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Driving LED with an N-Channel MOSFET I am driving a red LED with a forward voltage of approx. 2V as shown in the picture. The driving signal comes from a microcontroller with a logic HIGH of 1.8V. The LED is being sourced from an independent supply of 3.3V. The circuit works but there is something in my measurements that I cannot explain. I measured 3 mA through the LED branch. I need more current through the LED (about 10 mA would be ideal). Then I reduced R1 in steps until I had about 10 ohms. Yet the current in the LED did not change much. Sometimes I measured about 8 mA at the beginning of the measurement which after some seconds ended up in 4 mA. Why is this happening? I also reduced R2 but it seems to have no effect either. How could I increase the LED using a control signal of 1.8V and powering the LED with 3.3V? <Q> Vgs is probably not high enough. <S> I couldn't find the datasheet for the CMPD7002 (or any part named that). <S> First make sure the Vgs will give you enough current, find how much Vgs you need for the Rds on. <S> You'll need to do one or more of these things: 1) <S> If the GPIO's on the microcontroller are 3.3V tolerant (or if 3.3V is fine with a pull down) <S> Then use open drain on the GPIO and a pull up to 3.3V 2) Use a circuit or buffer to translate 1.8 to 3.3V 3) <S> Select a different mosfet that turns on at 1.8V and has a low rdson at 1.8V. 4) <S> Calculate the current vs LED drop and size R1 correctly. <A> Use a good ol' BJT, or find a logic-level FET rated for 1.8V. <S> I've called out a 2N3904 in the suggested schematic below, but <S> (A) you can use just about any small-signal NPN there, and (B) there are prebiased transistors that'll work and save you a resistor. <S> Design for an 0.7V drop from base to emitter, and a base current of about 1/10 of your desired collector current, to keep the transistor hard into saturation. <A> You need a MOSFET that turns on fully with 1.8V. Not many will. <S> Do you have a datasheet for CMPD7002AG? <S> I'm not getting any hits on it. <S> This one looks like it will turn fully with just 0.9V on the gate <S> https://www.onsemi.com/pub/Collateral/MCH3484-D.PDF <S> Digikey has them. <A> Although you have not linked any part numbers, I suspect you can drive the 2V LED Cathode directly from Vol with Anode to 3.3. <S> The Ron= Vol/Iol should be in the 50 Ohm ballpark. <S> Then when High , the LED will see 3.3-1.8 = 1.5V which will be very dim. <S> All you need is 10mA <S> so when ON, the driver will rise to 3.3-2V= 1.1V with a current of 1.1V/50R or 22mA so adding 50 R in series will reduce the current to ~ <S> 10mA and Vol to 0.6V . <S> For protection (opt.) in case of 3.3 rising before 1.8, just add a Schottky diode from output port to Vdd. <S> Then you need to invert the LED state in S/W. Proof with Sim and specs From Cypress datasheet p76 Iol = <S> 4 mA at Vdd=1.8V Vol = <S> 0.6 max,thus driver Ron= <S> Vol <S> /Iol=0.6V/4mA = <S> 150 <S> Ohms <S> I used a custom model of <S> Vf=2.1V @ 20mA <S> ( Isat=92pA, ESR 12 <S> Ohms, Emission coefficent =3.73, breakdown voltage= 2V) <S> The above simulates minimum current from an ARM Core data port using 1.8V supply. <S> Adding 150 OHms will reduce brightness in half to 5mA min.
| The problem is most likely that the mosfet does not have enough voltage to fully turn on.
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My push-pull converter uses a flyback transformer with an air gap. How does changing air gap size affect the secondary output voltage? My push-pull converter circuit uses a flyback transformer with an air gap. If that air gap is changed (increased), the primary inductance is decreased. How should I expect the secondary voltage and the output waveform to change after increasing the flyback air gap? <Q> All the energy is stored in the gaps between the magnetic core particles or gap in this case. <S> As the gap size increases, flux will start to 'spray' into the area of the windings in proximity to the gap. <S> This will generate high levels of eddy current losses in the wire. <S> To avoid this problem you can wind on a few turns of insulating tape around the middle of the coil former to exclude wire from the region. <S> The main advantage is it reduces the sensitivity to thermal runaway by moving the losses from the core towards the glass gap when increased. <S> There is an optimal gap to minimize the sum of all tradeoffs for increased winding loss, reduced core loss , increased gap coupling leakage and Eddy current losses. <A> There are other variables, you have to specify what you are keeping constant to get a valid comparison. <S> Is anything else other than the total output capacitance limiting the output voltage? <S> If not, then more energy stored in the flyback means a higher voltage developed into that capacitance. <S> Does <S> an increased airgap mean more or less energy stored in the flyback? <S> It depends what else stays constant. <S> If the primary voltage and pulse width stays constant, then the lower inductance will mean a faster rate of rise of current. <S> Fortunately, this increased current will not saturate the inductor, as the input volt.seconds are the same. <S> The current is higher by the factor that the inductance is lower, but as stored energy goes as \$0.5I^2L\$ , the energy is increased by this factor. <S> The output voltage will be higher. <S> If the primary is switched on until the current gets to a threshold, or like in a car coil the current is limited by the coil resistance and supply voltage, the current stays the same. <S> The increased airgap will also lower the secondary inductance, and may well reduce the primary - secondary coupling. <S> This means that the secondary waveform will tend to have shorter time constants. <A> My push-pull converter circuit uses a flyback transformer <S> If your converter is a push pull type then, although the transformer you used may have been called a flyback transformer when you acquired it, the new application (as part of a push pull circuit) puts it into operating as a conventional transformer. <S> If your converter primary voltage(s) remain as they were after the gap is increased, then the output voltage will probably not be altered but, bear in mind that enlarging the gap can mean a larger (usually unwanted) primary side current and, depending on other primary protection circuits, the secondary voltage may fall. <S> If you are running it as a true flyback circuit <S> then Neil’s answer is valid.
| With reduced inductance, the stored energy is lower, and the output voltage will be lower.
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Build a scale without computer I need to build a scale that is not attached to any computer. The goal is to weigh a dustbin and show the weight value on a small LCD screen. The weight I want to measure is approximately 10 kg, but I have to do the tare of the weight of the dustbin to weigh only the contents. The price of the equipment has to be cheap. I tried to look for an online solution and the only solution I found was with an Arduino. The problem is that the Arduino is connected and powered by USB to the computer and I can not use computer. How could I build the balance cheaply? Is there any way to run the program on Arduino without having to always connect it to the computer and feed it by stack for example? Example with Arduino: <Q> Flashing Once you upload your sketch to an Arduino, it does not need to be connected to a computer anymore; the Arduino IDE uploads the sketch into Flash of the Arduino, which will stay there until overwritten again, even when the Arduino is powered off and not connected to any power source. <S> The uploading process is called Flashing (done by the Arduino IDE normally). <S> USB power <S> For powering, instead of using the USB plug to the computer, use a socket adapter with a USB plug, same as a telephone/USB phone charger. <S> This will have the same behavior as connecting it to a computer, except that flashing is not possible. <S> Battery power (from comment below) <S> You can simply connect the battery to the Vin, because the Arduino already has a voltage regulator (which works recommended from 7 to 12 V and absolute maximum rating of 20 V). <S> The voltage output from e.g. the 5V or GPIO pins will be (approximately) 5 V. <S> The reason that you the minimum voltage should be 7 V of Vin, is that the internal voltage regulator is not 100% efficient (no voltage regulator is). <S> So if you connect a battery that is at least 7 V (e.g. a 9 V battery or 5 or 6 AA batteries you are ok). <S> Note that if you use 5 V batteries from 1.5 V, you get 7.5 V so the limit of 7 V will be reached quite soon as the voltage of the batteries drop. <A> The Arduino in your photograph has a barrel jack on it for powering from a battery. <S> A suitable battery pack is one like this. <S> Battery Holder - 4x AA to barrel jack connector <S> I teach Arduino robotics at the local university. <S> As others have stated, use the computer and USB to actually program the Arduino, then remove it from the computer and simply run it from the battery pack. <A> short answer: <S> yes. <S> The program is written on the controller and can run without computer. <S> I guess you will find quickly plenty of examples for that. <S> If money and size is of essence and you are willing/abled to handle a "bare" µC <S> outside the arduino eco system I can actually recommend the MI0283QT Adapter v2. <S> It is intended as an rpi/arduino shield but comes with a built in NXP <S> LPC1114 µC which can be easily reflashed to work without an external rpi/arduino. <S> Adding a simple Battery monitor system and li battery one gets a nice stand alone touch display mini "computer" with touch display. <A> Yup. <S> It is very easy. <S> There are 2 ways: Provide 5-12 V on the Vin pin of arduino and connect (-) or ground to the ground on the arduino. <S> You can also use an adapter whose output pin matches the black coloured jack on the arduino. <S> Connect the programming cable to a mobile charger (USB port on the charger). <S> For the 5-12 V source you can very easily use a small LiPo battery (depending on budget) or AA batteries in series should also do the trick. <S> (Note: 5-12 means 5 to 12 V, any voltage in that range would work.)
| Powering an arduino via USB charger is done frequently and using a battery is well within the possiblities.
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Do PWM solar controllers increase current? There are two kinds of solar charge controllers. PWM does simple pulse width modulation to limit intake voltage (E.g. 18V) to that appropriate for battery charging (e.g. 14V). The only thing it considers is correct battery voltage. MPPT controllers also "scan" the panel dynamically, varying their current draw to seek maximum watts. This is constructive because solar panels are not linear, and don't naturally seek watt maxima. Decreasing current draw 5% may increase voltage 15%, and that's a win. This MPPT strategy can get 30-40% more power out of a panel. What I want to know is, does the PWM naturally increase output current when it decreases output voltage, I.e. does it act like a buck converter? Or must the excess voltage be consumed as heat in the PWM charge controller, like an LM7805? If my Panel input is 14A @18V, is it possible for the PWM output to be 16A @14V? Or is it impossible for amps out to exceed amps in? I mention MPPT because folk wisdom in solar is PWM controllers act like 7805s, and you need an MPPT to get the buck conversion/current multiplication effect. That advice sounds off. <Q> Just the term PWM does not give enough information, it needs to be more explicit. <S> Some solar controllers use PWM but only to control the average power charging the battery, they do not attempt to store any energy during the cycle and are lossy Some of those operate by shorting the PV array so the unwanted power is dissipated within the array. <S> Others may disconnect the array in which case the array does not output any power when not required. <S> MPPT controllers use a DC-DC converter which itself uses PWM together with energy storage elements (inductors) to avoid energy loss. <S> The energy is stored during part of the PWM cycle and then sent to the battery during the second part. <S> The average output current, in this case, can be more than the average input current but the power output cannot be more than the power input. <S> Efficiency can be up to the high 90% region. <A> Thousands of pulses per second. <S> "Simple" Pulse Width Modulation of the current from a Photo Voltaic panel can achieve efficient power conversion at MPP (matched impedance) into a resistive load if suitable capacitance is connected in parallel with the PV panel & PWM is appropriately controlled. <S> This is due to the capacitor is charged during PWM off time from PV current & during PWM on time current from PV + Capacitor are supplied to load. <S> This is easy to test by applying a load with lower impedance than the PV panel(DC fan or Filament lamp etc...). <S> Using a simple motor PWM board & capacitance on input (eg 1000uF/5khz/5A) adjust the PWM for maximum output & measure PV Volt & Current. <S> Example: <S> ___ 100W 18Vmp panel, 100W filament 12V autolamp. <S> Below is a power plot of data acquired from PWM sweep from 100% -- <S> > 0% with 250W old panel & 1.5R load. <S> A PIC16Fxx provided the PWM.The horizontal axis <S> is the linear 1024point PWM sweep. <S> At MPP the effective load resistance(calculated) is 3.29ohm, data recorded 26.06V, 7.92A, 206.4W. @51% PWM. <S> point 520 on graph. <S> (conditions not STC). <S> I have a modified firmware solarPWM controller doing MPP tracking for my 30V HWS, no inductors (except the cabling), no cold showers for me. <A> No sane designer would implement a 7805-like voltage stepdown when trying to harvest max power from a panel, they'd always use a switch mode device, often a buck converter. <S> An ideal converter has the same power out as power in, so if the voltage is reduced, the current is increased. <S> A practical buck will deliver an efficiency in the 80s to 90s percent.
| PV PWM charge controllers don't perform voltage transformation, they just supply current during the PWM on time(% of PWM period).
