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How to step down from 4.5V to 3.3V? I'm trying to step down my source voltage of 4.5V to a 3.3V suitable for the MPR121 . The chip has at the best settings a typical current of 393μA. The 3.3V has to be quite accurate as to not damage the keypad. When researching, I've come across two possible solutions that could achieve what I want to do: Voltage Divider In this case, can the output be stable and accurate enough to be able to step down and voltage and if so, what value resistors would I use? Voltage Regulator This is then another IC just to step down the voltage (footprint wise), any suggestions on which regulator I would use (linear, switching)? <Q> 3.3 V is a common voltage, so there are many fixed linear regulators available at that voltage. <S> These things have only three pins and are very simple to use. <S> The pins are the input voltage, ground, and the output voltage. <S> You are dropping (4.5 V) - (3.3 V) = <S> 1.2 V. <S> You have to be careful to choose a regulator that can work with that headroom. <S> These are often called LDOs (Low DropOut). <S> The efficiency from the voltage drop will be 73%, plus a little more loss for the quiescient current. <S> At only 400 µA output, the overall wasted power will be very small. <S> For some linear regulators, that would add significantly to your 400 µA figure. <S> Others work with only a few µA. <S> Take a look at the MCP1700 series, but there are many many others that would be fine too. <S> Some older LDOs are not "0 ESR output cap stable". <S> Simply stay away from them. <S> They were designed before the era of small and cheap ceramic capacitors that could do a few µF. <S> The MCP1700 series I mentioned is 0 ESR output stable, requires a maximum of 350 mV headroom, has only 4 µA quiescient current, and can deliver up to 250 mA. <S> These are my "jellybean" LDOs, meaning that's what I use unless there is a good reason not to. <S> I don't see one in this case. <A> Do not use a voltage divider. <S> The problem here is that as the current in your load changes the divider voltage will change in proportion. <S> Sometimes this can be worked around to a suitable level of accuracy by making the current through the divider resistors be twenty or forty times the average load current but this solution is very wasteful of energy. <S> For your solution an appropriate solution will be to use Low Drop Out linear voltage regulator. <S> A normal linear voltage regulator will not work for this because the input to output voltage differential for those needs to be on the order of 3.5 volts or more. <S> A properly selected LDO should work for a input to output differential at the 1.2V as needed by the OP. <S> A small switcher may also be used but to convert the voltage levels needed here at the very low current requirement the LDO regulator will be the easiest and lowest cost to implement. <A> For this current (less than 1mA) and this low voltage drop (about a volt) <S> your best choice for higher efficiency is linear regulator. <S> There are LDO's with low Input Quiescent Current like this for example - MCP1701A - only 5uA!
| Since the voltage drop ratio is relatively low and the output current requirements also low, use a linear regulator. You will also need a 1 µF or so ceramic cap between input and ground, and between output and ground. You can also make a shunt regulator with TL431 - a bit cheaper, but power consumption will be more (at least 1-2mA). Also take a look at the quiescient current spec.
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Is FCC certification needed for receiver device? Is FCC certification needed for product acting as receiver but having transceiver chip inside? or there should specifically be a receiver only chip inside in order to bypass complicated certification process? The document by TI "ISM-Band and Short Range Device Regulatory ComplianceOverview" here states that "Receivers do not need a certification, but the vendor has to state in a Declaration of Conformity (DOC) that each device complies with the spurious emission requirements of unintentional radiators according to section 15.209." The frequency of operation for this device is 902–928 MHz ISM band. <Q> You should have it tested as an "Unintentional radiator". <S> The local oscillator in a superhet can end up radiating from the antenna or from the PCB, just for one example. <S> If you have a processor or microcontroller, that's another potential source of radiated noise. <S> It's a relatively inexpensive test, and if you've done you work properly you will pass. <A> A certification process is required in both cases. <S> It might be easier to pass it. <S> You will have to prove it is an RX only device. <A> For an exact answer, specify the frequency range and application of your receiver. <S> The presence of a transmitter chip has no bearing if the chip is not enabled and cannot readily be enabled by the consumer.
| Receivers can definitely radiate energy, and can violate the limits. The FCC regulations stipulate different conditions depending on the type of receiver.
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RC circuit as clock source I am not understanding something very fundamental here , I have read that an RC circuit can be used as a clock source. From this I assume a DC input. Playing around with my oscilloscope the only time I see an RC circuit oscillate is with an already oscillating input, throwing in a DC input there is no oscillating just the obvious blocking of DC by the cap, so how does an RC circuit work as on oscillator with a DC input? And I know you can make fancy ones with more components but I mean a very simple resistive/capacitive circuit? Thanks in advance. <Q> You have basically already answered your own question. <S> A simple R & C circuit configured as a low pass or a high pass filter circuit is not capable of sustaining oscillation. <S> For oscillation you need something that provides feedback from output to input of the circuit with 180 degree phase shift. <S> Almost always that requires the circuit to contain some type of active component that adds gain so that the oscillation can be sustained. <S> Without it any oscillation from a Time 0 application of the feedback will die out in a few cycles due to the lossy nature of the circuit. <S> If you do get an R & C circuit to oscillate with some active components (transistor, MSFET, opamp, comparator, tube, valve) you can then use the resulting AC waveform as a clock signal should its amplitude, frequency and average voltage level meet the needs of the clock consumer. <S> Voltage and amplitude characteristics are easily adjusted with some additional circuitry. <S> Frequency can be changed by either changing R & C values in the oscillator circuit or using counters and flip flops to divide the frequency to a lower value. <A> You're correct about oscillators being used at clock. <S> But you need more than just a filter to have an oscillator... look at the picture below (I got this from Electronic Tutorials). <S> As you can see, in order to have an oscillator, you need a feedback network. <S> Depending on how many branches you have with these capacitors and resistors, it can change the difference in phase shifting. <S> The output frequency can be defined as \$ f_r = <S> \frac{1}{2\pi RC\sqrt{2N}} \$, where \$ N \$ is the number of RC branches that you have. <S> This is one out of many solutions you can have for an RC Oscillator. <S> You may use an Op Amp if you desire for an another example. <A> Another Astable uses a CMOS Schmitt Inverter with a large R and small C. <S> This Astable Osc uses negative feedback for centre DC input bias and Hysteresis to determine transition time ramp (triangle wave) with a square wave out. <S> f=0.72/RC for input hysteresis= Vdd/3 . <S> simulate this circuit – <S> Schematic created using CircuitLab
| A lot of computing systems use RC oscillators to have a clock reading for a particular processor.
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OpAmp Buffer configuration with max allowed output I'm designing a circuit, and a section of it is as follows: simulate this circuit – Schematic created using CircuitLab What circuit can i use to allow OA1 to follow the Non-inverting input from 0 to 5V exactly, and anything above 5V in the input remains 5V on the output? (i thought of using 2 resistors or a trimmer to reduce voltage on the input, but i have no use of the 5-10V range, and i want to use the full 0-5V range on the output) (the 0-10V input is not available for redesign, the OA2 doesn't exists, but i placed it there to signify a high impedance circuit after the output.) Note: I'm trying to be accurate, i could use a resistor+zener diode on the Output to limit to the closer commercial value for zeners eg. 5.6V, but there must be a better and more precise way to do this.. <Q> Here's my solution: simulate this circuit – <S> Schematic created using CircuitLab <S> How it works is that the 3rd opamp will compare Vlim against a 5 V reference voltage. <S> As soon as the voltage rises above the 5 V reference, the output of OA3 will go low <S> this basically works as an ideal 5 V zener diode. <S> As soon as Vlim goes below 5 V, OA3's output will become high and the diode will be in reverse mode. <S> So the diode is to prevent OA3 interfering when Vlim < 5 V <S> The accuracy of the 5 V clamping level is mainly determined by that level itself and the DC offsets of the OA3. <S> The diode is in OA3's feedback loop so the voltage drop across it does not matter. <A> Just make the positive voltage rail +5V rather than 24V and use a rail to rail op amp. <S> The circuit may not have great ac characteristics if you're going to be slamming the op amp into the 5V rail. <A> A web search for opamp min-max circuits some time ago led me to a fascinating AD8036 <S> Voltage Feedback Clamp Amp from Analog Devices. <S> This device, when used in non-inverting mode can clamp the input voltage between an upper and lower limit. <S> Figure 2. <S> Clamp-amp block diagram. <S> (See page 16 of the datasheet.) <S> \$ <S> +V_{IN} \$ is applied to A1's non-inverting input until \$ +V_{IN} <S> > <S> V_H \$ at which point S1 applies \$ V_H \$ to A1. <S> Similarly when \$ +V_{IN} <S> < V_L \$ S1 applies \$ <S> V_L \$ to A1. <S> There are a few limitations to the device - particularly the supply voltages (+/- <S> 3 to <S> +/- <S> 5 V) and 6.3 V max \$ <S> \Delta \$V between \$V_H\$ and \$V_L\$ - <S> but you can work around those by, for example, working in the 0 to 1 V range on your inputs and putting a x 5 amplifier afterwards. <A> simulate this circuit – Schematic created using CircuitLab Might have to change the resistor values, but the ratio is there and you get the idea.
| You could get 5V from 24 with a linear regulator or build your own with a voltage divider buffered by another op amp and a pass transistor.
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How to create a zero crossing detector using a full bridge wave rectified circuit Currently 1) I have a full bridge wave rectified ac waveform of 10Vmax. 2) I also have a potentiometer adjustable DC waveform. 3) My rail voltages are +12V and -12V. 4) I currently supply my full bridge wave rectified wave in to the non inverting terminal and the DC voltage to the inverting terminal. 5) Using an opamp or a comparator, I want to compare the two signals and produce a square wave which varies from 0 to a positive voltage at the output. 6) At every zero crossing , my square wave should go to zero. In all the other times it must stay positive. I have created an incomplete schematic below and have also tried many different methods but I am having difficulties in achieving my result. Thank You simulate this circuit – Schematic created using CircuitLab EDIT New schematic Removed the Buffer, Positive Feedback resistor, Ac source floating and load resistor added. simulate this circuit <Q> Source link : <S> The falling edge of the output pulse happens at approximately 200 μsec before the zero crossing. <S> You can use the circuit to safely stop the triggering of a thyristor gate, giving it time to properly turn off. <S> The circuit generates short pulses only when the mains voltage is approximately 0V, thereby dissipating only 200 mW at 230V and a 50-Hz input. <S> simulate this circuit – <S> Schematic created using CircuitLab Rearranging the circuit you can get the same for your XFORMER input: simulate this circuit <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Modified OP circuit. <S> Your second circuit has no AC return path. <S> This modification fixes that. <S> 10 V RMS will peak at 14.1 volts which will overload the input to the comparitor. <S> Rearrangement of the two 1k resistors will drop this to 7 V. Figure 2. <S> The LM311 has the emitter of the open collector output available on a separate pin. <S> Using the LM311 you can tie the emitter of the output transistor to ground (or anything else) to prevent negative output excursions. <S> When I simulate this circuit in the circuit lab , it didn't give me any square wave at the output of the comparator? <S> Any ideas on that? <S> Figure 2. <S> Running the simulation on Figure 1 with the potentiometer K factor set at 0.15 results in the above simulation. <S> (Settings: 0 - 0.1 s, <S> 0.001 s steps.) <S> You probably had the pot setting too high in your simulation. <S> Don't forget that most comparitors have open-collector outputs <S> so I've added R3. <A> This is an old Elektor design. <S> First of all the zero detection is realized on the mains side. <S> The value of the capacitor C1 is important to obtain the desired output pulse. <S> I would strongly discourage the use of transformer since the use of it creates a zero crossing not in sync with mains.
| The circuit in this Design Idea generates a zero-crossing pulse off the ac mains and provides galvanic isolation.
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Linear power supply + zero-crossing detection on transformer secondary Having read this comment , I'm a bit concerned about whether one of my projects is going to work: simulate this circuit – Schematic created using CircuitLab The comparator is also inside the uC, but shown explicitly for clarity. The intent of this question appears to be an exact duplicate, but its only answer completely misses what I think is a critical point: The transformer is (almost) completely unloaded at the time that I want the ZCD pulse. I can understand that having both linear and reactive components, considering the transformer and the load together, can cause a load-dependent phase-shift, and that's what the other answer was about. But in my case, and in the case of the other question, the transformer becomes unloaded for all except the peaks, and we want to sense the zeros. Given that detail, is there still a phase-shift to worry about between the primary zero-crossing and the secondary one? <Q> Realize the zero crossing first without transformer. <S> Separate optical. <S> See this old elektor design. <S> The value of C1 is crucial. <S> The received puls can be send to the micro processor. <S> I have realized this circuit many times and it works without a flaw. <S> A short quote from the old Elektor magazine:Due to reactive, non-uniform loads, zero crossing points can be exactly determined on the secondary side of the transformer only under certain conditions. <S> The non-linear transfer characteristic often causes the secondary voltage to be deformed and offset in phase, so it cannot be assumed to be a clean, phase-aligned sine-wave signal. <A> In an ideal transformer, the primary voltage and the secondary voltage are identical (except for scale) at all times. <S> This is because it's the changing flux in the core that creates the voltages on all the windings. <S> When we introduce reality into the transformer, it disturbs that slightly. <S> The finite primary inductance means the transformer draws a magnetising current, in quadrature phase to the input voltage. <S> The finite winding resistance means that the magnetising current creates a small voltage in quadrature phase to the input voltage. <S> This creates a small phase offset between the primary and secondary voltages. <S> The load currents are in phase, and so irrelevant. <S> The phase error in a good transformer is small. <S> It will be significant if you're trying to make a precision lab instrument for measuring phase shift. <S> It should not be significant if you're triggering a zero-cross switching circuit, or phase-shift Triac dimmer, try it and see. <S> Note that transformers get better as they get bigger. <S> A 'good' transformer may have to be a big transformer, say 50VA and up. <S> It's likely that PCB mount matchbox-size and down could be too non-ideal for this use. <A> The transformer is never open-circuit. <S> The output is zero at zero-cross, if that's what you mean. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Simplified schematic. <S> D1 allows monitoring of the full-wave rectified signal for zero-cross detection. <S> With this arrangement your comparitor will switch slightly before and after zero-cross. <S> You can measure the time of this pulse, divide by two and apply that timing offset in your code to be close to the true zero-cross. <A> ;) <S> I'll simulate it. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The trick is to wonder what exactly sets the potential of points "AC1" and "ACD" relative to our post-rectifier ground? <S> The answer is that it depends on what diodes actually conduct, and if they are off, it depends on their leakage currents. <S> Fortunately, you made it... umm, simpler by including a second bridge rectifier. <S> Now, here, obviously R3 and R4 conduct some current almost all the time, which biases one of D10/D11 (and D12/D13) into conduction also. <S> The result is that AC1 and AC2 relative to GND become this: <S> Unfortunately, the voltage on the "COMP" node relative to GND (which is what your comparator sees) is the minimum of these two (minus a diode drop) and thus is not what you'd like at all to detect any kind of zero cross...... <S> This won't work, not because of phase shift, but because both bridges interfere and the waveform is just... wrong. <S> Now, what if we remove the second bridge ? <S> simulate this circuit Very nice! <S> Since the diodes constrain AC1 and AC2 to be between GND and VCC (modulo diode drops), all we have to do is to feed AC1 and AC2 to the inputs of our comparator, with adequate input protection resistors of course, and it will then detect the zero crossing. <S> Now, what is RLeak? <S> In the previous simulation, it was omitted. <S> But of course one diode will leak more than the others, which our neat simulation did not account for, since all diodes were exactly identical. <S> Adding RLeak to account for this means the waveforms on AC1 and AC2 will be more like... this... <S> Amusingly, a comparator between AC1 and AC2 would still work... <S> The Elektor solution posted by Decapod is much better, since it negates transformer phase shift. <S> But it needs more parts. <S> Your choice!
| This circuit won't work, but the reason is a bit subtle, and you guys all got it wrong
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How to deal with non UL DCDC converters? We have a product, and we're trying to get it MET marked for export from the UK to the US. It uses an APD S10 converter [ datasheet ] to generate +-250V rails from a 12V supply. It's an isolated converter, but we're not using it as such, we've tied the negative input and the centre output together. The relevant standard for our product is EN61010. Since the converter is not bridging an isolation barrier, I assumed that the only relevant part would be the flammability requirements. APD can provide us with datasheets showing V0 flamability for the PCB and potting compound, and the whole thing is in a metal case, so that seemed sufficient. And initially the test house agreed. The test house has now changed their mind. Apparently MET labs think the converter should be tested separately to EN 60950. This seems like a sensible choice of standard, but the converter does not have a UL number and has not been tested. The manufacturer says the norm is to just test it as part of a complete product, in the same way the rest of the circuit is tested. I realise we'd be better off with a UL recognised converter, but there aren't many at the right voltage and it's too late to change - we've already passed EMC tests with this one, and we need to ship ASAP. What am I missing here? What is the norm? APD is an American company, serving an American market, so there must be some way for them to sell these non-UL-recognised components for inclusion in a UL listed product. But now MET seem to be telling our test house that that is not possible. Yet APD are telling me that it happens all the time. I feel like I'm in the middle of a game of Chinese whispers, and want to understand the American system better. <Q> we need to ship ASAP ... <S> What am I missing here? <S> Someone that understands the certification process. <S> If you had such a person, they would have been involved from the start, and not allowed you to leave certification as something to do at the end. <S> Thinking <S> "We'll ship next week, just got to get this thing certified" shows extreme naivety to the point of incompetence. <S> Uncertified components <S> Ultimately, you want your whole product certified. <S> That can be quicker and easier if whole subsystems are already individually certified. <S> Your test house then can treat the subsystems as black boxes and not look inside. <S> If a subsystem is not separately certified, then they have to look inside. <S> This takes more time and therefore costs more, but does not by itself mean your whole unit can't be certified. <S> If you do go thru this process, it will cost less to get another product certified that uses whole subsystems from this one. <S> It is quite possible that the same testing house has already done this with your exact subsystems for another customer. <S> That will save them money, but they'll never tell you that. <S> From your point of view, if ADP won't provide a certification, then your test house will have to look inside that module enough to prove to themselves that it meets all relevant requirements for how you are using it . <A> You may wish to consider talking with your converter supplier about sharing in the cost of obtaining UL certification for the converter. <S> The testing regime will be about equal <S> but it is a win-win scenario. <A> First on all, it is important to consider that UL listing is not a matter of matter of submitting, passing, done. <S> To manufacture items with a UL mark, UL will review the manufacturing plant and its quality assurance system. <S> You must sign a contract that must be in force for as you as you are manufacturing items with a UL mark. <S> You must agree to ongoing unannounced UL inspections. <S> Many manufacturers have one of their own engineers assigned to assuring UL requirements are met. <S> Many complex products contain some components that are not UL recognized. <S> UL will review the risk of failure of those components, possible failure modes and the risk of fire, electric shock or other safety issues. <S> They are concerned only with safety not with products meeting any performance standards. <S> If a US company that wants to incorporate your product in a UL listed product, they may very well have reason to be confident that is manageable without your product being recognized. <S> Other potential US customers may have a different view. <S> If the UL inspector knows a company has reliable people of their own assuring the maintenance of UL standards, tow whole process is easier and less expensive.
| To some extent it depends on their relationship with UL.
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Does the impedance of a PCB track matter if the length of the track is far smaller than the wavelength of the signal? I have a PCB with tracks of no controlled impedance. The longest track is shorter than 1/5000 of a wavelength. Does the impedance of the track even matter? If not, then at what length would I need to start thinking about matching the track impedance to the source and load impedances? <Q> I have a PCB with tracks of no controlled impedance. <S> The longest track is shorter than 1/5000 of a wavelength. <S> Does the impedance of the track even matter? <S> No it won't matter. <S> If not, then at what length would I need to start thinking about matching the track impedance to the source and load impedances? <S> Well, not all scenarios like this require matching - for instance if you are designing a quarter wave impedance transformer you don't match on purpose. <S> If, on the other hand, you are transmitting data then it makes complete sense to match the impedances to avoid reflections and the possibility of data corruptions. <A> General rule of thumb: the impedance of the track begins to matter when its length is greater than 1/10th the wavelength. <S> The reason it doesn't matter when the track is short: although there will be reflections due to impedance mismatches at each end, those reflections can propagate across the line so fast relative to the rate at which the signal is changing that equilibrium is reached "instantly", or at least fast enough <S> it's negligible in most cases. <A> A controlled impedance trace is the same as a transmission line. <S> A quote from <S> electronicdesign.com : <S> A cable becomes a transmission line when it has a length greater than λ/8 at the operating frequency <S> So at 1/5000 of the wavelength you're still very far away from that point so a controlled impedance track would not even behave as a proper transmission line (at that frequency). <S> You don't have to impedance match, even if you are working with a transmission line. <S> It depends on the behavior that you want. <A> Back in the old world odor, I was configuring 1553 data busses. <S> The rule of thumb there is that anything less than 1/5 the pulse rise time is treated as a lumped impedance. <S> Our stuff flew just fine.
| It starts to matter (as a rule of thumb) when the track (or wire) length becomes about one tenth of the wavelength of the highest frequency signal of importance. If you want good power transfer and little signal reflections then yes you need impedance matching.
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Zener over voltage protection from DC hand crank generator I have a board powered by a hand crank generator. Under normal conditions the voltage peaks at 18-20 volts. However spinning the handle violently will peak higher at 30+ volts. The generator runs through a schottky bridge rectifier and into a 5 volt regulator rated at a max of 26 volts. The circuit as a whole pulls less than an amp. In trying to keep the simple circuit simple I was thinking about a 20 volt zener (D5) and a tiny 10 ohm series resistor(R1). When it comes to protecting a voltage regulator is this the most ideal method or should I be looking at something different? <Q> A linear regulator is inappropriate here. <S> Use one of the many many buck regulator chips out there. <S> Pick one that can handle a bit more than the max possible output after the full wave bridge, and you don't have to worry about that part at all. <S> A buck switcher will be smaller and cheaper than a linear regulator after you include the cost and space of getting rid of the heat. <S> Even at only 500 mA at 5 V out and 20 V in, a linear regulator will dissipate 7.5 W of heat. <S> That's not trivial to get rid of, and it's not the worst case either. <S> Use a buck switcher rated for 40 V, and you're done. <A> I don't think I can draw a schematic in the comments so I will put in another answer. <S> Follower Circuit: simulate this circuit – Schematic created using CircuitLab <A> Chances are you can put out more than an amp from your generator, but even if it is only 1 amp, your zener will get hot. <S> You will have 20 volts times one amp or 20 W. <S> This will blow any board mount zener. <S> If you want to use a zener, it will have to be a chassis mount with a heat sink. <S> It should work fine, however. <A> It will make more power from the generator available to your load and it will reduce the thermal dissipation. <S> You will find that the component count will approach that of your linear regulator design. <S> If this is indeed powered by a generator, you can also eliminate the full wave Schottky bridge.
| I would replace the passive regulator with a simple buck converter. Answer to the question in the comment below - If you know the current and voltage you will expect, and want to dump the energy into a resistor instead of the zener, build you can build an emitter follower or source-follower-type circuit running off a small zener with a load resistor.
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How can I test battery for true mAh rating? Is there a way I can test the capacity of a 3.7v rechargable lion battery like this at home? All I really have is a multimeter. ( Image source ) <Q> To really test a battery capacity requires a timed test with a constant current load on the battery. <S> Since the battery voltage changes as the battery gets depleted trying to use a fixed resistor type load leads to the load current getting smaller as the battery voltage goes down. <S> The constant current load varies its effective resistance dynamically to so as to keep the current the same over time. <S> For low current levels it is relatively easy to build your own constant current load device using an opamp, power MosFET, a few resistors and a capacitor or two. <S> Power the constant current circuit from an external power supply. <S> There are plenty of circuits on the web to go by. <A> A lithium ion cell is considered to be discharged when it reaches 3.0 volts. <S> The amp-hour capacity is over a one hour period. <S> Your picture shows 9900 mAh cells which means they should supply 9.9 amps for one hour at which time the voltage of the cell should be 3.0 volts or greater. <S> You can approximate a capacity test by connecting a resistor of 0.34 ohms across a cell. <S> You can make this value by paralleling 20 6.8 ohm resistors for example. <S> The resistor will need a power rating of at least 40 watts. <S> It will get quite hot, so take care. <S> Monitor the voltage of the cell during the one hour test. <S> If it drops below 3.0 volts, stop the test. <S> If it makes it to one hour, you can be reasonably sure of its capacity. <S> This is only an approximate test since as the test progresses, the voltage of the cell is dropping. <S> This means the resistor is not a true common current load. <S> But for some simple home testing, this should be close enough. <S> Anticipate that battery will explode so wear appropriate personal protection gear and have appropriate fire fighting equipment at the ready. <A> JUWEI 2in1 Digital Type-C USB tester. <S> Has a built in countdown timer and reset switch. <S> Tells watt hours as wells as mAh. <S> It's under $20. <S> Use a simple usb fan(one on a stand) and let it run until it goes off. <S> This will give you an accurate reading of usable the current needed to power a device from the battery!
| When testing batteries, always put everything in a fireproof enclosure.
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Excess energy from my off grid system I would like to know what would happen to the excess energy produced by my off grid solar panels. The scenario is that I have enough panels to run my appliances and fully charged my battery at same time and still have some energy left from the panels, I would like to know what would happen to that extra energy. The system is off grid <Q> Does a solar panel produce full power all the time? <S> Short answer : <S> No <S> First, let's take a look at a solar panel's characteristics, which are always some variation of this graph : Image source : http://www.pveducation.org <S> You can see current and voltage and current are linked through the red curve. <S> In practice, it means that if you do not pull current, your solar panel will quietly sit at Voc (open-circuit voltage) and pulling current from the panels lowers the voltage, but increases the total power output, until you reach the maximum power point (maximum of the blue curve), from where pulling more current will reduce the voltage so much that te total power available decreases. <S> Which means, as @mkeith pointed out, that you can adapt the solar panel's energy production at will. <S> Now assuming your power conversion works correctly, it should perform both these tasks : <S> If the battery is fully charged, provide exactly the power needed for your appliances <S> If the battery is not fully charged, provide the appliances power + charging power for the battery, while taking its maximum charge current into account. <S> What happens to the sun's energy when not turned into electricity? <S> Short answer : It is turned into heat As shown above, a solar cell does not produce its full power all the time, but still receives energy from the sun. <S> Conservation of energy tells us that it should go somewhere. <S> Basic logic tells us that there are only two ways : Light and heat. <S> This page from pveducation tells us that it is simply turned into heat, which will then be dissipated via convection, conduction or radiation. <A> If you withdraw less energy from the solar panel than it is able to produce there are only two things that are physically possible to happen: <S> The solar panel absorbs less light (i.e. it becomes more reflective; it gets brighter). <S> I have, however, never heard of such an effect. <S> The solar panel heats up more and radiates excess energy as heat radiation <A> There are no harmful side effects from excess potential solar energy. <S> The panels will not overheat, no wires will melt, etc. <S> It is simply lost energy that you cannot recover. <S> If this happens frequently, adding additional battery capacity will harness this energy as long as your charge controller is capable of handling the additional charging current. <S> But if you do not ultimately consume this extra stored energy, there is no justification for this added capacity and expense. <S> As you may imagine, it is hard to strike the perfect balance between energy production, storage, and consumption. <S> It often comes down to a question of economics.
| If your load (battery charging plus household) requires less energy than what your solar panels are able to produce, the excess potential energy simply goes to waste.
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get dynamic length packet from UART? I Write a program for receive data from Uart ISI, my Data Packet is something like : 2 first byte contain Device ID and type of command. Data is contain some bytes that could be have different length, minimum 1 byte to max 15 bytes, and the last byte represent the packet is finish. Uart ISI code is: volatile char UartFlag=Fulse;volatile unsigned char count=0;unsigned char coder[13];ISR (USART_RX_vect){ coder[count]=UDR0; if (coder[count] == 20) UartFlag=True; else count++;} That receive each byte and save in coder . As you see when receive number 20 , stop receiving process by UartFlag=True; and in main call a function DoSomthing(); to do something with coder as below: while (1) { if (UartFlag) { DoSomthing(); count=0; UartFlag=Fulse; } } But I have problem that sometime the Data section have 20 in their bytes and its lead to i dont get Correct packet. I try to put \0 (the last byte=0 instead of 20) but maybe have same error. I am looking for best way to find dynamic length of packet. What is the best way to find the dynamic length of packet? A way that maybe work is put the data length in first byte of Data section, but it add extra process in sender device. <Q> In my opinion, to be 100% that it will work in all cases, you have to do what you thought of yourself: Include the length of the data (the number of bytes you are sending) as an additional header field, for example after the Type field. <S> This is actually what the Ethernet protocols do. <S> About extra process on sender <S> I don't know how the sender is implemented. <S> But it doesn't seem to require much processing effort . <S> I mean the sender already knows how many packets will be sent, no? <S> So, as you fill in the other two fields (Device ID and Type), you can fill in this additional field as well. <A> The two common ways to solve this are the ones you mentioned. <S> Use a dedicated length field at the beginning of the package <S> Use an unambiguous end marker at the end of the package. <S> If you don't want to use length fields you have to use an end marker. <S> If you want to be able to use any of the 256 byte values in payload you have to extend your code by intrducing escape sequences in order to code more than just 256 distinct values. <S> E.g. Use 0x20 0x00 to code original 0x20 in payload <S> (i.e. replace any 0x20 by 0x20 0x00 in payload) <S> Use 0x20 0x01 as end marker. <S> Since escaping increases the size of the coded payload it is advisable to use an escape character value that is unlikely to appear in original payload data(so 0x20 wouldn't be a good choice if payload is ordinary ASCII text). <A> Another thing to consider is how the data field is encoded. <S> (Note that in applications that I create I also use a start marker that is unique and cannot occur in the data region. <S> This allows quicker re-sync to the command stream without losing one extra packet). <S> So if a data region is just numbers for example encode them in character format as opposed to binary format. <S> Then also select your device ID and packet Type as a readable character. <S> This scheme can leave you a series of bytes to be used as "control characters" two of which you use as the start marker and end marker. <S> The ASCII character set neatly reserves the encodings of 0x00 to 0x1F <S> as control characters. <S> This is MANY years old and so you can see that this problem was solved long ago. <S> There are even names given to some of them that can be directly leveraged such as SOH (start of header) and EOT (end of text).
| You can avoid a length specifier in the packet by ensuring that the end marker is not ever expected in the data field.
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Input voltage range for lm336z-5.0? I want to use lm336z-5.0 as a reference voltage for Arduino Uno. In the data sheet I see an example of the chip being connected to a 10V source via a 5k resistor. In Absolute Maximum Ratings section I only see that the max reverse current is 15 mA. What is the lowest and highest possible voltage at the input (I assume I will adjust the resistor to keep the current well below 15 mA)? <Q> The term "input voltage" is actually not exactly applied here, since LM336 is a shunt regulator. <S> You can see it roughly as a zener diode. <S> Check also Figure 11 in the datasheet . <S> As long as you provide a bias current to it that is at least 0.6mA <S> and at most 10mA (15mA absolute maximum) then the voltage across the regulator/diode will be around 5V. <S> So, the value of the supply voltage is not important in this application, as long as it is of course higher than the reference voltage. <A> The LM336 type you have is a shunt voltage regulator capable of regulating between 4 volts and 6 volts. <S> What is the lowest and highest possible voltage at the input <S> Because the LM336 is a shunt regulator it means that the input voltage is also the output voltage i.e. between 4 volts and 6 volts. <S> I see an example of the chip being connected to a 10V source via a 5k resistor <S> Providing you use an input resistor that adequately limits the current to 15 mA <S> then you will be ok. <S> Check that at higher temperatures the 15 mA figure isn't degraded of course. <S> For instance you could run with an incoming voltage of 100 volts but the resistor will need to drop 95 volts and, if you want an operating current of 15 mA <S> then the resistor value will be 6333 ohms and dissipate nearly 1.5 watts. <S> But you don't need to operate at <S> 15 mA - the LM336 will work at 0.6 mA and this means a resistor value of 158 kohm and a power loss of only 57 mW. <A> The data sheet for the part lists the operating current range at 0.6 to 10mA in the teaser specs section on the first page. <S> The specification itself gives this same range in the "Conditions" column where they specify the max voltage change over operating current: <S> So to answer your question you need to use a resistor to limit current to the shunt regulator to within this range if you want to achieve the specified accuracy.
| The size of the resistor determines the input voltage range.
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Getting 240VAC output from 220-230VAC input For lab testing at work I need to be able to test an appliance (often more than 3KW, e.g. deep fryers and waffle irons) at 240VAC. The problem is that our power grid only provides between 220 and 230 VAC. So I'm looking for the best way to get a steady 240VAC output from a slightly variable input (we're in an industrial area). A variable output would be nice, but is not a requisite. Any help/hints/direction would be more than welcome! <Q> I'd be considering using a decent UPS - it can provide a stable output and a decent sine wave with variable input voltage conditions. <S> You must choose one that always converts the input to DC and then uses an inverter to generate the output voltage. <S> Some UPS designs will pass through the input AC voltage and only "switch in" the backup system when there is a power fail. <S> These types are not sufficient for your needs. <S> Maybe one of these <S> and the giveaway is that they can output 50 or 60 Hz irrespective of the input frequency - this tells you that they are true inverting converters. <A> These devices are usually made for stepping down mains AC, but many have a slight "overdrive as well". <S> A common range seems to be 1 : 0-1.1. <S> They are available for reasonable prices up to several kilowatts. <A> The most convenient old-skool way to do it is to use an auto-transformer, followed by a step-down transformer. <S> As you want 3kW loads, it's worth doing this two-stage approach. <S> For instance, a 2A autotransformer followed by a 500VA 20V fixed transformer would allow you to vary the incoming mains voltage by up to +/- <S> 20v, at an output current of 25A. To do that directly from an autotransformer, you would need a 25A version. <A> There are variable AC power supplies around from several companies. <S> Most of them allow you to change the frequency as well, which might be a useful feature for some appliances. <S> They are quite costly (starting at 5k with 3kVA), but will regulate frequency and voltage very well. <S> Companies which should provide good stuff: <S> EA (Elektro-Automatik, not Electronic Arts), Elettrotest, Chroma, Keysight and surely some more.
| For these kind of tests, especially if you are doing verification and not just happy tests (so there is a test report and the result must be reproducible), I suggest buying professional testing equipment. Get a Variac (more accurately; a variable auto-transformer).
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Multilateration with TOA using lowcost transceivers (like NRF24L01) and uCs My problem is the following, I would like to localize an object (main node) via multilateration using a wireless sensor network of 4 nodes with less than half a meter accuracy. The equations are already worked out and what is left is to obtain distance measurements from the network nodes to the main node using TOA method (this is what I am working on right now). My hardware specifications are the as follows: 5 NRF24L01+ transceivers (2.4GHz, 2Mbps, GFSK, 16MHz inner clock) 4 ATTiny85 uC (for the wireless sensor network) (8MHz clock) PIC16f877A (main node) (16MHz clock) Specifically, what I want to know is: can I measure the time of arrival of a signal (in less than a meter), with this hardware? If not, point some way out for me, either acquire new hardware (not too expensive) or change the method implemented. PD: If possible look at the basic idea I have in mind before answering (and give me feedback on it): Narrowing down the problem time of flight between two nodes only (N1 and N2), the total time for a bit of data to go from N1 to N2 and back is:$$t = 2(t_b + t_c) = 2\left ( \frac{1}{R} + \frac{1}{c}\times d \right ),$$where \$t\$ corresponds to bit transfer rate (\$R\$ = 2Mbps for NRF24L01+), or the time the transceiver takes to send that one bit to the channel; \$t_c\$ is the time in the channel that is approximately, the length of the channel over the speed of light. The NRF24L01+ can send up to 32 bits in one go, this implies:$$t = 64(t_b + t_c) = 2\left ( \frac{1}{R} + \frac{1}{c}\times d \right ).$$However, for \$d=0.5\$ (as I want) this yields \$t_c = 0.106\mu s \$, with a 8MHz MCU this time is invisible (1/8MHz = 0.125\$\mu s\$). A posibility is to resend the message to increase \$t_c\$ to a visible value.Is the previous reasoning accurate? <Q> Why does everyone try to do this digitally with modules NOT designed for precision timing? <S> Think mixers, filters and oscillators, not computer datacomms, there will probably be a computer in there somewhere <S> but the high speed timing and phase shift measurement should be in hardware. <S> I would be looking at modules from MiniCircuits not Nordic. <A> The basic problem with your formula is that t b has nothing to do with the time of flight. <S> The start of the transmitted bit arrives at the receiver delayed only by the speed of light, not by the bit time. <S> The processing time of bits and messages is not related to time of flight either. <S> So increasing the message length does nothing with regard to time of flight resolution. <S> If you wish to get 1/2 meter accuracy, your solution will need to resolve a delay of less than 1.67 ns. <S> The notion of pinging a signal back and forth multiple times to reduce this burden will require that the turnaround delay on each end have a delay error orders of magnitude less than 1.67 ns. <S> This is probably not achievable with off the shelf radio modules. <A> Depending on what is legal (if you care) in the 2400MHz ISM band, you can implement instantaneous phase-reversals in your data-streams, generating broad-band bi-phase modulation. <S> Mixing that 180 degree phase flip in a receiver's mixer will produce a useful reversal of sign in a differential-output mixer, but the broad-band nature of the phase flip produces a SNR problem. <S> You can restrict the mixer's differential output using a common-mode capacitor, producing a controlled slewrate, and the greatly reduced noise bandwidth gives a greatly improved TimeJitter measurement. <S> But this approach requires custom design work, instead of gluing modules together. <S> Some RF DACs may suffice; use the DAC to produce low band (100MHz?) <S> Biphase, then upconvert to 2400MHz. <A> None of the Nordic RF ICs are suitable for TOA/TDOA measurements. <S> The best you can do with these devices is make use of RSSI (only available on some of the ICs) and even that is unlikely to provide the accuracy you desire. <S> For accurate positioning consider UWB hardware such as the DecaWave DW1000 (available as a module as DWM1000). <S> The DecaWave site also has many whitepapers describing the challenges present in real-time location systems.
| A different approach you could pursue is to triangulate on the device by using electronically steerable antennas.
