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I need to run a 12v fan (440w) from a 240v supply - what would be the best way? I need to run a 12v fan (440w) from a 240v supply - what would be the best way? Should of mentioned that this fan is attached to a car conditioning rad and therefore dc. - Sorry. Thanks to those who answered. <Q> At that power level, probably your best bet is to buy a suitable enclosed power supply such as the one pictured (600W) (despite the name 'enclosed' it will require an enclosure to cover the mains terminals etc.) <S> The exact minimum rating should be nailed down properly- <S> the starting current will be higher than the run current, and most switching power supplies can behave undesirably when connected to such a load. <S> It should not current limit when subjected to the motor stall current or you may have problems. <S> The rectifier/transformer that @Charles suggests would not have the startup problem (use a circuit breaker for protection), but it would be heavier. <S> The regulated supply might be a bit easier on the commutator, assuming it's a brushed motor rather than a BLDC motor. <A> For a $50 solution, consider using the +12V output on a computer power supply. <S> If the fan does indeed need 440W, then it needs to provide at least 36.6A on the +12V rail – let's say it needs at least 45A to allow some margin. <S> Most modern supplies will happily accept mains anywhere from 100-240V, 50-60 Hz. <S> In that case you could a part like the EVGA 600B , a 600W power supply that can source 49A on its 12V rail. <S> This is just an example of a part that would work; other vendors might as well. <S> Some PSUs have multiple +12V rails with limits on individual rails; you may to test if this is the case, and possibly use a load sharing circuit or better, find one that is above the limits. <S> As discussed in guides like this one , you will need to provide a dummy load on the +5V to stabilize the power supply's feedback loop, and a momentary contact mechanical switch to turn the power supply on. <S> Of course, the power supply will contain its own fan, which may not be suitable depending what you are trying to accomplish. <A> A transformer and rectifier would be sufficient. <S> The rectifier would need to be able to withstand a substantial but brief high inrush current that will occur when the motor is started. <S> The impedance of the transformer will limit the inrush current. <S> You shouldn't need any voltage regulation or filtering for the power supply. <S> For a complete purchased supply, look at automotive battery chargers. <S> Added re comments: <S> A battery charger that is rated to supply the required current should come with adequate cables. <S> For a long wire run, voltage drop needs to be considered. <S> Look for tables online or in electrical codes. <S> I have assumed that this is a commutator DC motor since a brushless motor requires a controller rather than a power supply and 440 W seems too large to have a built-in electronic commutator. <S> A battery charger with a built-in electronic charge control would probably not work very well. <A> step down transformer and rectifier is enough to run the fan. <S> Transformer will step down the voltage form 240v/12v and Rectifier will convert the AC to DC.
| You need a simple one that is just a rectifier and transformer or perhaps an electronic voltage converter. You might even be able to salvage one from an older computer, but make sure to check the maximum current on the +12V rail – this is usually written on the label. The wire size required is dependent on the type of insulation, the temperature of the environment, bundled or open-space wiring, and whether the wire is copper or aluminum. Exactly what dummy connections and the pinouts will depend on the class of supply you use; that guide is for ATX supplies.
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Crosstalk between Wires in a single Twisted Pair I was reading about the Twisted Pair Cables, and how they implement Balanced Line methodology to cancel out all the noise. All fine and good. However, I just couldn't grasp this one simple doubt: When we are twisting two wires (in a pair) close up, which are carrying equal but opposite signals, aren't they going to interfere with each other? Isn't this interference going to add noise? I couldn't find the answer to this simple doubt anywhere. Perhaps, I am missing something very fundamental. _____________________________ Note: The original question appeared to refer to wires in multiple pairs. The OP has advised that he is asking about interference between the two wires within a single pair. The original subject line and text has been edited to reflect this. <Q> Within the one pair the difference in voltage or current IS the signal. <S> If used as a single pair carrying a signal the signal is the relative difference between the two conductors. <S> What you do get is series inductance and resistance and parallel capacitance (and possibly interwire leakage but that is usually negligible) which causes losses. <S> This leads to increasing loss with frequency. <S> This can be passively addressed with "loading coils" which flatten the frequency response across a desired range with a consequent high frequency cutoff and low pass filter action. <S> This method was traditionally used with twisted pair telephone cables to get extended range at voice frequencies and a very sharp cutoff above the range of interest. <S> From dim memory 3.2 kHz was considered to be the top of the speechband for residential circuits. <A> When we are twisting two pairs of wires close up, which are carrying equal but opposite signals, aren't they going to interfere with each other? <S> Isn't this interference going to add noise? <S> The problem we're trying to avoid here is creating a loop antenna that will pick up interference from other sources. <S> We can do this reasonably well by running the feed and return conductors very close together but twisting the pair improves this significantly. <S> Each alternate twist forms a small loop but each consecutive loop is out of phase with the previous one and the net effect is to cancel out the interferece. <S> So, I hear you repeat, "aren't they going to interfere with each other?" <S> Figure 1. <S> The feed and return currents' magnetic fields do not oppose each other. <S> There's no reason to. <S> Because the feed and return carry the same current in opposite directions the magnetic field of each conductor goes through the loop in the same way. <A> When we are twisting two pairs of wires close up, which are carrying equal but opposite signals, aren't they going to interfere with each other? <S> Isn't this interference going to add noise? <S> Usually, the two individual pairs are twisted at a different spacing (say one twist per inch versus 1 twist for every 2 inches) to each other thus, over a length of cable, cancellation of one pairs signal onto another pair is fairly good. <S> If you look closely at the pairs below they have different twist spacings: - Alternatively two pairs can be twisted all together is a diamond, star or quad formation. <S> One pair will use wires at opposite corners of the "diamond" and this will theoretically cancel any cross-talk to the other pair due to symmetry: - <S> Given that the question has changed, here is my revamped answer. <S> Consider the following diagram showing two conductors with equal and opposite voltages present: - Lines of equipotential exist between the two wires and at exactly halfway is a zero volts equipotential plane extending to infinity. <S> It should take only a small leap of faith to imagine this plane is in fact a solid but this highly conductive earth plane. <S> This can be regarded as an electric field screen and the problem then reduces to one wire above a ground plane on the left and ditto on the right. <S> Thus, there is no interference between the two cables only a reduction in capacitance. <S> The fact that twisted pair cable is errrm , twisted does make the visualization more difficult but the truth is there is still effectively a 0V equipotential point between the two conductors.
| IF the pair was used as two circuits with voltages relative to eg ground then you would get 'interference' but that is not how a twisted pair is properly used.
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Proof that every circuit with diodes has exactly one solution Consider an electronic circuit consisting of linear components plus a number of ideal diodes. By "ideal" I mean they can either be forward-biased (i.e. \$v_D=0\$ and \$i_D\geq 0\$ ) or reverse-biased (i.e. \$v_D\leq 0\$ and \$i_D=0\$). These circuits can be calculated by arbitrarily declaring each diode either forward-biased or reverse-biased, and setting \$v_D=0\$ for every forward-biased diode and \$i_D=0\$ for every reverse-biased diode. After the resulting linear circuit has been calculated, we have to check whether at every forward-biased diode \$i_D\geq 0\$ and at every reverse-biased diode \$v_D\leq 0\$ is satisfied. If yes, that's our solution. If not, we have to try another set of choices for the diodes. So, for \$N\$ diodes, we can calculate the circuit by calculating at most \$2^N\$ linear circuits (usually much less). Why does this work? In other words, why is there always one choice that leads to a valid solution and (more interestingly) why are there never two choices that both lead to valid solutions? It should be possible to prove that on the same level of rigor with which e.g. Thevenin's theorem is proven in textbooks. A link to a proof in the literature would also be an acceptable answer. <Q> I assume this is for a contrived problem where there is a circuit with known passives and some <S> I's and V's given and spots marked for diodes of unknown direction. <S> My answer is: Hopefully the creators of the problems have constrained themselves to instances where their assumptions lead to their conclusions. <S> It could be theoretically unsolvable by having a diode be extraneous; consider grounding both sides of a diode. <S> There could be non-trivial cases using virtual grounds or other equal voltages that could be hard to spot. <S> There surely could exist valid circuits that only differ by the direction of a diode for any value of "valid circuit" that includes diodes. <S> Consider modeling switches using those ideal diode rules, how can you decide if a switch was meant to be on or off? <S> Hopefully the given currents and voltages give enough hints. <S> And hopefully they haven't given you conflicting hints. <S> This shifts <S> the question to "How can you tell if an instance has enough information to be unique? <S> " I remember the answer being something like you need one independent given for each independent unknown, but I'm sure I couldn't prove that or come up with a general test for the independence of either. <A> For ideal diodes, there can be multiple solutions. <S> Trivial counterexample: <S> Take any circuit containing ideal diodes which you have solved. <S> Now replace one of the ideal diodes with, if forward-conducting, a pair of diodes connected in parallel, or if reverse-biased, a pair in series, maintaining the orientation in either case. <S> How do you solve for the distribution of current or voltage between the two? <S> You can't, the ideal diode model leads to a convex hull of equally-valid solutions. <A> <A> From Wikipedia load lines entry <S> There is only one unique solution because of the nature of the problem. <S> This is best illustrated graphically, in the form of load lines. <S> The diode has an equation that describes the relationship between current through it (y axis), and voltage across it (x-axis). <S> Here, the x-axis is the voltage across the diode. <S> Look what happens to the current across the resistor as the voltage across the diode changes. <S> If the voltage is Vdd across the diode, then there would be no voltage drop across the resistor, as the voltage across the resistor and the diode must sum to Vdd), and there would thus be zero current across the resistor (Ohm's Law). <S> Similarly, if there were a zero voltage drop across the diode, there would be Vdd across the resistor, and the current through the resistor would be Vdd/R. <S> Now, we know those to be unrealistic situations, as the current in the diode and resistor must be equal. <S> Given the equation for the resistor (linear) and the equation for the diode (non-linear, but monotonic increasing), we can see on the graph that this can only happen at one unique point, the intersection of the two curves. <S> Thus, the simultaneous solution of three equations (the resistor, the diode, and the fact that the two currents must be equal) give use one unique solution. <S> This method will work for all circuit elements. <S> It's a bit different for reverse-current diodes, as the resistor current goes the other way, and a quadrant needs to be added to the graph. <A> I think ist quite simple: you can treat the forward biased ideal diodes as shorts and the reversed biased ideal diodes as open circuits. <S> So in any case you get circuits with only linear components (because all diodes either resolve to open circuits or shorts) and those linear circuits are known to have exactly one solution. <A> The 'proof' of this would only work for certain circuits. <S> If you have some gain and the only nonlinear elements are the diodes themselves you <S> can have multiple possible states. <S> For example (may not be the simplest possible example). <S> This circuit will work with an ideal perfectly linear op-amp and the output never goes off to infinity or saturates, yet with 0V in it can be about +6 or about -6 at the output, with one pair or the other of diodes conducting. <S> It will also work with 'almost ideal' diodes that have a forward drop when on and no other nonidealities. <S> (and of course tunnel diodes are a special case with their non-monotonic I-V curve). <S> The proof would probably have to require only passive elements such as resistors (no dependent current or voltage sources). <S> Or perhaps only with ideal diodes with 0V <S> Vf. <A> This is not a complete proof, but perhaps it will get you on track: If there are multiple solutions, there is at least one diode that can be either forward or reverse biased. <S> Consider one such diode. <S> In a given solution, it is either forward or reverse biased. <S> Let's define the voltages at its terminals, <S> Va and Vb, such that if it is forward biased, Va <S> >= <S> Vb, and <S> if it is reverse biased, Vb >= <S> Va. In either the forward- or reverse-biased case, the Rest of the Circuit (RotC) produces these voltages at the terminals of the diode. <S> Since you stated that the circuit consists of linear elements and diodes, either the RotC is a purely linear network, or it includes more diodes. <S> If the RotC is a purely linear network, it has only one solution, and the only solution to the constraints Va > <S> = Vb and Vb <S> >= <S> Va is that <S> Va = Vb. <S> If the RotC includes more diodes with multiple possible solutions, consider the next such diode. <S> Again, it is either connected to a linear network, or a network with more diodes with multiple possible solutions. <S> If we assume there are a finite number of diodes in the circuit...
| I don't have a rigorous proof, but the general idea is that as long as the components of a circuit have V-I curves that are single-valued functions (this includes diodes as well as linear components), there can be only one solution to the circuit overall.
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Will a fuse improve the safety of my non-isolated buck converter design? I have a power circuit that implements an AC-to-DC rectifier and a non-isolated buck converter to convert a 120 VAC 60Hz signal to 12V and then a linear voltage regulator to step the 12V down to 5V. The 5V will be used to power a micro controller and there is a possibility that people will touch the pins of the micro controller. I realize that it is not ideal to use a non-isolated power converter, but due to size constraints I do not want to put an isolated power supply with a transformer in the design. The question is: Will a 500mA fuse after the 12-to-5V linear voltage regulator make the design safe against electric shock since the buck converter is non-isolated? <Q> Will a 500mA fuse after the 12-to-5V linear voltage regulator make the design safe against electric shock since the buck converter is non-isolated? <S> No. <S> A fuse does not protect against electric shock. <S> What you are proposing is potentially lethal. <S> You must isolate the low voltage circuit from mains if there is any risk of contact with the circuit. <S> Just for reference, RCD / GFCI earth leakage protection devices generally trip at 30 mA as this is considered a safe limit for protection of humans against electric shock. <S> Put safety first. <S> Then worry about size and cost. <S> An isolated SMPS (switched mode power supply) with 5 V output won't be any bigger than what you are proposing. <A> No. <S> 500 mA is significantly greater than the current necessary to put a human heart into fibrillation. <S> The main potential failure mode here is if the device is plugged into mains wiring with the live and neutral conductors swapped (which is disturbingly common). <S> Then both your "+5V" and "0V" outputs will be superimposed on top of line voltage, and will be an immediate hazard. <S> Or if you're using a full bridge rectifier on the front end then the output will be floating at \$\frac{1}{\sqrt{2}}\$ of mains voltage even if the outlet is correctly wired! <A> There must always be a full isolation barrier between AC mains and anything a person might touch, and it must be arranged in such a fashion that no plausible single-point failure can breach it (certain components specified such that fail-shorted scenarios need not be considered plausible). <S> That point should be considered non-negotiable. <S> If you don't want to put an isolation barrier between your electronics and the mains, you must put an isolation barrier between those electronics and all persons nearby <S> , i.e. don't let anyone touch anything connected to the electronics . <S> It's unclear why you are expecting people to touch the pins on your controller, but if you are expecting that, you must use an isolated supply. <S> A fuse will reduce the likelihood that a failure in your device will start a fire. <S> If you're lucky, the fuse may reduce the harm caused by some other kinds of failures, but fuses aren't really good for much beyond fire prevention. <S> Devices that don't burst into flames are generally safer than those that do burst into flames, so in that sense fuses do improve safety, but they in no way affect the isolation-barrier requirements.
| Fuses are for limiting current in a fault condition to protect equipment, wiring and prevent fire.
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How to generate a -24V signal I have to interface a specific motor, it has a weird driving board, I was in contact with the technical team from the manufacturer, and they told me that I have to send a -24V on a pin to allow it to turn. I have a +24V power supply, several step down modules, on 5V and 3.3V for my microcontrollers and other things.So here is my question:What is the simple way to produce a -24V voltage source from a 24V or lower one?I have searched a lot, but found nothing...My first idea was an inverting opamp, but if I were able to power it down to -24v then I would'nt need it at all :) Any Ideas are welcome. My ideas are now oriented to step down ICs able to output negative voltage from a positive one... I guess something around -18V should be enough to be understood by the driver. <Q> Linear tech produce a number of inverting switchers like this: - Check the data sheet to see if it can produce -24V. <S> If not then use the search engine on this page here <S> There's also this one: - <A> Assuming the -24v is a control input, so doesn't need much current, and only needs approximately -24v, then a cheap, quick and dirty way is to use a charge pump. <S> Like this. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> What's drawn as an opamp is any circuit capable of making a rail to rail square wave. <S> The output is unregulated, and drops with increasing current. <S> If you do need exactly -24v, then you can use a 2 stage charge pump and regulator. <S> If you need more current than a few mA, then it's worth going to a magnetic solution, an inverting boost, rather than a charge pump. <A> An easy way is to use a DC-DC converter with an isolated output. <S> For example, an PDS1-S24-S24-S . <S> This particular one, while cheap, has a minimum output load requirement of 4mA so add an LED indicator or whatever if your load is insufficient (the consequence of insufficient loading is that the magnitude of the output voltage rises above the expected value). <A> What is the simple way to produce a -24V voltage source from a 24V or lower one? <S> You don't have to produce a new source. <S> Connect the "0V/GND" for your -24 Voltage to +24 and your -24V connector to 0V/GND.
| Because +24 and GND/0V is just a name for a potential we can switch the postitive and negative potential to get a negative voltage.
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Power Dissipation of Schottky Diode I am implementing a circuit as shown below :- Source is a 15V Solar panel & Load is the Battery. The intent is that if someone tries to connect a bigger battery on the output, Diode(Schottky is what coming to my mind) shall block reverse flow being reverse biased. In normal scenario, all the current will pass through forward biased schottky diode. The rating of the system is 15V/20A. Upon looking into few schottky diodes available, which have a forward voltage drop of about 400mV, the power dissipation is coming out to be Vf*If = 0.4V*20A = 8W which is a lot of power to dissipate. Am I using the right approach?? Any suggestions here. <Q> Am I using the right approach?? <S> Any suggestions here <S> Maybe you should consider the standard MOSFET circuit (but modified) <S> that acts as a very low volt drop diode i.e. it protects a circuit from reverse polarity but turns the MOSFET on so that forward volt drop is barely a few milli volts. <S> The basic circuit is this: - + <S> Vin is where you would connect the solar panel and +Vout is the battery connection. <S> If you add a bipolar transistor and base resistor you get this: - From a quick simulation, the BJT turns off the MOSFET when the battery voltage exceeds the SP voltage by about 730 mV (R3 is 10k not 100k). <S> Between the voltages being equal and the battery being 730 mV higher there will be some hundreds of mA flowing back to the panel. <S> However, if you are trying to protect the panels from the wrong battery being put in place (i.e. 24V instead of 12V) then it should do the trick. <S> It's just an idea and not proven other than by a quick sim. <S> Caveat emptor!! <A> Source is a 15V Solar panel & Load is the Battery. <S> This may work, somewhat, for awhile. <S> @DerStrom8 gave a link to a power Schottky which would work and be quite robust to both drop little voltage and prevent a "larger battery" from damaging the solar cell. <S> Namely, that the battery should have dedicated charging circuitry. <S> If none is used, the battery electrolyte may boil away, destroying it. <S> Other modern battery types are no exception - a dedicated charging circuit is a really, really good idea. <S> In the case of LiPo batteries, they can fail catastrophically if charged incorrectly. <S> Another issue you'll discover is that solar panels are quite inefficient most of the time, and to get the most power out of them requires a technique called Maxiumum Power Point Tracking or MPPT. <S> Now this isn't a requirement, but it will give a nice boost to the efficiency of the overall system. <A> but you could get into some problems discussed here <S> Is paralleling diodes a bad idea? <S> So my suggestion is that you take a look at Smart bypass <S> diode which is essentially a mosfet that contains all the necessary driving electronics to be used as a diode exactly but with much lower voltage drop and power loss. <S> you can also take a look at this article by infineon <S> Automotive MOSFETs Reverse Battery Protection and TI Reverse Current/Battery Protection Circuits
| However, if charging lead-acid batteries, a whole other gamut of issues comes into play . you could use multiple schottky diodes in parallel to reduce the current passing in each on them so you can manage the heat dissipation easier .
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safe way to reduce 200V to ~20 volts? edit2: After calling customer service, the impedance is 350k. Does this change anything? I am totally clueless when it comes to circuitry, but am trying to build a grid charger for my old hybrid car, using an online DIY write-up. This charges the battery at a constant current (350-600ma selected ) and the voltage is monitored as it rises from a low of ~120v absolute min to ~ 195 volt absolute max. I want to add to the setup by monitoring this with a voltmeter display that measures 0-33v, so a logical way to do this would be to divide the voltage by 10. It seems to be unsafe to just use resistors to achieve this, as I understand this will be dissipating some 30-40w of heat. What is a safe (and preferably efficient) way to do this? I need to use this particular display because of its accuracy precision, as I only know to stop charging when the voltage stops rising, and the voltage change will be extremely slight (0.1v/hr) at the end of an extended charge. edit: the intended display is a self powered Drok 5 digit DC . I did not see any info on impedance. Am I correct in assuming a self powered voltmeter has no or negligible impedance? <Q> What is the input impedance of your voltmeter? <S> If it's 1 Mohm, you could use a 1 Mohm/100k divider ( <S> the absolute accuracy would be poor, like 10%, but you care about trends, not accuracy) and the total power in the divider would be a max of about 36 milliwatts. <S> A 100k/10k divider would provide accuracy of about 1%, and only dissipate 360 mW. <S> You don't need absolute accuracy, since if nothing else the battery voltage will change with temperature. <S> What you're trying to do is detect small changes (or lack thereof), and the absolute accuracy is irrelevant to that task. <S> What you care about is stability, and with low power dissipation this ought to be good enough. <A> you've got one of those 3 wire LED voltmeters with red yellow and black wires <S> You'll probably find that an ordinary 100 to 220V AC to USB adaptor can be connected to the 100 to 195VDC and will produce 5V at reasonable efficiency which will be enough to power the voltmeter, (red and black wires) for the measurement side add a series resistor to get the appropriate scaling. <S> The impedance of the meter can be determined by hooking it up to a low voltage source (eg: 12V) measuring that and then adding resistors in series with the yellow wire until the reading halves. <S> (the added resistance then matches the meter impedance)You need nine times that to get a one-tenth reading. <S> Some meters have a solder jumper on the back of the meter that can be moved to shift the decimal point, but I suspect that yours is not one of them. <A> The simplest way forwards here is to use a passive potential divider network as you've already suggested. <S> If you make the total resistance of the potential divider = 1M, then the power dissipation will be very small. <S> In order to prevent the input impedance of your monitor affecting the output of the high impedance potential divider, feed the output of the potential divider into an OpAmp stage set up as a unity gain amplifier. <S> Then connect your monitor to the output of the OpAmp.
| if your meter has 1 megaohm input impedance a 9 megaohm resistor in series with the white wire will reduce 195V down to 19.5V, (connect the black wire also to battery negative) Check the data sheet for the resistor, some cheap resistors are not suited for continuous use at 200V, adding a fuse in series with the resistor is probably not a bad idea either.
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Reverse polarity protection for a very low voltage devce I want to add reverse polarity protection to a device which is powered by a single AA battery, so its voltage might go down to 0.8 V. Forward diode is too huge of a drop. I've already thought about adding P-MOSFET but what bothers me is a diode's forward voltage. It's usually around 1 V. What it means is that if battery is inserted the right way, MOSFET won't be open because Vgs is not high enough. As I understand, if after diode voltage is high enough, MOSFET will be opened and diode won't matter anymore. Am I wrong or should I consider other ways to implement reverse polarity protection? Just to be clear what circuit I'm talking about: I was considering Si2329DS which has a maximum Vgs of 0.8 V but diode voltage leaves it unusable. edit:So, I've been doing some research and I've found ideal diode controllers that enable me to use NMOS. I'm going to tell if I find something suitable <Q> Did you see (and understand) <S> Spehro <S> Pefhany's comment?I take the liberty of turning it into an answer because that's also what I'd suggest. <S> Add a diode in front of your circuit that shorts the voltage if the battery is reversed (=V2). <S> In that case the (resettable) fuse will break the current flow. <S> If the battery is not reversed (=V1) <S> there is almost no voltage drop (only a small one across the fuse because it won't have 0 resistance): <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The fuse, of course, must be able to sustain enough current for supplying the circuit in normal operation but must blow/deactivate if the battery is shorted via the diode. <A> You dismiss diodes very quickly. <S> Have you looked for Schottky diodes or specialized diodes with very low voltage drop? <S> Look for the TI SM74611. <S> It has a Vf of 26mV and is available for less than $5 on digikey. <A> Ok, I will add one more answer myself for more clarity for future. <S> Usually when trying to block reverse current diode seems like a nice way, nevermind if it's Schottky or regular Silicon diode. <S> The problem with it is a fairly high voltage drop which might be unacceptable in most of cases where device is battery powered. <S> The good thing is that you can use devices called Ideal Diode Controllers. <S> It drives N-MOSFET's gate so you can drive voltages as low as 0.6-0.8 <S> V. <S> Some of these controllers tend to work from 4 V or something onwards but there's LM74610 <S> which is suitable for such an application
| The diode must pass enough current at reversed polarity to deactivate the fuse.
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Are circuit boards necessary? Can I solder the components together? I just started to learn and I don't have a circuit board. <Q> without using a prototype board, as long as you are careful not to let any leads short together. <S> Unlike most cases, where you would twist wires together for mechanical strength, when prototyping this way you generally want the solder to hold the leads together <S> so it is easier to later unsolder and reuse the components again. <S> Instead of soldering, you can also just clip leads together using alligator-clip jumpers. <S> Either would work for a circuit of the complexity in your question. <S> Much more complicated, and you will want to get a solderless breadboard (however, see below!). <S> You can probably stick the other end of the resistor directly into the socket of the Arduino. <S> You will need to get some hookup wire for the ground and <S> +5V leads; 22-guage solid is probably the best size. <S> Before PCBs were common, consumer electronics were all hand-wired. <S> Components with leads were soldered together using terminal strips, like the one at the top of the picture. <S> This is the chassis of a 1948 Motorola Golden View 7" television set. <A> No, a pcb is not necessary. <S> Many people practice Deadbug style circuit soldering. <S> It's just not all ways professional looking or easy to duplicate in mass production (i.e. automation. <S> It's man hour intensive). <S> It's mostly limited to hobbyists, one off projects or prototyping/testing designs prior to a finalized version. <S> An example with SMT components: <S> An example with Through Hole components, casted in resin, for aesthetics reasons: <A> Can I [just] solder the components together? <S> Sure, the process is called dead-bugging ( wikipedia .) <S> I remember an article on hack-a-day, called volumetric circuits , where someone created some software to help design a circuit and create an assembly plan. <S> It should be noted that printed circuit boards (PCBs) have other advantages, and design concerns. <S> ex: <S> FR4 - flame retardant type 4 is used as a PCB substrate <S> It has favorable strength, insulating and fire resistant properties <S> Better control over noise and other transient behaviors <S> /phenomenon through design features such as ground planes, shielding and element placement (to name a few.) <S> Tractable designs for RF (radio frequency) and beyond (GHZ, etc); design constraints get stronger as the frequency of operation increases. <S> As long as the circuit you are building/designing/whatever has wide enough design tolerances to be dead-bugged, it should work. <S> Even a bare micro-controller like the ATMEGA could be dead-bugged into a working Arduino clone! <S> A breadboard for prototyping, or proto-board for permanent/semi-permanent circuits would generally be more productive and useful than dead-bugging; dead-bugging can be useful in a pinch. <S> The ability to repair/disassemble <S> a circuit is, however, not to be underestimated ; hence the saying "don't build something you can't take apart." <S> n.b. <S> sometimes because of design constrains some components are 'stacked' even on a PCB, ex. surface mount resistor soldered on-top of surface mount capacitor to reduce parasitic capacitance where it would be a significant fraction (10%+) of the caps capacitance. <A> This answer is for your particular circuit. <S> Other answers have covered the general question. <S> Yes, you can solder components directly together. <S> The main concern in your circuit would be the motor, which probably takes a lot of current. <S> I recommend giving the Arduino power and ground a separate connection to the 5V supply. <S> This will reduce the noise on the Arduino's power supply when the motor turns on and off. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> You're asking something similar to <S> "can I carve wood sculpture using a chainsaw?" <S> Of course the answer is that some people do, but it takes a lot of skill, and it's ridiculously easy for one careless move to destroy everything you've done instantly. <S> Given that you're asking this question, there is no doubt that you're a complete beginner to electronics. <S> As a complete beginner, your soldering and assembly skills will not be very good. <S> I would not suggest my grandmother should try chainsaw sculpture, and I would not suggest you try doing this! <S> What you want is called breadboard . <S> It lets you build circuits simply by pushing components into holes. <S> Save the soldering until you know what you're doing.
| The actual practicality of dead-bugging is rather limited. You can literally solder components together, i.e. one end of the resistor to the base of the transistor etc. If you can afford the components, you can afford the breadboard. You could - but you shouldn't !
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DC motor and encoder for pendulum with low static friction I would like to build a pendulum from scratch and model/control via Matlab. I don't know what DC motor should I use that has low friction (low static friction), so the pendulum can rotate by its self. Moreover, I would like to have an encode with "good" precision, less then 1^o.I want to control the position of the pendulum by using the motor to put energy in the system. It will be a closed loop control and the controlled variable is the voltage drop on the motor. How to check in a technical specification if the motor has low static friction? Eg. the technical specification for a motor include: Values at nominal voltage: Nominal voltage; No load speed; No load current; Nominal speed; Nominal torque (max. continuous torque); Nominal current (max. continuous current); Stall torque; Starting current; Max. efficiency; Terminal resistance; Terminal inductance; Torque constant; Speed constant; Speed / torque gradient; Mechanical time constant; Rotor inertia. <Q> The problem with motors is that while static friction and dynamic friction can be addressed by selecting the right bearings, many motors have other undesiderable characteristics: Brush friction is present in all brushed motors. <S> Torque ripple affects all electric motors to some degree, with the exception of the impractical homopolar motor . <S> The reluctance of the magnetic circuit changes as the rotor and stator poles step in and out of alignment while the rotor turns. <S> This causes a torque which tries to align the rotor at certain specific angles. <S> Cogging torque is a specific case of torque ripple which occurs in all slotted permanent magnet motors and persists even when the motor is at rest and unpowered. <S> Affected motors include permanent magnet brushed DC motors, brushless DC motors and stepper motors. <S> Coreless DC motors, induction motors, reluctance motors and electrically excited (field winding) <S> DC motors are immune. <S> Torque ripple can be alleviated by increasing the motor pole count or adding a transmission between the motor and pendulum, but this will obviously cause additional friction and add inertia to the system. <S> Given the fairly specific requirements, it might make more sense to use a voice coil rotary actuator (like what is found in hard disk drives) rather than a motor. <S> Not only does such an actuator have zero torque ripple, but there is no physical contact and the moving part can be made very lightweight to reduce its impact on the system. <S> The variations in torque in response to position can be fairly linear as well, and are easily compensated out: <S> The downside is a reduced range of motion. <A> Actually, you need to rethink your controller. <S> As stated, forcing the motor voltage to zero will cause the motor to provide a large damping force to the pendulum. <S> The reason is that any motion of the motor shaft will cause the motor to act as a generator. <S> Keeping the motor voltage at zero is effectively the same as providing an extremely small resistance across the motor (E = iR, after all), and this will resist any motor motion. <S> So simply controlling the motor voltage will not allow the pendulum to swing freely, regardless of friction. <A> You could start by looking at brushless motors. <S> " There are motors sold for laboratory or prototype building that are described in more detail by the sellers than motors sold for hobby or toy use. <S> The same is true for encoders. <S> WhatRoughBeast has a good point. <S> The controlled variable needs to be pendulum position as determined using the encoder. <S> The circuit that energizes the motor will need to "open circuit" the motor at the end of the pulse. <S> I think it will need to provide a trapezoidal shaped pulse. <S> Upon further consideration: I was thinking a trapezoidal pulse would minimize the back emf generated when the pulse is shut off. <S> Upon further consideration, I think that it is going to be more difficult to prevent any permanent-magnet motor from producing some braking torque when it is not energized. <S> You can shut it off without providing a path for generated current, but there will still be some reluctance torque. <S> An induction motor might be better, but you would need an inverter to energize it.
| Try looking for motors that are described as "high quality" or "precision.
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Noise problem on input of an analog to digital converter I just finished building and testing a board using an LTC2440 24Bit analog to digital converter. It functions, but there is a noise problem that I don't understand. The signal I am trying to digitize is very clean (for what I need.) When I connect it to the LTC2440, however, I get a very large amount of noise. It seems to come from the LTC2440 itself, although the noise on the input is larger than the noise on the 5V supply to the LTC2440. What I am trying to measure is a slowly changing DC from the MAX2015. I need the DC voltage accurate to 0.00001 or better. The voltage represents the level of 13GHz RF received by a satellite dish LNB. Right now, the circuit far outperforms the original solution I had that used a SatFinder to provide the level reading voltage. I expect to have to improve the circuit and layout in order to get the performance that the LTC2440 is capable of. This question is NOT about that, though. Here I am only trying to find the cause of the enormous peaks that you can see below in the picture labeled "Noise on LTC2440 Input." Information: Schematic: Noise on LTC2440 Input (pin 5): Signal (Output from U5 MCP6231 pin 6, when disconnected from LTC2440 pin 5): This picture doesn't do the real signal justice. The camera has a problem capturing the trace. If it is bright enough to be sharp, the camera captures the glare around it. If the trace is dim enough that the camera doesn't capture the glare, then it also doesn't capture the trace. In reality, the trace is a razor thin line with small, slow, excursions. If you look closely, there is a thin, almost white line down the middle of the fuzz. That's the signal. The rest fuzzy from the glare. Even so, that is a small fraction of 1mVolt as compared to the approx. 10mV peaks in the noise picture above. The MCP6231 was added during testing because I found that the MAX2015 puts out a lot more noise than I expected. The simple RC filter clears that up, but doesn't make the glitches go away. 5Volt to LTC2440: 3.3Volt to MCP6231: <Q> The main problem is that the voltage reference (LT1019) is not suited for that ADC. <S> The ADC requires substantial decoupling on the REF inputs because it is an unbuffered input and will glitch merrily at whatever sampling frequency you are running at. <S> However, the LT1019 likes to run without output decouplers. <S> It's a very old device and the data sheet is poor and just does not tell you how good the LT1019 is at dealing with high frequency switching noise caused by the DAC. <S> More modern devices show a graph of output impedance versus frequency to give you better facts to work with. <S> This is also true of the inputs but at least you are using a buffer amp (MCP6231). <S> However, that buffer amp has a very poor equivalent input noise (50 nV per root(hertz)) and, now that I've checked, does not specify what output impedance it has (versus frequency) either. <S> Poor choice of op-amp and poor voltage reference. <S> There may be other things wrong too but these stick out to me as bad design choices. <A> I see two things. <S> One is noise where there should not be, and two is an absence of bypass/decoupling capacitors at your LT1019 voltage regulator, and pins 3 and 4 of your ADC ref inputs, and minimal filtering at your op-amp. <S> I would add a 4.7uF across its power pins. <S> Add a 100nF and 4.7uF capacitors to pins 3 and 4 of your ADC. <S> Also your main power feed could use a 470uF capacitor at pin 2 of the voltage regulator. <S> Make sure the grounds leads from the RF receiver to the op-amp and ADC are as short as possible. <S> Get the power feeds clean first , then check for signal clarity again. <A> I think I know where the problem is coming from, and why it isn't affecting my measurements as badly as I would expect from the size of the peaks. <S> While putting things together to post this question, I had to dig out the LTC2440 datasheet again. <S> I also had good look at the pictures of the scope traces while cropping them. <S> The peaks occur at the sampling rate of the LTC2440. <S> That is 880Hz the way I am operating it. <S> Pages 23 and 24 of the LTC2440 datasheet, explain what happens when the LTC2440 samples, and it also explains the peaks. <S> The LTC2440 appears to need a very low impedance source to sample from. <S> The datasheet also explains how to drive the LTC2440 so as to keep the peaks from being a problem. <S> Basically, driving directly from a simple non-inverting buffer won't cut it. <S> You need a hefty capacitor on the input (1µF) and an op-amp configured to drive that capacitor. <S> The capacitor supplies the current needed by the LTC2440 input, and gets recharged to the correct voltage by the op-amp between samples. <S> So, my low power MCP6231 isn't up to the job of driving the LTC2440. <S> The interesting thing for me is that this explains why I'm not getting completely crap results. <S> I tested the circuit with a small program that just reads a bunch of values and calculates the peak-to peak AC. <S> The peak-to-peak values were in the range of 1mVolt, which wouldn't be possible if it were simply noise - the spikes are 15mV high. <S> I think what is happening is that my little MCP6231 manages to pull things within 1mV of the real signal before the sampling takes place in the flat spot between peaks.
| Your op-amp only has minimal filtering with a 100nF capacitor.
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What is the significance of "microcontroller based on microprocessor"? I am aware about the basic theoretical differences between the two. Like micro controllers have ROM RAM memory etc., but microprocessor has none. But I am still confused. Why is it said "A microcontroller based on microprocessor" ( the very first line ). If ARM is a microprocessor, then why STMs use it and call it as micro controller. What's the difference? Is eval board like "TI c-series launchpad" has a micro controller or a micro processor or both. Kindly correct me if I am wrong anywhere. <Q> ARM is the CPU core, which can be used to implement a microprocessor or a microcontroller. <S> The full sentence you referenced is: The STM32 family of 32‑bit Flash microcontrollers based on the ARM <S> ® Cortex®‑M processor is designed to offer new degrees of freedom to MCU users. <S> Clearly this is largely content-free marketing babble. <S> Don't pay much attention to it. <S> It actually never says "microprocessor". <S> It refers to the ARM core as a "processor", which isn't strictly right either. <S> It's just a core, which can be used to implement various kinds of processors. <S> The core is more like the engine of a car. <S> You license the design from ARM, but can heavily configure it to your needs, and putting the chassis and wheels around it is your job. <S> You can make the result a sports car, a pickup truck, or various other types of vehicles. <S> The marketing babble above is like saying "We've based this pickup truck on a sports car" . <S> No, they haven't. <S> They've base the pickup truck on the same basic engine technology others have used to make a sports car with. <S> Again though, the important point is that this is all marketing babble. <S> There is nothing useful to see here. <S> Move along everyone. <A> A microprocessor is effectively a computer without any peripherals <S> (I/O, ADC, timers, etc). <S> A microcontroller is a processor with those peripherals connected and all combined into one package. <S> Therefore, a microcontroller is based on a microprocessor, and adds peripherals to it. <A> When you consider purchasing microcontroller, you usually assume that it has many features required for your design, and main thing is to achieve a task, not the speed. <S> When you consider purchasing a microprocessor, you know that you are good in terms of speed, but you will have to put additional effort to integrate it to your system. <S> So, in your particular example, saying that a microcontroller based on certain microprocessor would imply that this microcontroller's computation power is fast compared to other microcontrollers. <S> So, think of it like someone used that microprocessor and built a system for you <S> so you don't have to do additional work. <S> That's my understanding, correct me if I am wrong.
| Microcontrollers are said to be "complete" systems(human body), while microprocessor is just a "brain".