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Vector Control in Induction Motors Can vector control in induction motors be considered as power factor correction ? From my researchs, it seemed to me angle between rotor and stator flux is 90 degree when load is purely resistive (power factor is unity).The goal of vector control is to make this angle 90 and get the maximum torque. Is that right ? Thanks in advance. <Q> Only in a very nonstandard sense. <S> Vector control of an induction motor pertains to the power (possibly the instantaneous power) going to the motor. <S> Yes, it's may be good in some cases if this power is, for the most part, real. <S> However, to most engineers "power factor correction" will mean the ratio of real to reactive (or harmonic) power at the point of connection to the power grid -- and that's not where vector control happens. <A> It's not about active/reactive power. <S> Vector control in a drive simply says your control loop is fed with the phase-corrected momentary I/U parameters instead of <S> I/U averaged over several periods. <S> The goal of vector control is having a finer and faster response to torque changes from the load, as there is no low-pass and averaging over several periods. <A> It has no relationship to what happens on the LINE SIDE input of ta VFD. <S> That said, a standard "6 pulse" VFD, by virtue of what it is doing, in effect will correct the displacement power factor of the power INTO the VFD rectifier to near unity, but then adds some distortion power factor in the form of harmonics. <S> If your utility is not capable of measuring the higher frequency distortion power factor, you may appear to be at .95 displacement PF or better. <S> But this has nothing whatsoever to do with the output being vector control or not.
| "Vector Control" is a terminology that only applies to the OUTPUT of a VFD in terms of performance response and accuracy.
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Sizing the pulldown resistors for 74HCT08 logic inputs I have come across some text or blog before but I cannot find the source that: when we use 74HCT08 AND gate IC and for the logic inputs if we use pulldown resistors to prevent floating then we need to use the resistor pulldown resistor values Rp low such that the input should not rise over 0.9V. Regarding a such input as hand drawn below: Why does the value of resistor matter and how could high values make the input higher than 0.9V when the switch is open? <Q> If you know nothing about CMOS, you need ohms law, a data sheet, and some common sense. <S> If you know something about CMOS, then just the common sense will do. <S> Per the data sheet , the input leakage current for a 74HCT08 is \$1 \mathrm{\mu A}\$ . <S> Per ohms law, this means we need <S> \$R < \mathrm{\frac{0.9V}{1\mu A} = 900k\Omega}\$ . <S> Common sense says that unless we're designing something for insanely low current consumption, and for which we can guarantee an electrically quiet environment, using the maximum resistance here is silly. <S> So choose something convenient, like \$\mathrm{10k\Omega}\$ . <S> A basic knowledge of CMOS says that for all practical purposes a CMOS chip's input current is nothing -- so you fall back on common sense, and you're done. <S> Note : <S> that this is not the case for a lot of modern microcontrollers -- they are often designed so that without the pins being programmed, they have built-in weak pull-ups, to reduce current, and for historical reasons (see my final note, below). <S> But sometimes they have built-in weak pull-downs on some pins, for practical reasons. <S> So always read the data sheet. <S> Final note : <S> If you were designing circuits 30 years ago, then the gate in question may well have been TTL. <S> In that case, you would want a pull- up resistor (because TTL inputs source a bit of current, but sink less). <S> Then you'd use the same \$\mathrm{10k\Omega}\$ , but it would be to VCC. <S> There is no reason not to do this for CMOS, and it's oh-so-slightly more comfortable for old circuit designers when they see it. <A> The basic rule is never leave CMOS inputs floating for EMI noise and ESD immunity reasons. <S> Thus a resistor to the opposite rail of the closed switch is used to shunt weak stray currents induced by external means. <S> The lower, the R, the greater the immunity when the switch is open. <S> Schematics do not not show the length of the path to the switch, so the trace or cable ESL inductance and the mutual coupling to external current transients (B field) or the stray capacitance to high dV/dt E fields is unknown. <S> You must know the environmental noise voltage and impedance to choose how to design for immunity. <S> The easiest way is a shunt resistor as shown. <S> 10k is better. <S> For lower micropower a better way is with a filter and shielding. <S> If the switch is not gold plated (normally done if rated <2A) then you would eventually see intermittent contacts from a lack of wetting current. <A> The input to a CMOS logic gate typically has a specified leakage current of up to \$1 \mu\$ A. <S> If the resistor is 1 M \$\Omega\$ <S> then the resulting voltage would be 1V, so in this case the resistor value is too large.
| A large resistor value will also cause a slow fall time at the input, which can cause undesirable behavior.
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How to best clean this sealed mechanical rotary encoder / volume knob? How can I best clean, replace or lubricate this fairly well sealed digital rotary mechanical encoder? The equipment operates, but the encoder is super finicky, often encoding in the wrong direction when spun. With patience it's possible to get it to the right value, but quite tedious. The equipment is about 15 years old, and the encoder has been unreliable for the last year or so. Question 1: if it can be replaced, how would I find a pin for pin match? Question 2: if it can't be replaced, what's the best way to clean or re-lubricate it? It has a crack, I can probably flood some chemical junk into it. Update: I opened the cleats: <Q> By the limited information you give and the (not too great) photo, it seems it is a mechanical encoder, not an optical one. <S> You don't give any information about the equipment it is mounted on, but 15 years of continuous operation may be quite a lot for a mechanical encoder. <S> Probably the contacts have worn out and there is no reliable way to fix them using any sort of chemical. <S> Best course of action is to replace it with a new one. <S> The cheapest crappy encoder you can buy on ebay (~1$) could work better than your worn-out encoder, at least for a while. <S> Of course that is not a suggestion for a long-term fix. <S> If you care about the equipment, you could probably find a suitable replacement on any major component distributor (digikey for example, or RS components), for a couple of dollars. <S> Just for example, I just did a quick search on digikey trying to find something vaguely similar: BOURNS PEC11L <S> Series - 11 mm Low Profile Encoder (datasheet) . <S> As you can see from this datasheet excerpt (emphasis mine): the expected minimum encoder life is 100k full rotations. <S> Assuming ( optimistically ) that the average life is twofold (200k rotations) and that the shaft is rotated on average 50 times a day (not uncommon in a control console in a work environment) you get 4000 days average life, i.e. about 11 years. <S> Therefore what I initially said about your encoder being at its end of life, is perfectly reasonable. <S> All this assuming it is not some specialized high-reliability encoder. <S> If you want better advice post more information on that encoder (model number, better photos, info about the equipment, etc.). <A> Figure 1. <S> Open the cleats. <S> I have successfully cleaned an encoder in an audio amplifier by squeezing the cleats on one side enough enable me to create a crack through which I squirted some switch / contact cleaner. <S> (The cans with the straw attached help direct the spray.) <S> Then the switch should be pressed together and the cleats spread once again. <S> The technique avoids the risk of parts flying off in various directions, not to mention the trouble of finding a suitable replacement, de-soldering, etc. <A> That encoder has a de facto -standard design, and replacements are cheap and readily available <S> (I've got a box with dozens of them, from multiple manufacturers). <S> (Basically, you'd use needle-nose pliers to squeeze the two split-and-flared flanges shown in the photo, as well as the ones on the other side, then lift off the top half from the blue housing.) <S> If you're reasonably adept with a soldering iron and with a desoldering braid and/or solder sucker, you should be able to replace it yourself without too much difficulty, though an iron under 40 watts or so may have trouble desoldering the anchors on the side. <S> Otherwise, any electronic repair shop can do it for you easily. <A> I simply open it up completely (no desoldering is necessary), then with a small screwdriver scrape up any clean lubricating grease to save for reapplication after cleaning <S> Then use alcohol and brush to clean away any left over grease on the rotating disk and the small wipers, but be careful, the wipers are very fragile. <S> Then take a pink eraser and clean the disk and tiny wiper pads, this will shine them right up, reclean with alcohol, then re-apply the saved grease. <S> This step is optional: very slightly bend the wiper contacts to give slightly more mechanical pressure. <S> Reassemble and test. <S> I have fixed countless mechanical encoders this way. <S> Good luck. <S> P.S. <S> This works on pots too, <S> if the pot or encoder is very worn, you can carefully shift the wiper contacts left or right to contact a new section, this works wonders, almost new again.
| While it is possible to disassemble and clean one, doing that without first desoldering it from the board would be tricky and could easily result in accidental damage to the board. The best way to clean it is replace it.
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Is the continuity test limit resistance of a multimeter standard? The continuity shows us the shorts, but many times I guess, even if the wire has a resistance of up to 50 ohm, the continuity will beep to indicate a short. Is it always 50 ohm, or what parameter should we look for in the manual? <Q> It is not a standard value from meter to meter. <S> Different models will beep at different resistances. <S> An example in the comments were the fluke models 175,177 and 179. <S> In the datasheet for these, you can see on page 12 that the meter beeps at <25 ohms. <S> The multimeter I use is a Tenma 72-7732A, and in the datasheet for this model, on page 35 it is stated that the beeper will sound for conditions less than 50 ohms. <S> In the popular EEVBlog multimeter, page 25 of the Datasheet states the continuity threshold is between 30 and 480 ohms. <S> These few examples are enough to determine that there is no standard value. <S> If the information cannot be found in the datasheet of your particular model, then getting a few resistors between 10 - 250 ohms and measuring them should be enough to tell you the threshold. <S> An easier way, as pointed out by @HarrySvensson in the comments is to turn a potentiometer/rheostat until the beeping stops/starts and measure the resistance. <A> However, I would personally say, there is an unofficial industry expectation that the continuity beeper should not trigger on a pn-junction. <S> Assuming the meter is using an excitation current of 1 mA and choosing a conservative IV curve for a pn-junction <S> (assume <S> Id < 100 uA <S> at say 200 mV - 300 mV). <S> By Ohms law, that would give you an upper continuity threshold of 200 \$\Omega\$ to 300 \$\Omega\$ . <A> The continuity resistance threshold is not standard. <S> The exact value is certainly quoted in the datasheet/manual, if the instrument comes from a reputable manufacturer. <S> If you have a cheap, unknown instrument then you can try a few resistors, and check which one beep, and which one do not beep. <S> Some expensive multimeters allow you to manually set the threshold, on my Keithley 2000 <S> the threshold can be set anywhere between \$1\ \Omega\$ and \$1\ \text{k}\Omega\$ .
| I'm not aware of any standard resistance values for the continuity mode of a multi-meter.
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Output capacitor in LDO regulator I want to use an Analog Devices LDO regulator ( ADP7104ARDZ-5v. ) When I checked the datasheet, I found out that it requires just a 1uF capacitor on the output pin. Then I searched more and I found an evaluation board ( EV-ADF4159EB1Z ) that this regulator had been used on. On the evaluation board, there is a 4.7uH inductor, a 0.33 ohm resistor, and a 100 uF capacitor rather than just the 1uF recommended in the datasheet. I want to know why they used a very low value resistor together with a high value capacitor. Here is the circuit from the evaluation board: <Q> The evaluation board is for a very sensitive circuit, that needs an extremely clean power source. <S> The additional components form a low pass filter at the output of the regulator. <S> The inductor and the capacitor make a low pass filter with a cutoff frequency around 15kHz. <S> I can't say for sure, though, since my phone has a problem downloading the ADP7104 datasheet. <S> The resistor may also be there to reduce oscillation (more correctly, "ringing") in the LC circuit. <S> EDIT Finally got to look at the datasheet. <S> As @Hearth mentioned in the comments, the ADP7104 is made for use with low ESR ceramic capacitors. <S> That makes it much more likely that the series resistor is there to reduce ringing in the filter. <S> In any case, you can follow the recommendation in the ADP7104 datasheet and go with just the 1uF capacitor - unless you know for certain that you need the extra filtering as used on the PLL evaluation board you were looking at. <A> Linear regulators have stability requirements for the output capacitors, you can find the requirements in the datasheet. <S> The ADP7104 datasheet specifies a minimum of 1µF effective capacitance and a maximum of 1R ESR. <S> It's also quite common to have a minimum ESR as well, it really depends on the regulator circuitry and you have to check it in advance. <S> This is the same principle as in the well-known SMPS feedback loop design but usually a lot easier to manage. <S> The large output capacitor improves step response, you can find in the datasheet an example of step response with the 1µF capacitor. <S> In your circuit 10µF is the actual regulator output capacitor, 4.7µH inductor + 100µF capacitor are a low-pass filter. <S> The 0.33R might be there to dampen the LC circuit spikes at resonant frequency. <A> Most EVM's will do everything to make their part preform the best (lowest noise, ripple, efficiency, etc). <S> So in your app you might get away with the minimum, a single 1uF output cap.
| The small resistor is probably there because the ADP7104 probably requires that the output capacitor have a small amount of series resistance to keep it from oscillating - many parts that were designed before low equivalent series resistance (ESR) ceramic capacitors depended on the innate resistance of the capacitor.