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Are radio waves naturally polarized like light or is that a function of how they are produced? I am thinking of let us say AM or FM radio being transmitted by a vertical antenna. Would the they be polarized vertically or horizontally and wouldn't the angle of the receiving antenna then determine the strength of the signal. <Q> Radio waves and light are both electromagnetic waves. <S> The only difference between them is the wavelength. <S> The polarization is initially determined by the way they are produced FM Broadcast radio is generally transmitted by circularly-polarized antennas in order to accommodate receivers in most any orientation. <S> AM broadcast towers are vertically-polarized. <S> The propagation tends to be by ground waves so retain their polarization for the most part. <S> You may remember the old transistor radios used ferrite loop antennas that had a particular polarization, and you could move them around to find the best angle for reception at the best strength for the receiver's circuitry. <A> So in principle if the transmit antenna were strictly vertical and your receive antenna were strictly horizontal, you would receive nothing. <S> But there are a couple of complexities: <S> Partially-aligned linear radio antennas can receive each other with modest losses. <S> A circularly polarized antenna can receive any linear polarization with modestly reduced efficiency, and vice versa. <S> Short-wave signals are generally received after bouncing off the ionosphere, which randomizes the polarization. <S> Similarly, Wi-Fi and other 2.4/5 GHz signals are often bounced off buildings or walls, which tends to randomize the polarization. <S> Signals that are not narrow-band can have complex mixtures of polarizations, and polarization can change very rapidly with time. <S> The key difference between radio waves and visible light is that most of the radio signals we are familiar with are produced by coherent emission processes, which (usually) produce fully-polarized radio waves. <S> More, almost all detectors of radio waves coherently detect just one polarization; radio astronomers usually use pairs of crossed dipoles so we can record both polarizations and reconstruct the input signal's polarization state. <S> Most of the visible light sources we deal with are incoherent and produce unpolarized light (an even mixture of polarizations) and our detectors mostly aren't sensitive to polarization anyway. <S> Lasers are coherent and indeed are polarized, but unless the laser is designed to have a stable polarization, you tend to get random jumping around on very short time scales, averaging out to unpolarized. <S> The human eye is in fact very slightly sensitive to polarization, though we don't usually pay attention, and there are processes - like reflection - that readily add polarization to light, hence the utility of polarized sunglasses (to preferentially block light reflected off horizontal surfaces). <A> The polarization of the emitted wave is a result of the antenna design. <S> Typically, antennas that primarily extent in vertical direction have vertical polarization. <S> However, it's often hard to see the actual antenna inside the protective cases they come in, so, if you see an antenna thing that higher than wide, please don't assume vertical polarization. <S> It might just be one case with many stacked smaller antennas inside, or a loop, or something else completely. <S> And yes, the polarization of the receiving antenna must match the polarization of the wave, or else you get worse reception.
| Radio waves emitted by an antenna have a specific polarization, and receiving antennas are generally sensitive only to a specific polarization.
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Electronic drum set powered with batteries I have a electronic drum set that I want to make portable . Is it safe to use couple of parallel joined 9v 500 mA batteries to replace the 9v, 500 mA AC adapter and could the batteries last for couple of hours? The drum set is consisted of total 7 elemens plus a simple controlling device.the connecting cablels are relatively long, total of cca 4,5 meters.. <Q> Is it safe to use couple of parallel joined 9v 500 mA batteries ... <S> Batteries are rated in mAh - not mA. <S> If they are 900 mAh then they can (ideally) provide 900 mA for one hour or, more likely, 90 mA for 10 hours with somewhat less at higher discharge rates. <S> To estimate run time you need to measure the current drawn by your drum set. <S> To do this you need to somehow wire a multimeter on <S> DC amps range in series with the power supply. <S> You can estimate your battery run time by \$ h = <S> \frac {battery\;rating}{meter\;reading} \$ where h is hours. <S> You can parallel the batteries if they are the same type. <A> DC is not AC <S> Is it safe to use couple of parallel joined 9v 500 mA batteries to replace the 9v, <S> 500 mA AC <S> adapter <S> It's an AC adapter and batteries produce DC so, without a full examination of the schematic, no. <S> 9 volt AC peaks at 12.7 volts and may indeed be used to generate + and - internal voltage rails of +12 volts and - 12 volts using half wave rectifiers. <S> This won't happen with a DC battery of course, <S> It's irrelevant asking about battery life because batteries should not be used to power this piece of equipment without a thorough examination of the circuitry. <A> If you take another look at your AC adapter, it should tell you Its specifications I.E. 120 volt AC, and then it says what it provides which is 9 volts, 500 mA, but it will also say whether it is AC or 9 volts of DC power. <S> Many appliances use DC. <S> If it is DC, then I recommend that you use 6 - AA batteries in series - the positive connected to the negative etc of each successive battery. <S> You would still get the 9 volts AND you would get closer to 2500 mAh or 5 hours of play time. <S> If you really want to get crazy, apparently Energizer D sized battery will give you 20,000 mAh, so 6 of them in series will let you drum until you get tendinitis! <S> The 9 volt 'square' battery is designed for low power products, more like <100 mA <S> or you get serious voltage drop, so <S> it is just not a great choice for your setup, assuming you even need DC. <S> Good luck
| You can find battery holders that will hold the 6 batteries togetherness in series for you making this a very simple project.
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Connect signals to N.C pins in diode array (for ESD protection) We created a circuit with diode array for ESD protection PN: IP4283CZ10-TBR (MFR: Nexperia) Data sheet As you can in the datasheet, there are some pins which supposed to be N.C, but we did connect them to the signal pins, as you can see here: We afraid that this connection might explain some problems we see in our samples, for example- one PCB didn't communicate with UART until we removed the part and another circuit failed during first watchdog test (again, until we removed this part). In the datasheet all I could find that is related to the N.C pins was this line that mention that the capcitance between N.C pin to signal pin is 0.07pF (sorry, can't upload another image) Correct me if I'm wrong, but it doesn't say much (just that the pins are seperated using small value CAP, nothing about not short between them). Hope you could help me understand if this what is causing the problem or is it something else.We also use another 3 identical diodes arrays with same "connection problem", but they seem to work fine (Connected to RS422 RX,TX and Debugger lines) The debugger we are using is J-LINK (using JTAG communication). <Q> or "We didn't connect this pin.". <S> This looks like the latter. <S> The 0.6 pF is just parasitics, and picofarads will not affect a UART or reset. <S> Your problems are most likely unrelated to these connections. <S> Since you're seeing differences between copies of supposedly identical boards, there are probably manufacturing or assembly defects (but this isn't 100% certain - could also be marginal design). <S> I would start with a thorough visual inspection under a microscope looking for soldering shorts and opens. <S> Check polarity / pin 1 position of all active devices. <S> Maybe when you removed this part from your defective boards, you also reflowed some other nearby parts. <A> From the datasheet I get the impression that these N.C. pins are really just not connected internally. <S> And this 0.07 pF between these pins and the signal pins are probably just parasitics (although I don't quite understand why they state the value of such parasitics). <S> An idea that may apply in your case is that perhaps this diode array that you removed from the boards that didn't work as intended was defective (perhaps destroyed during soldering) and for this reason it affected the signals connected to them. <S> So when you removed the defective part the circuit worked ok. <A> looks painfully familiar. <S> If i understand you correctly, your device occasionally would not run properly until you removed diodes, right?It seems to be tricked @poweup into JTAG/Debug mode by those diodes. <S> you need different diodes for esd protection
| When listed in a datasheet, "No connect" can mean "Don't connect this pin!" Anyway, my point is that I don't think you have any problem with the connections that you have made.
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4N35 opto-isolator for MIDI input does not work? I have the following circuit which I would like to use to connect a MIDI enabled device. I am not getting any signal on pin 5, could it be because the resistor from the base to ground in too small? I am not sure how to calculate it from the datasheet. Update #1 I have replaced the opto-isolator, tested the input using an LED and could see the LED light up indicating current flows in the MIDI loop. I have placed a scope probe on the output of pin 5 of the device. pin 6 is now not connected to anything. Normally (idle) the signal is pulled high, when a signal comes in the signal goes low but only down to ~2.8V which is not enough to register as a logic low. Update #2 I have replaced R17 with 360 Ohm. Now the signal goes as low as 1.6V. Update #3 - Final one I've swapped the 360 Ohm resistor for 1K. Now the signal looks much better and it reaches almost 0V when low. It is worth mentioning the scope claims rise time of 7us and fall time of 3us (probably on the low side). Considering the MIDI freq. with about 32us per bit this is short enough not to have false readings. The MIDI signal goes to an Atmel AVR and from testing I did right now it seems the messages are going through clearly. I agree that this is not the optimal device for this task though, and the circuit one would make for it (if he desires) looks very similar to what is contained by the PC900 optocoupler from Sharp. <Q> The 4N35 is too slow for MIDI. <S> With a current transfer ratio of 100 % (which is specified for 10 mA, so you get even less for MIDI's 5 mA), you cannot rely on getting more than 5 mA through the output. <S> This means that to be able to drop the full 5 V, you need a pull-up resistor of at least 1 kΩ. <S> And this means that you will not be able to switch at the MIDI baud rate of 31250 Hz (and <S> a digital UART signal requires much more bandwidth than a sine wave): <S> And with a resistor at pin 6, the CTR would become even worse: If you really want to use the 4N35, then it is possible to speed it up by adding a transistor as an amplifer with a lower input impedance: <A> I have used 4N35 OCs for MIDI. <S> Fast enough transition from ON to OFF needed ridiculously low series resistor for the output transistor. <S> The output voltage swing was about 0,5V. <S> I added a comparator to make the proper logic output. <S> Why 4N35 and not some recommended type? <S> Because I had a handful of 4N35's. <A> Check your input: <S> Replace U5 with a standard LED connecting the anode to 1 and cathode to 2. <S> It should blink when MIDI data is sent. <S> Test again with the LED in series with R18 and U5. <S> If it glows (probably much more dimly) it proves current is flowing. <S> If not it might be just that there is not enough drive voltage. <S> Check your MIDI <S> input signal polarity. <S> Check your output: <S> Disconnect R? <S> from the photo-transistor base. <S> It's there to make sure the photo-transistor turns fully off but should work for test purposes without it. <S> Put an LED in series with R17. <S> This should glow when data is being sent. <S> Failure with these tests may indicate that the opto-isolator is damaged. <A> Having the base of the opto transistor based to ground via a resistor is a double edged sword. <S> On the one hand it can help the transistor to shut off a bit faster. <S> But at the same time it makes it harder to turn on which is almost equivalent to lowering the current transfer ratio. <S> Particularly with the poor current transfer ratio of parts like these. <S> So as you have already started to experiment and has been commented in other answers here the base resistor needs to be made much larger or eliminated. <S> In addition the collector resistor needs to be made much larger as well. <S> People always select the 4N35 because it is cheap. <S> But, as they say, you get what you pay for. <S> There are better opto-couplers that offer way higher current transfer ration and have full logic output stages instead of just an uncommitted transistor. <S> There are some types of opto-couplers that will work better in a digital isolation scheme such as the 6N136/6N137. <S> These offer an improvement over the cheap garbage but can still can present an challenge. <S> In async serial applications I have experienced problems with 6N137's due to the problem of pulse shape distortion due to non-uniform on and off propagation delays through the part. <S> A robust solution is to look for parts that have full output drivers built in to then specifically for digital applications. <S> One example is the part number ACNT-H61L from Avago (now part of Broadcom). <S> These offer performance up to 10MBd and source and sink 3.2mA and maintain good logic swing outputs.
| But the simplest method to get the MIDI input to work would be to use a high-speed optocoupler like the H11L1 or 6N137. The small sized collector resistor is a huge challenge for a clunky old beast of an opto-coupler like a 4N35.
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Why does my GPS and Wi-Fi work inside a sealed copper clad box? I have just finished assembling a Raspberry Pi-based project into a DIY box made from copper clad FR4 PCB, with the edges soldered together and the copper surface connected to ground. I expected, when I put the lid on the box, the onboard Wi-Fi and the USB GPS receiver would stop functioning - that is, the Pi would drop off Wi-Fi and the GPS fix would be lost. Instead, there is no discernible effect. Wi-Fi and GPS function as if the metal lid is not present. Given the entire reason I put this project in a copper clad case was to shield it from RF (it will be operating in the near field of a 5W VHF FM transmitter), I could do with understanding what's going on here. Here's a pic and with the temporary lid on... (note the thin stripped wire is just to ensure the lid is making electrical contact, and the USB power bank on top is just providing some downwards pressure also to ensure contact) <Q> Even if you have good contact points every cm or so, if your antennas are very near the gaps (a few mm), the RF energy will come inside strongly. <S> Some months back, I answered a question about "why metal cages around IR receivers". <S> Why are many IR receivers in metal cages? <S> Reading Richard Feynmann's lectures, I found the attenuation is 1 neper (8.6dB <S> ) * 2 <S> * pi * wavelength/separation. <S> Thus a 3 mm wire grid (making the Faraday Cage) with the antenna 3mm inside the grid, has attenuation of e^-(6.28) <S> = <S> 1/533 or 54dB. <A> For the box to be an effective Faraday shield, the entire peripheral of the top lid must be in electrical contact with the rest of the box. <S> Failing this, RF can easily couple from the largely isolated plane of the lid to the internal electronics. <S> Also don't overlook the large opening you have on the side of the enclosure. <S> I doubt it would allow the GPS to work <S> but it could enable the ingress/egress of a 2.4 GHz signal. <A> There are two big gotchas here. <S> The first is the lid issue. <S> The lid needs to be well connected along the entire edge. <S> The easiest way to do this is to use some copper foil tape. <S> The second issue is that you have wires penetrating the cage. <S> Any wire or cable going through a hole will act as an antenna. <S> Holes by themselves are ok as long as they are small <S> it is the wire that is the problem. <S> Presumably you need wires. <S> There are basically three ways to get signals and power in and out of a Faraday cage. <S> First is to use optical fiber. <S> This is simple and effective, but expensive, non standard, and either slow or really expensive. <S> It is also hard to send much power over fiber. <S> Second is to add filters to every wire in the form of a capacitor or pi network to the shield right at the point of penetration. <S> This is good for power and low speed signals like serial ports, but not for high speed. <S> This is what you need to do for high speed signals like usb or Ethernet. <S> If you look at the IO panel on the back of a PC, you will see there is a metal piece with cutouts for the connectors and spring fingers. <S> The purpose of that piece is to electrically connected the connector ground shields to the chassis. <S> Without it, the shields would only connect to the motherboard ground and shielding would be compromised.
| The final way is to use shielded cables and connect the cable shield to the enclosure right at the point of penetration.
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How to solder surface mount chip I have to solder a chip in a TSSOP-20 package on a board but unfortunately I have no "pcb oven". During the design I had access to a such oven but it's no longer possible. After some search on internet, I found many post about DIY pcb oven. As far as I understood, the heat is monitored to fit the soldering curve on the datasheet. I don't have time to make this kind of oven or spare money to buy one. So, I wonder if it's possible to solder the TSSOP-20 chip thanks to an "unmonitored" oven? Note that I'm jus a hobbyist, I do not need a very high quality weld. <Q> I finally soldered the chip on my own without a reflow oven. <S> It was way easier than I was expecting, thanks to many comments and some videos found on YouTube. <S> Materials: <S> Soldering iron, 2mm head Solder wire with incorporated flux <S> Flux in a syringe <S> I first tinned the top-left pin footprint in order to be able to place the chip for further solder. <S> I then melted the tin again and placed the chip with a little clamp with accuracy to make sure that all the pins were on the corresponding pin footprint. <S> I then soldered the opposite pin. <S> After that, I was sure that the chip would not move during the rest of the process. <S> I started to solder each pin one by one until two pins got linked due to my lack of soldering skill. <S> To correct my mistake, I put a drop of flux on <S> and then I heated both pins with the soldering iron. <S> With much surprise, it appears that it perfectly corrected the mistake. <S> For the other side, I linked half the pins together on purpose and then use some flux. <S> It's way easier, faster, and produced a cleaner soldering result. <S> Here is the result of the process: <S> I did some research about the oven part of the post. <S> Since it's non-monitored, it is not advisable to solder sensitive chips. <S> However, it's pretty safe to solder every resistor and capacitor with the oven and then solder chips like this with a soldering iron. <S> (Now I know how to do it properly <S> so it's no big deal.) <S> But please, DO NOT use the oven you are using to bake food! <S> Solder wire/paste releases toxic fumes when it melts, so use a dedicated oven that you are sure to not use again. <S> I made this answer with the help of comments left by other users. <S> This post might seem useless for experienced people <S> but I think it can help other people like me who have no real experience in soldering. <A> You don't have to use a temperature profile, and the chips you solder (not weld) to the board don't have to work either. <S> If you don't follow the temperature profile, your not guaranteed that the chip will work. <S> I've seen some chips fail. <S> Skillets work in a pinch, so do toaster ovens, but these don't follow a profile and have uneven heating. <S> There are two bad things that can happen: 1) <S> The uneven heating causes parts to be misaligned (which is easily fixable if a soldering iron is accessible) <S> 2) <S> You get too much heat, and burn up parts, some chips are some sensitive then others, for example: mems parts are extremely sensitive to thermal stress or humidity. <S> It may be worth it to invest in a cheap oven if working with expensive parts or if time is valuable to you. <S> That being said, almost all chips are solderable as a hobbyist with only a soldering iron. <S> As long as the pins are on the outside of the chip an iron can melt the solder and get the job done. <S> The main idea is you have enough flux to help the solder to only stick to the metal and run a blob of solder past the pins, it will only stick to the pins. <S> QFN's with pins underneath are solderable especially when the pads are extended beyond the part. <S> Even exposed pads (the kind used for thermal sinking) that are underneath TSOPS or DFN's are solderable if you create a via for heating underneath the chip. <S> If you want to remove chips, however a heat gun may be required (or you could just clip it off if your careful and solder on a new one) <S> BGA's and LGA's need a hot air gun minimum. <A> Yes it's possible. <S> I have soldered 7x7mm QFN-48 package successfully using only heatgun. <S> I did this by pre-soldering the pins on pcb with iron. <S> Then place the chip. <S> And last carefully descent the tip of the gun towards target. <S> When the solder flows the chip should find it's place. <S> Just take your time as the soldering curve shows. <S> You can also use solderpaste but be careful about the amount of paste. <S> Check before that your heat gun is hot enough.
| It appears that it's possible to use an unmonitored heat control oven (like a kitchen oven) to solder an entire board at once.
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How to determine the nominal current of an LED? I salvaged the LED disk from a defunct LED light bulb. (The problem was an arced transformer, the LEDs are intact.) It has several unmarked white SMD LEDs soldered on it: 5 lines parallel, with 3 LEDs in series on each line. I want to use this disk in a strobe. My aim is to get the maximum brightness out of it without damaging the LEDs. Is there any way I can determine (vaguely) its nominal current rating, rather than guessing it? <Q> It has several unmarked white SMD LEDs ... ... without marking you cannot get a datasheet and <S> without datasheet there you don't have a chance of finding out the real current. <S> However there is a possibility of getting a rough impression of the current needed by the LEDs: <S> Slowly increase the current from 0A until the product of current and voltage is the power you estimate (the nominal power of the bulb multiplied by the efficiency of the transformer; maybe 5W for a 7W bulb). <A> VERY approximately: LED power rating = <S> bulb_ratinf/number_of_LEDs. <S> So if eg 5W bub with 15 LEDs. <S> LED power ~= <S> 5W/15 <S> = <S> 1/3 <S> Watt per LED. <S> LED forward voltage is typically ~= <S> 3V. <S> So LED <S> current ~= <S> LED_Power/3V = <S> 0.333/3 <S> = <S> 111 <S> mA. <S> This should not be too far from "about right". <S> Bulb Wattage ratings are expected to be ~= LEDs DC Watts in. <S> Modern white phosphor LEds usually have quite a low ratio between i_operating_max_typical <S> and I abs_max - often in the 10% - 20% range. <S> LED lifetimes increase for I_LED < Imax and usually are acceptably affected for I_LED <S> > <S> Imax by a moderate amount. <S> Operating on the conservative side seems wise. <A> I got two great answers from @MartinRosenau and @RusselMcMahon. <S> Both answer relies on the wattage of the bulb. <S> They both missed the fact, however, that this is the consumed power by the LEDs AND the series resistor. <S> One could find and measure the resistor, but I used another method. <S> What I did was using another working bulb of the same type. <S> I eliminated all other light scources, then measured <S> it's brightness from a given distance using a smartphone and this app. <S> Then I used an adjustable supply to find the voltage/current where the brightness of the led disk matches the bulb's from the same distance. <S> The result is rather accurately 150mA, or 30mA per LED. <S> Pretty beliveable, so the method seems to work.
| Attach a current source and a voltage meter to the LEDs (the current source may be a voltage source with a resistor in series).
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Why modern cellphones do not have an externally visible antenna? Old phones had either a fixed antenna inside a housing which was externally visible, or a retractable antenna, which was visible when used. I'm assuming this was needed to obtain a certain total length. In modern cellphones all the antennas are inside the case. The height dimension of older cellphones was similar to current modern cellphones (smartphones). Why now externally visible antennas are not needed? <Q> Short answer: <S> external antennas are unattractive to the user. <S> The consumer drives the market, and these days, consumers want thinner, lighter, faster, etc. <S> This is actually a huge problem for antennas because the physics are fundamentally limiting. <S> Most antennas in cell phones are some sort of variant of a PIFA (planar inverted-F antenna). <S> Antenna-Theory has some cool articles on PIFAs and the now-famous iPhone 4 antenna . <S> PIFAs are fairly pattern invariant, and provide a decent impedance match. <S> They have low gain but if you do a link budget calculation, you'll see that your noise floor is like -110 dBm <S> (I think... <S> I forget but <S> somewhere in that range). <S> The miracle that makes phones work is digital communications. <S> Advanced modulation, extremely robust protocols, MIMO, etc have all contributed to ensuring communication even when the antenna on the phone itself is a terrible radiator (-5 dBi gain). <S> If you do the link budget calculation to reach the nearest tower, you'd see that it works, so long as your link budget factors in all of the communication properties. <S> That is, you have to consider what signal level the radio (cell phone) needs to demodulate the digital data and get information. <S> Beyond analog gain (antenna, amplifiers, etc), you have digital gain through things like error-control coding (i.e. coding gain) and robust modulation techniques like OFDM (the standard for LTE and the future). <S> Multiple access techniques (CDMA, TDMA), and MIMO also contribute to this. <S> If you look at the "generation standards" like 3G and 4G, you'll see where on the protocol level things get super complicated. <S> All of this goes into making cell networks work. <S> One of the toughest challenges facing folks now is that cell phones need many antennas that all couple and communicate on the same bands. <S> So you have to be able to integrate multiple radios and antennas on a small platform and make sure everything works. <S> When you don't do it right, you get the iPhone 4/4S... :-) <A> Way back when antennas were long there was a shortage of cell towers and those that were there, were short. <S> Nowadays frequencies have changed, cell towers have become more powerful (taller and more efficient) as well as the technology in the phones has increased. <A> Older cellphones mostly used traditional monopole antennas, which (basically) consist of a straight wire with specific length requirements (as stated by you). <S> Some modern cellphones use microstrip patch antennas. <S> This type of antenna is obtained by strategically designing certain figures (like squares or others) and printing them on a printed circuit board (PCB), typically with a ground plane on the opposite face (See why this works here ). <S> This allows the antenna to be integrated with the rest of the cellphone's circuitry and to be fit inside the case. <S> The most common type of antenna in modern cellphones seems to be the planar inverted-F antenna (PIFA), which is a special type of monopole in a compact microstrip (patch) format. <A> In modern cellphones all the antennas are inside the case <S> That's clearly not the case. <S> The iphone for example is famously known for its visible antenna bands. <S> The same for many others phones as well. <S> Newer ones go to great length to hide their antenna bands making them less visible.
| There's been a huge hiring thrust in the mobile industries to hire antenna engineers to design functioning antennas for thin phones.
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Is my 2 relay h-bridge safe? I am trying to control a linear actuator using a cheap remote controlled 2 channel relay (it controls a gate for my workshop dust collector). I'm think it's as simple as connecting each relay the same way... NO: +12v from power supply Comm: one motor line on relay 1 and the other motor line on relay 2 NC: neutral (ground? Or -12v???) from power supply This gives me the following voltages on the common lines... No relays activated = NC / NC = neutral Relay 1 activated / relay 2 off = NO / NC = +12v Relay 1 off / relay 2 activated = NC / NO = -12v Both relay activated = NO / NO = 0V I feel comfortable I got everything above right except that last line. If I put 12v on each side of the motor then I should have no voltage difference and therefore no current running through. This thing sits around in one of these states all the time in a remot barn so I just want to confirm nothing is dangerous about my setup. Is there a better way to do this? If above is safe then I have to remember to shut off each relay once the actuators reaches its end position (closed or open). What I need is a circuit tho if I press 1 it closes (+12v) and if I press 2 if opens (-12v). The linear actuator has a switch that opens once it has rich its final position. So trying to press 1 or 2 more times won't do anything until it goes the other way. Thanks for any feedback or suggestions. <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> (a) <S> One way. <S> (b) <S> Another way. <S> Your question was a very hard read. <S> I think you are describing (b) which will work fine. <A> It could be a problem if you are using relays that are make before break. <S> This would cause a momentary short circuit on your 12V supply when switching. <S> From forward to reverse. <S> From reverse to forward. <S> From off to forward or reverse. <S> From forward or reverse to off. <A> This design will work, but it is likely to be rather hard on relays unless one includes a transient suppressor to absorb inductive flyback from motor. <S> An alternative would be to use four flyback diodes: one from each side of the motor to the positive rail, and one from each side to the negative rail.
| Normally when using a relay to switch an inductive load one would use a back-biased diode across that load for flyback suppression, but that won't work if the motor may run both ways.
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Electronic component that is like an LCD version of an LED? Is there a liquid crystal display version of a light emitting diode? I want a visual indicator that is a small and simple 2 pin device like an LED, but instead of emitting light it just turns black or white, or black and clear like how an LCD screen does. I want to make a solar electronic compass. Its a magnetometer wired up to a solar panel and then to some kind of indicator that turns on when you point north. LEDs would be perfect for the indicator except they cannot be seen in sunlight, and they drain several milliamps which is a bit much for a solar panel. LCDs can operate in the microamp range which is perfect, and can also be seen in direct sunlight because they reflect light. This seemed like an ideal fit, but most LCD modules I can find are several inches across which is kind of big for what I had in mind. They also involve a lot of pins and complex interfaces. Also the cheapest LCD modules I could find cost about 2 dollars. All of this is a bit much for a simple on-off visual indicator. LEDs cost like 2 cents each for comparison. After that I looked into 7 segment displays, but almost all of those that I can find for a reasonable price are LED versions that you would be unable to see in sunlight, and would again draw too much power. They make LCD versions of these too, but they seem to be much more rare and expensive. Again you have a lot of pins and large sizes and higher costs. All I want is a little plastic dot that can turn black or clear when you run some power to it. Does this exist? This doesn't seem like it should be so hard to find. Does humanity not have little black dot technology? We definitely have little glowing dot technology; you can get LEDs in huge packs for cheap in every color of the rainbow. I guess you could call this a "LCD pixel" or "single-segment LCD display" or something like that. UPDATE: A few people said I was underestimating the visibility of LEDs and overestimating the visibility of LCDs. I decided to test it just to make sure and took a few photos. Direct sunlight. Left LED is on, right is off for comparison. Outside but shaded. Left LED is on, right is off for comparison. Indoors. I really could see the green LED even in direct sunlight. It was faint, but I could tell that it was on. The LED on the right is completely unpowered for comparison. This LED drew 3.25 milliamps. The LCD I used to compare is a cheap eggtimer I got from walmart for 88 cents. Judging from the posts here it does not seem like this circuit component exists. The light valve was the closest one and they don't make them small enough to be used for a small indicator. Could we get this made? Based on the large number of views relative to other questions asked on this site it seems like people are interested in a component like this. Is there any chance that writing to some kind of LCD manufacturer could get something like this made? I know LCDs run on AC power, but I also know factories can build fairly complex circuity right into the plastic somehow. I have seen LEDs that have built in blinker mechanisms, where you just hook them up to a battery and they automatically blink once per second. The blinking circuit is built right into the base of the LED so small you can't even see it. It seems like they could do a similar setup with some sort of h-bridge to invert the power to AC so the LCD would work even just hooked up to regular dc power. Another possibility was a "Grating light valve" which uses microscopic metal ribbons instead of liquid crystals. As far as I can tell it runs on DC power. It also looks like the patent is just about to expire. <Q> It's a single piece of LCD glass with two electrodes on it. <S> If you've ever used a self-darkening welding helmet, this is what's used to make the window in those. <S> Most of these will probably be bigger than what you have in mind, though. <S> If you plan on manufacturing this product in quantity, contact an LCD maker. <S> They make custom LCD glass designs all the time <S> -- you could probably get a real "compass" design made, instead of just a single indicator. <A> A better solution for this may be a mechanical thing. <S> This could be a device that works similar to the pixels of a high contrast dot matrix sign where a colored dot is flipped between color or black. <A> They provide extremely low power consumption, can be seen under direct sunlight, and interface similarly to LCDs. <A> If you need low power consumption and ability to see in sunlight, I would recommend using an e-ink display. <S> The drawbacks are: you will probably need a controller IC, the price. <A> What you might look for is a single pixel, negative mode, TN (twisted nematic) <S> lcd glass. <S> It's black in powered or unpowered mode, depending on polarizers used. <S> Adafruit has one. <S> It looks like it's .75" sq <A> This is an old question but this might fit the bill. <S> A flip dot indicator . <S> It’s basically a small electromagnetic pulse relay mechanism attached to a fluorescent dot. <S> It requires considerable current to flip (a couple hundred milliamperes) but it’s just a 2ms pulse.
| You can buy relatively inexpensive E-Ink modules that have very simple displays (just a few dots / pixels). The closest thing I can think of to what you're describing is an LCD light valve .
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Problem with MAX756 Boost converter I'm using MAX756 as a DC-DC boost converter to provide 3.3V from of a single 1.5 AA battery. I've implemented the circuit on page 1 of datasheet with 1N5819 instead of 1N5817 and 144uF caps between Vin and Gnd instead of 150uF. Also Vin comes from bench power supply. I've tested three 22uH inductors with different current capabilities (all above the required 1.2A due to datasheet (page 6). ~SHDN and 3/~5 is connected to out . LBO is left open. The problem is the Vout is 3.3V in no-load. But as soon as I insert the 220Ohm resistors, Vout drops and in Iout = 60mA, the Vout drops below 2V. Here is the picture of the setup: lower rail in Vin and upper rail is Vout I've shown the other two inductors that I tested. Anyone knows where the problem might be?Thanks in advance. P.S. Glenn W9IQ says: Take a bench supply capable of 2 amps output and set it to 7 volts. Put a 4.7 ohm, 10 watt non inductive resistor in series with your inductor and put this circuit across the supply. Monitor the inductor voltage with your scope. You should see a nice exponential, asymptotic curve that heads toward zero when power is applied. This will happen within 1 us of applying power. If the beginning of the curve has a different slope, your inductor is in saturation at less then 1.5 amps. But your problem is more likely the ESR of your inductor and capacitor. Use the right parts - it makes a difference! – I've paralleled 5 , 2 Watts 22 Ohms resistor (so 4.4 Ohm in theory and 4.7 due to DMM). Here is the circuit: And here is the scope's result immediately after applying the power : <Q> The app note shows that the minimum input voltage is 2 volts but you are attempting to power it with 1.5 volts. <S> Look at page 6 for recommended suppliers and part families. <S> Your AA cell has an internal resistance and limited energy capacity. <S> You should monitor the input voltage to your switcher to make sure the cell is holding up. <S> Take note that the best efficiency for this boost converter is about 80% with proper component selection. <S> Building a switcher supply on a protoboard is never a good idea. <S> Dead bug / Manhattan style construction on a solid ground plane is a better approach. <A> {Added diagrams of SinglePoint VDD and SinglePoint GND} Circuit works with no load and with 220 ohm (12mA) <S> Iload, with 3.3 volts output. <S> Consider inductive bounce on the breadboard. <S> Assume 2" of loop (in Gnd or elsewhere), or 50nH. Assume 0.1Amp in the IC's internal FET; the IC timing circuits have a minimum ON time, and thus some minimum current; we'll use 0.1amp. <S> Those internal FETs can be very fast, so assume 10 nanosecond to turn OFF or ON. <S> Result? <S> Vbounce = <S> L <S> * dI/dT = <S> 50nH <S> * 0.1a /10nS = 0.5 volt bounce. <S> Nothing is well-controlled when GND or the GND rail is bouncing around with 0.5 volt. <S> Suggestion: build your circuit atop a sheet of copper (one-side copperclad). <S> Here is one physical setup that will generate this 0.5 volt upset <S> simulate this circuit – <S> Schematic created using CircuitLab <S> By the way, this is not the output swing, but inductive upset within (inside) <S> the GND wiring between the IC pins. <S> (for this example) <A> The output voltage on the MAX757 is set by a voltage divider between ground and the output which is connected to the feedback input (pin 2), the formula to calculate the required resistors is: VOUT = <S> (1.25) <S> [(R2 + R1) / R2] To get 5V I have used 30K for R1 and 10K for R2, for 3.3V you could use 18K for R1 and 11K for R2. <S> If you can find it an alternative part is the MAX756 which is a bit easier if you only need an output of 5V or 3.3V as the voltage can be set by connecting pin 2 low for 5V or high for 3.3V which negates the need for the voltage divider arrangement. <S> https://www.electroschematics.com/wp-content/uploads/2010/11/portable-5-volts-power-box.png
| The selection of the inductor and capacitor is quite critical to the operation. With the high currents and high frequencies involved, the protoboard introduces too many strays.
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How drift current is independent of bias voltage in semiconductors The equation J=nqv where n is a concentration of electrons or holes,v is drift velocity.also, v=uE where u-mobility, E is an electric field.SO as bias voltage increases, in turn, increases the Electric field and then results in an increase in Current Density.So how it is possible that the drift current is independent of bias voltage. <Q> The drift current essentially depends only on the rate of generation of the minority carriers. <S> It is because the minority carriers are generated much slower than they are swept due to the field. <S> Had the generation rate been higher than the speed at which they are swept due to the field, the electric field strength would have affected the magnitude of drift current. <S> Conclusion: <S> drift current is independent of bias unlike diffusion current <A> v which is the drift velocity depends on the electric field v = u*E where u is the electron(or hole) mobility. <S> Since the electric field E = V/d. <S> Depends on voltage,Then the drift current density Depends on voltage through the electric field dependence. <A> The drift velocity ( \$v_d\$ ) is a function of the electric field ( \$E\$ ). <S> However, the equation you provide, $$v_d = <S> \mu E$$ is a simplification that is only valid for low electric fields. <S> At high electric fields, interactions with optical phonons cause the drift velocity to approach the "saturation velocity" \$v_s\$ $$v_s = \sqrt{\frac{8E_p}{3\pi m_0}} \approx 10^7 <S> \text{cm/s (Si and Ge)}$$ <S> Where \$E_p\$ is the optical-phonon energy for the material and \$m_0\$ is the mass of an electron.
| The equation of drift current density J = nqv actually shows implicitly that drift current Density depends on voltage!!