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Efficient way of controlling many LEDs I am working on a project which is going to have approximately 50 LEDs. These will be status LEDs to show whether an appliance is On/Off. I will also need a brightness control for the LEDs so PWM is a requirement on the pins. What will be most efficient way (low cost, low PCB area and low component count) to accomplish such a thing? I had a couple of ideas but then found some issues with all of them: 1) Get a micro-controller with too many GPIO - Controllers with 55 GPIO appeared to give the best GPIO/price ratio. I have selected one with 55 GPIO but even that controller might not be able to handle everything. Putting two micro-controllers complicates things. 2) Use i2c port expanders - These are very costly. I'd rather buy another micro-controller. 3) Use shift out ICs (595) - Lack of PWM. Implementing PWM on code side doesn't seem to be a good idea to me (maybe I am wrong but it feels like I am unnecessarily making things difficult this way). <Q> Another possibility is to use addressable RGB leds, based on the WS2812 or similar. <S> This is an RGB led with a PWM controller built in. <S> They can be bought in strips, or individually. <S> These are very popular in the Arduino community. <S> (Search on NeoPixel or APA102.) <S> You can string a very large number of them together serially, and then control them with two output pins on a microcontroller. <S> But you have multi-color capability, which might help for indicators. <S> A possible downside is that if you microcontroller goes off in the weeds, the LEDs will continue to display their last data, which could be a safety concern. <A> You could use dedicated led driver chips like 7216 or their modern equivalent. <S> 3) <S> Use shift out ICs (595) <S> - Lack of PWM. <S> lack of pin-independent PWM. <S> you can apply pwm to the OE pin and control the brightness of all of those leds. <S> not individual leds, however. <A> Your solution 3 (use shift registers with latch) is quite cheap and absolutely feasible for the given number of LEDs. <S> You will need 7 shift registers and therefore have to send 56 bits followed by a latch pulse (use the SPIs Chip Select signal) <S> x times within T, where x is the desired PWM resolution and T is the period. <S> Both values determine the necessary SPI baudrate. <S> Of course the micro has to be fast enough to calculate the bit streams and transmit them to the SPI. <S> Wich micro to choose depends on your PWM performance requirements. <S> 250Hz LED dim frequency at 8bit PWM resolution should not be a big issue for any standard 32 bit device - just to give a number. <S> You may have a look here for some example implementation. <S> However solution 1 may be the better choice. <S> For instance, the Renesas RH850 F1L (100 pin housing) can generate up to 48 individual PWM signals, and I'm sure there are some peripherals left to generate the remaining 2. <S> If your desired controller doesn't have enough dedicated PWM outputs, it should have a DMA unit and be able to write at least 16 bit wide output ports. <S> It depends on your application, but letting the CPU do all that transmission stuff may overload it.
| Depending on what you get, you may need an external 12v supply. You can use an SPI interface for generating the bit streams.
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PCB routing T joint alternative Whenever I come across this situation: I feel a strong urge to make a T - joint like this: However 90 degrees are usually frowned upon. As such what's a better alternative to this T-joint? If I follow Tyler's suggestion, I'll get this: However this contains an acute angle which is even worse. <Q> <A> Why not take the track point from the resistor pad? <S> Apologies... <S> my KiCAD is broken (recompiling now) <A> You never want to have acute angles, as you seem to know already. <S> There really isn't anything wrong with T-junctions, especially if you use "teardrops" (or fill in the 90-degree corners) <S> : <S> Personally I just use "Y" junctions, which only have one 90-degree corner: <S> Take your pick, but don't use Tyler's suggestion. <S> Acute angles can trap etchant which can eat away at the copper over time. <S> This isn't as much of a problem nowadays but it's still best practice to avoid such routing. <A> Have an acute angle (or 90°) on a PCB is a problem if the junction drive high frequency signal (like communication buss) or a great current amount. <S> This two cases will cause some ECM issue. <S> Otherwise, it's okay to have T junction. <A> However, this is considered bad practice <S> and I recommend avoiding it. <S> The other answers are correct; route through pads or add fillers in the inner angles of 90-degree intersections. <S> Another thing I do is to place an appropriately-sized via at the intersection. <S> Make sure the ring around the via has enough copper to handle the trace's current. <S> Oh, and don't do this with high-frequency signals :)
| You can probably get away with 90-degree intersections until you need really thin traces or are routing rf signals. There's nothing wrong with a T joint (with two 90° angles). But if you really want to avoid it, it might be possible to route the entire line through the pad: Do not use Tyler's suggestion.
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Is it possible to use a coil to create sustained displacement/preasure? without a permanent magnet? So, as far as I understand, speakers are basically coils coupled with permanent magnets, then when a current is run through the coil it creates a magnetic field, pushing out against the permanent magnet, moving the diaphragm that's attached to the permanent magnet and creating sound waves. What I'm interested in is using a coil to do something similar but instead create human perceivable pressure. For example, something to lightly but noticeably push against your finger at a constant rate (not something that constantly moves back and forth like a speaker diaphragm). If I just make a small coil out of some magnet wire and hook it up to a battery, I think that the magnetic field created by the wire would create a constant pressure against the magnet with the current being proportional to displacement. Is this correct? If that's right then my real question is if this is possible to do without using a permanent magnet? I know you can create an attractive force to metal with a coil but is there a way to create a repulsive force? <Q> You can either use two coils of wire, or one coil of wire and a permanent magnet. <S> Both will work. <A> What you are describing is commonly known as a "solenoid". <S> It's a readily available electrical components something like a relay. <S> Go to www.mcmaster.com and search on "solenoid" and you will see lots of commercially available varieties, though most are probably too large for your specific use. <S> The solenoid is a coil of wire (actually many turns) formed around a circular tube. <S> An iron, but un-magnetized, rod slides into the tube. <S> When the coil is energized the rod (often called the "plunger") will move in or out of the coil depending on the specific design. <S> So, yes, the mechanism you describe is indeed possible. <S> However, one problem you may run into is battery life. <S> Solenoids are typically current "hogs". <S> In order to maintain the magnetic force on the plunger, the current must remain flowing thru the coil. <S> While current is flowing battery power is being consumed. <S> If battery size and life is an issue you may have to apply some cleverness to your design and come up with some kind of latching solenoid, probably with two coils - one to push the plunger into the finger and another to pull it back. <S> This way you will only need to energize the coil, and consume battery power, when the plunger is moved, rather than the entire time it is engaged against the finger. <A>
| One way to create a repulsive force is to use a diamagnetic material for the coil's magnetic field to push against, another is to use @user96037 's excellent double coil suggestion.
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Timing warnings for functional model I am writing a controller for a low power/mobile DDR module on my FPGA. To allow debugging, I use a functional model written in Verilog. In it, the setup and hold time for some signal is set to 1.5 ns. If I understand everything correctly, this means that the signal can not change 'within' 1.5 ns of a rising clock edge. However, the RTL that I've written does not include timing, so the signal appears to change instantly, yielding hold time warnings. On the one hand, I am not too concerned; I'm only getting warnings, and I think that during a project for my university, we were told to simply ignore these errors. On the other, I don't like to ignore warnings. The manufacturer would not have implemented these warnings if they have no purpose. Since Xilinx ISE is able to check timing constraints, I feel like it should be possible to route and map my design, and use the generated timings somehow (but maybe I'm making things too simple here). I'm sure there are more people with the same problem. What is the proper way to deal with these warnings? Edit: On this page , I found some more information. You can generate a post-map or post-place-and-route simulation model. I suspect this includes the timings. However, it seems only modelsim can actually perform the simulation. Clarification: Ideally, I would be able to synthesize (or at least get as far in the process of generating the layout as possible) my part of the design (I have the RTL and I have specified the board, so I think this should be possible), then combine it in a testbench with the functional model to test if my design has the proper timing delays. However, I can't make this work in Xilinx ISE 14.7. <Q> Setup and hold timing checks only make sense with post-layout information. <S> A century ago, you could do timing analysis without layout structural information because the the device delays were overwhelming compared to routing delays. <S> Some models are written to be used in both RTL and structural simulations, so you can ignore these warnings. <S> Even better would be to turn of the timing checks which improves RTL simulation performance (maybe not relevant on your project, but that what people do in industry). <A> As dave_59 pointed out, timing checks on RTL are not very meaningful. <S> That said, here are 4 possible approaches: <S> Don't compile the timing checks with RTL <S> Use a compiler directive macro ( `ifdef ... <S> `endif ) around your timing checks; timing checks disabled with RTL Inject hold time into RTL behavior by changing all flop assignments ( <= ) to <= <S> `D , where `D is a your delay (ex `define D #2 ) <S> Idea from NBA with Delays , by Cliff Cummings <S> For synthesise define <S> `D as empty (aka `define D ) <S> Hack your checkers to to have a masking atribute. <S> ex: <S> $setuphold( D, CLK &&& actiming_enable, tSETUP, tHOLD); <A> There should be some way to input a timing delay. <S> This could be done a couple ways I can think of (maybe more): <S> Modify the launching flip flop to have a clock->out delay which is greater than the hold time of the capture flip flop. <S> Modify the flip flop model to have a -.1ns hold time. <S> Modify any gate (like a buffer) to have a fixed delay of 1.6ns and put this before each capture flip flop. <S> Shift the clock signal at the capture flip flop (second flop which experiences the hold violation) by at least 1.5ns relative to the launch clock <S> (goes to first flip flop). <S> Most of these will involve modifying the representation of some gate or buffer to include a delay. <S> If you are using .libs for the characterization, you could just set the entire load/slew/delay table to a default value of 2ns.
| You can no longer do accurate timing analysis with RTL code.
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what do I call this relay? I want to build a circuit to switch a 1 1/2hp pump ON and OFF. ON to fill a large tank. OFF when the tank is full. I want to use a mechanical start (push the button) with a switch to cut the power. I see this as a mechanical latching relay application. I prefer to have a 24vac coil to match other devices in the system. I want latching so that the coil is not powered during the 3 hour fill time but only gets momentary power when the full float completes the coil circuit. After that, the coil and the pump lose power until a finger pushes button. The catalogs I have read suggest the mechanical portion of the relay is an accessory. How do I spec this relay? I know I need to have contacts for two hot legs to the pump and contacts to control the power to the step down transformer for the coil.I am a hobbyist. Thank you for any clues. <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Control circuit. <S> Your question is a little confused <S> but I think this meets your requirements. <S> XFMR1 is the mains to 24 V AC transformer. <S> RLY1 provides power through its NC (normally closed) contact to the rest of the circuit. <S> When the float switch contact closes RLY1 is energised and power to RLY2 is cut. <S> RLY2 runs the pump when it is energised. <S> Pressing START will cause RLY2 to be energised and run the pump. <S> If this isn't quite what your require then please edit your question to clarify and post a comment below to notify me. <A> Panasonic makes a line of latching relays with mechanical actuators (which can also be electrically controlled, of course). <S> http://www3.panasonic.biz/ac/e_download/control/relay/power/catalog/mech_eng_djh.pdf <S> A float switch on the tank triggers a one-shot which drives the reset coil; and your pushbutton triggers a one-shot which drives the set coil. <S> In either case, using latching relays ensures that the state survives power loss. <S> https://www.electronics-tutorials.ws/waveforms/monostable.html <A> Since you are switching mains power, Code requires that you must use commercial off the shelf mains control equipment, not random components from Mouser. <S> The obvious thing to me is to use exactly what you propose: a latching relay. <S> The General Electric RR7 comes to mind. <S> Anyplace you see light switches which are centered and you push them up or down to turn the lights on/off <S> , that's an RR7 or competitor. <S> As such, the RR7 is well rated for inductive loads such as motor loads. <S> It has two 24V coils, one for on, one for off. <S> You smack one momentarily and it throws and stays. <S> RiB also makes some relays which mount in the knockout of a junction box. <S> You could also use a common off-the-shelf air conditioning relay, about $12 and fits handily in a 4-11/16 square deep box with extension. <S> DPST, 30A throw and motor rated. <S> This has a 24VAC coil, and you would need to hold that coil on at all times, it won't latch. <S> You could do the latching in your control electronics, which can be low voltage, and therefore electronics-tier components from Mouser are allowed.
| Or, you could get a dual-coil latching relay and drive it with one-shots. The relays are called 24 V AC control relay. A second NO (normally open) contact bridges out the start button and keeps RLY2 energised until STOP is pressed or the float switch contacts close or the power fails. Make sure the contacts are rated for voltage and current at least as great as your load. It is made to control large banks of discharge lighting such as those found in the ceiling of any big-box warehouse store, typically HID lighting with transformer based ballasts with a wicked inductive kick.
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What are the manufacturing difficulties for a large Li-Ion battery pack? How is a large Li-Ion battery pack, such as in a car, manufactured? What makes it difficult to produce, if anything? This question is born from the high cost of Tesla Model S , which is around $70k. In contrast, the average 4-door car is more like $20 or $30k despite having an internal combustion engine. I don't understand that. I would have thought that a big battery, which has no moving parts, would be much much cheaper than a 4-cylinder engine. So what's the expense in making a large Li-Ion battery? Is lithium much more expensive than steel? Are there major (expensive) safety precautions in manufacture? Is the battery pack very intricately "woven" with cells? Something else? <Q> Like horta said, battery grade lithium (~$5.00 USD per pound) is quite expensive compared to steel (depends, it can be less than $1 USD per pound) or aluminum (also usually less than $1 USD per pound) used in an engine block. <S> Lithium likes to react with the atmosphere, forming lithia/lithium oxide. <S> There are quite a few regulations regarding the shipment of lithium. <S> Large Li-ion batteries or cells, especially in cars, can pose a safety risk when they are punctured or exposed to atmospheric oxygen. <S> Just google, "exploding li-ion battery". <S> Having smaller/individual cells like 18650 batteries could help minimize the amount of lithium that is exposed when the is battery damaged. <S> However, having a greater number of smaller cells can increase the cost of manufacturing as they could easily be more sophisticated (cooling, wiring, assembly of cells etc...) <S> In conclusion: Main costs of large Li-ion batteries: Cost of transport (special requirements) Cost of battery grade lithium Cost for manufacturing large numbers of small cells Cost for assembling small cells into a large battery <A> It is not a large single battery, it is build from a lot of small cells. <S> But you could not only connect a lot of those cells in series and lot of those serial connected strings in parallel, you have to add temperature control and charge control. <S> Fast charging produces a lot of heat, but the temperature of each cell has to be kept within limits. <S> Charging must be controlled to balance all cells and to stop charging before any single cell goes over the limit. <S> If one or some cells fail the large cell array should still operate with only a small reduction of capacity. <S> The complete battery should work in an automotive environment with low and high temperatures, vibration, fast acceleration and deceleration. <S> Even a crash should not cause a fire. <S> A protection against internal and external short circuits is necessary. <S> The battery should work reliable over many years and a lot of charge and discharge cycles. <S> The cruise range after some years should only be a little smaller than the range of the new car. <S> No owner of an electrical car wants to find his car burning at the charge station due to a battery failure. <A> Construction of the battery pack: https://teslamotorsclub.com/tmc/threads/pics-info-inside-the-battery-pack.34934/ Engines without fuel are entirely passive components. <S> Li-ion batteries are a fire hazard if shorted or punctured; see Samsung's recent troubles. <S> They start life with charge in them, so while assembling the 400V pack there is an electrocution hazard. <S> The design is based around considerations of cooling and crash safety. <S> Hence the use of lots of individual standard cells rather than giant cells. <S> They can be kept surrounded by coolant, and they're able to survive if the pack is deformed. <S> In order to protect the pack there's a titanium underbody shield, which is a very expensive material. <S> The raw battery materials are also quite expensive as mentioned in other answers: lithium and cobalt. <A> As Mike stated, it's made out of a lot of smaller 18650 cells. <S> Just the cost of the number of those alone accounts for a lot of the cost. <S> If you were to buy those batteries alone, they generally run 4-8 dollars for one battery. <S> Tesla gets the benefit of scale and gets them somewhat cheaper, but still, it's an exorbitant amount. <S> Tesla has created the best cooling/charging/discharging and safe battery packs on the market and so charge accordingly. <S> The Tesla's have the ability to pull power from the batteries better than any others on the market without the batteries spontaneously combusting. <S> Lastly and probably most importantly, lithium metal is far more expensive than steel. <S> A basic google search shows that lithium costs anywhere from $5000-$20000 per ton of lithium metal. <S> The price range for a ton of steel is $429-$569.
| Refined lithium or just lithium in general will likely require special handling, also increasing costs. If the large block of cells overheats they may be destroyed and even burn.
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Should I use this powerstrip with different voltage on it? pic of label: https://i.imgur.com/gKfhmZ3.jpg See how they marked off something there? The voltage i think. here in this country the normal voltage is 220. The powerboard is imported from germany, and they have it at 230 V. Would that difference affect anything? I bought this without no box, its how its sold. Its the only brand of powerboard sold here thats not generic. its cheaper than APC or alike, for some reason its cornered the market. It also has its plastic pealing off in places for some reason. The other alternative to this from same company is an aluminum one, i guess that one comes without the plastic pealing. pic of powerboard: https://i.imgur.com/npP4CeY.jpg <Q> Furthermore, if you plug it in to your wall, your wall socket will supply 220V and that is what will be delivered to your devices. <S> The powerboard will not change the voltage that is supplied to it. <A> No, it will not affect anything. <S> The voltage of the power strip is a recommended maximum voltage that it's designed for. <S> The power strip will work with any voltage up to that (and probably several tens of volts higher at least). <A> There is no practical difference between a mains power strip running at 220V vs. 230V (or even 240V) <S> The utility mains voltage is not all that precice.
| For normal electronics, 230V and 220V are the same thing (these are AC voltages anyhow and your devices likely convert them to DC using a regulator circuit). It would not be unusual to find mains voltage varying 5% or more in many parts of the world where those things are used.
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Which kind of regulator/battery setup should I use for 5 volt microcontroller and transmitter circuitry for optimum performance? I have a project which uses microcontrollers and a radio transmitter, both running at 5 V. The circuitry consumes less than 1 mA from the 5 V rail when idle, and 1 A when transmitting data using the radio transmitter. The transmission phases do not happen often, as the device spends most of the time in idle mode, usually transmitting only once every 24 hours for a few seconds (maximum 20 seconds). Now I'm designing a battery input for the circuitry and I cannot decide whether to use 11 V battery packs (LiPo) with switching regulators, or LiFe batteries instead where I can get a low regulating voltage difference between input and output for the regulator (output of 6 to 7 V) and use a linear regulator with those lower voltage batteries. Which of these setups gives more efficient results for regulation circuitry in this kind of use? switching regulator with 11 V input, dropping the voltage down to 5 V; or a linear regulator at (mostly) low current and maximum of 2 V voltage drop over it? One of the main interests in addition to the battery life, is the simplicity of most linear regulators versus switching regulators, which reduces time spent on designing the product and saves space from the board. Switching regulators can also cause some radiation that might interfere with other circuitry and they are also more expensive. <Q> This will mean that in idle mode only the low power regulator's current consumption is discharging the battery. <S> Of course this means an extra IO line to "enable" the higher power regulator AND some shortish period of "waiting" whilst the radio becomes "ready" for the data transmission from the microcontroller. <S> The problem with NOT enabling/disabling the higher power regulator (feeding the radio) is that its quiescent current consumption might be hundreds of micro amps or even low milli amps and this will certainly deplete the battery. <A> Interesting question because I can see the answer going either way, depending on environmental circumstances. <S> Some aspects of the design which may not be immediately obvious... <S> 1) <S> Switchers are notoriously poor at a tiny fraction of their load - they may consume several mA internally, or they may lose regulation and deliver 7V below some value, say 1% of rated load (10mA in your case) without special care in design. <S> 2) <S> One answer could be a linear regulator during sleep, (even from 11V <S> but there's nothing wrong with a 2S Li-Ion - nominally 7.4V max 8.4V) and the MPU has to wake up a switcher before transmitting. <S> If the linear regulator only supplies a few mA, you can probably find a SOT-23 to do the job, or SOIC-8 at the largest, so I don't believe size is the issue 3) <S> A linear regulator for 1A will need some heatsinking even for 20 seconds ... <S> if there's a convenient chunk of metal, use it. <S> Linear may be more reliable from its simplicity. <S> But what happens if the TX gets stuck "on"? <S> Running the battery flat is one thing, destroying the equipment is another... . <S> 4) I would not, personally, change battery technology simply as a way of tuning supply voltages. <S> If you need lower fire hazard, or greater charge/discharge cycles, or some characteristic of <S> LiFePO4 that's a reason for using them - otherwise stick with commodity batteries for economics and simpler servicing.. <A> Look at the datasheets. <S> This really should have been obvious. <S> The datasheets of linear regulators will tell you the quiescent current. <S> The datasheets of buck regulators will tell you the quiescent current, and give some guide to likely efficiency. <S> From these, you can figure out the overall efficiency and average power draw from the battery. <S> You also need to do some basic math. <S> One obvious thing to determine is whether your biggest problem is the occasional but high power RF transmission, or the constant but low power idle current. <S> There are 86,400 seconds in 24 hours. <S> (1 A)(20 s)/(86,400 s) <S> = <S> 230 <S> µA. <S> That's the average current draw due to the radio transmissions. <S> This means the 1 mA idle current dominates the total by over a factor of 4. <S> There is no substitute for looking at a few plausible alternatives and doing the math to see which one is more optimal. <S> However, my hunch is a buck switcher with good idle characteristics. <S> This would be something that has PWM/PFM switchover capability. <S> Put another way, it doesn't just change the lengths of switching pulses at a fixed frequency, but at low power lengthens the time between pulses too. <A> With 7V input and 5V output <S> the linear regulator will have to drop 2V, thus achieve an efficiency of about 62%, no matter the load. <S> Switching regulators have very good efficiency (>90%) at high loads, but will be very bad at lower loads. <S> You’d have to look up the datasheets for real numbers but at a fraction of their maximum load they can easily drop below 50%. <S> Their quiescent currents are often also quite high.
| There are some good switching regulators out there, you might find one which fits your requirements perfectly. I would estimate that your most efficient solution is to use a low power regulator for your microcontoller and let the microcontroller activate a buck regulator to feed power to the radio transmitter.
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Better options for induction coils than water-cooled copper In an induction heater one of the main problems is that the coil carries a large current. This causes the coil to become hot. There is a limit to how large diameter the wire can be, because it needs to have multiple turns (I think). The standard solution to the problem is to use hollow tubing, then run water through the tubing, thereby cooling it. Is there a better option for making the coil? The overall goal is to make a coil that can transmit large amounts of current without getting hot. One obvious improvement I can think of is to make the coil out of silver . Silver is both more electrically conductive than copper and also more thermodynamically conductive, so it will be able to transmit electricity better and be less susceptible to heating up. Is there an even better way, maybe superconductors? <Q> Commercial Induction Heating Equipment is all about the economics of the equipment and the process of heating. <S> There are commercial units using super conductors Seen in this post and in <S> This Post <S> Fabricating exotic induction coils for a one time sale might not make sense. <S> Induction Heating metals of low conductivity (aluminum and copper) would be candidates for using super conduction coils. <S> Induction heating of copper components using copper coils results in efficiency less than 50%. <S> Heating ferrous objects (below the curie temperature) is probably not a good candidate for exotic induction coils since the efficiency could be quite high (85% to 90% efficient). <S> However in multi-megawatt systems running continuously could save lots of power and $. <S> Flux concentrators , see Fluxtrol also help with efficiency of the heating process. <S> (not an endorsment for fluxtrol). <S> Flux Concentrator taken from Induction Presentation <S> The wall thickness of the copper tubing can be made thinner, so as to increase cooling. <S> But, decreasing wall thickness begins to harm the efficiency (skin depth limitation). <A> One obvious improvement I can think of is to make the coil out of silver. <S> First of all conductivity tables: - Silver conducts about 5% better than copper so not much of an improvement here. <S> The better option is Litz wire because this counteracts something called skin effect. <S> Skin effect happens when AC current passes down a conductor - magnetic fields induced in the conductor tend to force current to flow at the periphery of the wire and so individually insulated multi-strand wire improves things. <S> Skin effect: <S> - As you can see, most of the AC current flows at the periphery - this is why copper pipes work so reasonably well <S> - the centre of the copper would hardly conduct anything so it can be filled with water for cooling purposes. <S> Here's a table of skin depth versus material type versus operating frequency: - This table is the backbone for designing Litz wire - depending on operating frequency, the individual strand diameters can be determined i.e. at 10 kHz <S> there is little point in choosing a copper conductor that has a radius bigger than ~0.6 <S> mm. <S> Here's a good example of how increasing the number of strands is a benefit: <S> - So for the higher graph the construction is 16 strands of 0.2mm diameter. <S> This is a cross section of copper of 16 x pi x 0.2 x 0.2 / 4 = 0.502 <S> sq mm. <S> For the next lower cable the cross sectional are of copper is the same at 25 x pi x 0.16 <S> x 0.16 <S> / 4 = 0.502 <S> sq mm and the same all the way down to the wire made of 255 strands of 0.05 mm diameter wire (0.501 sq mm). <A> The heat in the coils comes from the radiant heat off the work object not from any resistance is the copper coils themselves. <S> Water run thru the coil is the superior way for all but larger industrial units. <S> Water via a pump that cycles thru a radiator with a electric fan attached. <S> Balance the radiator size volume to the work size and power of the unit to get a 100% duty cycle. <S> No need to get any fancier than that.
| Also, many uses of Induction Heaters are one of a kind for a specific product. The efficiency gain of using silver rather than copper would likely still require water cooling. In conclusion, Litz wire is probably the best way to go but by how much depends on the operating frequency.
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How can I make connection on pcb (circuit board) holes without solder (for prototyping)? I have a Pro Micro board (very small -- see pic below) and it has connector holes in the printed circuit board. Best Non-Solder Connection (Easy and Removable) I'm wondering the best way to make non-solder connections for doing my prototype work. By best, I mean easily removable while maintaining solid electrical connection. Are there pre-made connectors for this type of connection? What Part Is Most Conductive / Best Electrical Connection? Also, how can I know I have a good connection? Is it the inside of the hole that has the conductive metal, outside of hole? Front / Back Separate Traces? And, are the front and back of the holes electrically connected normally or do they have separate traces normally? One Idea: Would It work Very Well? What if I had header pins pushed into a breadboard and then up through those holes? Then I connect my wire to the top of the header pin? Would it be a solid enough connection? Or would it not make enough electrical connection? ie - would header pins make connection inside of holes and would that be enough electrical connection? Edit -- I Wish They Made Banana Plugs That Fit This Wouldn't it be cool if you could use a banana plug type of connector. Then just plug in each one and put wire in hole and clamp it down? UPDATE 11-11-2017 Interesting that in the time since I've posted this someone came up with a solution similar to what I was thinking with banana plugs: Hammer Header Male - Solderless Raspberry Pi Connector It's really for use on a RPi Zero but it's the type of snap-on header I was interested in at the time. However, the installation is not easy so it may not be practical.Take a look at what you have to do to install it : https://learn.pimoroni.com/tutorial/sandyj/fitting-hammer-headers Probably easier to just solder on the header pins. <Q> Solder a pin header <S> Typically you solder pin headers to these boards. <S> Either male or female. <S> With male pins, you can solder them with the pins pointing down , so you can put them in solderless breadboards. <S> 1 <S> 1. <S> This may ruin the breadboard strips <A> Use grabber test clips, which are basically like smaller alligator type clips. <S> I have some really mini ones. <S> Yes, the pads should be plated on the inside as well as the top and bottom, electrically connecting them. <S> No, just pushing a standard 0.1" header into the hole won't work. <S> They are not offset holes, or push fit tight. <S> They are generously loose. <S> Some people have used rubber bands but that's not very secure. <A> For me I would rather use these PCB terminal blocks, which you can easily remove and connect your wiring without worrying about loose connections every time. <A> These jumpers are available in m-m, f-f, and m-f configurations. <S> You could get the connectors and make them yourself, but that requires a special crimping tool. <A> A search turned up an option called "press fit" pins and headers. <S> Here is a picture of one from the whitepaper <S> my Google search turned up: <S> The problem with this is you still need to connect to that header somehow. <S> You might as well solder a single-row pin header in and use jumper leads as in other answers, though. <S> It really is the best way. <A> Need: Normal size paperclip and one standard jumper (plenty on your old hard drives)1. <S> Crimp paperclip with vise grip pliers, and adjust if needed to just fit snugly into the hole in the PCB.2. <S> Cut paperclip to length and insert long end into jumper.3. <S> Insert crimped end of paperclip into hole in PCB.4. <S> Use other side of jumper for standard male prototyping cable connectors.5. <S> Leave in place, which has the advantage of not constantly inserting/removing a connector and thus wearing out the hole. <A> What Part Is Most Conductive / Best Electrical Connection? <S> Usually you solder to the circular 'pad'. <S> Front / Back Separate Traces? <S> No, normally they're plated through. <A> If your module doesn't have headers you can have someone add them (when you're purchasing them) or you can add them yourself. <S> This goes along with what @pipe suggested above but rather than placing into a solder-less breadboard one would make connections like you can see below. <S> There is not an easy way to wrap the wire into the holes themselves. <S> Possible but prone to user error since you'd have to do it yourself with no tools. <S> This does require you to purchase wire-wrap wire, a decent stripping tool, and a wire wrapping tool. <S> If you look on eBay they're dirt cheap compared to what you can get them for new from Digikey (or the like) <S> The connections are robust and I've made some serious product prototypes before ever spinning a board. <S> I wrote an article and made a video on wire wrap for prototyping on Circuit Dojo . <S> Feel free to check it out.
| Normally you solder a male header to the board, and if you aren't plugging them directly into another board, you connect them with either ribbon cables or jumpers that have individual sockets on the end. Though regular alligator/crocodile clips may work, depending on the size or how many side by side ones you need.
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What is the best sensor to use in a wearable device to monitor body temperature on a continuous basis: Thermistor, RTD or a semiconductor device? Most wrist-worn wearables on the market today don’t seem to feature body temperature sensing which makes me wonder if all these sensors are inadequate in some ways. <Q> In general, thermistors are the sensor of choice for measuring body temperature. <S> They are sensitive, cheap, and their accuracy is adequate. <S> After all, body temperature does not need to be measured to better than about 0.1 degree <S> F. <S> With that said, measuring body temperature at the wrist is a losing proposition. <S> There are simply too many chances for wrong temperature readings. <S> The wrist, after all, is out at the end of the arms, with lots of opportunity for heat loss, and who knows what your hands have been holding on to. <A> The problem of monitoring body temperature in a wearable device is were to place the sensor. <S> The skin of arms and legs are known for delivering bad results. <S> The temperature there is some kelvin lower than the body core temperature. <S> The difference is not constant, if you feeel cold, the body reduces the blood flow thru arms and legs in order to keep the body core temperature. <S> The only place for good results is the ear canal. <A> While I’m no expert in bio-monitoring, there are some basic things that leap to mind that may help you decide: For continuous monitoring to be useful as a diagnostic tool, the sensor must be accurate, fast (therefore: low thermal mass) and low power because you want as small a battery as possible; <S> To minimize the overhead of corrective circuitry it should have as linear an output as possible over the measurable temperature range (which is actually not that large for body temperatures); Since conduction, rather than radiation or convection, would be the most efficient heat transfer mechanism in this use case <S> , you’d have to have close and consistent contact with the skin. <S> I’m not sure that’s possible with a loose fitting bracelet or watch. <S> All of the above leads me to think in terms of an analog semiconductor sensor (over a thermistor or RTD) as part of an adhesive body patch. <S> The packaging may be the most important aspect of the design after the choice of the semiconductor sensor. <S> I did a little bit of research and found a great example of a bio-patch kit with which you could experiment: https://goo.gl/vWrVJq
| Another way to look at it is that all available sensors are inadequate - they measure the actual local temperature rather than the body core temperature.
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Audio playback design issues with WM8731 I am using WM8731 Audio Codec taking hints from the Audio Board PROTO with a 1 W Audio Power Amplifier LM4889 to drive a speaker from CUI INC: CDS-13138-SMT . The setup works but not as expected. I am using the recommended circuitry given in LM4889 Datasheet for high gain amplifier. My application requires a speaker to output frequencies in 17 to 20 KHz. But when I pass a wave audio of pure 19 KHz(from here ) to it, the speaker outputs a frequency at around 9KHz and 15 KHz.(Why?) The playback is slow too.(I tried playing a song. It ran in slow motion)(can't understand Why?) (I am using Spectrum Analyzer . It shows the two peaks at 9 and 15). The whole setup is interfaced with Raspberry Pi through I2S. I don't have much experience with audio applications. Is my approach wrong? Am I using a wrong speaker for my application? Can anyone provide any alternative? What all can be improved in the design? I am sorry if I come across as a complete noob. Thank you. <Q> Or maybe a circuit design problem where you are hot clocking audio data into the D <S> /A converter at the proper rate. <S> The other problem you describe is most likely the result of "aliasing" which is a common side-effect of analog to digital and digital to analog conversion. <S> The analog audio must be properly bandwidth filtered to eliminate false alias signals. <A> Well, if the input frequency is 19 kHz and you sample it at too low a frequency like 27 kHz you will output an aliased signal at 9 kHz: - Your sampling rate needs to be greater than twice the highest frequency you wish to reproduce. <S> This could also explain why when you play a song it plays back too slow. <S> If the sampling is at 34 kHz <S> then you could also produce 15 kHz due to aliasing. <S> I suspect you have your output sampling rate set too low. <S> Also, as another idea, if your source audio is digitized at a different sampling rate compared to your playback rate then you will get similar problems. <A> This sounds like a software issue of the I2S bus on the Raspberry PI. <S> In these cases I would be very tempted to break out the scope to probe the BCLK and DACLRC lines of the DAC board. <S> The LRCLK signal should tell you the sampling frequency of the DAC. <S> Also make sure the BCLK is correct. <S> You can verify that by looking at the bit depth of the DAC and the sample frequency. <S> The product of these 2 should give you the BCLK frequency. <S> It seems like the WM8731 is pretty flexible in the audio format it accepts. <S> You should be able to send it 48kHz 16 or 24-bit I2S data without any issues. <S> Make sure your RPi software is configured to output those exact settings to the DAC. <S> Maybe the software is still configured to 44100Hz. <S> I am not sure if this DAC supports that sample frequency (because it does not fit nicely between 8, 16, 24, 48, 96 kHz)
| The playback speed issue is most likely a software problem. It could be the audio sounds like lower/higher pitch or distorted, because an incorrect sample frequency is used.
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Why do these two single supply noninverting amplifers not work but the third does? Why don't Circuit 1 and 2 work? Circuit 1: I don't get gain of 4 but of about 1 like buffer amplifier. Circuit 2, I just connected R2 to ground and the Vcc/2 generated using R4 and R5 to noninverting input of opamp: More mysterious; here the opamp is stuck at vcc. Circuit 3, similar to circuit 2 but now there is a capacitor between R2 and system ground: This one works. EDIT: OK, thanks for the answers. I want to reach a conclusion now. I conclude that the following circuit from SLOA058 page 6 is wrong: Since I use a voltage divider to created Vcc/2, this circuit simply will not work. Do you agree? It seems that this application note has wasted a lot of my time. <Q> Circuit 1 doesn't work mostly because your 100k pair voltage divider is not appropriate to set a bias point for the inverting (-) feedback network. <S> The currents flowing through the 4k & 1k will easily pull that so-called voltage divider waaay off where you thought it would be <S> and this is what gives you a gain of approximately 1.1 Also, <S> your input to the non-inverting (+) input of your opamp has no DC bias set - it depends entirely on the opamp's biasing/leakage currents. <S> Circuit 2 doesn't work because you're applying a DC bias to your non-inverting input (good), but then amplifying it by the gain of your opamp circuit (bad). <S> So your opamp is trying to drive its output to 25V, but it only has a 10V supply. <S> This circuit has an AC gain of 5, but a DC gain of only 1, so your input signal appears on the output amplified by 5 and riding on a 5V DC level. <A> First circuit: <S> There is no DC bias path for the (+) input. <S> Second circuit: <S> The (+) input is biased to +5v, the feedback is amplifying with a gain of 5 from ground. <S> 5 <S> * 5 = 25, which is greater than 10, so the output is pegged to the rail. <S> Measure DC before worrying about any any signals! <S> See also Single Supply Audio Amplifier <S> Edit: <S> This figure 3 circuit is wrong because there is no DC bias to the (+) input, not because of anything with the voltage divider. <A> I'm not sure what the problem with Circuit 2 is, but the problem with Circuit 1 is clear. <S> Circuit 1 doesn't work because <S> the dual 100 <S> kΩ divider you're using to try and bias the op-amp's +IN to ½Vcc is also half in series with the ground leg of the feedback divider, so that the gain isn't calculated as 4 kΩ <S> ÷ 1 kΩ + <S> 1 = 5 (not 4!) <S> but as 4 kΩ ÷ <S> (1 kΩ + 100 kΩ) + 1 ≈ 1. <S> The virtual ground biasing trick you're attempting to use only works with the inverting op-amp configuration.
| Circuit 3 works because you've separated the DC gain from the AC gain by adding that capacitor.