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Why are THP 1/2W resistors bigger than 1/4W while 1206 resistors are all equal? How come THM (through hole mounting) 1/2 W resistors are bigger than their 1/4 W or 1/8 W counterparts, while SMD 1206 1/2 W and 1/8 W resistors are equally sized? <Q> SMD 1206 defines the flat/projected size of the component / in other words, it defines the footprint. <S> But they do have differences when their wattage is different. <S> How much heat the component can handle not only depends on size, it also depends on other factors: component matérials used, board layout, ... . <S> I could find SMD 1206 resistors that can handle 2W. <S> The manufacturer indicates that the actual power depends on how the device is actually mounted. <S> The 2W device has an aluminium core which helps conducting the heat to the board and possibly a heat sink. <S> The SMD resistor is directly on the PCB <S> so it is "easier" to transfer heat from the resistor to the PCB than it is for the THT resistor. <S> In addition, the PCB layout can be designed to improve heat dissipation. <S> There is of course also the issue of pricing - <S> when you are using THT you typically care less about space occupation, and increasing the size of the THT resistor to cope with double the power, is a relatively small increase in the 2D space taken on the board. <S> (The axial resistor is cooled by the air flow, unless you put a heatsink on it using a clamp for instance). <A> Thus these variables define the Pmax. <S> I found that% tolerance is the biggest factor due to max temp R error. <S> The exceptions increase cost dramatically for high temp, low tolerance error, low temp. <S> coefficient. <S> They are not all the same. <S> So you may find 1/8W 0.1% to 1/4 1/3 to 1/2W 1% and many options in between with 1/4W as the standard rating for this size. <S> You will find a 1/2W rated 1206 part with <=0.1% tolerance <S> and it will have exotic materials and cost >100x as much while operating much hotter at max power. <S> These are the standard sizes. <S> Code Length (l) Width (w) <S> Height (h) <S> Power US Metric inch mm inch mm inch <S> mm <S> Watt 0201 <S> 0603 <S> 0.024 <S> 0.6 <S> 0.012 <S> 0.3 0.01 <S> 0.25 <S> 1/20 (0.05) 0402 1005 0.04 1.0 0.02 <S> 0.5 0.014 <S> 0.35 <S> 1/16 (0.062) <S> 0603 <S> 1608 <S> 0.06 <S> 1.55 <S> 0.03 0.850.018 <S> 0.45 <S> 1/10 (0.10) <S> 0805 <S> 2012 0.08 2.0 0.05 <S> 1.2 0.018 <S> 0.45 <S> 1/8 (0.125) <S> 1206 <S> 3216 <S> 0.12 <S> 3.2 0.06 <S> 1.6 0.022 <S> 0.55 <S> 1/4 (0.25) <S> 1210 <S> 3225 <S> 0.12 <S> 3.2 0.10 2.5 0.022 0.55 1/2 (0.50) 1218 <S> 3246 <S> 0.12 <S> 3.2 0.18 <S> 4.6 0.022 <S> 0.55 <S> 1 2010 <S> 5025 <S> 0.20 <S> 5.0 <S> 0.10 <S> 2.5 0.024 <S> 0.6 3/4 (0.75) 2512 <S> 6332 <S> 0.25 <S> 6.3 <S> 0.12 <S> 3.2 0.024 <S> 0.6 <S> 1 <S> ** <S> Read more http://www.resistorguide.com/resistor-sizes-and-packages/ <A> It's simple, the through hole resistors have the leads connected to the board, the smd resistors have pads which are directly in contact with the body of the resistor and the body itself is pretty much in contact with the board. <S> That allows them to dissipate heat by transmission way more efficiently, plus when they are black they also dissipate by radiation fairly good, due to the smaller surface area and the fact that they are directly over the board they dissipate a little less by convection <S> now the transmission depends on the material of the body too, so if it's made lets say of aluminum it will be able to handle more heat than one made of ceramic, also if the film that actually makes the resistor can handle higher temperatures the heat dissipation capacity will be also bigger, as le_top said though it will depend on the layout too, its not the same to have a trace 1mm wide connected on each side than a 0.1mm one in a through hole you mostly dissipate by convection, meaning that you must increase the surface area
| The power rating is a combination of higher tolerance , lower tempco and higher max operating temp.
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Why does my Inverter overload? I have a battery pack consisting of 4 10 amp-hour 12 volt batteries wired in parallel. We recently bought a 2000 Watt rated HammerDown power inverter to run a 500 watt window air conditioner. The air conditioner is supposed to draw 5.6 amps. But when I connect the AC to the inverter the air conditioner always shuts off as soon as the compressor starts running. I measured the current going to the inverter and as soon as the compressor started it shot up from around .3 Amps a whopping 26 Amps before turning off! Now I understand there is a surge of current needed when starting an appliance but I didn't think it would be this much. I also noted that the voltage dropped a little too (from 13.5 to 12.9 v). Given all that though, I don't understand why the inverter failed to supply the power to the AC. With 40 amp hours shouldn't it be able to provide 40 amps of current at any moment (more than needed for surge)? Another interesting thing to note is that sometimes the inverter would read OVERLOAD after one of these tests. <Q> John D has answered the inverter overload issue. <S> I'll answer another part. <S> With 40 amp hours shouldn't it be able to provide 40 amps of current at any moment (more than needed for surge)? <S> To understand the maximum current from the battery you need to consult the datasheet. <S> 40 <S> Ah is an energy rating (sort of - a better energy measure would be Wh, watt-hours which is V x Ah = 12 <S> x 40 = 480 <S> Wh in your case). <S> The Ah rating is usually measured on a 10-hour discharge so a 40 Ah battery will typically supply 4 A for 10 h. <S> At lower discharge rates the time will be higher than the rating would suggest while at high discharge rates it will be lower. <S> i.e., at 40 A discharge, if the battery can handle it, the discharge time will be considerably less than the 1 h the calculation might suggest. <S> The <S> 40 Ah rating does not tell you that the battery can actually supply 40 A. <S> It might not or might be able to supply 400 A. Again, the datasheet is your reference. <S> I also noted that the voltage dropped a little too (from 13.5 to 12.9 V). <S> Try measuring at the compressor. <S> If it drops more there then your cable is inadequate. <S> Notes: <S> SI units named after a person have their symbols capitalised but are lowercase when spelled out. ' <S> V' for volt, 'A' for ampere, 'K' for kelvin, 'Ω' (capital omega) for ohm, etc. <S> Meanwhile 'k' is for kilo. <A> Your compressor is likely run by an AC induction motor, and they take very large surge currents to start. <S> Your inverter is rated at 2kW, and no matter how much battery capacity you have it will cut off when it reaches its overcurrent limit. <S> If your mains voltage is 120VAC 40A would be 4080W, well above the rating of your inverter. <S> (Even worse if your mains voltage is 220VAC). <A> Motors when starting draw from 3x ( soft-start 3 phase type) to 5x typ to 8x with Start-Cap surge current to 12x for high-efficiency BLDC motors. <S> (10x typ) <S> Your inverter is unable to handle the startup load. <S> It is intended for Blenders, Vacuums and Power Tools and not an air-conditioner compressor which has a much longer startup time. <S> For industrial use, they may choose a VFD instead of an inverter. <S> This allows the user to choose a voltage a slow voltage ramp with a rising frequency and fixed V/f and thus it is possible to limit the surge current much lower values with high torque, instead of 8x as in your case.
| Your inverter can likely handle short surges above rating, but starting an AC compressor is a notoriously hard thing to handle.
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Level shifter for WS2811 led strip I'm trying to control a WS2811 Led strip with an ESP32.I'm using a logic level shifter to convert the 3.3V output of the ESP32 to 5V.I bought this level shifter (txs0108e) https://fr.aliexpress.com/item/32404824354.html but I cannot manage to use it properly. The leds are flickering and are not the good color.What is strange is that if I touch all the pins of the level shifter with my fingers it works perfectly. It works also perfectly when I touch the "data input" line of the strip with one multimeter probe (if my explanation are not clear I can try to upload a video). Here is the wiring diagram : Do you see any problem with this wiring or without this level shifter for this application ?Do you have any idea why it works only when I touch the level shifter whith my fingers ? Thanks <Q> This is entirely a signal integrity issue using relatively high series impedance of the level shifter which is not good for driving inductive long cables with a Vol/Iol equivalent to about 800 Ohms max for port B. <S> I would buffer them with HCT CMOS 5V logic that is closer to 50 <S> ~75 Ohms and has a threshold of TTL = 1.4V or 74ACT series CMOS which is closer to 25~33 Ohms driver impedance instead of a bidirectional level shifter. <S> This uses a 5V supply. <S> Then you have far better immunity with 100 Ohm STP CABLE or similar 120 Ohm twisted pairs. <S> You can glue your fingers to the data line or add shunt caps to attenuates stray capacitance noise or as suggested load it with pull-up/down resistors but this depends on the frequency of the radiated noise. <S> If due to fast risetimes in the cascaded WS chips or due to low frequency line noise. <S> IF YOU can , add the BUFFER. <S> If not, scope ]the signal properly with two balance probes in differential mode to see the signal integrity issue then a fix of ACTIVE LOAD R , passive C or cable routing, shielding with STP CABLE OR <S> SOME OTHER EMI PREVENTION SOLUTION BECOME OBVIOUS. <S> ( sorry BT keyboard caps lock error (CROSSTALK WITH THE LETTER A and fat fingers) <S> Why is a bidirectional (unbuffered) level shifter not the best solution for a line driver? <S> Examine Vol @ <S> Iol . <S> this ratio is the series resistance in two FETS used as autosensing bidirectional drivers actually “ NON-ideal dual diode switches” due to low Vgs/Vt on the low V Port A side inside the chip. <A> A genuine TXS0108E has pull-ups on all pins (see Functional Diagram on page 18), so leaving data pins unconnected or touching unconnected pins should not have an impact. <S> Considering the shop you got it from, it could be a counterfeit. <A> thanks all for your answers. <S> I managed to fix the problem just by grounding all unused inputs. <S> I also re-soldered everything <S> so maybe there was a faulty one <S> and I also used only 1 ground pin of the ESP. <S> If I have time I will try to do more tests in order to add more details to this answer.
| Your meter inductance and finger capacitance is adding overshoot and suppressing stray common mode noise respectively. Ask the seller to include a bunch of spare fingers with your purchase as the IC you got doesn't seem to work without them.
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why is transformer not considered as an amplifier? I know that in a transformer the power is conserved on both sides, and hence there's no power amplification as such. That's fine! But if I instead want only a voltage amplification, then choosing proper turns ratio and connecting the load across the secondary would serve my purpose. So can you consider it as voltage amplifier? If yes, then why is it not used as one? <Q> It's just a subtlety of the terminology. <S> As others have pointed out, there are other functional differences between transformers and amplifiers, one being that power can flow both ways through a transformer but generally only one way through an amplifier. <S> Finally, we already have a word for a device that can change voltages without providing power gain, so we don't need to call these things amplifiers . <S> We can just call them transformers . <A> Only for ideal transformers, power is conserved. <S> Real transformers have losses, which may be load dependent. <S> So, in that case, the 'output' voltage is not the gain times the 'input' voltage. <S> For voltage amplifiers, you also want them to have unlimited (or at least large) current supply. <S> This unlimited current is needed, so the output is not dependent to the load, i.e. the voltage should not collapse because of the load. <S> For transformers, this 'output' current is dependent and limited. <A> Amplifiers typically provide input-output isolation. <S> Changes in output energy need not require more input energy (tho the VDD rail must support the output demands) <S> A transformer of course does not provide isolation.
| If there's no (capability of) power amplification, then we don't call it an "amplifier" even though the output voltage might be higher than the input voltage.