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How to find SMD resistor values and size? I want to do reverse engineering of a PCB, in which some SMD resistors are used. On top of each resistor it has marked with 1R0, 150 , 0 etc. Is that represents the value of the resistor? If so, how to find the size of the resistor. Say, a 2512 resistor SMD package has a width of 3.2mm and a length of 6.4mm, will it be available with all resistor values like 1kohm , 2kohm etc. Thanks in advance. <Q> The size of a SMD resistor affects it's thermal dissipation, and maximum voltage rating (although other factor are also likely to be limiting for most values in a range). <S> Different types will also have different tolerance and temperature performance. <S> There is no significant interaction between the available resistance values, and the package size. <A> Yes. <S> Or, to be more precise and exceed the minimum required character count for a cheap answer - Yes. <A> It is gennerally in a form of \$ABC\$, where this means \$AB \cdot <S> 10^C \$. <S> For example, 100 would be 10 Ohm, 473 would be 47kOhm. <S> As others have said, different sizes give you different power and voltage ratings. <S> However, other factors also influence this, such as the construction of the resistor (thick/thin film, wirewound, ...), as well as the PCB and mounting (sometimes you can get higher ratings of power by having thicker copper pads).
| The number on the resistor is the value.
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what is d-sub connector metal shield made of? I want to solder a ground wire to the d-sub connector's metal shield( the bright color metal part of the connector), using tin-lead solder 60-40. I just wonder what kind of metal it is. I searched the web and couldn't find the answer. This kind of connectors <Q> Taking one connector at random on digi-key, http://www.assmann-wsw.com/fileadmin/datasheets/ASS_4884_CO.pdf <S> you can find that the shell is SPCC. <S> So soft steel. <S> Not so fun to solder on. <S> Some (most?) <S> shells are plated though, with copper, tin and/or nickel, which are much nicer to solder on. <S> The other factor is that the shell is basically a HUGE heat sink (in terms of soldering electronic components) <S> so you might have a hard time getting it hot enough to have a good solder join. <S> Good luck. <A> It is commonly nickel-plated steel. <S> Soldering to this requires an immersion flash-gold plating process, which is probably beyond your means. <S> Otherwise, HCl acid with ZCl and NH4Cl and cleaners with toxic fumes when soldering is also not recommended. <S> The recommended solution is to use a thin ring-term. <S> lug on outside screw connection to chassis ground and crimp wire connection to lug. <S> One example might be to terminate a shield at one source only to circuit 0V to prevent ground fault currents yet provide 0V coupling to reduce noise. <S> The shield area has a 1/4 wavelength of several GHz so it practically won't see any egress in normal applications but ingress might only come from arc welders nearby or more likely ESD discharge. <S> In which case I would prefer 10kV insulation over a low impedance current path for ESD to go around a plastic shell screws with heat shrink. <S> But you may have different requirements. <S> Others might terminate to pin 7 ground, but then ESD tests to shell may induce a false start bit. <S> This may be irrelevant to you but are real issues for commercial equipment. <A> I have also seen some that are die cast zinc or nickel, too. <S> The actual manufacturer will generally give you this information in their datasheets. <A> In general, the d-sub connector housing material is SPCC, but there are also some SUS301 and SUS304 stainless steel.
| They tend to be steel, sometimes plated with nickel. Finding the datasheet or mechanical drawing of your part would be the best way of getting the answer.
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How to convert from binary to hex display in verilog? The question needs some explanation: Suppose I have an 8 bit value, say 8'b00000001 (1) Suppose I have the module as follows: module hex_decoder(hex_digit, segments); input [3:0] hex_digit; output reg [6:0] segments; always @(*) case (hex_digit) 4'h0: segments = 7'b100_0000; 4'h1: segments = 7'b111_1001; 4'h2: segments = 7'b010_0100; 4'h3: segments = 7'b011_0000; 4'h4: segments = 7'b001_1001; 4'h5: segments = 7'b001_0010; 4'h6: segments = 7'b000_0010; 4'h7: segments = 7'b111_1000; 4'h8: segments = 7'b000_0000; 4'h9: segments = 7'b001_1000; 4'hA: segments = 7'b000_1000; 4'hB: segments = 7'b000_0011; 4'hC: segments = 7'b100_0110; 4'hD: segments = 7'b010_0001; 4'hE: segments = 7'b000_0110; 4'hF: segments = 7'b000_1110; default: segments = 7'h7f; endcaseendmodule to output on HEX0 and HEX1, I can do something like this: hex_decoder h0(.hex_digit(value[3:0]), HEX0);hex_decoder h1(.hex_digit(value[7:4]), HEX1); and this will display 01 on HEX1, HEX0. The only problem is that after 9 , the values will become in hex letters. I want it so that if I pass in binary 10, ( 8'b00001010 ), then HEX1 HEX0 should be 1 0, not 0 A (as hex works) How can I convert it like this? <Q> The problem you are having is quite a common one - how to convert a binary number to something called "Binary Coded Decimal" (BCD). <S> In BCD each digit is 4 bits, but those 4 bits are only used to represent the numbers 0-9 <S> (hence the decimal bit). <S> This is an ideal format for outputting to 7-segment displays, screens, in fact anything that needs a decimal number to be displayed. <S> (both names for the same thing). <S> Essentially the approach is to scan through the binary representation, then any time you see a number which is 5 or higher in each group of 4 bits, you add 3 to it. <S> This approach basically is a way to overflow any values greater or equal to 10 in a digit into the next one without too much hardware. <S> Here is an example, stolen from this Wikipedia Page : <S> Double Dabble Conversion of 243 <S> Hund Tens Unit Shift <S> In 0000 0000 0000 <S> 11110011 <S> Initialization <S> 0000 <S> 0000 0001 <S> 11100110 <S> Shift <S> 0000 0000 0011 <S> 11001100 <S> Shift <S> 0000 <S> 0000 0111 <S> 10011000 <S> Shift <S> 0000 <S> 0000 <S> 1010 <S> 10011000 <S> Add 3 to ONES, since it was 7 0000 0001 <S> 0101 <S> 00110000 <S> Shift <S> 0000 0001 <S> 1000 <S> 00110000 <S> Add 3 to ONES, since it was 5 0000 0011 <S> 0000 <S> 01100000 <S> Shift <S> 0000 0110 0000 <S> 11000000 <S> Shift <S> 0000 1001 0000 <S> 11000000 <S> Add 3 to TENS, since it was 6 0001 0010 0001 <S> 10000000 <S> Shift <S> 0010 0100 0011 <S> 00000000 <S> Shift <S> 2 <S> 4 <S> 3 BCD Building this process in Verilog is relatively straight forward. <S> I'll leave it as an exercise for you. <A> The problem with Double-Dabble is that it was written for software programs, so the solution is inherently serial. <S> FPGA's and ASIC's are parallel in nature: we need to use their strengths to our advantage. <S> While Double-Dabble works, the example given here uses 11 clock cycles after initialization to arrive at a 3-digit result from an 8-bit input. <S> For an 8-digit result from a 27-bit input Double-Dabble requires 35 clock cycles. <S> A reachable goal for an N-digit result is N-1 clock cycles. <S> How? <S> By doing long division in parallel. <S> For an eight digit result from a 27-bit input, start by doing 9 subtractions in parallel, input1 <S> minus: 9000_0000, 8000_0000, 7000_0000, ..., 2000_0000, 1000_0000 and implicit 0. <S> Take the remainder from largest factor where the result is not negative (test the top bit of the result), record the factor as the top digit and use the remainder as the input to the next stage, which is again 9 subtractions in parallel. <S> Input2 <S> minus 900_0000, 800_0000, 700_0000, ..., 200_0000, 100_0000 and implicit 0. <S> Repeat this procedure down to input7 <S> minus 90, 80, 70, ..., 20 and 10. <S> The remainder from the last subtraction is the one's digit. <S> Using this method requires seven clock cycles after initialization for an eight digit result. <S> Depending on the FPGA, this might be implemented with nine DSP's and 63 constants. <S> Without DSP's, the number of bits of subtraction go down by three or four bits on every cycle, so it does not take as much fabric as you might expect, and it still only requires seven clock cycles to complete. <A> A rom table could work if you like using up FPGA gates.... <S> Say you have a typical FPGA dev board with four 7 segment displays, and you want to show unsigned values. <S> That's 10000 total characters, 7 bits each, round up to 16K by 32 bits is a 64 Kbyte rom.(The 7 seg beside me also has a decimal LED, so 8 bits per character..) <S> Something like a "character generator" from the stone age of ascii terminals. <S> You can also use up one 7 seg and get from 9999 down to -999. <S> Filling the rom table can be done by C or Python etc using one of the above and converting the output to the FPGA rom file format.. <S> At FPGA speeds this is slightly (!!) <S> more than a bit faster than needed and thus a waste of gates, but if you have space why not? <S> Definitely fast enough for Persistence of Vision games.. <S> Typically though, if you already have a small embedded microprocessor, use one of the sequential routines above and let the micro handle it as one of its' tasks..
| The simplest way of converting from binary to BCD is an algorithm called "Shift-Add-3" or "Double-Dabble"
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Does the electrical impedance of a headphone depend more on the cable or on the headphone itself? I'm currently considering buying a defect headphone with a built-in broken cable which would be quite expensive otherwise. I'm quite sure I'm able to solder a new cable onto the headphone to fix it. My question is the following: Does the impedance of a headphone noticeably change if I use a cheap standard AUX-chord or does it mainly depend on the internal electronics of the headphone? <Q> The cable does have a finite resistance, but that is small compared to the impedance of the headphones. <S> Unless the cable is unusually long or thin, the cable is not a issue. <S> A few feet of even thin cable should be fine. <A> A standard set of headphones has no electronics. <S> The sound is produced by miniature electro-magnetic loudspeakers. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The impedance of the speakers is typically > <S> = 8 Ω and the wiring would be less than one tenth of that for good control of the speakers. <S> Be careful to get the phasing of the speakers right. <S> If you swap them the sound will be out of phase and will give a weird hole-in-the-middle-of-your-head effect. <A> The impedance of a headphone or speaker is determined by the speaker. <S> The resistance of the wire has almost nothing to do with the impedance. <S> There are cases where using the wrong wire or type of wire can cause problems. <S> Most of those involve high power, high frequencies, or high power and high frequencies. <S> Your headphones certainly aren't in the realm of high power. <S> Audio doesn't come under the heading of "high frequency" either. <S> So, use any cable that is long enough, flexible enough, and easily worked with and soldered. <A> In short - the speaker. <S> The cable is usually neglectable in comparison.
| The impedance of headphones is a function of the headphones themselves, not the cable.
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Can 12v dc circuit spark and start a fire? I'm new to electronics and electricity. I have the following scenario: I am wiring a LED light switch and mounting it into wooden cladding. My concern is, can the wires spark unexpectedly and start a fire? We are talking about 12V DC, with three, 2W LEDs. The circuit is properly fused. <Q> Yes, 12V can certainly cause sparks if there is a connection that is opening and/or closing. <S> Your 12V power supply likely has some reasonable current limit, and It's hard to ignite wood with sparks. <S> If you were around a flammable gas or liquid, however, any sparks could be devastating. <S> Once I was stranded at night (with my malfunctioning car) on a road in the wilderness. <S> I had some camping gear with me, but no method to light my cookstove. <S> I used jumper cables from the car's 12V battery to create sparks and light the stove. <S> I don't recommend this! <S> It took a while for the hair on my forearms to grow back :) <S> P.S. Make sure you note @Trevor's important point about strain relief in his answer. <A> Assuming the connections are sound and no extraneous bits of wire etc. are around it "should" be safe. <S> However, when burying things like this it is prudent to ensure that no stress can be exerted on the connections, either as part of the install, or later. <S> As such you need to relieve those cable turns and add appropriate cable tie downs to make sure nobody can pull on the cable later breaking the integrity of the connections. <S> Ultimately, there is ALWAYS a risk of fire, so nobody on here will stand by a blank "Yes, it's fine and won't start a fire! <S> " statement. <S> However, as a side note, your design may not be a great idea anyway. <S> If at some point that switch fails you will need to replace it. <S> I have a feeling you do not want to rip the trim out of the wall to do that either. <A> Sure, if you bring them close together enough, or if you introduce a liquid, or if you connect them to a high voltage or something. <S> But as long as you properly terminate or insert them, with no loose strands or nicks or cuts, you should be fine.
| However, I expect you are safe: Since you are fused, any high-current short should blow the fuse before anything bad happens.
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Linear regulator GND pin MUST be connected to common ground? I'm trying to make a constant current source of 2.1A which will operate around 1.7V. I did some web searching, and I decided to use linear voltage regulators; simply put a adjusting resistor between the voltage outputs, and connect my load between the cathode and the ground: TPS757 constant current source with output voltage of 1.5V. Note that GND pin would be at 1.7V with respect to input ground terminal, which is response voltage of my load near 2A. This means there is 3.3V different between VIN and GND, as long as my load keeps constant voltage across. However, I found that some regulators have GND pin, while others have ADJ pin; I believe that the output voltage indicated in data sheets refer to the voltage difference between OUT pins and either of those pins. For those which has GND pins instead of ADJ pins, must I connect GND pin to the same ground of the input voltage? I've seen some examples with regulators with ADJ pins that had the ADJ pin voltage above the common ground voltage (ex LM317), but I'm not sure it would apply the same to regulators with GND pins. If so, how would I make a constant current source out of them? Thanks in advance! <Q> However, what the "third" pin does is rather less defined in some devices than others. <S> When using it the way you indicate, you must remember there is a bias current that emanates from the "ground" pin. <S> Basically the quiescent Icc of the device. <S> This bias current will of course be driven through the load in addition to whatever you have set your regulator circuit to produce. <S> You must therefore compensate your values to take that into account. <S> How much ambient current that is will depend on a the device and a number of variables that may or may not be well specified, or even controlled in the device itself. <S> Furthermore, if the load is a fixed load, using a voltage regulator may work for you, however, I'd be cautious if the load varies, or is switched on and off, since this will also affect the bias point of the regulator and it may become quite unpredictable. <A> LM317 has a large drop and will get hot. <S> Consider a 3.3V source from ATX supply and this or equiv. <S> I'm trying to make a constant current source of 2.1A which will operate around 1.7V. <S> THis is equivalent to a 3.6W load. <S> If it is an IR diode it will over-heat unless pulsed. <S> Otherwise you can use an LC filter with a series drop from 3.3V of (3.3-1.7)/2.1A = 0.76 ohm or use a Darlington emitter follower biased at 3V from 3.3V with equivalent load at 1.7V/2.1A= 0.8 ohm and Rb=1K equiv bias to 3.0V simulate this circuit – <S> Schematic created using CircuitLab <S> Depending on what parts you have avail, there are many ways to make a CC source or sink. <S> and exactly what your load equivalent cct. <A> Please consider these two equivalent circuits: simulate this circuit – Schematic created using CircuitLab <S> A voltage source is a short with a voltage across it. <S> See how you can move the LED across the voltage source without changing anything to the current through it? <S> From the view of the regulator, it stays all the same. <S> It's just the LED isn't in the return path any more. <S> Does it make a difference? <S> No. <S> What the regulator does is maintaining the set voltage across its OUT and GND terminals. <S> That results in a fixed current through the OUT pin. <S> The current through the GND pin has to be negligible at the same time — <S> the whole idea of using a voltage regulator as a current regulator depends on that—. <S> So $$I_{in} <S> \approx I_{out}$$ <S> Are your doubts about GND gone?
| Generally linear regulators can be used in various modes which are not quite as the original developer envisioned, including as current sources.
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Does a high voltage electrostatic precipitator power supply provide a balanced or unbalanced output? Does a high voltage electrostatic precipitator power supply create a balanced or unbalanced electrical output in relation to the primary side/ground? If the device provides an unbalanced output, does the power come off of the negative or positive side?Here is an example of such a power supply: http://www.ebay.com/itm/High-voltage-electrostatic-precipitator-power-supply-with-400W-60kV-/152550032861 . Also, here is a schematic for the power supply: <Q> Joshua Guertler - Please stop messing around with this stuff. <S> You are going to get someone hurt. <S> Why do I say this? <S> Because your questions indicate that you have no idea what you are talking about. <S> Does a high voltage electrostatic precipitator power supply create a balanced or unbalanced electrical output in relation to the primary side/ground? <S> No. <S> Or rather, maybe. <S> But in light of the rest of your question, no. <S> Also, here is a schematic for the power supply: <S> And again, no. <S> That is not remotely the power supply for the unit you linked. <S> So, you don't understand what balanced or unbalanced means, and you don't realize that the schematic cannot possibly apply to the eBay unit you linked to. <S> (Or maybe you do, since in your previous question you referred to the schematic as being "generic". <S> But in that case it cannot be "the power supply". <S> Either way you are hopelessly confused.) <S> Marcus Miller obliquely pointed this out in comment, but it apparently passed you by. <S> It's perfectly possible to make a HV supply which is either balanced or unbalanced, it's just that there is no reason to think that either term applies to the unit under consideration. <S> Please find somebody who has a clue about this stuff and take his or her advice. <S> I'm afraid you're going to hurt yourself or someone else. <A> The source is balanced differential output from a HF transformer but the load is unbalanced with HVDC ionizing particles may detonate more on the + terminal since electrons emit freely from conductors and transfer charge to dust which become -ve and discharge against +ve terminal attraction. <S> Thus dust might detonate in the centre of the stack rod and be converted to a benign gas. <S> ( just a hunch) <S> Thus since arc impedance is low compared to the capacitance and high R of the stack electrodes. <S> The imbalance is due to the earth grounding of the stack and polarity of discharge on one side relative to the series <S> current limiting impedance of the transformer. <S> The WYE 3 phase may be earth grounded at the Neutral or it may be floating Delta windings, but common mode coupling to ground at RF is makes it relatively low impedance high capacitance coupling compared to the earth grounded stack capacitance and inductance to the HVDC source. <S> Thus high DM noise but shielded by the stack somewhat and more important, the CM noise of the feed wires , so attention to EMI must consider CM balancing at RF from 50kHz to 1GHz and discharge currents via the stack, which is not visible from the logic diagram.. <A> There is not enough information on that eBay ad to determine whether the output is AC, -DC, or +DC. <S> You need to ask the seller about that, because those four flyback transformers might very well have built-in diodes that output +HV that are impossible to change! <S> That was the case for many of the last CRT-type television sets , which were +DC and could not be changed. <S> I have a whole box of those here somewhere, and they are destined for the garbage. <S> If there are no embedded diodes --and the output is AC-- then it might be possible to add some HV diodes so that the output is -DC. <S> Maybe you'll get lucky and find that the HV output is -DC, <S> but I seriously doubt it.
| It is unbalanced pulse current which creates large EMI unless a HiV CM choke is used with twisted pairs low ESL wiring.
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Alternatives for EMG muscle sensors? I would like to measure muscle tension with a sensor. When I'm sleeping sometimes a shoulder muscle cramps and I would like to make something (for example with an Arduino board) to wake me up. The problem is to measure the muscle tension. A EMG sensor uses three pads. The Myoware EMG board can be used for DIY projects. I have not tried it yet. I think it is not easy to attach it every evening and I don't known where to attach the three pads to select a specific shoulder muscle. But so far this is the only thing that comes close. [ADDED] I wonder what happens if I attach the pads to a muscle that has already a high tension. Would the EMG be able to measure the absolute tension ? A ultrasonic liquid density sensor is big, they can not stay attached while sleeping. It could also be a safety risk. Perhaps the ultrasonic could harm tissue and blood cells. The best way to measure muscle tension is to pinch or press it. Perhaps something small that mechanically pushes into the muscle and measures the strength could work. I have no idea how to make this. With dry needling , a relaxed muscle does not react, but with a muscle that cramps or has a high tension, then the muscle reacts a lot. However, I can't push needles into my body while sleeping, that would be silly. If you know a way to measure muscle tension in a practical way, please let me know. A similar question was asked in the Health section as well: How can I quantify muscle tightness? [Update] First tests with a EMG sensor . I have been testing a EMG sensor. The Myoware 4th generation EMG sensor is easy to use for DIY project. To prevent grounding or noise, I used a battery and it is not connected to the computer. Sadly, it is not near to what I want. With my fingers I can feel how much tension a (shoulder)muscle has, when I move my back/head/shoulder a little I can feel what is going on. The EMG sensor is more like a lie detector. It detects the tension of a muscle, but that is mostly relative tension and it depends a lot on where the pads are located an how well they are attached. When I use it the next day, I want the same tension to result into the same numbers. That is not possible. <Q> Although my answer goes against being electrical engineering in nature I have to say you are going about this all wrong. <S> What you need to consider instead is a holistic approach. <S> If you are having muscle seizures and cramps while you sleep be assured that is almost always caused by something going on in your life or physical self. <S> It could be stress from any one of dozens of factors. <S> It could be the condition and environment you put yourself through during the day and reflecting that in through the night. <S> There will be added tension and stress due to the anticipation of being awakened and will certainly lead to less sleep and lower quality sleep when and if you do get to sleep. <S> You may want to try some of these things to try to being more calm and reduced stress and tension in your being: <S> Try to reorganize your day - get up earlier <S> - do your most energetic activities <S> early and then by evening make sure there is a quiet wind down time for a couple of hours before trying to go to sleep. <S> Try learning yoga and take a class. <S> The benefits to overall body stress relief can be amazing. <S> Even one session will show you what total relaxation can be like. <A> I agree with Michael. <S> Although you could probably use any metal foil that is compatible with your skin would work. <S> The gap between the epidermis dielectric and electrodes creates a galvanic skin voltage and any gap modulation also changes capacitance so when moving, a false reading can occur, which may be what you want ir if not and just muscle tension, it will be a pulse noise envelope < 1kHZ BW. <S> But you will need extra pads which normally use snap buttons. <A> You might try a piezo strip sensor taped to the skin. <S> The are inexpensive and don't require an external power source, but generate an electrical signal when flexed which could be used to trigger your Arduino or measured by the Arduino ADC. <S> You could poll the ADC and look for voltage and keep a log or initiate some other action like the alarm you suggest.
| I would suggest that even trying to hook yourself up to a gadget that wakes you when these conditions happen will make things even worse. So special low-k dielectric constant partially-conductive adhesive gel pads work best as perhaps provided with that instrument. Consider a treatment of whole body acupuncture. Even if you could somehow make a gadget how can you be sure that trying to measure tension in one muscle is even going to be the right one when the next time it may be in completely different area.
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How do I desolder these specific points? I'm currently attempting to replace some bloated caps on an old PC motherboard but am currently having problems with some specific points. Here's a pic of the motherboard: . The area outlined in fuchsia are the series of caps I am trying to desolder. Notice they are in pairs of square and round solder points. The round solder points are no problem to desolder. It's the square ones that won't desolder. No matter how much I heat them up, the solder on the points just won't melt and so I can't pull the caps off! Is there a special technique or tool I need to perform this task? Thanks in advance for any help you can provide. <Q> The square connections are in the ground plane of the board. <S> That is the large yellow area they are embedded in. <S> That is a large piece of copper, and there are probably also large copper surfaces on the internal layers of the board. <S> Copper conducts heat very well, and it also radiates it away. <S> The large copper areas are basically sucking up all the heat your iron can provide and radiating it away fast enough that it can't get hot enough to melt solder. <S> The solution is an iron that can put in heat faster than the board can dissipate it. <S> So, you need an iron with more power. <S> Many irons are only around 30 watts. <S> You'll need much more than that. <S> When I've had to do that kind of thing, I borrowed a huge 150 watt iron from my father in law. <S> It isn't intended for electronics, but it has the raw power needed for large copper surfaces. <S> As for technique, high wattage irons often have wide tips. <S> I apply some extra solder to the heavy joint with the iron heating just the ground connection. <S> When that finally melts, I rotate the tip of the iron to heat both pads for that part. <S> The solder melts pretty quickly, then I can pull the part out. <S> Afterwards (if you need to to replace the part) you can clean the holes with a solder sucker or solder wick. <S> While you are removing the part, you actually want as much solder as possible on the connection. <S> Removing solder makes it harder to get the part out, not easier. <A> Is there a special technique or tool I need to perform this task? <S> Nothing special but try to get hold of a rework station with blower. <S> Using the blower adjust the blow speed and temperature (such that the board doesn't burn up). <S> Apply flux, heat the desired location and with the help of tweezers plug out the caps. <S> I have also experienced the same problem before, so I plan on dividing my ground plane into smaller sections during routing. <S> Edit: In Eagle, there is an option of enabling thermals for a pad(enabled by default). <S> Enabling this keeps a bit of gap between the pad and the surrounding plane. <S> This helps in quickly heating the pad before the heat dissipates. <A> Most likely both pins are connected to planes. <S> The ground plane on bottom layer is obvious, but there is probably a power plane on the other pin as well. <S> So the solder will melt half way through the board (and visually, from your side, it looks melted) <S> however there will still be solid solder inside. <S> Have someone grab the cap from the other side with a pair of pliers and pull GENTLY. <S> Apply one >90W soldering iron with flat wide tip to each pin, adding leaded solder to lower the melting point. <S> Yes you need two irons and two hands. <S> Another method is: Cut a piece of thick (like 4mm2) copper wire long enough to span both of the cap's pin. <S> Place it across the cap's pins. <S> Heat it with a powerful iron (like an instant-on 100W soldering pistol). <S> The copper will spread the heat to both pins. <S> Apply a generous helping of leaded solder to lower the melting point. <S> This method only requires one iron (albeit a beefy one) and two hands, so it is more practical than the previous one. <S> With a bit of practice, desoldering QFPs without damaging the board is rather simple. <S> This are the simplest "ghetto" method. <S> If you have a hot air rework station, you can preheat the board, which will make your job a lot easier. <S> Don't use something like a 2000W hot air gun, this is very effective at burning and delaminating boards (after all it is intended to strip paint...) <A> Butcher the capacitor off and solder the replacement to the stub of the leg that remains. <S> This is known as the RAG technique (Rough As Guts).
| A cut piece of copper wire bent and shaped properly also works wonders to heat all the pins on big SMD chips, connectors, etc, and makes desoldering a lot easier. You need a bigger soldering iron, as in "more power." As pointed out by others, the square ones are indeed ground plane which require a bit more time to heat up due to the large surface area. The solution is: Clamp the board to something solid.
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Correct calculation for voltage drop of transistor I'm new in electrical stuff, but I know that a transistor has a voltage drop of around 0.7 V. So I have the following circuit: (Hope it's correct): I want to trigger the transistor if the base get's flow from 3.3 V.My question: How do i calculate the voltage drop? $$\frac{5~V - 0.7~V}{50~\Omega} = 0.086~A$$$$\frac{3.3~V - 0.7~V}{500~\Omega} = 0.0052~A$$ So total at the ermitter: $$0.086~A + 0.0052~A = 0.0912~A = 91.2~mA$$ Is this correct? Or it is only $$\frac{5~V - 0.7~V}{50~\Omega} = 0.086~A$$$$\frac{3.3~V}{500~\Omega} = 0.0066~A$$ or $$\frac{5~V}{50~\Omega} = 0.1~A$$$$\frac{3.3~V - 0.7~V}{500~\Omega} = 0.0052~A$$ I don't know where exactly the voltage drop should be in my calculation. <Q> For the circuit you show, 0.7 only applies the base-emitter junction. <S> The emitter-collector junction will attempt to affect current, rather than voltage. <S> So, the base current will be (3.3 - 0.7)/500, or 5.2 mA. Calculating the collector current is harder. <S> You must assume a value for the current gain, hfe or beta. <S> Let's start by assuming an hfe of 100, which is a decent starting value for small-signal transistors. <S> 5.2 mA times 100 is .52 amps. <S> If this were true, the voltage across R1 would be .52 times 50, or 26 volts. <S> And this obviously won't happen. <S> So it's likely that the transistor is not operating in the linear region, but is close to saturation. <S> (The difference between linear and saturated operation is determined by whether the collector-emitter voltage is greater or less than the base-emitter voltage.) <S> A good rule of thumb is that, in saturation, the gain of the transistor is less than 10 or 20, and the collector-emitter voltage will be in the range of 0.2 to 0.3 volts <S> Let's try 20. <S> Then the current is .104 mA, and the voltage across R1 will be 5 volts. <S> This is consistent with the circuit, and is a reasonable solution. <S> The exact collector-emitter voltage cannot be determined easily, but in general the details won't matter. <S> The current through R1 will be a bit less than 100 mA, and the voltage across the transistor will be about 0.2 to 0.3 volts. <A> The Collector-Emitter current (limit) is beta times Base-Emitter current. <S> Beta is the gain of the transistor. <S> Collector-Emitter voltage is (5V - 0.7V) / 50 <S> Ohms = <S> 86mA. Base-Emitter current is (3.3V - 0.7V) / <S> 500 Ohms = <S> 5.2mA. <S> If beta were 200, then that Base-Emitter current could support driving up to about 1A of Collector-Emitter current. <S> What you have to do is size your base resistor so that, for the beta of your transistor, at least enough current is flowing through the Base-Emitter junction in order to satisfy the demanded / desired / designed Collector-Emitter current. <S> This might be a good reference for you to take a look at. <A> This is a grounded-emitter amplifier. <S> That is, a common-emitter amplifier with RE=0. <S> The gain of such circuit would be -RC/re, where re is the intrinsic emitter resistance, which is 25mV <S> /Ic. <S> For example, for a quiescent current of 1mA <S> the gain is -Rc <S> /re=-RC.Ic/25mV=-50*1/25=-2. <S> But this is heavily dependent on the collector current. <S> For your example, the gain will vary from -200 all the way down to zero. <S> So in practice, it's better to avoid an emitter-grounded amplifier except when very small input signal swings are involved. <S> People use emitter-follower with degeneration, where the voltage gain doesn't depend on the collector current.
| You can generally assume a 0.2V drop across the Collector-Emitter junction.
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BJT or Mosfet as Cascode I was going through this lecture pdf. If I want to have maximum output resistance which cascode configuration I should use. BJT cascode or Mosfet cascode? <Q> You want to use MOSFETs if you want maximum output resistance. <S> Draw the small signal model and see for yourself. <S> Try stacking N BJTs vs N MOSFETs to form Cascodes and see the limiting case for Rout. <S> Because of a BJT's rpi, maximum output resistance is limited as you stack more and more stages. <S> For MOSFETs the only practical limit is the voltage rails. <S> Edit: Clarification for the incorrect downvote For BJT's forming cascodes by stacking ideal transistors it not that beneficial as the stack grows because the resistance rpi compromises the resistance boost of the cascode structure. <S> So as the BJT cascode stack tends to infinity the output resistance tends to rpi and NOT infinity. <A> Often you don't have that choice because many designs are made with CMOS processes where (good) BJTs are not available. <A> The pdf you quote is carelessy written, with too many ambiguities and unidentified cicuit references. <S> The problem you are asked to solve could be tackled by using simple three terminal models for the circuit elements. <S> Simple inspection will generate the answer you are looking for.
| Of course you would want to use a BJT because it has a higher intrinsic gain and would give the better (higher) output resistance.
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What do 3-digit reference designators mean? While looking at many circuit boards and schematics, I often notice that 3-digit designators are used, for example R101 instead of R1. What do these digits actually mean? The numbers are not in sequence, for example only a single relay, K101 may be present, still numbered 101. <Q> You should be able to figure out it looking at the device as whole. <S> You can find such numbering in the devices with multiple boards, multiple physical or logical blocks in it. <S> First digit may designate block #, other two (usually) designate component # in the block. <A> For example, R101, R201, R301... will each be the "same" component, but in channels 1, 2, 3... <A> I personally usually use 4 digits: first two are the sheet number and last two are the unique number for this component. <S> That means I can use up to 99 components of each type on each sheet! <S> Of course sheets 1-9 have 3-digit reference designators. <S> I suspect this or something similar is the reason you see 3-digit numbers. <S> I find this method quite convenient especially during layout. <S> When you are in the process of placing components, this is a good way of more or less knowing where does each component belong to. <A> There is also a common system for reference designators that follow a coordinate zone system for large boards that have thousands of components. <S> (Think PC or server motherboard for example). <S> The board will have zone numbers along the sides that may range from 1 to 9. <S> Along the top/bottom the zone numbers may range from A to K (skipping I) on the primary side and ranging from L to V (skipping O) on the secondary side. <S> Reference designators then become something like R7N16 for the 16th resistor in the "7N" zone region. <S> This system makes locating the physical resistor on the actual board easier if the reference designators are actually all included in the silkscreen of the board. <S> Particularly useful when testing and qualifying the first or second fabrications of a new board. <S> In the advent of small components like 0402 and 0201 SMT parts it is getting more common to not even include the reference designators on a board and engineers and techs need to keep their laptop handy right on the lab bench to be able to look up part locations! <A> Reference designators are arbitrary. <S> Very generally, the first character denotes the type of component (R for resistors, C for capacitors, U or IC for integrated circuits, D for diodes, Q for transistors, Y or X for crystals, J or P for jumpers and/or connectors, F for fuses, K for relays, etc... <S> The numbers are just as arbitrary. <S> You shouldn't be confused by the number of digits. <S> I typically re-annotate <S> my boards so the designators "flow" according to how they're placed, with 1 being in the upper left and increasing as you move across and down the board. <S> I've tried numbering 1-99 for components on the top and 101-199 for components on the bottom, but over the years my boards sometimes got to be more complex <S> and I opted for just starting at 1 at the top left and letting the numbers just naturally progress as they transitioned to the bottom of the board. <S> Of course, when you re-annotate based on physical location then the numbers on the schematic are all over the place, but that's fine; trying to find R75 when you see R73 is much, much easier when they're numbered by location. <S> The schematic organization is much easier to observe and doesn't need numeric "locality". <A> Different people assign reference designators differently. <S> Some use location on the PCB, some location in the schematics. <S> Some the order in which components were added to the design. <S> One reason to use 3-digit designators even when not strictly needed is to make the length of the designator independent of the value. <S> That means that when you re-annotate the board or copy and paste a subcircuit you don't have to go round re-positioning all the designators on the silkscreen layer.
| Often, in multi-channel circuits (same circuit repeated several times), the first digit will indicate the channel, while the remaining digits indicate the specific component.