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Alternative for arduino for simple requirements I want to capture new email in inbox. I want something very tiny to accomplish this task. Arduino is quite big in size. Is anything available out there which I don't know by name. Please help. <Q> There are other options that you can try opting to in Arduino like their Nano, micro or Fio boards. <S> They are quite more comfortable as their footprints are even provided in the link, you can actually design what kind of components you want in the board. <S> Maybe you can try them, they are not as small as the lilypad <S> but they can provide a better function as the fio board comes along with the wifi compatibility in the circuit. <S> https://www.arduino.cc/en/Main/ArduinoBoardFio <A> You could use Arduino Lilypad, mainly used for wearable electronics because of its small size and for creating sewable electronic pieces. <S> Check out the existing projects done to get a good idea of its processing capabilities. <S> It has 9 I/ <S> O pins and an ATMEGA32u4 at its core. <A> I'm not quite sure of your requirements, so I'll just answer based on my understanding. <S> There's tons of examples out there <S> so I'll jut link a few for you. <S> Arduino UNO on breadboard - Arduino <S> webstie <S> Arduino UNO on breadboard - Instructables Mini Arduino <S> using ATTINY85 <S> (Some reading required) <S> If you've already got an arduino lying around to program the chips then you can make the circuits pretty small and also teach yourself some useful skills in the process. <S> Less buying, more making I say. <S> Edit: After some clarification in the comments and it being email as opposed to physical mail you will either need an arduino with an Ethernet/Wireless shield. <S> Or you can buy premade tiny Ethernet/Wireless controller as you will need network connection to use POP3 protocol
| If you want a smaller version of an arduino then you can build an equivalent circuit on breadboard/protoboard.
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Is there an instrument to measure/uniquely identify gadget's RF signal strength I'm not a technical person. Apologies for not being able to express myself clearly. I have a base station(blackbox audience response system bought from manufacturer) used during events that transmits 2.4GHZ RF signals across the halls. (E.g Huge Convention Halls) I want to measure the signal strength from this gadget at different parts of the hall where it is placed in order to see which area has weak signal. However, in public places, there may be other RF signals from unknown devices. Are there instruments that I can use to measure/uniquely identify the power/signal strength of my base station? Thanks in advance! <Q> You have already identified that the crucial part of this instrument is to only measure the signal strength of the voting system base station. <S> If that unit used a standard protocol such as WiFi there are a number of instruments available that can do exactly that. <S> Even an Android phone can run a WiFi Analyser app to give you an uncalibrated measurement of signal strength. <S> But it sounds like your voting system uses a non-standard protocol. <S> the modulation scheme the format of the data packets so that you can decode the packet and identify that it is from the voting system base station. <S> The only people who will know all of those things are the manufacturers of the voting system. <S> It is possible that they already have a signal strength meter for use with their system. <S> I would contact them. <A> An RF power meter can be used to measure the power level of a specific RF signal. <S> That signal will be the transmit signal of the RF channel. <S> If multiple RF devices use the same channel/s (like wifi), then you may need a different device to decode the RF signal identifier for each device. <A> Normally you use a field strength meter to measure continuous or near continuous transmissions. <S> Probably doesn't suit your scenario as you need to uniquely identify the correct transmission, and it may be a very short transmission. <S> It might also not be transmitted on a single frequency. <S> A radio receiver can serve as a field strength meter. <S> Some (single frequency) receiver chips have a received signal strength indicator (RSSI) <S> A spectrum analyser is the ultimate tool to, well, to analyse a spectrum of frequencies. <S> Unfortunately it scans from one frequency to another and can miss very short transmissions. <S> The handsets have very short transmissions to prolong battery life. <S> The "base station" may or not transmit at all. <S> So, we don't really have enough information to give a fool proof answer. <S> You could get a qualitative answer by reducing transmitter power. <S> Say by half, and wander about the room to see if it still works everywhere. <S> You can reduce the signal by half by altering the antenna. <S> Or cutting the feed to the antenna and inserting 3 resistors (Pi attenuator). <S> Or maybe software configuration somehow. <S> Ask a radio ham for help. <S> RegardsIan Mccrum __
| To be able to measure its signal strength you need to know: the frequencies or channels in use
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What is the exact purpose of this 2.2kOhm resistor in the circuit? My best guess is that it's there to drop the voltage down to 0V by the time the current reaches the negative terminal but I'm not sure if that's the case; and if it is, is it absolutely necessary to be there and why exactly? Also a side question: Why is the voltage dropped to 5V in the beginning of the circuit, wouldn't it work if it remained 9V? <Q> What is the exact purpose of this 2.2 kOhm resistor in the circuit? <S> [...] is it absolutely necessary to be there <S> and why exactly? <S> Its purpose is to form a voltage divider with LDR (photoresistor). <S> The resistance of LDR varies with the light intensity. <S> The resistor turns the variable resistance into a variable voltage , which is then compared to a reference voltage that you can control with the potentiometer. <S> Yes, it's necessary in this circuit, and most other circuits using an LDR. <S> Why is the voltage dropped to 5V in the beginning of the circuit, wouldn't it work if it remained 9V? <S> You're right here. <S> Due to the circuit's construction of only relying on the relative resistances, the circuit would work fine (even better) without the regulation. <S> You would only have to change the resistor after the LED, because the output from the operational amplifier will be closer to 9 volts than 5 volts. <S> One minor benefit of having a 5 volt regulation is that you will only get at most 5 volts out of the operational amplifier. <S> This could be useful if you want to connect it straight to a digital input. <S> Note that the μA741 amplifier chosen here is a pretty bad choice, but why it is so is another question . <A> This question wasn't asked but I thought it might be useful information for you. <S> Why is it a 2.2K resistor? <S> Can I just use any resistor? <S> I've plotted a little graph so you can see the effect of using different fixed resistors for the LDR in your circuit schematic <A> The sensor is a LDR, which changes resistance when exposed to light. <S> The circuit is a operational amplifier, used as a comparator, that will turn on an LED when a certain level of light is reached. <S> Comparators respond to voltage, not resistance, so somehow, the resistance value of the LDR must be converted to a voltage. <S> There are a few ways to do this. <S> One, which is not real common, is to use a constant current source, and then the voltage across the LDR would be proportional to it's resistance. <S> Another way to is put the LDR into a voltage divider. <S> The 2.2 KOhms forms the lower half of the voltage divider.
| It's all a matter of sensitivity, the LDR will be the most sensitive to light changes in both light and dark if your fixed resistor is a similar resistance to the midway point of the LDR.
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Electrical specification for minimum bias current available from 3.5mm cellphone microphone input? If I understand correctly modern microphones integrated with headphones with 4-contact 3.5mm jacks for use with cell phones are condenser microphones, and usually contain a small, integrated amplifier - consisting of at least one FET and a few passives for bias isolation. These are very low current, but require a small DC bias available from the microphone input to operate, and usually a resistor divider there requires and draws at least a small current. Question: Is there a standard, or generally and widely accepted minimum current that a designer can expect any modern phone to make available at the 3.5mm 4-contact plug - or voltage plus maximum internal series resistance (within the phone)? I'm guessing it's going to be somewhere in the ballpark of 1 to 100 micro-amperes and between 1 and 2 volts, but I don't actually know this for sure. Here is just a random example, the TPA6166A2 3.5-mm Jack Detect and Headset Interface IC was the first I found in a search, and it seems to be rated to supply a maximum of 1.2 mA when it detects a need for it and the internal bias resister is set to bypass. However it is no evidence by itself of a minimum standard . There are plenty of examples, and other questions within this stackexchange, but so far I haven't found the actual published, or industry accepted minimum current, or minimum voltage plus maximum internal(phone) series resistance. Hypothetical use example: This is not what I'm doing but it's a good working example for the question. "pico-LEDs" - small surface mount LEDs attain useable brightness at only a few milliamperes. A power harvesting-like circuit could collect enough charge from say 50 microamperes over one minute to "blink" the LED for 100 milliseconds. above: screenshot from an example picoLED-eco surface mount LED. Comments on personal experience are welcome, but for an answer I need a link to a real, relatively reliable specification from which design something with the reasonable expectation that it should work with any modern cellphone with the 3.5mm 4-contact plug and draw a small but useful amount of power from it (microwatts). <Q> (Not enough reputation to comment on @Kevin Reid's answer <S> so I have to do so in another <S> answer.)The Android docs are a useful source of info regarding this question. <S> I interpret them differently though: I think the 1.8 V minimum bias voltage in the spec is an open-circuit value. <S> It's hard to be certain of course, because no output impedance for this voltage is stated. <S> However, the text also talks of a "2.2V mic bias applied through 2.2 kOhm resistor" which one might infer to be the reference method of providing the bias supply, and would allow for no more than 1 mA to flow (into a dead short). <S> Given that mics which need bias tend to draw tiny currents, this would make sense. <S> Finally, the TI chip referenced in the question defaults to 2.0 V for the mic bias voltage and 2.2 kOhm for the mic bias resistance. <S> This would give just over 0.9 mA into a dead short. <A> I doubt mobile phone companies use an outside standard if they can use an improved mic. <S> They would have internal standards which are proprietary. <S> Then they have to choose compatible external devices. <S> To look at the most sensitive devices, consider new MEMs microphones Impedance <S> 4.5 kOhm <S> Voltage - Rated - Voltage Range 0.9 <S> ~ 1.3V Current - Supply 17µA <S> This might be your worst case for the near future. <A> It's not <S> a <S> universally applicable standard, but Android publishes specifications for headset <S> plug and jack . <S> These documents (as of June 14, 2017) specify that the jack's mic bias voltage shall be in the range “1.8V - 2.9V”, and that the microphone shall have a DC resistance of “1000 ohms or higher”. <S> \frac <S> VR = \frac{1.8\,\mathrm{V}}{1\,\mathrm{k\Omega}} = 1.8\,\mathrm{mA}$$ and at most 2.9 mA. Note that to follow this specification <S> you must design your device's power supply so that it presents an input impedance of at least 1 kΩ. <S> In particular, if it is less than 100 Ω then it may be considered a short and detected as a 3-pole headphone plug with no mic.
| While it is not stated outright, from assuming that the two devices are compatible, we can thus conclude that if presented with a load of 1 kΩ, the jack must apply at least 1.8 V across it, resulting in a current of at least $$I =
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Choosing transistor for this circuit I have a problem: I have a circuit with a MSP430 MCU.I have a sensor which is supposed to be powered in 3.3V. My main power rail in my circuit is 3.3V. I need to power on and power off this sensor using a GPIO to pilot the power, so I suppose I need to use a transistor BJT being controlled by GPIO to let go the current in on/off mode into the sensor. Problem is power rail value = sensor value and my transistor got VCEsat = 0.7V (or I can also change it to -0.7V) So how do I manage with the Resistor value (R208) ? <Q> simulate this circuit – Schematic created using CircuitLab <S> The sensor will always be connected to ground on one end. <S> The power supply can be enabled or disabled by the MCU. <S> this is just the basic FET based switch. <S> Advantage here is that, the sensor can always have ground reference. <S> FETs can be from TI . <S> the threshold voltage is well with in 2 V to turn on the FET. <S> Correct me if i am wrong. <A> Get rid of R208 and use a low on-resistance N channel MOSFET instead of the BC817. <S> Make sure the MOSFET is turned on enough with a 3V3 IO line - look at Alpha and Omega - they have some low gate threshold devices. <A> I second Andy's answer. <S> Here is how the LEDs on my CC3200 Launchpad are connected. <S> You could use a similar arrangement and replace the LEDs with your sensor. <S> VCC_BRD is 3.3V and the GPIO lines are active high at 3.3V. <A> With a collector current of 100 mA, the BC807 would have a typical saturation voltage of about 50 mV: <S> However, even if you can live with this 50 mV, this would require a base current of about 10 mA, which is too much for your MSP430FR2311. <S> This requires that Q1 is a FET with an R DS(on) <S> value of much less than 500 mΩ; some widely-used models would be DMP2066, IRLML6401, or FDN338P.
| A P-channel MOSFET can give you a smaller voltage drop:
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Balancing copper on inner layers of PCB A similar question was asked - What is Copper Thieving and why use it? And my question originates from the comments of the answer in the linked question. I have read and understand that to help the etching process, you balance the copper so that it effectively eats away at the board at the same rate. One of the answers/comments makes a note that its for the outer layers. Thieving is added to the outer layers in order to help a more balanced chemical process for the plating. What about the inner layers ? I have a 8 layer stackup where some of the internal layers routing is not as dense as others. This is a relatively large board (15" x 8" ). My gut/understanding tells me, that I should be balancing the copper on all layers and not just the outer layers, but I would like to be sure its needed and that my understanding is correct. <Q> When a multilayer board is produced, the copper structures of inner layers and outer layers are manufactured in a different way. <S> If an inner layer should be 35 µm thick, the take a thin board with 35 µm copper on both sides, add fotoresist, expose thru a film with the structure, develop the resist and etch all unwanted copper away. <S> The outer layer are done different, because the drill wholes should be plated with copper. <S> They use a copper foil with only 17 µm for the outer layers and add another 17 µm of copper by galvanic deposition. <S> But all areas where you want no copper are covered with resist. <S> The galvanic deposition of copper is done only at uncovered traces and copper planes and inside the drill holes. <S> But you want all traces to get an equal deposit of copper. <S> If there is a single trace in an area of the board only, all copper from the galvanic anodes goes there and this trace get much more copper than only 17 µm. <S> At other areas of the board there may be a very dense population of traces. <S> The copper from the anodes inside the galvanic bath spreads to all traces there and the deposition of copper will be less than 17 µm. <S> If the outer layers are structured with a good balance, you will get an equal deposition of copper to the whole area. <S> But you should ask the board manufacturer about his recommendations for inner and outer layers. <A> While the chemical process is important the primary issue with unbalanced boards are the thermal properties during production of the PCB itself. <S> The PCB manufacturing process involves baking the boards to cure the epoxy between layers. <S> Copper expands more with temperature than the underlying board material (typically FR4). <S> This means that with an unbalanced board the PCB can warp during production. <S> I've seen some wonderfully curved pcbs that weren't flat enough to mount BGA parts to due to an unbalanced layer stackup. <S> Since inner layers are by definition closer to the centre of the PCB they will exert less torque on the board and so are less critical <S> but they can still matter. <S> Generally as long as your power planes are symmetrical you will be close enough to balanced that things will be OK. <S> If some of your tracking layers are very bare in comparison to others then you can always consider moving some of the tracks to a different layer. <S> A good quality fab house will normally be able to check and warn you if they think there will be a problem with your board. <A> For sparse routing inner layers you can leave as is. <S> There should be no need to make a big deal about those because those layers are typically just etched and then bonded to the adjacent layers without plating. <S> On the other hand for PWR and GND planes you want to try to fill those out to full fill. <S> Even on split PWR planes where you may be creating islands for several different voltages you want to fill out the whole board with copper (except for the spaces between islands). <S> There are a number of reasons for this including: <S> It is good to match the GND plane cover fully with PWR plane cover. <S> There are potential thermal expansion issues where boards with unbalanced PWR/GND plane copper coverage can lead to board becoming warped. <S> The one main exception I can think above for the above is that on designs where there are high current switcher supplies the 'switching node' between the switch FET, inductor and diode (or 2nd FET for a synchronous switcher) want to have a component side island to interconnect the node. <S> This node should have a copper cutout in the PWR/GND layers <S> so the <S> it is not able couple noise into the power and ground.
| Balancing the copper is recommendable only for the outer layers of the multilayer if the inner layers are structured by pure etching.
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Can I use ST-Link programmer for non-ST chips? Can I program all kinds of SWD-programmable chips (ARM-MCUs) with ST-Link? <Q> You could also flash the st-link and convert it to a Black Magic Probe . <S> The same image will also convert a Blue Pill into a BMP. <S> I've done both. <S> The Blue Pill has the advantage that the usb-rs232 bridge the BMP exports is easily available. <S> The BMP supports a range of chips to include but not limited to: ST Microelectronics STM32F0, STM32F1, STM32F3, STM32F2, STM32F4, STM32F7, STM32L0, STM32L1, STM32L4 Atmel SAM3N, SAM3X, SAM3S, SAM3U, SAM4S, SAM4L, SAM D20, D21 Nordic nRF51, nRF52 <S> (These are why I use the BMP) <S> The BMP is open source, can be used for commercial programming and the hardware can be cheap(The <S> "real" BMP costs around $60, <S> a blue pill 5 pack from amazon was less than $20 shipped). <A> In particular, if your target is from the Atmel SAMD21 family of MCUs, you are likely to run into weird issues, where the processor is detected, but any attempts to erase flash sectors fail with an error. <S> The reason is rather involved , to quote: <S> AFAIK the problem is in half word (16-bit) write to NVMCTRL->CTRLA register. <S> STLink does not implement half word memory access and OpenOCD <S> emulates is as two byte operations. <S> Unfortunately CTRLA register comprises from key and command an have to be written atomically. <S> If STLink writes two bytes, NVM controller sets PROGE bit in STATUS: " <S> An invalid command and/or a bad keyword was/were written in the NVM Command register" <S> The workaround involves recompiling OpenOCD with code patches. <S> Not fun. <S> However, if you don't mind flashing your ST-Link (through another ST-Link), you can convert it to a CMSIS-DAP adapter , which works just fine with the SAMD and should also work with STM32 and other Cortex-M chips. <A> Yes I believe this is possible, although I haven't tried it. <S> You would likely be in breach of the ST Licence agreement, if you tried programming devices other than those from ST <S> As an example <S> Segger (One of the market leaders in debuggers and programmers) supports converting your ST-Link to J-link (essentially overwriting the ST-Link chip with the Segger code). <S> This is also reversible, so if you want to restore your ST-link device back to its original form you can. <S> There are several version of ST-Link, however. <S> Take a look at the following links which provide further info and a guide on how to do this. <S> https://www.segger.com/jlink-st-link.html <S> https://www.segger.com/jlink-ob.html <A> Yes, its possible. <S> i am using stlink v2 (original, as well as Chinese usb stick type clones) with Coocox IDE. <S> Coocox has also standalone programmer app - CoFlash , Which could be used to program a lot of chips. <S> I have experience with only two series: LPC176x and LPC175x, over SWD interface. <S> And i haven't any issues with those chips. <S> I also haven't made any specific changes or updates for the programmers (except that original st link required the high voltage level at TVCC (PIN1), otherwise, it didnt allow to program chip, so i connected directly pin 19 (3.3vcc) to pin 1 on stlink v2). <A> An STLink V2/V2.1 or V3 with recent firmware can also be used with the PC-Hosted Blackmagic pc-stlinkv2 platform. <S> Compile with "make PROBE_HOST= <S> pc-stlinkv2", start the debug server with "blackmagic_stlinkv2" and connect to :2000 in gdb or directly load binaries with "blackmagic_stlinkv2 ". <S> This works with all targets that BMP knows. <S> However V3 activly denies working on non-ST targets, while V2 does not care. <A> Sure, any MCU can be programmed with STlink/V2, I use it for NXP, Silabs and for STM32, well.. for all my cortex-m devices. <S> E.g. For Silabs (or others) <S> EBlink a device independent tool for STlink/V2
| To add to the existing answer; some chips are known not to work properly with ST-LINK and OpenOCD.
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Can I make a one-speed AC motor variable-speed by modulating the power input? simulate this circuit – Schematic created using CircuitLab I'm using a Raspberry Pi with a SainSmart solid-state relay. If I connect the relay to an AC motor (a box fan in this case) and run the following Perl script with different timing values, the motor speed does change accordingly. use Time::HiRes qw(usleep);for(;;) { system("gpio write 24 0"); usleep(50000); # 50ms system("gpio write 24 1"); usleep(50000);} The question is, is this safe and efficient? If not, is there a way to make it so with additional inexpensive hardware? EDIT: On further observation, this might not be a good idea after all. I attached a meter to the AC supply and the fan actually draws more power on a 95% duty cycle than it does at full speed, even though the fan turns noticeably slower. Presumably that extra power is just heating up the motor. <Q> Can I make a one-speed AC motor variable-speed by modulating the power input? <S> You can not modulate power. <S> You may be able to control the voltage. <S> The motor and the load working together will determine what current will be drawn and how much power will be taken from the source. <S> If you control the voltage to a box fan, it is likely that the speed can be controlled over a limited but useful range. <S> This will work with a permanent-split-capacitor or a shaded pole type of induction motor driving a fan or centrifugal pump. <S> It will not work with other types of induction motors or other types of loads. <S> Even under the above conditions, there is a possibility that the motor will overheat eventually. <S> The device used to control the voltage needs to have appropriate voltage and current ratings. <S> You must be able to trigger the voltage turn-on in mud cycle to get a waveform like this: <S> The question is, is this safe and efficient? <S> Safety depends the design and implementation. <S> Part of that would be determining if the specific motor and load work properly and that the motor does not overheat. <S> To be completely safe, the motor probably needs a thermal fuse imbedded in the winding my the manufacturer. <S> This will probably not reduce the efficiency very much as compared with constant-speed operation. <S> Reducing the fan speed will reduce air flow and require less energy than full-speed operation even if the motor is slightly less efficient. <A> Pulse width modulation works well with small DC motors and voltages, but I don't believe it will work well on AC. <S> I can see that relay and motor Will be under a lot of stress due to high current draw. <A> You may want to look into how variable frequency drives work. <S> This is generally accomplished by rectifying the AC line to DC, then putting a variable-frequency inverter on the DC to convert it back to AC. <S> I've never seen it done, but the same principle should apply to single-phase motors. <S> (I think modulating the voltage is more likely to reduce the torque more than the speed, though they'll be related to some degree.) <S> A VFD inverter usually runs a high frequency PWM signal, 2-6 kHz carrier being typical to generate a ~60 Hz fundamental. <S> But if you don't care about detailed control (or the lifetime of your motor) you may be able to use a relatively simple square wave or modified square wave inverter. <S> Most solid-state relays I've seen have switching times measured in milliseconds, so trying to shape any AC waveform with them will be problematic. <S> Power transistors (IGBTs or FETs) are probably the way to go. <S> Of course, designing even a single-phase square-wave inverter to drive a motor is non-trivial! <S> And from my experience, off-the-shelf inverters are vastly cheaper than anything you could ever hope to build for any significant power level. <S> Makes me wonder if you could take an off-the-shelf inverter and modify its references in some way to give a variable frequency... <A> AC induction motors Do Not Work That Way. <S> Here's what's happening inside an induction motor. <S> It's a cylinder, with 3 field windings around the edge. <S> It uses either 3-phase power, or single-phase with an extra "start" winding to get it spinning in the right direction. <S> The three windings make a magnetic field which rotates around the cylinder at line frequency, so 60 cycles/second (3600 RPM) or 50 cycles (3000 RPM). <S> It's like there's an imaginary spinning magnet with 2 poles: north and south. <S> Yes, you could use 6 windings and have the imaginary spinning magnet have 4 poles. <S> That would rotate half the speed. <S> The rotor (spinning part) is nothing but a hollow cylinder that catches the rotating magnetic wave and is dragged along by it. <S> It resembles a hamster wheel and is actually called a squirrel cage. <S> There are no permanent magnets in it, this all happens via eddy-current induction, hence the name of the motor. <S> By nature, this has a small amount of "slip". <S> So a motor whose field spins at 3600 rpm is rated for 3450 RPM. <S> The slip varies slightly by load (if you've ever worked in a wood-shop, you know the sound of a motor "loading up") but that is the one speed it goes , assuming the input frequency is 60 Hz. <S> The entire motor is carefully tuned to be efficient (avoid making internal heat which would have to be actively cooled) when fed a sine-wave. <S> Well. <S> Now you know the score, you can see where altering the 60Hz waveform will do absolutely nothing to change the speed of the motor, but break the sinewave it needs to be efficient (coolable). <S> Also, you were getting a lot of inductive back-EMF at your switch, which I'm amazed <S> didn't fry it. <S> And now you know how to control the speed of the motor <S> : feed it a sinewave at a different frequency. <S> Yeah, that's harder. <S> And not really thinkable until silicon power switching got pretty good. <S> Which is what a Variable Frequency Drive is: the "magic bullet" to solve this problem. <S> While you're doing VFD, you might as well also do proper 3-phase, to increase efficiency, allow reversibility and avoid the hokey "start winding" found on single phase motors. <S> By the way, an interesting work on this is Don Lancaster's "Magic Sinewaves".
| For three-phase motors, you generally control the speed by modulating the frequency of the AC voltage.
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How can I build a micro USB hub? EDIT: @EugeneSh. I ment how can I build a hub Double edit: Sorry everyone, I mentioned how can I build a USB hub... I'm looking to build a very small USB Y splitter for my Nexus 7 2nd Gen so it has 2 USB ports.The reason why I'm not buying one from eBay (ect.) is because i don't want a cord and I want to build something like the "sugoi hub" mentioned below.I WAS going to buy a sugoi hub but those things are nowhere to be found! Anyways, do any of you guys have an easy wiring diagram for a (micro) USB 2.0 Y splitter? <Q> All these y splitters are, are a usb hub ic, a few passives, and the usb ports. <S> It is an active device, not just wires. <S> Some all in one USB hubs like GL850G are used. <S> You could get a cheap dollar hub with one of those, then modify it as you want. <A> You can't split the wires from a USB 2.0 port to make two or more ports. <S> A USB port can connect directly to (a) a USB hub i.e. an active electronic junction box or (b) a USB device, such as a printer or flash stick. <S> (There's a mountain of stuff underneath (b) to explain further <S> but it won't help the point. <S> Wikipedia's USB article will give you enough.) <A> You can start with this reference design, offered by one of manufacturers: http://www.microchip.com/Developmenttools/ProductDetails.aspx?PartNO=EVB-USB2422 <S> Then remove components that you don't need, replace Type-A connectors with micro-AB receptacles. <S> Then you can use any free design software (e.g. https://www.expresspcb.com/ ), and manufacture a mini-prototype, it is cheap, under $100 for a 4 layer board. <S> Make sure your are using traces with 90-Ohm controlled differential impedance for USB D+/D- wires. <S> Then you probably need to design a custom enclosure for your hub, by modifying some small plastic case. <S> Good luck.
| Unless your USB socket carries more than one pair of USB connections, you need a USB hub to go from one USB port to two or more. One challenge will be to accommodate a micro-B plug, there is no connectors made for printed circuit board. Or there are services that can do this for a couple of thousand bucks.
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Does the inductance change in an LC filter based on frequency? I sort of understand the formulas for inductive reactance. Do they apply to a LC filter? I have this, inductor at 700nH and caps at 120pF for a 20MHz cutoff emi filter: simulate this circuit – Schematic created using CircuitLab The specs for my EMI filter say max 8 ohms DC resistance. If I solve for inductance in the reactance formula for 8 ohms I get much higher inductance levels than 700nH. Is this happening, or am I missing a key concept here. <Q> Not so fast! <S> Also they possess some capacitance due to the windings being side by side in close proximity. <S> These parasitic quantities are not inconsiderable at high frequencies - indeed an inductor will become self resonant at some frequency - this is often quoted by the manufacturer for RF inductors. <S> Also, you clear state '8 ohms at DC' - there is zero reactance at DC; ZL = 2 <S> * PI <S> * f <S> * L (DC equates to f = 0) <S> Shown below is a model of a typical inductor (source - http://www.coilcraft.com ) <A> This question is a logical falacy. <S> You can't solve for inductance after you fix the inductance value. <S> The overall circuit reactance may change, but that is due to the frequency change OR relationship between the capacitors and the inductor, not the inductance changing. <A> Inductance is, in general, a fixed quantity dictated by the component construction, etc. <S> Under some cases as pointed out by @N. G. near there can be parasitic effects that change this value. <S> However, at 20 MHz, I doubt you will see any of these effects on your 700nH inductor. <S> My guess is that the EMI spec is written so that at DC you don't have an inductor of more than 8 Ohms. <S> Such a resistance would come from the wire windings since they are not perfect conductors. <S> As long as your inductor of choice has a DC resistance lower than 8 Ohms, your filter should pass the EMI requirement you mention.
| Inductors do exhibit some resistance due to being wound from wire that has a finite conductivity.
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8-bit shift register 74LS164 not working I am trying to use an SN74LS164 8-bit serial-to-parallel shift register with my Raspberry Pi, but I have some problems. I connected: VCC, A, B and CLR to +5V GND to 0V CLK to Raspberry Pi GPIO simulate this circuit – Schematic created using CircuitLab When I power-on the circuit, all the LEDs are off. Then I send a clock pulse from the Raspberry Pi (go high for 100ms then go low). However instead of having the just first output (QA) go high, the first 3 or 4 outputs go high (QA, QB, QC and QD). What am I doing wrong? I took some photos of the scope, the first is the clock from the GPIO (without nothing) and the second is the clock connected to the register. -- EDIT -- I added capacitors and MOSFETs to respect max output current. I re-added pull-down resitor on clock line because of the natural +1V offset (this offset come from the register not the pi) <Q> I resolved my problem. <S> The register was clocking multiple times because the clock signal wasn't enough clean. <S> So I added a schmitt trigger between the raspberry pi and the register clock input, and now it shifts exactly the number I want. <S> Thanks for all your advices and suggestions, have a nice day ;) <A> A Low power Schottky input typically sits at about 1.3V when open circuit, and sources ~0.25mA when pulled low (<0.8V). <S> The Pi should have no trouble driving this load, but in your case it only managed to pull the clock input down to ~1.1V. <S> You got it down to 0V with a pull-down resistor, which suggests that the LS164's input is OK and your Pi's GPIO line is not pulling down properly. <S> Cleaning the pulse up with a Schmitt trigger got it working, but is masking the real problem - you <S> Pi's GPIO line is either not being controlled correctly by your program, or it's blown up. <A> Take a peek at the data sheet . <S> Thus exceeding operating conditions, all bets are off.
| This chip can't drive the kind of current you want it to.
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Help on ID for SMD component (probably diode) I recently smoked a DC motor control board and need help identifying the wounded component. before picture (H2 and sideways 3) and here it is, wounded It looks to me like a resistor SMD; but, of what type/value? Thanks and regards <Q> Judging from the location on the board and the marking H2 , it's a Zener diode, to protect the 12 volt rail from overvoltage. <S> I'm not quite sure about the package. <S> Maybe SOD-323 or SOD-523. <A> Looks like it is a 12V zener diode. <S> http://www.diodes.com/_files/datasheets/ds31038.pdf <S> The linked document gives codes for various zener diodes from Diodes Incorporated. <A> You could remove it to test the rest of the board. <S> It would be best to test with a load consisting of LEDs or a small 12 V lamp as these would not give an inductive kick on switch-off (which the diode may have been there to absorb / shunt). <S> If that works you could replace the dead diode with a regular through hole type - if that's all you've got. <S> If the pads are damaged beyond use then piggy-back onto the adjacent component. <S> Others are suggesting that the component is a zener diode. <S> Without seeing more of the circuit I couldn't comment on the configuration.
| The value would be something a little above 12 volts. Yes, there seems to be a stripe at the left side of picture 1 indicating a diode.
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What does the 1k resistor do in this schematic? I'm very new to circuit design and have been doing lots of reading and research in order to understand everything going on in this schematic. I think I understand all of it except for the purpose of the 1k ohms resistor going from the control signal to the emitter. Can someone explain what that does? This circuit is part of a bigger project outlined here: http://victorbush.com/2015/01/arduino-rfid-door-strike/ <Q> The 1k resistor is a pull-down resistor. <S> Not the weakest one of that, but that's okay. <S> It is meant to pull the base of the transistor to a known state (ground) when the control signal is missing/open/high input. <S> It does form a small voltage divider, but not enough to really affect the transistor performance. <S> Due to its value, it will waste 0.5mA to ground when the control signal is high, <S> so it's value may need to be changed for battery applications. <S> A 10k to 47k may be a better value. <A> I'm not saying this is what was in the designer's head, but it's possible that with the Arduino microcontroller not powered a powerful RF signal could be rectified by the transistor base and cause the door to unlock or heat up the transistor and solenoid. <S> It also reduces the leakage current if the transistor was hot and the Arduino was not powered. <S> It doesn't do much of anything useful if the Arduino is powered and the output is configured to pull down. <S> The 1K takes 0.7mA of the available base current only when 'on' (a few percent probably, and less than 0.3% of the total current including the solenoid) <S> so it's reasonable enough. <S> If replaced with a larger value resistor, a capacitor from base to emitter would eliminate the RF concern. <S> To pick a reasonable value say that Xc = (say) 1000 ohms at (say) <S> 100kHz <S> so a few nF would do the trick. <S> However there appears to be a flaw in the ointment (sic) here- <S> the solenoid inductance is a powerful storage reservoir of magnetic field energy that will cause the voltage to rise to hundreds of volts and thus break down the TIP3055 when the power is shut off to the strike. <S> To avoid this, a reverse-biased diode should be installed across the solenoid. <S> It's possible there is one already inside, but I see no mention of it on the datasheet or install sheet. <S> A 1N4004 or similar will do the trick. <S> The TIP3055 is a fairly robust device and it will probably survive this for a while, but it would be a major pain to have it fail (probably on)- it appears to be rated at 30 seconds maximum 'on' time- and burn out the expensive strike mechanism, whilst leaving the door open for rascals to ransack the place. <A> This resistor performs two functions. <S> 1) <S> 2) <S> It is part of a voltage divider. <S> This allows the threshold voltage to be set to the desired value. <S> For example, it is possible to pick resistor values such that the transistor doesn't turn on until say, 120 Volts is present at the circuit input. <S> Almost any voltage greater than Vbe can be selected. <S> This allows you to improve resistance to false or low-level signals. <S> Note that the threshold voltage is sensitive to both temperature changes as well as variation between devices. <S> Nonetheless, it is an important design element.
| To prevent a floating signal or even leakage current from turning the transistor on, to make sure the strike doesn't open or engage randomly . It is a pull-down resistor.
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How to deal with the high current flowing into your wires? So i've got this 1000 watt 36V with a full-load current of 35,6 amps. In the circuit I want to wire a switch that can handle this high current. Unfortunately it seems that I cant find a switch (specific reason?). What can I do to deal with the high current. I am making an electric gokart btw. Ps: In a circuit, does the motor pulls 35,6 amps into my wires or is it the battery? <Q> As said already you need to take a step back and start learning basic DC electronics and some automotive electrical. <S> This means large wires don't have to detour through the cab (shorter runs so less voltage drop) and you can use smaller more intricate switches among other benefits. <S> Think of your headlight switch on your dash or indicator stalk, this then triggers a relay under the bonnet. <S> So the loom to power your headlights just has to travel from battery to fuse box and relay then on to your headlights. <S> Another problem will be protection of wiring. <S> Not sure if they use the chassis of a go kart like in a car and bond it to the negative terminal but if they do there is potential for shorting on the frame or each other due to wires chafing, rubbing, melting and so on. <S> Lastly, how is the motor actually driving the go kart? <S> I am no expert on motors by any means but is it direct drive by a chain or belt ? <S> How are you going to do 'throttle' and control the rpm of the motor ? <S> Unless you are purely asking about a manual on/off switch ? <S> Any gearing, transfer case or clutch of some sort ? <S> Are you going to want to do regenerative braking, have the motor brake dynamically or just free wheel? <S> Motor driving one way or capable of reverse ? <S> How are you going to monitor charge and battery status ? <S> Sounds like a microcontroller or similar could be handy ? <S> Even if you only want this at the back I would imagine you may need a kill switch accessible to the driver up front? <S> You need to flesh this out more. <A> The magic word for high-current switching is a "contactor": http://www.te.com/usa-en/products/relays-contactors-switches/contactors/mil-aero-contactors/dc-contactors.html?tab=pgp-story <S> (Obviously those are the expensive ones, but you should be able to find a e.g 50A contactor somewhere) <S> However, since this is apparently for an electric go-kart, you might want to think about speed control before going much further ... <A> Solar disconnect switches are designed for high currents like this. <S> Here is one that would appear to meet your stated requirements... <S> http://www.industrialcontroldirect.com/dc-disconnects-316/enclosed-dc-disconnects-317/dc-disconnect-4088.html <S> But I would echo @jbord39's comments - this is enough current to weld contacts and vaporize conductors <S> so you should probably try to find someone with experience to help out.
| Typically in a car for example you are switching remotely a relay or solenoid on and off. So you will need to know how to select the right gauge of wiring and protection for the wiring, typically a fuse and where to locate it.
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Can the /OE pin of a buffer be left floating? Here is a buffer from OnSemi which has an /OE pin that should be tied high for high impedance mode. For all active modes the pin value is stated as low. Would my buffer still work if I were to leave this pin unconnected or do I explicitly need to ground it? <Q> In general, if there is no explicit mentioning in the data sheet that a logic pin is internally pulled up/down, you have to provide a defined logic level. <S> If you leave it floating, it may appear low, but it may also be susceptible to noise. <A> Either drive it from another CMOS device's output, or connect it to Vcc or Gnd. <S> If you leave a CMOS logic input floating, that input will "drift" into the no-man's land between a legitimate logic-low voltage and a legitimate logic-high, which will cause the output PMOS-NMOS pair to enter the linear zone of operation. <S> In the linear zone both transistors will partially turn on, causing a large current to flow between Vcc and Gnd within the device. <S> When this happens the device will misbehave in several ways. <A> The /OE pin is an input just like a regular input such as INA on the device. <S> The device's data sheet says that the maximum input leakage current is up to +/- <S> 1 <S> uA. <S> This means that it may drag down (to 0V) <S> the unconnected input in trying to draw 1 uA from an open circuit OR <S> it may drag up the input (to Vcc). <S> This means you cannot leave it floating because you CANNOT know what state it will move to.
| This is a CMOS logic device so all inputs must be tied to a legitimate logic level.
