valid state in less than half a clock cycle, in which case gating /OE with the clock may avoid such contention. <S> Some other devices, however, can't drive the bus so quickly and would need to have closer to a full cycle to drive the bus. <S> Devices which drive the bus more slowly, however, are less apt to pass excessive current during momentary periods of device contention. <S> If a device is slow to release the data bus, it may be necessary to control its gating signal so that it gets released early to ensure that by the time another device tries to drive the bus the first device will have ceased driving it. <A> There are at least 2 issues here: thermal surges, and VDD transients. <S> Lets put some numbers on this. <S> Assume transistors with silicon-extent of 20 micron by 20 micron, and 10 micron depth. <S> The volume thus is 20*20*10, or 4,000 cubic microns. <S> Older technology bipolars, with collectors under the base-emitter region, are approximately this size. <S> The specific-heat of silicon is 1.6 picoJoules/cubicmicron/°C. <S> Our device is 4,000 <S> * 1.6pJ <S> = 6.4 nanoJoules/°C. <S> How much temperature rise can we generate, in 10 nanoseconds of thermal spike? <S> Use 5 volts, and use 100 milliAmps (a rather good spike, between 2 opposing bus driver transistors). <S> The power is 0.5 watts, and energy is 0.5 nanoJoules per nanosecond. <S> In 10 nanoseconds, the energy is 5 nanoJoules. <S> Now just divide: <S> 5nJ <S> / 6.4nJ == 0.8 <S> °C rise. <S> Assumed uniformly distributed inside the 20*20*10U volume. <S> Given the majority of bipolar volume is the buried collector, "uniform" is valid assumption. <S> Thus 1 <S> °C is answer, per buss driver. <S> If 8 drivers are in one package, does the 1°C number change? <S> No, because the heat sources are spread around in 8 different regions, and the transient occurrence is low duty cycle. <S> Now for that second issue: the VDD ringing. <S> Early buss driver ICs held 8 circuits, with only one GND and one VDD. <S> Rail collapse was a big problem. <S> Why? <S> Assume 10nS GND+VDD inductance. <S> Assume charging 8 loads, 50pF each, with Trise of 10ns. <S> Or 2 volts/nanosecond slewrate. <S> Given 1pF at 1v/ <S> ns needs 1mA, our single output needs 100mA. <S> The eight outputs need 800 mA. Assume the charging surges rises from ZERO current to 800mA, in half the time, or 5ns. <S> What is the rail bounce? <S> V = <S> L <S> * dI/dT = <S> 10nH <S> * 0.8amp/5ns = 1.6 volts. <S> Thus GND moves up by 0.8v and VDD moves down by 0.8v. <S> Because I assumed a triangular current pulse (rising and falling in 5ns), the charging rate is less than needed. <S> To meet the full-charge pulse timing, we need to double the peak currents, and the bouncing becomes 1.6 volts for both GND and for VDD.
| If there was only a conflict for a very short time, this would cause no problems at all. A couple-nanosecond difference in when devices receive enable signals isn't apt to matter as much as the timing with which the devices themselves respond to enable signals.
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Non-linear torque vs speed response for DC motor I have a Faulhaber DC motor (model 1524T012SR) that I coupled to a cheap DC motor in order to characterize the cheap motor to use as a generator (and get RPM and torque from voltage and current). The Faulhaber motor uses a motion controller (MCDC 3006S) that was purchased with the motor years ago. I am using the motion controller to step through various motor speeds from 1000 to 9000 RPM. I expected the current output from the Faulhaber motor to be linear for a linear change in speed but it is not. The current increases until about 2500 RPM and then starts to drop as the RPM increases, similar to a stepper motor response. The problem that I have is that the datasheet provided with the motor only gives a single torque constant (11.5 mN-m/A) so I have no idea how to get the actual torque supplied to the coupled motor. Is this a problem caused by using the speed controller for the Faulhaber motor? What can I do to determine the actual torque output by the Faulhaber motor? Here is my setup: Here is the current drawn by the driving motor for a linear increase in speed: (it says torque but it's just the current times Kt.) <Q> The Current required by the driving motor is not necessarily a linear function of speed. <S> The current into the driving motor depends on the torque required of it as well as a other factors. <S> The torque depends upon its own friction as well as the output torque to drive the load. <S> There will also be current required to account for electrical losses. <S> These other terms may not be linear with speed. <S> To determine the torque being absorbed by the generator first plot a curve of the motor current without the generator attached - <S> this will give you the current required just to spin the motor. <S> Then repeat the experiment when driving the generator. <S> The difference in current should be a function of the torque to drive the generator. <S> You should perform the test with various loads on the generator at each speed to get a complete characteristic. <S> Subtract the plot with no-load from the ones with a load. <S> That will represent the additional current required to drive the generator. <S> Multiply that current by your torque constant to get the actual torque required by the generator. <A> A brushed DC motor has a linear current vs torque curve. <S> The upper boundary of the curve is the stall torque and current. <S> The lower boundary is zero torque, zero current. <S> The slope of this line is the proportionality of motor current to torque (A/Nm), and is called the current constant. <S> The reciprocal of this slope is the torque constant of the motor (Nm/A). <S> The motor spec sheet lists the stall torque as 6.52 mNm and the torque constant as 11.5 mNm/A. <S> From this we can compute that the stall current is 6.52 mNm/11.5 <S> mNm/A or ~0.57 amps. <S> You can confirm this experimentally by locking the rotor and measuring the current. <S> So in summary, simply measure the motor current, multiply by 11.5 mNm/A, and you will have the mNm of torque developed by the motor. <S> If you wish to calculate RPM, first measure the resistance (R) of the motor (take several readings, rotating the shaft each time, and take the most commonly read value). <S> The spec sheet lists this as 19.8 ohms. <S> The Back-EMF constant, K e , for your motor is listed as 1.21 mV/min -1 . <S> Now monitor motor current (I) and voltage (V O ) and calculate RPM as: RPM = <S> ((V O -(I*R))*1000)/K e <A> On the Faulhaber website there is available for download very extensive information about the motor including definitions of all of the constants, equations of the relationships among the constants and data items etc. <S> Complete analysis of your set-up and the data obtained will probably explain everything. <S> Each applied motor voltage creates a different speed vs. torque characteristic. <S> Each speed creates a different generator voltage. <S> Each generator voltage creates a different generator load. <S> All of that can be calculated. <S> The Faulhaber motor current doesn't need to be controlled. <S> To characterize the motor that is being used as a generator, it is sufficient to control the driving speed and the load resistance. <S> Note that the Faulhaber motor motor efficiency is only 37.3% at rated speed and load. <S> Rated speed = <S> 4130 <S> RPM <S> Rated torque = <S> 2.9 mNm Mechanical output power = 4130 <S> X 2.9 / 9549 = <S> 1.25 <S> Watts <S> Rated voltage = <S> 12 <S> V <S> Rated current = <S> 0.28 <S> A Electrical input power = <S> 12 <S> X 0.28 <S> = 3.36 Watts Total losses = 3.36 – 1.25 <S> = 2.11 Watts Motor resistance = 19.8 ohms Losses at rated speed and torque: Copper loss = <S> 0.28^ <S> 2 X 19.8 <S> = 1.55 Watts Friction torque = 0.08 mNm Friction loss = 0.08 <S> X 4130 / 9549 = 0.035 <S> Watts 2.11 – 1.55 – 0.04 = 0.52 Watts windage and other losses
| Each generator load creates a different motor torque.
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