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Are Adafruit RGB 30 LED/m flexi-strips suitable to wear in large numbers? I am working on a project that involves using a large number of LEDs on a single garment. According to my initial design, I would need as much as 1800 RGB LEDs. I have seen these LED strips from Adafruit https://www.adafruit.com/product/285 . Since the strips draws 60mA 600mA per meter, and I would need roughly 60 meters, this amounts to a total of 36 amps. Using them would imply (I guess) either using dangerously high amps or distributing the current by using unacceptable numbers of battery packs. My questions are therefore: I am correct in my assertion that using the LED strips linked to above is not realistic (i.e. 1800 ish LEDs on a single garment)? Or can I overcome this problem in any other way, as for instance by setting software limits to the PWN controlling the LEDs, thus limiting the current flowing through them? Are there any other types of LEDs that would be more suitable? They need to be easily controlled by a microcontroller without additional circuitry, and ideally RGB. (Updated text to show the maximum "current per meter" value, 600mA per meter, quoted on that Adafruit page, instead of the originally written value of 60mA per meter; this change has affected answers.) <Q> Edit: Sam caught an error by the OP which changes my answer. <S> You will need to supply a total of 36 amps if you turn everything on a once. <S> But this might be a bit boring because you will just look like a large white light bulb. <S> Consider adding a small programmable controller that cycles through various colors and turns sections on and off to make it more interesting. <S> This will also greatly reduce the energy that your battery will need to supply. <S> To help you with your battery calculations, if you had 1/3 of the LEDs turned on at any one time on the average, you would then only need ~12 amps. <S> Battery life is expressed in amp-hours so a 12 amp-hour battery would power your LEDs for 1 hour. <S> If you used a commonly available 7 Ah battery, this would power them for less than 1/2 hour. <S> The app notes indicate that the strips will work with 9 volts. <S> If you used a 12 volt battery with a 12 volt to 9 volt buck converter, you could extend your battery life by at least 15%. <S> The LEDs would be slightly dimmer but it may be worth it for the extra life. <S> You could also consider Li Ion cells that add up to a voltage lower than 12 volts. <S> Do break your LED strings into strips that draw no more than ~4 <S> amps each and put a 5 amp fuse in series with each such strip from the battery to head off a potential clothing fire. <S> Monitor your battery temperature when in use to make sure it is not getting hot. <S> Always charge your battery away from the clothing in a safe fashion. <S> Have fun with the project. <A> Actually, the Adafruit datasheet says, " <S> Max 0.6 Amps per meter (all LEDs lit full brightness)" <S> So the fear of OP might be valid. <S> However, if the controller is smart enough and limit light patterns, maybe more reasonable consumption can be achieved. <S> In my experience, a high-efficiency LEDs can output pretty intense light running at 2 mA. <S> If the strip really is useful at 60 mA per 30 LEDs (as Adafruit says), this project will take 3-4 <S> A total, as Glenn noted. <S> Then, with a 12V boost converter, this will translate into 10-12A at a single-cell Li-Ion battery. <S> So even a single high-discharge (~5C) <S> 18650 battery can lit the thing up for 10-20 minutes. <S> Three such cells in series will power the project directly for about an hour. <S> There are flat batteries , 5-7mm thick, up to 20000 mAh capacity and 8C discharge rate, they will last longer. <S> ADDITION: if money and design effort time is not a problem, there are tiny FPGA chips designed specifically to drive a single RGB LED, 1.4x1.4mm in size, 16-ball WLCSP, iCE40 series from Lattics Semi . <S> You can use some serial interface and control everything in any desired way. <S> ADDITION2: <S> To illustrate how overstated the 20 mA "requirement" is (or how low quality the strip LEDs are): in one project I was using an OSRAM RGB LED, in PLCC-6 package, 3.2x3.2mm, P/ <S> N LRTBGFTG-T7AW-1+V7A7-29+R. <S> To get a light that would not blind an eye, I had to use ~0.7 mA <S> (!!!) on Green, and 1.6-1.7 mA on Red and Blue. <A> I am correct in my assertion that using the LED strips linked to above is not realistic (i.e. 1800 ish LEDs on a single garment)? <S> Or can I overcome this problem in any other way, as for instance by setting software limits to the PWN controlling the LEDs, thus limiting the current flowing through them? <S> Yes, you could PWM them and so reduce both the brightness and the power consumption. <S> Keep in mind that the maximum power is for them on full brightness white, other colours will be lower power since they won't require all 3 colour LEDs to be on. <S> So a 50% PWM in red would be about 1/6th of the power consumption. <S> However you're still looking at a lot of power consumption no matter what you do. <S> Are there any other types of LEDs that would be more suitable? <S> They need to be easily controlled by a microcontroller without additional circuitry, and ideally RGB. <S> 1) <S> Those need additional circuitry, they are 9-12V <S> so at the very least you need a FET controlled by the microcontroller. <S> 2) <S> Not really. <S> Less or dimmer LEDs will reduce your power but ultimately LED power isn't that variable, for the same brightness you'll be looking at roughly the same power level no matter what type of LED you use. <S> You could look at a completely different technology like Electroluminescent wires here or for an animated effect here , whether it's suitible depends on the effect that you are trying to produce. <S> It's often easier to incorporate into clothing then LEDs but isn't as bright, not as easy to vary the brightness of and the only way to get multiple colours would be to run multiple threads next to each other.
| Yes, that power consumption is unrealistic assuming you want them on for any length of time rather than brief flashes.
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What is the most efficient way to effect a voltage drop in a dc cct? I have a 24V supply and need to drop the supply voltage to 19V to power a TV set. The plug pack that powers the TV states the output as 19V and the current at 2.1A I could use a resistor to drop the voltage but it would need to have a high power rating. Can anyone suggest a cct using easily obtained components to solve this problem? <Q> You can't really use a resistor because voltage drop depends on current draw. <S> If you choose a resistance value to drop 5V at 2.1A then if at any time the TV draws less than 2.1A the output voltage will rise above 19V. <S> So you need a circuit that regulates the output voltage <S> so it remains stable when load current varies. <S> The two main classes of regulator are linear and switching . <S> It suffers the same power dissipation problems as a resistor, but has the advantage of a relatively simple circuit. <S> Switching regulators pulse current through an inductor to 'transform' the voltage down. <S> They can be over 90% efficient because the inductor stores and releases the pulse energy rather than wasting it as heat. <S> The disadvantage is a more complex circuit with critical component specifications and circuit layout. <S> Using a linear regulator to drop 24V down to 19V at 2.1A would waste (24-19)*2.1 = 10.5W. <S> You would need a fairly large heat sink to remove this heat. <S> However the circuit efficiency would be 79% which is not bad. <S> A 90% efficient switching regulator would only dissipate 4.4W. <S> The LM2596 is a switching regulator IC that only needs a few components to complete the circuit. <S> Here's an example from the datasheet:- <S> However this apparent simplicity is a bit deceptive because for best performance the inductor characteristics have to be carefully matched to the input and output voltages and current. <S> You could try building a switching regulator using the LM2596 with the datasheet to guide you, or just buy a ready-made module that uses a similar chip. <S> These are cheap and readily available from eBay and electronic hobbyist suppliers (eg. <S> IC Station ). <A> simulate this circuit – Schematic created using CircuitLab <S> Here's a rather hacky alternative to Bruce Abbott's answer. <S> A silicon rectifier diode drops about 0.6V <S> when forward biased, If you wire 8 of them in series, that's pretty close to the 5V drop required. <S> Make sure the diodes are traditional silicon rectifier diodes, and rated 3A or more. <S> Edit: This is "efficient" in the sense of cheap and easy. <S> It does not have a good power efficiency, as the diodes will all get warm. <A> The simplest answer is to get a 19 V supply and stop trying to make the round peg fit the square hole. <S> That's rather a lot for a adjustable 24 V supply, but it's worth a try. <S> Otherwise, use a buck converter. <S> There are many chips out there, and it's not too hard to roll your own either. <S> The TV will have its own internal regulators, so a little ripple on the 19 V supply shouldn't matter. <S> A simple pulse on demand system should work plenty well enough. <S> A linear regulator is also not a good idea. <S> One of those would burn up (24 V - 19 <S> V)(2.1 A) <S> = <S> 10.5 W in heat. <S> Getting rid of that heat would be more complicated, costly, and larger than a buck switcher.
| If you really want to use the 24 V supply, then first see if you can adjust it down to 19 V. A linear regulator acts like a variable resistor which is continuously adjusted to keep the output voltage constant.
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GND of 9V batteries in a circuit I have a very simple question. I'm building a circuit that uses some integrated circuits, some of them need 5V and GND, and some like opamps need symmetrical power supply.In summary, I will need 3 outputs for GND, + 5V and -5V to power the ICs. So I'm going to use a kind of linear source with 2 batteries to convert the 9V signal to + 5V and -5V.I will use 2 batteries in series as follows: My question is: Can I use the output I have named "GND" to power the ground pins of my ICs? Another question: On the ground "REF GND", do I need to put some external GND reference? Note that if I remove this "GND REF" from my simulated circuit in Proteus, I can not get + -5V at the outputs.I am using the arduino too, if necessary, I can use the arduino GND reference, and put in "REF GND". But is this really necessary? Last observation. I will not use the 5V pins of the Arduino, because my circuit will need a lot of power, so I have the 2 batteries to power it. Thanks! <Q> You won't get "a lot of power" out of two 9V batteries because they don't have it. <S> Outside of that, the circuit will work. <S> The "REF GND" is particular to the simulator, and you don't need it in real life. <S> You connect GND to the ground of the rest of the circuit, and that's it. <S> Back to the 9V batteries. <S> They have a capacity of a few hundred mAh. <S> If you draw more than maybe 100mA, then they will be drained in a very short time. <S> They are not designed to deliver large currents, and cannot do it. <S> Further, you are using linear regulators. <S> They work by throwing away power as heat. <A> simulate this circuit – Schematic created using CircuitLab <S> Do you have any "ground symbols" in your parts drawer? <S> These three schematics are exactly the same circuit. <S> All the ground symbol means is <S> all the things attached to it are part of the same node in the circuit. <S> It simplifies the drawing, but has no relevance to the circuit. <S> You can use any node you want to power your ICs, <S> as long as whatever is connected to "Vcc" is 5V more (or whatever the datasheet specifies) than "GND". <S> Typically the datasheet also specifies all the other pins must be at voltages between the limits set by the power pins. <S> You could, if you wanted, use "Vout -5V" in your schematic as "GND" for an IC, and "GND" in your schematic as "Vcc" for the IC. <S> Why not? <S> There's a 5V difference between them, and as long as the other inputs to the IC are within the limits in the datasheet, why not? <A> The GND is only a used for the simulation and indication that this is a "zero potential" point (aka Ground https://en.wikipedia.org/wiki/Ground_(electricity) ).When <S> designing the board, this point would probably be called a “floating ground” as you are not able to discharge to the actual ground so some layers of the PCB are used as GND. <S> To lower the voltage, you can either use a "Buck" or a voltage divider (the second is a more efficient way)- Why not use a 5v battery? <S> I am not entirely sure how flexible is designing an Arduino board is.
| Using linear regulators, you are wasting a large part of the little energy that a 9V battery can deliver.
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Why does the voltage double in a doubler-circuit? I’m having a hard time understanding how the final voltage is achieved. I understand that in the first half of the sine wave the first capacitor gets charged to 1V.In the 2nd half the first one is already charged so the 2nd capacitor gets charged to 1V as well.But how does this account to 2V? Looking at it, it seems like these capacitors are in parallel, so I’d expect them to only provide 1V.Where’s my mistake? <Q> The output of the transformer (A) is a waveform that rises up to some peak positive value and down to some peak negative value relative to 0 volts. <S> Let's say it rises to +100 volts and drops to -100 volts (200 volts peak to peak). <S> Typically, the frequency would be 1 kHz to hundreds of kHz and this means that the capacitor closest to the transformer will pass this peak-to-peak voltage virtually without attenuation. <S> However, passing this AC voltage without loss doesn't mean that it will still be + and - 100 <S> volts at point (B) <S> - the capacitor blocks DC so point (B) <S> could be raised or lowered by some DC amount with the same AC content superimposed. <S> However, at point (B) the negative excursions are trapped and restricted (by the diode) to being no more negative than -0.7 volts. <S> This means that the voltage at point (B) now only moves between -0.7 volts and +199.3 volts i.e. there is still the same peak-to-peak voltage but virtually the whole range is now positive. <S> This is the hard bit done because what feeds (C) is a diode peak detector and at (C) is smoothing capacitor thus, you will get a DC peak voltage of about 199.3 volts minus another 0.7 volts making it 198.6 volts. <S> If you take current from (C) there will be a ripple voltage superimposed on the DC output just like any conventional half wave power supply rectifier circuit. <A> Imagine it in half cycles. <S> During the first positive half cycle the capacitor C1 will pass the voltage and C2 get charged (since D2 is reverse biased). <S> During negative half cycle, D1 is open and D2 is conducting charging the capacitor C1. <S> And now when the next positive cycle comes, C1 also acts as source adding up to the value supplied by source. <S> Below is the simulation of your circuit. <S> You can also simulate the same on your end using LTSpcie and probably get a better perspective. <S> Simulation files <A> But how does this account to 2V? <S> Charge transfer. <S> The left most capacitor is charged up via the left most diode during the negative cycle.
| That charge is transferred to the output capacitor during the positive cycle (minus the voltage drop over the diodes).
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Reducing voltage in power supply with diodes Hello there i am building a variable power supply as a summer project. I have ordered some parts, and want to begin building it. I start with the transformer power supply: I've ordered a transformer (230 V - 36 V) and i expected it to give a output voltage of 36 V peak (silly me), and then found out the output will be in RMS, i. e. an peak outout of 50,92 V. I can work with this, but the capacitors i've orderes is only rated at 50 V, so this will give me some smoking problems i guess. The rectifying diodes will give a voltage drop of minimum 0,6 V, which is not quite enough to go under 50 V. Is it possible to just use two diodes extra? As seen in the simulation: The diodes added is of course D5 and D6. Their only purpose is to lower the voltage, to under 50 V. This will theoretical be enough. But will i need to go lower in voltage, as the overhead on the WV of the capacitor is too small? And is the "extra diodes" way even a viable way of doing this? <Q> As a rule of thumb, it is always desirable to select capacitors which are 1.5 times (2times is even better) the desired voltage. <S> So, in short, the answer is NO. <S> Theoretically(ideally) <S> it should, but practically it will be a bad design. <S> An Even slight increase in input voltage or any small variation in the circuit than ideal <S> and you might see the circuit smoked. <A> What is important for the capacitor selection is the rms current it can accept and at what temperature. <S> This is the major cause of failures in power supplies as manufacturers often select cheap capacitors which quickly overheat due to poor characteristics (high equivalent series resistance or ESR). <S> Current ratings are usually given for a 120-Hz ripple so you should find the adequate information in the data-sheet if you operate from a 60-Hz network. <S> So from your simulation results, extract the rms current in worst case (max output current and min input voltage for instance) and base your capacitor selection including some derating factor (safety margin). <S> Then, voltage is of course important but not as sensitive as reported by some of the answers. <S> A capacitor can operate close to its max voltage without problems. <S> For instance, you can find 420-V bulk capacitors in hi-volume PFC-based adapters in which the nominal output voltage is 390 V (7% derating). <S> However, using diodes to reduce the peak by a few volts is not reasonable especially considering a voltage drop depending on the absorbed current plus variations in the input voltage (+/- <S> 15% in Europe). <S> If you are stuck with the caps you have and providing they can accept the current, the best is to stack them in series and add equilibrium resistances so that they equally share the voltage. <S> This is often done in hi-volume power supplies. <A> this will give me some smoking problems i guess. <S> No point in guessing. <S> Capacitor ratings are magic and it wouldn't explode just because the voltage is a tiny bit off when unloaded <S> Yes you can use more diodes to further reduce the voltage on the capacitor but it isn't the best solution <A> At best, you'd be running the electrolytic capacitors right at their maximum voltage. <S> That is not a good strategy. <S> It will decrease their lifetime, and it will still leave you vulnerable to voltage spikes or a little higher than usual line voltage. <S> You should aim for running electrolytic capacitors at around 75% of their rated voltage most of the time. <S> Then it's OK for spikes to occasionally go up to the rated voltage, but never above it. <S> Do it right. <S> Get the right transformer, or get the right capacitors.
| No, adding a few diodes to drop the voltage is not a good idea.
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Recommended isolation paper for hobby circuitry I'm having a hard time finding out what type of insulation is enough for my projects. I will be building this: STM32 OLED Soldering Station Case Temperature Controller T12 KR BL ILS BC2 Handle Electronic Soldering Iron Tips 220v 70W I'm wondering what type of insulation paper is needed for the metallic body, to protect the internal transformer and other bits. This video shows what I'm going to be doing: https://youtu.be/25rww-pXqr0?t=5m39s He states any insulating material is enough (0.3mm?). i'm wondering what is best bang for the buck so to speak. Looking at Aliexpress/Alibaba that is. I'm also planning to use the paper in a few other projects! <Q> For a hobby circuit, if in doubt, wrap it in electrical tape. <S> Paper is not a good insulator since it can be abraded thru easily, and doesn't insulate well when wet or damp. <S> Make sure the electrical tape isn't adding thermal insulation to things that need to dissipate significant heat. <A> As shown, that kit is a death trap waiting to kill an inexperienced buyer. <S> Like you. <S> You do not need paper. <S> You will need insulated hookup wire to make connections. <S> You will also need 4 longer screws in order to mount the power supply. <S> Drill the bottom of the case to fit the mounting holes on the supply. <S> Purchase 4 standoffs about 1/4 inch long which have a center hole (either threaded or not) which fits the screws, along with 4 nuts and lock washers which fit the screws. <S> Push the screws through the bottom of the casing. <S> Put a standoff on each (using threaded standoffs will make this easier, since you can screw them on and not have to worry about the screws falling out). <S> Then put the supply board on the screws, then use the nuts/lock washers to hold the supply in place. <S> Tighten everything down. <S> The standoffs will hold the underside of the supply away from the metal case and no insulation will be needed. <A> Air gaps needs to be 3mm from smooth AC surface to metal earth ground, Paper needs to be stiff so solder tips won't cut thru. <S> I prefer Mylar or Polycarb sheets but you might be able to cut from PVC packaging material 0.3mm min. <S> As long as it wont short out if you step on it or spill coke on it. <S> ( Official UL safety test is to use a hammer and coke) <S> if there is clearance, check if nylon washers will work with TH wires sheared.
| Hot glue can also be useful in the short term.
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Is connecting a 12V circuit to +12V and GND the same as connecting it to GND and -12V? After reading What is negative voltage? and the answer given by kellenjb I wondered if I understood the topic sufficiently, and started theorizing. If the + or - sign is just a convention and the voltage is the potential difference to ground, and my PSU has a +12V, -12V and GND connector, would it make no difference wether I connected a circuit to +12V and GND or to GND and -12V? In other words, say I had a 12V LED, could I connect the anode to GND and the cathode to -12V and still have it work same as when I connect the anode to +12V and the cathode to GND? <Q> the voltage is the potential difference to ground Voltage is the potential difference between two nodes and has nothing specifically to do with ground unless one of those nodes happens to be ground. <S> You can connect a circuit that requires 12 volts between ground and -12 <S> volts for sure but make sure the polarity is right i.e. positive terminal of the target circuit connects to ground and the negative terminal of the target circuit connects to -12 volts. <S> In other words, say I had a 12V LED, could I connect the anode to GND and the cathode to -12V and still have it work same as when I connect the anode to +12V and the cathode to GND? <S> Yes. <A> Short answer: <S> yes. <S> Longer: think, would the device know the difference. <S> If it only has two wires on input, only voltage between them matters. <S> But in practice sometimes it is not that simple. <S> Sometimes this GND is somehow still in system as GND. <S> For example as enclosure potential. <S> Then you may have limitations. <S> Bottom line- convention may be almost as strong as physics, especially if you have no tools or knowledge to check, like it happens in complex systems. <A> Yes, the LED would work. <S> Think of it like this <S> So if you have the anode to +12V and the cathode to GND (which is basically our 0V reference point) <S> then the LED will see a voltage of +12V-0V=+12V <S> But then, if you have the anode to GND and the cathode to -12V <S> the LED will see a voltage of 0V-(-12V)=+12V. <S> As you see, there is no difference in practice. <S> Also you say: the voltage is the potential difference to ground <S> That is true for an absolute value of voltage, like the +12 or -12V of your PSU; they are referenced to its GND terminal. <S> But as I said before, voltage is actually the potential difference between two points. <S> In this sense, the absolute voltage of the anode and cathode of the LED is not important. <A> As long as you have a potential difference there will be a current flow. <S> In your case the potential difference is (0 - (-12)) <S> volt = <S> 12 volt <S> so your 12 volt load will work.
| : Voltage is basically the difference between the potential of two nodes.
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What is a good alternative for the UA741 opamp in hobby projects? Related to What's the uA741's appeal? , a lot of schematics use this rather dated opamp. My follow-up question would be: What is a good alternative? Of course if you have specific requirements you just search for the correct part. What I'm talking about is, what should be the default opamp I stock in my parts bin? Imagine the average hobby project with a breadboard and an Arduino, single-supply, probably 5V, frequencies in the audio spectrum, and accidental shorts. It should just be a nice, single-ended, well behaved (no phase reversal) opamp with a large voltage swing (preferably rail-to-rail), and reasonable performance for audio. Some comments on the linked question mention LM358 and LM324, but I've read the low slew-rate may be bad for audio, which is what most of my analogue signals are. <Q> In complement to Andy's list of criteria, I'll add that most of the time, one of the usual suspects will do just fine. <S> For example: MCP6002 is a good one. <S> Dirt cheap (bag of 10 is 3€). <S> It's your crummy CMOS opamp. <S> It is slow so it won't oscillate on your breadboard, has Rail to Rail <S> In <S> /Out, the input common mode even extends a little bit beyond the rails. <S> Low power too. <S> Performance is nothing to write home about of course, but it is good enough for many uses. <S> Microchip makes opamps? <S> Well, yeah. <S> They have a whole line of CMOS opamps which do absolutely nothing spectacular besides being cheap, RRIO, low-power, low-voltage. <S> These work on 3.3V and 5V so a good match for Arduino, Pi, etc. <S> If you want faster, try MCP6292 (10MHz). <S> Good thing with these cheap crummy CMOS opamps is they work well on battery voltages, and they don't have the LM358 gotchas, like the input common mode going to GND but not VCC, or the output being "kinda able to go to 0V <S> but only if it never has to sink any current" and the like. <S> Yeah, there are other $0.50 RRIO opamps which would fit the bill. <S> Do I want to spend 2 hours selecting a 50c opamp for a hobby project when I know this one will work? <S> Probably not... <S> Also, cost. <S> For amateur, hobby stuff, you gonna spend hours laying out that board, so when performance is needed, it's not worth it to skimp on the parts! <S> Get a $3 opamp if it saves you a headache <S> , it's well invested... <S> For example, if you want to filter and buffer a DAC or a PWM from an arduino and get a 0-5V output, there's no excuse for using an opamp which needs a negative supply voltage when you got 30c RRIO opamps which will do the job with only a 5V supply! <S> Audio, for example: if you need cheap, NE5532. <A> what should be the default opamp I stock in my parts <S> bin? <S> It's a failure on your part to think that there is a default <S> op-amp that suits most applications. <S> It's as simple as that. <S> It's also a failure to limit what you think are the important parameters to those that you have stated. <S> There are many important parameters that may be important to a particular target application <S> but, these parameters may be of no-consequence to a different application. <S> Examples are, but not limited to: - Input <S> offset voltage Input offset voltage drift with temperature Input bias and offset currents <S> Input common mode range <S> Input impedance (resistance and capacitance) <S> PSRR <S> CMRR Phase margin Input equivalent voltage and current noises <S> Ability to operate down to 1.8 volts on power supply Operating current Output impedance versus frequency <S> Open loop gain <S> Your question is asking about the uA741 <S> yet you quote a general application that would run on 5 volts - are you aware that the minimum recommended power supply is sometimes as high +/-10 volts with some hints in some data sheets <S> that <S> +/- <S> 5 volts will be OK. <S> The ST uA741 is quoted to run at 5 volts but this is the exception rather than the rule. <A> Ne5532 is a good all purpose opamp. <S> Tda1308 is another one for good low voltage application. <S> It has poorer DC performance however. <A> Because you mentioned Arduino (low power, low voltage): LM358N - 0.5mA <S> , min 3V 1MHz <S> MCP6001 - 0.1mA, min 1.8V <S> 1MHz <S> LM324 - 0.7mA, min 3V 1MHz <S> (quadruple) <S> NE5532a - ? <S> mA min <S> +-5V <S> (10MHz!) <S> TDA1308 - 3mA, <S> min 3V, (class-AB stereo headphone driver, <S> circ boards already available on ebay for $1) <S> LM193, LM293, LM393, LM2903, <S> min 2V <S> Note: <S> * min is minimum voltage required to run <S> * mA is typical current consumption <S> * I am going to prepare a RAR file with the data sheets for all the above ICs and upload it somewhere (for lazy people like me).
| If you need high quality, OPA1642, OPA1652, LME49720 and others in the same families are very good.
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Strategy to balance a Wheatstone bridge for strain measurement The purpose of my project is to build a sufficiently accurate "Strain indicator" based on the use of a Wheatstone bridge using an Arduino. Currently I am using a HX711 as load cell amplifier which is working well since I use it with a full bridge. Currently I want to connect only one strain gauge to a wheatstone bridge built using 3 resistances of 120Ohm (0.1%). I am able to measure something but I want to balance the Wheatstone bridge. My inexperienced question is it is possible to add an accurate balance system to my Wheatstone bridge? I know that classical solution use potentiometer but the principle is based on a choice of a variable resistance with similar value of the others 3 resistances on the Wheatstone bridge (in my case 120Ohm). Unfortunately, finding potentiometers around this value is not possible. In addition the potentiometer must allow very small change of the resistance around 120Ohm. I am from the mechanical field, sorry if my question seems to be very common... <Q> The way I do it is to use an instrumentation amplifier (TI, Analog devices etc.) <S> that amplifies the difference signal and then apply a variable voltage to the reference input of the InAmp to balance the output to zero. <S> An InAmp basically comprises three op-amps as shown below: <S> - Note the "reference input" bottom right and the potential divider that sets the offset voltage on the InAmp output. <S> This can be a potentiometer or even the output from a DAC. <S> It all depends on what you want to do and how automated you want the zeroing process to be. <S> Some applications will allow a slow integrator to be fed from the InAmp output to the reference input and in this way you are always auto zeroing but at such a slow rate that the measurements you make are largely unaffected. <S> This may or may not be of interest to you of course. <S> There may be a way of adjusting the HX711 (load cell amplifier) to do what I suggest above <S> so if you want help in that direction, please link to a proper data sheet for the chip. <A> There are two ways of using a Wheatstone Bridge. <S> The first classical way is to balance it, that is, adjust the arms until you get a zero output. <S> Then you can assert that the ratios of arm impedances are equal, and calculate your unknowns from your knowns. <S> The second way, which is (I think) a slight abuse of the name Wheatstone Bridge, is to use the bridge configuration, often with strain gauges, Hall sensors, resistance thermometers, where there is a small variation to be measured in the face of a large resistance. <S> The bridge is approximately balanced, to mostly remove the effect of the big static resistance, and then the residual output is measured with a sensitive voltmeter, like an HX711. <S> There is no need to balance the bridge precisely, and in fact, as the sensor changes resistance, this will unbalance it. <S> The purpose of using the bridge is to reduce the dynamic range between the sensor static resistance, and the small measurement range. <S> Once it has been reduced enough to match the voltmeter, and the system noise, there is little point in improving balance further. <A> There is no need to use equal resistances in each half of the bridge in most cases. <S> You can use something like this: simulate this circuit – Schematic created using CircuitLab <S> This gives you the equivalent of about +/-5 ohms adjustment in the reference resistor. <S> Pick R1/R2 to control the current through R3 depending on your excitation voltage. <A> Unfortunately, finding potentiometers around this value is not possible.
| For what you do the bridge needs only to be roughly balanced , not perfectly balanced.
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Measuring power consumption in Arduino and raspberry-pi system I have a system that consists of 1 Arduino, 1 Raspberry-pi, 2 power banks, and a set of different sensors. Each of the Arduino and Raspberry-pi is powered by a separate power bank. I need an accurate method to measure the total power consumed by the entire system. I also need this power measurement as feedback input to the Arduino to control some sensors. Is it better to tap some power measurement circuit on each power bank or is there a better way to measure it by Arduino/Raspberry-pi. <Q> Linear make a range of high-side current monitoring devices. <S> Figure 1. <S> Linear <S> LTC4151 . <S> 0.02 Ω is probably for a high current. <S> This particular one seems to use serial transmission. <S> Analog versions are available too. <S> That will give you the current. <S> For power calculation you will need to measure the voltage too - probably from before the voltage regulator so that you measure the losses there too. <S> You then calculate power from \$ P = <S> VI \$. <S> You will have several problems to address including how you average out the current if it varies rapidly. <S> ... <S> feedback input to the Arduino to control some sensors. <S> Generally you will be reading sensor rather than controlling them. <A> However, if there are significant load-associated ripples on your voltage then the accuracy of this method is reduced. <S> So, if you have a reasonably ripple-free terminal voltage that only exhibits slow droop as the battery becomes discharged, then a simple two-quadrant multiplication of voltage and average current should be acceptable. <S> I would do this using the circuit shown in Transistors answer or with external ADCs with appropriate filtering. <S> Multiply the two digital numbers representing voltage and current and average a few values to get average load power. <S> Even if the load is turning on and off at a high rate you can still average current to get true power providing the battery ripple voltage is low. <A> Since your system is cobbled together from ready made subsystem components you may want to consider adding one of these low cost USB power measuring devices inline between your power bank and the rPI and/or the Arduino board. <S> There are several models available from places like eBay and Amazon. <S> These devices do take some power so you would not want to leave it in place permanently. <S> Use it to profile your system power through the various usage scenarios and then take it back out. <A> Instead of measuring the power consumption interactively, power the whole system from a fully charged battery for couple of hours/days and then measure how much mAh were needed for the batter to be fully charged again. <S> This can give you a good picture of much power was needed over time and also how long can your system run powered on this battery. <S> When changing components you can then compare two measurements if this change was towards power-saving or towards power-consumption.
| The average current taken by your "load" can be used to estimate power by numerically multiplying it, at regular intervals, with the battery voltage.
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diode selection for large current reverse bias, small current forward bias I'm designing a circuit where I want to pass a small test current from one leg, but block a large "firing" current (rocketry) coming from another leg in reverse bias. I have a 5vdc indicator line (with about 1-2 vdc and <.5 amp once it would get to the diode) I want to pass through a forward biased diode to indicate the fuse is still viable without burning the fuse, but need that same diode needs to block 12 vdc at as much as 30 amps in reverse bias for a short duration (maybe as long as 5 second until the igniter lets go and opens the circuit and the "firing" voltage is turned off manually). How do I select the proper diode? What am I looking for on the diode spec sheet to be sensitive enough to allow the low current forward bias and strong enough to block the large current reverse bias. I know that is what diodes do, but how to select the right one escapes me. <Q> What am I looking for on the diode spec sheet to be sensitive enough to allow the low current forward bias and strong enough to block the large current reverse bias. <S> I know that is what diodes do, but how to select the right one escapes me. <S> Just use a high speed diode like a 1N4148, BAS16 or 1N914. <S> It will certainly block 12 volts and won't conduct any appreciable reverse current when the heavy duty 12 volt supply is activated. <S> I said use a fast diode because reverse recovery time for those stated above and in the single figure nano second range. <A> Andy's answer is simple and correct, I only wish to add to it (since you asked about how to select a diode and what specs to look at). <S> Standard diodes conduct when forward-biased and block (mostly) when reverse-biased - you already know this. <S> To use the water-in-pipe analogy, without sufficient pressure a one-way valve will stop backflow indefinitely, no matter how much water is in the pipe. <S> When choosing a diode for general use, you'll want to look at the following: <S> Diode type - standard "silicon" diodes have high threshold voltages (~0.7V) and high reverse breakdown voltages ( <S> >50V). <S> Schottky diodes have much lower threshold voltages (~0.2-0.4V) and faster switching times, but also generally lower reverse voltages (~30-60V). <S> Zener diodes conduct in reverse when reverse-biased above their "Zener voltage", essentially operating in both a normal forward mode and a precise, controlled breakdown mode. <S> Forward voltage - determines the needed voltage for the diode to be fully conducting (real diodes conduct some current when they're forward biased at all). <S> Reverse voltage - determines what amount of reverse-bias the diode can repeatably withstand before it enters breakdown and begins to conduct in reverse. <S> Real diodes allow a very small amount of leakage current (on the order of up to a few micro-Amperes) when reverse-biased, but generally not enough to cause problems. <S> Continuous forward current - if you need a specific current throughput when your diode is "on", you'll need to make sure it's rated to allow that much current. <S> Reverse current - the amount of "leakage" current when the diode is reverse-biased. <S> In your case, you aren't overly concerned about a few uA slipping by, but some applications may have a maximum allowable reverse current. <S> "Low-leakage" diodes may have a reverse current on the order of pA. Recovery time - the amount of time a diode needs to switch between "conducting" and "blocking". <S> High-speed diodes can make this transition in nanoseconds, Schottky diodes faster still. <S> Temperature ratings <S> There are other specs for diodes that specific applications may need to be in certain ranges, but for fairly basic applications these specs should suffice. <S> In your specific application, a Schottky diode or high-speed silicon diode would work perfectly to block the 12V reverse. <A> There are two parameters for that diode you need to consider. <S> Forward <S> current <S> You'd have trouble finding one that would not do that. <S> Reverse voltage <S> At first glance, it looks that 12v would be sufficient, or 7v looking at the difference between the 12v rail and the 5v rail. <S> However, the fuse current could be up to 30A, and we don't know what the inductance of the wires from the 12v source to the fusible link is. <S> This large current could store a lot of energy in the inductance, which when the current is broken rapidly, could generate hundreds of volts. <S> The 75v PIV of a 1N4148 might not be able to cut it, or it may. <S> You could hope and use a 4148, or aim higher and use a 1000v diode. <S> You might expect that the fusible link would arc well below that voltage and keep the voltage below 1000v. <S> A transzorb placed across the fusible link would certainly limit the high voltage.
| Bias is expressed entirely as a voltage, so a diode can block quite a bit of current if the voltage is low enough. You need to be able to light the LED through it, say at least 20mA. Cheap 'cooking grade' diodes like 1N4148 will easily do that.
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What seems to be the issue with this simple diode bridge rectifier with smooth capacitor? I don't know if it is some limitation on the simulator, or my mistake, but I am trying to build this circuit: But this is what I get: For some reason, the voltage at the marker 5V_DC only gets one positive half-cycle, at a peak of 34.80mA . I was expecting 50mA 5v / 100ohm or 36mA [5v - 0.7v - 0.7v]/100ohm , not 34.80mA. And the current at the LOAD is not smooth, despite the capacitor being there. <Q> You have grounded the wrong point in your circuit. <S> If you monitor the bottom of your load resistor in the simulator you will get another surprise. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> (a) <S> In the positive half cycle D4 and D5 are in play. <S> (b) <S> In the negative half D6 and D7 are in play. <S> Ground at the bottom of C1 instead. <S> Normally for a ground referenced DC supply like this <S> we let the transformer "float" and ground the DC-. <S> And the current at the LOAD is not smooth, despite the capacitor being there. <S> You can get a good approximation of the discharge time by the RC time-constant: <S> \$ \tau = RC = <S> 100 <S> \times <S> 100n = 10 \ <S> ; \mu s \$. <S> The capacitor will be discharged by 63% at this time. <S> For 1 kHz (as in your schematic) you would need at least 100 times your C value. <S> For 50/60 Hz you need 1,000 to 10,000 times that depending on your load. <A> There are two significant problems with your circuit: <S> 100 nF <S> with 100 Ω load has a time constant of only 10 µs. <S> You connected one side of the AC input to ground, and you then used that ground to measure the positive output against. <S> Your AC frequency is 1 kHz, so there are 500 µs between peaks. <S> The labels on your graph are too small to read, so let's say the AC voltage is such that you get 15 V out of the rectifiers without load. <S> If you want the drop to be no more than 1 V with 100 Ω load, then the capacitor needs to be (150 mA)(500 µs)/(1 V) = <S> 75 <S> µF. <S> So 100 µF would be a good value for these made-up numbers. <S> Note that this is 1000 times more than what you used. <S> Don't connect either side of the AC source to ground. <S> Instead, connect the negative output of the rectifiers to ground. <S> Then you can measure the positive output with respect to ground and get the output of the rectifiers. <A> Move your Gnd to V- as your 0V reference and ensure source is isolated(XFMR) <S> otherwise it is offset from Vin 0V by Vf. <S> V and I then will be relative to the correct node then for load to Gnd. <S> My rules of thumb <S> $$\text{For <S> <10%pp load ripple use } T= <S> RC=5 to 10 <S> \times \text{input cycle time}$$ <S> I prefer 8x for higher voltages where Vf is negligible. <S> Thus <S> $$C>=8/(1\text{kHz}*100R)=80 \mu F <S> $$ for 5 Vp sine input 1kHz , R=100, <S> C=80µF <S> Vdc = <S> 5-2 <S> *0.8V <S> you can expect, around 0.7 to 1V drop per Si diode depending on power rating, where ESR above 0.6V <S> (Si) <S> $$\text{ESR(diode)[Ω] = <S> k <S> /Pd[W] for constant k=0.5~1 typ. <S> tolerance on diodes.}\tag4$$ <S> You should intuitively expect 10% Vpp out ripple , causes at least 1/10% (=10x) <S> input current ripple compared to Iout. <S> RC values affect both load ripple V and input ripple I, also both affect diode peak voltage drop. <S> $$ V_{ dc(pk) } <S> ~~~= <S> ~~V <S> _{ dc(avg )} <S> ~~~+~~ \frac{1}{2 <S> } V_{ o(pp)~}~~ \text{ ripple} ~~\tag5 <S> $$ or... <S> $$ (V_{ dc(pk) }~-~V <S> _{ dc(avg)}~~~~)~~*2~~=~V_{ o(pp) } <S> ~~ <S> \text{ ripple} ~~\tag6$$ <S> \$V_f=0.6V+ \dfrac{I_{ i(pk) }}{ k P_d}\tag6\$ <S> In order to charge quickly and discharge slowly with low ripple voltage ripple current is very high (inverse related). <S> If Vf reaches 1V at peak current, consider higher power rating diodes with lower Vf. <S> Now you can try using 80uF. <S> I get 5Vin(pk) <S> - 2Vf + Vo(pk) <S> and I assumed Vf=0.8V <S> thus Vdc=3.4Vdc with 10% of Vdc = <S> 340 mVpp sawtooth ripple <S> This guideline gets worse at lower voltages and low frequency as it implies huge caps. <A> Your expectation of exactly 0.7V drop per diode isn't realistic. <S> 0.7V is just an often used approximation for standard silicon diodes, so having a little more or a little less isn't unexpected. <S> Your capacitor is only 100nF and your resistor is 100 Ohms. <S> This gives a time constant of 10 microseconds, where as your 1khz waveform has a much longer half cycle period of 500 microseconds. <S> The capacitor needs to be much larger for the smoothing you seek for the 1kHz frequency sine wave input. <S> Try making it 100uF instead of 100nF.
| The capacitor is way too small.