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TV screen in still photo: Why are there dark, blackish bands? I was trying to capture the scanning process of a CRT TV screen in a photo. It gave me these dark, broad bands, shown below: Fig. 1 a,b,c I'm trying to understand the cause for this banding. However, I couldn't understand it clearly in very similar questions here , here , or here . All web sources I visited very briefly explained it as a difference between the scan-rate of the TV and the camera, and none provided a diagram. Fig. 2 a,b,c I assume what is going on is in Fig. 2. The blue lines indicate the scan lines where cathode rays hit most recently (phosphorescence occurring), and black lines indicates where the cathode ray has not been in the image, so phosphorescence is all that is left and a dark band appears. I assumed this, because out of 10 photographs I took, I found only the above 3 types of photos, and none of the type shown in Fig. 3 or Fig. 4: Fig. 3. an Fig. 4. I also assumed that the camera is much faster than the TV scan rate, so it could take picture in an instant. My question is, Have I assumed correctly? or is something else going on? Additional information: TV: Videocon 120 W colour CRT, Electricity power supply 50 Hz. Camera: Nikon Coolpix L24 digital camera. (please also explain other possible situations, such as this one in Fig.5. I didn't experience it myself. It is similar to Fig. 4. Fig 5. From this question I'll be grateful if any explanations use diagrams. Update: I've obtained several fig3 conditions, and though they are less frequent, they're not very rare. File properties show exposure time for this still-photo is 1/125 second . I've also took some videos (in PAL and NTSC mode... both gave same result), file properties show frame rate is 30 frame/sec ; and run them on slowest motion. I found similar, but much brighter and quite unclear bandings, and it seemed from slow-motion video that the alternating bright and dark band rising on the upward direction. From that videos I screenshoot some frames as-successive-as-possible. .Red small bars added to indicate the margin of each banding <Q> Yes, you're basically on the right track. <S> The bands appear because the camera shutter is not synchronized to the vertical scanning of the CRT. <S> A fast shutter speed will only capture part of a scan. <S> A shutter speed exactly equal to the vertical scan period will capture a full scan, but there still might be a narrow band (either dark or light) somewhere in the image if there's any mismatch in the timing. <S> To minimize the banding, use a shutter speed that spans several vertical scans, but this only works if the image on the CRT is static. <A> The only reason you didn't see "Fig 3" (bright stripe in the middle) is that your shutter speed is too slow. <S> You have a much higher chance of capturing a dark gap in the middle, because those pictures all include a vertical blanking interval (the time when the beam is offscreen and "moving" back to the top) as well as all of the vertical part of the picture hidden by overscan. <S> You can see from Fig 1b that your shutter time captures all but 1/3rd of the visible part of one field. <S> To get "Fig 3" would require pretty precise timing, and your bright stripe would be nearly the full height of the display. <A> Your scanning beam description seems right on, to me. <S> But my experience would be with a film camera, not digital. <S> Your more complex images make me think of scan-on-scan of reading out the digital pixels vs cathode ray trace, presuming the readout speed might be similar to the scan rate on the CRT. <S> My first SLR had a left to right shutter. <S> My better one did top to bottom. <S> If you had an iris shutter, the results would not show such skewing because it opened from center out and back. <S> The standing rule for taking a photo of a TV screen was to set the speed to 1/30 or lower to get the full interlaced image. <A> Not as a confirmed-answer, but hypothetical one, the other-2 situations could be this: 2 . <S> when Camera having speed synchronous with TV Fig. <S> 6 <S> Whether the camera (i)starts scanning with the TV, or (ii)start scanning out-of-phase (above picture); in both-cases we should get the accurate picture... <S> neither dark band nor bright band. <S> 3. <S> when camera is slower than TV Fig. <S> 7 <S> a, b, c, d, e from left to right. <S> conditions like fig 1 <S> a, b or c , but <S> with an extra shade of brightness could be obtained. <S> So we could get a medium-bright band and a very-bright band (no dark-black band) <S> Fig7 a (leftmost) indicates ultimate photograph that should be obtained . <S> fig7 b -> c -> d- <S> > e indicates step by step formation of 7a. <S> (gradients would take place if I consider the camera can detect the gradual loss of phosphorescence.) <S> (feedbacks welcome)
| In film with a planar shutter, the choice of shutter speed could result in skew for the CRT image against the film.
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One particular part number not showing up on Eagle board layout I finikshing up the silkscreen for a board about to be sent out for fab, and for some reason, I have two diodes whose part numbers are not showing up on the board layout, while all my other parts numbers are displayed just fine. Here is one of the parts; it is a diode which should have a label "D2": The part is from the SparkFun DiscreteSemi library (SparkFun-DiscreteSemi.lbr), and has a >NAME label defined in the appropriate layer (95 Names): When display the properties for the part on the layout, everything seems ok including the name: Any idea why the label for this one part isn't showing up, while all my others are? <Q> The ">NAME"/">VALUE" designators on the "Symbol Editor" refer to the Name and Value on the Schematic. <S> For them to be displayed on the PCB they also have to be set on the "Package Editor": <A> It could be that the library PCB footprint is missing the designator. <S> In the footprint editor the designator for silkscreen should appear as >NAME . <S> If you have other footprints of the same type, and none of them have the silkscreen text for designator, that would suggest that it's missing in the footprint library. <A> The easiest way to resolve this issue is to hit the reposition attribute button , then the component. <S> The reference designator will return. <A> I would try a small test schematic and board with the diode as the only part. <S> I would check all layers for the missing part label. <S> The label may be placed far away from the part origin. <S> The text size of part label may be way too small.
| If the problem is not found, I would try editing a new symbol, package and device for this part under a different name.
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Zener diode selection in a led boost ic circuit I'm wondering how to select a correct zener diode value for this led driver I am using RT4533AGJ6 (link to datasheet). My understanding of boost converters is limited, although I understand that there is an internal switch that charges up the inductor and when switched off, the inductor discharges onto the load...through the zener diode, which is to my point. The datasheet does not give any values for breakdown voltage, let alone why its required.. i did some further searching on other led driver datasheets and ran into an additional issue, the reaction/reset speed of the zener diode, which is of course dependent on the switching speed. To make matters worse, I had contacted a customer service representative from a electronics warehouse, he said that it didn't matter what value diode I choose....which just seems silly and must be wrong in some way. <Q> This is a boost converter topology. <S> The diode should be Schottky, not Zener. <S> The reference design for RT4533 has a bill of materials, where D1 is a Schottky. <S> The symbol in the datasheet is for a Zener. <S> I blame lack of proofreading. <A> That is not a zener diode, that is a schottkey diode. <S> A Schottkey diode is basically just a fast diode -- honestly, in my experience almost all "diodes" are shottkey diodes now. <S> So when the guy said "it doesn't matter what value diode you use", he meant it: just use any old diode that can handle the current you need it to handle, and is called a schottkey. <S> Just don't use a zener <S> : they generally can't handle the amount of current that might be necessary for a LED driver. <S> Schottkey diodes are faster than "bog standard diodes", and should be used instead of something old and useless like the 1Nn00x series. <S> Most diodes that you find someplace like digikey will be shottkey, and will be able to switch fast enough. <S> Older diodes (like the 1N400x series), struggle to keep up. <A> You can select a 1N5819 (there are SMT versions) which is cheap and will handle 1A/40V, adequate for the maximum voltage of this chip. <S> You should not pick Schottky diodes with excessive PIV voltage capability because the forward drop goes up with voltage capability until they are no longer much better than conventional diodes. <S> For example, for a lower voltage circuit you could use a 1N5817. <S> For higher voltage (more like mains) an ultrafast conventional diode is called for such as a UF4005. <S> The additional forward drop even in your particular circuit will not be that important because the voltage of those series LEDs is relatively high. <S> Do not use a regular 1N400x diode- <S> the switching frequency of this converter is 1MHz and that is about two or three orders of magnitude higher than is suitable for a diode with several microseconds of reverse recovery time.
| As others have mentioned, the diode is a Schottky not a zener.
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In automotive electronics ECUs, why is an external EPROM IC often used instead of the MCU's internal EPROM? I am a bachelor of vehicle engineering.It's easy to see many Automotive class MCU own interior EPROM,for example S9S08DZ,S9S12XS and some PIC 8-lead chips.But,I have seem many Automotive electronics ECU use a EPROM IC(for example M95320).Why they don't use MCU internal EPROM directly?Is it because of lifetime?I know some PICF12 8-lead watchdog MCU own interior EPROM,they won't lose power.(Beacuse they are watchdog MCU) <Q> I used to work in automotive, and there could be any number of reasons to speculate on. <S> Next is qualification. <S> If a SW team has already written a driver for an EEPROM and HW has qualified that it meets AEC-Q100 and other internal tests, it might be "cheaper" to use the external device initially to meet schedule, and revise the design at a later date to optimize for cost. <S> Additionally, in the old days, the EEPROM was a socketed device that was programmed externally and then placed on the boards during assembly. <S> Often you'll see aftermarket modifications that change the contents of that device; modern tuning companies just reflash the MCU itself, though. <A> But if the program is external, then you can use any available MCU, with or without program memory <S> Second reason is cost. <S> External memory ICs are almost invariably cheaper than the same amount of memory integrated into the MCU. <S> This cost difference comes from the semiconductor process differences or bonding+packaging+testing workflow complexity. <S> Third reason might be reliability; you might not want to put programmed ICs into a reflow oven. <S> I'm not sure about this, but I would feel safer to program the IC externally with a gang programmer or order the ICs from the vendor already programmed, and then push them into sockets after production so that the programmed ICs are never heated up. <S> Fourth reason might be product differentiation. <S> Perhaps the same ECU is used with different software in different models. <S> Using an external IC for the program allows the manufacturer to make all units in a single lot and choose the software version at a very late stage in production. <A> I'm sure the combination of it being a small size (1K x 8) and not re-programmable made it worthwhile designing the ECU with external EPROM.
| First and most obvious is size -- if the internal EEPROM (or other non-volatile media) isn't large enough, an external part is obviously needed. I can think of four reasons: one is component availability: if you design for a particular IC with integrated program memory, you basically need to use that (exact) MCU. On very old ECUs, the micro-controller has a write-once ROM that can only be programmed at the factory. The Volkswagen Digijet ECU has an Intel MCS-48 P80A48H, using an Intel M2716 2K x 8 UV-Erasable EPROM.
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What kind of component lets me measure distance in a small space? I have bought a beautiful old radio , which I'm trying to convert into a DAB radio ( there's no future for FM radio in my country ). Of course I could just use a DAB receiver, plug it into the old radio's speakers, and be done, but there's no fun in that. Instead, I want to be able to operate the old radio, turning its knobs and pressing its buttons. So that when I turn the tuning knob, I'm switching between DAB stations instead of tuning between FM frequencies. The first step (at which I'm stuck) is to translate the position of the tuning knob (A) into a digital value I can use to control a DAB receiver. I'm a programmer, with only limited experience in electronics, so I'm not sure how to best accomplish this. [ ] Originally, the radio displays the frequency with a vertical bar (B) that travels over a frequency band (C). My idea is to attach something to that bar (or to the carriage that the bar is attached to), and measure the distance (D) between the bar (B) and the inside wall of the radio . If I could do that with some electronic component, I could get a signal that I could translate into a number that again could be used to select a DAB station. The problem is that space inside the radio is limited. Images This image shows a horisontal, round bar on which the frequency bar moves along. When the radio is fully assembled, the speakers are located on top of this, leaving about 1cm space between the horisontal bar and the speakers. It must also go clear of two lightbulbs (for illuminating the frequency scale). The frequency bar is sandwiched between the glass dial scale, and a metal plate. Here's what it looks like from above. Here's the back of the tuning knob and its shaft . As can be seen, there's not too much space here. The knob is dual shaft - the outer ring is the speaker selector . Here is a birds-eye view of the interiors , when taken out of the enclosing cabinet. And a close-up of the variable capacitor's shaft , as per Transistor's request. Oh, and here's the schematics , if there are any Norwegian speaking radio enthusiasts out there. Here are the options I have investigated/tried so far: An ultrasonic rangefinder (like http://letsmakerobots.com/node/30209 ) These are too big to fit. Also, I doubt I would get any sensible signal from it, as there are many parts inside the radio that would reflect the ultrasonic waves. The same goes for infrared rangefinders . A slide potentiometer (like these: http://www.potentiometers.com/select_slide.cfm ) The ones I have found are difficult to fit, and there are not many that are longer than 10cm. The radio is ~50cm wide, so that would leave most of the band unused. A SoftPot (like https://www.sparkfun.com/products/8681 ) These works by reacting to mechanical pressure somewhere on the surface. They come with an "actuator", which is basically a screw with a round, plastic tip. This was promising - I attached the actuator to the part moving the bar, and let it travel along the softpot membrane. However, the tuning knob is carefully designed so that when the bar reaches one end of the scale, it stops moving, because the wire (E) that drives it will start slipping on the tuning knob to avoid damage. The pressure needed to activate the softpot was enough to cause too much friction, so that the wire was slipping, and the bar did not move. A string potentiometer (like these: http://www.unimeasure.com/ ). I could attach the string to the bar, and the stringpot housing to the inside wall. However, most stringpots I have found seem to be meant for heavy industrial use, and are priced and dimensioned accordingly. The added friction could also be an issue. What other options do I have? <Q> My suggestion is that you forget about trying to figure out where the needle is at and build additional circuitry that interfaces with the radios tuned circuit. <S> You could select the AM band which tunes the internal oscillator from about 500kHz to 1.7Mhz. <S> If you got your hands on the schematics for this radio you could figure out where to put your "tapping" point. <S> The problem is not to load the circuit in a way that would alter the frequency, but that's doable. <S> It could even be as simple as reading the value of the variable capacitor that the dial is altering. <S> Doing it this way, you are less likely to deface a beautiful old radio and you'll know a bit about how radios work when it's done. <A> OK, assuming you can detach the variable capacitor from the surrounding circuitry, then it should be a simple matter to read it into a digital system. <S> On your (rather nicely hand-drawn from before the days of CAD) schematic, it's the parts 3326/3327 at the top left just below the antenna. <S> If you incorporate it into a loop of Schmitt trigger buffers, you can turn the capacitance value into a frequency which can then be counted in software on a microcontroller. <S> Adjust values of R1 until the frequency range is "sensible". <S> simulate this circuit – <S> Schematic created using CircuitLab <A> The tuning knob will be driving a variable capacitor. <S> Replace that with an ordinary potentiometer and you should be able to read its value trivially using analogue to digital conversion. <A> Quick answer: use a rotary encoder, axis mecanically connected to the knob axis; leave the frequency scale mechanism as it is. <A> You could consider making your own linear pot using a piece of nichrome wire and a sliding contact, preferably precious metal <S> (cannibalize it out of something). <S> A straight 40cm piece of AWG 30 Nichrome wire would have a resistance of about 8 or 9 ohms. <S> If you put 100mV across it (use a voltage divider from your ADC reference and an op-amp buffer), that's only about 11mA. <S> Then amplify the wiper voltage with a decent op-amp <S> and you're done. <S> Linearity should be excellent (in the 0.1% class most likely). <A> I would put a grid of Gray Code on the BACK of the dial scale, and a duplicate of the "cursor" on the back to read the gray code with reflective opto sensors. <S> That would tell you the ABSOLUTE position the cursor along the dial scale. <S> And it would be completely STATIC so that the position could be retrieved even from cold power-up. <A> Then you can use that to position the cursor at the direction of the microcontroller. <S> You could even use one of those "filament LED" strips to make a glowing cursor behind the glass window.
| Another alternative would be to use a tiny stepper motor to a length of threaded-rod ("all-thread" as it appears to be called in Europe).
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Accidental short circuit of lead acid battery, can I still use it / charge it? Just a minute ago I accidentally short circuited my lead acid battery. The wiring touched, glowed bright, the plastic around it melted and smoke got released, but I'm not sure from the battery or from the plastic. It smelled like plastic burning, so I'm hoping the latter. The battery discharged quite a lot and the voltage dropped from ~12.80 volt to around ~12.55 volt. However, this short circuit lasted only a fraction of a second, the wires only touched and they almost immediately lost contact. The battery seems to still be working, but I'm not sure if it's safe to charge. Is it safe to assume I can still use it properly? It's the first time I short circuited something like this and it scared me a bit ha. <Q> Probably okay. <S> Consider using a fuse, ideally a fuse right at the terminal or inline as close as possible or a short length of "fusible link". <S> A good practice to get into. <A> If you used such a thin wire that you can connect to a breadboard <S> then I think that incident isn't that "short" circuit, from the battery's point of view. <S> The wire resistance is enough to limit the current, perhaps 10..20A or so flown for a fraction of second, it is pretty normal for these SLA batteries. <S> Short circuits not only happen mechanically by accidents, anything wrong can happen in the circuit you experimenting with. <S> This is especially true in prototyping. <S> It is also a good safety practice to always disconnect any power if you leave the area. <S> Always. <S> The sole exception is long-term testing, but that should be done after thorough pre-testing and already thought out and operational safety systems. <S> The bigger the power you working with the importace (and complexity, cost...) of the safety systems gets more and more significant. <A> You're ok to continue using the battery. <S> Typical 12 volt lead-acid car batteries can be discharged to about 9 volts and be recharged, so you're in the clear. <S> Cars pulls hundreds of amps and their batteries aren't exploding. <S> I'm guessing <S> your wires didn't make very good contact otherwise they would have welded together. <S> This can be dangerous and you should be careful of this. <S> If the wire is thick enough, it will weld and stick. <S> If it is thinner, it will weld and then melt away, like a fuse. <A> Well, a such wire can draw at most 50 amps for two or three seconds before melting, but a car starter can draw from 90 to 150 amps, depending of the type, so for me the only risk is that molten metal touches the battery case and penetrates the battery, making an internal short circuit.
| Next time it is a good idea to add a fuse as close as possible to the battery terminal, and use a fuse that not just blows when the wires already red hot, use one which is rated for the application. Discharging a lead-acid car battery below 9 volts reduces the battery's capacity but it doesn't cause explosion or anything dangerous like that.
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How does current work in hubs? Okay, I'm a total noob and don't know if I'm even wording this question correctly, but I'll try. My understanding is that the power supply (an USB power supply in my case) determines the current. I have a 2.5A 5V supply, and a USB-to-DC cable that connects that power supply to an USB hub (5V as well). Now that hub would get a total input of 900mA from its data USB 3.0 connection to the PC + 2500 mA from the power supply. Does it just always get that current, regardless of whether it actually needs it? In other words, if I just have one little USB drive in the hub, will that USB drive get all the 3400 mA and burst into flames? <Q> The power supply works as a voltage source up to 2.5A current. <S> You cannot combine the current of the externals supply and the PC's port - that would require expensive adaptive circuitry. <A> Your power supply can deliver up to 2.5 Amps at 5 Volts, but does not force 2.5 Amps into a load. <S> Any load will only "ask" the power supply for the current it requires. <A> Most power supplies regulate voltage. <S> And can supply enough current to maintain that voltage up to a certain limit. <S> In math terms voltage is equal to the current times the resistance of the load. <S> For a given voltage, something that need a lot of current to run will have a low resistance. <S> The usb power supply's job is to make sure that the load's resistance times the supplied current is always 5 volts. <S> So, remember this equation and you should be able to solve these types of questions from now on using math: <S> V = <S> I <S> * R
| The insight needed here is to uderstand that only enough current is supplied by a voltage regulating power supply to maintain the voltage.
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USB to TTL communication problem I'm trying to connect a device to my laptop using a TTL to USB converter like this one but it doesn't works so I tried the following configuration (USB to Serial, RS232 to TTL and then to the device) And works perfect. Both (USB to Serial and USB to TTL) are using the PL2303 driver. The question is why the USB to TTL didn't work? it uses a different protocol? <Q> The Prolific driver download site has a warning and also says how to detect the fakes. <S> Basically, check the Windows Device Manager. <S> If your Serial/USB device shows a yellow warning triangle and has error code 10, then you have a fake. <S> It will not work with newer Prolific drivers. <S> You might find an old Prolific driver that will work with your device. <S> There are lots of sites on the internet discussing this, and lots of advice on finding old drivers. <S> You might also just save yourself the time rooting around the internet and buy or order an adapter from someplace reliable so that you don't get burned by another fake and don't have to worry about the drivers. <A> These USB-TTL converters with PL2303 chips are crap. <S> I've got many problems with these sticks, i.e. one described here . <S> The problem lies in PL2303 chip - 99,9% of these terribly cheap sticks have counterfeited ICs. <S> They may look like original ones from Prolific, but obviously they do not. <S> Differences are very subtle, for example mine have invalid production code. <S> As Artūras <S> Jonkus said You should use stick with original chip, these from FTDI are great (but not cheap), and there is a small chance for buying counterfeited FTDI chips. <S> Original Prolific chips are also very good, but there are plenty fake ones on the market. <S> Also good USB-TTL chips without many fakes are from Cypress. <A> There may be several problems. <S> For begining, connect your TTL to USB converter to computer and measure if there are voltages that are supposed to be there (3.3V and 5V). <S> Some converters might need additional jumpers to enable voltage rails on output. <S> If everything is fine, proceed, if not - it's likely that something in pcb is broken. <S> I don't think that's the case this time. <S> You might try switching TX and RX wires because that's usual mistake and making sure that you've chosen right communication configuration (baud rate, stop, parity bits). <S> If you have oscilloscope, I highly recommend to send something from your computer and see if it's transmitted; send something from device and make sure it's received. <A> Connect conterter TX to its RX and type something in the terminal <S> , you should see you characters, if you don't than the converter is definitely broken. <A> Had the same issue, i just downgraded from 3.8 to 3.0 drivers. <S> (This works for cloned PL2303, original works with new drivers)
| You may well have gotten bit by a fake Prolific 2303 chip. Check the device manager on your windows under Ports if you see a yellow warning icon next to your device it is because of drivers. You just need to download older drivers for Usb to TTL prolific and it works.
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Transmit a signal using RF transmitter/reciever without using microcontrol or a decoder/encoder Objective: I am trying to build a transmitter that transmit an on signal when a button is push. The transmitter will need to be small enough to wear or hold in a pocket. I prefer not to use any microprocessor or encoder for the transmitter. For the receiver, it have to be able to receive the signal from the transmitter and play an alarm. 433Mhz RF transmitter and receiver: https://www.amazon.com/UCEC-XY-MK-5V-Transmitter-Receiver-Raspberry/dp/B017AYH5G0/ref=sr_1_1/163-3850928-5773553?ie=UTF8&qid=1470685976&sr=8-1&keywords=rf+transmitter+module Question: I am planning to use a 433Mhz RF transmitter and receiver. Since I am only sending an on signal, I wonder if I can just send a voltage as a signal or data through the data pin on the transmitter chip. Would that be possible for the transmitter to transmit such data and for the receive to interpret it as a 1 or an on signal? <Q> Would that be possible for the transmitter to transmit such data and for the receive to interpret it as a 1 or an on signal? <S> If you want a somewhat reliable system you cannot just send a high or a low voltage and expect that to be adequately decoded with any measure of reliability. <S> Think about the receiver - it's very sensitive (by design) <S> so, in the absence of a proper signal, it's trying to demodulate spectral noise and producing basically garbage on the output. <S> It will have an AGC (automatic gain control) that allows it to receive and demodulate anything that ranges from a proper signal to noise. <S> If it receives a non-random-noise signal it backs-off the AGC to properly receive and demodulate the signal. <S> However, that signal could be from someone else down the street opening their car doors. <S> If you are looking at cheap radio transmitters and receivers you have to: - Send some form of data preamble that allows the receiver to adjust its AGC or detector to properly deal with the amplitude of the received signal. <S> Send an address so that the MCU looking at the received data has some idea that it is from the transmitter it wants to focus on. <S> Send data and this could be high or low, one bit to any number of bits Send a checksum so that after all is received you can get even more confidence that the signal is valid <A> The problem is that without a pattern to the transmission and gaps, it is very hard to reliably recognize, especially in the presence of noise and other authorized users of the same frequency. <S> In effect, to simplify the transmitter you end up having to make a receiver sophisticated enough to try to "guess" if a sudden increase in received power at a given frequency is due to your transmitter or something else. <S> And however you do it that algorithm will have false positives, or negatives, or both. <S> It is far simpler to use a fixed function encoder or a tiny micro controller to on-off-key (OOK) the transmitter in a recognizable and somewhat unique code pattern. <S> That is how most inexpensive products in that range work. <S> If you are concerned about space, you merely need to pick your parts carefully - look at the size of an 8-pin SOIC MCU for example - there are smaller devices, but that has few enough large enough connections to be approachable for an initial project. <S> If size is a concern you may also want to look at a higher frequency - for example 2.4 GHz could mean smaller antennas, and you can get a radio and MCU combined in one package. <S> But you wouldn't get the true size benefit until you moved away from modules and made something custom with the chip and tiny supporting capacitors, inductors, and resistors on your own board. <A> Two very simple circuit ideas that don't require a micro controller and haven't been discussed: Transmit a narrow band CW tone, build a narrow band filter at the receiver, then set a power threshold that triggers a 555 timer to make the alarm sound. <S> Could add RC circuit as Andy mentioned to improve reliability. <S> PROS: simple transmitter and receiver, silent while not transmitting (ideally) CONS: requires a sharp filter to reject interference, unreliable in several ways. <S> Transmit the alarm sound and demodulate it directly. <S> PROS: still fairly simple and removes need for a sharp filter. <S> CONS: required demodulation (AM sill simple), <S> noisy unless a squelch circuit is added increasing complexity. <S> Neither circuit uses 'bits' per se, and would act like a wireless switch. <S> It could be a fun project as well. <S> But I agree with the other answers: it's probably better overall to go digital. <S> A garage remote circuit would be perfect, pre-built, and cheap.
| It would be technically possible to build a transmitter with a button which merely enabled the RF circuitry and thus transmitted an unmodulated carrier as long as it was held down (perhaps with some RC timer limit).
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USB on STM32F107RCT and Stm32CubeMX I have STM32f107 MCU soldered on custom board. I would like to use USB peripheral in CDC mode. I connected pins PA11 (D-), PA12 (D+) and GND directly to a USB cable which leads to the computer. The device is self-powered, thus I don't use VDD pin. In STM32Cube I created a new project, configured only USB_OTG_FS device and USB_DEVICE in middleware section. Also I configured High-Speed Clock in RCC section so that USB peripheral has 48 MHz clock. Then I generated the program skeleton and uploaded it to the MCU. Unfortunately it does not work. I think the main problem is that there is no voltage on D+ pin so that the computer is not able to recognize a new device. When I execute the CDC_Transmit_FS function, the processor falls into the Hard fault. The strangest thing is that when I configure STM32F429 MCU on Discovery kit in the similar way, the USB works out of the box. I tried to diff the source code, but I haven't found any crucial difference (except the STMF4 MCU uses USB High speed OTG in full speed mode). Please does anybody have some experience on using USB with STM32F107 and STM32CubeMX program skeleton? EDIT: When I run the MCU in bootloader mode, the D+ line is pulled to 3V3 - therefore I think the MCU really does not need any external pull-up (also datasheet says this). EDIT2: Does anybody know what Activate VBUS option in CubeMX exactly do? I can understand it in Host mode, but what does it do in Device-Only mode? <Q> Even though the device is self-powered, you still need to connect the USB's Vbus (pin 1) to the microcontroller's OTG_FS_VBUS (pin PA9). <S> This isn't to provide power, but to allow the microcontroller to know when to start up the USB subsystem. <A> Thank you for both answers, you are right. <S> To sum it up I will make another answer. <S> I hope someone will find this useful. <S> USB connection requires GND, D+ and D- lines. <S> VBUS sensing can be disabled at least on some MCUs. <S> In this case, VUSB wireis not necessary. <S> When you enable USB OTG peripheral in CubeMX, it does not light up the VBUSpin <S> even it is necessary (unless you check the Activate VBUS checkbox). <S> I cannot say what does Activate VBUS option do in Device mode. <S> I triedto generate sources both with and without this option and both projects required VBUS line to make USB working. <S> When you generate code from CubeMX (I am using HAL version 1.31) with USB CDC in Device mode, the generated code really works on both <S> stm32f1 and stm32f4 cores out of the box. <S> Even if you don't implement send/receive handlers, theMCU is still recognized as VCP. <S> I wanted to disable VBUS sensing on STM32F107 by writting zero to VBUSBSEN bit in GCCFG register, but I didn't succeed or the value has been overwritten by the usb library (the library is quite complicated, thus I don't know where exactly put the register assignment). <S> Also it is interesting that reference manual on the page 926, section 28.17.3 says that in Device mode the VBUS sensing must be on. <S> STM32F107 MCU does not require any pull-up resistor to make the USB working. <S> If I have stated anything incorrect, please correct me in the comments. <S> Klasyc <A> That pull-up R on D+ (or D-) is used to indicate the USB speed of the device. <S> While host can detect the plugged device by this pull-up, in your case your have USB Device (Slave). <S> And since it is self powered, as @bitsmack answered it shall detect connection by monitoring VBUS and start USB operation. <S> If you configured USB peripheral as dual role/OTG, you have to also connect USB_OTG_xx_ID pin to ID/OTG pin in your micro-USB connector. <S> For CDC, in CubeMX you can configure your device as Device_Only, connect VBUS (connector <-> MCU). <S> If using micro USB connector then also wire connectors ID/OTG <- <S> > GND. <A> One way to register a high level on the VBUS pin (PA9) in software (or without a physical connection), is to uncheck the "Activate_VBUS" checkbox in CubeMx and make PA9 a GPIO output, with the default level being high. <S> This worked for me with a STM32F105 which has the same USB_OTG_FS Stack as the F107. <S> CubeMx will highlight a warning, but this can be ignored. <S> The device enumerates as it should. <A> VBUS sensing is mandatory for self-powered devices (slaves). <S> USB specification demands that a device can't source power to D lines in the absence of VBUS. <S> A self-powered device must disconnect the pull-up resistor as soon as VBUS disappears. <S> For example, a self-powered device may be physically connected to a PC which is powered down. <S> As the device still has power, it sources power to D line unless it can detect the absence of VBUS. <S> I'm not sure what happens if you fail to satisfy this requirement. <S> I guess your device may work well with some hosts but not with others. <S> Or it works most of the time but may cause some strange host behavior sometimes. <S> It's best to follow the USB specification and not take risks. <S> VBUS sensing is not required for bus-powered devices, as there is no risk of driving D lines when the VBUS is gone. <S> When VBUS is gone, the device will be unpowered too, obviously. <S> OTG_FS hardware of STM32 manages its internal pull-up resistor automatically, if VBUS sensing is activated. <S> The other USB_Device hardware, which is present in some other STM32 microcontrollers, doesn't have this feature and needs additional software and external pull-up.
| VBUS is used only for USB cabledetection.
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Can I control a load of 12v DC with a solid state relay 24-380VAC? I need to control a load of 12v DC and 30A, Can I control the load with an AC solid-state-relay 24-380v and 40A ? If it isn't possible, what should I use?Thanks. <Q> No you can't. <S> Figure 1. <S> The SSR-40 DA. <S> I need to control a load of 12v DC and 30A ... <S> The device clearly shows that: The output (pins 1 and 2) are AC. <S> The output will switch 24 V to 380 V AC. <S> Even if it could handle DC your voltage is too low. <S> If it isn't possible, what should I use? <A> what should I use? <S> You can control a 30A DC load of any reasonable voltage with a suitable MOSFET. <S> Your options are wider if the gate control voltage is 10 volts or above. <S> For switching 12 volts, I'd look at 20 volt rated devices with an on-resistance in the milli ohm range. <S> For a 10 milli ohm on-resistance at 30 A, the power dissipation is going to be 9 watts <S> so, some form of heatsinking will be required. <S> You can get a few MOSFETs that are around 1 milli ohm of course. <S> To make this work with reliability you need to have an under-voltage lock-out system should the 12 volts drop to below 10 volts. <S> This prevents the MOSFET being turned on when it can only achieve an on-resistance of maybe 100 milli ohm to avoid burn-out. <S> If you are switching an inductive load you will need to use a flyback catch diode rated potentially at full load current. <A> Use mechanical relay. <S> Input AC and output AC or DC
| A relay or a DC SSR with voltage and current ratings to suit. It can not switch DC.
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LM2596 buck converter overheats converting 36DC -> 5DC at 600mA In my circuit I'm using an assembled LM2596 converter module (those which are like $2 on ebay). The input voltage is 26V AC (current 140mA), which is rectified (making it ~36 peak DC), and feeds the buck converter. The output is 5V DC, 600mA. All seem to be within reasonable ranges for the converter, but it overheats pretty fast to the point I can't touch it anymore. And if left overnight for testing, it burns out. Both the LM2596 chip and the 330 induction heat up; the inductor seems to be hotter. So far I tried to drop the voltage at the converter by inserting the capacitor (up to 100mF, non-polarized) into the AC circuit for capacitive resistance. This drops the input voltage to ~9V but the capacitor itself overheats really fast. I also tried with different modules, and they all do the same. My questions here: Is it normal? If yes, what could I do? Is it reasonable to chain the converters, like using the first buck to drop something like 36 -> 20, and then the next one to drop 20 -> 5? Any side effects, besides the cost of two converters? Or is it more effective to put them in parallel? Edit: photo of the module. The dark chip says LM2596 -ADJ: Conclusion: I have tried a bigger capacitor (2200uF), a different inductor, and a heatsink. Nothing worked, except stacking regulators. <Q> These converters are often advertised on eBay as working up to 40V and 3A with 92% efficiency. <S> Don't believe it. <S> But even if it's a 'good' fake, what kind of performance can you expect? <S> I simulated the LM2596 in TI's WEBENCH®. <S> Here's the circuit... <S> And here's the simulation result... <S> Only 74% efficiency at 600mA! <S> That equates to 1W of power loss, which could make a small board quite hot. <S> However in the simulation the inductor only dissipated 0.16W, much less than the LM2596's 0.54W. <S> Your inductor is getting hotter than the IC, which suggests it may have higher resistance and/or magnetic core loss than the simulated component. <S> The combination of low inductance, high voltage drop and low current results in high ripple. <S> A circuit designed for low output current might achieve higher efficiency by using a larger inductance value, but then it would be worse at high current. <S> I am guessing the original designers of this converter wanted to get the greatest current and voltage range they could out of it, so they used the minimum inductance value they could get away with. <S> Then someone else copied the circuit, but substituted the inductor for a physically smaller part with higher resistance and lower saturation current. <S> And if the LM2596 is a fake... Is it reasonable to chain the converters, like using the first buck to drop something like 36 -> 20, and then the next one to drop 20 -> 5? <S> Yes, this should help. <S> Efficiency improves at lower voltage drop, so you could try cascading converters (eg. <S> first from 36V down to 12V, then from 12V to 5V) so the voltage differential that each one has to handle is reduced. <S> Total efficiency may be worse, but each individual converter is more efficient so they should run cooler. <A> No, it's not ok. <S> I would blame switching and iron losses, but the frequency is fairly low and current is far below rated. <S> Input voltage is OK, don't touch it. <S> I would replace the module at this stage. <S> Are you sure about your load current? <S> By the way, do you have a scheme of the module, or at least a picture? <A> You say that input voltage is 36V, current 140mA, output voltage 5V and current 600mA.So, input power is 5.04W, output power is 3.0W. Efficiency ~60%. <S> (That conforms to datasheet graphs) <S> This is efficiency for 3A load. <S> I can't find efficiency for lower loads, but efficiency is usually lower for currents lower than rated current. <S> You dissipate 2W. <S> There is almost no heatsinking, so you can't be surprised that it gets very hot. <A> Suggest trying substituting a larger value inductor from a reliable supplier. <S> It is probably saturating. <S> Saturation will also cause the LM2596 to heat up. <S> You could also pull one or two inductors off the ones you burned out and connect them in series with the existing one. <S> If inductor saturation is the problem, converting in two stages may not be of any help. <S> I suspect it's the high input voltage with a relatively small value inductor that is better suited to a 12V input. <S> The LM2596 is certainly fake <S> (a real recently purchased one would say TI on it), but may still be okay.
| The inductor getting hot probably indicates it is undersized for the relatively high input voltage you are feeding it. The 'LM2596' may be a fake. That is just poorly designed and/or not very suitable for your application.
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Why does the opamp output voltage saturate? simulate this circuit – Schematic created using CircuitLab Vout = V in (1 + R2/R1) According to this positive feedback topology - made an error here, it's Vout = Vin(-R2/R1) (edit) Now I read in a tutorial - http://www.electronics-tutorials.ws/systems/feedback-systems.html , that If the input voltage Vin is positive, the op-amp amplifies this positive signal and the output becomes more positive. Some of this output voltage is returned back to the input by the feedback network. Thus the input voltage becomes more positive, causing an even larger output voltage and so on. Eventually the output becomes saturated at its positive supply rail. How does this happen? Shouldn't Vout be constant at a particular value, because we aren't changing Vin or the resisitors right? Shouldn't the equation Vout= Vin ( 1 +r2/r1) always hold? Error Vout = Vin(-R2/R1) (edit) So how is the Vout increasing iteratively? <Q> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Non-inverting amplifier configuration. <S> Your configuration is non-inverting but has positive feedback. <S> This will give a Schmitt trigger effect. <S> Remember that the op-amp output will be \$ (V_+ - V_-)A \$ where A is the open loop gain and <S> typically > 1M. <S> If we apply V1 = 1 V as shown in your schematic when the output is zero <S> then \$ V_+ \$ goes to 0.5 V. <S> The output will then try to go to \$ (0.5 - 0)1 <S> M \$ \$ = <S> 500,000~V \$! <S> It will get as far as the positive supply rail and stop with <S> \$ V_+ \$ <S> half-way between 1 V and the supply voltage. <S> This will hold the opamp in that condition indefinitely. <S> Schmitt trigger , ʃmɪt/, noun , ELECTRONICS: a bistable circuit in which the output increases to a steady maximum when the input rises above a certain threshold, and decreases almost to zero when the input voltage falls below another threshold. <A> Just look at what you've done - you are mixing up the circuit of an op-amp with positive feedback with the theory of an inverting op-amp with negative feedback. <S> Look at the two scenarios in the link you gave - in the negative feedback scenario <S> the gain is -Rf <S> /Ri (the article says Rf ÷ Rin <S> but there should be a negative sign). <S> In the positive feedback scenario they say: - <S> Eventually the output becomes saturated at its positive supply rail <S> And this of course is true when the input signal is positive enough to exceed the schmitt threshold imposed by the positive feedback resistor and input resistor (otherwise it will be saturated negative). <A> " So how is the Vout increasing iteratively? " <S> It is not possible to demonstrate the saturation effect as a sequence of several steps ("iteratively"). <S> The input voltage of 1V (at first, without taking the feedback path into consideration) will appear at the non-inv. <S> opamp node and will bring the output immediately into saturation (pos, supply voltage Vcc). <S> Now - we have two voltage sources at both ends of the resistor chain: <S> At the left Vin and at the right Vout. <S> That means: <S> The voltage at the non-inv. <S> node will even be larger than 1V and there is no reason that Vout changes its value - it remains at Vcc.
| Vout= Vin (1 +r2/r1) is the equation for a non-inverting op-amp with negative feedback.