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How to remove this component from PCB Anyone have any idea how I could remove this component? I've tried blowing hot air, used solder wick, tried to poke the holes with two soldering irons, cut the pins to near the PCB, but it won't budge. It would be ideal if I had a rectangular piece of metal that could get up to around 380 degrees celsius and that I can just place on the pins all the while pulling to remove the component. <Q> You need a lot more solder, so as to cover all the eight contacts and heat them to their melting point . <S> Think "solder pool. <S> " <S> As the comments attest, this will make the part very hot. <S> Silicone gloves have come to the rescue when desoldering large parts, for me. <A> With big through-hole parts, I attack one pin at a time with a solder sucker (or "solder pump"). <S> I prefer this type because they're big and cheap! <S> It can remove more solder per use than the smaller, more refined pumps. <S> You want to keep adding heat until the entire volume of the PTH is molten. <S> It can help to add solder to improve heat flow. <S> Then, use the pump to remove the solder. <S> Don't remove the iron from the pin until after you use the pump. <S> If the PTH looks empty, move on to the next pin. <S> If there is still solder left over, fill it up with new solder (and flux) and try again. <S> There will always be just a little solder left over holding one edge of the pin to the inner edge of the PTH. <S> Here's the trick: grab each pin with the tip of some needle-nosed pliers and wiggle it until it breaks free. <S> Once all the pins are loose, pull out the chip. <S> Touch each hole with a clean soldering iron to reflow the remaining solder. <S> There should be so little solder left that this action effectively clears out the holes. <S> In this specific case, you may have a problem with Step 4 since you have clipped the pins. <S> Perhaps there will still be enough to grab onto. <S> And, of course, this makes a mess of the device's pins, so it shouldn't be done if you want to re-use the part. <A> Dremel off the body, and get the pins one at a time with an iron. <S> Clean up with a solder sucker and braid. <A> For something like that, conventional soldering tools are the wrong answer. <S> Possibly, a very high powered hot air rework station would do it, and could be the right choice if you need to work around other components that shouldn't be dismounted, but unless you already have it or are seeking an excuse to buy one, it's probably not the solution. <S> For something large on a sparsely populated board what you probably want is a hardware store type heat gun. <S> If you don't particularly care about the board, the old school method was a propane torch; beware the board will probably catch fire, and you don't want to breathe that. <A> It looks like you have a hot-air gun. <S> This should make the process fairly painless. <S> (Unless you touch something hot, of course!) <S> I would try to suspend the board to ensure the component isn't touching anything. <S> Then heat up the pins on the back of the board with your hot air gun. <S> When the solder melts, the component's weight should cause it to fall away from the board. <S> One pin looks like it's tied directly to the ground plane, and a few others might be tied to a copper fill on the component side of the board. <S> These will require a lot of heat: the whole plane needs to get hot before the solder joints will melt. <S> If the solder melts but the chip doesn't fall off, I recommend poking the pins with a thin wooden dowel (or a toothpick). <S> The wood might start to smolder, but I haven't had one catch fire yet :) <A> You can also try Chip Quick, it's a low temperature solder. <S> It stays hotter longer. <S> I use it a lot specially when I want to save the component I'm removing. <S> It costs a lot but it works really well. <A> I would just use a hot air solder machine. <S> May not be the conventional way but you will find it 100x easier. <S> Just make sure to keep the hot air moving <S> so you don't leave a burn mark on the pcb. <S> If you don't own a hot air solder machine I would recommend buying one since most components these days are smt based and then you would have the ability to fix both styles of components. <A> Here's an idea and one that works for electronic companies. <S> Keep raising it until the board it too hot to touch. <S> Get everything prepared, flux, braid, sucker, gloves and what ever else you use. <S> If the oven is close to you, the better. <S> Try and find a small gap under the component and run flux and begin desoldering. <S> It may take a few goes and you may need to place it back in the oven as the board will cool down. <S> But make sure when the pins have desoldered that they move in the holes and then do the top. <S> It's all about keeping a constant heat. <S> You'll find your own way and technique, but I find it so invaluable. <S> I'm sorry I can't help you if your oven is too far away. <S> This is an idea that seems to work <S> and I hope it helps you and anybody else. <S> Have fun......
| First, you heat up the pin and melt the solder around it. Once you pull the component, then you can remove the solder pool using whatever means you have at hand. I use my electric fan oven and put it on its lowest heat and bake the board so it reaches a good temperature, roughly between 80 and 120 degrees.
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Filtering for USB VBus to 3.3V I have a circuit taken from an FTDI app note that takes USB +5V VBus and filters it into 5V Vcc . This circuit works well. I now want to modify this circuit to output +3.3V and I am unsure exactly what to do about the filtering. Should the filtering be placed before or after the 3.3V regulator? Is it best to filtering the 3.3V after the regulator (bottom drawing)? Or is it better to filter the 5V and then run that through the 3.3V regulator (top drawing)? Below are circuits for both configurations. <Q> If you want to be conservative, though, you should do it as shown in your first circuit. <S> (Probably with a small electrolytic on the output for stability). <A> Since I can't see how to post a picture in a comment, I will do it this way.diegogmx,I incorporated your suggestions. <S> The TI datasheet indicates that the highlighted is needed. <S> So I placed a 10UF and ferrite as the incoming filter, connected that through TI's required circuit, and added a 100nF to the output. <S> Diegopmx, is that similar to what you were suggesting? <S> And thanks to all who contributed. <A> The approach i would use is to put an LC filter before the regulator with generous capacitance lets say 1-10uF with a cut off frequency in which the regulator has a really good psrr (datasheet for that) and then in the output side a couple of ceramic capacitors, one around 4.7uf and the other one 100nf, so you have low impedance over a wide frequency range
| You likely can do away with the filtering all together, as the linear regulator will filter out the noise sufficiently.
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Dim LED lamp with constant current power supply using resistors I own a really bright led lamp that I want to dim. It is powered by a constant current power supply that can supply 37mA between 11V to 17V. If my lamp is connected the voltage settles at 13.58V. Please note, that there are more than four LEDs in unknown wiring in this lamp. I only drew four of them as illustration. With a bench powersupply I determined, my desired brightness is reached at 1mA where the voltage drops to approximately 12.7V. To limit the current to 1mA while still using the constant current source I've added a parallel resistor like this: For some reason this wont work. The voltage drops to 2.7V, the current on the branch with the resistor is 7.28mA, and the current on the branch with the led lamp is immeasurably small. Is there something I'm missing? Or is there another possibility to dim the LEDs besides PWM? I thought abouth changing the sense resistor of the power supply but I don't really want to mess with the power supply. <Q> Figure 1. <S> LED I-V curves . <S> According to my Figure 1 a typical white LED could be expected to drop about 3.5 V at 37 mA. Four in series would drop 14 V which is in pretty close agreement with your 13.6 V measurement. <S> Accurate figures for low currents aren't often published but if my graph is any way accurate you should see a voltage drop of about 1.5 V at 1 mA or 6 V across the four in series. <S> You're measuring 12.7 V so your LEDs don't match the graph. <S> They appear to have a much more vertical I/V curve. <S> Your calculation looks pretty close as far as I can see <S> so I don't know why it doesn't work. <S> A more efficient alternate solution would be to connect a series resistor to limit the current. <S> The CC PSU will then rise to 17 V and you'll need to just find the series resistor value as you would with any constant voltage supply. <A> PWM may cause current measurement errors and long wires can introduce inductive voltage/current errors. <S> I suggest you use a TO-220 transistor <S> so it doesn’t get hot, and bias the base current with a Voltage control Rseries or a pot and series R to limit the base drive, in order to dim the LEDs . <S> This is using it as an active shunt load like your resistor except with short path control between source, control and load in a small loop area to minimize wire inductance. <S> It is the rise time that can cause issues not the PWM rate which might be >>1000x higher bandwidth if running at 2kHz for e.g. <S> If we had some idea of the circuit, I could be more exact. <A> This isn't really exactly what I asked for <S> but I solved this by getting rid of the constant current power supply and used an of the shelve power supply with fixed voltage and a current limiting resistor which made it really easy to met my requirements. <S> I've chosen a 15V Power supply and used a resistor in series with the LEDs: R = <S> U / <S> I = <S> (15 - 13.58) <S> / 1mA = <S> 1420 Ohm
| A voltage divider to 0.4V will turn off the transistor then bias to 1V with 1mA should shunt the collector and turn off the LEDs by choosing the right R ratios and values from Vdc.
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Meaning of 'Scatterer' and 'point scatterer' as opposed to just using the word 'target'? https://earth.esa.int/web/guest/missions/esa-operational-eo-missions/ers/instruments/sar/applications/radar-courses/content-3/-/asset_publisher/mQ9R7ZVkKg5P/content/radar-course-3-the-radar-equation "The geometry of scattering from an isolated radar target ( scatterer ) is shown in the figure, along with the parameters that are involved in the radar equation." I'm a bit confused about usage of the words "scatterer" and "point scatterer" in radar. The above makes it sound like 'scatterer' is exactly synonymous with target. That can't be right- then we would just say "target." There must be some connotation with the above that makes us use the word scattterer. My googling efforts: I searched "scattering" on wikipedia but am confused how it relates to the above https://en.wikipedia.org/wiki/Scattering "The types of non-uniformities which can cause scattering, sometimes known as scatterers or scattering centers, are too numerous to list, but a small sample includes particles, bubbles, droplets, density" <- this doesn't sound like anything related to radar. A scatterer in wikipedia seems to be anything that can deflect radiation There is also the term "backscatter" (something to do with bouncing according to this site http://www.radartutorial.eu/07.waves/wa51.en.html "“bouncing” or “backscattering” radio waves off the ionosphere" ) but I can't tell if it is related to "scatterer" <Q> I found this in your link: <S> The radar equations (eq. 8 and eq. 10) are general equations for both point and area targets. <S> That is, the scattering cross-section s is not defined in terms of any characteristic of a target type, but rather is the scattering cross-section of a particular target. <A> I'm a bit confused about usage of the words "scatterer" and "point scatterer" in radar. <S> The terminology here has a specific meaning. <S> A "point" scatterer is one that is much smaller than the wavelength, and so scatters isotropically in all directions. <S> A non-point scatterer is approximately a wavelength or larger, and so has high anisotropic scattering (usually forward and backward peaked). <S> As the object becomes many wavelengths wide <S> it begins to reflect (redirect the beam without scattering). <S> In the extreme case of a large, smooth object you have a mirror, which does not scatter at all (ideally). <S> The distinction is relevant here because they are looking at the object with widely separated TX and RX angles. <S> A point scatterer will look the same for all angles, while a larger scatterer will look much dimmer when the RX channel is at a high angle (as depicted). <S> There is also the term "backscatter" (something to do with bouncing according to this site <S> All scatterers are backscatters. <S> Backscatter is relevant if you have a combined antenna for RX and TX, since that is the only type of signal you will receive. <S> Backscatter tends to be highly efficient when the object is about a wavelength wide. <S> Too small and energy is lost to side scattering. <S> Too large and it is lost to reflection. <S> I searched "scattering" on wikipedia but am confused how it relates to the above <S> The page on Mie scattering (object about 1 wavelength wide) has a lot of helpful background, including plots of the radar cross-section of an object vs. size: https://en.m.wikipedia.org/wiki/Mie_scattering <A> There are both technical and semantic reasons that can make a "target" and a "scatterer" mean the same thing, or refer to different, but potentially related, things. <S> For example, one can think of a "target" as a generic object that will in someway or another reflect energy we are interested in. <S> In the case of a radar, this object will reflect some of the power that we have transmitted. <S> Here, the target can be thought of as "a scatterer". <S> This is the most appropriate when the target is modeled as a point-target. <S> A target with a complex geometry <S> can me modeled as being composed of multiple scatterers. <S> Here, the scatterers themselves produce a return but together make up the entire "target" return. <S> They are treated in many cases as isotropic to simplify modeling. <S> You can then use multiple scatterers to model a more geometrically complex object. <S> Using these terms really comes down to context: <S> A target can be modeled as a single scatterering point. <S> Here a "target" and a "scatterer" are one in the same. <S> A target can be composed of multiple scatterers. <S> Here, the two terms refer to two theoretically different things, but are obviously related since the scatterers compose the target. <S> The term "target" usually refers to an object of interest. <S> There can be many scatterers in the environment, but none of them is your "target".
| In practice, the term "scatterer" is used to refer to point or near-point objects that will produce some kind of return.