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How would I make a circuit that takes a voltage (V) as input and outputs min(V, Vref)? Simple question, how do I turn the following block diagram to a real world working circuit? I know I could use the following: 1) A comparator, but that will give me only either +Vcc or -Vcc. 2) I could use a rail-to-rail comparator and set V- at 0 Volts and the maximum allowed voltage at V+. 3) I could use a Zener diode... But I've zero experience with using them... simulate this circuit – Schematic created using CircuitLab <Q> It depends on how accurate you need this to be. <S> Here is a simple concept: D1 and D2 provide the MIN function. <S> The voltage at the top of the diodes is the minimum of Vin1 and Vin2, plus a diode drop. <S> It should be obvious how to expand this to any number of inputs to take the minimum of. <S> D3 tries to compensate for the diode drop, so that Vout is the minimum of all the inputs with the diode drops cancelled out. <S> If within a few 10s of mV is OK, then this might do. <S> Vhigh and Vlow are voltages you have to supply. <S> Vhigh must be a bit higher than any input voltage of interest, and Vlow a bit lower. <S> The impedance of the input voltages need to low enough to overcome R1. <S> The output impedance at Vout is higher due to R2 needing to be high to not interfere with the signal. <S> That's the basic concept. <S> The next step is to realize that BJTs can be thought of as diodes with gain. <S> Here is the same concept carried out with BJTs using their gain to advantage: Look carefully <S> and you'll see its really the same thing, using the B-E junctions of the transistors in place of the diodes in the previous circuit. <S> The advantage is that the input signals don't have to supply anywhere near as much current. <S> The B-E diodes still do the MIN function, but most of the current comes from the negative supply. <S> The gain is used the other way around so that the B-E junction of Q3 loads the signal much less. <S> Due to the gain of the transistors, this circuit has much lower output impedance while having higher input impedance. <S> One drawback of using the B-E diode of BJTs this way is that the max input range is narrower. <S> This is because the reverse voltage characteristics of the B-E junction is usually fairly low. <S> If you're only doing this over a 5 V range, then there should be no problem. <S> Of course, always check the datasheet of whatever transistor you plan to use. <S> Choose ones with high B-E reverse voltage capability if you want a wider input voltage range. <A> The AD8036 Voltage Feedback Clamp Amp from Analog Devices may suit. <S> This device, when used in non-inverting mode can clamp the input voltage between an upper and lower limit. <S> Figure 2. <S> Clamp-amp block diagram. <S> (See page 16 of the datasheet.) <S> \$ <S> +V_{IN} \$ is applied to A1's non-inverting input until \$ +V_{IN} <S> > <S> V_H \$ at which point S1 applies \$ V_H \$ to A1. <S> Similarly when \$ +V_{IN} <S> < V_L \$ S1 applies \$ <S> V_L \$ to A1. <S> There are a few limitations to the device - particularly the supply voltages (+/- <S> 3 to <S> +/- <S> 5 V) and 6.3 V max \$ <S> \Delta \$V between \$V_H\$ and \$V_L\$ - <S> but you can work around those by, for example, working in the 0 to 1 V range on your inputs and putting an amplifier afterwards to restore the signal level. <S> See my answer to OpAmp Buffer configuration with max <S> allowed output for further details. <A> In addition to the many good answers provided already. <S> The concept is applied heavily in power electronics, allowing multiple feedback loops to overdrive the output (typically with priority to lowering output power). <S> A sample schematic is the following, <S> Here there are two op-amp buffers with there outputs diode or'd. <S> Neither opamp can push the output high, only pull the output low. <S> So the minimum voltage overdrives the output and the result is the Min(A,B). <S> A sample simulation,
| You can diode-or the feedback path of a set of buffer amplifiers.
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Maintain state of GPIO pin across reboot I have an ESP-12 soldered onto this adapter board . I am using NodeMCU Lua based firmware to code the logic. I have scheduled node.restart to execute periodically. 1 of the GPIO pins is used to drive an external component using HIGH or LOW state. I can restore the GPIO state (HIGH/LOW) after node.restart but during restart GPIO state is undefined. I want to maintain that state especially if the state was HIGH. 1 approach I have in my mind is to have another micro-controller such as ATTinyX which is connected to ESP-12 via UART. ESP-12 can give a string/char to ATTinyX to maintain the state (ESp-12 GPIO pin and ATTinyX pin is connected to the external component via an OR gate) and then go for a reboot. Once ESP-12 reboot is completed it can tell ATTinyX not to maintain the state because ESP-12 can maintain it now after the reboot. This adds a bit of cost and components, is there a different way to achieve the same? I'm more on the software side, have less electronics knowledge. Any help is appreciated. Maybe it is similar to 1 bit memory which is set/reset by ESP-12, can I use a flip flop in this case? <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> State holding capacitor. <S> Wire up a spare GPIO as shown. <S> On reset read the state of the capacitor and set the output appropriately. <S> This will give you a short-term 1-bit memory. <S> //Pseudo <S> code to go early in boot <S> sequence.pinPullup(pin) = false; //Turn off the pull-up.pinMode(pin) = input; //Set the pin to input mode, if required.pdState = pinRead(pin); //Read <S> the input to get the power-down state.pinMode(pin) = <S> output; //Configure as <S> output.pinWrite(pin) = pdState; //Restore <S> the power-down state. <S> You may wish to swap the order of the last two lines (depending on micro) to avoid a momentary blip. <S> Note that in this configuration the pin can't be used for anything else. <A> Try using a capacitor that would provide the required voltage for the reboot duration. <S> The following link may help you to find the appropriate capacitor value. <S> http://www.electronics-tutorials.ws/rc/rc_2.html <S> edit: simulate this circuit – Schematic created using CircuitLab <S> To be more specific, I am assuming that the time required for reboot is less than two seconds <S> so now the RC network's time constant must be nearly twice the reboot time. <S> Choosing a time constant of 4.7 seconds. <S> Time constant T = <S> R <S> * C <S> ie. <S> , 4.7 s = <S> 4.7 <S> K OHMS <S> * 1000uf <S> The reason why we should have the time constant to be twice the required time is, when a capacitor starts discharging, at half of the time constant the capacitors voltage <S> would be about 60% of the voltage while fully charged, which is 3 volt at 2.35 second, since the reboot time is less than that, a high level will be maintained. <A> Have a look at an I2C expanders like MCP23008. <S> Basically the same setup as your ATtiny minus another firmware. <S> It will just keep any state you wish regardless of ESP reboots. <A> If the state of the pin prior to power off is known, you can use pull up or pull down to maintain that state while the my is booting up. <S> The issue is that the pins typically default to in out or non GPIO functions so you look their states in reset. <A> Try using a simple latch, but you might need three GPIO pins. <S> While running, if the output at the GPIO that you are using is HIGH, provide a high-low pulse at SET, and if LOW do the same at the RESET.During power on check for the state of Q, if HIGH maintain the required GPIO pin as HIGH else make it LOW
| Large capacitors can also help.
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How can I add a logo to PCB with Altium Designer? I am trying to add a logo to my PCB using Altium Designer 16.1 via "PCB Logo Creator" software. However it was not so neat. Do you have an idea to prepare better logo which would be taken from a photo or picture? <Q> A bitmap image can be copied and pasted into Altium designer if it is a true 1-bit monochrome image. <S> I normally use MS Paint. <S> It is not too bad if you start with a massive image and then shrink it in Altium - you can re-size it in Altium after it has been pasted in. <A> If it's a black/white logo which can be rasterized, you can also go the "Font" way. <S> This works for both Schematic and PCB with the added benefit that it doesn't blow up your PCB in size (which the bitmap option does). <S> Disadvantage: Wherever you want to have your logo displayed, you need that font installed. <A> If you have your logo in vector form, you should export it as dxf which Altium can import. <S> Might not work well with pictures, but since most Logos also come in "simplified" vector forms, it might still be an option. <S> The procedure of creating a "solid" logo from the DXF lines is a bit lengthy, but very reliable and gives high quality results. <S> Here it is: Import your dxf file (File>Import>DXF). <S> Choose the correct units. <S> Use the Inspector panel to set the line width of the DXF geometry to a very small value, <S> e.g. 1 mil Draw a polygon pour around the shape (Place>Polygon pour), select pour over same net objects and disable removal of Islands. <S> Set small values for neck width and arc approximation. <S> Adjust the clearance for the polygon to the DXF lines to 0 by adding a specific design rule Repour the polygon, then select it, <S> right click on it and use Polygon Actions <S> > <S> Explode selected Polygons to Free Primitives <S> Now your polygon consists of several copper regions. <S> You can delete all regions that should not be filled, as well as your original DXF lines. <S> Group your regions by selecting them and creating a union from them (Tools>Convert>Create Union from selected objects). <S> This approach works best if you put your logo into a PCB library, so you can simply place it on any PCB and edit it at a central location. <S> There is also an official Altium Video explaining this process. <A> Per the official documentation , there are a few methods: TrueType Font Embedding images as glyphs in a TrueType font allows you to take advantage of vector graphics and their inherent scalability. <S> I won't cover how to do this here, but there are many tools such as FontLab or a free online tool such as Fontstruct . <S> This isn't an endorsement nor guarantee of these tools working.) <S> Pasting from Clipboard Altium supports metafile formats such as 1-bit bitmaps, lines, arcs, and TrueType characters. <S> (Copying a 1-bit image from your favorite editor won't necessarily work unless it supports Windows metafile data; as I discovered attempting to use Irfanview.) <S> The documentation recommends pasting first in Microsoft Word, for example. <S> The current layer will be the target for pasted data. <S> Placing as OLE Object <S> This works through the Place <S> > <S> Object from File command in Altium Designer. <S> Only the BMP format is mentioned. <A> The basic issue is that your PCB image is going to be 2 tone (black/white, yellow/red, etc) and most source images are not. <S> PICK <S> A GOOD SOURCE IMAGE <S> First you need to pick a good image as your source. <S> The best images are going to not have fine details and will be made of large, solid colored, blocks. <S> PREPARE THE IMAGE <S> You'll need to import the image into a tool like the GIMP (GNU Image Manipulation Program). <S> Once in GIMP you can use their tool-set to recolor and threshold the image so that it becomes a 2-tone version of the original. <S> Save it as a BMP. <S> IMPORT INTO ALTIUM <S> Download the Altium Example Scripts and extract them to the Examples folder under your Altium installation. <S> Run PCBLogoCreator from within Altium <S> On the PCB Logo Creator dialog box click Load and select the BMP file. <S> Click <S> Board Layer and choose Top Overlay . <S> Click <S> Scaling Factor and experiment with sizes (1000 mils = 1 inch). <S> Click <S> Convert and wait. <S> Once the conversion is complete hit the Exit button. <S> Now you have a logo on the Top Overlay <S> and you can use it with Copy/Paste onto <S> a .PcbDoc file. <S> MAKING <S> THE LOGO REUSABLE Highlight <S> the whole logo and hit copy Create (or open an existing) .PcbLib file and create a New Blank Component Paste the logo into the blank component and save the footprint Create (or open an existing) .SchLib file and create a New Puppet Give the new Puppet a shape you can drop onto your schematics, a designator and name you like, and the select the Footprint you created above in the .PcbLib . <S> At this point you can drop your new Puppet onto any schematics and your "footprint" (which is actually your 2-tone image/logo) will be imported into the associated .PcbDoc file. <S> You can read a writeup with pictures and more details here: How to put a Logo on a PCB in Altium .
| Use a Font creator tool (e.g. FontCreator, costs a little) to create a Font with a single letter (you can easily import vector graphics); then, assign that logo to a letter (e.g. L); then install that font in Windows and use it in Altium; To import an image (especially a complicated one) onto a PCB (in Altium) you're going to have to do some work.
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How determine the number of LEDs that I need? So I am trying to do some calculations for a UV curing system that the place I am interning for is working on. One of the tasks that I am assigned to is to determine how many LED does the system need given the energy needed to cure 1 cm square of UV ink. So I have done some research on my own and find some formulae online but I am still not quite sure which unit I should depend on to determine the amount of LED that the system need. So far I have found a formula that gives me W/m2 and one that uses Lm (Lumen).Thanks in advance guys. <Q> Luminous flux (lm) is not what we want- <S> it incorporates the sensitivity of the human eye. <S> For UV curing we need irradiance in W/m^2 and an exposure time to get a total integrated energy that is sufficient. <S> We also need to consider the peak wavelength of the UV LED and the sensitivity of the material you are polymerizing to that particular part of the spectrum. <S> Different materials are more or less suitable for a given wavelength, and LED light sources that provide shorter wavelengths tend to be more expensive and perhaps shorter life. <S> You can always increase the exposure time to increase the total dose, but be careful about trusting the claimed output power of 'no name' light sources, and be careful about running them at the full rated current. <S> It would be best to derate the latter by at least 30% and to keep them cool via a heatsink and fan or closed-loop liquid cooling). <S> It may also be somewhat difficult to estimate the actual exposure with an adhesive or thick material (or if the material is partly blocked by something like glass which absorbs some of the UV) vs. a simple overcoat or UV ink cure, so you may have to experiment. <A> You want to cure 1 square centimeter of ink. <S> Look in the ink datasheet to find what UV light intensity is required for curing. <S> That should be in units of power per area, like W/m 2 . <S> You know the area, so you know can calculate the power required for the area you have (1 cm 2 ). <S> If you do find some, then you look at their datasheet to see how much power they can put out. <S> Use the output power and the radiation pattern to determine the power one LED can dump onto your area from the distance you can arrange. <S> From that you know how many LEDs it takes to dump the necessary power. <S> Always derate a bit. <S> For example, if you find that one LED provides exactly 1/5 the power you need, don't use just 5 LEDs. <S> Add a few extra so that the system still works when the inevitable stuff happens. <A> I would use the formula for lm, because LEDs have different luminous efficacies, i.e. the lumen output per watt depends on the LED you use. <S> To do this, you need to know the characteristics of the LED you are using. <S> In the picture below, obtained here , you can see the lumens as a function of current. <S> As an example, assume that you select a drive current of 1A for the LED. <S> This will generate 400 lm per LED. <S> If you need 20000 lm in total, you will need 50 LEDs.
| Next you look at what wavelength the ink needs and see if you can even find LEDs that emit within the suitable wavelength range.
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How to command a 220v device through a 12v output I have a vending control board whose output is 12v-DC but the problem is that with this device I have to control a device whose input is 220V-AC. Can anybody help me make this possible? Sorry for my dumb question but i am relatively new to electronics. <Q> Relays are commonly used for this type of application. <S> These are available for panel mount or as PCB modules for incorporation into a project case or machine cabinet. <S> Figure 1. <S> Industrial relay and panel or DIN mount base. <S> Figure 2. <S> PCB relay on board. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 3. <S> The schematic symbol is(?) <S> self explanatory. <S> The coil pulls the contact closed when energised. <S> Pick the relay mechanical packaging to suit your application. <S> Choose one with a 12 V DC coil. <S> Choose one with contact current ratings greater or equal to your load current. <A> Use something called a relay . <S> These are basically electrically-controlled mechanical switches. <S> The input side is a electro-magnet. <S> When the magnet is activated, it causes something to move, which changes the state of a mechanical switch. <S> 12 V is a very common relay coil voltage. <S> You can easily find relays that can switch several Amps at 220 VAC, and that are controlled with 12 VDC. <A> I have used their 110VAC module, which comes with the plugs already attached, in my own home. <S> Their 220VAC versions look like you must add the plugs yourself, but that actually offers much needed flexibility for your application, seeing as how 220VAC supplies differ in their connections. <S> Also Google for guides to set these up - many people have used them in home automation projects.
| If you are new to electronics and looking for a solution with parts already picked for you, check out the PowerSwitchTail for 200-240VAC: http://www.powerswitchtail.com/240vac-kits
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Simple wireless transmitter for notification when circuit turned on I have a simple home project I'd like to implement. Being a electronics newbie I'm not sure exactly how to go about it. Basically, I'd like to build a device that is attached to a (12V) power supply line that simply sends a wireless signal/alert to my smart phone when the power is turned on i.e. current running through the wire. The signal only needs to be transmitted roughly 1-3m, so I assume some sort of Bluetooth transmitter spliced into, and piggybacking off the power running through the 12V wire will suffice? The device does not need to receive signals i.e. one-way communication only. <Q> I would suggest using a Raspberry Pi with a wifi board. <S> Feed the voltage (after regulating to 3.3 V) to a GPIO pin. <S> Monitor the pin, and send yourself an email when it triggers. <S> Alternatively, power it directly from the 12 V so that it is off until it's needed. <S> Have the message sent automatically after boot. <S> Then you can program it to shutdown. <A> - You'll need a power supply and you'll need to be able to put some firmware into it. <S> Anyway, the application will likely be outside, with no wifi, and at times no data/internet of any kind. <S> Then there is the ESP32S combined blue-tooth and WiFi module: <S> - There are quite a few options for blue-tooth/WiFi <A> I suggest you to try NRF51 based boards. <S> which is cheaper and uses BLE to communicate.
| If you have WiFi then use an ESP8266 :
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How to convert a device pluggable to a car lighter, changing to a regular home socket I bought an air compressor that is powered by the car lighter in my car. It does the job when I want to inflate my car tires, but if I want to use for something else like there are 3 bikes in our house and they all have to be regularly check, it becomes a bit of a hassle. I'd like to change the adapter of the air compressor from a car lighter to a regular adapter, but I'm not sure what should be the voltage and amperage? Or will this depend on the air compressor specs? <Q> Get yourself a cheap power supply with a built in 13.8 VDC port. <S> I have an old one from Radio Shack that I bought 20 years ago, and still use for exactly this purpose. <S> Your cigarette lighter plug will socket directly into the supply - into the round hole just above the terminals. <S> Another option is to get a supply like this: iSaddle, 12 VDC, 10 A . <S> It's more restrictive, but great price. <A> The easy thing to do would be to purchase one of these. <S> https://www.amazon.com/dp/B01579KBFW <S> Then you don't have to build anything that could potentially be dangerous <S> and you don't have to modify the compressor. <A> In general, car "cigarette lighter" outlets are fused for 10 amps <S> so I'd doubt any device designed to plug into one would be designed to draw more. <S> The car voltage is nominally 12 volts (though would generally be higher while the car is running) <S> so you would be safe with a power supply that supplies 12 volts and at least 10 amps. <S> You can easily get a "cigarette lighter" outlet, which as this at Walmart . <S> The center pin (red wire) is connected to the positive terminal. <A> Your device should have Input label on it. <S> Use that to determine what kind of power supply you will require.
| Buy an off the shelf bench power with the correct Voltage and >Current Rating.
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Is it a good PCB design practise to choose components with the same footprint? Most of the components I have selected for my project have 2012(mm) footprint while some others have 1608, 3225 and 6432. Does it matter that some components have a different footprint? What is a good design practise? <Q> Generally you will have some choice for many parts- for example, resistors that are not critical in value or dissipating a lot of power can be (inch) 0402, 0603, 0805 or 1206. <S> Ceramic capacitors that are not near the limits of what can be made in a given size may be available in several case sizes. <S> There are a few other factors- the size of the part affects the overall size of the PCB, which may point towards smaller sizes. <S> You may have reels and reels of parts on the shelf and wish to use them. <S> The price vs. size curve is typically bathtub shaped with the smallest available parts expensive and few suppliers, and the larger parts (which are used less and less in high volume) more expensive and harder to source. <S> If you are hand soldering the board, (inch) <S> 0402 <S> = <S> metric 01005 and smaller are harder to handle than larger parts. <S> They also come more to a reel. <S> For high power dissipation resistors, or very high accuracy resistors (especially of higher values) <S> larger sizes are advantageous or necessary. <S> For capacitors- you may be able to buy the value and voltage you need in an 0603 from one supplier at a high price and with nasty voltage coefficient, but there are a plethora of suppliers making 0805 parts which are better and cheaper. <S> Assembly wise, it doesn't really matter too <S> much- <S> provided <S> you don't go nuts on making the parts ultra-tiny. <S> Most PCB assembly suppliers can handle the various sizes (though at the very small end some may have older equipment that can't handle really tiny parts well). <S> As a practical matter, larger resistors are often marked, which can be an advantage. <S> Parts smaller than 0603 <S> (inch) are seldom, if ever, marked. <S> You may notice that 'glue' logic chips come in quite a wide range of packages, and logic families, just as the market is narrowing due to increased use of CPLDs, FPGAs, and incorporation of similar (very small scale, usually) functionality on processors. <S> So I would suggest being careful about picking packages without multiple sources if the product has a long life cycle. <S> It is not always the largest packages that become discontinued, sometimes the smaller ones don't catch on. <A> I see a small benefit of sticking to the same default size if you are assemblying yourself (in-house), in that you don't need to constantly check if you have resistor X in form factor Y in your inventory. <S> This way, you can have sufficient quantities of the usual E values in one specific form factor, and simply stick to it. <A> As Luke Gary said -- it is also a DFM (Design for manufacturing) concern. <S> First is visual uniformity -- which helps with visual inspection of assembled boards. <S> Second one is assembling capabilities of your PCS assembler. <S> Many PCB assembly houses got few assembly lines -- for example two lines with older Pick-and-Place machines and one with newer. <S> The older ones might not be able to place the smallest component. <S> So if you design your board with mostly 0805 or 1206 components and sprinkle, here and there, a few 0402's (without the need for them) your board will have to be assembled on the newer machine just because of those few parts. <S> And the assembler might not be happy about this (or simply it might be more expensive). <S> [But this probably only applies to smaller/medium assemblers.] <S> Third one is BOM (Bill of materials) reduction. <S> If you have, for example, 1uF capacitor in 0603, 0805 and 1206 packages on one board, it would be good idea to check if you really need them in those packages. <S> Maybe you have the space to use 0805 version in all places? <S> Etc. <S> But at this point it becomes a trade-off between size/parameters/assembly cost and price of the component itself.
| In my opinion, it's a good idea to keep the footprint sizes in the same ballpark for a few reasons.
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Extract 12V from USB port I wish to drive a 12V stepper motor with a USB port as the source of power. However, the USB port only outputs only 5V as output. This will obviously not work. I had thought up of several ways to overcome this. 1. chain multiple USB ports in serial to bump up voltage. 2. Use some sort of circuit ("booster") to increase the 5V to 12V I have tried searching this issue up online, however I have seen very differing answers. Some people say that using 2 USB ports in serial will create a short circuit while other people say that this will work and is the basis of Y cables. Could someone clarify and suggest the best option to increase a 5V supply to 12V? <Q> You cannot put the power connections of two USB ports in series. <S> This will cause a short circuit. <S> You are limited to the power available, though. <S> USB is limited (with exceptions) to 0.5A at 5V. <S> That's 2.5W. <S> If you boost the 5V to 12, the available current goes down. <S> The power stays the same, but the voltage goes up. <S> Therefor, the current has to go down. <S> At 12V, you would have less than 200mA. <S> I doubt that is enough to power your motor. <A> You'd be much better off powering the motor with a 12V power supply and designing some type of interface circuit to control it via the USB port. <S> But it doesn't sound like you have the level of knowledge designed to build something like that. <A> I think the best option for you would be to use a Boost Converter <S> (There are many available and with example circuits in the datasheet). <S> Make sure you know :-Current RequirementsNoise Requirements (if any) <S> If you dont feel confident enough to design and build a 12V boost converter then go to Ebay <S> there are many cheap options availbale which work of the shelf.
| You could use a boost converter to get 12V from 5V. Y cables put the power connections of the USB port in parallel and allow you to get more current.
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Impedance matching on DC voltage line to remove ringing/reflections I'm generating a 6 MHz signal in an Altera FPGA. I am feeding this 1.8 V signal out through a level shifter to 3.3 V to a connector on the board. This is the signal at that connector: I am then sending this 3.3 V signal back into the FPGA (stepped down to 1.8 V). It comes out of the FPGA (stepped back up to 3.3 V) and is brought out on another connector on the board. This is the signal at that connector: There is a lot of ringing from reflections on the line which I am trying to dampen. I have put a 33 ohm resistor in series on the track, but it didn't help. What do I do to match the impedance? I only have hobbyist level electronics. <Q> At 20 ns/division, edge rates of 2 ns, you need to dampen; use 50 ohms and 33/47/68/100 pF in series from trace/wire to GND. <S> Or try 100 ohms and a small capacitor. <S> By the way, your ringing frequency of 100 MHz is where 15 pF probes and 100 nH scope-probe ground leads (6") resonate. <S> Bypass your ICs, right at the package. <S> Using SMT capacitors. <S> On ground planes. <A> Make sure to use the best possible measuring methods with the scope to avoid the introduction of false impressions. <S> Short GND lead referenced right at the load measure point. <S> Make sure the series resistor in the signal line is as close to the driver output as possible. <S> The series resistor value may have to be adjusted up or down to improve the line match. <S> For particularly critical applications you may need to place a pullup and pulldown on the very end of the line to match line impedance and prevent reflections. <S> The match impedance is the parallel equivalent resistance of the two resistors. <S> If you have this all built on a proto plug board with flying wires everywhere <S> the pictures of the waveforms may be as good as it gets. <A> From the output pin of your FPGA <S> you may adjust the slew rate or at least get the information of the slew rate. <S> If there is parasitic inductance on your wire, this ringing will be generated due to the voltage at an inductance given byU = <S> di/dt at each switching. <S> If possible, try to use a ground like in the picture. <S> Maybe your signal is much more beautiful in reality without the probe. <S> Link
| The ringing could occur due to extra inductance, brought into the system by your probe.
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Reducing voltage drop over 7805 regulator? I have several projects (clocks) that require 12V and 5V supply. The 5V is for TTL ICs and the 12V powers a SMPS(Switching Mode Power Supply) to give HV(High Voltage) for a tube. The 12V pulls less than 100 mA, the 5V about 400 mA. The power source is a 15V 1A wall wart. For my first project, I wired both the 7805 and 7812 with heatsinks to the 15V supply. The 7805 heatsink got to around 43.3C (110F). After the first build, I discovered the the heat produced by the 7805 depends on the voltage drop, and the current drawn, and I want to reduce the 10V being dropped over the 7805. For my 2nd build, I was going to wire the 15V to the 7812, and run the 7805 off of 12V instead of 15V (giving 7V drop instead of 10V). I was wondering if there was a more efficient way to do this? Maybe using a Zener diode, or a voltage divider? I already have the 7805 and 7812 soldered, as well as the output caps, but I haven't soldered the inputs yet. For the next clock, I am open to suggestions for using other regulator ICs. <Q> Simply use an SMPS replacement for the 7805.These are readily available ... <S> here's a commercial variant that plugs into the TO 220 pinout you have with no heatsink required: <S> There are many variants available on Ebay or from Digikey/Mouser. <A> you thought think about a smps for the 5v output. <S> alternatively, you can use a pre-regulator to off lay the bulk of the heat. <A> Choose the resistor value so that at maximum load the 7805 input is just above the dropout voltage level. <S> Taking it a step further, put a resistor between +15V and +5V to supply the minimum load. <S> You'll still get the same power dissipation but the 7805 will be cooler. <A> The same question came up recently, here are a few 5V out buck DC-DC converters which will do the job. <S> https://electronics.stackexchange.com/a/317550/13616 <S> They are readily available from the usual suspects (farnell, digikey, mouser...) <S> Also you could use a 12V supply and skip the 7812 altogether.
| The simplest way is to put a resistor between the 7805 power input and +15 volts.
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Reason for Phase lead in RC high pass filter I'm struggling to understand the phase response of a high pass RC filter. At low frequencies, does the capacitor charge/discharge so quickly that there is a phase lead? But then we should have seen a similar behaviour in the low pass filter. At high frequencies, in a low pass filter, the capacitor is unable to keep up with the frequency of V IN and charges/discharges at a slower rate resulting in a phase lag. Should we not see a similar behaviour in the high pass filter? On the contrary, V OUT actually gets more and more in phase in the high Pass counterpart which I don't understand. <Q> If you want to understand phase relationships of passive filters you have to buy into the right terminology else you will forever be going back to first principles and that first principal is that current flow is determined by capacitance multiplied by the rate of change of applied voltage <S> If you then apply sinewaves to that first principle you start to see and understand the phase relationships. <S> So, for ac applications we just remember the simple acronym CIVIL. <S> It stands for Capacitor:I leads V and V leads I in an inductor (L). <S> We also remember that the impedance of a capacitor reduces as frequency increases and that the impedance of an inductor increase with frequency. <S> All proven from the basic formulas for capacitors and inductors but conveniently converted to different pictures when thinking about AC analysis and, if you talk about phase shift you are buying into AC analysis. <S> To talk about charging and discharging in the samesentance as phase shift is missing the point. <A> The current through a capacitor charges and discharges it. <S> A positive current charges the capacitor, the voltage across its terminals rises. <S> It can be shown that a sinusoidal current produces a sinusoidal voltage across a capacitor. <S> The positive half wave charges the capacitor and produces the rising edge of the resulting voltage, going from the minimum to the maximum. <S> So, when the current reaches its peak the voltage is in the middle of the rising edge. <S> It can be seen that for a sinusoidal waveform the current leads by 90 degrees. <S> The same behavior can be seen when analyzing a high pass filter. <S> One decade below the cutoff frequency most of the input voltage drops across the capacitor and only a small fraction of it across the resistor. <S> The current through the capacitor is leading its voltage. <S> The output voltage is just the current through the capacitor times the resistor and therefore has the same phase as the current through the capacitor. <S> Therefore the (small) output voltage is leading the input voltage. <A> It is easier to understand the phase shift using complex numbers. <S> By Ohm's law: \$ V = Z \times <S> I \$ <S> , Z is the impedance, I the current and V the voltage. <S> The impedance of an inductor is noted \$ <S> j <S> \omega <S> L \$ , with L the inductance value, <S> \$ \omega \$ the frequency in radians per second (i.e. \$ 2 \pi f \$), <S> j is engineering notation for imaginary number \$ <S> i \$ and is equal to \$ 0 + 1i \$, i.e. a vector of amplitude 1 and 90° phase shift. <S> Therefore voltage is ahead of current in an inductor . <S> The impedance of a capacitor is noted \$ \frac 1 {j \omega C} \$, which equals \$ -j \frac 1 {\omega C} \$ <S> (*) . <S> Pay attention to \$ -j \$ here, which is equal to \$ 0 - 1i \$, i.e. a vector of amplitude 1 and -90° phase shift. <S> Therefore voltage lags behind current in a capacitor . <S> (*) <S> \$ \frac 1 i \$ equals \$ -i \$, just multiply both the numerator and the denominator by i . <A> AC current always lags voltage in a capacitor regardless of its use in series or parallel. <S> THe impedance (f) always determines the current flowing thru it with an AC voltage across it. <S> But the impedance ratio as a function of frequency determines both transfer function and the phase shift relative to say a resistance where the breakpoint is 45 deg when the R=Zc(f) with lagging current <S> then approaches 90deg depending on the position of the cap Series or Shunt and of course freq. <S> It is easy to visualize with Phasors and math if you understand it. <S> Perhaps harder is the intuitive understanding. <S> Transfer function of XY out vs in is shown for both HPF,LPFand Transfer function of each part is shown XY=VI where a resistor is always a 45 degree linear line for V/I slope , but Caps always 90 deg lagging current. <S> Thus the position of the Cap determines if the phase of the output is partially leading or lagging when the voltage across the cap reduces near 0 above the same breakpoint. <S> We consider Caps as short circuits intuitively when Zc<< R and thus when attenuating in a LPF it is always 90 deg. <S> at high f and in a HPF it is always 0 deg shifted or out= <S> in for a transfer slope depending on scales for XY looking like ~45 <S> deg more or less IF they were equal scales. <S> So in summary each R and C the XY plots or Y/X= I/V on the Y/X plots and transfer function <S> X/Y= <S> Out/in <S> So I made a simulation here. <S> Although sweeping both filters in slow motion, they must span almost 3 decades (50Hz to 20kHz) to show the entire 90 phase shift of each filter, where it shifts most rapid at the break point f-3dB or 0.707 of input. <S> Note <S> only the HPF transfer function reaches vertical or 90 deg phase shift at max f to match the phase shift of I/V in the cap when it has near 0 voltage across is. <S> Thus the load V(R) shows the Cap current I(C) going thru it leading by 90 deg relative to the input X or + 90deg. <S> because the cap current is ALWAYS lagging current or Leading Voltage by 90 deg.
| For frequencies higher than the cutoff frequency output and input voltage are almost in phase since the capacitor is nearly a short circuit.
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Buck Converter for 59V to 58V? Some batteries last twice as long if they are kept to 80 percent of their maximum charge. It's good to add a small adjustment to undervolt a charger output by 0.5 - 1V. Myself i have a CC/CV charger that is .5 volts too high for a given BMS. How can I take the voltage down by .5 - 1V at the output? <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. \$ V_{OUT} = <S> V_{IN} - 0.7 \$ . <S> A silicon diode will drop about 0.7 V when more than a few milliamps is running through it. <A> At your very small input /output differential you will find it hard to get a chip to do this .You want to drop 1 Volt out of 59 .Even <S> the most efficient buck converters would waste more power than a simple linear solution. <S> When you have a big drop in percentage terms there is more to gain from the buck .Say <S> if you want 12V from 24 then the linear will be at best 50% efficient and you would be winning with if the Chinese Alibaaber buck is 78% efficient .In <S> your case a linear solution which could be diode based as Transistor suggested will be cheaper and waste less power . <A> Put a power diode in forward series with the positive supply. <S> Each diode will drop the voltage roughly 0.65 V. <S> A diode forward voltage increases only logarithmic with the current. <S> Also make sure there is adequate heatsinking provided for the diodes. <S> The amount of heat generated would 0.65*I, where I is the maximum current expected to pass.
| Make sure that the diode is rated according to the current you wish to pass, or if not, put several diodes in parallel so that the ratings add up.
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Can you continuously flash a high power LED bulb or will it break? I want a bright flashing display light. Imagine the sort of thing on a one armed bandit machine, that flashes a couple of times a second. And it would run for days on end. So either one of these which are rated at 5W:- or:- but the former is preferred as it contains all the necessary current control and is very easily replaceable without soldering. Are these suitable? I'm concerned that as they run hot they'll be no different than an incandescent, suffering continuous thermal shock and mechanically breaking. Is there a more suitable high power LED for flashing or might I just as well use a traditional filament bulb? There's Strobing very high power LEDs but that seems to deal more with frequency than resilience. There are some other questions too but again they seem to deal with implementation issues. <Q> The first one is a LED bulb, the second is a bare light emitting diode (LED). <S> The LED bulb may contain a semiconductor current driver or may use a single resistor. <S> If current is limited only with a resistor you can safely drive it with pulses with constant voltage amplitude. <S> Otherwise you should check how the internal electronics will work in such mode. <S> In most cases this mode will be bad for the bulb's electronics and may shorten it's life. <S> Pulsing the bare LED with constant current amplitude is best, but you have to be sure of the current waveform trough it <S> and how much current overshoot occurs during turn on & turn off. <S> This requires precise selection of the current driver and the switching off technique - breaking the circuit or shortening the LED. <S> Not any "off the shelf" driver will work good enough. <A> The Phillips is not made to be blinked. <S> It is rated at 50,000 switching cycles. <S> Also it has a 0.5 second turn on delay (warm up). <S> It is powered by 12V DC or AC. <S> LINK: <S> CorePro LEDcapsuleLV 2-20W 830 G4 <S> The second is a cheap Chinese LED. <S> LEDs are made to be turned on and off. <S> You do not want to use either of these. <S> Use a Cree XP-3G, the most efficient LED available today. <S> In white or blue. <S> Use a Microchip MIC4802 or MIC4801 CCR driver with the enable pin driven by an inexpensive ultra low power µController. <S> Very simple driver circuit. <S> Microchip MIC4802 Datasheet <S> This combination will give you the most efficient, very flexible, low cost, and very bright blinking LED. <S> It is easily battery powered by a Li-ion 3.6V battery such such as Panasonic NCR18650B. <A> The first one is a packaged consumer product that is made for direct application of a fixed voltage like 12V. <S> It has drivers and controls embedded. <S> It is not made to flash . <S> If you search elsewhere you can find "consumer product LEDs" made of nothing but LEDs and resistors, typically 12V or 24V, notably in automotive replacement lighting, as well as many LED strips, puck lights, etc. <S> These are perfectly fine with being flashed, in fact, flashing them too fast for the flashing to be visible is how they are normally dimmed. <S> * <S> * <S> The second one is a raw LED emitter device. <S> It responds very well to being pulsed, modulated or flashed at any imaginable frequency, but needs a driver circuit which is correct for this application, nominally-constant-current, but in your case, modulating current. <S> ** <S> But this is not the best way to dim LEDs. <S> They can be dimmed simply by regulating their current flow. <S> In this mode, they are the most dimmable lights ever made, happy to perform from "barely perceptible" to their max rating.
| LEDs can be driven in pulses and in most applications they are.