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TVS for 48v system I have a 48v battery powered system is charged by normal lead acid charger that might reach up to 14.5v and maybe 15v per battery ( total 58v to 60v ).The circuit contains components rated for 75v and 100v . As everyone else i get confused by the various ratings and how they are related to my needs: 1. reverse standoff voltage 2. break down voltage 3. maximum clamping voltage I would like to know which TVS fits 48v circuit (voltage rating wise): from littel fuse 1.5KE series I cut this image : my first impression is to use the 1.5KE68A because the reverse stand-off voltage is approximately equal to maximum Vcc, but the maximum clamping voltage is way above the 75v rating of the regulator. Even the 1.5ke56 has clamping voltage of 77v. Question:Which TVS should i choose ? or is this a limitation of TVS and i should consider other type of clamping ? Note for whom might consider this a duplicate: i check 3 posts on this site but have not found the right answer. TVS Diode selection Choosing TVS Diodes TVS specification understanding <Q> Since OP added information on the actual circuit, here's one solution. <S> LM5118 is an external switch controller <S> so there's no need to use the same input voltage as you're using for switching. <S> I'd put a 22R series resistor to the switcher input and clamp it with a 68V 5W zener diode. <S> This would comfortably protect the circuit @ 88V input voltage. <S> At least in theory. <S> In practice the zener diode would overheat and die unless you had metal clad PCB or perhaps a solder-on heatsink to dissipate all that heat away. <A> If I interpret the OP correctly, you're concerned about your battery charger output stage. <S> That shouldn't require a high-power TVS at all, the regulator chip is responsible for keeping the voltage withing boundary conditions. <S> Main concern would be sparking on connecting the plugs that is not a high energy source. <S> But on TVS chips: Here's an appnote from Microsemi that explains it in a fairly straightforward manner: http://www.microsemi.com/document-portal/doc_view/14650-how-to-select-a-transient-voltage-suppressor <S> Simply put, the part you want does not exist. <S> Standoff voltage should be equal or higher than the max normal operating voltage, in your case 60V. <S> So the 64V part is the way to go. <S> On the other hand, max breakdown voltage on that part is 79V which may or may not keep the 75V regulator alive. <S> You would probably get away with using that 58V part but the min breakdown voltage of 64.6V is awfully close to the actual operating voltage. <S> Finally, the clamping voltage applies to sudden transients, think lightning and other spikes in the power network. <S> Spec says that if you have transient current of 14.8A, the part is guaranteed to keep voltage below 103 volts. <S> "Other clamping solution" is probably pretty hard to find. <S> You'd need a comparator to trigger the clamping part (probably a mosfet) with enough precision. <S> But how will you power that part? <S> Yes, you could use a 9V zener diode or something <S> but then you're dissipating 51V on the series diode(s).. <A> This picture should help: - Stand-off voltage relates to the maximum normal working voltage your system might see and, according to the detail in the question, this will be as high as 60V. <S> However, in your specific situation I would look at the break-down voltage specification. <S> In the Littel Fuse table it tells you that Vbr is measured at 1 mA and this is a really insignificant current for a large battery <S> so, maybe you could start to consider the 1.5KE56. <S> However, the maximum clamping voltage could be 77 V so this puts in jeopardy components only rated at 75 volts. <S> The next device down has a clamping voltage of 70.1 volts and this would be suitable for use as a protection device <S> but how much current will it take when 60 V is applied across it? <S> You cannot make a decision on the information given. <S> Maybe 10 mA leakage current (Ibr) is acceptable and maybe the 1.5KE51 could be chosen through manual select-on-test procedures.
| Since the system is battery powered there should be no chance of actually exceeding 60V on steady state operation to start with. I would consider trying to get components rated at only 75 volts up to a rating of 100 volts.
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How do you shift a square wave down? I have a square wave the oscilates at 25khz from 0 to 30v. I would like it to oscillate from -15v to +15v. Is there a SUPER DUPER SIMPLE circuit to do this? Keep in mind it's going up and down at 25khz. Here's the circuit. The square wave I want to shift is going into the PiezoSpeaker at the lower right of the circuit. <Q> There are several simple options. <S> Rail-to-rail op-amp <S> If you have a rail-to-rail op-amp, this can be done using four resistors as follows: <S> This will divide the signal down to 0-15V which is in the range of the supply of the op-amp, and then compares it with half of 15V (which should be the new midpoint). <S> There is no negative feedback which means the output will be either of the supply rails. <S> If you supply +/-15V to the op-amp, this will output a +/-15V signal. <S> Transistor Level Shifter <S> A second options is to use a PMOS transistor/resistor level shifter followed by a CMOS inverter to correct the inversion. <S> You need to make sure the transistors are happy with 30V+ on the Vds, and also the Vgs. <S> This is quite high for MOSFETs, but you can get 40V rated ones, so not impossible to find. <S> AC coupling <S> The capacitor acts as a high pass filter which removes the DC bias. <S> You have to be careful however that the capacitor is rated for at least 45V. Also if you use a polarised capacitor, make sure the positive terminal goes to the input signal as that will always be the higher potential. <S> The downside to this approach is that if your signal ever goes away, the output will float to the mid supply (0V) which may or may not be ok for whatever you are connecting it too. <S> Additionally, a 30V zener diode is required on the output to prevent it rising to 30V during the initial transient (while the capacitor effectively "measures" the DC offset of the input signal). <S> You can see from the simulation waveforms how the initial transient looks with the zener diode in place (top trace) and without (bottom trace). <S> Notice how without you get a 30V signal initially. <A> How do you shift a square wave down? <S> Just an RC will do it: - If fed with a sine wave (for the convenience of the attached picture), it will reproduce the sine wave at the output but with the DC level removed thus maintaining the correct peak-to-peak amplitude. <S> Here is what happens in the first few cycles as the DC level is removed: - It will work in the same way as a square wave (DC or average value = <S> 15 <S> V will be removed) but, please ensure that 2\$\pi\$RC << than 40 us to ensure the following doesn't happen: - If C = 10 uF and R = 1 kohm, the cut-off frequency will be 15.9 hertz and miles away from causing the droop problem seen above. <S> However, if you in fact have a non 50:50 duty cycle square wave you may need a more complex circuit because the DC level won't be at the midpoint of the waveform. <S> On the other hand, if you are driving a simple piezo transducer, why bother shifting it to equal positive and negative levels - it's purely capacitive and it won't care so, a really simple circuit becomes "a wire". <A> A simple circuit using a few jellybean BJTs: <S> It won't get quite to the rails, maybe <S> +/-14V with 15V rails, depending on the load. simulate this circuit – <S> Schematic created using CircuitLab <S> No fuss, no startup transients, can't go beyond the rails, it will just work. <S> Edit: If you are really only concerned with removing the DC bias on the piezo, simply add a series capacitor such as a 1uF ceramic and a resistor of, say, 100K across the piezo.
| If your signal is a continuous waveform, you can use a simple AC couple and rebias circuit - which consists of just a capacitor and two resistors as follows:
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Dynamo phone charging leads to discharging I am trying to make my bicycle dynamo charge my phone. Currently, I've got dynamo (outputs 6V AC), connected to a rectifier, then capacitor (2200 microfarad), regulator (outputs 5V), capacitor (2200 microfarad), and usb extension cord to which I plug my phone. simulate this circuit – Schematic created using CircuitLab The diagram is wrong Regulator: On the output I get stable ~5.2 V When I start pedalling, my phone begins to (agressively/quickly) discharge. Even with the turned off phone, ~5 min of pedalling discharges the phone by 30-35%.Having said that, the charging icon is ON on the phone, so it detects the current Phone model: LG Nexus 4 What may be the problem? Tested on bunch of lamps. All work fine P.S. my background is not in electrical engineering <Q> Okay to make this a bit more clear. <S> To prove that the phone is discharging through the UBEC. <S> Install a diode in series on the BEC output. <S> Ideally a Schottky like 1N582x <S> although a general purpose diode like 1N540x should be okay. <S> Picked those purely as easy to obtain parts. <S> Check that the voltage is okay, as there will be drop across the diode. <S> Hence the preference for a Schottky. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This is a crude fix, but if the phone is no longer discharging this will prove it. <S> Your UBEC is a buck DC-DC converter <S> , this means it drops voltage down but does not work on voltages lower than the output with some dropout, in this case the voltage in has to be 5.5VDC or higher on the input. <S> My concern which Transistor has proven with maths is that your smoothing capacitor after the bridge voltage is potentially dipping below this voltage on each cycle. <S> Essentially what is mentioned with ripple. <S> ( http://www.electronics-tutorials.ws/diode/diode_5.html ) <S> So essentially those 'troughs' <S> as the capacitor discharges is the ripple, so the voltage may be dipping above and below the 5.5VDC input for your UBEC. <S> Also my other concern was the dynamo voltage output may be dropping below that required for the 5.5VDC on the UBEC, <S> say if not pedalling, is there a flywheel or anything on the dynamo ? <S> It would be useful to measure the voltage output of your dynamo, after the bridge rectifier and into the UBEC. <S> Have you tried this circuit with a standard DC power supply such as a plugpack to confirm functionality, the UBEC is rated for 5.5-26VDC so 12 and 24VDC are both very common, could even use a battery ? <S> As has been mentioned, you may need to look into resistors for pull-up or pull-down or across the data lines. <S> These various tests should help determine exactly where the problem actually lies. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> The corrected schematic. <S> Please check that you have wired as shown in Figure 1. <S> If you can confirm this we can proceed to the next stage of debug. <S> You should probably use something other than an expensive phone to test the circuit. <S> A 6 V bicycle lamp might be a good choice. <S> Figure 2. <S> The output voltage and frequency of Shimano NX-30 hub generator. <S> (Wheel diameter is not specified.) <S> Source: Electronic Power Management for Bicycles . <S> Figure 2 shows that at normal cycling speeds the output frequency of the generator is considerably lower than that of the 50 or 60 Hz mains. <S> This needs to be taken into consideration in sizing C1 (Figure 1) smoothing capacitor to keep the ripple-voltage low enough for your circuit to work. <S> Ripple voltage is calculated as \$ V_R = \frac { <S> 1}{2fC} \$ where f is the frequency. <S> If we decide we want less than 0.8 V ripple at 30 Hz (17 kph) and 1 <S> A load then we can calculate C1's value as \$ <S> C = <S> \frac {1}{2fV_R} = \frac { <S> 1}{2 <S> \cdot 30 <S> \cdot 0.8} = <S> 20,800~\mu F\$. <S> Any lower capacitance or lower speed will increase the ripple and the UBEC output will drop out. <S> Your C1 value is one tenth of this and may be the first problem with your circuit. <S> Read Electronic Power Management for Bicycles for more details and graphs. <S> I don't understand how the phone is discharging through the USB port as you claim as it shouldn't be sourcing power. <S> Can anyone enlighten me? <A> Purely for testing. <S> Try putting a diode in series out of the UBEC. <S> Ideally a schottky but general purpose rectifier may be okay, just check the typical and max current rating. <S> Just check the voltage on the output under load first. <S> If too low maybe do a quick and dirty string of several diodes in series and change the output to 6V and check your output agsin first. <S> Should prove a theory, don't have the design finesse to fix it though. :) <S> Edit: <S> My theory is based on this manual: <S> https://www.google.com.au/url?sa=t&source=web&rct=j&url=http://www.hobbywing.com/products/enpdf/UBEC3A2-6S.pdf&ved=0ahUKEwiXhce4pr3OAhVJjZQKHWgnCFMQFghuMA4&usg=AFQjCNF7nim20wiBn4Fb73q7jZok91T5Cg&sig2=FaTgMlymjIC_Aeb7NFd7sg <S> It requires an input of 5.5VDC. <S> I'm a tech, not an engineer <S> but I think that the smoothing cap voltage is dipping down enough every cycle and the phone is feeding power backwards into the BEC. <S> Just a thought. <S> Something vaguely along those lines. <S> Seems to be a buck only. <S> Might be an idea to make some measurements on the input side, before and after bridge and after the cap. <S> Also have you tested it with a regulated 6VDC supply? <S> Maybe also long term consider a battery or other storage as well as part of your design.
| Could also be that your dynamo voltage is lower or not consistent.
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Splitting voltage and current unevenly I want to make my lab variable power supply from an old laptop power adapter (19V 6A) with an LCD to show Volts and Amps. The LCD needs an independent powersupply. Can I make a circuit to give me the power from the laptop supply in a way that is 'independent' or should I find another solution? This is what I have in mind. Just a couple of Buck Converters, one of them will have the POT's moved to the front of the case and the other small one to power the LCD, fixed at 5v. simulate this circuit – Schematic created using CircuitLab <Q> You need to clarify what you mean by "independent power supply" <S> One of them with low current (assuming LCD draws milliamps) and fixed voltage. <S> The other a variable voltage, where the laptop's 19V are reduced using either a linear regulator (simple but difficult to construct for larger amperage), or a buck regulator - much more difficult to implement but can handle larger current without an excessively big heatsink and fan. <S> If you mean the LCD needs to be galvanically isolated from the voltage provided by the lab supply - it is going to be very difficult. <S> You need to convert DC to AC, pass it "through" a transformer, get it back to DC. <S> This is a tall order for anyone who doesn't specialize in PSUs. <S> Maybe you are actually asking on how to measure the voltage / current of your supply, and display those on an LCD. <S> You should improve your question by adding a circuit you are considering for the actual "supply" part, and then we could help about the LCD portion. <A> Figure 1. <S> Battery powered LCD voltmeter. <S> If you're stuck with only your 19 V laptop supply I would recommend that you purchase two LCD voltmeters and power each from a 9 V battery. <S> This means that you can connect them on the high side or low side of your bench power supply to measure voltage and current as required without risk of damage through common grounds, etc. <S> These typically have very low current consumption and the battery will last for a long time. <S> Alternatively you could use a couple of 6 to 9 V wall warts to provide independent power to each unit. <S> Most of these meters read full-scale at 199.9 mV <S> so you need to add a voltage divider for the voltmeter and a shunt for the ammeter. <S> Since your PSU is rated for 6 <S> A your ammeter will read 6.00 at 6 A. <S> This will occur at 60 mV <S> so your shunt will need to be \$ R = <S> \frac <S> {V}{I} = \frac {60m}{6} = <S> 10~m\Omega <S> \$. <S> If you can find a meter such as shown in Figure 1 you can enable the 'V' and 'A' units on the display which is a nice touch. <A> Knock the 19V down to a regulated 12V with an LM7812 and then use an isolated DC-DC converter module to give you +5 or +/-5 or whatever your meter modules require. <S> An LED meter might use 0.5W <S> so the LM7812 will only dissipate a couple hundred mW <S> and no need for a heat sink for powering one or two DC-DC converters. <S> LCD meters will probably use even less, depending on the backlight consumption. <S> You could also make the DC-DC converter from scratch and wind your own transformer, say on a pot core, (perhaps omitting the LM7812) <S> but it's not too cost effective for a one off-- the core, bobbin, wire and a suitable chip will likely cost more than the module.
| If you mean the LCD needs an appropriate voltage to run, and that voltage must be fixed, and independent from the voltage output by your power supply, then yes - you can just use your laptop adapter to generate two separate output voltage lines. The only alternative to power from your laptop supply is to use an isolating DC/DC power supply which may be more trouble than it's worth.
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Is there a relay to digitally check if mains is available I am working on a system that will remotely turn on/off an appliance (RaspberryPi running a web server, sending input to a relay). One problem that I still need to get sorted is that, how can I check if the electricity is available. Is there some kind of a relay (or other component) that I can use to tell my RaspberryPi that electricity is available for the appliance to be turned on/off. <Q> Yes. <S> You can use a mains relay and use a contact of it to indicate that mains is present. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Mains present indication. <S> Enable the GPIO pin internal pull-up resistor. <S> There are other ways using opto-isolators, etc., but since you are already using relays it makes sense to use another. <S> The isolated or "volt-free" contact provides isolation to prevent mains getting near your Pi. <S> The relay coil should be AC type with voltage selected to suit your mains voltage. <A> There's many ways to do this, some more dangerous than the others. <S> How about just taking a 3 volt wall DC supply and connecting the output to your Raspi? <S> When there's power available, the output will be at 3 volts. <S> And when not, it will be at zero. <A> These are called "27" relays because the ANSI function number for loss of voltage is 27. <S> You will not have to worry to much about the specs for the contact on the relay as this will be feeding an input to your GPIO pin on the Pi, so will be 3.3V at very low current. <S> However, to separate the GPIO pin and the mains power you will want the contact to be "dry", meaning you need to supply it with voltage. <S> So I think the circuit diagram in transistor's answer should look like: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> A link to a common relay here in the midwest is: A-B relay <S> You can probably find cheaper relays, but this is similar to what I think you would want. <S> Of course, you will have to match the type of relay for your installation, i.e., flange, socket, panel mount, etc. <S> but this is just to give you an idea. <S> Note, I used what's called a form "C" relay, meaning you have normally open and normally closed contacts - if voltage is good, the normally open contact will close and the normally closed contact will open. <S> That way you can wire it up whichever way you want (good mains voltage provides a high signal at the Pi, wire to the normally open contact; good voltage provides a low signal at the Pi, wire to the normally closed contact.) <S> As transistor also stated, you will need to configure the pin as an input and make sure you have the right pull-ups configured.
| Any standard relay should work, just match the coil voltage to the voltage you are working with as transistor stated.
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Can We connect LED on AC line I see that LED light is replacing nowadays every old style light. I open one LED bulb and found that had a little SMPS inside it. Those convert AC line voltage to DC to supply the LED. Wondering we can design to connect LED's on directly on AC line . It will improve efficiency As per attached circuit. If we can connect LED on Direct AC line Will it work? Will it be safe and reliable? Whether it needed any limiting resistor, since all AC line voltage will be drop on LED . (Vline = 230 V AC, Vp = 230 * 1.414 (+15% tol) = 360 V approx. Total no led is 120 (changed 76 to 120), each led has 3V forward drop). How much line fluctuation will effect on LED? <Q> 230V is RMS value. <S> Peak value is 325V. <S> So you would need 109 LEDs, not only 76. <S> And 230V is not precise. <S> It can be ±15%, so peak voltage can be between 290V and 360V. <A> 230V AC means ~325V at peak. <S> So 325V/76 <S> = 4.27V. <S> At that voltage and with virtually unlimited current LED would not have a chance to survive. <S> :) <S> No, look explanation above. <S> No. <S> They will change light intensity depending of the voltage. <S> Keep in mind <S> they are current driven devices, so very low light if voltage is below at least 2.5V. <A> If an ideal diode is given 17 mV more voltage, it draws twice as much current. <S> Only very non-ideal diodes would survive in the AC connection shown, because an AC line may (even if there's local protection) have occasional peaks, nearly 1Vacross each diode. <S> Even a line-voltage allowed variance, of 10%, would makea normally biased (3V) <S> LED see 3.3V, and that causes a current excursionof a factor of 2**18 (about 250 thousand times) <S> the 'normal'. <S> What one CAN do, is put a series inductor, similar to a fluorescent lampballast, into the circuit. <S> Then, instead of 10% voltage deviation causing25 000 000% current increase, it causes 20% current increase. <A> Will it work? <S> It will work for some time, perhaps. <S> Figure 1. <S> An graph of LED current versus voltage. <S> Note the huge variation in current for a small variation in voltage. <S> Will it be safe and reliable? <S> No. <S> The current will vary with voltage. <S> A voltage spike or rise will cause a disproportionately large change in current - possibly to the point of destruction. <S> Also a reduction in voltage will result in a large dimming of voltage. <S> Whether it needed any limiting resistor, since all AC line voltage will be drop on LED . <S> (Vline = 230 V AC, Vp = 230 <S> * 1.414 <S> (+15% tol) <S> = <S> 360 V approx. <S> Total <S> no led is 120 (changed 76 to 120), each led has 3V forward drop). <S> Yes, current limiting is required. <S> How much line fluctuation will effect on LED? <S> See above. <S> The big problem is that the LEDs will only blink briefly at peak voltage. <S> You can see from the graph that if we were working at 2.75 V per LED at peak mains voltage that we would get 30 mA. <S> When the voltage drops by 20% the LED would effectively be turned off. <S> There would be severe 100 or 120 Hz flicker. <S> The SMPS supplies current limited DC which solves all these problems.
| Yes, at least some current limiting resistor is needed. It is not very good to rely on mains voltage.
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Sending 120VAC to a device for a certain amount of time after a single button press I am trying to use a push button switch to send 120VAC for a certain amount of time after it has been pressed. I want it to only be pressed once, and not have to be held in the closed position. Is there some sort of timer/timed relay I could use to accomplish this? <Q> You can buy buttons that do exactly that . <S> They are often pneumatic, holding themselves in until enough air has leaked in to release them. <S> They're used for lighting stair wells, etc. <S> Alternatively a 555 timer in monostable mode driving a relay, or a small microcontroller doing the same job (easier to make configurable and get more precise times). <A> Purely as an example: http://au.rs-online.com/web/p/products/8285161/?grossPrice=Y&cm_mmc=AU-PLA- -google- -PLA_AU_EN_Automation_And_Control_Gear-_-Timers_And_Counters&mkwid= <S> scbgdPzdy_dc|pcrid|99325705834|pkw||pmt||prd|8285161 <S> This may not be exactly what you are after <S> but you can generally get various timer schemes or multifunction units in various voltage ranges. <A> There are industrial timing relays that can do this, in addition to the pneumatic ones that Majenko mentioned . <S> From the outside, they appear to work similarly to traditional relays, except that there's a (usually programmable) delay between the "coil" receiving power and the contacts moving. <S> Here's one possible way to wire one: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> When you push the button, the load and both coils get power. <S> The normal relay closes immediately and maintains power when the button is released. <S> The timer relay does nothing for the programmed time. <S> When it finally moves, it removes the normal relay's ability to hold the circuit on, and so everything resets.
| You can also get various DIN and panel mount timer relays that do this.
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Does fixed point number processing require ASIC hardware? I've recently started diving deeper into DSP and have come across the term 'fixed point number'. The idea of a fixed point number is simple enough and makes since to me, however, I'm somewhat curious as to how fixed point operations are carried out. Are fixed point operations carried out by processors with instructions to perform fixed point operations? Or is there usually a dedicated chip? Thanks! <Q> Fixed point operations are carried out as integer operations, possibly with bit shifts to get things lined up. <S> The whole point of fixed point is to get some of the benefits of floating point math without the performance penalty, especially on CPUs that do not support hardware floating point math. <A> I was using fixed point back in the Intel 386 days when processors didn't have floating point hardware. <S> It's a very effective way of using integer units to represent numbers with fractional parts, ideal for low-accuracy 3D and audio processing. <S> Let's do it in base 10. <S> Suppose we have a fixed arithmetic unit capable of handling 2x four decimal digit inputs multiplying to eight, and you want to compute 4.2 <S> * 3.14 (\$\pi\$ rounded). <S> We move the decimal point two spaces to the right, and our operands are now 0420 and 0314. <S> Do the multiply, to get 00131880. <S> Shift the decimal point back four spaces to the left to get 13.188. <S> It's exactly the same in binary: if you can operate on 32-bit numbers, then you can do fixed point with a shift of the binary point by 8 or 16 bits. <S> Every time you do a multiply you end up incorporating the shift from both arguments, so you need to shift the result back. <S> Bitshifts are fast. <S> Addition and subtraction are transparent and don't require extra bitshifts. <S> (Don't do division, division is hard) <A> Fixed point on normal processors works perfectly well, provided (a) <S> the existing word widths (typically 8,16,32,64 bits) satisfy your needs, and either (b) your programming language supports it directly <S> or (c) <S> you keep track of the binary point yourself. <S> If you need custom word lengths, the compiler (if it supports fixed point) will use the next largest type (or return an error if there is no suitable type) <S> Option (b) - a language which supports fixed point - is by far the easiest way. <S> The code it generates just uses the integer instructions, which can be more efficient than floating point instructions (and especially, much faster than FP emulation on CPUs without an FPU), and keeps track of the point, shifting where necessary. <S> Alternatively, you can implement your own hardware on FPGA, there is no need for ASIC. <S> VHDL-2008 incorporates synthesisable fixed and floating point packages .
| No special hardware is required to use fixed point math, certainly not an ASIC unless you're trying to do something very specialized.
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Why is kerosene stopping my red LEDs from illuminating? I initially posted this on chemistry.stackexchange but didn't get any answers, so I'm reposting it here. Long story short - we have an electronic product that is submerged in fuels (kerosene being one of them) and uses an RGB LED ( click here for datasheet ). Due to a sealing problem in the enclosure, kerosene has managed to get in and cover the PCBs. What's interesting is the effect that has had on the PCB. The PCB's functionality has been completely unaffected, apart from the fact that the red LED in the RGB LED module has completely stopped illuminating. We've replicated this ourselves manually by submerging 2 new PCBs in kerosene for a day and then taking them out and powering them up and seeing that the red LED stops illuminating entirely. The green and blue LEDs continue to illuminate just fine. Examination of the failed boards shows that there are no other electrical faults. It is just the red LED that completely stops illuminating. We measured the forward voltage across each of the LEDs in the failure condition, but didn't notice any significant difference that would explain the fault. After leaving the PCBs to dry, the red LED starts working again. So the problem is not permanent. Looking at the last page on the datasheet, the LED material is listed as AlGaInP / GaAs . Is there any obvious reaction between kerosene and these materials that would explain why just the red LED stops working? Update 1 : I've carried out the following experiments: Dripping kerosene on to the LED. Submerging the PCB+LED in kerosene while running. (Videos to follow up later on today, hopefully) In both cases, there was no perceived effect on the LED - it continued to operate just fine. This would seem to indicate that the problem is not purely an optical problem between the kerosene and the LED. So far, the problem has only occurred after soaking the LED in kerosene for some time. Update 2 : I've taken a fresh PCB with LED on it (haven't done any tests yet with just the LED) and soaked it in kerosene. I've taken some close up photographs of the LED before soaking, after soaking while it's not working and after it resumes working after it's been left to dry. What the photos show is that there is a very obvious bulge in the LED lens during the period when it's not working. Once the bulge recedes, the LED illuminates again. Unfortunately, I don't have a camera set up on the PCB to see the exact moment that it stops working. I'd let it soak for about an hour before it stopped working. I checked on the LED every now and then and noticed no change in the LED brightness. I came to check it once and it was just off. My suspicion is that the change is sudden. Judging by the swelling, I'm going to guess that there is some mechanical damage internally that's moving something and once the swelling recedes it springs back to position. Left: Kerosene-soaked LED; Right: Normal LED LED in failed state after soaking Normal LED Left: Kerosene-soaked LED after being left to dry and in the working condition; Right: Normal LED Kerosene-soaked LED after being left to dry and in the working condition <Q> We measured the forward voltage and didn't notice any change. <S> Physically, I'm pretty sure <S> this means the semiconductor interface is still producing photons at the same rate and wavelength as before. <S> So, something happens to those photons. <S> What you should do is get a working source of red light of the same wavelength (e.g. another one of your LEDs), extract the "lens" material from a "donor" LED: e.g. by cutting it off with a razor blade, testing transmission of red light before and after having soaked that material in kerosene. <S> Since that lens is tiny, you should probably use something like a piece of cardboard with a hole punched through it with some kind of needle <S> (don't let the hole get to small, lest you want to have much diffraction...) and put the lens in front of that hole. <S> My guess is that soaking the material in kerosene leads to a drastic change in optic properties, and that might very well mean that either <S> your lens is now absorbing red light or your lens is now not focussing red light, but spreading it. <S> To rule out 2., you'd need a very very dark room and some way to guesstimeasure the distribution of light. <S> So in effect, without optical design lab equipment, either way, Kerosene contains a mixture of different hydrocarbons, and those are soluble in other hydrocarbons, such as the transparent material used to protect the actual LEDs and act as a lens. <A> My 5 cents: <S> Most LEDs are potted with silicone today. <S> Silicone has a good permeability for VOCs (volatile organic compounds, e.g. alkanes and their isomeres), which are part of kerosene. <S> VOCs entering the silicone can interact with the silicone matrix, changing its optical properties. <S> Often observed damage: the potting/lens can get milky or diffuse, and yellowing can be observed. <S> Certain VOCs will be broken up by the blue light of an LED which usually leads to a blackening of LED potting/lenses. <S> Those effects are known to be (partially) reversible. <S> I.e. the discoloration of the lenses will disappear if the VOCs are able to gas out again. <S> This happens faster if heated under the LED's operating conditions. <S> So my explanation is:Edit: Highly speculative Large amounts of kerosene may contain also aromatic compounds, which are known to be optically active <S> (e.g. See azo dyes pigments ). <S> Van der Waals forces can change the resonant behaviour of aromatic compounds, which is possible when the VOCs enter a silicone rubber matrix. <S> This could explain why fractions of the kerosene achieve a red filtering behaviour when entering the potting. <S> Edit:I cannot rule out interaction of VOC with the semiconductor itself, but I habe difficulties imagining how this could work. <S> The crystal is nearly impermeable for anything at room temperature hence the interaction can only happen at the surface of the dice. <S> Because light emission happens everywhere near the pn-boundary <S> I doubt that kerosene components can prevent the generation of photons. <S> IMO only absorption and filtering are the effects to look after again. <S> Another culprit in LED deterioration is hydrogen sulfide, which can be found among other sulfur compounds in kerosene, too. <S> But sulphur corrosion in LED isn't reversible AFAIK, so this can be excluded IMO. <A> So you have the double effect of absorption and dispersion of the red photons. <S> There is also the possibility that at some point, the heat produced swelling of the plastic, creates a high resistance connection, which goes back to "normal" after the LED dries.
| My guess is that the kerosene absorbs the red photons and heats up the plastic lense causing it to bulge, which in turn causes dispersion of the photons.
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USB powered and battery powered on the same connector I need to make this Arduino Nano-like board with one single connector: a mini USB type B which will of course be connected to the PC for programming and interface. Additionally, as it will be used away from the PC, I need to be able to connect a 6V+ battery pack using the same connector. There's no way I can fit a second connector for this. Usually the FT232 (or similar) won't take more than 5.5V, so I've thought of placing an LDO between the USB-VCC input and the rest of the circuit. The LDO would drop about 0.3V but solve my overvoltage problem. It could take 9V easily. My question is: will the voltage drop cause problems with the USB communication (since the FT232 will get power from a slightly lower voltage source)? <Q> If you are using a FT232R chip, this should work, provided you have a minimal voltage of 4.0V at the output of your LDO with all conditions. <S> Other FT232 chips might have similar properties, check the datasheets. <S> The datasheet of FT232R states for VCC on page 11: +3.3V to +5.25V supply to the device core . <S> However, in Note 1 (page 12), they precise that to use the internal oscillator (I suppose you want to save the space for a crystal) you need a minimum VCC of 4.0V. <S> Thus, the manufacturer ensures that the FT232R will reliably work, without an external crystal, with a VCC between 4.0V and 5.25V. <S> This of course includes the USB communication with the USB host, which is of course an essential capability of the chip. <S> If you look at Chapter 6.2 of the datasheet (self-powered configuration), you can see that the self-powered circuit can be supplied with any voltage between 4V and 5.25V and still reliably communicate with the host. <S> There's no reason this would not remain true if the power is derived from the USB supply voltage instead. <S> Although not mentioned in your question, another point you need to consider is the power supply voltage of the ATMega chip (I guess you will be using a ATMega168 or ATMega328, if you want to be compatible with Arduino Nano). <S> Everything will be fine with your 6V external power supply as you will end up with 5.0V power. <S> However, if you power from USB, the LDO will introduce a drop on the V BUS voltage, which is 5.0V nominal. <S> If you use a 0.3V drop LDO, you will have 4.7V on the microcontroller, which is still fine. <S> However, if V BUS goes below 4.80V, your VCC will drop under 4.50V, which is the minimum required voltage to operate at full speed (20 MHz), and the maximum frequency decreases with the voltage (see Chapter 33.3 of the datasheet for the ATMega 168P) <S> In this case, you will have to operate at less than 20 MHz. <S> If you use the 16 MHz typical of Arduino boards, you should always be in the safe area. <A> Usb Power is typically 5V +- 5%, or 4.75 to 5.25 volts to be with spec. <S> USB data is 3.3V. <S> The FTDI has an on board regulator to drop it down, as does the Arduino board. <S> As the FTDI and ATMEGA both have very wide latitude in voltage range, dropping it to 4.7 (6%) wouldn't be an issue, as the FTDI uses a good LDO internally. <S> Many usb devices simply use a LDO and run off 3.3V instead. <S> That said, the solution is simpler. <S> Add the LDO to your battery pack before the usb cable you will be cutting up. <S> Alternatively, use a USB power bank . <S> 5V out, rechargeable, come in a variety of shapes, chargeable by any usb power supply or computer, etc. <S> (Mind you, some have a minimum current draw that a simple Arduino wouldn't draw, so you need to shop around) <A> According to FT232 specifications, VCC pin can hold +6V. <S> If you decide to use a LDO between VBUS and FT232, you can safely set the LDO output to +4V, which should be perfectly fine, all in accord with specifications. <S> Or even to +3.3V if you are using external crystal for FT232. <S> In both cases USB D+/D- signaling should meet USB signal specifications. <S> Addendum: to be on a safe side, an inexpensive in-series diode can be used to meet both ends of the particular voltage conditions.
| So you do not need to do anything, except maybe you need to limit the VCCIO pin to a somewhat lower level than 6V.
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Are SPI slave select lines hardware enhanced? I have a question regarding SPI communications. I feel like I have a good understanding fundamentally of how SPI works. However, I'm often confused when implementing the slave select line of SPI. Is the slave select line on a microcontroller, in general, enhanced via hardware? That is, is there anything different between a slave select output and an output controlling an LED? Do microcontrollers allow certain I/O to be toggled faster when used for SPI? <Q> It depends on the microcontroller. <S> I tend to use the GPIO interface in these cases as it allows for more flexible routing. <S> But I've also seen SPI controllers that do more advanced stuff like managing the SS lines per transfer packet. <S> That comes in very handy if you communicate with multiple slaves on the same SPI bus via DMA. <S> You just send out the data via DMA and the controller will do all the switching between slaves at the correct time without needing any CPU attention. <S> Other notable things I've seen: <S> Slave Select logic which just latches the state-change and delay this request it until the internal transfer FIFO is empty. <S> This is very useful if you're just doing writes and don't care about the MISO data. <S> You just write out your bytes to the SPI bus, de-assert the Slave Select and you're done. <S> No risk in breaking the last byte because you've switched Slave Select to early. <S> Slave Select logic that allows for various SPI related protocols like TI-SSI, I²S etc. <S> Slave Select logic that automatically inserts an idle-bit between each transfer unit. <S> Not sure why they do this as it makes SPI unusable for lots of SPI flash chips <S> (Yes, I'm looking at you, NXP). <A> No difference, as far as drive strength or transition speed are concerned. <S> Some uC's I've worked with <S> don't even have a dedicated SS pin. <S> You can implement it in code using whichever pin is convenient. <S> However, some microcontrollers will toggle the SS line for you (without you having to toggle the pin in code). <S> This can reduce the dwell time between SPI transactions, decreasing the total time elapsed during multiple-transaction transmissions. <S> Also, if you are designing an SPI slave device, it is very convenient to use a uC that has a dedicated SS pin, which is used by the uC's internal SPI module. <S> The STM32F1 ARM-based microcontrollers, for example, have a dedicated SS pin for each of their SPI busses, with the option to disable the SS functionality and free up the pin for general use. <A> As others have said, typically there is no difference in the pins electrically. <S> However some have a dedicated SS pin, and some of those require the use of it if they have capability to be a slave device. <S> The SS pin may be linked internally to some function of the SPI module. <S> ATMega and ATTiny devices that have SPI hardware do indeed have an SS pin which has special purpose. <S> In this case if the hardware SPI module is enabled, it can only be a slave if the SS pin is set to be an input. <S> In slave mode the SS pin is used to tri-state the MISO pin to allow it to be used in a multi-slave system, so the pin must be an input during slave mode. <S> For master mode in this case, the pin direction requirements vary between parts. <S> On some parts it must remain an output (but can be user controlled as GPIO) in order to stay in master mode. <S> On other devices, it can be either an input or an output - however in these devices, if configured as an input, the pin must be held high in order to stay a master. <S> The latter option allows for multiple masters by switching into slave mode the moment the SS pin is pulled low. <A> This allows for a faster SPI transfer (and saves some CPU cycles). <S> But the select is one-per-slave, and needs to be changed only twice during a transaction, so it makes more sense to use aq GPIO for this purpose, and that is how the uC's I have seen work, <A> For more advanced protocols using the SPI hardware (most notably I2S) <S> the SS pin is generally controlled by hardware. <S> In the case of I2S it is often used as the LRCLK. <S> The SS pin is quite often specially linked with the SPI hardware in such a way that when running in slave mode the SS pin can automatically control reception of data, and even raise interrupts when it changes state.
| Microcontrollers often have special hardware to handle CLCK, MISO and MOSI. In most SPI peripherals that I have seen the slave-select management is doing the exact same thing as a GPIO.