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Paralleling mosfets to reduce Rds On? This if a fairly simple question. I have some APT43M60L MOSFETs with a fairly high on resistance. They get quite hot and waste a lot of power in my induction heater. They work great but get really hot really quick. I tried IRFP250M MOSFETs, they also worked great, but I just needed a little more power. We all know that putting 2 identical (or near identical) resistors in parallel halve the resistance. For example, two 1k resistors in parallel would make 500 Ohms. So does this concept apply to parallel MOSFETs? Could I simply put 2 of these APT43 MOSFETs in parallel to lower the on resistance without any drawbacks? <Q> Yes. <S> It does. <S> Note that in general you can't blindly parallel transistors. <S> You can parallel MOSFETs without special measures since as they get hotter they conduct less well which distributes the load more or less evenly in spite of individual component differences. <S> Positive temperature coefficient. <S> BJTs conduct BETTER as they get hotter so the BJT that conducts best conducts even more in a positive feedback loop until it is conducting the entire load and fries while the other parallel BJTs conduct nothing (unless you take special measures such as load balancing resistors). <S> Negative temperature coefficient. <S> but if you don't run it at the right operating point it will fry (unless you take special measures). <S> The drawback is there is now more total gate capacitance/charge so the MOSFETs will take longer to turn on and off for the same gate drive which matters if you're switching at high frequency. <A> So does this concept apply to parallel mosfets? <S> Yes, it is quite common to do this in high current DC DC converters. <S> The other nice benefit is you get double (or whatever number your paralleling) <S> the heat dissipation while lowering the resistance because of the additional devices. <A> Most but not all MOSFET have a low PTC which permits current sharing easily,. <S> ALL CMOS logic has a PTC effect on Ron as well. <S> All BJT <S> ’s and IGBT have a NTC effect which requires a small series R (high power) to share current. <S> It does this by the added resistance so that the NTC effect never causes thermal runaway with the rise in current with voltage <S> yet drop in voltage with temp rise. <S> So the net effect is to limit the current and share the current by linearizing this net resistance. <S> This is usually just greater than the Rs of the diode, LED or Power transistor often defined as Rce. <S> However MOSFETs in the Triode mode are not safe to current share as the heat is not evenly shared in nanolayers of HEXFET junctions on many parts designed as switches. <S> In GW switches used on large power sources, this must be carefully done so that the linear changing Ron transition does not experience a commutation burst of power and be prone to failure and self heating. <S> Deadtime is crucial based on L/R + RC time constants of the network load. <S> Is there a design you would like to share for heat reduction opportunities? <S> Sure you had thrown lower resistance at it in parallel FETS and if resonant expect higher Q, current amplification and resonant currents. <S> But instead, I would choose better FETs such as Silicon Carbide instead. <S> SiC devices have 1% of the RdsOn (1mΩ/cm²) for same chip size, and 10x times higher breakdown voltage Vds max than the IGBT silicon devices for the same chip size. <S> The drift region is also 10% of the Si FET which is \$W_{drift}≈ \dfrac{2V_{BR}}{E_C}\$ with this smaller region as 10µm vs 100µm. <S> Here are some photo's of 10kW induction heaters.
| Then there are IGBTs that where the temperature coefficient changes between positive and negative so if you run it at the right operating point, you can parallel them
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Terminology confusion with sensitivity used with gain Regarding force transducers like strain gauges, the resolution is the smallest change in the load that represents a change in the output values of the measured forces and moments. The smaller the resolution of a sensor, the larger is the sensitivity of the sensor. Lets say we have a force transducer outputting 1mV/N; if then we amplify this by gain of 10 it will output 10mV/N. Does that mean we increased/changed the "overall sensitivity"? Or the term "sensitivity" should only be used when no amplification is introduced? <Q> I would actually call the \$x\$ <S> V/N number the sensitivity . <S> You can add amplification to make that 10 \$x\$ V/N <S> but that does not change the sensitivity of the sensor itself. <S> The amplification only increases the output level. <S> What then defines the lowest value we can measure? <S> If we would make the applied force lower and lower at some point we would not be able to measure a sensible number anymore due to noise . <S> This can be thermal noise or induced noise (or radiated signals) emitted by surrounding equipment. <S> The equipment we can remove or shield, the thermal noise we cannot avoid. <S> Since noise is a random signal it can be "averaged out". <S> A noise signal will have an average value of zero (actually it is a limit function converging to zero) <S> so the longer you measure, the more accurate you measure and the more resolution you get. <S> Then obviously you can measure slowly changing forces (more time to average) more accurately than quickly changing forces (less time to average). <S> So: resolution depends on how fast you want to measure. <S> An additional amplifier would amplify both noise and the signal (voltage) <S> you actually want to measure so an amplifier by itself does not increase the resolution at all. <A> The terms Resolution, Accuracy, Sensitivity and Sensitivity Analysis may have different definitions in Test Engineering depending on how many input variables are known. <S> To know the sensitivity of any formula, we analyze the signal or control system or component tolerance or sensor measurement system, and simply take the derivative of the system output over input. <S> this incremental rate may not be linear over the whole range but at least over some range like ESR is Effective Series Resistance, the small changes of V/I or the error due to supply load variations. <S> That is a supply to load regulation % error sensitivity. <S> S{F(x)}= <S> dF/dx such as in your example. <S> If uncertainty includes environmental variations, quantization levels, non-linearity, offset, gain error, Thermal noise, “other” stoichastic noise , unknown transients, repetitive electromagnetic wave interference (EMI), causal errors, non-causal errors, human errors, equipment or component drift out of calibration, etc etc. <S> Then we have a measurable “uncertainty % or std dev. <S> level” on the absolute values. <S> That is a defined NIST traceable accuracy. <S> Resolution is simply the smallest incremental change that can be measured. <S> Often it is misused as the inverse of Sensitivity when other error factors are unstated or unclear . <S> It may be used as inverse correctly with constraints in context to limits or with a sensitivity affecting resolution due to other factors like noise, temperature, calibration error, supply voltage, human error. {?} <S> * <S> * <S> Gain may improve both resolution and sensitivity IFF (if and only if) the signal to noise ratio is adequate. <S> Where noise can be DC, AC, random, EMI/RFI, AC hum. <S> Such as your example : Av= 10 does increase the rate or derivative of two linear variables or Sensitivity , for the measured parameter and thus lower your resolution. <S> But if the error or noise also increases in the signal, then there is no gain in sensitivity <S> Resolution can also get better with gain but Sensitivity get worse when it is not done properly, which is a major source of questions in this forum. <A> Just to give a different perspective, there is a formal definition. <S> "sensitivity of a measuring system: quotient of the change in an indication of a measuring system and the corresponding change in a value of a quantity being measured". <S> This is just to point out that, according to the formal definition (which is frequently not used precisely in Test Engineering documentation) sensitivity should be applied to measurement systems and not to sensors .
| That being said, all other things kept unchanged, if you increase the gain, yes, you would increase the sensitivity of your measurement system .
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Do Computer keyboards have ICs? I have always assumed that computer keyboards do not, in the typical case, contain integrated circuits. And also, now that I think about it, there must be a set of standard electrical (digital?) signals defined by key that a keyboard sends. I understand that some fancy keyboards may have their own macro capabilities supported by ICs, but basic keyboards don't use ICs, or? <Q> Even the earliest mainframe computer keyboards had circuity! <S> Keyboards buttons are usually placed in an X-Y matrix such that any inputs need to be converted to some serial code (scancodes or keycodes) and sent to the computer. <S> The matrix reduces the number of individual buttons to scan which in a keyboard, saves a lot on wiring and processing power. <S> The scanning procedure typically involves iterating through every combination of rows and columns to check if button press: <S> The use of the matrix created several well known problems such as rollover. <S> When multiple keys are held down simultaneously, this circuit is unable to register all the inputs. <S> Modifications in the both the layout of the keys and software of the microelectronic later improved these conditions. <S> Also, if you might have heard of the connection types <S> PS/2 or USB <S> (and of course the AT Connector) which send serial communication through these ports. <S> These standards regulate the communication protocol between the keyboard the computer, necessitating the microcontroller. <S> Image Source: https://en.wikipedia.org/wiki/Computer_keyboard#/media/File:FunctionalCircuitDiagramOfKeyboardNumPadScanningProcedure-small.gif <A> Do Computer keyboards have ICs? <S> Short answer. <S> YES <S> You CANNOT have a PC keyboard without IC's ( <S> and they ALL have microcomputers as at least one of the ICs in the keyboard). <S> The PC keyboard has a serial protocol of events sent to the computer it's attached to. <S> The serial protocol has little to do with scanning the matrix of keys on the keyboard, which is only a sub task of those required by the keyboard controller. <S> The computer expects in the serial protocol to be advised about keydown/keyup events so that the status of keys may be maintained. <S> Read this document to get an idea of what is going on. <S> Nothing is simple. <A> Yes. <S> Most keyboards arrange the keys into a matrix to reduce the number of signal lines connected to the keys, but that still leaves an inconviniently large number of signals to route over the cable from the keboard to the main system board. <S> For example a PC keyboard has about 21 columns and 6 rows (though not all positions are used) so to wire up the keyboard matrix directly would require about 27 wires. <S> Therefore all full size keyboards i've seen (small keypads are a different matter) contain some form of active circuitry to reduce the amount of wiring. <S> Historically different computers handled the keyboard differently, The BBC micro keyboard for example contained multiplexer and demultiplexer chips to reduce the number of signal lines, but left the active scanning process to the main processor. <S> IBM PCs and compatibles on the other hand have the scanning of the matrix handled in the keyboard itself. <S> The keyboard uses the results from the scans to generate key down and key up events which are sent over USB or PS/2 to the computer. <A> but basic keyboards don't use ICs, or? <S> What makes you come to that conclusion? <S> Modern keyboards use USB, in order to communicate with a PC over USB some protocol has to followed. <S> Implementing this protocol without using any IC would make for a very large and expensive and power hungry keyboard. <S> When the keyboard is communicating with the PC it sends codes . <S> Also making these codes without ICs is theoretically possible but your keyboard would require a humming box next to it similar in size to a refrigerator, it would consume a lot of power and it would be expensive. <S> In an IC we can integrate all the functions above for a few $$ or less. <S> I challenge you to find a computer keyboard that does not have any IC inside. <S> My bet: you will be unable to find one. <A> Yes, of course keyboards have ICs. <S> Think about it. <S> On one hand, you have several dozen switches. <S> On the other, you have a serial data stream representing the closing and opening of those switches. <S> How do you propose getting from one to the other without ICs? <A> They were heavy, slow, noisy and required considerable maintenance.
| The only keyboard in my experience which didn't use ICs would be that of a Teletype, which used motors, gears, cams and switches to send serial data to a computer.