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Driving a MOSFET with fastest rise time possible I'm working a circuit to send very short pulses to a load, with the requirement that the pulses have the fastest rise time possible (~10ns or less). I've picked out a driver that has a fast rise time, the UCC27524P ( http://www.ti.com/lit/ds/symlink/ucc27525.pdf ). I have a question about choosing a MOSFET, is it the driver that determines the rise time or the MOSFET's rise time in it's datasheet ? I couple of MOSFETs I'm looking at has rise times of around 20ns, are they simply unable to switch any faster? I've attached a simple circuit to illustrate. <Q> It's both the driver and the mosfet that determine the rise time. <S> The net rise time is going to be about twice if both driver and mosfet are similar. <S> You might also be interested in fall time and this can be harder to achieve especially if the gate isn't driven negatively when removing gate charge. <S> If the data sheets say 20 ns <S> and there isn't some obvious improvement in the example circuit used to measure rise time <S> then you are probably out of luck. <S> Keep plugging away and look for better mosfets. <S> I think there is a new type out referred to as something like silicon-carbide technology but don't quote me on it. <A> Here's your solution. <S> 400V, 4 ns. <S> https://www.avtechpulse.com/medium/avrf-4a/ <S> The fact it costs $15k could hint at the fact that it is not a problem that can be easily solved with 2 active devices... <S> hmmmmm... <S> Now, if you really insist... <S> Okay, you want to turn on a MOSFET really fast. <S> The MOSFET gate is a capacitor. <S> So you need to pump lots of current into it really fast. <S> This means lead inductance is not your friend. <S> This means leaded packages like TO-220 will only result in more pain. <S> So, I searched DigiKey for a suitable part, and whaddya know, found one! <S> Tada! <S> (It is also available in TO220 , if you want to take chances). <S> Now, this is not your average MOSFET. <S> It's actually a cascoded GaN-FET. <S> Ths cascode handles your other problem, Miller effect of Cgd capacitor, which at 400V is going to be truly humongous. <S> Notice how the gate charge is ludicrously low compared to your IRF840 (like 10x less). <S> Also the rise/fall times are very fast (like 5ns). <S> And the TPH3206LSB version has the Source on the big fat cooling pad on the back, which means you can solder it to your ground plane to cool it. <S> (If it had been the drain, extra capacitance on the output would have been introduced). <S> Note the TO220 version also has the Source on the tab, so you can ground the heatsink and even remove the insulator. <S> I've never used a GaN-FET, but damn, this part looks like it means business! <S> With a solid layout and a bit of luck, it could work. <S> Maybe... Well, you can always try... <S> Although you should really work on your specs first, since we don't know anything about the load yet, so this is really a shot in the dark. <A> The problem is not the gate charge (you chose a good gate driver) or intrinsically the MOSFET itself or even the stray inductances (assuming 10nH gate-source) but it is the drain-source charge combined with the load impedance. <S> A quick LTSpice simulation with the STP8NM60 (650V MOSFET that is built in) shows 50ns rise time with 600R load, but just under 10ns with a 60R load (agreeing with the datasheet 10ns figure). <S> The output capacitance for the STP8NM60 is 100pF. <S> A GaN transistor such as GS66502B (650V part available at Mouser for £8.19) has an output capacitance of 17pF, which means it could just about manage your required 10ns with a 600R load. <A>
| You can use ISL55110 driver with GS66502B GaN Mosfet, they should allow you < 2nSec fall time
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How much can I overcharge a 4.35V LiPo? I'm building a charger for a set of LiPo cells. My circuit needs to make sure they get fully charged to their maximum level of 4.35V. At the same time, the circuit should protect the cells from overcharge. The charging stops when the 4.35V level is reached. Tolerances on components might cause a tiny overcharge, for example 10mV. Can this be harmful for the battery? From what overvoltage does the battery get damaged? <Q> The charge (termination) voltage level of Li-ion batteries is specified by manufacturer. <S> The specification is based on reasonably accepted number of cycles ("battery service life") <S> a battery can withstand, say, 500, or 1000. <S> This parameter depends on particular cell chemistry, internal construction, charging current, and is picked up by manufacturer for the best marketable value. <S> The manufacturer's recommended voltage is a trade-off between these two parameters. <S> Contrary to urban myths of "damaged" batteries, the dependence of "battery service life" on charge voltage is a smooth continuous curve. <S> Certainly the lifetime dependence ends at some point with catastrophic failure, but fears of 10 mV overcharging are grossly overstated. <S> However, 100 mV over 4.35 V (for Li-Po battery) might cause a problem, see, for instance, this publication from Texas Instruments , page 3-5. <S> So, overcharging of 150 mV over the nominal 4.2 V leads to about 10% more capacity for first 50-100 cycles, but the service life shrinks from 500-1000 cycles to about 200. <S> Extrapolating, the another 100 mV will result in maybe 30-50 cycles life. <S> This means that 50 mV over the spec won't kill the battery. <S> The page 3-7 is also fairly informative. <S> It says that 70-80% of capacity is coming during CC stage, while the tail (CV stage) makes up only 20-30% of capacity, so there is no much reason to wait down to 0.03C. <S> Most TI chargers defaults to 256 mA to terminate the charge process. <S> For more insights and correct application of chargers, one might want to examine other materials as THIS ONE . <A> Charging Li-Poly and Li-Ion is not as simple as just pumping a voltage across it. <S> It's a two-stage operation involving both constant current and constant voltage. <S> Apply a constant current of (say) 1C (so for a 1200mAh battery that's 1.2A) until the voltage rises to 4.35V Switch to constant voltage applying 4.35V until the current flattens out (usually around 0.03C). <S> Here's the charge curve from a 1200mAh battery I charged the other day using my bench power supply: <S> As you can see most of the time it's waiting for the current to drop. <S> The longer you wait with constant voltage the greater the capacity. <S> I waited until exactly 0.03C. <S> Increasing the voltage over and above the charge voltage isn't going to force any more into the cell. <S> Waiting longer for the current curve to become flatter is what you are interested in. <S> Operating any device outside of the published specifications can lead to undefined results. <S> When operating something as potentially explosive as a battery outside of the published parameters can lead to explosive results. <S> Usually in the form of an absolute maximum specification. <A> At 4.35 V cutoff, the battery is already being damaged for an ordinary LiCoO2 cell. <S> You need to cutoff at 4.15 V or 4.2 V maximum. <S> The amount of energy capacity between 4.1 and 4.2 V is only 1.2% of the total energy capacity. <S> Between 4.2 V and 4.3 V it is even less, probably 0.6% of total capacity. <S> There really is no point to go above 4.2 V, given that you only increase the energy content by 0.6% and it will quickly and permanently reduce the energy capacity of the cell, even if you do not cycle your cell often. <A> simulate this circuit – Schematic created using CircuitLab <S> This is a simple constant current constant voltage design using 2 LM317. <S> I have to build one myself because it's very hard to buy a 4.35V charger in the market. <S> Circuit is over simplified, do remember to add the caps and protection diode. <S> Set the U2 output to 4.35V during no load using R3 pot before charging. <S> charging voltage. <A> Constant current switching to constant voltage and then shutting off at 4.2, as chargers work, isn’t necessary. <S> All that matters is u <S> don’t put in too many amps st a time and don’t go over or under voltage. <S> Even the amps u can safely put in can greatly vary depending on soc. <S> If it’s getting warm charging back off and will last longer. <S> Ironically blasting w very high current can actually heal cells increasing energy density and destroying dangerous dendrites. <S> But that’s going a step further.
| At full charge it will not exceed 4.35V. Also do check the manufacturer spec on the battery, mine mention 4.35V +- 0.03V Higher charging voltage leads to slight increase in battery capacity, but it shortens the battery life. If it were safe to charge with a voltage above the specifications the specifications would show that.
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Transistor Collector Heats Up I used power transistor to drive high power LED (used in streetlights) I added resistor between the Transistor Collector and the LED to verify which part of the transistor causing too much heat, and found out that it is really on the collector part. Originally, the LED is powered up by 42 DC Volts so I wanted to keep that source level. I added Photoresistor to turn ON/OFF the LED The problem is that the transistor heat much. I also wanted to keep the luminous/power dissipated by the LED as much as possible. UPDATE: I don't know the specs of the LED but I know the original power source of it: 24V - 40V ; 0.45A - 0.8A ; Maximum 42V <Q> Oh, cripes. <S> I'll add a BJT circuit that uses your existing TIP122. <S> You don't specify the LDR, so I can't be sure I've got the hysteresis thresholds right. <S> But I think these will do. <S> I'm offering this because it's probably the "least change" to what you have now <S> and it may work okay for you, assuming that I read you correctly when you updated your question. <S> (But I lack enough information to know for sure.) <S> I'm also assuming here that it is easier for you to grab a couple of jelly bean PNP BJTs than it is to grab a specific comparator IC. <S> Here it is, below. <S> I'm going to stay with your drawing format which buses the power supply around. <S> I don't like doing that, because it can distract from understanding the circuit. <S> But I suspect it may communicate better in your case. <S> So I'll stick with the format you are currently comfortable with. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> I kept the \$200\:\textrm{k}\$ resistor near the LDR in your circuit, because I don't know what caused you to use that value. <S> I felt safer keeping it in place, for now. <S> You can adjust \$R_1\$ to change the light level threshold, a bit. <S> Also, \$R_3\$ and \$R_5\$ have useful effects when you change them. <S> But I think I really need to know what LDR you are using (datasheet?) <S> before I can offer much better. <S> At least, this provides a very versatile topology that can be adapted to your needs when more information is available. <S> The basic idea is solid and it uses a key part that I know you already have (the TIP122) adding only a minimum number of cheap parts to get the rest done for you. <A> The LED has an exponential current-versus-voltage IV characteristic. <S> Its a diode. <S> The transistor has more of a constant current, with variable-collector voltage, behavior. <S> Result? <S> Most of the VDD appears across the transistor, ensuring the transistor overheats. <S> Insert 1Kohm resistor in collector, in series with LED, to limit the current. <S> That resistor will dissipate (42-2)*(42-2)/1,000 = <S> 1.6 watts, if the transistor is saturated. <A> UPDATE <S> : I don't know the specs of the LED <S> but I know the original power source of it: <S> 24V - 40V; 0.45A - 0.8A; <S> Maximum 42V <S> It seems to me that if you are trying to dim the LED by dropping 18 volts across the transistor you will get significant power loss <S> The LED will now have 24 volts across it and presumably want to take 0.45 amps. <S> That current will flow through the transistor and dissipate 8.1 watts in it <S> i.e. 18 volts at 0.45 amps is an internal power dissipation of 8.1 watts in the transistor. <S> If on the other hand you only want the transistor to turn on or turn off (based on the photoresistor) <S> you need to use a comparator so that there is no in-between half state that causes the transistor to dissipate all this power. <S> You need to go down this route if you want this to work properly. <A> For a maximum base current of only \$\frac{42{\mathrm V}}{200{\mathrm kΩ}} = <S> \bf 0.21{\mathrm <S> mA}\ \$you may need another transistor stage. <S> The typical current gain of the TIP122 is 2500. <S> From \$I_B = 0.21{\mathrm <S> mA}\ \$, you get \$I_C = <S> 525{\mathrm <S> mA}\ \$. <S> Not the 800mA you may require. <S> This will result in the TIP122 only "half-opening" and dissipating a lot of power as you are using it an analogue amplifier instead of driving it into saturation. <A> darlingtons are poorly suited for switches as they have very high Vce at saturation. <S> Go with a mosfet or bjt + driver.
| Check the specs of your lamp.
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Operating an adjustable voltage regulator below minimum Vout-Vin I used an LM350 adjustable voltage regulator https://www.onsemi.com/pub/Collateral/LM350-D.PDF and to keep the wattage dissipated by the device to a minimum I have opted to make the output voltage 24VDC. Max current drawn is around 500mA. The input voltage can vary from 19VDC to 32VDC but will be 24VDC under normal conditions (24V Vehicle Supply). When the input to the circuit "Vin" is below 25.1VDC the regulator Input-Output differential will be too small and the regulator will not regulate the voltage. I cannot find any information if this will have any negative consequences eg increased noise or limiting of current flow? The circuit is intended only to limit the voltage when it goes above the normal 24V and some noise filtering as a by-product. Is there any reason not to use the regulator in the "non-regulating" / "drop-out" state most of the time? <Q> Let's have a look at datasheet Figure 10 "Dropout Voltage". <S> At 500mA it will be a bit above 1.5V. <S> When operating in this mode, the output pass transistor will be fully on (saturated in this case, since it is a NPN bipolar). <S> If this was a PNP pass device LDO, you would expect excess ground current, as the regulator attemps to saturate its output device. <S> However, this one uses a NPN, so the excess base current will simply go into the output. <S> No problem here. <S> The regulator will not regulate anything though, this means it will act either as a resistor or as a couple of diodes in series, so output voltage may vary depending on current draw. <S> Also, output voltage will follow input voltage, so input noise will not be suppressed. <S> In this case it would simply act as a voltage limiter. <S> If you also want to filter noise, then something like a capacitance multiplier with an output voltage limit would be more suitable. <S> EDIT: <S> Example simulate this circuit – <S> Schematic created using CircuitLab <S> This is a simple capacitance multiplier. <S> It lowpass-filters the input (RC network) to filter out noise. <S> Zener limits the voltage. <S> I put in a CFP (double transistor) for lower output impedance, so you can say that's the "luxury" version! <A> I'd definitely be considering a buck-boost regulator like the one below: - They're not particularly cheap but they do what they say on the tin and with everyone trying to stop the planet heating, this is a fairly green solution in the longer run. <S> There may be cheaper and similar solutions from TI of course. <S> I cannot find any information if this will have any negative consequences eg increased noise or limiting of current flow? <S> This would be my concern and it will happen on most LDO regulators if taken too close to the limit i.e. there will be progressively more output noise as you approach the regulation voltage from a higher voltage. <S> Whether your circuit might cope I cannot say. <S> My philosophy is that if there is a more expensive solution that offers peace of mind then it is worth considering. <A> If noise is important, then consider the PSRR of LDOs versus switchers. <S> At high frequencies (and "high" may be 100Hz and above for some 1uA Iddq LDOs), the input noise comes through un-attenuated. <S> Examine the datasheet. <S> Switchers have their own added synchronous noise, with lumped Ls and Cs to filter; input noise lower than the switcher's frequency will be somewhat attenuated; examine the datasheet. <S> Part of your tradeoff will be the standby current. <S> You can purchase 1uA LDOs that use Pchannel FET regulation. <S> You cannot purchase 1uA switchers.
| If your load works on 19V, and you have 24V input, and the load can tolerate the regulator not rejecting input noise, then you're fine.
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Voltage regulator for Stepper and Microcontroller I'm designing my first PCB for a personal project : I'm planning to run a Stepper motor (3.8V 670mA/phase) with the DRV8834 driver and an Atmega328P from the same power source. The power will come from a classic USB charger (rated at 5V, 2A) plugged in my PCB. Do I need a voltage regulator on my PCB as the power coming should be already regulated to 5V by the USB charger ? If so, would I need two voltage regulators one for the Stepper that will draw higher amounts of current at times, and another one for the stable microcontroller, or can I use one for both ? I believe if needed I will go for an LDO 5V - 2A regulator like those found here . Would it be a good choice ? <Q> Stepper motor coils have high inductance , which resists current changes by inducing an opposing voltage proportional to the rate of current change. <S> In practice this means that to get the motor to step faster you have to apply higher voltage. <S> Current limiting is then applied to account for the lower voltage drop at lower speed. <S> This can be achieved by putting a resistor in series with each phase, or applying PWM <S> which switches the power on and off rapidly to reduce the average current draw. <S> 3.8V is the voltage drop across each phase of your motor when passing a constant (DC) current of 670mA, not the design supply voltage. <S> To run it on 5V you just have to limit the phase current to 670mA. <S> The DRV8834 uses PWM to reduce current draw. <S> The Pololu DRV8834 breakout board has a potentiometer for adjusting the current limit. <A> Be careful with the regulator. <S> Ensure you use a package that will dissipate the power. <S> One of those looks like a SOIC8 and would likely not suffice. <S> Current Drawn <S> x <S> (5-output voltage) gives power lost in regulator and the datasheet will tell you temperature raised per watt dissipated. <S> If your stepper motor can't take 5V (a quick glance simply says it's rated at 3.8V), you will need a regulator for it. <S> You can probably use the same regulator for the AVR too, it depends on the frequency you're running it at. <S> (Check the frequncy vs voltage limitations on the AVR datasheet). <S> Also, it might make interfacing between micro, stepper motor driver and the motor itself easier if all are running at 3.8V. <S> You might want to run the supply from regulator to the AVR though ferrites to each power pin and, most importantly, make sure you have the biggest possible uninterrupted ground plane along with thick, thick power tracking. <A> You may try to add a resistor in series with 5V regulator where that supply goes to stepper motor. <S> To supply 3.5 V to stepper motor whose current rating is 670mA/ phase, you need to add a resistor of ~2.24 Ohm. <S> It drops 1.5 V supply before Supply goes to that stepper motor. <S> Calculation: Voltage (drop) <S> = 1.5 <S> V Current <S> = .67 <S> A. Resistor = 2.24 <S> (approximately)
| High performance stepper motors are usually operated on a higher voltage than their coil rating.
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Can I recharge a battery that I don't know anything about? I like electronics but I basically don't know anything about it (electricity is too complex), the thing is that I've opened a battery from an old laptop that reads "10.8 V 4300 mAh Li-ion" The battery-pack contains other 6 batteries and a circuit inside, I suppose that the circuit is just for showing battery info in the computer, charge indicator and for preventing 'overcharge'. Batteries shows 2x 0.0 V, 2x 3.11 V and 2x 3.17 V (with a tester), but that's just their current charge, there's no info on them more than a serial number so... There's a way to SAFELY recharge them? I'm not in a hurry so time is not a problem. I thought about put them 0.5 V higher than they currently have over and over until they don't get any higher voltage, but I'm not sure if that will work nor if that will even recharge them. PS: I thought that Li-Ion batteries didn't contain a pack of standard batteries inside... <Q> If you know nothing about them then be aware that some batteries are not designed to be recharged and can explode if charging is attempted. <S> For rechargable batteries there is a different charging scheme for each technology. <S> Do it wrong and the battery is likely to explode or catch fire. <S> I strongly urge that, since by your own assertion that you know nothing about these batteries, STOP experimenting around before you end up hurting yourself, others or destroying property. <A> And yes, all laptop batteries are made of standard cells, from variants of 18650 or 17650 cells. <S> Or flat proprietary cells. <S> As Simon noted, you have one dead section. <S> While the chances to recover full functionality are small, you can try. <S> You need to try to "pre-charge" the dead section, which can be done with a small, ~200 mA current applied to this section alone. <S> You don't need to desolder anything, just find a way to connect external leads temporary. <S> The best would be a benchtop power supply, set it to 4.2 V and 200mA current limit. <S> Alternatively you can try a USB charger with a 22 - 33 Ohms resistor in series. <S> After one-two minutes check if the cell builds any voltage. <S> If the voltage goes above 2.5-3 V, you can stop the pre-charging and try to charge the whole battery inside your laptop, using the laptop's AC-DC adapter. <S> I had some successful recoveries using this method. <S> ADDITION: for those who spread myths about explosiveness of over-discharged batteries. <S> Here is a scholar research on the subject, "Effects of overdischarge on performance and thermal stability of a Li-ion cell" No explosive effects or thermal run-away <S> instabilities were found, other than a loss of capacity (sometimes drastic). <S> NOTE OF CAUTION <S> : The cells inside laptop batteries don't have built-in protection as many stand-alone cell have nowadays. <S> Be careful not to short them with thick wires. <S> Although no Li-Ion cell explode, they can rapidly "vent", which is sometimes accompanied by flames. <A> It is not safe to charge them anymore. <S> That small board that you have seen has many utilities. <S> It cuts the current to ensure that you never discharge your battery more than this level plus a security margin. <S> It also prevents you to charge it again in case it has discharged naturally below that level, and it also prevents you to charge it over its limit. <S> If one of your batteries shows 0V <S> it means that it has gone below the non-return level, so don't play with it (unless you want to make it explode).
| If Lithium batteries drop below a certain voltage, it is dangerous to recharge them. You have a standard 2P (two parallel) and 3S (3 in series) laptop battery.
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multiple LEDs in series I had a 2sets of LED series circuits which runs with 12V DC, 1A as shown below . Each set will be of 3 series of four LEDs. My problem is that when i connected to 12VDC,1A source, the set1 brightness is higher than set2. how to overcome this problem ? <Q> There will be some voltage drop in the wiring, depending on both the current and on the length of the wire. <S> In your circuit, set 2 will receive a lower voltage than set 1 due to the additional wire between the sets. <A> You don't specify the length of your wiring nor the gauge of wiring. <S> But your diagram shows that you are daisy chaining the power supply connections. <S> This adds more resistance for the second strip which could lower its brightness. <S> Try wiring each LED strip directly to the 12 volt supply using the same length and gauge of wire for each. <S> Increasing the gauge wire for both would also be helpful. <S> Your diagram shows that you are connecting the LEDs directly to the power supply without any form of current limiting. <S> This is generally not a good idea. <S> But to improve the situation, you would need a higher voltage power supply. <S> If you have one, indicate in the comments what you have <S> and I can add to this answer. <S> Finally, consider that LEDs of the same part number come in different brightness grades. <S> It is possible that your strips are simply two different grades. <S> By lowering the current on the brighter strip, it can be made to match the other strip. <A> The culprit here seems to be the power supply. <S> If each set needs 12V 1A and you're connecting 2 set in parallel, their current demand will be 2A. A single 12V 1A power supply won't be able to supply enough current and will start dropping voltage. <S> As a result of all this, the LEDs will start to show an impedance higher than what is desirable, due to insufficient bias. <S> Because of manufacturing tolerances, the LEDs in each set will have different impedances at those ill biasing conditions and, as a result, one of the branches could draw more current (and its light be brighter) than the other. <S> Try the following test: connect each set individually to the power supply, one at a time. <S> If each set shows full brightness tested like this, then the culprit is the power supply.
| Wire each set individually to the power supply. Increasing the wire gauge (diameter) will also reduce the voltage drop in the wiring.
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What is the difference between characteristic impedance and input impedance in transmission lines? What is the difference between characteristic impedance and input impedance in transmission lines? When are these quantities are equal? <Q> What is the difference between characteristic impedance and input impedance in transmission lines? <S> Characteristic impedance (\$Z_0\$) depends on the transmission line and its physical properties. <S> Mathematically it can be shown that if you know the inductance (L), capacitance (C), resistance (R) and conductance (G) per unit length, \$Z_0\$ is: - \$\sqrt{\dfrac{R+j\omega L}{G+j\omega C}}\$ <S> And of course these quantities can be deduced from the physical dimensions, dielectric properties (including dielectric losses) and conductivities of the materials used. <S> when these quantities are equal? <S> If you have an infinite line then input impedance = <S> \$Z_0\$ <S> If you terminate a non-infinite line in \$Z_0\$ they are equal If you terminate the line in an impedance not equal to \$Z_0\$ then, providing you choose the correct line length, the impedances can be made equal. <S> This last bullet point makes use of the relationship between input impedance, load impedance and \$Z_0\$ in the following way: - \$V_P\$ is velocity of propagation as a ratio to speed of light and, for normal coax cables is about 0.7 but is \$Z_0\$ value dependent. <A> The characteristic impedance is a function of the line only. <S> The input impedance of a line is a function not only of its characteristic impedance, but also of its loading impedance and electrical length (or physical length and frequency). <S> They are equal when the line is loaded in its characteristic impedance. <S> A quarter-wave line will present an input impedance of \$\frac{Z_{char}^2}{Z_{load}}\$ <A> These current and voltage values are the result of interference between incident and reflected signal. <S> If the line is infinitely long, the reflected signal from the load will take some time to go back to the beginning of the line. <S> Hence, the current and voltage values will only be due to the incident signal. <S> If you take the ratio of this voltage and current due to only the incident signal, you will get the characteristic impedance of the line.
| The input impedance of the line is the ratio of the voltage and current at any point.
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I've read that 18V li-ion batteries are over 18V when charged. How do I make sure this won't harm my motor or anything else? I've read that a fully charged 18V li-ion battery is over 18V when fully charged. Is this true? If so how do I ensure that I don't damage my motor? Here is my setup: 18V Li-Ion Batter -> PWM (Properly Rated) -> 18V DC Motor <Q> You don't tell us anything about the motor, but brushed motors are pretty tough. <S> Brushless motors tend to be even tougher, but their controllers may be another matter. <S> A moderate overvoltage won't normally kill a motor, just slightly overspeed it. <S> Now at hundreds of volts, you have to worry about exceeding the insulation breakdown voltage, but not at 21V. <S> And if it's a particularly high performance motor, that overspeed can overstress bearings or add to brush wear, or destroy the rotor through centrifugal force, but otherwise it's unlikely to do any harm, unless the motor is heavily loaded. <S> Now in combination with a heavily loaded motor, that's a different matter. <S> Loading a motor enough to slow it down, makes it draw a high current, which dissipates a lot of power in its winding resistance, and the high current can also burn out the brushes, melt the commutator, etc. <S> If the motor's this heavily loaded, then the overvoltage supplies excess current, resulting in more heat and more damage. <S> So, don't do that. <S> And if you're still worried, limit the PWM controller to 80% duty cycle when the battery is fresh. <A> The voltage across a Lithium Ion battery (and most other technologies for that matter) is not constant. <S> It depends on several factors, the main ones being residual charge, temperature, age and the instantaneous current flowing to/from the battery. <S> A Lithium-Ion cell may be rated 3.6V, but it will be at 4.2V when fully charged, and keep functioning without causing irreversible damage down to 3.0V (lower, in fact, but there is really no point bringing a battery that low as the slope of the discharge is really steep at that point). <S> When making a battery back, you can arrange cells in series to increase the voltage. <S> If you have an 18V Lithum-Ion battery pack, it sounds like 5x3.6V nominal cells in series. <S> This is a pure assumption from my part based on the info you provided. <S> In such a scenario, if your charger can fully charge each cell to its maximum voltage, you could potentially reach 21V. <S> Don't take this information for granted - try to find the datasheet of your battery pack and look for the discharge curve, or maximum charge voltage. <S> As for whether your motor supports that higher voltage, again I would suggest digging a datasheet. <S> It's likely it supports higher voltages and simply output more power in a less efficient manner. <S> It could also heat up beyond the normal levels if operated beyond the nominal voltage for a sustained amount of time, only the datasheet can help us. <A> You can also try a Buck-Boost or SEPIC converter to maintain fixed voltage at 18V if that is you want. <S> This will make the motor/PWM power sort-of independent from battery power. <S> If you battery is discharging and going below 18V the regulator should give you more reliable performance as well. <A> The voltage output of batteries changes by as much as 30% depending on the load and how charged the batteries are. <S> The controller and motor designers take this into consideration when specifying the voltage rating. <S> If you go above the maximum voltage of your controller/motor, you'll damage it. <S> If you go under the minimum voltage, it won't work because the controller thinks the batteries are almost dead (low voltage shutoff). <S> Voltage differences don't affect a motor much. <S> Just make sure the motor controller and motor can handle the maximum voltage of the batteries fully charged. <S> A motor is just a coil of insulated wire. <S> Too many amps and the wire melts. <S> Too many volts and the voltage jumps through the insulation like a small lightning bolt. <S> I have even gotten away with doubling the voltage to a Phoenix Racer electric bike motor and reduced the amps by 25% to make it spin twice as fast. <S> But doing something like that requires special high voltage controllers. <A> If the cells are LiFePO lithium-ion, then there is little to worry as its maximum voltage will be 3.6 V per cell, and assuming you are 5S, that will be the 18 V that you desire. <S> If it is non-LiFePO, then the voltage can be as high as 21 or 21.5 <S> V (assuming again that the pack is a series of 5 cells). <S> When the voltage drops to 18 V (non-relaxed) then only 19% of the energy remains in the battery. <S> So most of the time, you are operating at above 18 V. <S> With a voltage above 18 V it would be the PWM that may complain before the motor does, as motors generally take a wide range of voltages, as long as they are not overstressed. <S> If the overvoltage is still of a concern then string about four diodes in forward series with the battery output. <S> Each diode will drop the voltage by 0.65 V. <S> So if the battery is at 21 V, the PWM will receive only 18.4 V. <S> Make sure the diodes have sufficient current rating and have adequate heat sinking. <S> Once the voltage drops to 18V, you can bypass the diodes with a switch or relay.
| If the motor's lightly loaded, (determine that by measurisg its current) it'll probably be fine.
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If Conventional Current is wrong, how can I trace the flow of current in a schematic? If Conventional Current is backwards, how do I look at something like this schematic and trace the flow of current? I'd like to follow the design linearly from positive to negative. When the battery is on, does current go to 8, then 4, then through R1 and so on? Does the current get "back around" to the side of the capacitor, 1, and the speaker eventually (though fast)? How can this be so if in reality electrons flow from negative to positive? Edit: If the current were water here, how would I see that physically on the wiring, as if they were pipes? <Q> if Conventional current is wrong <S> Conventional current: used by engineers and physicists everywhere. <S> It's what's measured by ammeters! <S> It applies to all circuits, including the non-electron flows in dirt, nerves, acids, plasmas, etc. <S> "Electron current:" used by technicians during WWII ...and by several generations of students taught by them. <S> Applies to solid metals, and especially to vacuum tubes. <S> It cannot explain nerves and batteries, semiconductors and plasmas, or any situation where proton-flows or mobile ions are paramount. <S> In other words, electron current is wrong. <S> Don't use it. <S> If you have a textbook which employs backwards current (electron current,) just throw it away. <S> Heh, or perhaps chop it up, so it cannot harm anyone. <S> how can I trace the flow of current in a schematic? <S> To understand circuits, we don't trace the flow. <S> (After all, batteries don't spit out constant current.) <S> Instead, we study the schematic as a whole, and determine the pattern of voltages across the various circuit points. <S> Then, knowing the voltages, we can figure out the current in any component. <S> "Tracing the flow" doesn't work, since it isn't based on Ohm's Law, and it leads to mistaken thinking. <S> For example, at any "Y" junction, how do the charges know to split up? <S> How do they know which path to take? <S> They don't. <S> Instead, the voltages far downstream are determining the current in the entire circuit. <S> If tracing the flow doesn't lead to understanding, then what does? <S> Ah, that would be voltage, and the voltage-dividers scattered throughout the circuitry. <S> The goal of electronics students is to learn to "see the voltages" all throughout a schematic. <S> One education site which goes into the voltage-divider viewpoint is NCSU/Williamson: <S> http://williamson-labs.com/transistors/transistors-main.htm#animations Also check out Falstad's java circuit simulator, where voltages are shown as colors. <S> Here's the animated guts of a 555: <S> http://www.falstad.com/circuit/e-555int.html <S> ALso see his entire circuit-animations index. <A> Conventional current being backwards from electron flow has no effect whatsoever on circuit analysis. <S> If you really wanted, you could swap the signs on all your currents and voltages and the math would work out the same, but people would be confused. <S> If you're doing an ad-hoc intuitive analysis without any math, well, it still doesn't matter whether you start at the positive end or the negative end of things, or neither. <S> The causality , the ways changes propagate around a circuit, is also symmetric. <S> If you close a switch or make some other such change, the changes in voltage and current propagate away from the switch along both connected wires — with opposite signs, but otherwise completely identical, at the same speed. <S> Specific components (diodes, ICs, capacitors, tubes…) may have polarity , so they require or only allow a flow with a specific direction/sign, but it does not actually matter at all for understanding the behavior of the circuit which one matches the actual flow of electrons — only if you want to understand why the components do what they do does that begin to matter. <A> If Conventional Current is backwards, how do I look at something like this schematic and trace the flow of current? <S> It worked in all their calculations then and still works now. <S> In your schematic you have correctly drawn the positive rail at the top and the negative at the bottom as is favoured by most electronics designers. <S> Current then (generally) flows from top to bottom. <S> Figure 1. <S> Conventional current vs electron flow. <S> Both move simultaneously with equal effect - just in opposite directions. <S> Source: Electrical Market Plus . <S> To analyse your circuit you would need a diagram of the internals of the 555 to understand how it switches when voltages change at its terminals. <S> Start with basic circuits and build up. <S> It's a fascinating subject that you can spend a lifetime studying and continue to learn new things every day. <A> In your schematic, conventional current flows from the battery + terminal to the 555 pin 4 AND 555 pin 8 AND R1 simultaneously. <S> Any further analysis of current flow in that circuit requires a study of the internal workings of the 555 IC. <S> Forget the "water analogy" (and any other analogy you find), as it is only useful in very limited situations, and can be confusing or misleading otherwise.
| The behavior of currents flowing in wires and through components can and should be be understood symmetrically — positive and negative voltages/currents, or electrons and “holes”, being equal and opposite to each other. The convention that current flows from + to - was agreed on long before the discovery of the electron by J J Thompson in 1898.
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How do I power/wire something that is rotating constantly WITHOUT extra batteries? So I want to make a Lazy Susan that is rotating constantly in one direction. I also want to have some other things on top such as neon signs, moving things, and etc. Problem is I want this to all run off of one DC source. Problem is I cannot figure out how I wire it so that things don't get tangled since the entire thing is constantly in motion. <Q> You are looking for a slip ring . <S> This is a device designed to do exactly what you describe - transmit power and/or signals to a rotating object. <S> Generally, they work by having a rotating circular contact on one side and a spring-loaded pin which pushes against it on the other. <S> They are not super reliable long term or in harsh conditions, but should be fine for your project. <S> More expensive ones are generally better and last longer. <A> An alternative to slip rings is the rotary transformer. <S> This consists of two cup-shaped cores that face each other, with the windings inside the cups. <S> If you drive the primary at a high frequency (we used 25 kHz), the cups can be ferrite and the whole thing can be quite compact (we did about 100W in a unit that was about 1" thick overall and about 3" in diameter). <A> How about using two rings of ball bearings and use each ring as the contact. <A> The other option that you may not have considered is to put all of your power consuming items, including the motor, and the batteries or other power source onto the lazy Susan. <S> No electrical connectors to the outside world required then. <A> It has occurred to me that if you are driving the lazy Susan at a reasonable rate you are also in a situation where you can generate power on it. <S> If either on the central spindle or round the edges you could mount fixed magnets you could use appropriately placed coils to generate power on the moving platform. <S> If you were looking for really low tech you could even use an old push bike dynamo on the rotating platform running against a stationary surround of some sort. <A> Slot cars get their power from braid running on conductive tracks. <S> You might be able to glue flattened coax cable shield (braid) in two concentric circles on your base board and then use braid "brushes" to pick up the voltage. <S> Or use one circle and brush and pick up the other side of the voltage via the central pivot. <S> Or use the central pivot as one contact, run the braid around the outside of the platter and use a horizontal brush. <S> Or think about the good old Dodgem cars where the floor was one contact and the wire mesh ceiling the other. <S> Cover the whole of the base board with heavy duty domestic aluminium foil (leave a hole in the centre). <S> Attach a brush to anywhere on the platter as one contact and use the central pivot as the other contact. <A> I looked into this recently building a zoetrope with LCDs and a microcontroller on the spinning top. <S> I ended up making a DIY slip ring assembly from copper sheet and some carbon brushes designed for drill motors. <S> It worked out surprisingly well (still good after 100,000 or so revolutions). <S> Before I went down that road though, I investigated wireless power transfer and that looked really hopeful using a cheap tx/rx pair of modules like this: <S> Passing the drive shaft through the transmitter and attaching the receiver to the underside of the lazy Susan worked pretty well if you can keep them close together. <S> It didn't provide enough current for my application (I needed about an amp and these give about 500mA comfortably). <S> For my application I also needed to pass very low-rate uni-directional data from the base to the top, rather than add to my slip ring assy, I used IR LEDs and receivers, that worked really well. <A> Nicola Tesla said you could have a toothed ferromagnetic wheel on the rotor and stator each with windings. <S> A combination transformer and synchronous clock motor. <S> That avoids the legal expenses that you will have with liquid mercury, ozone poisoning, or electrocution. <A> But you could have a set of several smaller coils around the edge with similar coils coupling to them on the rotating platform. <S> The power from each would be intermittent, but if they were properly spaced, you might meet your needs. <S> Do expect nasty 60 Hz vibrations as the coils pass each other... unless you drive them at 100 kHz and use lightweight ferrite cores... <S> If you want to go to high frequency, you might do some kind of capacitive coupling, as suggested by that great Scottish poet, R.F.Burns.
| Tweed's rotary transformer is probably the best solution when it is possible to wrap a coil around the axis and have a ferromagnetic core for the axel. The other alternative is inductive power transfer/signalling, but this will be harder to get right first time, so might not be ideal for a hobby project. No need to add pins and contacts... Just to add, after taking note of the comments, that I was thinking of a low voltage supply here 5v or 12v just for leds or somesuch - I was not intending to suggest a 110VAC or 230VAC solution.