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"Dumb" power from USB I'm developing a small micro-ohm meter that I will be powering from USB, due to the ubiquity of USB power bricks and computer power. I've run into a few questions: What do I do with the data lines of the USB? I'm not using them, I'm just interested in USB power, should I tie them to ground, to each other, to +5V? Do I need to switch from "low power mode" to "high power mode"? This seems to indicate that you need to ask for high power using something called "enumeration", but the same article also talks about a 2.25W max... which I know is no longer true. Max capacitance for the USB 2.0 specification according to the above is 10µF. I've got significantly more than that. This leads me to believe that I should be using something like the NCP380 , MIC2545 , or TPS2112 . These are high side switches that can limit inrush current, allowing me to have more capacitance after the switch. Is this still necessary? Or has the capacitance limit increased since the above linked document was written? <Q> For your particular purpose, you don't need to do anything, just connect GND (black wire) and VBUS (red wire) to your design. <S> (1) leave D+/D- <S> unconnected; (2) don't worry about any "low" or "high" power modes. <S> These modes are for host to supervise and police overall power that host believes is available to supply. <S> It never worked correctly. <S> If you don't pull any data wires up, the host will not even know if your cable is even plugged in. <S> (3) Do not worry much about 10uF limit. <S> In worst case you might get "port overcurrent" message on a host, but it is rare. <S> Or you might disrupt other USB devices that might be working in adjacent ports. <S> But there are plenty of DC-DC converters that use input caps of 10uF or less, so it is advisable to use a better DC-DC converter in your design. <S> If you are not using any DC-DC de-coupling from VBUS supply line, it is a very bad idea for a sensitive analog design. <S> Regarding the BC1.2, forget this completely. <S> All you need to care is not to exceed 500mA DC limit. <S> Clarification for (2): <S> The OP is not building a USB device. <S> The 100mA is a requirement for USB devices , not hosts. <S> The OP wants to use the port capability of a USB host. <S> Every host port must have an ability to supply 500mA, regardless if there are "low" or "high" powered devices. <S> So 500mA is always fine. <S> If the port will be behind a bus-powered hub, there could be a different story. <A> USB supplies come in a few flavours and standards. <S> Basically low power (portable hosts), negotiable power (most smart hosts and professional mains adapters) and brute power (DIY and many car type adapters). <S> If you do not plan to negotiate for extra power you can ignore the data pins altogether. <S> Some devices use a 5th pin to do other clever tricks sometimes in barely standard ways. <S> There were 3 common ways of signalling a desire for more current from a host when I was looking at this a while back. <S> The old way, the new way and the Apple ways. <S> If you look at IC datasheets that support these standards you will see what your device would be required to do to trigger the power boost if this is what you need. <S> If you have a post regulator with generous capacitance (before and or after) you should be able to use any type of supply and use what it can give. <S> EDIT: <S> Most of the brute force chargers supply full power all the time and while they are very practical they may not be as safe with short circuits as an intelligent supply. <S> The low power hosts are usually incapable of providing more than the USB minimum by design to conserve their own battery life. <S> http://www.nxp.com/files/analog/doc/data_sheet/MC34825.pdf?pspll=1 http://www.st.com/content/ccc/resource/technical/document/datasheet/5b/c8/25/15/fa/83/4d/66/DM00086386.pdf/files/DM00086386.pdf/jcr:content/translations/en.DM00086386.pdf <A> USB port provides power by default - this is necessary for the device enumeration. <S> Thus, you could simply draw the power from the USB power pins. <S> The standard unit load available is 100mA (for USB2). <S> No device is permitted to take more than this before it has been configured by the host (according to the standard). <S> In case your power requirements are higher than the default (100mA USB2 and 150mAUSB3), you would need to enumerate the device indicating higher power needs. <S> This requires the whole USB enumeration protocol, where your device will be communicating using data lines. <S> The capacitance problem is related to hot-pluging your device while other devices are connected to the same hub. <S> Thus, high capacitance device may case a voltage drop/ reset of other devices. <S> If this is not an issue, you should be fine. <S> So, if your device just uses a little power and does not require communication with the host - I would not worry about any of your questions.
| If your multimeter will not use USB communication, then you probably do not need to do anything with data line.
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Downconvert high voltage to match dry contact rating I have a few dry contacts on a device where the ratings are MAX 30V, 300mA.I want to connect LED indicators (24V, 20mA) to these dry contacts and was thinking to setup an external 24V power supply to feed the LED indicators when dry contacts close. However, there seem to be a reason for me to use 230V power supply, so I was wondering if it works using same LED indicators (24V, 20mA) and a resistor of 10kOhm in series? $$U = U_1 + U_2 => U_2 = 230 - 24 = 206V $$$$I = 20mA => R = U_2/I = 206V / 20mA = 10kOhm$$ I just get confused given the rating of dry contacts are given in volts... How should I know what voltage there is across the dry contact?(Consider all components and supplies being AC.) <Q> When you switch a circuit involving 230 VDC the contact voltage when open circuit will be 230 VDC, which is greatly in excess of 30 VDC. <S> Contact voltage rating must be >= <S> the voltage that will appear across the contacts when open circuit. <S> If you are using 230 VAC mains as the source of HV DC and rectifying it then Vdc will usually be around 1.414 x V_AC_RMS = <S> 230 <S> x 1.414 <S> ~= 325 VDC. <S> Your resistor will need adjusting accordingly.eg (325-24)/20 <S> mA = 15k <S> Dissipation in a series resistor = <S> V <S> x <S> I = <S> 206V <S> x <S> 20 mA = 4.12W in the 230 VDC case or <S> = 301V x 20 mA = 6W in the 325 VDC case <S> A 10 W rated resistor is required at minimum in the 4W case and > 10W rated ideally in the 6W case. <S> The resistor MUST be rated for at least the full DC voltage used, regardless of power rating and any other losses. <S> Numerous capacitor dropper supply examples are available. <S> Use of HV DC in the manner you describe MAY make sense in some cases, but use of a transformer r similar to provide a low voltage supply usually will make more sense. <A> Your circuit does not down convert enough. <S> The reverse voltage needs to be blocked with a diode or bridged with 4 diodes for less flicker. <S> The peak voltage exceeds 230Vac by 40% and the switched voltage far exceeds 24V. <S> A small opto-Thyristor would work with a 5W resistor. <S> EDIT <S> A doorbell or old furnace thermostat transformer @24V would be a better match. <A> You could drop 206 VRMS across a 10k ohm resistor <S> but you'd be dissipating a power of 0.02 amps X 0.02 amps x 10,000 ohms = <S> 4 watts. <S> You might also break your LEDs because if the power supply is AC, you would likely exceed the maximum reverse voltage of the LED. <S> Also, it's highly unsafe doing it this way - consider using a transformer to provide the correct voltage and a bridge rectifier and capacitor if a DC supply is needed for the LED. <S> However, the bottom line is that if the contact rating is 30 V it SHOULD not be used in a circuit greater than 30 V.
| IF AC is used as the 230 V source power dissipation can be greatly reduced by using a series capacitor to drop most of the voltage.
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Why does the voltage of vinegar batteries in series not equal the sum of the individual voltages? I'm working on an in-class demo / hands-on for my sons primary school class, and I've made some small batteries with New Zealand 10 cent (copper coated) coins, and zinc washers, and vinegar soaked cardboard. Each individual cell is measuring about 0.96 volts, but when I put 4 of them together, I only get out about 2.6 volts. I'm wondering if there is something I'm unaware of about the nature of these batteries that makes them not add up. Also, even at 2.6 volts, the same voltage as I'm getting out of a pair of eneloop AA's, the LED is not very bright at all -- compared to hooking it up to the eneloop AA's, where the LED is quite bright. Is this because of the low amperage of the vinegar battery? Would putting more in series make it better (or do I need to make a second one and hook them up in parallel?). I'm a bit of a noob with electronics, mostly learning it now as my son is very interested, so having some fun learning it with him. Thanks for any tips. I've attached a picture below showing the intended final product (I squeeze the top and bottom of the led wires to complete the circuit, as a simple switch). On the right is what I'm using for my cells (minus the vinegar, and without the cardboard being cut to fit the coin.) <Q> As well as demonstrating some basic electricity generation you will be demonstrating why we don't generate commercial electricity using NZ coinage and vinegar! <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Every battery has an internal resistance which causes a voltage drop as current increases. <S> The effect of internal resistance is to cause voltage droop as current increases. <S> A simple demonstration of this will be to hook up as shown with the multimeter set to mA when in the AM1 position (make sure you have the leads in the right sockets) and set to DC V in the VM1 position. <S> Disconnect the LED. <S> Measure the open-circuit voltage. <S> There is no need to measure open-circuit current. <S> It will be zero. <S> Reconnect the LED. <S> Measure the current and the voltage. <S> Using this information you should be able to calculate the total internal resistance of your battery. <S> \$R = \frac {V_{OC} - V_{LOADED}}{I} \$. <S> It's hard to say why you're not getting four times 0.96 V when connected in series but a photo may tell us more. <S> Looking at the photo I suspect that you may have some leakage in your individual cells. <S> If the sides of the washers or coins get wet there will be a partial short on the cell. <S> To test this make four separate cells connected by wire rather than the stack. <A> When any battery is connected to a load its output voltage will drop a bit due to the internal resistance of the battery. <S> If you are constructing a battery then there are four simple things you can do to reduce the internal resistance of the battery: <S> Make the electrodes wider. <S> Generally most of the current will travel in a straight line directly between the electrodes. <S> The rate at which the chemical reactions can occur is proportional to the area of the electrodes. <S> This is equivalent to wiring multiple batteries in parallel. <S> Reduce the spacing between the electrodes. <S> The resistance of the electrolyte is going to be proportional to the distance that the current has to travel through it. <S> Using thinner cardboard, or using paper towels should reduce resistance. <S> Increase the concentration of ions in the electrolyte. <S> The conductivity will be somewhat proportional to the concentration of ions. <S> For example, vinegar contains acetic acid. <S> Using another brand of vinegar with a higher percentage of acetic acid could help. <S> If you use citric acid rather than vinegar, you can purchase pure citric acid powder from your local grocery store. <S> It is usually in the baking isle. <S> Use an electrolyte that has higher ion mobility. <S> The resistance of the electrolyte is going to be inversely proportional to the ion mobility. <S> Citric acid (found in lemon juice), and sodium bicarbonate (baking soda) <S> both make good electrolytes. <A> In addition to internal resistance phenomenon which lowers the battery voltage under load, you should pay close attention to @Jasen comment about internal short circuits. <S> Take a look at industrial-grade batteries, and you will notice that individual cells never share electrolyte . <S> I suggest you try to put a non-wettable film (like pieces of a plastic bag or candy wrap) between each coin from one cell and the washer from the next cell. <S> Since you still need them to be connected electrically, put a small wire across your insulation layer. <S> That should improve your battery output and bring the total voltage closer to the sum of voltages from individual cells, at least without load.
| The resistance of the electrolyte is going to be inversely proportional to the cross section of electrolyte that the current travels through.
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What is the 1000V - 200mV AC range for on my multi meter? This question draws upon elements of using a DVM for AC . My multi meter has a 1000V - 200mV AC range. Since there seems to be a consensus that multi meters can only reliably read 50Hz AC wave forms, why would you have this range on a meter? Very few are going to use a 50Hz signal for testing their kit. It might not even get through some small coupling capacitors. It doesn't really get to /from my sound card which I'm using as an oscilloscope. So unless you're measuring the mains, what's the point of such a huge AC range on my meter? In what circumstance would you be needing to measure for example 100 mV AC @ 50Hz? <Q> I have owned and used a large number of DMMs during my engineering career, and have calibrated the AC range against frequency on most of them. <S> The suggestion that the AC range on DMMs only reads 50/60Hz has elements of truth. <S> Not surprisingly, the cost of the meter has some correlation with the frequency range it can achieve. <S> Also note that an audio range DMM will always specify this, as it's a cost-adding feature. <S> Given that <S> to calibrate the frequency response, you need access to test gear that most amateurs do not have, the safest course of action is that, if your meter doesn't specify otherwise, only trust it at mains frequencies. <S> So, given that all meters will measure AC mains frequencies OK, why do you need 1000V to 200mV? <S> You need 1000V to measure the mains coming in at the wall. <S> Caution: this is not enough to be safe, you need a CAT2 or CAT3 protected meter to safely withstand (that is, fail safe) when hit with the all too common over-voltage spikes on mains supplies. <S> People have died using a non-CAT meter on mains, even on the 1000V range . <S> You need 200mV to measure the voltage drop on a wire, across closed contacts, or a current shunt. <S> Thanks to Chris for pointing out that there are plenty of intermediate AC voltages, like transformer secondaries, for which the 2v, 20v and 200v ranges are appropriate. <A> DMM's typically measure any sinusoidal AC audio frequency accurately, not just 50/60Hz. <S> But correction factors must be used if you know the waveform is not a sinusoid as they are often converted from average or sometimes peak to RMS. <A> I suspect you're thinking this because you're in the US, and it seems like too large a range for what you'd need. <S> Nominal mains AC in the US is 115V RMS. <S> In Europe it's 230V RMS. <S> If you want to test a three-phase supply then you've got 415V RMS between any two phases. <S> Suddenly that 1000V range doesn't seem too over-specified, does it? :) <A> British Standard 7671 defines low voltage as: below 1000 volts AC (rms) or below 1500 volts DC between conductors; below 600 volts AC (rms) or below 900 volts to earth. <S> your low-voltage multimeter is designed to accurately measure across this wide range of voltages. <S> if you require very accurate measurements at extra-low voltages, you need a specialist meter
| A more accurate summary is that the AC range on all DMMs will read 50/60Hz, many DMMs will get to 1kHz, a few will cover the full audio range, and it's a rare beast that will read accurately above audio.
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The potential difference between primary and secondary wiring of an isolating transformer Lets say we have an isolation transformer powering our circuit. What voltage is a multimeter going to show when we put one probe into mains ground, and the other one at the output of the isolated (secondary) transformer wiring. <Q> What voltage is a multimeter going to show when we put one probe into mains ground, and the other one at the output of the isolated (secondary) transformer wiring. <S> Theoretically 50% of your mains AC voltage, providing the multimeter has infinite input impedance and the secondary is disconnected from any load. <S> This value also assumes that there is no grounded shield between windings and that the net electric field is evenly distributed in one layer of windings. <S> All manner of other subtle issues can raise or lower this figure because it is based on inter-winding capacitance. <S> For instance, because it is likely that the primary and secondary are wound using multiple layers, the main electric field imposing on the secondary will tend to be more towards ground potential or more towards live potential. <S> This is dependent completely on which of the mains wires connects to the layer closest to the secondary. <S> Swapping live and neutral on the primary could easily give different results. <S> Should this measured voltage be a worry? <S> No because it is sourced through the inter-winding leakage capacitance and this will be in the order of 100 pF to 1 nF (hand waving alert!). <A> If it's a typical low cost isolation transformer then there is at minimum an electrostatic shield between input and output windings so the output winding has a capacitance to earth/frame. <S> If it's a high quality isolation transformer the output winding will have top/bottom and side shields (called triple shielded). <S> The capacitance to frame/ground/earth in a professional grade isolation transformer is likely in the 10-20pf range from output winding to ground and perhaps 1pf or less to the input winding. <S> Here's a diagram of a double shielded transformer, the two shields capacitance is in series: <S> transformer diagram http://electrical-engineering-portal.com/wp-content/uploads/construction-of-shielded-two-winding-transformer.gif <S> Here's a commercial grade transformer as an example of the input to output isolation that can be achieved: http://www.consulneowatt.com/products/transformers/ultra-isolation <A> Isolation depends on the ratio of coupling capacitive impedance from primary to secondary and , mutual impedance to ground such as line filters and the quality of the shield to reduce the coupling capacitance inside. <S> If you were to analyze the transfer impedance, you would find it quite high relative to the load but leakage current may be as much as your load's common mode filter to ground (0.5mA) if that is connected. <S> If you are looking for RF isolation, then a CM choke of suitable size may be added. <S> One test you can do is put a large plastic cap across the meter and measure the voltage and <S> if zero, then remove and measure AC current in mA scale to compute the effective impedance coupling of V/I for the frequency being used. <A> I agree with Andy aka, and also agree with the "nothing useful" comment. <S> BUT, there could be static charges that would add or subtract from the capacitance divider. <S> I sometimes have seen an intentional resistance of several megohms across an isolation barrier, to limit the voltage due to these extraneous charges. <S> The input resistance of the multimeter would accomplish the same thing.
| Assuming a real world multimeter (~1MOhm), and a professional grade isolation transformer you will measure zero between either output wire and ground if the output is not bonded to earth/neutral.
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what exactly backup fuses are? Rather basic, but I can't figure out what exactly backup fuses are! I was studying a Weidmuller PRO ECO 120W 24V 5A power supply's datasheet (accessible via the catalogue page ). It has recommended three types of backup fuse for input, but I can't find out what backup fuses are and why to connect them. I searched for it but it didn't help me that much! <Q> The datasheet states that the PSU has an internal fuse. <S> Figure 1. <S> Extract from Weidmuller PRO ECO 120W 24V 5A datasheet. <S> The datasheet gives you three options. <S> Choose one to suit - perhaps to match fuse or circuit breakers in the rest of you control panel. <A> The instruction manual states: <S> This unit is equipped with an internal fuse, so no additional external device protec tion is required. <S> The recommended backup fuse is listed in the Technical Specifications. <S> Notice: <S> When the internal fuse is triggered, the probable cause is an internal malfunction. <S> The device must then be inspected in the factory! <S> I don't see any limit on the fault current capability of mains supply to which the product is suitable to be connected. <S> However the standards that apply may state limits for such products and require backup protection with higher interrupting capacity if the product is connected to a supply that is capable of supplying a higher fault current. <S> Local electrical codes may also require external protection under some circumstances. <A> Redundant fuses or breakers would be a convenience for additional fault protection if there is frequent power cycling. <S> Normally PSU's in large companies such as Weidmuller are tested for reliability with 10k power cycles but the surge currents can cause stress on components. <S> If your line voltage is not well regulated and turn on occurs during peak voltage <S> this would cause the maximum surge current such as power restore in a building with imbalanced phases for startup with surge drop on one phase and rise in another phase. <S> Inrush current is 40A max. <S> Power cycling is a reliability test and the design of soft start would reduce the need for redundant tighter threshold of surge currents , if the protection inside is a "slo-blow" type fuse. <S> Consult with OEM tech support for additional reasons for choices and to confirm my above assumptions if you have doubts on power cycling or slow blow internal fusing. <S> Internal faults can be externally caused from unfiltered lines and transient conditions, so every situation is different, but I would expect at least minimal line filter is included inside this design. <S> If you plan on buying in large quantities, I would suggest a rated load power cycling test plan to confirm reliability.
| The backup fuse is the external fuse that you supply.
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X2 capacitor alternative component or circuit? My computer monitor (Samsung Syncmaster 732N) started to fail and it seems some capacitors have gone bad. They are all electrolitic but one is a CARLI MPX40 .47K275V-X2 capacitor , which where I live (Argentina) turns out is very hard to find. I didn't know what an X2 capacitor was so I researched a little bit and it seems they are capacitors designed to be used as filters in the power line, and because of that, they have to meet higher standards of endurance regarding temperature, voltage surges, etc. Is that correct? My question is, suppose I cannot get one of these where I live. Is there a suitable replacement component or circuit that would meet the same security standards? What could happen if I just put a common .47 electrolytic capacitor? Could it blow up or catch fire or something? Thanks! <Q> Replacing an X2 capacitor with a non-safety type of capacitor is not recommended. <S> The extra ruggedness built into them are needed when you're dealing with the unpredictable nature of input AC. <S> Furthermore, if you put a non-recognized capacitor in there, plug the monitor in and your house burns down, good luck getting any insurance money. <S> - I've been working in the switching power supply industry for more than 15 years <S> and I cannot say I've ever seen one. <S> If you cannot find the exact part, a less dangerous substitution would be another X2 capacitor with the same ratings from a different manufacturer (Panasonic, Elna, Wima, Kemet, etc.) <A> As @Marko implies in his comment, the chances of an X cap failing are very low. <S> Electrolytics, on the other hand, have a definite lifetime that is relatively short when hot. <S> Presumably you are shotgunning and have limited test equipment- <S> even so you can check the X cap with a regular multimeter. <S> If it measures open on ohms and has close to the desired capacitance <S> it's okay. <S> If you don't have a capacitance range, use the highest ohms range, touch the probes until the meter goes off scale, then reverse them. <S> There should be a short delay before it goes off scale again (because of the capacitance). <S> Anyway, a bad X cap (unless shorted) won't prevent your monitor from operating. <S> Ideally you check electrolytics with an ESR meter, but you can just replace them all. <S> The biggest ones and the ones close to hot stuff are most likely to be bad (and anything with a suspicious bulge or fluid leakage, of course). <S> Don't confuse the 'gunk' (typically white, off-white or brown) that manufacturers often use to mechanically stabilize through-hole electrolytic caps with dried electrolyte leakage. <A> The X2 Caps only serve to reduce conducted noise getting back on the line and not to provide power to the unit. <S> Check voltage capacitance and ESR of each part number and choose equivalent or better. <S> I believe this is the schematic for input power to your monitor. <S> Replace the X2 cap if it reads bad with a similar part as deemed necessary. <S> A series choke also reduces inrush currents from external transients.
| Failures of X2 capacitors are extremely rare in my experience (when used as an input line filter)
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At what UART baud rate is hardware flow control required? At low baud rate such as 9600bps, I do not think hardware flow control CTS/RTS is necessary. I believe at higher baud rates, CTS/RTS will be necessary. What is this baud rate? Can one still do without CTS/RTS at 115.2kbps? <Q> If a receiver cannot empty its buffers quickly enough, then it should deassert CTS (in response to which, the transmitter will assert RTS if there is more data to transmit). <S> Note that the transmitter may experience a buffer <S> underrun in which case it should deassert RTS <S> (because there is no valid data to send). <S> So it comes down to how fast the buffers can be moved <S> - that is more likely to be an issue at higher data rates but it is completely implementation dependent; there is no single speed. <A> You require some form of flow control (be that hardware or XON/XOFF) when you are transferring data at a speed greater than you can process it. <S> It's as simple as that. <S> Flow control is used for one end to tell the other end "wait a moment, I am still thinking". <S> Usually it is tied to a buffer being almost full (known as a "high water mark"). <A> Hardware flow control allows the communicating pieces of equipment to synchronize with each other. <S> It allows the receiving piece of equipment to indicate that it is ready to receive the data that is being sent. <S> If data is sent when the receiver is not 'listening', this will cause data errors. <S> The data rate that this occurs at will depend on a number of other things, including the type of device and software on the device. <S> If you are connecting two PCs at 115.2K, they will likely be able to run fine. <S> If it is two microcontrollers with other software load, they may not. <S> If you specify what type of devices are sending and receiving, you will receive better advice. <A> The need for flow control on an asynchronous serial interface is fully dependent on the application needs. <S> Sometimes a slower baud rate can lower the effective data rate to alleviate the need for flow control but that is again application dependent. <S> There are some techniques that can allow an application to function at higher baud rates without the need to use flow control handshaking. <S> One of these is to use an interrupt driven UART send and receive technique and queue the data flow between the application mainline code and the interrupt routines through circular FIFO buffers. <S> The use of FIFO buffers has to be made based upon whether the application processor can handle UART interrupts at the character rate (i.e. baud_rate / bits_per_transmission_unit). <S> If it cannot handle that interrupt rate plus the small overhead imposed by the FIFO buffer handling then it would become necessary to lower the baud rate enough to permit the interrupt rate to function. <S> Sometimes your application processor will have a UART that includes a hardware FIFO built in. <S> If these are used in conjunction with a interrupt handled FIFO buffering scheme it can be beneficial for the interrupt service routine design to empty receive hardware FIFOs or fill transmit hardware FIFOs completely at interrupt time to/from the software handled FIFO buffers which would typically be larger in size. <S> Configured and coded properly this can lower the net interrupt rate that an application processor has to handle at any particular baud rate. <A> I remember when we used to use "dumb" terminals, and you would type Ctrl+S (XOFF) to pause transmission so we could read the screen before the data scrolled off it. <S> Then we would type Ctrl+Q <S> (XON) to resume output. <S> The screen might be paused for 10 minutes. <S> It was nothing to do with the baud rate. <S> Similarly, a serial printer which was out of paper might assert hardware flow control to pause output while the paper was replaced. <S> Again it would be nothing to do with the baud rate. <S> Can one still do without CTS/RTS at 115.2kbps? <S> I routinely send data from my ATmega328P* to a serial monitor at 115200 baud. <S> That works fine. <S> * Arduino <A> Do you care about your data getting dropped in the bit bucket? <S> If yes, then use flow control. <S> Do you have a secondary parity/ECC/flow-control like TCP has? <S> If yes, then you are covered. <S> However, using TCP/OSI as a model, multiple layers do error checking and control to ensure the delivery of data and to keep errors to as few as possible. <S> Let's say your are running TCP/IP, without hardware flow control at 115.2kbps, your effective rate due to hardware flow issues could be terrible. <A> There are many considerations. <S> How big is your receive buffer? <S> How quickly can the receiver empty said buffer? <S> Will the receiver do things that cause it to stop looking at the buffer for some time? <S> if so is the buffer big enough to let the receiver ride over those cases? <S> If you do lose words due to receive buffer overflow how bad is it? <S> If the source experiances "back-pressure" due to flow control <S> how bad is it? <S> Can one still do without CTS/RTS at 115.2kbps? <S> Many systems do, see for example the serial console ports on nearly every embedded linux board. <A> If target device has buffer of 16 bytes but fast interrupt is not guaranteed , then RTS/CTS is always preferred and also include odd parity bit for reliability. <S> Without all the details, it is hard to know the threshold of target buffer overflow, so it becomes trial and error.
| It is not data rate that matters; we use CTS / RTS (or XON / XOFF for software flow control) where a possibility of a receiver overflow exists, although admittedly that is more likely at higher data rates. There is no magic number for "when do I need hardware flow control", it is a question that must be asked and answered in the context of a specific system.
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Determining motor speed and torque given a power and constant voltage I'm trying to determine at what speed my brushed dc motor will run at, given a certain power dissipation. The background is I've built a lathe, using this motor , and for a given depth of cut and feed rate (which is dependent on motor speed and how fast I turn the z axis handle), there is a certain power requirement to remove that much material. Calculating the power is easy, if I make an assumption about the motor rpm, but if I'm running the motor at 12v, I don't have direct control over the speed, just the range of speeds, correct? So, if I set the voltage, then attempt to stall the motor (huge depth of cut or very high feed rate), I will draw max current, and have zero speed, or I can take no cut, and have essentially the no load speed. For a given cut though, I can find the power, but there are two points along the torque speed curve (other than max power) that give that power, so how do I know which point I'm at? The high torque low speed point, or the high speed low torque point? Thanks! <Q> Looking at the datasheet, while it doesn't lie, it is definitely on the edge of misleading. <S> Note that the "max power" stated occurs at almost exactly half the unloaded RPM and half the stall current. <S> This is indeed the "max power" point for such a motor, but the datasheet fails to mention that it is also the nominal 50% efficiency point, thus dissipating 12V*68A-337W = <S> 479W in that tiny motor - destroying it, probably in minutes. <S> (Ideally, exactly half the power would be delivered, about 400W shaft and 400W heat, but the motor isn't ideal). <S> The motor is probably suitable for 100-150W continuous output and 200-250W short term. <S> So practically you must operate the motor at the upper end of the speed range, and if the speed falls below (say) 70% of the unloaded speed (or the current rises to 30% of the stall current) then - unless this is strictly temporary, like starting a heavy load or hitting a chilled spot while machining a cast iron surface, you need to cut the current and protect the motor. <S> Then the question of which side of the torque speed curve doesn't apply - unless the protection has tripped, you should be on the high speed side. <S> You can get circuit breakers that will allow short-term overcurrent. <S> These are "motor rated" or Class C breakers for the AC motors used in most machine tools. <S> I don't know of anything suitable for 12V DC though. <S> I'd be looking for a 12V DC supply that can be set to trip if its output exceeds 40A for more than a couple of seconds. <S> And as Olin says, if you want to monitor it yourself, measuring the current is definitely the way to go. <A> Can you sense and filter the voltage spikes or current switching on the motor to trigger a One shot and then make a tach speed speed output for servo control. <S> Or limit the current pulses , LPF and input to a typeII PLL and monitor the VCO control,voltage for tach speed? <S> 1st option is better if you can detect back EMF with a clean pulse conditioner. <A> You can measure the stall current and no-load currents up front. <S> Any actual operating current should fall within that range. <S> Since you apparently have a fixed voltage source (12 V in your case), the power is directly proportional to the current. <S> In fact, the power in Watts is the current in Amps times the voltage in Volts. <S> Note, however, that this is the power going into the motor. <S> The mechanical power coming out will be different. <S> One obvious way to see this is to consider the input and output powers when the motor is stalled. <S> The electrical input power will be maximum, but the mechanical output power will be zero. <S> In that case, all the power goes into heating the motor, which the motor may not be rated to take indefinitely. <A> For a given cut though, I can find the power, but there are two points along the torque speed curve (other than max power) that give that power, so how do I know which point I'm at? <S> The high torque low speed point, or the high speed low torque point? <S> You can tell by the smoke :) <S> At the high rpm low torque point the motor should be working relatively efficiently (delivering more power than it dissipates) but at the low rpm high torque point the opposite occurs. <S> Most motors are not designed to dissipate a lot more power than they output, so running them in this region will cause them to overheat. <S> A permanent magnet or shunt wound DC motor has a linear torque vs current and inverse rpm vs torque response to varying load, delivering maximum power output at around 50% of no-load rpm. <S> Your motor has the following specs:- Voltage: 12 volt DCNo load RPM: 5,310 <S> (+/- <S> 10%)Free <S> Current: 2.7 ampsMaximum Power: 337 Watts (at 2655 rpm, 172 oz-in, and 68 amps)Stall Torque: 2.42 N-m, or 343.4 oz-inStall Current: 133 amps <S> At maximum power output it is doing exactly 50% rpm, producing 337W mechanical output from 816W electrical input for an efficiency of 41%. <S> At this operating point it has to dissipate 479W, which is probably already above its continuous capacity. <S> Any higher load will rapidly increase power dissipation, up to 1596W at stall! <S> You do not want to operate the motor in this region.
| Probably the easiest way to tell the motor operating point is to put a ammeter in series with it.
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Is an ungrounded earth wire worse than no earth wire? The electrical installation in my house is old and have no earth wire, with the outlets having only two holes. However, the new equipment come with three pins plugs, so I installed some new three hole outlets without any earth wire connected. The problem is that if I measure the voltage between neutral and the chassi of a device that has a three pins plug, I get almost 60V. I think this voltage is being induced by the nearby live wire inside the power cord. In such a situation, would it be better to disconnect the earth wire inside the equipment, so the induced voltage coming from the earth wire doesn't energize the chassi? I know this is all far from ideal, but I'm just trying to figure out what would cause less harm. <Q> A 3-hole outlet installed in a wall is a promise that the third pin is grounded. <S> Suppose someone who doesn't know the truth plugs in a faulty piece of equipment and gets shocked because the ground is not there to protect them. <S> You would be held liable for creating an installation that doesn't meet the building code. <A> Electrically, it doesn't matter. <S> While it is illegal to do what you have done, it is not more physically dangerous (as long as everyone using the outlet KNOWS that it was modified) than using a 3 to 2 adapter. <S> While it may fool a building inspector, it creates a dangerous situation if there is a fault/break in the neutral line. <A> Yes, it's worse. <S> About the voltage you get when you measure between chassis and neutral, the chassis is not grounded so it has not to be any voltage in this case. <S> So I would like you to check the voltage between phase and chassis, and to check if there is a continuity between phase/neutral and chassis to have a clear situation. <A> The REASON that it is illegal is because it IMPLIES that the ground pin is actually grounded (else it wouldn't be there). <S> Your problem with seeing 60V (1/2 the mains voltage) on the chassis is because the device was designed with the assumption that the ground pin would be properly grounded. <S> And since you did not actually ground the safety ground pin, you are probably seeing low-current leakage of some kind of filtering circuit on the device. <S> This is just one example of WHY it is wrong (AND illegal) to leave the ground pin floating. <S> DON'T DO IT! <A> Time to take a chance. <S> For this you need a length of green/yellow cable, so that everybody can see that it is earth, and rated 13 Amps or higher so that if ever there is a fault it will blow the fuse before this wire. <S> Also probably a drill to get from the earth <S> 3rd pin in your indoor wall to a copper water pipe such as under the sink, some sandpaper to scrape the copper to shiny, and a bit of metal often called a "jubilee clip". <S> Even if the earth 3rd pin wiring in your house is incomplete, it is best that it really is earthed to your water pipes. <S> If you have a multimeter you can also check for earth leakage current from faulty or old or incorrectly wired appliances BEFORE you make this connection, mA from 3rd pin to earthed pipes. <S> If you get anything more than 0.01mA <S> then you have a faulty appliance somewhere in the house to find and disconnect. <S> If you get less than 0.001 mA then adding this earth wire will fix your fault. <S> If not then you have a faulty appliance. <S> For example I have a HP980 Deskjet printer whose earth leakage current crept up as its power converter board aged. <S> Eventually I took out the mains to 31V DC power converter and run it direct DC (only two wires) off solar panels or batteries and a "boost" DC-DC converter-regulator. <S> I make sure that the chassis of that appliance converted to two wire <S> direct DC is crocodile clipped to a green and yellow wire which goes to the same earth as mains earth. <A> What would cause less harm is wiring in the EARTH to all the sockets. <S> the appliance is "requesting" an earth connection for a reason
| If a leakage current passes to the chassis of any electrical device by fault, and someone touches that device, he will be shocked if it is not grounded. Yes, it's worse, in the sense that it is quite illegal in most jurisdictions. What you absolutely should not do, is connect neutral to ground inside the outlet.
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Two circuit voltage 5V, 12V efficiency with one supply and convertor Currently i'm bulding a circuit that need two voltages. 5V to feed the microprocessor, sensor, leds, LCD and others 12V to temporary feed small water pump (12V 4.8W) and a Electric Solenoid Water Valve (250mAh) [Activated by a MOSFET from MCU] 5V Will always run, but with a timeout that disable lot of sensors, leds, lcd, etc to save power, when user click any button it awakes. From x to x seconds it will auto awake to update sensors and values and sleep again. 12V will only be needed to turn on pump and valve by user interaction (Click a button) to fill a glass of water for example. I have a 5V 5A brick power supply with stabilization that i run some of my projects. The main question here is if i use this power supply with some or one boost convert 5V->12V will be better than run 12V and buck converter to 5V to main system Note: 5V Circuit can reach max 2A when its all ON and at Max Brightness <Q> There are lots of power supplies that deliver +12V, 2A and +5V, 2A; thatis a common power requirement for external hard disk drives. <S> I'd use oneof those, they're assembled and tested and available whenever an externaldrive dies... <A> Your 5V drain appears to be significantly more (over an extended period), which would imply that it is more efficient to take this rail directly from the primary supply, and generate the 12V rail on demand. <S> If you start at 12V, you will have an additional penalty from the 5V switcher efficiency. <S> However, without comparing all of the components you are planning to use, it is hard to be sure. <S> For example, it is possible that the low power state is dominated by leakage in the primary supply if it is rated for the peak load. <A> I prefer to use a separate power supply for each device I make, unless it is only used occasionally and I don't need my bench power supply for something else. <S> In general a buck converter is more efficient than a boost converter. <S> A 12V supply stepped down provides an extra level of stability on the 5V supply, so the 12V supply could be run hard out to operate the pump and valve without glitching the MPU. <S> This may be quite difficult to achieve, particularly if your pump has a high startup current.
| If you boost from 5V to 12V then the 5V supply must maintain stability over a wide power range. What's 'Better' depends on the situation. Whichever way you go the primary power supply must be sized to handle both loads (including the loss in the converter).
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Adding capacitors prevents car peltier cooler from making low battery light flicker, but why? I have a Black and Decker Heater/Cooler that I want to use in my Camper as a little drink caddy for in front of my Futon while I have company over and we are playing games, and I just switched to a new Camper and I stripped the old camper of some of it's parts like the converter So I wired the cooler up to the Converter and the "Vehicle Battery Low" Indicator light turned on and started rapidly flickering like it would if it was plugged into the car and the battery was dead. So I knew this solution would work, but I don't quite know why, (I guess I was thinking of Automotive Stereos and Amplifiers where you can sometimes buy a big Capacitor and it will prevent the AMP from drawing too much power from the battery and alternator and making the lights flicker during times of High BASS and Volume.) can anyone explain to me why this would work though? My solution was to take an electrolytic type 10,000UF 16V Capacitor and put it between the Positive and Negative with respect to appropriate polarity, but all it did was make the cooler blink faster, So I put another Identical Capacitor in Parallel with the first one and ZING the cooler turned on and operates as though it is on a 12 Volt Battery now. I am great at experimenting with electronics and I have been wiring circuits since I was a little kid, since I can remember anyways, but I am not an engineer. Can anyone possibly explain to me why or how this worked? I also know that sometimes if you over volt or over Amp a capacitor it will get hot and it will vent and that will destroy the capacitor. Could you help me to ensure this will not happen while running this device I have created? Cooler~Black & Decker: TC212FRB (Peltier Cooler) NOTE: My Multimeter says it is only pulling about 3.60 Amperes at 12 Volts 43.2 Watts Converter~Phillips: Model Number PC-201-A1, It puts out DC 12 Volt Maximum 20 Amps or 240 Watt Capacity 2X capacitors JRB 10000 Micro farad at up to 16 Volts Vented Large Canister type 105C in Parallel The Cooler is getting cold, but the Converter is getting slightly warm (there seems to be a Wire wound resistor inside heating up for some reason, but if I remember right it does this normally as old as it is, so IDK if this is a great idea or not,but it seems to be powering it well when it couldn't power it without the Capacitors, and I am really not sure why it couldn't if it can now. The capacitors are not adding voltage or amperes are they? Capacitors just store and discharge DC electricity right? This is baffling to me. <Q> First of all, TMI of the useless sort, not enoughof the useful sort. <S> That converter is made to operate lights and DC motors. <S> It probably has little to no filtering of the output, and may use only a half wave rectifier. <S> It seems to be pretty old. <S> What that means is, is that it doesn't put out a nice, clean 12VDC, but rather a pulsating voltage that approximates DC. <S> The low points in the pulses are low enough to trigger the low voltage warning on your cooler. <S> Adding the (big) capacitors smooths the pulses and holds the average voltage high enough to stay above the low voltage alarm. <S> The peaks on the pulses may well be over 16VDC. <S> You have no way to measure them, and I can't measure it from here. <S> So, no guarantees that it won't kill your (presumably) expensive capacitors. <S> Cheaper and more certain would be to use a modern switching power supply rated for 12VDC and 5A. Check the tags on the cooler, or look it up online to be sure about the voltage and current. <S> Or, just use a cooler made for 110VAC when you've got an outlet to connect to. <S> The recommended converter puts out 6A at 12V. <S> The model number is in the cooler hand book. <S> Buying one of those is probably cheaper than playing games with expensive capacitors, and safer besides. <S> Have you considered what could happen if you short those big capacitors? <A> Not entirely sure what this 'converter' is that you're referring to, but my guess is that your cooler unit draws a very sharp transient when it turns on and the converter can't put out enough current to keep up so the voltage droops and the cooler resets itself. <S> Then it tries to turn in again and just ends up resetting itself again, and so on. <S> With with capacitance, the drop is small enough that the cooler doesn't reset itself and it runs normally. <A> The external load terminal of that converter outputs raw, unfiltered, pulsating DC, which may be OK to operate incandescent lights and motors. <S> The Peltier element in your cooler may also be OK with the pulsating power, but the dips to zero volts twice a cycle are confusing the low voltage warning circuit. <S> Adding capacitors prevents the voltage from dipping to zero, as they are charged on the AC peaks, and discharge (relatively) slowly between the peaks. <S> Adding the second capacitor further slows the discharge, preventing the low voltage warning circuit from causing an alarm. <S> On boats, we normally run all 12 volt equipment, including lights, from the battery at all times. <S> When Shore Power AC is available, we just use a battery charger to replenish the batteries and power the DC loads - the battery maintains the voltage between AC peaks, so your cooler would have no cause for alarm. <S> I've never understood why RVers use these "converters" to switch some DC loads off the battery when AC is available.
| When you add the gigantic capacitors, this surge current will be drawn from the capacitors instead of the converter and the voltage won't drop as much.