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Why aren't non-isolated DC-DC converters made for high wattage applications? I am new to electronics. I am presently studying DC to DC converters. One thing I noticed is that most of the people are using isolated stepdown converters like half bridge, full bridge, or resonant LLC converters for conversion such as 400VDC in and 28VDC out at 1kW. But there are no non-isolated converters like buck converters. I am just curious to know why this is so?I am trying to build one AC to DC converter at 1KW, where my boost PFC gives 400VDC output. I think 7% duty cycle is normal to drive given with the technology we developed (as D = 28/400 => D = 0.07), still i am unable to find any manufacturer who makes such PMIC with that small duty cycle at high power such as 1KW. <Q> Like Aaron said, a topology with a transformer allows for better control of the step down/step up voltage. <S> A duty cycle of 7% seems fine on paper, but in reality you should aim for closer to 30%. <S> In an application with changing load that 7% nominal duty cycle can drop down to 1 or 2% with low or no load. <S> When using a PFC, especially a boost PFC, the 400V(ish) output is chosen for the AC line. <S> A universal input converter (90VAC-264VAC) will have a rectified voltage of 128VDC to 374VDC. <S> If you are only doing low line (90-130VAC or so) then it doesn't need to boost to 400. <S> Then there is safety concerns. <S> Any company is going to have to get UL/CE/TUV/KC or whatever other safety agency you can think of. <S> Think about a buck converter, if that top switch fails, how are you going to protect the output from that HVDC? <S> A transformer/coupled inductor provides galvanic isolation. <S> Then you start getting into efficiency concerns. <S> There is nothing you can do about it. <S> A resonant topology like a PSFB or LLC can lessen or eliminate the turn-on losses. <S> Look up Zero Voltage Switching. <S> Phase Shifted Full Bridges are really cool and a topology I work with often. <S> You can use a buck converter for a 1kW source, there are just better options. <S> Of course, better is the enemy of good! <S> For a medium power source like that, you can look into interleaved buck converters. <S> These are essentially 2 or more buck converters connected in parallel. <A> With a simple buck converter, you only have one degree of freedom, the duty cycle. <S> And as you are noticing, the duty cycles for those input:output ratios are very small. <S> With a transformer design like a flyback, you get two degrees of freedom, the duty cycle and the transformer winding ratio. <S> Usually you design the winding ratio so that you can be in the sweet spot of the duty cycle. <A> For consumer use, it is not popular, for light industry it is not common, but for heavy industry (MVA generation) as long as somewhere in the system design there is isolation and safety earth ground, there is no restriction for isolation if no service access is needed or safety hazard imposed in use. <S> You are likely never to see the latter unless you are involved in high power system designs such as Megawatt PV Grid Tied Inverters (GTI) powered from Solar Farms. <S> All I can say is , they exist in your power range, but you must be aware of the consequences of lightning strikes and lack of safety certification on home insurance.
| A buck converter will always have turn-on, turn-off and conduction losses in the switches.
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How to avoid voltage drop when using full bridge rectifier as reverse polarity protection I found a couple of old bridge rectifiers. After reading some docs and tutorials about reverse polarity protection decided to give them a try. The problem is nobody offered a solution against the voltage drop (and the power loss) after the rectifier, which is usually mounted at the load side not at the supply side. What is the way to avoid this drop - using higher voltage as input or additional circuit to overcome this at the load side? Note that I chose this way of protection so the protected device will continue to work <Q> If you want to avoid the drop, you need a different device that doesn't use as diodes for current control. <S> Diodes have a drop, the best you can do is switch to a diode that has a lower voltage drop, usually 0.2 is as good as it gets. <S> Mosfets have low resistances, you can either deisgn your own active rectifier by matching mosfets, or buy an active rectifier IC. <S> Another good option is to use a gate driver IC like the LT4350 <S> Source: <S> Storing the charge from a MOSFET Bridge Rectifier <A> What about this simple solution from www.ti.com/lit/an/slva139/slva139.pdf <S> You should put a zener and a large resistor to protect your MOSFET if the load voltage is larger than Vgs max: <A> The board won't work if power is applied in reverse, but it won't go up in smoke. <S> One place I worked used a fuse, and a diode on the other side of it that would crowbar the supply to one diode drop in reverse. <S> If the board was connected backward the fuse would blow -- so technically it would be "broken", but it would be an easy fix. <S> You need to take care if you do this -- you need to size the diode so that it doesn't get damaged before the fuse blows, and you need to make sure the system isn't going to be damaged by the brief short across the power leads.
| If all you want to do is avoid damage you can use a single Schottkey diode in the power lead (as opposed to ground) and end up with less drop.
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Ringing on SPI square wave and other SPI issues I've just started learning this material not too long ago and need some help. I'm trying to communicate with SPI from an AD7124-4 ADC IC to an ESP32 . The first problem I'm having is I'm getting a lot of ringing on the square waves which is shown in the photo on an oscilloscope. The ringing is about 1V and greater than 2V on the top signal and I believe it is interfering with the high and low signals, causing read errors. I've already compensated all four probes perfectly. See the schematic and also pic of the circuit on a breadboard (disregard parts not seen in the schematic) if anything stands out. Any ideas on what may be causing the ringing or how to fix? I'm using x10 probe and settings on the scope are for x10. I've set the SPI settings on the scope according to the manual (correctly I think). Everything looks great on all solder connections and the chip flashes just fine. EDIT: I realized I didn't have the decoupling caps on the AVDD and IOVDD. Here is the schematic from the circuit note CN0381. The second problem is the logic levels don't seem the correct voltage. ------------------------- I have wired as follows: ADC ---- ESP32 DIN --> MISO DOUT --> MOSI SCLK --> SPICLK CS# --> SPICS ------------------------- On the oscilloscope: PROBE 1 -> MISOPROBE 2 -> MOSIPROBE 3 -> CLKPROBE 4 -> CS ------------------------- On the ADC: IOVDD = 3.3V LOGIC INPUTS:LOW = 0.35 × IOVDDHIGH = 0.7 × IOVDD LOGIC OUTPUTS:LOW (MAX) = 0.4VHIGH (MIN) = IOVDD − 0.35 ------------------------- On the ESP32:VDD = 3.3V LOGIC LEVELS: LOW = 0.25 * 3.3 = 0.825V HIGH = 0.75 * 3.3 = 2.475V ------------------------- I have 3.3V powering the ESP32 and the ADC so the logic levels should match I would assume according to specs, but why are MOSI and CS lower than the other 2? And MOSI and CS are also look like they are on 4 different levels which is not good at all I would guess. All probes are at 2V / division. I'm trying to decode the signal also on a Siglent SDS 1104X-E but the signals don't look readable yet. Thanks and I would really appreciate any help or suggestions. ====================================================================== EDIT: Well I got the levels right, I had the CS on the wrong (non SPI) pin and I swapped the MISO and MOSI, I mistakenly thought the docs said that ESP32 was a full time slave (I realize it doesn't make sense now). Here are some screen shots: You can see that the length of the whole communication chain varies at different readings on the same time scale, that doesn't seem right but I'm new to SPI. I used a dev kit ESP32 board instead of using the ESP32 mounted on a breakout board and the noise seems low enough to work correctly, I tried twisting grounds around the signal wires and it helped but not a whole lot. Here's the pic. But still can't get it talking correctly, I'm using this code: https://github.com/epsilonrt/ad7124/tree/master/examples/ad7124-thermometer In the debug output it just prints "FAIL". I've successfully used ESP32 for UART and I2C but not with SPI yet. Any more ideas or pointers? My main goal again is just to successfully read the signals and I'll clean them up after that. ============================================== FINAL EDIT: I got it working finally, I could only get the SPI working with the AD7124 library when using this line of code in the setup() function: SPI.setClockDivider(SPI_CLOCK_DIV4); And it would only work using CLOCK_DIV4 or CLOCK_DIV2, no other speeds would work. Here's a scope shot of CLOCK_DIV64 which did not work: The signals look pretty clean but could not communicate properly over SPI. Here's a scope shot of CLOCK_DIV2: It was a bit noisier but worked. I believe If I'm counting right, the DIV64 signal is about 260KHz and the DIV2 signal is about 4MHz, it says in the ESP32 datasheet that default SPI speed is 80MHz so I don't know how dividing that by 2 gets 4MHz. The pic might have been from DIV4, don't remember. But 4MHz is the default speed on Arduino and I tried the AD7124 code on an Arduino Pro Mini and it worked great with no changes to the code. I also got rid of the long SPI transmission lines and the noise went right away. Also in the first scope shots at the top, you can see the noise on the yellow trace is much worse and that wire is longer than the others and wraps to the other side of the board. Twisting with a ground wire did help and reduced noise by about 20% but it was still bad. Also switched to an ESP32 dev kit board instead of the breakout board which had longer wires connecting it to the breadboard (I just didn't have the right length with the connector ends needed). Here's the new setup: But anyhow I was pretty happy to get it working and thanks to all for the help. <Q> ADC ---- ESP32DIN --> MISODOUT --> MOSISCLK --> SPICLKCS <S> # <S> --> SPICS <S> This is part of your problem -- "DIN" on the ADC means data into the ADC. <S> "MISO" means master in , slave out (or it's a kind of soup) <S> -- if you've got the processor configured as master, then you've got input to input and output to output ( DOUT --> MOSI ). <S> Also, you need to provide a decent ground return for the SPI signals, and you're getting cross-talk between your clock and the other signals. <S> I can't see from your scope <S> shot what the time base is set to, but <S> you're not providing much of any ground return for the SPI, much less a good one. <S> Get your innies and outies matched up correctly: make sure inputs go to inputs, and outputs to outputs (and double-check -- getting serial connections hooked up out to out and in to in always seems to happen). <S> Then make a ground return for each SPI line. <S> You should be able to get away with twisting each SPI line with a ground wire. <S> Do this for each wire carrying SPI between the controller and the ADC. <S> You want an extra ground wire twisted with the original wire (and connected to GND on each side). <S> This may seem excessive, but for high-speed signals it is not. <S> Start there, and let us know how it works. <S> (Next steps are twisted pair taking care to make the twist lengths different, and after that try to find some high-impedance coax -- that's probably unnecessary, though). <A> Spikes could be due to Oscilloscopes ground clip wire length as well. <S> Check that 1st. <S> Also check your power supply to the CHip, if it getting bounced, you need to provide proper decoupling at the CHIP VCC. <A> New answer, because I think you have the signal integrity and connection problems fixed. <S> At this point in the process, I would be carefully comparing the bits going by on the scope with the ADC data sheet. <S> You're running this through driver code, which complicates things -- but I suspect that you're just not "telling" the ADC the right thing. <S> You should also carefully review the SPI mode and polarity. <S> Everyone does their SPI a bit different, basically involving the polarity of the clock and the relative timing of clock and data. <A> watch out for the terminations, you don't seem to have termination resistors in place, also, you have a lot of crosstalk, not just ringing, if you are using long wires to connect an spi bus i would suggest you put 2 ground wires on the sides of the clock line
| You need to check the ADC data sheet, and make sure that the processor is set up to match.
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Why is conduit for 120V building power metallic? I saw wire conduit being installed in a building, and it was metallic. Wouldn't you want to use an insulating material instead? What if the live wire sheath started to fail? It would short to the metal conduit. Isn't that something worth avoiding? <Q> The conduit will be connected to earth. <S> If the insulation breaks down, the live cable will short to earth, drawing a large current that will trip a breaker. <S> If the conduit was an insulator the fault would go unnoticed. <S> The cable in the conduit may well be a steel reinforced cable that will have the conductors surrounded earthed steel. <S> If the live insulation did break down it would would then short to the steel and again trip a breaker, so in this case the insulated conduit would be unnecessary too. <A> It can be metal or plastic as offered in your local hardware store. <S> I believe all exposed wiring in commercial settings are required to be in conduit according to code. <S> There is no specification of type of conduit <S> I believe so there is no requirement for it to be metal at all. <S> Also I think plastic is used for more weather resistant applications vs metal. <S> Metal conduit is always grounded to prevent it from being live and will usually trip a breaker if an exposed live wire were to touch it. <S> (One reason why wiring is always sized for the breaker or higher loads) <A> Outside and underground Distribution wires may be coaxial <S> shielded XLPE cables with an outer layer of insulation to impede lightning current induced. <S> Fault protection gear will protect the customer entrance 1st and/or the switchgear and transformers to determine the optimal OCP time constants. <S> So it is often, something worth having, rather than avoiding. <S> Buried entrance is better for lightning avoidance but worse for winter thaw moisture creepage breakdown and often causes power failures in my residential region (maybe once a year) <S> every spring on a warm day after frost is leaving the sub-ground. <S> Power is often restored quickly but these are the tradeoffs for buried in moist clay that freezes in Canada vs above ground entrance. <S> Yet we have one of the best power utilities in the world. <S> So these types of interruptions are added to the maintenance tasks of replacing 30y old XLPE underground cables.
| Building entrance cables may reduce radiated noise from say arc welders and also improve shunting stray lightning arcs to earth ground by this means of earth ground shielding or sheath in the conduit. I believe contractors would prefer metal due to it's ease of use because it can be bent to fit around corners, etc where as plastic requires fittings for every bend.