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How do I prevent bridges while soldering SMD Components? I have noticed that whenever I try to solder a component with many pins close together (like an IC), I get tiny bridges of solder, which shorts the pins. Is there any easy way to solder SMDs without solder bridges? <Q> When soldering by hand, it is easier and faster to just ignore the bridges while you solder the part and then clean it up. <S> Just put some desoldering wick on the bridge, heat with your soldering iron (may need a bigger tip or more heat) and the excess solder will happily flow into the wick leaving perfect solder joints. <S> With practice the cleaning of even large parts just takes a few seconds. <A> There are three golden rules in hand soldering: 1. <S> Use a lot of flux; <S> 2. <S> Use more flux; 3. <S> Use even more flux. <S> And, of course, right iron power, clean tinned tip, sharp tip (I prefer), and right iron temperature for the solder you are using. <A> Forget the old soldering iron method. <S> Solder paste (and a heat gun) is your friend. <S> This method is MUCH easier and faster than soldering each connection! <S> Assuming you're soldering components on one side of a bare board: Put a small (appropriate) quantity of solder paste on each pad; for ICs, you can put a continuous line of paste across all the pads on both sides. <S> Carefully place all the SMD components over the paste Heat the underside of the board with a heat gun. <S> All the components will be simultaneously and neatly soldered in seconds, with no solder bridges and much better looking than with an iron! <S> I bought mine from http://kd5ssj.com/solderpaste . <S> The technique . <S> A video there showing how it's done , but there are much better videos on YouTube. <A> this extrapad will take all the excess tin. <S> In addition you should use lot of flux. <S> I would recommend you to try diffrent brands and find the best one for you. <S> You can also try to use gull wing soldertips. <S> I have one by myself and with the right flux i rarely have to clean up with solder braid. <S> But when i have to, it isn't the end of the world. <A> In addition to the excellent suggestions already provided. <S> Solder bridges are a fact of life even for the most experienced tech. <S> I find that cleaning the iron tip and then drawing the bridge away in a perpendicular motion to the IC (parallel to the pin/pad) is most effective for me. <S> Oh and regardless of how much flux you use, always clean it afterwards, even the "no clean" flux. <S> Your boards will be much more reliable and long lived. <A> Tip will pick up some excess tin and deposit it onto the next one which could use more. <A> As pointed out, you can clean that up afterwards. <S> Just be sure to put some solder on the desoldering wick first. <S> Lead-based solder has much lower melting point <S> so it's easier to work with. <S> That's obviously quite toxic <S> so don't chew on it or leave it where kids have access to it. <S> Solder joints are obviously pretty inert so handling it is ok. <S> The fumes from soldering that normally reach your nose are not dangerous, it's mostly flux. <S> Heavier than air part is the bad stuff so do not use a fan to blow the smoke all over the room. <S> You would likely benefit from a decent magnifying lamp. <S> 5x lamp makes everything bigger so you can actually see what you're doing. <S> They're not that expensive if you shop around a bit. <S> Don't touch soldering iron tip with the solder wire. <S> Heat the pin/pad and touch solder wire to another part of the pin/pad. <A> Besides flux, I'd also recommend using a soldering heat gun for this type of component. <S> Here are some steps you could follow to solder with a heat gun: <S> Apply flux on component's pads from PCB. <S> Apply tin solder to all pads with soldering iron. <S> Place component. <S> Apply heat with a soldering heat gun (Typically around 15 seconds with 270°C but this depends on the type of tin solder and heat characteristics from the component, which can usually be found on the data sheet).
| Clean the tip from excess solder, add flux and just slide iron over all the pins in one slow motion. Apply flux (Again). Rosin-based mildly-activated flux (RMA type) helps a lot. If you design the PCB by yourself you could make a "solder thief pad" also used in wavesoldering.
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How can RPM be measured in 3 phase induction motor? I was thinking measuring magnetc flux from embedded magnets on the shaft via hall effect sensors but I am not sure. What would be the easiest way to measure speed of the rotor? I am talking about mechanical speed so frequency of the phase currents is not the way to do it. <Q> An induction motor is asynchronous and it is therefore hard to measure the rotational speed based on flux. <S> Because the magnetic flux on the stator does not match the speed or the rotor. <S> It has slip. <S> Motor drivers are able to do it by pulsing short burst of DC on the stator windings whilst analyzing what comes back. <S> This only works when it's not driving. <S> This method is used for flying start. <S> Perhaps you can measure it based on the harmonics measured by the flux sensor if you mount it on the rotor. <S> But this is inconvenient since you'd need slip rings to get the signal out the electronics on the rotor. <S> The reliable industry method is to use a induction pickup on a trigger wheel. <A> The turbine industry magnetises one of the shaft fixing nuts and detects the changing field ... <A> Depending on the accuracy needed, there are plenty of available technologies that already exist; tachometer, encoder, resolver etc. <S> They all involve external components attached to the motor shaft. <S> In the VFD world, "encoderless" speed detection for the purpose of achieving Sensorless Vector Control is done with highly sensitive current detection systems that monitor and filter the stator current flow signals looking for the anomalies that represent the rotor bars passing through the stator magnetic fields and count them to determine the actual rotor speed. <S> You cannot use that for absolute position (i.e. when stopped), but with enough processing power in the drive, it can be highly accurate, as in .001% <A> You can calculate speed based on slip and input frequency if you know the number of poles the motor has. <S> I was at college <S> and it was part of an assignment <S> and it was in the late 1970s <S> so I don't know if this technique will work with more modern machines but it did back then. <A> The only way is to use an encoder mounted on a rotor shaft. <S> The induction motor has a slip, therefore measuring the stator flux won't give you any valuable information. <S> Further, the stator flux frequency is equal to the grid frequency, so you don't even need hall sensor to get the flux frequency, you can measure directly the line frequency.
| I once wrapped a "several turn" coil around a medium sized 3-phase induction motor and low pass filtered the output to get slip frequency i.e. I measured the flux produced by the rotor due to the asynchronous speed of the rotor compared to the fixed stator frequency.
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Splitting strain gauge signal before amplification I have a strain gauge full bridge force sensor with 3mV/V sensitivity and power it with 10 V excitation, giving up to 30 mV output. Normally the output is connected to a 3rd party device's strain gauge input (call it device "X"). That device can't plot the force during measurement, so I want to show the real time data in a different device. That is why I need to get the signal. I have 2 ideas: Option 1:Split the strain gauge's raw output signal, for example with a device like this . Splitted signal 1 goes to device X. Splitted signal 2 goes to my own device. Option 2:Get an external strain gauge amplifier, scale signal to 10 V, and split 10 V signal afterwards, as in option 1. The device X also has a +-10 V input, so no problem here. I think option 2 is better. But I really want to understand HOW the strain gauge output is affected by splitting the signal as in option 1. I know that the splitter is pure analog, no A/D conversion is happening.How does a splitter possibly work? Is it creating a parallel circuit? Then I assume that it would not increase the load on the strain gauge output, hence having no bad effect. But if the splitter would be somehow connected serially only, I think that the total resistance increases and I am going to measure a lower voltage in device X. Is it conceptually much different splitting a strain gauge signal in contrast to a potentiometer signal (for example from a travel sensor)? Could you please explain in detail? <Q> If you just have the one strain gauge and you are alright with whatever gain your selected instrumentation amp affords you <S> I would go with Andy aka's suggestion. <S> There are some really nice old Burr-Brown <S> (TI bought them a long time ago) instrumentation amps for this sort of thing that have adjustable gain via a resistor and can get you a nice large scaled single-ended voltage to measure instead of the small differential voltage. <S> For example: <S> https://www.digikey.com/short/3vpvh7 <S> If you have more than one strain gauge, or you just want more design detail control, you could instead try something like what's suggested here: https://eewiki.net/display/Motley/Analog+Bits+-+Analog+Combinator+Circuit+for+Load+Cells <A> An instrumentation amplifier (InAmp) has a high impedance input and is the turn-to device when measuring strain gauge bridges. <S> It takes no appreciable measurement current from the bridge and you would hardly know it is there. <S> More than likely the 3rd party device uses an InAmp and, there is no difficulty in you using an InAmp in parallel i.e. both measuring the same bridge output. <S> You just have to make sure that the bridge output voltage swing (a few milli volts centred at 5 volts) is compatible with your InAmp <S> i.e. you need to choose an InAmp that can work from your 10 volt excitation supply a bit like this configuration: - You should be able to power your InAmp from the reference 10 volt supply and 0 volts. <S> Most InAmps I use can do this such as the AD8221. <S> I have 2 ideas <S> If you can provide more technical info then great, but if you can't just go for an InAmp. <A> Actually with the splitting device you won't get any good result. <S> You would need a differential input signal DAQ. <S> The best you can try is to use a two channel scope and subtract both signals to get the difference. <S> Keep also in mind that these devices output very small voltage, they do need an amplifier or even more stages of amplifiers. <S> Temperature stability, drift, etc can be an issue. <S> With attaching a parallel measuring device, you will certainly add some noise to the cell output. <S> As for second option: A good stable na fast amplifier is expensive, a good solution is some front end with ADC tailored for the application. <S> Adding external amplifier will certainly make your signal acquisition worse than that you have. <S> You can however use an instrumentation amplifier for strain gauge and second for the displacement sensor and hook to scope to get X-Y curve. <S> When testing press/displacement you should get almost identical results for scope and your DAQ.
| Without digging around looking for specifications for splitter it's difficult to say.
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DC Voltage Polarity and Current Flow? So I'm getting pretty familiar with Electrical Theory but one issue keeps bringing me problems. In a DC circuit, there is a negative (-) and a positive (+). I am aware that in electron flow theory, current flows from the negative (-) to positive (+). See Figure 1 . But in a car battery, the positive terminal is considered the hot lead and the negative terminal is ground. If you take a wrench and connect the negative (-) terminal to the metal chassis of the car, it won't short, but when you short the positive (+) terminal to the metal chassis, it does. See Figure 2 .My question is.. How is it possible for the positive terminal of the battery to be hot when, there is positive charge on the terminal? ? If electrons are supposed to flow from negative(-) to positive(+), wouldn't the negative terminal be considered hot and not at ground potential? Also, when the positive(+) terminal is shorted to ground, is that positively charged terminal just allowing negatively charged electrons to flow from the ground? Please Help! simulate this circuit – Schematic created using CircuitLab <Q> Electrons are flowing from negative to positive when there is a connection for them to flow. <S> You can also think of it as "holes" (or the absence of electrons in an atom) flowing from positive to negative. <S> Because of historic reasons, we generally think of conventional current, or the "holes", rather than the actual electron current. <S> There is also no magical node called "ground. <S> " We simply assign a point that is easy to keep track of to refer to as ground. <S> In a circuit, you could refer to the positive terminal of the battery as ground, and simply be working with negative voltages (though it would likely cause some confusion for others interpreting your work). <S> Likely in the case of shorting the battery to the chassis of the car, the negative terminal of the battery is already connected to the chassis. <S> Possibly this is done for some sort of EM shielding (vehicles are inherently very noisy environments). <S> When you make another connection from the negative terminal of the battery to the chassis, no current flows. <S> When you connect the positive terminal, current flows through the chassis of the vehicle to the negative terminal. <A> In your figure 2, there will be no current flow, as there is no complete circuit - the negative terminal of the battery is not connected to anything. <S> "Ground" has no magic properties - it is (for most circuits) <S> just the point in the circuit that we choose to call "Zero Volts". <S> Please try to forget about electron flow <S> - most people speak in terms of conventional (positive) current, although we are aware that, in most materials, current is actually a movement of negative charges. <A> Ok Thanks! <S> , I get it now. <S> The reason why the positive terminal would be considered "hot" is because the negative terminal which is connected to the metal chassis at 0 volt potential will short to the positive terminal if something metal were to allow electrical continuity. <S> AND yes, I now understand that positive voltage is just the potential difference between a positively charged area and a negatively charged source with the voltage "pulling" the electrons from the negatively charged source to the positively charged area. <S> simulate this circuit – <S> Schematic created using CircuitLab
| You are correct that the positive terminal just "pulls negatively charged electrons."
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Choosing a PCB Manufacturer/Assembler for the first time I've just finished designing my first PCB, and I'm about to send out a batch of emails for quotes from manufacturing/assembly shops. My design is super simple: the only components are 7 identical SMD resistors, a 5v voltage regulator, a decoupling capacitor, an ATMEGA328P-PU and 16 pads for off-board wires. I'm looking to do an initial run of 50 boards, with the possibility of ordering more later. I'd also like to receive a prototype to verify that everything works properly before putting in a purchase order for 50 more boards. I'm very new at this, and Google provides an overwhelming number of shops to choose from. This post seems to have some good advice for ensuring quality, however it seems that such a small, simple board wouldn't fall prey to many quality issues. Beyond the advice in the above thread, does anybody have any recommendations for what I should ask in the email? And further, if anybody has any suggestions for particular shops to reach out to, I'm all ears. Less expensive is better, but not at the risk of receiving defective boards. Edit: Also, would you anticipate that it would be dramatically less expensive to just solder the boards myself? And if it's worth it to have the PCBs assembled, is it more or less expensive to let the shop source the parts themselves? I'm not sure if they would get a better deal on the parts, or if they would be overcharging me for them. <Q> Depending on how time-critical this is, go to China. <S> There are loads of Chinese places that will do this for a much better price than anywhere else. <S> The only drawback is it takes more time. <S> I have personally used iTead and fusion and have had no problems whatsoever with their finishing. <S> The only issue I had was if the silk screen is too small, it gets a bit illegible, but that has only ever happened when doing a space critical PCB with 0402 components. <S> It is up to you to do the math for if it is worth doing the soldering yourself or not. <S> As for what you should ask them, ask how much it is for different size batches (some do 'price per PCB' reductions every 50, some every 100 etc) and ask them what quality checks they do, and what process controls they have in place to avoid problems. <S> Ask minimum drill size, for vias, trace to trace width etc. <S> You will have to ask them how much it is to source components, they should be able to give you a quote on that, and then you can check yourself to see if you can get it cheaper. <S> It is up to you whether you want to go local for speed of delivery, or afar if you can wait, but do expect quite a price difference if you decide to keep it local! <S> EDIT <S> Both of the places I recommended also do very cheap prototyping, usually in small batches (10) where you can verify your design. <S> For prototyping, it is most likely worth assembling yourself so you can do anything you want to do during assembly to help testing. <A> You can get really fast quotes from OSH park <S> I believe they are an American company which should make things simpler for you. <S> They are good for simple boards. <S> Worth a try <S> and you can also get a small initial batch of 3 to test out for fairly reasonable cost. <S> I would say to source the parts yourself if you know your requirements and are confident enough. <S> PCB companie will most likely charge a premium for the service. <S> The type of questions to ask: Their minimum via size, minimum track to track clearance etc. <S> most PCB manufacturing firms have this sort of information on their website (design rules). <S> Ask them directly if you can do anything to reduce manufacturing cost, one PCB manufacturer helped me with this before. <S> If the PCB is fairly simple I would assemble my self for cost reasons if you are confident enough. <A> I have recently been in same situation. <S> I reached out to my network and some frequent name came back: <S> PCBway, seedstudio, eurocircuits and dirty PCB. <S> Now the initial cost of PCBA is quite high, so what i did for prototyping was buying PCBs and stencils at PCBway. <S> Then I did the soldering and assembly my self. <S> Now I'm moving to the next step which is production of a batch <S> and I'm probably gonna go with either PCBway or seedstudio. <S> Sparkfun has a great tutorial on this https://www.youtube.com/watch?v=WDIqtGMROjM Regarding the turnkey solution. <S> If you have a large amount of chips and buy the directly from the manufacture you often have a personal contact person which can help you a lot. <S> In my opinion if I had low volume - just get the full turnkey solution. <S> If the volume grows I would supply the key components such as MCU's my self and let them deal with resistors, capacitors etc... <A> If I'm prototyping, I usually use someone like itead.cc or oshpark.com . <S> I see you're from the US <S> so i recommend oshpark as they are based in the US and have pretty quick delivery. <S> I order a couple of bare boards (depends on size, but you can get 3 for ~$15 usually including shipping) and hand solder them myself just to check that my layout is correct and that I've not made any silly mistakes. <S> When it comes to emailing the manufacturer. <S> You need to make sure you have a BOM ready for them to and to send it along with your gerber files. <S> When making a BOM i usually list all the important parts with RS or Farnell numbers to indicate that I want that specific part. <S> As for unimportant things like SMD resistors and caps I just list the value and size I want and include a couple of lines at the bottom of my BOM: <S> SM Components: <S> Resistors to be TE Connectivity CRG series. <S> Ceramic Capacitors to be KEMET X7R, or equivalent RoHS compliant parts, unless otherwise stated. <S> For generic IC parts any equivalent RoHS compliant part may be used. <S> The reason I do this is that PCB manufacturers typically do what you tell them, <S> if I list an RS or Farnell part number, they'll just buy that at whatever price it is. <S> Leaving it open to them means they'll try to obtain the equivalent part from their own suppliers at a much cheaper price usually. <S> Also when asking for your quote ask for the 'bare board' price as well as the assembled price. <S> You can then decide for yourself if it's worth soldering them yourself or not.
| For something that sounds quite simple with few components, and only 50, it may be worth doing it yourself, and if you do larger batches in the future, then look at getting somewhere else to do it for you.
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Opamp input voltage divider creates wrong voltage - what could be the cause? I'm using above 1:2 differential opamp, when 72V applied to the 30:1 voltage divider, voltage across R1 is only 1.727V instead of 2.4V, opamp output is around 3.4V. Opamp is LMC6062IN/NOPB http://www.ti.com/lit/ds/symlink/lmc6062.pdf I think it's because of bias current, any suggestion on solution? Follow up question: Same circuit below but now this time 0V does not connect to circuit main GND, so there is not second voltage divider but i am seeing around 1.5V across R1 simulate this circuit – Schematic created using CircuitLab Follow up question #2: simulate this circuit Switch on both side of V1, top and bottom circuits are identical differential opamp circuit, bottom one is before switches and connect to v1 directly, top one is after two switches. When SW2 is open, SW1 is closed: voltage across R3 is 1.4v, output 2 is 2.83v.voltage across R23 is 0.532v, output 1 is 1.08v.Seems like R3 is in parallel with a 60K resistor, does that mean two op amps' inverting and non-inverting pin are tied together? I am using a dual op-amp. <Q> ... <S> when 72v applied to the 30:1 voltage divider ... <S> It's not 30:1, it's 31:1. <S> \$ V_{OUT} = V_{IN} <S> \frac {10k}{200k + 100k + 10k <S> } \$. ... <S> voltage across R1 is only 1.727 V instead of 2.4 V. <S> But you have a second voltage divider in parallel with R1. <S> "R1" is now effectively 7.5 kΩ. <S> The voltage at R1 will be \$ V_{OUT} = <S> V_{IN} \frac { <S> 7.5k}{200k <S> + 100k + 7.5k} <S> = 72 <S> \times 2.44\% <S> = 1.75 \; V\$ and the voltage into the non-inverting input will be 2/3 of that = <S> 1.17 <S> V. <S> Im (sic) <S> thinking its because of bias current ... <S> I'm thinking it's because of biased thinking ... <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> OP's second attempt. <S> In the standard inverting and non-inverting op-amp amplifier configurations the output will adjust until the difference between the two inputs is very close to zero. <S> For practical analysis they are connected together as shown in Figure 1. <S> Now it is clear that you have 20k in parallel with your 10k of R1 giving 6.67k. <S> Running the calculation again the voltage at R1 will be \$ V_{OUT} = <S> V_{IN} \frac {6.67k}{200k + 100k + 6.67k} <S> = 72 <S> \times <S> 2.17\% = 1.56 \; V\$ <S> Your question 2 circuit is a bit of a mess and I don't think I could analyse it for you. <S> I doubt that this is necessary. <S> Figure 2. <S> Just use the differential amplifier itself. <S> For example, setting R1 and R2 to 100k and Rf and Rg to 1k would give you a differential amplifier with a gain of 1/100. <S> Your 72 V input on the original circuit would then give an op-amp output of 0.72 V (if it can swing that low). <S> Why complicate things? <A> Another problem you may run into with your circuit is the input common mode range of the opamp. <S> For this device the input must not go closer to the positive supply rail than 2.6v to guarantee correct operation. <S> If the supply rail is 5v this is equivalent to 2.4v relative to ground - if your supply rail is slightly low (it won't be exactly 5v) <S> the max voltage may be lower. <S> A typical device will work better up to about 3v. <S> From the calculations done by @Transistor you do have margin but it is something to be aware of. <A> This is a classic mistake, as you are assuming that your difference amplifier circuit is acting as a high impedance load to your signal source and front end circuitry. <S> It doesn't, there is a (relatively) low impedance path through R4 and R9, as well as through R3 and R8. <S> This is similar to how non-inverting amplifier circuits (such as simple voltage followers) act as high-impedance buffers, but inverting amplifiers do not. <S> Transistor already did the resulting additional voltage divider math for you, but here is another example with an LTspice and Mathcad calculation comparison for an inverting amplifier circuit. <S> https://eewiki.net/display/Motley/Analog+Bits+-+Analog+Combinator+Circuit+for+Load+Cells#AnalogBits-AnalogCombinatorCircuitforLoadCells-SufferingFromaLackofBuffering <S> Also, I'm not sure why you are using a difference amplifier at all since you have one side connected to GND anyway. <S> I of course don't know all the details of what you are trying to do and why, but at first glance I would think you could just use a simpler non-inverting amplifier circuit.
| I think you are trying to combine a traditional voltage divider on an non-ground referenced source and feed this into a differential amplifier. There are a number of devices with improved input common mode range - often referred to as "Rail-to-Rail inputs" - I quite like the LMP7701 range that have very good offset voltage, extremely low bias current and can operate with a supply up to 12v.
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Help to Identify this IC chip sorry this is my first post on this forum. I've spent a fair while combing the internet looking for advice and have not been able to find any information. The problem:So I've converted the engine in my bmw e30 to a 6 cylinder engine from a 4 cylinder engine and the tacho for the instrument cluster is reading incorrectly. It is to my knowledge that the signal needs to be multiplied by 3/2 to achieve an accurate result. Originally my plan was to implement some kind of frequency divider to achieve this. However, I found out that the coding plugs for 6 cylinder cars were identical in construction to the 4 cylinder cars and when plugged into a 4 cylinder cluster would provide the desired result. Unfortunately, these coding plugs are out of production and are very difficult to source. The chip itself consists of one 8 pin IC chip. I've tried searching the model number printed on the top but it returns no results. The chip can be seen below: My goal is to clone the chip by having a board printed and finding an appropriate IC chip to build the circuit upon. I am looking for advice on how I can identify this IC chip and if someone can identify it whether they can point me in the direction of an off the shelf solution. Thanks, appreciate it ^^ Here are two images, however, these images show the 4cyl chip and not the 6cyl. Yeah, I know there's thousands upon thousands of IC's. I was just hoping that someone had come across a similar situation. I've worked on black box circuits in the past and have been able to identify various filters from generating bode plots of the circuits at different frequencies but I'm not sure how to go about this for an IC chip. I was going to start by probing the connections on my car to figure out the inputs and outputs of the chip as well as constant power and ground. I could also look up the original wiring diagrams from BMW in some hope can figure out what each pin does by tracing back its connection. Any other advice on how to go about identifying this IC would be greatly appreciated. UPDATE So I managed to get in contact with the seller of the clone chip and he was kind enough to point me in the right direction, really awesome guy. Here's what he said: The 8 pin IC I used is a microcontroller that acts as a coder - a device that uses a data set for scaling instrument cluster gauges. It is not an analogue signal converter. Does anyone have any experience with coders for use with analogue guages? <Q> I think probably a custom fabrication from Hughes - back when there existed such a manufacturing. <S> While identification from this schematic isn't very likely, it can help you to at least rule out possibilities: <A> Maybe you should think about emulating the part instead of finding a exact replacement. <S> If it's just a memory chip of some sort <S> and you have both parts to compare, then maybe you can simply drop a logic analyzer on it, figure out what the protocol is, and program a small microcontroller to emulate it. <A> Could be a clone of http://www.alldatasheet.com/datasheet-pdf/pdf/177121/TI/TL087.html . <S> If it is an op-amp (probably is, considering you're talking about some sort of amplification), you could try it out. <S> (I'm assuming your tacho is for indication purposes only) <A> The shown pictures show the BMW E30 Coder Stick. <S> By changing the engine from 4-cyl to 6-Cyl, it's sufficient to replace the coder stick to a version for BMW 320i (part # 62111381635) or 325i (part # 62111381683). <S> Both are still available at BMW. <S> The IC <S> HML-087 is a serial 8-Bit EEPROM which are very very difficulty to re-program.
| From what I have found, which is not much, there seems to be some BMW-Refurbishers-Community consensus that HML 087 is an 8-pin EEPROM.
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Why VDDIO is more than VDDcore supply in VLSI/ integrated chips? I have a question on different power supplies in a integrated circuit. I have seen VDDIO supply is more than VDDCore. If the input signal is more than the power signal, won't it affect the device?? Explain the purpose of this variations..?? <Q> I have seen VDDIO supply is more than VDDCore. <S> This is required, as the smallest transistors (and other structures) in modern chip technology cannot support higher voltages than VDDCore limits. <S> These structures are so tiny that they would break when supplied with VDDIO. <S> But you need these tiny transistors in the core because they are the "fastest" and consume the least amount of chip area. <S> The higher VDDIO is usually required to be able to connect to some standard busses (say RAM or mass storage). <S> The internal chip structures around the IO-pins are designed for these voltages and are thus quite a bit larger compared to the core. <A> Large integrated circuits usually have multiple power planes to meet the requirements of different parts of the circuitry on the chip. <S> Splitting core and <S> I/O supplies onto separate rails is quite common. <S> This is done so that the core can be built with smaller transistors that offer the best possible performance characteristics (higher density, smaller parasitics, lower power consumption, faster switching speed, etc.) <S> while the I <S> /O circuitry requires larger transistors and higher voltages to drive signals off-chip and interface with the outside world. <S> It also enables independent adjustment of the core voltage, enabling power-saving features such as DVFS (dynamic voltage and frequency scaling), which is common in low power devices such as cell phones. <S> Using an adjustable voltage regulator for <S> the core voltage enables lowering both the core voltage and core clock frequency to save significant power when high performance is not required. <S> Separate power supplies can also be used for noise reasons - sensitive analog circuitry including PLLs, serializers, deserializers, RF interfaces, etc. <S> will usually also get dedicated power pins to reduce noise either into or out of these components. <A> Core voltage and IO voltage are driven by different design considerations. <S> Core voltage is driven by achieving the desired performance while avoiding damage to the tiny transistors and consuming as little power as possible. <S> As process nodes get finer the optimal core voltage gets smaller. <S> Power consumption is still a concern with IO, this is why some high speed IO does use low voltages. <S> However considerations of compatibility and noise immunity are often equally if not more important. <S> Having multiple voltages on an IC does indeed complicate the design. <S> When going from a lower to a higher voltage region care is needed to ensure the voltage is high enough to correctly switch the transistors. <S> When going from a higher voltage to a lower voltage region care is needed to ensure the input transitors will not be damaged. <S> The implications of startup and shutdown also need to be considered (for this reason some ICs have pretty specific requirements for power supply sequencing). <S> But ultimately that complexity is the price we pay for progress.
| It is also possible to have multiple different IO voltages connected to different sets of IO pins, enabling a chip to interface with other circuitry that is running with different voltage levels.
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Using a 220ohm resistor in a mains circuit? I work in a charity shop where our display cabinet with a rotating glass shelf has stopped working. The electrics are clearly visible, and the fault is that a 220 ohm resistor between the live lead and the motor has broken. This is a 'normal' small resistor with no markings other than the resistance. Can I replace the broken item with a standard 220 ohm resistor? <Q> Yes, But... "Normal" resistors do come in different sizes, voltage ratings, tolerances, and power dissipation ratings. <S> You need to replace the burnt out one with one which matches. <S> Resistance <S> : Replace 220Ω with 220Ω. Easy <S> Tolerance : A resistor which is used in series with a motor is unlikely to have a fine tolerance. <S> To be on the safe side, you should replace with the same or better. <S> E.g. 5% could be replaced with 5%, 2%, 1% etc. <S> Look at the last colour band to see what the tolerance is. <S> Voltage rating : There is no easy way to check the voltage rating of the resistor. <S> Unless you have a particular reason not to, just choose a voltage rating well above the mains voltage in your country. <S> Power rating <S> : There's no easy way to check the power rating of the burnt out part either. <S> Most resistors of a given size have the same power rating though, so if you replace like with like, that should be fine. <S> Even better, you could go for a larger resistor with a higher power rating, that way it might not burn out again. <S> Size : <S> Provided you can fit it in, there's probably no harm in picking a larger resistor. <S> The main change is the power rating. <S> Be careful about clearances though - the resistor may have lots of space around it to make sure it doesn't arc to another component or to the case. <S> Don't put a larger resistor in if you can't maintain that spacing. <A> If you are sure it is the resistor that's broken, you can. <S> Remove the broken resistor and replace it with a new one. <S> However, a resistor breaking seems odd to me. <S> I would suggest measuring the actual resistance of the resistor with a multimeter. <S> Do this when it is removed, otherwise you will measure other resistances of the circuit too. <S> If it is broken, and you replace it, it could be it will break again soon, if the issue is caused by another component. <S> Since resistors are really cheap, you can certainly replace it to see if it fixes your problem. <A> Can I replace the broken item with a standard 220 ohm resistor? <S> The broken resistor can still be a symptom of more serious failures in the motor or something else. <S> Even if from your description, it appears to be a simple mechanical problem. <S> This goes with the usual disclaimers: Tampering with mains-powered electric and electronic devices is always hazardous. <S> Depending on where you are located, it may be illegal to operate devices that have been repaired by anyone else than a certified professional, and bring about nasty liability issues for you and the shop owner if the gadget starts a fire.
| You should replace it with a resistor that has the same rating (resistance and power, voltage) as the damaged component.
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How to find status (on/off) of any applicance ( light, fan, etc.) connected to an electrical circuit There are 4 appliances connected to an electric circuit, how to determine which applicance is on/off and pass that information to an electronic device in binary form. The idea is to make an electronic device which when attached to the electrical circuit determines whether the applicance is switched on or not and pass the information in binary format <Q> Figure 1. <S> CR-Magnetics CR2550 current indicator. <S> The CR2550 series Remote Current Indicators are a low cost method for providing a visual indication of electrical current flow. <S> The indicators are factory calibrated to provide a preset turn-on point. <S> The value of the turn-on point is determined by the customer and specified in the part number. <S> Attached to the transformer is a high efficiency, bi-polar LED that illuminates when the current is above the turn-on point. <S> The indicator is available as standard with 11 inch long leads and a red LED indicator. <S> designed as a low cost method for providing a visual indication of electrical current flow. <S> The current-carrying wire is routed through the window opening in the current sensing transformer. <S> Attached to the transformer is a high efficiency, bi-polar LED that illuminates when the current is above the turn-on point. <S> The indicator is available as standard with an 11 inch long lead and a red or green LED indicator. <S> Turn-on Point: 0.75 Aac RMS <S> Maximum Continuous Rating: 20 Aac RMS Frequency: 50 to 400 Hz <S> Operating Temperature Range: <S> -30 <S> C to +60 C Storage Temperature Range: <S> -55 C to +85 <S> C <S> LED Indicator Type: <S> Bi-polar <S> , Red/Red or Green/Green LED Mounting Hardware: <S> Plastic one-piece, press-in lens supplied From the spec you can see that for this model you would need to get the ampere-turns through the coil to between 0.75 and 20 A when your load is on. <S> For example, with a 0.25 <S> A load you would need four or five turns of your current carrying conductor through the transformer to ensure reliable switch-on. <S> There are a few application notes on the linked page. <S> You can glue the LED to an LDR to signal to your micro-controller or remove the LED and replace it with the input of an opto-coupler. <S> Figure 2. <S> If converting to an opto-isolator it would need the same LED arrangement internally. <S> Note that the spec says the LEDs are bi-polar which means that there are two back to back in the one package for AC. <A> You could try a Hall Effect sensor attached to the appliance power inputs. <S> These are non-invasive sensors so you would not have to wire them into the circuit. <S> Here is a popular one from Sparkfun . <S> Many of these are analog, so you would have to set up an analog-digital converter or some sort of logic circuit to convert to binary. <S> EDIT: as pointed out in comments, you'll need to pull out one of the mains in order for this to work, otherwise the currents will cancel each other out. <A> You could use a Non-Invasive Current Sensor like the one sparkfun has: https://www.sparkfun.com/products/11005
| The CR-Magnetics devices use LEDs as shown on the right.
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What is the name and purpose of the following component? I'm trying to repair a small electronic door bell. I was wondering what is the name and purpose of the white box component ? It seems to be just a wire with some white body around. Is it a resistor or a fuse? My guess is that 4W22R means 4 watts 22 ohms but I'm not sure. How can I test that component is still working ? AFAIK the ground (the black wire) goes through the 400V capacitor first, then it has to go through that white component before going to the bridge rectifier. EDIT : I found out the issue : the zener diode is dead and acts like a wire (it's always closed). Because of this, the main 400V capacitor is never charging and don't provide power to the rest of the circuit.I disoldered the diode and tried to put 5.0V directly between the solder joints using a external power supply, it works again !Thanks Marko Buršič for pointing out the zener diode. <Q> I was wondering what is the name and purpose of the white box component ? <S> It seems to be just a wire with some white body around. <S> Is it a resistor or a fuse ? <S> It's a resistor, perhaps wirewound. <S> Look at the PCB where that component is soldered. <S> Its component designator will probably be a number prefixed by "R". <S> You can read more about that type of power supply in this previous question and elsewhere. <S> My guess is that 4W22R means 4 watts 22 ohms <S> Yes. <S> How can I test that component is still working ? <S> With power completely removed and the capacitors on the PCB discharged , you could try measuring its resistance with a DMM and see if you get a sensible result. <S> Since I expect it to be in series with C25 and no other components in parallel with it, then you should measure 22 ohms +/- <S> 10% 5% <S> The lack of discolouring gives us no indication of overheating (although doesn't disprove that it has occurred). <A> Its a resistor of 4 W having 22 Ω resistance and the letter J means a tolerance of + or - 5%. <S> A multimeter will give its resistance and if it's faulty it will give no resistance. <S> The capacitor and the resistor drop the voltage (transformerless power supply) before it's converted to DC by diodes. <A> 22R 4W as already established <S> WARNING <S> This sort of power supply can kill you many seconds before you realise that you are dead. <S> ALL parts of the whole circuit MUST be treated as if they are art mains potential when mains connected as sometimes they may be. <S> Zener failure indicates that the circuit was "designed" about as badly as such circuits usually are and as all should be considered to be. <S> They may die due to mains surge, component under-rating or Murphy, and 'other things'. <S> Had the zener failed o/c, as happens less often than s/c, but often enough, modified mains voltage would have been applied elsewhere and das spitzen sparkem, bad smelzen and flashing lights maybe. <S> Briefly. <A> You got it. <S> Try to measure the voltage on zener diode (glass package in the middle position). <S> It's a capacitive PSU, not an SMPS, so be careful it's all live. <S> You can swap L and N terminals to make the measuring a little more safe. <S> In one combination the entire circuit is live, use the voltage tester tip. <S> In the other combination of L,N wires connection, only the input capacitor, resistor,... are live, all the other components are a Neutral potential. <S> So find a safer combination, first. <A> I am not that familiar with the markings and conventions thereof, but these ceramic pieces usually indicate resistors that need to withstand and dissipate a substantial amount of heat. <S> The latter of course speaks about a high wattage characteristic. <S> 4Watts is a high power rating for this small resistor.
| In combination with that capacitor C25 , this is a " capacitive dropper " from the incoming mains voltage.