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Powermanagement of a minicomputer in an automotive vehicle by engine start/stop and voltage threshold detection Revison 1.3 (New circuit, more background information added, PCB made and tested, it WORKS) Background: I would like to use a BeagleBone Black (=Minicomputer) for data acquisition in an automotive vehicle. It will be plugged into the OBD interface and will therefore be supplied by the 12.0 V - 14.0 V battery power / alternator. The following specifications were defined: The minicomputer has to power on when the engine starts. The minicomputer has to stay alive when the engine goes out (e.g. new start/stop system in new vehicles at traffic jams) A clean shutdown (i.e. not cutting the power) should be made when no CAN or OBD messages have been received for 2 minutes. After the BeagleBone Black shutdown the power has to be cut to the automotive battery. (i.e. no further power consumption) Actual progress/solution (Thanks to Olin and Dave!) Two separate circuits. The trigger voltage indicates if the main switch is open or closed. Voltage sensing by TL431: When a voltage > 13.25 V was detected, the main switch will be opened by a P-Mosfet. A NP-MOSFET switches the power supply for the BeagleBone Black. When the battery voltage is above a specific threshold (e.g. 13.25 V) it closes the P-Channel (engine started, vehicle battery ~ 14.0V, the BeagleBone starts up). When the BeagleBone Black starts up, the internal 3.3 V closes also the N-channel MOSFET (This will keep the BeagleBone on, also when the car engine is out) When the engine is out, and the BeagleBone receives the poweroff system call, the BeagleBone powers down. The connection to the vehicle battery is cutted (No trigger voltage and no 3.3 V of the BeagleBone). The circuit is shown below. Of course there is a protection circuit before and a step down converter after that part. MOSFET used: VISHAY SI4564DY-T1-GE3 +12.0V = Vehicle Battery TRIGGER = When vehicle battery is > 13.25 V, this voltage will be pulled down VDD = Supply voltage for the BeagelBone Black (Goes into a step down converter) <Q> What you are attempting sounds backwards. <S> You want to detect higher than idle voltage, not lower. <S> When the engine is running, the alternator raises the battery voltage to the float charge level. <S> This is usually around 13.6 V, but can vary quite a bit. <S> A threshold of around 13 V or a little higher is probably right. <S> The rest comes down to detecting this threshold voltage using very little current. <S> Something like a TL431 and a couple precision resistors may be good enough. <S> Carefully compute the error band to make sure it is acceptable. <S> Then turning on power to something else when the TL431 triggers is trivial. <A> Assuming you really want the computer to start up on cranking rather than running, I would propose a circuit like this: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The zener diode D2 determines how low the vehicle power must "droop" before switching on the computer. <S> This circuit draws no quiescent current. <S> If you replace D2 with something like a TL431, there will be some current through its bias network. <S> The general idea is that the computer will fire up on any cranking event (or other transient load), and then it can evaluate conditions and decide whether or not to keep running. <A> I strongly suggest that you do a little bit of searching around automotive power supply design. <S> There's a lot more going on than just a slightly (12-14V) power supply. <S> A good starting point is Little Fuses's AN9312 app note which gives a good overview of what's going on. <S> For example a figure of 14V is mentioned in the original post but <S> the alternator on my TDV8 L322 is regularly outputting 14.5V <S> because there is so much demand on the battery from all the electrical systems. <S> Similarly cranking a 3.6L V8 draws quite a bit of current, particularly over the winter, and it will easily drop to 6V, and these are not abnormal figures. <S> The main thing to be aware of is the rating on various devices. <S> For example Q1 in the schematic above is connected "directly" to the battery and noisy circuitry. <S> Care must be taken to chose a FET that is appropriately rated for the environment. <S> NB: there are devices (Linear produce some for example) that are specifically designed as automotive protection circuits and power supplies.
| Measure your battery voltage with everything off and with the engine running, then pick a voltage in between. The computer itself determines when to shut down — by pulsing the base of Q3 high.
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Put 1 of 3 batteries the "wrong way" to prevent accidental turning on? My headlamp accidentally turns on inside my backpack. If it used two batteries, I would put one the "wrong way" to prevent it from coming on. However, my lamp uses THREE AAA-type batteries. Can I invert one to get the same effect? <Q> If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. <S> However, batteries aren't like that. <S> The slightest difference in voltages mean that current will flow. <S> Not as much as normal, but some will still flow. <S> And that means, if the lamp is turned on, dead batteries - just not as fast. <S> Introduce a big imbalance, which is what you do with 3 batteries (2:1 voltage ratio) <S> you end up with considerably more current flowing round your circuit, which means flatter batteries quicker. <S> Add to that <S> the fact that you have current flowing backwards through one battery (even with just 2 batteries) <S> and you are really asking for trouble. <S> So all in all turning one battery around is a ReallyBadIdea™ and should be avoided. <S> Instead remove one battery entirely, or insert an insulating material between two batteries or between a battery and its contact. <S> Alternatively fix the switch so it doesn't keep turning on. <A> Reversing two batteries accidentally could damage the unit. <S> Also a headlamp that's badly designed so the power switch can be accidentally pressed is less likely to have protection circuits <S> (the main purpose would be to protect a possible rechargeable battery or possibly the circuit). <S> If you are willing to open the battery case anyway, you can just take the cells out and put them in a plastic holder for transport. <S> I use ones like this one: Or, devise some mechanical shield for the power switch. <A> Turning one of three backwards will not prevent it from being turned on. <S> I had batteries fail and drain to the point of reversal, yet could still turn the flashlight on. <S> It was dimmer, but still visible. <S> Eventually the reversed battery will leak, or worse explode (unlikely for alkaline, but possible). <S> The simplest option is adding a plastic or cardboard tab between the battery and the terminal in the battery holder, like they do in any number of products or toys that ship with batteries. <S> You break the connection, so the switch does nothing. <A> Don't play around with reversing the battery polarity just to keep the device form turning on. <S> You could ruin the batteries (or at least drain them) and risk potentially damaging your device. <S> Your "ideal" options: 1) Take all the batteries out and rubber-band them together in a cluster to prevent losing any. <S> 2) <S> Tape/affix your device's "on" switch to the "off" position to prevent it from turning on unexpectedly. <S> 3) Stick a piece of nonconducting material (tape? <S> cloth? <S> depends on what you have on hand. <S> electrical tape would be easy) between any two of the batteries and that should keep them from conducting in a circuit. <S> The easiest thing to do of these is the third since it doesn't involve opening the battery compartment every time you want to use the device.
| Whilst it is possible that reversing one battery would cause the low voltage lockout (if there actually is one) to actuate and not turn on, this solution is fraught with risk.
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Data exchange over 5 meter cable I have a microcontroller which needs to communicate over a cable, which will be approx. 5m long, with another very small microcontroller. The devices will be used outdoors in various conditions. The data rate will not be very high. When you connect the small uC with the cable to the big uC, a identification string and periodic (each minute) battery updates are sent from the small uC. I thought about I2C but I'm not sure if it is designed for such needs? CAN is also possible but I think it´s a little bit overkill for my needs. Does anyone have good ideas? EDIT:I want to use as few pins as possible for the cable. <Q> I'll also vote for RS232, you should be careful about one thing that no-one else has mentioned though. <S> You don't say what microcontrollers you are using, but the following is true for most that I've used... <S> Your microcontroller will have at least one data transmit line (TX) and at least one data receive line (RX). <S> These lines will toggle between low & high as you transmit/recieve between voltage levels of 0V & ~5V or 0V & 3.3V depending on the device. <S> Assuming both microcontrollers use the same voltage levels then, in principle, you could just connect TX from one micro to RX of the other and vise a versa. <S> This works for short distances & low data rates, but I think 5m is way too long for this <S> and I also think it is bad practice to do this between systems. <S> Better to use a level converter (such as the MAX232 family for 0-5V systems or the MAX3232 for 0-3V systems) on each end. <S> These convert the logic 0 to ~+12V and a logic 1 to ~-12V so that your comms line conforms to the RS232 standard electrically. <S> Hope <S> this helps, Joe <A> While CAN might be a bit "overkill", it is always to prefer if you have that option. <S> Brief comparison or CAN versus RS232/RS422 <S> /RS485: <S> CAN advantages: <S> Far more robust and tolerant to EMI. <S> But in terms of the technology itself and in terms of built-in protection in a standard transceiver. <S> Will work just fine without shielded cables at lower baudrates. <S> Built-in error handling, CRC and frame synchronization. <S> So there's no need to invent yet another obscure, custom UART-based protocol. <S> Meaning CPUs don't have to waste time encoding/decoding, calculating CRC etc. <S> Easy to maintain and easy to re-design from point-to-point into multi-node system, should the need arise in the future. <S> External CAN controllers are burdensome and a thing of the past. <S> Might have slightly higher current consumption than UART-based solutions. <S> The cost of CAN transceivers versus RS-xxx transceivers should be about the same (except if you pick old crap transceivers like MAX232 that need 5x 1uF decoupling caps). <S> Signal voltage levels don't matter, there's CAN and RS-xxx transceivers for both 3.3V and 5V. <S> The number of wires for a semi-duplex system will be 3 in either case. <S> In the case of CAN, you have CAN H, CAN L and signal ground. <S> In the case of for example RS-422 you have T+, T- and signal ground. <S> Skipping signal ground is really not recommended for either case, don't listen to people telling you otherwise. <A> Ignoring the question of best alternatives (I don't really recommend it, but it can work), I have sucessfully run I2C long distances (between buildings) in production and it worked for years. <S> You can reduce I2C clock rate right down to 10kHz, and put extra filter caps on the lines to filter RF out. <S> Many of the chips are perfectly happy at higher currents than 1.5mA run 5,10 or 20mA <S> depending on what your chips can handle <S> Make sure you are using I2C chips (cmos schmitt thresholds) not SMBUS (ttl levels) as they are far more noise prone. <S> feed the power down the same wires, have suitable bypassing. <S> twisted pair/shielded cables. <S> http://www.i2cchip.com/i2c_connector.html#Crosstalk use bus switch to isolate this segment from local I2C parts realise that errors are to be expected and design software around it. <S> (i.e. do repeats e.g. update your display every few seconds, rather than leave it for hours between updates) <S> Beware of I2C bus deadlock, which can occur from noise injecting an extra clock pulse, and make sure your master detects and deals with it. <S> (I bet arduino stock code doesn't) <S> See section 19 I2C Bus Deadlock in http://www.i2cchip.com/pdfs/bl233_b.pdf
| CAN disadvantages: Not really a sensible choice unless your MCU has a CAN peripheral on-chip.
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How many pull up resistors per I2C bus I am designing a PCB wich I need to talk with 13 IC's on a I2C bus. I am wondering how many pull up resistors should I put on the bus? The chips are MAX9611 and here is the current layout of the I2C board area (continues to right side): As you can see R17 and R18 are current limiting resistors and R21 and R22 are pull up resistors (1 for SCL and one for SDA). The SCL trace goes over top layer and SDA goes in bottom layer). I just saw some examples in TI I2C bus guide , and I am wondering if what I have already designed is wrong, because it seems for each device there is a local pull up resistor on SDA and SCL lines: I would really appreciate your suggestion, before I fail miserably with cost of a bad PCB layout! <Q> Look closer... <S> Your diagram shows four separate I2C buses. <S> The multiplexer and the repeater isolates the segments. <S> Thus, since you only have one bus, you only need two resistors: One for SCL and one for SDA. <A> One thing I think I am seeing is that all of your chips look like they have the same footprint which is making me think they are all the same chip. <S> If they are the same chip then all of them need to have a unique address, otherwise all of them will try to respond to the same commands given. <S> This is somewhat fine if they all need to behave at the same time, not so great if they need to act independently. <S> If they have the same address though, if one of them breaks, is missing, didn't get a command, etc, then your code will be unaware of this. <S> As for pull-ups, pipe is absolutely correct, you only want one pair of pull-ups used per bus. <S> The picture you posted shows four completely separate busses (micro to multiplexer, multiplexer to IO expanders and LED blinkers, Hub repeater to more IO expanders, and multiplexer to data converter, eeprom, etc). <S> Section 7.1 of the I2C standard (www.nxp.com/documents/user_manual/UM10204.pdf) shows how to size the pull-ups. <S> Table 10 <S> in section 6.1 specifies the rise times for the different speeds. <S> One trick I have learned for correctly sizing the pull-ups is specify a mid-range resistor (~5k) as the pull-up, then scope the waveforms on the prototype unit. <S> The actual resistor required will be equal to CurrentResistor <S> * (TargetRiseTime / MeasuredRiseTime). <S> A resistor smaller than the number specified will give you the rise time to meet the specifications. <S> If the resistor is smaller than the minimum calculated pull-up (from section 7.1) you may need to break your bus up into multiple segments using repeaters, multiplexers, or something else. <S> There are two advantages to measuring and then calculating, the first is that I feel it is quicker than trying to calculate your bus capacitance especially since many factors can affect it, the second is that you actually get to see the waveform and it can raise a red flag if the signal looks bad. <A> There are only pullups needed at the master. <S> (I2C is not a fast bus, and does not use termination resistors, and is not impedance matched.) <S> BUT You must ensure that bus segments after the mux/switches remain pulled up when off. <S> Depending on how you implement muxes, you may need pullups on the isolated bus segments. <S> These can be high R, they are only to keep the lines high, not pull them up during operation, e.g. the master pullups could be 1k5, and the segment pullups 100k. <S> You will see this in our bus switch board: http://www.i2cchip.com/pdfs/BusSwitch_MUX3.pdf <S> There is a second reason for segment pullups: One isolated segment is longer/high capacitance, and needs more current i.e. lower R. BTW, this case - one one segment, is a very good reason to use bus switches, as it isolates problems (eg noise) to the chip/s on that segment, while the rest of the system will be reliable. <S> see also: http://www.i2cchip.com/i2c_connector.html#Crosstalk
| There is only a single pair of pull-up resistors for each bus .
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10 pcs 10amp rectifier Diode in parallel for 14.v 90amp DC input I'm thinking to have 10 pcs 10amp rectifier Diode in parallel like in my drawing so I can have 100amp diode for 14.v 90amp DC input voltage. I tested the voltage drop is minimal from actual and simulation. Is this ok? please help <Q> It's not a great idea, there's nothing to guarantee each diode will do 1/10th of the work, and in fact that is quite unlikely. <S> You could balance it better by adding a small resistor in series with each diode, but then you'd be dropping more voltage and making more heat. <A> It is extremely unlikely ( <S> as both Colin and JRE have noted) that the forward voltage of all the diodes will be the same; that implies that one of them will have a slightly lower forward voltage than the rest. <S> Now we have a diode with a large amount of current and that is heating it up, but as it heats up the forward voltage will reduce, further increasing the amount of current through the device (it is now hogging current significantly), causing further heating and so forth until it gets so hot that it fails. <S> This nasty situation is known as thermal runaway and will very possibly make the victim part become spectacularly pyrotechnic; then the next lowest \$V_f\$ part will be subject to the same treatment until the remaining parts cannot handle the load and fail anyway. <S> Why do you need the diodes? <S> To catch inductive kick or are you rectifying a source? <A> Paralleling diodes: <S> Definitevily not a great idea but can be done with good thermal coupling, wise wiring or pcb layout to help current share and, of course, allowing for some derate to single diode specification. <S> This is true since one-diode current positive temperaure coefficient leads to run-away only if overall loop again is above unity. <S> Also positive diode series resistance temperature coefficient helps a while. <S> This AN from ST may help going deeper.
| Let's put it down simple: take no math and go for some 20 off 10A diodes paralleled to carry 90A. This is not a good idea, due to current hogging.
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Putting electronics in sand to keep them cool? I am running some RTL-SDR software defined radio USB dongles. They have metal cases, and can get pretty hot. I saw it mentioned that they can be cooled down by burying them in sand. Since i won't need to touch them after they're set up, is this a good way to go? I could attach a heat sink to each of them but it would require some work (cutting, etc.) <Q> Sand is a pretty good thermal conductor (k=0.2) compared to air (k=0.02)and good clean silica sand is an excellent electrical insulator. <S> Vegetable Oil or mineral oil is about the same used for transformer cooling (k=0.3) . <S> and only has a dielectric constant of 2... <S> but messy if your case leaks. <S> But forced air can improve thermal resistance by 60% or more depending on surface area+ roughness with just a low velocity of 3m/s just over a large flat area or a heatsink. <S> (my own test results) <S> The limitation of heat transfer could possible be to the outside surface area for cooling the sand thru the case but it would do well in spreading the heat. <S> As an example of heat transfer, I recall seeing a youtube video of middle eastern coffee brewed with extra fine coffee in a small BBQ container filled with sand. <S> The aluminum pot for a few small cups takes several minutes on a 2kW stove top. <S> But when the small thin pot is dipped in propane heated sand pit, it takes only one half a second until it foams up., which is great for mass production at an expresso party. <S> Although it might be worth trying, I would use fins with a 30 mm fan. :) <A> (this might be an interesting research) <S> If the device comes as a module like that and has no specifics on cooling, I'd expect it to run fine in normal conditions (up to 35°C air temperature or something). <S> Thinking of an USB dongle with an antenna connector and sand just made me imagine how small particles end up in all those connectors which you will probably take apart and put together at some point. <S> The scratching noises make my hair stand up even before you've done it <S> (yeah <S> you said you don't need to touch them, but things change). <A> This sounds like a bad idea to me. <S> Sand is a temperature INSULATOR, not a conductor. <S> You would end up CONCENTRATING heat right where you don't want to. <S> "Hot" isn't necessarily "bad". <S> If you can still touch it, and keep your finger on it, it probably isn't too hot to function correctly.
| Sand is a better heat conductor than air (like an order of magnitude better), but with sand you negate the effect of convection which might cool the device more than the conduction of the sand.
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What security risks does the Test Access Port (TAP) introduce? Trying to get my head around the JTAG world I looked at how the Test Access Port (TAP) works and although I get why we need it, the diagram seemed to me (as a software developer) introducing some kind of backdoor to the architecture. In the software world such mechanisms can be used in testing some parts of the software, however they get removed (hopefully) before the software is ready for release. So the questions are: Doesn't TAP introduce security risks to the hardware? Are there standers to mitigate such risks if they exist? <Q> Yes, TAP introduces hardware security risks. <S> However many chips provide a JTAG disable configuration that disables the JTAG after programming. <S> As far as risk mitigation, if you need such low level physical security you might consider choosing an IC that doesn't have this feature, or allows you to disable it. <S> Some disable signals are only a flash bit, and a full chip erase can restore JTAG access, others permanently alter the IC so JTAG will not be available after disabling it. <S> There are also implementations of encrypted JTAG, where you cannot use JTAG without the proper encryption and keys. <S> The 2009 paper, "Attacks and Defenses for JTAG" provides significantly greater detail in the various means and ways that JTAG can be attacked, and defenses available. <A> First, you're under the false impression that SW debugging mechanisms get removed from final products. <S> For example check Windows debugging API : it has broad potential for abuse (reading other processes' memory and such) and none of it is needed by the final users. <S> Second, in order to exploit a hardware interface like JTAG, the attacker must have physical access to the system which usually means "game over" in terms of security. <S> Even if the customer version of the product didn't have JTAG, nothing would stop the attacker from swapping the PCB with a custom one, which has whatever interfaces the attacker needs. <S> About the only real security issue arises when the developers want to protect themselves from malicious users who could abuse JTAG (for jail-breaking, firmware cloning etc). <S> Relevant mechanisms for such protection are mentioned in other answers already - fuse bits, encryption keys etc. <S> Of course, removing the programming/debugging interfaces completely would be more secure, but it would also be more expensive (e.g no in-system programming) and reduce the possibility of quality control. <S> JTAG in particular is used in End-Of-Line testing of PCBs, which makes sure that defective devices are not shipped to customers. <S> That's the big difference compared to software world: if you have released a software, you don't have to test every copy of it before download. <S> If you release hardware, you have to test every copy or assume the risk of production failures. <A> This is one of the ways in which users jailbreak phones: through the JTAG port. <S> It does indeed let you access all the processor state and memory, and in some cases the contents of any internal ROM. <S> This makes it very useful during production as a debugging feature. <S> Many systems will provide a "fuse" in the chip, which is a one-time-programmable device, to turn off JTAG access in such a way that it cannot be turned back on.
| If you have access to the JTAG port you may be able to introduce false input/output.
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Is it safe to remove earthing from a scope? I need to make some differential measurements on a device, but both nodes need to be floating from earth. I found out the hard way that the negative clip of my oscilloscope voltage probes are connected to the mains earth conductor, but I don't have access to a differential probe. I was wondering if there is an easy solution to this, and I figured this: inside the scope I have seen that the mains are only connected to a power, old style transformer, while the earth lead is screwed to the metal chassis. Since of course the transformer output is isolated from its input, I think that if I just remove the earth screw from the chassis and isolate the cable I can now connect my device to the scope, avoiding the pesky ground current that destroyed my first board. The negative terminal I would hook to the probe clip is about 60V over earth, the maximum current it can source is just above 30mA. The input transformer can surely withstand a 60V offset, right? I think that this is safe because the shell of my scope is all made of plastic, the only metallic parts are the BNC connectors and a couple of blade connectors that I use to calibrate my probes, but they should all be safe. Is it ok to remove the earth connector from my scope to make such a measurement? I have seen this suggestion all over the web, so I think it must be safe, but I want to ask a board of professionals just to be extra sure. <Q> The best answer is to isolate the equipment you are trying to measure. <S> When it is isolated, you are free to ground any one point, which will be done thru the scope probe ground lead. <S> That keeps the scope, its chassis, and anything it may be connected to or touching from becoming live. <S> If isolating the circuit you are trying to measure is not possible, then a portable battery-operated scope is the next best option. <S> These are intended in part for this sort of thing, and are usually well insulated, unlike scopes intended to sit on a bench. <S> If you are really really careful and can ensure everyone else around you or that might enter the room will be really careful, then you can put the isolation transformer on the scope. <S> Before you do that, very carefully make sure the scope isn't touching anything conductive, and there are no connections to is to elsewhere, like a USB connection to a PC or something. <S> This way you get what you suggest without modifying the scope. <S> I just can't recommend what you suggest. <S> You may think you found the only connection to ground, but the scope wasn't designed to be modified like that, so you don't really know without testing. <S> A Hi-pot test would tell you, but may also blow up the scope in the process. <S> Do it right. <S> Use a isolation transformer, preferably on the equipment under test, not on the scope. <A> The more common way I've seen "poor-man's" differential probes implemented is by using 2 channels, then using the math A-B mode. <S> The two ground clips are attached to proper ground. <A> It's possible, and we used to have a 'special' cord sets with the ground pins broken off in the old days, but it's not particularly safe, especially if the voltages are relatively high. <S> In this millennium, I have a Tek scope with four (mutually and from ground) isolated channels for this kind of work. <S> It also has a Li-ion battery pack for when more AC isolation is required. <S> Differential probes can work, but their common mode rejection ratio isn't all that great so if you are trying to look at a 10V waveform on a noisy 300V bus <S> the results may not be all that representative of the truth. <S> As well as the internal power transformer you <S> may also need to worry about the isolation (or lack of isolation) of communication ports such as Ethernet, IEE488, USB or RS-232. <S> I have an isolation transformer with an internal electrostatic shield (two shields would have been even better) <S> that is also useful in these situations. <A> The line filters to ground would become open and a charge accumulation from line voltages on the primary coil may result. <S> It would be better to use an isolation transformer if the internal design is unknown between neutral and ground. <S> Then verify on the same point in A-B mode for flat line. or better yet , make a diff amp board to output the voltage and put near DUT test points. <S> Then attenuate inputs within CM range and amplify with sufficient gain.
| Keep in mind that there will be (sometimes large) capacitance across whatever isolation device you insert to isolate the measuring or the device under test and that may load the signal you are trying to measure and distort the results. Normally using two matched and calibrated 10:1 probes can produce reasonable results using no ground leads and tip removed with adjacent test pins for V+, current sense V and earth ground. It cannot be recommended, mostly for safety reasons.
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Should I switch the transfomer input or output? I am installing a remote receiver switch into my outdoor lighting circuits and wonder should I leave the 230-12volt transformer (500va) permanently supplied & install the switch module on the 12 volt output or would switching the 230 volt input be best. <Q> I would aim to switch the high voltage. <S> A 500 VA transformer will waste power even when unloaded. <S> No-load losses are caused by the magnetizing current needed to energize the core of the transformer, and do not vary according to the loading on the transformer. <S> They are constant and occur 24 hours a day, 365 days a year, regardless of the load, hence the term no-load losses. <S> They can be categorized into five components: hysteresis losses in the core laminations, eddy current losses in the core laminations, <S> I 2R losses due to no-load current, stray eddy current losses in core clamps, bolts and other core components, and dielectric losses. <S> Hysteresis losses and eddy current losses contribute over 99% of the no-load losses, while stray eddy current, dielectric losses, and I 2R losses due to no-load current are small and consequently often neglected. <S> Thinner lamination of the core steel reduces eddy current losses. <S> The biggest contributor to no-load losses is hysteresis losses. <S> Hysteresis losses come from the molecules in the core laminations resisting being magnetized and demagnetized by the alternating magnetic field. <S> This resistance by the molecules causes friction that results in heat. <S> The Greek word, hysteresis, means "to lag" and refers to the fact that the magnetic flux lags behind the magnetic force. <S> Choice of size and type of core material reduces hysteresis losses. <S> Source: Copper.org . <S> For example, if the 500 VA transformer no-load losses are 5% (and I have no idea if that's reasonable) <S> then continuous power waste is 25 W. <S> There are 8760 hours in a year <S> so energy loss is 8760 h x 0.025 <S> kW = <S> 219 kWh*. <S> You can save energy in all the off hours by switching the primary. <S> Do the right thing! <S> Reference: <S> Transformer Losses and Efficiency . <S> Carroll-Meynell . <S> * <S> To generate 219 kWh of electrical energy with a 40% efficient oil or coal-powered generating station would use about 50 kg of fuel (based on 10 kWh per kilo of oil). <A> Check the voltage ratings <S> oh your input. <A> My answer would be switch the lower voltage / DC side, as I think it's less susceptible to arcing and therefore safer... <S> but I think there's going to be some varied opinions here. <S> The flip side of that argument is that you're going to be switching smaller currents at the higher voltage, but my gut says voltage is the primary variable when deciding what to switch in power circuits. <S> I'm prepared to be down-voted by people better versed in these matters, though. <S> It's really somewhat more of a question for an electrician than an electrical engineer... <A> I would switch the mains side. <S> As pointed out already, this will prevent losses when the circuit is idle. <S> However you should ensure that your switching circuit is adequate to deal with a 500VA inductive load. <S> Due to the inrush current and the tendency of an inductor to sustain the flow of current when voltage is removed (that's what inductors like to do) it must be rated to carry these currents and/or use strategies to reduce them. <S> A relay (the simplest solution) should have an inductive load rating (which will be much lower than its resistive load rating) and you should still have a generous margin with that. <S> 500VA at 230V is little over 2 amps, so look for a 4 or 5A inductive load rated relay. <S> If your switch circuit is electronic (e.g. triac) you'll need to likewise rate it for these momentary switching currents. <S> You can reduce these currents with considered design. <S> This application note for instance http://www.st.com/content/ccc/resource/technical/document/application_note/2f/f7/75/f5/9d/3d/44/04/CD00003867.pdf/files/CD00003867.pdf/jcr:content/translations/en.CD00003867.pdf <S> discusses switching (with a triac) and is a good read. <S> It shows how you should (a) switch at the peak of the mains cycle (when the current flow due to the magnetic field on the inductor is crossing the zero point) <S> (b) always switch on at the opposite polarity to the last switch off. <S> This is due to the remnant magnetisation of the core. <S> If you can do that, you can seriously reduce the surges. <S> The application note actually shows the measured currents on a 200VA 12V lighting transformer. <S> But that's probably more than you want to get into, so just use a nice big hunky relay with a nice big inductive load rating on the live connection of the transformer. <S> Switch the coil of that with your electronics. <S> Also make sure you install a normal switch for safety isolation purposes on the transformer supply. <S> A relay or electronic switch does not provide safety isolation (for when you want to work on the transformer or the loads it supplies).
| Check the operational voltage range. If it has an inbuilt adapter to convert 230 to 12, you don't need the transformer to do the job for it.
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Pull-up vs Pull-down on enable pin I am examining a schematic which concerns the TPS61030 boost converter IC and I have a question regarding the logic behind pulling the enable pin (EN). The schematic is from Adafruit and can be found here: https://learn.adafruit.com/adafruit-powerboost-1000-basic/downloads#schematics General information: The TPS61030 boost converter ( datasheet ) utilizes a signal on its EN pin to turn the converter on or off. If the EN pin is set to 0VDC, the converter is put into shutdown mode which effectively isolates the load from the input. When the EN pin receives an input >= 1VDC (but less than the 7VDC maximum), the device is put into operation mode and starts to function. The circuit is designed for a 3.7VDC (nominal) LiPo battery. Under ideal conditions, the circuit boosts the 3.7VDC from the battery to ~5.2VDC required by many devices. I believe the +0.2VDC is meant to compensate for low-grade wires. In their diagram, Adafruit pulls the EN pin high and uses a switch to GND to turn the device off. The schematic resembles the following: simulate this circuit – Schematic created using CircuitLab When the switch is open, the converter will enter operate mode. When the switch is closed, the device converter will enter shutdown mode; however, the circuit will become a parasitic drain on the battery. It seems odd to me that the circuit was designed to draw current when the converter, and by extension the device being powered, is meant to be off. I know it will only be ~0.02mA but I do not understand why they wouldn't pull the pin low and use a high side switch instead such as in the following schematic: simulate this circuit The above circuit will form a voltage divider which will output ~2.4VDC to the EN pin which is more than enough to enable the device. When the switch is open, the EN pin will be pulled to GND and the converter will enter shutdown mode. This would eliminate the parasitic drain when the converter is in shutdown mode. Questions Why they are pulling the pin high instead of low to eliminate the parasitic drain? I am a novice in electronics so I assume Adafruit has a good reason for doing so and I'm just too inexperienced to understand why. Any input would be greatly appreciated! <Q> Possible reasons: part count, reliability, lack of concerned about that that level of leakage, avoiding complexity, default-on behavior for user convenience (from rioraxe in the question's comments). <S> To expand on leakage current being of little concern: check the specification of the TPS61030: 20uA (typ). <S> Then 1uA (max) in shutdown. <S> What will another 20uA of leakage through the pull-up do? <S> Expanding on laptop2d's calculations: 21uA leakage from a 1000mAh battery gives 47,600 hours of "standby" time (discounting battery self-discharge). <S> Over 5.5 years! <S> The self-discharge of the attached secondary cell and the usage of the device are certainly of greater power-loss concern than shutdown leakage! <S> Leaving shutdown then trades pull-up current for the converter's quiescent current. <S> Thus, the expected use of this board is not greatly concerned about leakage currents in comparison to the 100 to 1000 <S> + <S> mA loads <S> the battery will see in normal operation (e.g. phone charging, running an rPi). <S> If you were using this for something other than a USB power bank, you might be concerned about battery life. <S> However, long-lifetime battery operated devices usually don't come equipped with 4A-switch boost converters. <S> Note: <S> the proposed 200k pull-up and 400k pull-down with switch circuit would not work. <S> It is put into a shutdown mode when EN is set to GND. <S> This looks to be a normal logic input, so it is intended to be driven close to the rails. <S> It is not a comparator input like the LBO pin or certain other regulators that have precision-threshold enable inputs. <A> Will it become a parasitic drain? <S> Depends on the definition of parasitic. <S> The answer is read the datasheet again and again, if you don't understand something look it up. <S> From the datasheet: <S> Shutdown current VEN= 0 V, VBAT = <S> 2.4 V Current of enable pin = <S> 0.1 to 1 <S> μA <S> 1uA of current from a 1000mA*H battery will last you 1e6 hours, I hope thats enough for you. <S> Thats why we love FET's A bigger worry than the enable pin <S> is the part itself, it has 25uA of Quiescent current. <S> That's only ~38k hours on the same 1000mA*H battery. <S> A better way to handle this in your schematic is have your entire device switched from the battery if possible and tie the enable to Vbat. <A> Actually looking at this again, they are using it as a pull-down and the resistor is simply a load resistance so as not to have a dead-short. <S> Disregard: <S> So this may not be the entire concept, as I haven't had a chance to look at the datasheet in detail, but you've pretty much addressed it with your insight about the voltage divider acting in the same way. <S> You could use this 400ohm resistor as a current limiting resistor, but the voltage drop is undesirable and this may change the "loading" of this pin when the switch is closed. <S> Given that this is looking at a voltage, I'll assume it's a very high impedance so this would prob be negligible, but it should be noted that this will add resistance to ground which can cause looping issues. <S> In both of your configurations the resistors are acting as pull-ups, just one is a stiff voltage divider which can introduce a ground differential.
| The job of a pull-up is to ensure that the voltage level is correct when switched, this seemingly being an "active low" you want to make sure that it stops in good time, so pull-up, flip the logic and you may need a pull-down. From the datasheet, The device is put into operation when EN is set high. And the pin varies from part to part or is dependent on the voltage so it could even be lower than 1uA.
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How did smacking electron tube TVs help? In old movies (or new movies playing in these times), I often see people smacking the top of electron tube TVs or screens. Somehow it seems to help to stabilize/sharpen the picture. But why? Are there any reasons an electronic circuit could go from not working as expected to working as expected from a sharp mechanical jolt? What are the conditions that cause such an issue in the electronics? It would appear that the Atari ST suffered from a similar problem that could be fixed by performing what became known as the "Atari drop" (as discussed in this section on wikipedia ) although this is described as being due to loose connections, are there any other failure modes that could be remedied in a similar manner, and why? <Q> This practice is generally known as 'Percussive Maintenance'. <S> Any touching contacts, for instance in connectors, valves and their bases, and between the wiper of a potentiometer and the track, have a tendency to build an insulating film between the contacts. <S> This happens most readily at higher temperatures, in high humidity, and when there is airborne contamination, especially sulphides. <S> This can introduce higher resistance, non-linear behaviour, or break contact altogether. <S> It can produce intermittent behaviour, changing with humidity, or voltage across the junction. <S> Sending a mechanical shock through the equipment can move the contacts with respect to each other, disrupting the film, and restoring contact. <S> In the case of a TV with a loudspeaker in it, sometimes the vibrations from the audio will change the contact state. <S> High contact pressures, and gold plating, improve the reliability of contacts against this sort of problem. <S> TV valves, because they got hot, were especially vulnerable, and of course they aren't used these days. <A> It's worth noting that older TV's were constructed with point-to-point electronics soldered by hand, lacking a firm place for the components to be anchored to, such as this image showing the underside of the chassis of a 1948 Motorola VT-71 7" television. <S> (image taken from the wikipedia page for Point-to-point construction ) From the image alone it is clear why a solid whack to the box could provide enough energy to move loose components and bring them to rest in such a state they go back to working correctly. <A> Bad connections, corroded vacuum tube sockets making poor contact, cold solder joints and so on could sometimes be temporarily mitigated by the judicious application of 'percussive maintenance' techniques. <A> This was not just a tube-era practice. <S> In transistorized-era displays <S> (ie, where the only remaining vacum tube was the CRT itself) <S> the target of percussive maintenance that could occasionally get more use out of a failing monitor was most likely magnetic components such as transformers, chokes, the flyback, etc. <S> These tended to be physically large and massive <S> so could develop intermittent connection to the PCB as a result of temperature cycling or shock, and also to be composed of parts such as core laminations which could loose mechanical fixation if their cementing compounds failed and audibly vibrate, often to the detriment of image stability. <A> [ <A> I am no expert, but I remember when I was a kid, doing this to my PC 486's 14" monitor (brand was Leading Edge). <S> at some points of time, first appeared to be randomly, the display was going bananas, showing something like C64's loading screen colors but with much higher frequency <S> (it did not matter whether I was in the black DOS prompt or a colorful game!). <S> The only solution I knew as a kid was to punch the damn thing, and after few punches everything was back to normal. <S> Later in time, even kicking the damn thing was not helpful, so I called my dad to have a look, he found out that the 15pin RGB at the monitors end is lose from inside the cabinet, soldered it back again and it was fine and dandy again. <S> I have no experience with old CRT TV's, but my guess would be the punching momentary fixed the position of some lose passive components e.g. caps/resistors/inductors or wires. <S> not sure even if a calculated punch could tell electrons whether or not show clean or distorted picture on the screen.
| It could be down to exposed conductors touching, or a poor solder joint being loose and the movement provided from a solid hit moves the offending components back into a position that they work again - by either breaking connections that shouldn't be connected, or by doing the opposite to the components that should be connected. The contacts in the drum/turret tuners were a big contributor.