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Run electric motor CW & CCW with DP-DT switch I have a Harbor Freight drill press 1-1/2 hp, running on 120 volt, capacitor start. Is it possible to run CW & CCW rotation, using a DP-DT switch? Motor has four wires coming out of it, all with color and numbers.black #1/gray #2/red #3/yellow #4 The picture is from the motor, if this is possible please show the wiring diagram. Thanks. <Q> As the motor is capacitor start, the answer may be yes, if the four wires give you access to the main and start windings. <S> The main winding stays connected as normal. <S> As the circuit diagram tool isn't saving for me at the moment, let's see if ASCIIart will do. <S> DPDT cap ______________________ <S> |-----||---/ <S> | <S> ----| <S> | ____|--start winding--|----------/ <S> | <S> ----------------------| <S> To identify if you do have all the wires for all the windings, measure the resistances between the wires with a meter. <S> You are looking for two isolated windings. <S> The main one will be lower resistance. <A> The diagram below shows a likely internal motor circuit. <S> There may be other possible internal circuit configurations. <S> The green and red lines are the 115 V and 230 V connections. <S> The start circuit is always connected in parallel to just one run winding, so it receives only 115 V regardless of the power connection. <S> If you can get access to the start-winding connections between points 5 and 6, you can swap the connections to the start coil to reverse the motor. <S> That should only be done when the motor is stopped. <S> I would not expect a good outcome if you attempt to reverse the motor while the shaft is turning. <A> The simple answer is NO. <S> This is a single phase AC motor. <S> There is no polarity to switch to make the motor turn in reverse unlike a DC motor.
| Try starting the motor with the start winding reversed, it may start in the opposite direction. The four wires most likely refer to Neutral, Ground, 110, 220.
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Voltage signal and Current Signal Please give an intuitive explanation of difference between how a signal or data is carried using voltage and current? I'm wondering even though current and voltage inevitability co-exist, why do we use the term for one concept(current or voltage)? If there is a signal out there somewhere, it is both current and voltage at the same time. But we name only one of its property. We see in some circuits like heart rate signals or some other small signals which have to be in form of voltage in order to transfer the whole signal it to an amplifier (which has high input impedance). Even if it is voltage signal, we still have current passed to the amplifier.Why it is mentioned that voltage is transferred to input of the amplifier and why not current (when voltage causes current to pass)? <Q> Usually it has to do with what the author thinks is the best way to sense the signal that's on that pair of wires. <S> An example of a current signal is a 4-20mA current signaling loop. <S> A device that uses this expects to be fed a wide range and possibly varying voltage, and it will impose a current proportional to the quantity it's measuring on the power supply. <S> An example of a voltage signal would be the output of a typical op-amp. <S> The op-amp imposes a voltage which remains largely unchanged regardless of variations on the current, at least up to the point where the amplifier can no longer coerce the output to be correct. <A> There can be benefits to both in different situations. <S> And in reality, as you said, we never have a pure 'voltage' or 'current' - you have both. <S> The key is the impedance: A high output-impedance source will have it's output current change very little if the load changes, but the output voltage will change a lot - you want your load (say an amplifier) to measure the current. <S> A low output-impedance source will have a very small change in output voltage when the load changes, but the output current will change a lot <S> - you want your load to measure the voltage. <S> Take for example the inside of an integrated circuit. <S> If we have a voltage signal, what we actually mean is 'the voltage between two points' - usually this is between the signal wire and the ground. <S> However, the ground is not necessarily stable - other currents are also flowing through this ground and might cause voltage drops, which will lead to errors. <S> A current signal will not suffer from this moving ground - a current of say <S> 1 mA is still a current of 1 mA if you have a bit of difference in ground. <S> (this is incidentally the same reason why industrial sensors like using current mode sensors - the long wires can cause a significant voltage drop that leads to errors if you were to use voltage-mode signalling). <S> There is also the matter of impedance. <S> Some sensors have a high output impedance, such as a photo-diode. <S> Because of this, these sensors are usually used as a current-source, and not a voltage source. <S> On the other hand, at low speeds, a MOSFET appears like a high-impedance input voltage-to-current (the 'correct' name is transconductance) amplifier. <S> You want to drive it with a voltage, as it will turn this voltage into a current at its output. <S> So when driving a MOSFET, you want a voltage signal. <S> Things change when you go to systems with matched characteristic impedances at higher frequencies. <S> At that point, every device has the same impedance, and you don't really think about voltage and current anymore, and instead consider power. <A> We use the different terms because in the case of a voltage signal, the observed quantity is voltage, and in the case of current signals, current. <S> That's really all there is to it. <A> Data transfer especially with controlled impedances is often done with either voltage or current sources with source and load impedances matching the cable transmission line. <S> Voltage source amplifiers are true for single-ended or differential drivers like RS-485. <S> In Current Controlled Logic ( ECL,PECL,CML) <S> gain in speed by using Emitter Couple speed advantages and also uses differential line currents to reduce EMI and skew. <S> So both Voltage mode and Current mode drivers are used depending on the link speed ( or rise time in ps,ns or us )and distance. <S> The overlapping bit rates commonly used may be in the 100MHz to 1GHz range but are not hard limits but for >>1GHz , CML is preferred and is standard up to 10Gbps with higher levels possible. <S> 1Gbps Ethernet links use differential voltage drivers with controlled impedance lines and the current into the matched load is used by voltage amplifying receivers. <S> But it also uses a bandwidth compression form of modulation and compensation to make the cable less sensitive for signal integrity/ quality. <S> They also with BALUNs and transformers to magnetically improve the Common Mode noise rejection needed from line noise and ground noise differences. <A> If a signal is 5V and 1uA (or maybe even 1nA), would you rather be inspecting the large 5V? <S> Or the miniscule 1uA or 1nA? <S> You probably want to look at the voltage since it is larger and easier to read, so we call it a voltage signal since that's the component you are treating as carrying the information. <S> But if your signal is 10mA being pushed down wire by the transmitter applying 2.5V, which would you rather inspect for data? <S> They both seem to be magnitudes that are reasonably easy to inspect. <S> But wait... <S> that 10mA is a pretty high current that would have caused a voltage drop along the wires so if you inspect the voltage it won't be accurate to what is on the other end of the wire. <S> The 10mA current, on the other hand, remains unchanged down the line and all the extra power was spent to push it down the wire <S> so it would be a waste if you didn't take advantage of the current somehow (you could easily push 2.5V down the wire into a high impedance input so the current is much lower and therefore consumes much less power with much less voltage drop for a more accurate voltage). <S> Why wouldn't you inspect the current for the data? <S> So we call it a current signal.
| If they say "voltage signal" then they (probably) feel that it should be applied to a high-impedance amplifier that responds to voltage; if they say "current signal" then they feel that it should be applied to a low-impedance amplifier that responds to current.
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Is it possible to use 7805 regulator without capacitors at its input and output pins? I have to drive a magnetic buzzer whose current is 50mA and max. voltage is 6V. I will use 7805 regulator to drive it as my supply voltage is variable 9V to 24V DC. There will be another small load connected on this 7805 whose current will be 100mA. So total current drawn from 7805 will be less than 200mA. As the current is small so is it ok to use 7805 without the capacitors at its input and output pins? I want to avoid capacitors to reduce the component count to minimum. If instead of the two capacitors I only use 1 capacitor then which one to use.. input or output capacitor? and what would be its sufficient value for a current of less than 200mA? <Q> Can <S> a 7805 regulator be used without input/output capacitors <S> Short answer: <S> YES .... <S> but there are conditions <S> First you should read a datasheet for the regulator such as this one . <S> The datasheet clearly lays out: If your DC in supply is not too far away <S> (distance/length not specified) <S> you don't need a capacitor on the input. <S> If your input supply is a small AC/DC walwart it will have a filter capacitor on its output. <S> With a meter or so of cable you could probably do without the input capacitor on the regulator. <S> You don't need an output capacitor BUT the transient response of the regulator will suffer. <S> In your case you have a load that is quickly turning on/off (your buzzer, so it may well be advisable to add an output capacitor of at least 0.1uF. <S> so if your loads turn off then you should consider a resistor of 620 Ohms on the output for a 5volt regulator. <A> That has not really anything to do with the maximum current drawn. <S> The 7805 just needs the 100nF at input AND at output to run stable. <S> Without these capacitors there is a high probability of your output voltage to oscillate. <S> Don't save on these two cheap components, in the end it will more likely cost you even more because of failed parts you have to replace. <S> Also be aware that the regulator will have to dissipate a maximum power of (24-5)V * 0.2A = 3.8W. <S> So you definitly need to use a heat sink. <A> The simple answer is, the capacitors are not required as long as the input is a clean DC and/or the load does not demand strict transient regulation. <S> According to the datasheet, the two capacitors are required for the following reasons: C(in) is required if regulator is located at an appreciable distance from power supply filter. <S> C(out) <S> "improves" stability and transient response. <S> The above conditions tell us that the capacitor is optional in most cases, as long as the input is a clean DC. <S> But since it's cheap and easy to add them, it's a good practice to include them. <A> Capacitor is used for reducing voltage regulation.
| But if you have a good power source you may avoid the capacitors. It's also worth noting that there is a MINIMUM load current specified for the regulator (5mA)
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How to match a power amplifier and a loudspeaker array? I am trying to use a power amplifier that is available to me to actuate an array of loudspeakers. The array has a total number of 30 loudspeakers (ways of connection can vary). And I would like to know how I should connect the loudspeaker array to the amplifier to make sure there's enough power for the array and also not overload the array. Below are some parameters of the amplifier and the loudspeaker. Do you think serial connection in this case will be fine? If possible, would you please show the chain of thought to reach the conclusion? So will help me when I want to change the no. of loudspeakers in the future. Power amplifier: Power output capacity: 75 VA into a 3 ohm resistive load Input impedance: 15 kohm Output impedance: 0.04 ohm: [10Hz-5kHz], 0.08 ohm: [5kHz-20kHz] Max output current: 5A or 1.8A rms according to selected value Max output voltage: 15 V RMS Loudspeaker Input power: 2W (rated), 3W (max) impedance: 8 ohm Array The array will be composed of the loudspeaker above with a total number of 30 . <Q> Power output capacity: 75 VA into a 3 ohm resistive load <S> So your power amplifier can deliver the highest power into a 3 ohms load. <S> If you would connect all speakers in series you would get 30 x 8 ohms = <S> 240 ohms, then the maximum power to the speakers is limited by the voltage that the amplifier can make. <S> This is about 15 V, into 240 ohms that means P = <S> U^2 / R = 15^2 <S> / 240 <S> = 0.94 <S> Watt which is not much. <S> If you would connect all speakers in parallel you would get 8 ohms / 30 = <S> 0.27 ohms, then the maximum power to the speakers is limited by the current that the amplifier can deliver. <S> This is about 5 A, into 0.27 ohms that means P = I^2 <S> * R = 5^2 <S> * 0.27 <S> = 6.8 Watt which is more than the 0.94 Watt but still not much. <S> Also, this quite low impedance of 0.27 ohms is bound to cause problems, the amplifier simply might not like it. <S> Your best bet is to combine series and parallel to get close to 3 ohms. <S> My guess is that 4 Ohms would be good enough as well and somewhat easier as two 8 ohm speakers in parallel equals 4 ohms. <S> Then use series and parallel again to add more speakers but keep a total of 4 ohms, like so: simulate this circuit – Schematic created using CircuitLab <S> There are 8 speakers, between the top and bottom there's 4 ohms. <S> You can combine 4 of these in series and parallel <S> and then you'd need 32 speakers. <S> If you must have 30 speakers maybe you can use 5 <S> x 6 = 30 meaning make sets of 5 speakers in series, connect 6 of those sets in parallel, that would result in a total of 8 ohm * 5 / 6 = 6.7 ohms. <A> Your speakers are rated at 2W and 8Ω, which means that they can take a maximum of \$\sqrt{P\cdot R} = <S> \sqrt{2 \cdot 8} <S> = 4\$ <S> V RMS . <S> The amplifier produces up to 15 V RMS , so you need to connect the speakers in strings of 4 in series across the amplifier terminals in order to make sure that the speakers are not overdriven. <S> You can put any number of such strings in parallel. <S> Each string consumes <S> 15 V/32 Ω = <S> 0.47 <S> A RMS , so you can have up to 5 A/.47 <S> A = 10 strings before you overload that particular amplifier. <A> If MPT impedance is 3 Ohm load ( due to thermal IC limits) and you have 30 x Zo=8 Ohm loads, that would wish to arrange as a 3 Ohm load, then you have several options. <S> For an array of xSyP where x is the qty in a series string and y is the qty of parallel strings <S> ; \$Z_{equiv.}=3<= \dfrac{x}{y} \cdot Zo\$ <S> where x*y=30 and Zo=8 Solve this : 4S10P=3.2 Ohms which is > 3 Ohms and gives you closest to MPT with <S> a 75VA source gives you 75VA/30 = <S> 2.5 VA per speaker below its MAX limit of 3W peak and rated limit of 2Wrms.
| Another approach is to match impedances for maximum power transfer out without overdriving all your speakers.