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Two 12v batteries in parallel 7ah and 20ah I have two 12v batteries, 7ah and 20ah. Can they be connected in parallel without damaging the 7ah battery? They will be charged separately. <Q> If you simply parallel connect the two batteries for discharging, then the combined Ah capacities do not add. <S> This will typically cause the higher voltage 20 Ah battery to begin charging the 7 Ah battery. <S> Whichever way the current draw occurs, the action of one battery charging another battery wastes energy in the form of heat. <S> Since part of the energy of one battery is being used to charge the other battery, the combined capacity will likely be more than 7 <S> Ah but less than 27 <S> Ah. <S> The exact answer is not known without an extensive analysis of the load characteristics and the specifications of the batteries. <S> By adding some electronics to the system, it may be possible to realize an overall higher Ah capacity compared to simply paralleling the batteries. <A> Yes they can be connected in parallel. <S> Different Ah of batteries means they have different ampere hours, thus when used alone a 7Ah battery will discharge sooner than a 20Ah battery when connected to the same load. <S> They can be charged from the same source of ~14V simultaneously. <S> If voltages were different then the higher voltage battery would act as a source while the lower voltage battery would act as a load which is not recommended. <A> As a general rule I never put batteries (of same nominal voltage of course) in parallel: they might have different State of Charge (SoC), different internal resistance, etc. <S> and the situation will turn out into a battery charging the other, rather than having the current all going where it is required to be (... your load). <S> To replace batteries on the spot and to protect against reverse polarity, I always use a diode in series, possibly Schottky to reduce voltage drop and heating. <S> simulate this circuit – <S> Schematic created using CircuitLab D1 and D2 may be different components, or the two legs of a common-cathode 2-diode component (see below). <S> With a normal PN diode and 10A flowing, forward voltage drop of 0.9-1.0 V is common, with 10W dissipated across the diode (!), and 1V less on your load. <S> Schottky diodes have a much lower voltage drop (about 50%): 1 leg of a common-cathode 2-diode MBR2045 at 10 A has a 0.58V voltage drop; with the 2 legs in parallel (5A each) and some self-heating that brings temperature around 50-60°C, 0.45V can be achieved = <S> > <S> 4.5W. Vishay VS-48CTQ060S-M3 can reach 0.37-0.38V, thus 3.8W! <S> A connector suitable for current rating is always a good idea
| As current is drawn from the paralleled batteries, the voltage of the 7Ah battery will typically drop more quickly than the 20 Ah battery. When both are connected in parallel they become one 27Ah battery of 12V.
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(How) can IEC-specification miniature circuit breakers be interlocked? A relatively common way of implementing a transfer or bypass/isolation function in North American electrical practice is to use mechanical means to interlock the handles of multiple circuit breakers or molded case switches so that they can't be both on at the same time. This can be done using a piece of metal retrofitted to a panelboard/loadcenter, or as part of a factory transfer switch assembly that uses circuit breakers or molded case switches as the electrical switching means. However, the devices used to do this are designed around the North American busbar-mounted types of breakers. Breakers designed to the IEC standards typically mount to DIN rail instead, using a cable-in/cable-out construction, with a single row of breakers vs. the opposing-columns construction typical in North America. Is it possible to mechanically interlock this style of breaker (i.e. breaker A and breaker B cannot be on at the same time) without violating its approvals? If so, how is it done in practice? <Q> I <S> m German these are called Lastumschalter (loaded-operation selector switch). <S> Smaller ones are available for DIN rail. <S> They are never combined with overload or short circuit protection. <S> Hager HIM304 40A <S> Lastumschalter <A> breaker A and breaker B cannot be on at the same time <S> This is a peculiarly American arrangement. <S> In the UK, I can't think of any reason I'd ever need to ensure that switching off my upstairs 240V lighting circuit also turned off my 240V cooker circuit. <S> If you have a backup generator or a solar installation, you have a transfer switch to handle that separately from your main panel. <S> If you need to isolate both neutral and live, you generally have dual-pole switch near the appliance (e.g. electric shower, alarm systems, etc). <S> Thats a switch not a breaker and it isn't in the main panel but typically wall mounted like most switches. <S> You can get dual-pole MCBs <S> but they are for isolating both live and neutral (not two phases) <S> and I believe are only needed under special circumstances (e.g. TT system ?) <S> I'm not familiar with industrial arrangements, presumably they use multi-pole breakers where needed. <S> Perhaps these are designed as such rather than having linking kits added on? <A> I wouldn't expect to see this sort of mechanical interlock in a "normal" North American breaker panel. <S> It is more likely to be seen in a custom assembly housing two breakers, and the mechanical interlock. <A> You can use motor reversing contactor. <S> It has mechanical interlock, additionally you can make also an electrical interlock with the separate contacts wiring. <S> You have to add a selector switch and circuit breaker, that is the same for both sections. <S> EDIT: <S> In a similar way, you can get two contactors + additional mechanical interlocking kit (for that series of contactors) and you wire them on separate circuit breakers. <S> Motor reverse contactor is simply a kit with two contactors, mechanical interlock and jumper busbar. <S> IMO, you won't find the device you are looking at, because the circuit breakers are protective devices, they are not meant for switching purposes, while switches, contactors are switching components.
| As long as the breaker handles are easily accessible, and move a reasonable distance between "Off" and "On", it should be possible to arrange a mechanical interlock - may need some special brackets around the breakers to support the interlock slider. I suspect this requirement most often arises in USA homes due to the use of split-phase supplies where alternating breakers are on separate phases and you can combine adjacent 120V breakers to produce a 240V circuit.
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Connection of dusl 12v NiCd battery packs I have the 12v battery (it is actually 10 Ni-Cads in series), and, of course, it has a red wire coming from "+" and a black wire coming from "-". However, on the schematic I am confused about whether the red wire should go to terminal 24, or the black wire. (It's the top line, with terminals 24 and 20.) I was changing the batteries and didn't take a photo of the connections first, and didn't see this battery was -12v, not +12v. Stupid... Appreciate any help. This is probably very simple for anyone here, but I don't want to make mistake and bolax this instrument. <Q> Black wire need to be connected with terminal 20 for negative -12V battery. <A> For whatever reason, your device requires a dual-rail supply. <S> This is typical in, for example, audio power amplifiers. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Audio amplifier with dual supply allows generation of AC signal for the loud-speaker without any DC offset. <S> However, on the schematic I am confused about whether the red wire should go to terminal 24, or the black wire. <S> (It's the top line, with terminals 24 and 20.) <S> Figure 2. <S> Coloured-up battery connections. <S> Hopefully the explanation of the reason for the dual supply above will give you the ability and confidence to finish the job. <A> (24) = <S> 0V (GND) = <S> Red + (20) = <S> -12V <S> = <S> Black - Check cell balance too or cell temp. <S> be careful.
| Since it's a -12V battery and based on the schematic, red wire should be connected with terminal 24 which is connected to the ground terminal.
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Is it possible to overcome a transistor's current limit by running it at lower voltage? Let's say I have a transistor which is rated for maximum 800mA current at 50V. Is it possible to have it running at 1000mA & 5V for example? Edit : please provide me some details in your answer by using the BC337 as sample. Here is its datasheet https://www.onsemi.com/pub/Collateral/BC337-D.PDF <Q> Not only can't you do that, but very likely you can't run it with 800 mA and 5 V at the same time either. <S> Typically the voltage rating is the maximum voltage the transistor is guaranteed to block in the cut-off state. <S> And the current rating is the maximum current in a fully saturated state (BJTs) or in the triode region (MOSFETs). <S> When operating in the forward active region (BJTs) or saturation region (MOSFETs), the maximum power limit is likely to give a lower limit on current & voltage than either of those limits. <A> If it's rated for 800mA, that's at any voltage (within the rated voltage limits). <S> The specifications are the conditions under which the manufacturer guarantees that it will work. <S> If you're willing to exceed the specifications, having a better heat sink is going to do more for allowing the transistor to run at higher current without damage than a lower voltage will. <A> Based on the edited question, now that we are talking about the BC337, I will make comments specifically about that part. <S> First of all, the section at the beginning labeled "Maximum Ratings" should not be used for design. <S> In other words, if Vceo maximum is given as 45V, you should avoid exposing the IC to 45V. Maybe stay at 40V or below. <S> Also, note that you should avoid all of the maximums. <S> You have noted the voltage limit and the current limit, but you should also note the power limit. <S> In this case, there are two power limits, which give different conditions. <S> One applies when Ta (ambient temperature) is 25C, and the other applies when Tc (case temperature) is 25C. For our purposes, we can ignore the Tc limit because it is essentially impossible to keep the case at 25C when the transistor is dissipating a lot of power. <S> The second thing to note about the Ta limit is that it only applies when ambient is 25C. See the note which says the limit must be derated by 5mW/C. <S> This means that for every degree above 25C, you must subtract 5mW from the limit. <S> For example, if the ambient temperature is 45C, that is 20 degrees above 25C. <S> So the limit will be 625mW - (20 * 5mW) = <S> 525 mW. <S> The power dissipation is Ic <S> * Vce + Ib <S> * Vbe. <S> So you will have to calculate that with your specific operating conditions and see if it works out. <S> But if you actually need something like 800mA, I suggest you look at using a MOSFET. <S> The DC gate current on a MOSFET (analogous to Ib) is negligible, and the voltage drop from drain to source (analogous to Vceo) is typically much smaller, so the MOSFET doesn't need to dissipate as much power. <S> Hope this helps.
| There is of course a chance that any given 800mA-rated transistor may work for a while, or possibly even forever, at 1000mA, but you can't count on it.
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What are the risks removing sleeve off the aluminium electrolytic capacitor? I have a device with very strict dimensions; the space electrolytic capacitor will be installed in is exactly 5 mm. I have tried several manufacturers, and learned that some tightly fit, some do not at all. Looking to various datasheets I see dimensioning declared as 5 + 0.5 mm, or 5 ± 0.5 mm. While -0.5 mm would be great, so far I have a bunch of caps which are more than 5 mm. Capacitor is mounted horizontally on the board, thus board at one side, and plastic casing at another (with space of 5 mm). I am looking to ways how to put these caps in. There're actually two ways - Increase space where cap would be installed, somehow cutting/using abrasive cloth/heat to melt or anything else on the plastic casing. This plastic casing is transparent, thus working on it will be visible from the outside of the device, and most probably will look ugly; decrease diameter of the capacitor, and the obvious idea is to remove sleeve off it, it must save several tenths of the millimeter. Is the second option a good idea from your experience, or I would better work on casing? Update: thank you for suggestions so far. The capacitor is aluminium electrolytic, here's the set of datasheets: Panasonic , Jamicon and Hitano . Boards are already produced, and must be of 1.6 mm thickness due to edge connector specification. The peculiarity of the design is that capacitors are mounted this way from the both sides, but only one side does not fit. Caps are 10 uF 25 V bi-polar , not available in 4 mm diameter body. Here's the pic. Update: I use bi-polar capacitors. The question is NOT about finding appropriate part , 5 mm diameter caps are already ordered and on their way. Question is about if I can remove black sleeve with marking off the cap on the picture without adverse physical and electrical effect. <Q> Removing the heat shrinking from the cap has no electrical downside <S> but it is labor intensive. <S> I would first consider routing the capacitor body shape out of the PCB, protrude the cap through the annular, and solder the cap from the bottom of the board - gull wing style. <A> TDK makes through-hole variants like " FG series " to satisfy the board aesthetics. <S> This is not something like input mike amplifier, it is the headphone output stage, so any possible piezo-effects would not matter. <S> The one <S> 25V <S> 47uF <S> FG26-type has 5mm pitch and only 3.5mm tall. <A> Solving by proof by counterexample: Caps are 10 uF 25 V, not available in 4 mm diameter body. <S> Not true. <S> Use a 4mm or 4.5 <S> mm diameter cap instead , e.g. Panasonic EEA-FC1E100. <S> Generally, octopart.com is a nice search engine.
| You need to substitute your electrolytics caps with ceramic.
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How Long Will This Battery Keep My Fridge Powered For? I have an interesting mathematical/electrical dilemma and I'm not sure where to start. I am considering buying this fridge (specs on page). I will be hopefully buying a battery pack with it to support a long-road trip (a few months) wherein I'll keep medication in the fridge (powered by the battery pack when parked and by the car when driving). The medication needs to be keep between 2 and 8 degrees C at all times. Here's the battery pack I'm thinking of getting: Can someone please help me understand what I need to do in order to calculate whether or not this fridge + battery combo will work, and if so, how much time will I be able to power the fridge for. Ideally, I need a combination of fridge and battery which will allow me to have the fridge constantly powered, assuming I sleep in the car for approx. 8 hours a night and drive for the rest of the day (although, there will be times which I'm away from the car during the day for several hours + times where I hope to have access to a wall outlet). Specs: Fridge: Input voltage (AC 220-240 V Input frequency 50/60 Hz Rated input power (AC) 64 W Rated input power (DC) 46W DC12V/50W DC24V W Input voltage (DC) 12/24 V Battery: Capacity: 453Wh Battery Type: Lithium-ion Input Voltage: AC 110-220 V, 50/60 Hz; Solar Power: 18 V; Car Input: 12 V Output Voltage: AC 110V 60 Hz; DC Output: 4 * 12V 4 A (Max.10 A); USB Output: 4 ports: 2* 5V 2.1 A and 2* 5V 1 A Power Output: Rate 500 W Peak/Surge 1000 W <Q> I presume it draws this power when the cooler is running. <S> But it is not running all day. <S> So you should check for how many hours the cooler will run in a day, how much energy it will use... <S> This depends on outside temperature, the quality of the insulation, etc... many variables, and many unknowns. <S> Say you have a box <S> and you want the contents to stay below 5°C. <S> So you install a cooling system. <S> You also install thermal insulation. <S> What is the compromise? <S> You could install a foot thick insulation, and then you'd need a tiny cooler to compensate for the losses. <S> Or you could use just a little bit of insulation, like an inch of foam, and use a more powerful cooler. <S> I'm trying to tell you that your question is a little bit more complicated than you think. <S> Fridge manufacturers usually do not quote an average power, because it depends on ambient temperature, and also it depends on user habits, like how often you open the door and let warm air in, if you place a warm beer inside and expect the fridge to cool it, etc. <S> assuming I sleep in the car for approx. <S> 8 hours a night and drive for the rest of the day <S> In this case, the most cost effective thing to do is not to store electricity, but to store cold. <S> If you have a power supply available all day except during the night, then the best option is to put a few gallons of bottled water (as thermal mass) in your fridge next to your sensitive medicine. <S> When you have power, the fridge cools the thermal mass. <S> When you don't have power, the thermal mass absorbs the heat flowing though the insulation. <S> The good thing about a few gallons of cold water used as thermal mass is that it doesn't break down, doesn't need maintenance, it can't fail. <S> Even if your car dies, your fridge dies, and everything stops working, you still have time to find someone who owns a freezer, or buy some ice, whatever. <S> It is low-tech and foolproof. <A> Just looking at the essentials: <S> Fridge : <S> Rated input power (DC) 46 W, DC 12 V /50W <S> DC24V <S> W <S> Battery : <S> Capacity: 453 <S> Wh , Output Voltage: <S> DC Output: 4 <S> * 12 V , <S> 4 A . <S> Power Output: <S> Rated <S> 500 W peak <S> Checks: <S> 12 V fridge and 12 V power supply outlet. <S> Pass. <S> 46 W fridge and 500 W power supply. <S> Pass. <S> Fridge current: \$ <S> I = \frac { <S> P}{V} = \frac {46}{12} <S> = 3.8 \; <S> A \$. Socket: 4 A. Pass. <S> Run time (100% on): <S> \$ t = \frac <S> {Wh}{W} = \frac {500}{46} = <S> 10 <S> \;h \$. Since the unit takes 7 - 8 hours to recharge it is putting about 50 W + losses. <S> Let's say it's about 75% efficient <S> so input power will be 50 <S> * 4/3 = 66 W (approx). <S> We can calculate the approximate charging current on DC as \$ <S> I = \frac { <S> P}{V} = \frac {66}{12} = 5.5 \; A \$. <A> Worst case your fridge draws 3.8amps (46W/12V) continuously which would be 3.8Ah per hour or 92Ah (3.8*24) in a day. <S> Depending on ambient temperature, how often you open it <S> etc <S> you could do a lot better than that but impossible to quantify. <S> Which would power your worst case fridge consumption for less than 10 hours (37.75A/3.8A) without a recharge from the car. <S> A running car could easily power the fridge and recharge the battery. <S> I couldn't easily find how long the battery takes to recharge on DC <S> but it says 7-8 hours from the wall <S> so let's assume that for DC too. <S> If you were to end the day with a 7-8 hour drive and then stop for 8 hours the battery would keep the fridge running worst case. <S> You did not mention alternative batteries but a 12V AGM battery could be discharged to around 20% and would recharge to around 80% off the car alternator. <S> A regular car alternator may provide around 40A of charge. <S> A 160Ah AGM would be quite a bit cheaper (potentially still cheaper if installed in your car, close to the alternator) and would recharge from 32Ah to 128Ah in less than 3 hours. <S> It could then supply your fridge for 24 hours before the car would need to be run again. <S> There are lots of lead acid choices beyond AGM but AGM is a good choice because it can take a very high charge in it's bulk charge phase and can be discharged well beyond 50%. <S> There are also three way fridges which run off propane or battery. <S> In your case you could run a fridge just off propane for close to a month with a 9kg bottle or between the car and propane probably even longer.
| There isn't enough information to answer this, because your fridge's specs only tell the maximum power draw. Your battery can supply 37.75Ah.
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Protecting a laptop from USB connected devices In the past couple of months I've managed to destroy 2 perfectly good laptops through programming development electronics through USB. I've been wondering what could be done to prevent this. A couple of ideas I've had (not sure how practical): USB connector / hub with clamping 5V zener diodes on power and data lines USB connector / hub with opto-isolators on data lines supplying the usb connection with external power source plus previously mentioned idea In combination with these ideas perhaps resistors on both 5v D+ and D- lines to limit current to say 200mA (enough for most of the devices I use). Just wondered what some more experienced engineers than myself thought about these ideas and their practical implications? <Q> You can solve transient overvoltage with clamping diodes, but these won't help if you power supply is strong enough – they will just fail, and then you're in the same situation as before, only milliseconds and the smell of burnt semiconductors farther. <S> Your problem is a bad one, for a lot of reasons, and your laptops are the least of that: <S> USB is meant to be handled manually (that's redundant wording), so if this fault kills your laptop, I don't have the strongest confidence that it's inherently save for human interaction. <S> There is, for good reason, design criteria for circuitry that involves switching higher voltages with lower voltages. <S> Generally: you need galvanic isolation between whatever any human (even a service technician) could touch during operation and the dangerous voltages. <S> Hence: strictly separate your USB controller and the things it switches. <S> It's common practice to drive inductive or high-voltage loads using optocouplers, whose secondary side is driven by a separate power supply. <S> The board layout must make the high-voltage regions separate from the 5V/MCU environment. <S> Only optocouplers, transformer cores, and relays can be allowed to cross that boundary. <S> No compromises. <S> A typical problem is that your device's power supply ground has a completely different potential than the USB ground – though that should not be an issue for a laptop, which itself should be galvanically separate from anything else, there's a lot of cases where you run into trouble with that (e.g. laptop ground ends up on ethernet ground, audio ground, RS-232 ground...). <S> The strict separation (isolation) between controller and controlled makes that an inherent non-issue. <A> I mainly work with Arduino and PIC based development electronics for controlling washing machines and vending machines <S> (It wasn't the same device that destroyed both laptops). <S> As its mainly just the risk of high voltage would the use of zener diodes be a simple cheap solution or would that mess with the data side of things? <S> Good. <S> Since you mention the "use of zener diodes" to protect against an ill-defined "high voltage" , we now have one reliable data point: you have no idea about what you're doing. <S> Therefore, you need a USB isolator to protect the laptop port from yourself. <S> I feel sorry for that laptop, btw. <S> I mean, laptops are isolated from ground due to their power supplies being not grounded. <S> So, blowing a laptop's USB port is like... difficult... <S> How exactly did you manage to do that exactly? <S> Did you send mains voltage down the USB port or something? <A> The Arduino development environment fits nicely on a Raspberry Pi which you can power from your embedded environment and talk to over wireless from your remaining laptops or perhaps a desktop. <S> This won't prevent you from frying a Raspberry Pi, but it will reduce the cost of a circuit failure to about $35. <A> Ok, after multitude of clarifications and various presented cases, let me offer a solution to the problem how to protect a laptop from frying in the environment of field servicing: <S> Always connect your test/diagnostic equipment to the service object first, and power the thing up. <S> Before connecting the equipment to laptop, check for any suspicious voltages between USB shields using a low-impedance (standard 10-20k) <S> DMM, between DUT port, and laptop port, both in DC and AC mode. <S> If a significant voltage is found (5 - 10 - 50 V), use a true isolation transformer on one of the ends. <A> Using a Pi/Beaglebone/Tinkerbox over WiFi would be cheap and easy. <S> CHIP looks good, see <S> https://getchip.com/pages/chip . <S> 5-9$ per device. <S> You can murder 7 CHIPs for the price of 1 Pi3. <S> Obviously you still need to address the issue that kill-voltages reach USB ports, but at least your workstation will be safe. <S> Don't connect it to your LAN with ethernet, though. <S> Transferring the files is simply a matter of using key-based passwordless ssh access to scp files to the Pi/other and a remote shell to upload. <S> For arduino, there is the arduino-mk package, and there are similar tools for PIC. <S> No doubt you can also just save your files on an exported SMB/CIFS share, and use the remote shell to make and upload it to your boards. <S> If a kill-voltage occurs again, you will only lose 30-35$. <S> I still haven't found out how these voltages ever manage to traverse your MCUs and enter the USB port. <S> Is it a short? <S> Where does it come from? <S> I have destroyed many arduino clones, and am sincerely curious. <S> Also, the USB ports are usually polyfused against moderate voltages. <S> You actually need to place a negative voltage on the 5V pin to kill the port (or positive to GND pin), controller and perhaps motherboard. <S> How are you managing murdering two laptops? <S> Could it be the magic of alternating current at work?
| The real problem is not the unprotected USB port, the real problem is that your device puts you and your devices at risk of being connected to higher-voltage, relatively high-current sources. Since you are using Arduino as a base, the solution is easy ....use a disposable processor for programming and debugging.
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Does the output bypass capacitor of an LM7805 double as a decoupling capacitor? I'm using a 5v linear regulator (specifically an LM7805) that outputs directly to an ATMEGA328P. According to the LM7805 datasheet (page 23) input and output bypass capacitors should be used, as seen below, to tame peaks and ensure stability. It is good practice to also include a decoupling capacitor in front of an IC, in this case an ATMEGA328. Does the 0.1μF capacitor on the output side of the LM7805 act as a decoupling capacitor if the regulator feeds into the IC directly after the output bypass capacitor? <Q> You should have capacitors at the output of the 7805 and at the ATMEGA328. <A> The most important thing about about decoupling capacitors is that they are placed physically close to the device they are decoupling, to minimize the trace inductance. <S> The actual capacitance is often chosen by rule of thumb. <S> This implies that two chips can share a decoupling capacitor if their power supply pins are right near each other. <S> Or, in other words, if two identical decoupling capacitors end up in parallel right near each other, you can drop one of them. <A> If the ic is close to the regulator, one capacitor might be sufficient. <S> However if the input to the regulator is loaded by some other circuit, there is a possibility that the regulator output can discharge quickly than the input. <S> So I'd suggest putting a diode between the regulator input and output for reverse polarity protection. <A> THe Zout of 7805 is 0.016 Ω @ <S> 1kHz <S> * <S> but since internal feedback gain like an Op Amp reduces with rising f, so does Zout rise with f thus at 10MHz <S> it is out of bandwidth and limits to the load regulation = <S> 100mV/5V=2% @ <S> 1.5A <S> 0.1V/1.5A=67mΩ <S> Then add any trace inductance <S> and you get... <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The location of low impedance ceramic caps affects both what the regulator sees with the Q of the input RLC cct as well as the attenuation of a step C CMOS load dump on the voltage. <S> Thus as Peter says both locations become necessary for long traces with about 10nH per cm for traces 10:1 length /width up to 30nH/cm for 100:1 ratio as I recall for typical traces. <S> so 50nH is 5cm or 2" for 0.5cm or 5mm wide power traces for 0.035mm thick. <S> But for power/Gnd planes this reduces to ~ 1nH/via <S> ( depending on L/D ratio) and 2nH/cm for path length for any square plane and thinner dielectric also increases nF/cm^2 with low ESR inverse squared but limited by dielectric breakdown and defects for burrs shorting the supply. <S> Commercial solutions for this exist. <S> p.s. C2 is the equivalent C for the uC and it also has ESR not shown. <S> C causes dynamic power rise with clock rate. <S> or delta Ic=CdV/dt <S> * delta <S> f. <S> thus C can be estimated. <S> Where dV/dt slew rate is assumed constant but rises with T ['C] thus C becomes the ratio of changes for ΔIc/ <S> Δf * 1/ slew rate. <S> ESR is harder and depends on number of FETs switching each about 25 Ohms in parallel. <S> Thus a final ripple current depends on very low ESR*C = <S> T values <= <S> and >= rise time for load regulation of switched C from Coss of CMOS. <S> This is my technical analysis of our Rule of thumb to use low C values as close to both the source and the load. <S> as smaller C have lower ESR values limited by smallest size. <S> Tantalum and ultra-low ESR alum e-caps can as low 1us or <1MHz effectiveness, and Ceramic <S> << 1us to <1ns for microwave caps with low ESL.
| Decoupling capacitors should normally be placed as close as practical to the power supply and ground pins of their associated IC.
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Opamp detection of extremely tiny voltage I am working on a project that requires measuring absorbance of a solution. The circuit consist of an emitter led, a photodiode attached to an opamp with \$10^7\$ gain (TIA).The solution absorbs light from the led and causes different voltages corresponding to different concentrations. This works well for large concentration solution. For example the voltage obtained with solution of: 0 mol/l (no solute) is 9.95V 100 micromol/l is 8.92V But for concentrations <=5 micromol/l the voltage is also 9.95V, which indicates that the change in voltage is so small that I cannot measure it. How can I improve sensitivity? I considered increasing the gain, but it can lead to a maximum of 100V detected. An opamp's saturation is around +/-15V. <Q> Instead of a Log-Amp, I'd use a 24 bit sigma-delta ADC. <S> Here are a few examples . <S> These achieve extremely high precision for voltage measurement. <S> You also need an accurate voltage reference, though. <S> Also, keep in mind that your LED and photodiode are temperature sensitive, therefore you will need calibration. <A> Spectrophotometers often use a reference beam and a sample beam. <S> The idea is that rather than having to detect these small differences directly in the presence of noise and other (e.g. temperature) fluctuations, only the difference with respect to the reference beam is measured. <S> So, build two identical sets of your apparatus, and in one of them put a cuvette containing only the solvent. <S> Then use an instrumentation amplifier with large gain to amplify the difference between the TIA outputs of the two devices. <S> Instrumentation amplifiers are fully differential amplifiers, so suitable to measure this small difference in the presence of a large d.c. offset. <S> Of course, since the two absorption cells will not be completely identical, you will still have a d.c. offset to deal with. <S> But it will be much smaller than otherwise, and you can probably either calibrate it out, or compensate it by adjusting the gain and offset for the reference TIA. <A> Posting a schematic of your circuit may help in providing more insightful answers. <S> One approach would be to reduce the current to the illuminating LED and increase the op amp gain when the solution concentration is low. <S> The reduction in light reaching the photodiode will increase headroom for amplification before the ouput approaches the rail. <S> The additional gain can be obtained with a basic op amp following the TIA. <S> This will keep your costs down. <A> Implement an optical chopper, switching between 2 paths with standard and the test specimen. <S> Then use AC_coupled amplifier to boost the squarewave output. <S> This, as experiment, lets you determine what is possible.
| This allows you to effectively increase your low concentration resolution. This will of course require a second calibration procedure for the lower concentration range.
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Question about coil emf Hi I am a Korean student studying physics. Recently I was solving a physics problem about coil emf and I have a question. As you see in this circuit and the graph, when you close the switch the emf of the coil reaches '-E'(voltage of the battery).However when you open the switch the emf reaches far more than 'E'Why does this happen? I already asked my physics teacher and he said that in theory when you open the switch the emf should be 'E' How does the voltage reach far more than E? <Q> The inductor stores energy and the energy stored is used by the inductor to try and keep the current flow through it constant. <S> So the instant the switch opens, there is a current flow that the inductor tries to maintain <S> but, the only circuit path includes the fractionally open contacts. <S> This can be as high as several hundreds or thousands of volts. <A> Theoretically, the inductor follows V = <S> L <S> * di/dt. <S> If the switch is opened instantaneously, then di/dt is infinite, so V is also infinite. <S> In reality, there are small stray capacitances which will cause the Voltage spike to be less than infinite, both in amplitude and in rise time. <S> Also, as noted elsewhere, an arc will likely form across the switch. <S> This will also affect voltage spike. <S> I have no idea why your professor thought the emf should be equal to V when the switch is open. <S> Theoretically, when the switch is open, the battery current is zero, and the battery does not really enter into the analysis. <A> When the switch is opened, all the energy gets transferred into the parasitic capacitance across the switch contacts. <S> Given Energy = 0. <S> 5 <S> * <S> C * V <S> * V, a very small cap requires the V to increase as squareroot.
| Therefore, to maintain current flow it will generate a voltage that can form a small arc.
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Measuring water level inside a drinking glass Background: I am creating an automatic tea brewing machine using anArduino and an electric water kettle from Amazon. My family and Idrink lots of tea and we have many different sizes of glasses. I was going to use a peristaltic pump to transfer the water from the kettle to the tea glass (mainly because it's food safe and self-priming). Task at hand: I need a way to know when to stop pumping the boilingwater from the kettle as to not overflow the tea cup. I was thinking of 3D printing a bracket that will hang over the lip of the cup (like on the side) that will have two wires facing down towards the bottom of the cup. One wire would be connected to ground and one to an input on the Arduino. That way, as the water fills to the top and eventually touches the exposed ends of the wire, the Arduino would know to stop pumping because the wires are now conducting to Ground. My question is: Is my method listed above safe to do in drinkingwater (exposing a path to ground in the water for a few seconds), oris there a better/safer/easier way to accomplish my task? <Q> A peristaltic pump delivers a set volume per revolution so relate that volume to the size of cup and have a select button : size1 or size 2. <S> Count the revolutions and you deliver the correct volume . <A> You could use a rangefinder mounted above the cup to measure how much fluid you are adding to the cup, and have the cup set up to automatically add a specific "height" of fluid, eg. 4 cm. <S> You would even adapt Solar Mikes suggestion and have presets (1 cm, 2 cm, etc.) <S> You could look into items like an ultrasonic sensor (Google arduino ultrasonic sensor, or https://www.sparkfun.com/products/9495 ), or an infrared rangefinder ( https://www.sparkfun.com/products/8958 ). <S> I've used the ultrasonic sensor before and it has a resolution of about 1 cm, which might be a bit low for this application. <S> I believe the infrared sensor has a better resolution, but I've never used one. <S> EDIT: <S> After a bit of thought, one clever, but arduous and tedious solution would be to attach RFID tags to each cup bottom, and a reader on device, then program the desired amount manually. <A> One wire would be connected to ground and one to an input on the Arduino. <S> That technique should work for you. <S> It is commonly used by visually impaired people for that task. <S> Figure 1. <S> Liquid level sensor alarm for cup / mug. <S> Because you are feeding into a micro-controller GPIO you will need to make some test measurements. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Probes and pull-up. <S> Without the micro-controller connected run some tests with a volt-meter connected between 'GPIO' and GND. <S> Adjust R1 (a pot might be handy for this) until you can guarantee that GPIO gets reliably pulled down to about 1 V or so when the probes are covered to the required depth. ' <S> GPIO' will read 5 V when probes are dry. <A> I'd go a completely different route: <S> have a device that identifies cups/glasses calibrate your containers <S> The second part is really easy. <S> Have a mode in your firmware where you push a button until the cup is sufficiently full. <S> Your firmware then memorizes that amount of pumping and applies it next time it encounters the same container automatically. <S> The first part might be harder, but on the other hand, cups sound relatively easy to tell apart when they're always placed on the very same spot in front of a camera. <S> I'd recommend a webcam, a white LED to illuminate the observed site of the cup, and something like a raspberry pi. <S> Apply OpenCV's classical object classification, or experiment with the fancy deep neural networks (DNN) <S> everyone and their mother plays with these days – tensorflow might be a good tool to start with, and they have a tutorial where they recognize digits that are horribly scrabbled; well-placed, optically precossed cups (edge detection?) should be easier. <A> Answering your actual question, instead of proposing alternatives: of course it would be safe. <S> Why would it not be? <S> @Transistor shows an off-the-shelf product that you could adapt for purpose. <S> There are others, like water sensors to put on the floor near a water heater to sound an alarm if it starts leaking.
| I was thinking of 3D printing a bracket that will hang over the lip of the cup (like on the side) that will have two wires facing down towards the bottom of the cup.
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How can I make battery supply power when wall adapter supply is disconnected? Hello, I have a question about battery charging and power supplying. I'll go with picture for better understanding Starting from using 12V 1A wall adapter, Power will be supplied to My Raspberry PI through 5V rail. My plan is as follows : 1. Down convert 12V to 5V.2. The converted voltage will be connected to both 5V rail and VIN for battery charger (LTC4067)3. When wall adapter is connected, output of 12V to 5V regulator will supply power for RPI directly through 5V rail.4. When wall adapter is disconnected, stored energy in 4.2 LI-PO battery will be supplied to RPI. I know that recommended input voltage for RPI is 4.75~5.25 and expect that 4.2V LI-PO battery supplies quite insufficient voltage. My questions are here :a. Is boosting output voltage of LTC4067 to 5V necessary? b. If so, is it okay to just connect the boosted voltage to 5V rail?( in aspect of backward protection) I'm really really unskilled novice and I don't know much about constructing circuit. I would be appreciated if you answer this question. <Q> Re-layout the circuit as above will give better efficiency and ease of design. <A> a. Is boosting output voltage of LTC4067 to 5V necessary? <S> Well, obviously the Pi will not operate with voltages under 4.75V. <S> That means you indeed have to boost the voltage of the battery when the main supply from the wall adapter is not present. <S> b. <S> If so, is it okay to just connect the boosted voltage to 5V rail? <S> ( in aspect of backward protection) <S> Here I suppose you mean if it is ok to connect the boosted 5V from the battery to the 5V rail. <S> But I don't understand the backward protection concern. <S> Anyway, that should be OK to do, but somehow you have to make sure that the boost converter is disabled while the main voltage from the wall adapter is present and only enabled when the battery has to provide current. <S> But what do you need the OUT pin of the LTC4067 for? <S> Alternatively, you could connect your main 5V power rail ONLY to the OUT pin of LTC4067 (and the output of the boost converter, as described previously) but NOT to the 12V to 5V converter. <A> Take a look at the Adafruit Powerboost 1000c . <S> It does everything for you. <S> It is open source and you can take the schematics and put it on your own PCB. <A> Then buy a DC power "barrel connector," the mate of the connectors commonly used on wall-transformer supplies. <S> These have an internal switch and three terminals. <S> The switch is used to connect the battery whenever the wall transformer is unplugged from your barrel connector. <S> (Of course, this won't work if you unplug the wall transformer from the wall.) <S> Another common method is to pair a 5VDC regulated wall transformer with "diode AND logic," a pair of diodes, one each in series with the battery and with the external DC supply. <S> In that case the higher voltage powers your device, while the diode to the lower voltage battery gets reverse-biased. <S> But diodes waste power and also lower the input voltages by 0.7VDC. <S> That's why the usual solution is the switch on the barrel-connector
| The ancient oldschool method: Use a 5VDC regulated wall transformer, the type with an internal switching supply. You should connect the boost converter directly on the battery, no need to use the OUT pin.
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voltage matching between two ICs I am using a processor which operates at 5 V and a MUX with 3.3 V. If I want to give selection lines from the processor, I have step down the voltage level. how can i do this by simply placing a resistor across it. How can I find the value of the resistor? <Q> There are several options, e.g: <S> Use a voltage divider <S> Use a series resistor. <S> This is used to limit the current into the protection diode of the input pin. <S> This has to be done with care, the input characteristics and ratings have to be considered. <S> The resistor must be bigger than (5.0V-3.3V)/MaxInputCurrent . <S> The maximum input current is normally listed in the datasheet of the mux. <S> If you are lucky, the selection lines of the mux may be 5V-compatible, check the datasheet. <S> Depending on your circuit and type of the mux, you can supply the mux with 5V. No level shifting needed. <S> Please keep in mind that the resistor-based solutions are generally relatively slow due to the time constants involved. <S> If you want to change you selection lines very fast, it may not be the appropriate solution. <A> For an almost complete guide on 3.3V to 5V (and vice-versa) interfacing, see http://ww1.microchip.com/downloads/en/DeviceDoc/chapter%208.pdf <A> Use a voltage level translator if they are only digital lines. <S> Something like this might work: TXB0104
| Use a level shifter or a logic buffer/inverter with 5V compatible inputs like the 74LVC series.
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