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Solder paste not wetting at all I have an issue with solder paste, I would like to know its origin so I can fix the problem and solder by components properly. I use a lead free Sn42/Bi57.6/Ag0.4 solder paste, manufactured by ChipQuick. Here is the datasheet. The syringe I'm using was opened three weeks ago and stored at ambient temperature until now. (I close it with the protective cap between each use, of course.) I ran some tests before actually using it to solder components. I simply deposited several bits of it on a copper board, which I previously wiped with alcohol. I have at my disposal a soldering oven (not some salvaged toaster, a real oven designed for this application).) However, it works like regular timer-ovens: set a time with one button, and set a temperature with another. This is the process I used so far: I put the board in the oven, at ambient temperature I start the oven at 90°C and I wait one minute I set it to 140°C and wait for two minutes I set it for 180°C and wait for the solder paste to "melt" and get transformed to actual solder Finally, just after the activation, I turn off the oven and open the door to allow a quick return to ambient temperature. Problem is, I always end up with a nice sphere instead of observing spread solder across the copper face.Exactly like this : I want to know if I am doing something wrong during the process, or if this is linked to the storage conditions of the solder. Note that the manufacturer indicates a good "shelf life" but I don't know if it implies that the container should not be opened. <Q> My guess would be that the copper board is not being given enough time to heat up. <S> Due to its thermal mass the copper heats up much more slowly than the solder, and the solder melts before the board reaches the correct temperature. <S> If you choose a smaller piece of copper, or an etched PCB with less copper on it, or leave the copper board in the reflow oven longer, the solder will eventually flow as expected. <A> I've heard bad things about desktop ovens like this. <S> They don't necessarily have the oomph to get the job done correctly. <S> Saying "I turned knob X to temperature Y and waited Z minutes" does not mean that you have any idea what's happening to your board. <S> The only reliable way of knowing would be to measure, maybe with a thermocouple in contact with the board (not perfect, but likely close enough). <S> You're obviously reaching an adequate temperature, because the solder is melting. <S> It's certainly possible that your oven doesn't provide enough oomph to actually heat the board, and the solder is melting on top of cold componets. <S> You may also be having problems with the flux. <S> Either the flux in the paste is past <S> it's prime, or the heating profile you're actually getting isn't giving the flux enough time to do its job, or the flux is activating too long before you bring your solder past the melting point, and a new oxidation layer is forming. <S> My advice is actually to forego no-lead solder unless there is some regulatory reason why you need to work with it. <S> It's just harder to use- requires higher temperatures, which makes it tougher to come up with the right temp profile short of using real equipment. <S> You may still have problems with lead, but probably less so. <S> Just as an aside, regardless of the nature of your oven, if it doesn't have heat-ramp-soak control with feedback, it isn't "meant for this purpose". <S> Update -- given the low temp nature of the Chipquik, the comments on no-lead solder don't apply. <S> I think it might highlight the issue of premature and prolonged activation of flux though, if the oven is a very powerful one. <S> No real way of telling whether its that or a cold board, though, without measuring. <S> Temp crayons might shine some light on this. <S> Lead solder might actually help. <S> Flux activation temps are better documented, so the soak profiles can be tweaked to slow things down before activation to avoid oxidation. <A> I suspect a problem with the copper board. <S> What about the surface, is it bare copper only, or is it covered with tin? <S> I would try a conventional soldering iron and lead rosin-core solder to do some test joints. <S> If the solder does not flow well, there is something wrong with the board. <S> The surface may be oxidized or the copper areas are to large to get heated. <S> Wipping the board with pure alcohol does not remove copper oxide from the surface, very fine abrasive paper does. <A> There's a problem with your solder paste. <S> May be <S> the paste you are using is expired. <S> If you mix it with flux it will look like mud and if put it in oven it will never melt. <S> You can verify that paste with heat gun or solder iron. <S> You will get the same results as in the oven. <S> In the past I suffered the same problem and that was resolved by changing the solder paste. <S> Hope this will help you
| It's probably just that the large thermal mass can't heat up enough before the solder melts.
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Is there a way to adapt a TSSOP pad to a SOIC-16 IC? So I messed up and printed a circuit board with 16 TSSOP pad when I actually need a SOIC-16 pad!!! Is there a clever way of somehow adapting the TSSOP pad to a SOIC-16 chip? <Q> The fastest way to deal with this is to "Dead Bug" your IC. <S> Find a clear place on your board to glue the chip to it, leads up in the air (hence "dead bug"). <S> Solder 32AWG wire wrap wire to the pins, and to the appropriate pads. <S> It will take some practice and patience. <S> Obviously, if you're doing more than a few, redoing the board is the better approach. <A> Adafruit makes breakout boards for 16SOIC and 16TSSOP parts on a single board, with one side SOIC: <S> And the other side TSSOP: <S> So if you had enough room around your chip (a big if, understandably) you could use a heat gun/reflow oven and solder paste on the TSSOP side to attach it to the main board and then solder your chip to the SOIC side. <S> The pins on both sides are connected electrically by those through-holes, which are normally used for a row of pin headers. <S> If that's not an option <S> then I'm going to echo the sentiment of previous answers and say that, for more than a few boards, you should just correct the layout and get more made. <A> Unfortunately there's not really any kind of adapter or way to convert the two package sizes. <S> Although the pin pitch is: TSSOP pin pitch: .635 <S> mm SOIC pin pitch: 1.27mm which would make it possible to bend the TSSOP pins onto the SOIC pad <S> , I really wouldn't recommend doing that. <S> Your best bet unfortunately would be to correct the design of the boards and get some more printed.
| Another solution could be to solder Kynar wire onto the Pins of the TSSOP and then in turn solder that onto the pads, but again it's not really recommended as it's a fiddly job and is not really practical for more than a few boards.
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Use of a single cell lipo for VIN enough to power motors and board I successfully powered two tiny brushed motors and an MCU using a single 1S Lipo (3.7V) battery 300mAH. I did this by supplying battery power directly to the motor controller and VIN in parallel. Per datasheet, the external power supply inputs voltage range for VIN should be 7-12V. The motors spun even though I am below that range, why? In this case, using just 3.7V for VIN. This makes me believe that 3.7V can be used for the +5V input as well to run the motors. If that's the case, then I won't even need the LD1117 voltage regulator if I plan to use a 1S to power everything, correct? I would still need the LD39050 voltage regulator if I plan to add logic devices. Questions:Is there a more efficient, less power drain, or reliable way to share a battery between MCU and motors than in parallel as done above? I read that I may need to decouple the shared supply for noise. Is there anything I need to be aware of when using a small 1S Lipo to power everything as a standalone unit besides less battery life? A 5V voltage regulator is unnecessary if my power supply is not above 5V, correct? I am trying to make something like the jumping parrot drone without the jumping, but make it smart enough that it doesn't just die because of terrible design flaws. I have plans to incorporate wifi, sensors, and battery management IC on top of the motor controller. My goal is to have it drive but only using the 1S. The brushed motors I used are taken from a quad and they have no specs on them that I know of other than 5.5mm diameter and 18mm length. I just know they use 1S Lipo. STM32 Nucleo-64 boards page 60/66 <Q> So yeah, providing the board with 3.7V on the VIN pin will result in a massive under voltage on the supply rails due to regulator dropout. <S> The LD1117 used on the input has a dropout of around 1.2V which is then followed by a diode which will have around 0.5V dropout - as a result the 5V rail will end up being 2.5V or so assuming a fully charged LiPo (4.2V - 1.2 - 0.5) or closer to 2V for a nominal charge (3.7 - 1.2 - 0.5). <S> The 3.3V rail will then be closer to 1.8V based on the second regulator which is only around 0.2V dropout. <S> I'm honestly surprised the MCU actually runs correctly at that voltage, though it is probably just about within spec. <S> Bypassing the 5V regulator would allow the 3.3V rail to correctly regulate as it has sufficiently low dropout voltage to do so <S> - you need at least 3.5V in order to maintain the output voltage of that regulator. <S> As the LiPo discharges it will get down towards 3V, by which point the 3.3V rail will have begun to drop off again. <S> So it is still not ideal, but should be workable for a large portion of the battery charge. <S> To bypass you would need to connect the 3.7V to JP5 pins 2 and 3 <S> leaving pin 1 disconnected. <S> You will still not have a 5V rail, however all parts of the board use 3.3V rails which are regulated using ultra low dropout regulators from the 5V line which should be able to maintain regulation even with a LiPo on their input. <A> I have a few concerns/comments with your approach. <S> 1) <S> Do you plan on putting a battery controller on this so your lipo does not go into deep discharge? <S> This is a potential fire/safety hazard. <S> You might be able to get a controller/regulator in a single package. <S> 2) <S> Non-isolated logic/motor supplies can cause logic dropouts. <S> 3) <S> If you need 5V from a lower power source (3.7V), a charge pump may be useful. <S> 4) <S> It should be reasonable to to power an LED from 3.3V, check specs of LEDs. <S> Some need higher voltages. <S> 5) Consider using an h-bridge or a motor controller, <S> that way you can get at least forward/reverse control. <A> LiPo batteries are rated as nominally 3.6V or 3.7V but have a working voltage range of about 3.0V to 4.2V. <S> I the battery was above 80% charge <S> it would have been at 4.2V on light load. <S> A motor spinning and spinning usefully may or may not be quite different things. <S> Only you can tell (as you've given us no data) whether the motors will meet your needs at 3V. LiPos will discharge to below 3V if you make them do so <S> but wise use is to stop discharge at about 3V minimum. <S> Whether your processor will run reliably at 3V at the clock speed you intend to use depends on whether your processor will run reliably at 3V at the clock speed you intend to use. <S> ie until YOU can state exactly what your design is nobody can tell you what the undisclosed system will do. <S> Changing Vbat across 3V - 4.2V range may be OK or may not. <S> You need to tell us more. <S> At Vbat = <S> 3V you do not want to waste ANY voltage drop - no series diode, no usual regulator - a super low dro out LDO of\r a controlled FET switch. <S> Just connecting them is likely to have predictable and undesirable results. <S> Tell us ALL about what you want to do and not HOW you intend to do unknown things <S> and we can help.
| Your motor may or may not meet your need at 3V. Unless you DESIGN what you do, your approach invites sorrow and disaster, alas. This should ensure power is provided to all parts of the board which need it, and also mean that if you connect the USB port it doesn't start to charge the battery incorrectly. You CAN successfully parallel motors and processor and more from the same liPo cell IF you filter and design correctly.
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Purpose of voltage amplifier (VAS) in three stage amplifier design Can someone explain me benefits of this design (found in Douglas Self's book) in comparison with "classic" npn-based long-tailed pair? What second stage actually does? <Q> The second stage obviously provides additional voltage amplification. <S> However, it has an additional function as well. <S> The capacitor from collector to base provides the dominant pole compensation that makes the amplifier stable (in general from unity gain to very high closed-loop gain.) <S> Since this is likely an integrated op-amp, where capacitance take a lot of area, the compensation takes advantage of the Miller effect to multiply the effective capacitance by approximately the gain of the stage. <S> The Miller Effect <S> Also, the PNP input stage allows the common mode voltage range to get much closer to the negative rail. <S> Could be useful for single-supply application. <A> It also rolls off the high frequencies (the miller capacitor) so that negative feedback applied externally does not cause instability or oscillations. <S> An op-amp should have lots of open-loop gain, be stable (closed loop), provide a low output impedance and a high input impedance thus, the 2nd stage greatly enhances the gain whilst providing stability when negative feedback is added. <S> Regards the use of PNP transistors, this is of no consequence to what stage 2 does. <S> An NPN input pair will still have a stage 2 <S> but, it might be a PNP transistor. <A> There is an additional task the second stage must fulfill: Correction of unsymmetry caused by the first differential stage. <S> The problem is that the npn current mirror cannot be fully symmetric (cannot provide two equal DC currents). <S> The current through the most right pnp transistor (with the "-" sign) is larger than the current through the left pnp transistor. <S> The difference is caused by the two base currents (IB1+IB2) of the current mirror npn transistors. <S> However, balancing the currents through the diff. <S> amplifier is important for the performance of the diff. <S> pair. <S> That means: Without any input signal, the quiescent DC currents of the pnp´s should be equal! <S> Now - this problem can be solved when the current gain and the operational point of the second-stage npn transistor is selected properly: Its base current must be identical to (IB1+IB2). <S> In this case, the quiescent DC collecor currents of both pnp transistors can be made equal. <S> Then, the diff. <S> stage is fully symmetric and delivers a signal to the second stage which depends on the diff. <S> input signal only. <S> EDIT (comment): There is an additional aspect - as far as the second stage transistor is concerned: <S> Further improvement of symmetrical properties of the differential stage. <S> The collector-emitter voltage VCE1 of the left <S> npn current mirror transistor is identical to VBE of the second stage transistor (app. <S> 0.7V). <S> Hence, we can assume that VCE1 is practically identical to the coll.-emitter voltage VCE2 of the other ( right ) current mirror npn transiustor <S> (VCE1=VCE2=VBE2). <S> Therefore, the quiescent DC voltages VCE of both diff. <S> stage pnp transistors are alo equal. <S> This symmetry improvement eliminates variations that would result from larger voltage swings within the diff. <S> stage as well as at the collector of the second stage transistor <S> (VCE only ap. <S> 0.7V).
| The 2nd stage takes the output of the diff pair and applies a lot of voltage gain.
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Can all USB Type-C to Type-C cables themselves handle 100 W? I am designing a machine and trying to make it look aesthetically pleasing. Could I use off the shelf USB Type-C to Type-C cables to handle both power deliver and encoder data to/from a motor? The DC motor is about 80W (@24 V) and uses 2 leads for power and 8 for the encoder. Could I technically use just one USB type cable to achieve this? I would not use the type-c connector, just the cable. The encoder has differential wires so hoping interference would not be an issue. I understand that passive type-c cables are rated for only 3A but is that limitation due to the cable or the connectors and electronics in them? <Q> Don't do this. <S> Always think about the user. <S> The user sees a USB port and WILL connect it to another USB port of his PC. <S> Instead, go for exemple with some kind of M8/M12 connector. <S> Those can be nicely integrated into your housing, you get it with all kinds of connector configurations and the user is unlikely to plug it into a USB port. <A> If you are using them for none usb spec applications, then all you are really interested is in the gauge of the power wires. <S> Keep in mind that usb C is not rated for 3.5+ Amp at 24V, and even at spec rates, not all manufacturers of Usb C cables will meet spec. <S> Many Usb 2.0 cables won't even meet Usb Dedicated Charging Port specs. <S> Some barely meet Usb 2.0 data specs of 2.5W <S> (5V +- 0.25V <S> [2.5%] @ 0.5A). <S> You pay for what you get. <A> Can all USB Type-C to Type-C cables themselves handle 100 W? <S> That's really not the right question since the power (hopefully) isn't consumed in the cable. <S> When choosing a cable you need to be mindful of: <S> Current carrying capacity . <S> This is affected by wire material (Al, Cu, etc.), cross-sectional area and its ability to dissipate heat in the working environment (free-air, conduit, etc.) <S> without melting the insulation. <S> Voltage rating . <S> How many volts can the insulation withstand between cores and between cores and ground. <S> High frequency requirements . <S> This is beyond the scope of this answer but includes EMI, cross-talk, shielding, etc. <S> Acceptable voltage drop . <S> Voltage drop will be proportional to the product of length and current. <S> To meet the maximum allowed voltage drop requirements the cross-section area of the wire may have to be increased. <S> So a better question would be: <S> Can all USB Type-C to Type-C cables handle 3.5 A at 24 V. Figure 1. <S> Hybrid power and data cables for servo applications. <S> Source: <S> MacRAE'S BLUE BOOK . <S> Lapp, for example, do a wide variety but you may have difficulty buying in small quantities.
| Depending on the size of your project it may be worth searching out suitable quantities of hybrid servo cable.
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relay or switch that remember previous power status after power failure I have application which need a push button on/off switch feature. But i need to do something buy which it can remember or store in memory the previous status after power failure. same like home air conditioner which save previous status if power cut and return. One of my suggest do it with IC 555 but i don't think it can remember the previous status if power cut... please give me a cheap solution.. my application consume only 12V and 2A. Edited :Purpose of this solution is to power on/off remote application which dose not have a UPS or power backup. I am using Arduino with Ethernet module to remotely on/off system power but the problem during long power outage.. application shutdown and i am needed to power on it again.. I looking for a Solution to store state of application even if power reset. <Q> A latching relay is designed for this very purpose. <S> "A latching relay is a two-position electrically-actuated switch. <S> It is controlled by two momentary-acting switches or sensors, one that 'sets' the relay, and the other 'resets' the relay. <S> The latching relay maintains its position after the actuating switch has been released, so it performs a basic memory function. " <S> Unlike ordinary relays the latching relay has two sets of coils which move the contacts into one of two positions. <S> In a power outrage they stay in the last position set. <S> A simple toggle circuit could be used to enable the use of a single push switch to operate both coils (first push ON (coil 1), second push OFF(coil 2)) <S> e.g. http://www.azatrax.com/latching-relay-circuits.html <A> Offhand, I'd say that an EEPROM is the obvious solution for this kind of thing. <S> They store state even in the face of a total power failure. <S> They're fairly inexpensive, as little as $0.29 each in small quantities <S> (ex: https://www.jameco.com/z/24C01-Major-Brands-2-Wire-Serial-EEPROM-1K-128-x-8-DIP-8_276592.html ). <S> The biggest downside is that it means adding extra circuitry to the system, to read/write the EEPROM. <S> I'm pretty sure you could get by without adding a full fledged micro-controller, but it would definitely require at least a few extra components. <S> Using an SD card could also be an option, but I think this would be even more complicated and more expensive, which might make it less than ideal. <A> I have application which need a push button on/off switch feature. <S> But i need to do something buy which it can remember or store in memory the previous status after power failure. <S> Figure 1. <S> A typical toggle switch. <S> Infinite memory duration. <S> Independent of supply voltage. <S> Status can be read even with power disconnected. <S> Very high reliability. <S> Low cost. <S> Multiple suppliers. <S> Tactile feedback during switching. <S> Uses no power. <S> Available in a wide variety of voltage and current ratings. <S> No electronics required. <S> Edit after remote control requirement explained. <S> Figure 1. <S> The Digital Loggers series of devices is an example of a web-controlled power switch (for the North American market). <S> There are a range of devices which perform the functions you require. <S> These are used in a variety of applications such as server rooms where they are particularly useful in forcing a reset of an unresponsive device by cycling the power. <S> The device above has switched and unswitched sockets and power status memory. <S> Search for "web-controlled power switch" and you should find something suitable for your country (India) at a good price. <A> Maybe your Arduino version has a "brown-out" detection feature, as some others microcontrollers do. <S> In which case, an interruption would be fired, so in your ISR you could store that event and more data, from where to read when the Arduino boots up again. <S> Regards. <A> Arduino Uno has 1KB of EEPROM which is prety enough to store status of 1024 devices. <S> There is no need to use external EEPROM for this. <S> Here is the link to examples and tutorials. <S> I suggest you to write the code in a such a way to execute minimum read/write operation from EEPROM, <S> since EEPROM has a life time depend on number of read/write cycles.
| A toggle switch, instead of a push-button switch, meets all your stated requirements and more.
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How to integrate a signal in LTSpice? Is it possible to integrate a signal in LTSpice and plot the result? <Q> <A> Summary Yes, it is possible to integrate a signal in LTSpice. <S> Use .meas directive. <S> To plot the integral use .step directive. <S> Details Below is an example of how to calculate average charge supplied by a switched capacitor circuit as well as an average current flowing in it. <S> To calculate the charge we integrate the actual current over one period. <S> To calculate the average current we further divide the charge by the time period value. <S> Suppose, there is the following switched capacitor circuit. <S> The period of non-overlapping clock signals \${\varphi _1}\$ and \${\varphi _2}\$ is 10n (clock frequency f is 100 MHz). <S> Let’s first calculate analytically the average charge and current flowing in the system: $${Q_{one\_period}} = <S> C <S> \cdot <S> (VDD - {V_{test}}) = <S> 100fF <S> \cdot <S> (1V - 0.5V <S> ) = 50fC.$$$${R_{eff}} = {1 \over {C \cdot f}} = <S> {1 \over {100fF \cdot 100MHz}} = 100k\Omega <S> $$$${I_{average\_one\_period}} = {{VDD - {V_{test}}} \over {{R_{eff}}}} = {{1V - 0.5V} \over {100k\Omega }} = <S> 5\mu A$$ <S> Now, let’s get these analytical results in LTSpice. <S> Build the following circuit: <S> Code for the LTSpice directives: .param <S> P=10n.param t=5n.step param t 10n 40n <S> 10n.meas tran Charge_one_period INTEG I(Test) <S> TRIG time VAL= <S> t <S> -P/2 TARG time VAL= <S> t+P/2.meas tran <S> Average_Current_one_period <S> INTEG <S> I(Test)/P <S> TRIG time <S> VAL= <S> t <S> -P/2 TARG time VAL= <S> t+P/2 <S> Notes: <S> a) P stands for integration interval, t are measurement points (10n, 20n, 30n, 40n). <S> b) <S> Download link for the file cmosedu_models.txt (PMOS model P_50n used). <S> Now run simulation to see the actual current flowing in the circuit: <S> Not really enlightening, isn’t it? <S> However, if we use our directives: View -> Spice Error Log - <S> > <S> RClk -> Select “Plot .step’ed <S> .meas <S> data” -> <S> In an opened window RClick -> Visible waveforms and select the calculated data: As one can see, the simulation result for the charge supplied and average current correspond to the analytical calculation above. <A> Without stealing @FakeMoustache's suggestion (and since the OP didn't specify), I'll just add, along his lines, that definite integration (the basic moving-average) is also possible: <S> Here, integration from 0 to 1 for a 1Hz raised cosine. <S> Note that V(b) is plotted with a 0.1V DC offset, for better comparison. <S> As seen, both methods yield the same result (V(a) and V(b)). <S> Despite the larger number of elements and nodes, the second suggestion may come faster and more orecise, as the tline has a fixed delay that is not dependent on the sampling rate (i.e. the simulation timestep). <S> Of course, given the conditions, a simple behavioural source may be more convenient -- <S> this is, entirely, a user's choice. <A> Simply use a 'B' behavioural voltage source with the integral function. <S> V=idt(V(sig)) <S> Where sig is (in this case) <S> a node voltage signal you want to integrate. <S> You can gate the integral using either time or some system state.e.g. <S> Using a node voltage called 'trigger' V=idt( u(v(trigger)) <S> * v(sig) ) or, using time as a gate <S> V=idt( (time > 1m & time < 3m <S> ) * v(sig) ) <S> Or, using the reset parameter to reset the integral after 5 milliseconds V=idt( (time > 1m & time < 3m <S> ) * v(sig), 0, time > 5m ) <S> To plot the result, simply give the B source output a name and plot that name. <S> If you want you can divide the 'voltage' by '1V' to make it unitless. <S> From the manual, there are two forms of integral: 1 idt(x[,ic[,a]]) <S> Integrate x, optional initial condition ic, reset if a is true. <S> 2 <S> idtmod(x[,ic[,m[,o]]] <S> Integrate x, optional initial condition ic, reset on reaching modulus m, offset output by o.
| Sure, you can use the idt function, for example with a behavioral voltage source, we see the integral wrt time of a sine wave with an offset.
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How to use IO pins in PIC microcontroller as PWM? I am trying to use a PIC18F4520 microcontroller to drive 8 servo motors. For this I need 8 PWM signals, but the PIC unit has only two built in PWM. So, I decided to use some normal I/O pins as PWM, and searched how to do that. I found some hardware based solution like using extra IC's or software based solution as using interrupt based PWM. But why can't I just use this code segment: while(1){PORTA.RA0=1;delay_ms(10);PORTA.RA0=0;delay_ms(10);} The pin is high for 10ms and low for 10 ms.So the cycle period is 20ms. Shouldn't this create a 50Hz signal with 50% duty cycle? Or am I missing something? <Q> The delay() <S> function causes the PIC to loop and prevents any other activity other than interrupt triggered routines. <S> In your example it will turn on RA0 for 10 ms, pause (preventing servicing any other output) and then turn off RA0 for 10 ms with the same inhibiting action. <S> The pause() function should really be hidden from beginners as it encourages this kind of poor programming practice. <S> Instead you should run a base timer for your PWM outputs and do something like this pseudo code. <S> while(1) { timer += 1; // increment the timer. <S> if (timer > tmax) { <S> timer = 0 <S> ; // reset } PORTA.RA0 = <S> (timer < t0ontime); // <S> Output on until ontime exceeded. <S> PORTA.RA1 = <S> (timer < t1ontime); PORTA.RA2 = <S> (timer < t2ontime); PORTA.RA3 = (timer < t3ontime); PORTA.RA4 = <S> (timer < t4ontime); PORTA.RA5 = <S> (timer < t5ontime); PORTA.RA6 = <S> (timer < t6ontime); PORTA.RA7 = <S> (timer < t7ontime);} It's been many years since I've used the PIC compiler <S> so this may not work as written, but you get the idea. <S> You should make the timer increment on a fixed timebase rather than the way I've shown it so that it won't vary if you change the chip oscillator, etc. <S> One other improvement you could make on this code is to avoid starting all the pulses at the same time. <S> This would smooth out the demand on the power supply and may help reduce noise a little. <A> You have bigger problems than not enough hardware PWM outputs and your software attempt not working. <S> You misunderstand the pulses such hobby servo motors require for control. <S> These units work only on the pulse length. <S> Usually the 0-100% travel range results from 1 to 2 ms pulse width. <S> Each pulse tells the controller in the unit where you want the motor position to be. <S> This position command should usually be repeated every 20 ms or so for smooth operation if the commanded position is changing. <S> There are several ways to use hardware PWM to do this. <S> Another is to multiplex the output of a single PWM generator. <S> Use the CCP module in one-shot mode. <S> First, switch the output multiplexor to a particular servo unit, generate the pulse for that unit, switch the output mutiplexor to the next servo unit, etc. <S> Since the entire travel range is encoded in 1 ms of pulse width, this is really something you want the hardware doing to get the resultion without jitter. <A> Shouldn't this create a 50Hz signal with 50% duty cycle? <S> Or am I missing something? <S> It <S> does (approximately). <S> The problem is that RC servos require a pulse width of 1 to 2ms, not 10ms. <S> The signals you should be generating look like this:- <S> The channel pulses can be output one after another, using a simple software delay timer if you don't need to do anything else at the same time. <S> However since the total time for each 'frame' may vary from 8 to 16ms depending on the individual pulse widths, you need some way to bring it up to 20ms. <S> Another way is to set up a hardware timer that fires every 20ms. <S> You could also have all this in an ISR so that the gap time is available for other 'background' processing which is not timing sensitive. <S> NOTE: <S> the channel pulse widths need to be produced precisely with low jitter (<=2us), but the frame time doesn't. <S> Most RC servos are designed to handle any frame time between 16ms and 22ms, and and an even longer time won't hurt (the servos will just move slower).
| The first obvious option is to get a PIC with more PWM outputs. One way is to add up all the pulse widths and subtract from 20ms to get the extra time delay required in the gap.
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Strain Gages Measurment Error and Saturation I have an experiment composed of 16 strain gages, mounted on a steal beam. The steal beam is horizontally placed under a compression cylinder for a stress test. the test took around two hours to complete and composed of different stages and levels, the maximum load achieved on the was of 160KN. The stress machine is a YLE compression machine with National instruments PXI modules for data acquisition After retrieving the measurement results, most of the gages have stopped changing in value and become somehow saturated as shown in figure 1 while the remaining strain gages did not preserve a linear variation, the installation used a quarter bridge and the results were calibrated at the start. The strain gages used in this experiment is the Vishay strain gages general purpose linear module 125BT http://www.vishaypg.com/doc?11196 So the questions are:- How did the strain gages get saturated? and how to choose the best strain gage?-what are the causing errors/noise which can cause such behavior? Does the amplifier gain limits affects the resulting values? if so why didn't it affect all of them?-what are the most common errors and noise generated when installing and using strain gages (the strain gages where soldered to a 10m shielded cable and connected to the machines inputs) Thank you <Q> Figure 1. <S> Strain gauge differences. <S> It seems to me that there are a few problems. <S> Gauge 1 seems to be bearing the brunt of the load with 2 and 3 following. <S> All three seem to have reached some limit and deformed permanently. <S> The traces around 4 seem to be behaving normally with one (green) exception. <S> The traces at 5 are going the opposite direction which suggests that they are in tension rather than compression. <S> The first and last points make it appear that the setup is mechanically faulty, that the load is not balanced between the cells and some leverage is occurring causing some gauges to go into tension. <S> This will put further stress on the sensors in the pivot area - probably 1, 2 and 3. <A> Which gage series, exactly, are you using? <S> Most of the series listed in the doc you point to have a max strain of 1.5 or 2%, but the EP series goes to +/-20%. <S> You might try using that one, if the saturation is mechanical. <S> A quarter bridge, I assume, is just a constant current running through your gage. <S> If the saturation is because of the current supply reaching its voltage limit, you could try turning down the current through the strain gage. <S> We have no way of knowing the answers to your more detailed questions unless you provide more detailed info. <A> The strain gauges have not saturated, actualy the metal body on which the gauges were applicated has bent. <S> According to Hook's law you entered in a region of no return, causing a plastic deformation of the body.
| If you've over-strained the devices, I don't know that they're still reliable.
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RF over sampling to increase SNR Using a 32MHz pic processor, and over sampling the demodulated analog output of an RF receiver (instead of the data slicer output) is it possible to use averaging to reduce increase the SNR of the system? The data is NRZ OOK at 4.5kHz. Update I found this article from Chipcon on the subject, but it addresses Manchester encoding, though somewhat applicable to NRZ. <Q> Assuming the PIC has an analog input capable of operating at 32MHz or significantly higher than 9kHz (Nyquist Frequency), then yes it is possible to use averaging to increase the SNR. <S> It is important to consider the issues with respect to your SNR. <S> For example, if the SNR is low due to transmission distance (attenuation), then averaging may not be very effective, whereas if transient spikes or emi are the source of your issues the averaging could be very helpful. <S> In both cases, you may decide you need a hardware solution such as a pre-amlifier or a bandpass filter respectively. <S> I suggest you use an oscilloscope to look at the eye diagram , this will help reveal the specific problem with your received signal. <S> Silabs have a useful note on oversampling and averaging here . <A> You want to improve upon bit error rate of the receiver's internal data slicer. <S> In on-off-keying (OOK), internal data slicers do a pretty good job in the face of signal strength that varies greatly. <S> The programming effort in implementing an adaptive data slicer is considerable. <S> OOK decoding requires different algorithms than Manchester as well. <S> For example, your NRZ OOK modulation bandwidth extends to zero Hertz, whereas Manchester bandwidth does not. <S> A truly excellent reference is Steven W Smith's e-book " The Scientist and Engineer's Guide to Digital Signal Processing ", where he gives a good explanation of how averaging reduces broadband noise proportional to square root of number of samples. <S> Chapter 15 is relevant to your situation. <S> Yes, there's math, but also many examples and illustrations. <S> Free PDF download. <S> Not much help with modem algorithms though, and how your receiver's on-board filtering already does some signal averaging. <S> If on-board analog filters are optimal, you'll not improve much on them with DSP oversampling. <A> By "averaging", you apparently really mean low pass filtering. <S> Averaging, or more specifically, box filtering, is a form of low pass filter, but not a particularly good one for most usages. <S> Yes, low pass filtering to reduce the high frequencies <S> you know aren't relevant increases signal to noise ratio. <S> You can look at this in the frequency domain and try to eliminate everything above the highest frequency you care about. <S> Or, you can look at this in the time domain and throw as much filtering as possible at the system while making sure the step response is still sufficient within the time of the shortest possible level. <S> True averaging is really only useful if you know when the bit times are. <S> Then you can average with as many samples as you can manage during one bit time. <S> The average is then the best indication of the measured bit level. <S> Put another way, this is really synchronous averaging, and can be quite effective. <S> Unsynchronized averaging is just silly. <S> If you don't know where the bit boundaries are, then some IIR (equation based) low pass filter will be better.
| Yes, you can oversample the receiver's analog RSSI output, and likely improve error rates, but you may be expecting too much.
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Diamond symbol in schematic I am having a lot of Diamond symbols in my schematic named as Pxxxx. I am not able to find what it means as there are no legends mentioned in the file. Also, are there any general conventions that govern the usage of symbols in schematics? <Q> Those look to me like testpoints. <S> On an actual board they would either be empty pads or they may actually be testpoint components. <A> I'll hazard a guess that these are test points . <S> These provide an convenient access to a signal. <S> Test points are useful for debugging, quality control testing, calibration, troubleshooting. <S> A signal can be probed at a test point. <S> Or, a test signal may be injected into a test point to exercise the downstream circuitry. <S> On the PCB, a test point can manifest itself differently, depending on the expected type of test. <S> It can be a throughole of surface-mount component for clipping a probe to. <S> It can be throughole for sticking an oscilloscope into. <S> It can be a surface-mount pad for automated test equipment. <S> As far as I know, there is no single standard schematic symbol for test points. <S> Here are a few symbols that come with schematic software. <S> (from OrCAD) <S> (from Altium) <S> The standard designator for test point it TP ( source ). <A> <A> The test points could document locations for probing on a bed of nails functional/ICT fixture defined by a Test Engineer working with the Design Engineer for high fault coverage for fixture design.
| It's either a net label or a test point.
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Purpose for NOR, NAND flip-flop I am currently studying flip-flops. I understand the purposes of flip-flops, and that they can be made by using NAND or NOR gates. But what are the reasons for making them these two ways? Does either way work for storing bits? Or is this just for cost consideration? <Q> There are a number of ways of building a flip flop depending on the style of logic and the circuit requirements. <S> One method is to use pass transistors and build something that somewhat resembles SRAM cells. <S> Another method is to use dynamic logic where the stored value is actually stored as charge in parasitic gate capacitance, however this results in a minimum clock frequency requirement to continuously refresh the stored value. <A> Even though the comments and the answers are quite reasonable, i would like to add some information that i assumed you didn't know based on your question [CMOS] <S> Why NAND and NOR gates are the building blocks? <S> Because all the CMOS circuits are complementary circuits <S> you can`t build a CMOS AND gate, you will end up building a NAND followed by NOT gate to form this AND gate, this is because PMOS is used as pull-ups while NMOS is used as pull-downs. <S> For more information you should look into This question .The <S> figures below shows the CMOS implementation of the NOR,NOT and NAND gates <S> What is the difference between NAND and NOR versions? <S> Let me give an example below are the NAND and <S> NOR implementation for this logic function One of the differences here is that A input in the NOR implementation is connected to only two transistors [NOT gate], while in the NAND implementation is connected to 4 transistors [2x NAND gates] <S> which means that in the NOR version the A input have HALF <S> THE CAPACITIVE LOAD than the NAND version <S> So in this example both versions differ in the loading on A input <S> you might start worrying about this capactive load if for example <S> A is connected to many other circuits <S> The gist is: depending on many factors one can decide which version is suitable for a circuit but both do the same function <A> I used to use S-R Flip-Flops made from NOR or NAND gates, depending upon what I was trying to accomplish with each particular design. <S> The simple way to think of it is this: a FF made from two NAND gates has active LO inputs. <S> A FF made from two NOR gates has active HI inputs. <S> Ignore the technology behind those gates <S> - this didn't matter for low-speed industrial systems. <S> You get 4- dual-input gates in a 14 or 16 pin package and the goal was to minimise the package count while maintaining extremely-good reliability. <S> You could get your choice of AND, OR, NAND, NOR, XOR, XNOR. <S> They all cost about the same per package. <S> The goal was to minimise the total number of packages in the design. <S> Many of my designs were used in industrial settings. <S> I would use FFs made from NAND gates so that the inputs were active LO - this allowed the use of standard "contact closure to ground" signals. <S> One other advantage to using NAND gates for FFs fed from external inputs: I would use cd4093 CMOS Schmitt-Trigger NAND gates with RC filters on the external inputs. <S> This gave me extremely good noise filtering and the series resistor provided significant ESD and transient protection. <S> Some of those designs were used for decades with no failures. <A> both NAND and NOR are universal gates because any circuit can be derived from these. <S> so you can use either one but <S> NOR is the better choice.
| They both do the same logic function, but sometimes one of them is better than the other. Well, in practice flip flops are usually built at a transistor level with a highly optimized circuit as they are used all over the place and need to be as small and as efficient as possible.
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230V 50Hz to 15V Switching Transformation I have an audio circuit which is supplied by a voltage of 9V dc and 12V dc. These voltages are produced by means of a conventional transformer which transforms the mains 230V ac 50Hz to 15V ac 50Hz. After the transformer there is a rectifier (diode bridge and capacitor) followed by two regulators, one for the 9V dc and one for the 12V dc. My problem is that the transformer produces a heavy magnetic field, which couples with my circuit producing a low hum. I am thinking about substituting the linear power supply with a switching one, thus avoiding the 50Hz magnetic field. Now: Is there around on the market a switching power supply with the requested characteristics? Is it better to use a power supply giving directly the 9V dc and 12V dc, or is it possible to Step down the 230V ac to 15V ac (without using a Conventional transformer) and keep in place the rectifier and regulators? <Q> Stray E or H fields are called Common Mode , meaning common to both {signaland ground} or {power and ground} and when floating or high impedance can cause the unbalanced input to convert this to a differential voltage. <S> This may be conducted or radiated interference. <S> Each can require similar or different solutions. <S> In some cases an unregulated DC supply such as you describe injects conducted noise into the output stage which draws from this voltage. <S> The most common solutions are; <S> Identify if it is magnetic radiated noise by moving the source. <S> Identify if is conducted by measuring the AC ripple voltage on the DC supply. <S> Shunt the common mode field by earthing or ground connection to a nearby ground. <S> Add ferrite CM and differential chokes to the DC input. <S> ( Many wall supplies have CM ferrite clamshells molded inline on the DC cable) <S> shielding of high impedance signals may be necessary in some cases. <S> A linear low drop out regulator (LDO) offers the lowest noise rejection but low in efficiency. <S> SMPS are common and 50mV ripple is common as well but more efficient. <S> A PC PSU has 12V. THis can be used for external applications from an unused HDD Molex plug. <A> I am sure the hum is caused by the magnetic field of the transformer coupling to the coil of the inductance inside the wha wha pedal - I made many cross-checks. <S> In fact, even when powered by a battery, the wha picks up the magnetic field of the transformer when the latter is connected to the mains. <S> I'm sure it is the wha because the noise vanishes when it is turned off. <S> That makes perfect sense. <S> Figure 1. <S> The popular Dunlop CryBaby wah-wah pedal. <S> Image source: ElectroSmash . <S> Figure 2. <S> The Cry-Baby PCB with the inductor clearly visible in the centre of the PCB. <S> Image source: ElectroSmash . <S> You seem to have a preferred relative positions for your setup. <S> If everything else is OK (and assuming the construction of your pedal is similar to this one) then one option is to demount the inductor from the PCB, solder on extension leads and reorient it to minimise hum. <S> Once the optimum position is found you can fix it in position using a silicone sealant or similar. <A> I left in place the rectifiers and regulators. <S> This completely eliminates the hum - a toroidal transformer produces much less magnetic field in its neighbourhood than a conventional one. <S> For those who are interested, I bought a transformer produced by Nuvotem Talema.
| Solution: I substituted the conventional transformer by a small toroidal transformer, supplying 14 V ac at full load (more precisely, it has two secondary windings at 7 V ac, which I connected in series).
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