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Adjustable Power Supply with LM317 I want to create an adjustable power supply for my electronics projects, and the cheapest/fastest option seems to be using an LM317.However, I do not want to blow things up by doing something wrong (what is pretty possible, considering my little knowledge in electronics), so I want to have a few questions cleared: What power should R1 be able to handle, considering a maximum drop of around 30v (from 32v to 1.25v) and, let's suppose, 1A of current (I have no idea of how much current I'll use, but I'm setting the value high to make sure)? If the calculation I did are correct, the power dissipated by the LM317 is of around 30w (I have a huge heatsink, heat on the LM317 shouldn't be a problem), but what should be the rated power of the resistor? 30w resistors aren't cheap, and the store I buy components from only offers resistors up to 10w (that are also very expensive). Do I need a capacitor between the input of the LM317 and ground? If I don't use one, what could happen? If I wanted to put a LED to indicate if my power supply is on, how could I transform my input voltage of 32v in the voltage of the LED without having to dissipate a ton of heat, and without expending much? How can I protect my circuit from a short circuit, so I don't blow everything if mistakenly connect something wrong? <Q> 1) As has been stated in the comments, you will never see 30V across R1 nor will 1A ever pass through it. <S> The voltage across this resistor is about 1.25V. <S> 2) <S> The datasheet should tell you whether, or under what circumstances, you need a capacitor to ground. <S> This information is in Figure 1 of the datasheet I found online. <S> 3) Heat is not measured in tons, and I don't know what you mean by "expending much". <S> Choosing a resistor to use with an LED has been discussed over and over again, on this site as well as others. <S> 4) <S> What does the datasheet for the LM317 say about short-circuit protection? <S> You should find all of the information you need. <S> Finally, I will say again that you need to revisit your plan to dissipate 30W in this regulator. <S> I don't think this will be possible without "expending much". <A> This is how it adjusts Output Voltage and when across a power shunt resistor becomes a current limiter instead with no ground reference or voltage regulation. <S> To test it , use another power transistor on the heatsink as an active load and simply bias it with 0.1V or less max current drop for sensing current with Vout . <S> Then put any sine wave on your DC bias and use XY on a Scope for current vs voltage to see your results. <S> Use a 300Wac or 30W <S> 24V Halogen light bulb or motor for current limiting your active FET to reduce the heat. <S> But warning, inductive loads it has no flyback diode, to Vin for back EMF clamping. <A> Power rating for R1 and R2 in a simple LM317 regulator <S> Since the voltage across R1 is a constant 1.25V, the power rating is 1.25^2 <S> / 240 = <S> 6.5mW so a 1/8th W resistor will be fine. <S> The voltage across R2 will always be that developed by the sense current through R1 (5.2mA, <S> and I'm ignoring the 100uA IAdj of the regulator). <S> So at the highest voltage you want out of your regulator (you said 29V) you would need R2 to be (29 - 1.25) / <S> 0.0052 <S> = 5336 Ohms and the power dissipated would be 0.0052^ <S> 2 <S> *5336 = 145mW. <S> I'd recommend that you use a .25W potentiometer for this purpose. <S> Do I need a capacitor between the input of the LM317 and ground? <S> YES. <S> While technically (and under ideal conditions) you might not need it, it's simply good practice to ensure stability with all the variations that can occur in wiring and on PCBs. <S> (You don't mention what generates your Vin … <S> .if <S> it's a rectifier <S> then you certainly need a large filter capacitor) <S> Can I use an LM317 to supply 1A load current over the whole voltage range? <S> MAYBE. <S> From the LM317 datasheet you need to look at one particular graphic to understand how the LM317 deals with current limiting (it's very novel, and most folks don't take time to understand). <S> Notice that when the input to output differential is small (<5V or Vout of about 27V) or large (>30V or Vout of about 1.25V) then the current limit is reduced. <S> It looks like you can get 1A at low output voltages, but above about 27V you may not get 1A. <S> Do notice that in the mid range (from about 7 - 24V) you will easily get your 1A output. <S> The LM317 is more than capable of dissipating 20W on a decent heatsink (altered from comments), though you'd be better considering a switching regulator to avoid large heatsinks. <S> Providing an LED power indicator <S> A simple series resistor and LED is all you require. <S> This can be run from the Vin voltage (32VDC), though it is of course a bit energy wasteful. <S> So your final circuit would look something like this: <A> LM317 won't be able to run 1.25A load fully from 32v input due to thermal waste. <S> max load without seriously illying the LM317 would near 0.8A. <S> specs say you're able to effectively reach 1.5A only when Vin/Vout difference is about 1/3. <S> not the case for variable regulation below 19V for this case. <S> it has a pretty bad thermal efficiency. <S> Its thermal efficiency rate has a lobe you could interpret by yourself later.
| R1 MUST have 1.25 V across it while the other R adjusts the output voltage.
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DIY generator. Simple and small I have read many posts on this site and everyone seems very competent, so here I am... very new to this. I'm trying to learn how to convert a motor into a generator. I'm just getting a bit overwhelmed with details and specs. Sorry I'm just a carpenter that likes to mess with stuff... I'm trying to basically build a diy generator, nothing big, mostly conceptual. I won't waste your time with my failed attempts and research. Could anyone just give me a rundown on the type of motor (bush less induction perminate magnet). Where to commonly find one (fans drills microwaves)and how to regulate the power if nessary. If I can even run a simple led or maybe a constant 5v to charge a USB device that would be fun. I know this has been posted before but again I'm just a hobby guy getting confused. And tips or rule of thumbs or advice or links is much appreciated. Thank you all for your help. Update: I just found a washing machine motor if that works. Researching how to convert it. Stupid question how can I tell if I need ac or dc, sorry I work with wood... Thank you for responding I am researching your advice <Q> How to convert a motor into a generator? <S> Easy motors <S> If it's a permanent magnet motor, it's easy. <S> Spin the motor, use the power, profit. <S> Brushed permanent magnet DC motors. <S> Often found in toys, and old-style cordless drills and screwdrivers. <S> These will produce DC. <S> The gearing is often reasonable to simply spin an electric screwdriver via the chuck with your turbine. <S> If you buy a BLDC (brushless DC motor) <S> these produce AC, which you can just 3-phase rectify from the output. <S> These produce a (to me) stupendous amount of power per volume, as they use very strong magnets. <S> This is what I'd use for a DIY windmill or turbine. <S> Mains synchronous motors like microwave turntable produce single phase AC, but at flea-power level, maybe worth a play. <S> Hard motors <S> If there are no magnets in it, it's hard. <S> Spin the motor, provide excitation voltage, or specific starting and running conditions, waste of time if you're not already a motor/electronic engineer. <S> Induction motors - fans, washing machines Wound motors - mains electric drills, food mixers, car starter motor <A> These motors are typically stiff and coggy when unpowered. <S> They also are magnetic, unlike induction motors and universal motors which spin freely. <S> There are basically 3 types of permanent magnet motors. <S> synchronous motors - <S> these motors require AC for operation and when used as generators produce AC when the shaft is rotated, they have only 2 power terminals brush motors - these motors require DC for operation and when used as generators pruduce DC when the shaft is turned. <S> electronically commutated motors - these motors have an electronic circuit that controls them, the circuit will need to be bypassed to use them as generators. <S> when used as generators they produce AC. <S> For an easy first success, the platter motor in most microwave ovens is a synchronous motor with a gearbox. <S> When the shaft is turned by hand, an AC voltage will be produced. <S> It can be directly connected to AC LED lamps of suitable voltage so that turning the shaft by hand produces light. <S> For a second project, perhaps a pulley or power drill spinning a DC brush motor (found in ink-jet printers, tape players, many toys) powering individual LEDs. <A> To light an LED, look for "ultra simple generator" made of cardboard and wire. <S> And magnets. <S> All of the little bitty DC motors are also DC generators. <S> But to create a few volts, you'd have to figure out how to spin them fast. <S> Or, just buy a little DC motor which already has a gearbox. <S> Turn the shaft slow with pliers, and the motor spins fast, as a generator. <S> Search: DC gearmotors, on hobby surplus sites like allelectronics.com, sciplus.com, <S> surplussales.com Also search: toy crank generator <A> As people have been saying pretty much any electric motor is also a generator. <S> The type of current it produces is generally the same as the type used to drive it (other than 'brush-less DC' motors which are actually AC). <S> That leaves you with two problems. <S> How do you turn the motor and how do you do something with the current it produces. <S> Given you want to make a 'generator', that probably means either turning it by hand or using some sort of combustion. <S> Turning it by hand is the most simple, either using a gear box or maybe a flywheel and just getting it turning really fast. <S> Now you have your mechanical power being converted into electrical. <S> It's not going to be great for using directly though, since the output is probably fluctuating quite a bit if you're turning it by hand. <S> You need something to regulate the voltage. <S> Voltage regulation and conversion is done all the time, so there are lots of options. <S> First you need to measure what voltages you are producing and also figure out if it's AC or DC. <S> With that knowledge you can choose the appropriate electronics to regulate your output into something you can use. <S> A buck/boost regulator is probably going to be your best bet <S> and it's possible to get them fully assembled on a 3 pin package. <S> That makes them as easy to use a linear regulator, just more expensive (but they are better). <S> If it's producing AC you will of course have to convert that to DC first, but this can be as simple as putting a diode on the outputs. <S> In summery: Get pretty much any motor and come up with a way to spin it. <S> Measure the out put of the motor while spinning it. <S> Use the measurements to get a buck/boost regulator with those values and the ouput you want (5V in the case of USB) Profit from your 'free' power! <S> From that point you could start looking at adding storage in the form of capacitors or even charging up a lead-acid battery. <S> And of course there is a lot you can do with at the input end. <S> Add a model car/plane petrol motor to turn it, etc. <A> To keep things as simple as possible, check which speed your mechanical power source has which power your mechanical power source has which voltage your generator should deliver Select a permanent magnet brushed DC motor with the matching characteristics in motor mode. <S> If the speed from your mechanical power source is way off, you need a gearbox. <S> Select a 5V DC buck-boost regulator with a roughly 3V…10V input range and the matching power rating.
| The easiest generators are made from permanent magnet motors.
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What would be the most efficient way of wiring a solar power bicycle? I have an electronic bicycle; 350 motor that runs off a 17.5 Ah battery at 36 volts. I've just ordered three 12 volt, 150 watt solar panels, that I'll wire in series to total 450 watts; giving 12.5 amps at full rating. I've got an MPPT controller to boost the charge voltage. However, the battery is limited to 2.5 amp charge current, meaning that the 10.5 will be wasted I suppose? Would it be better to wire the panels direct to the motor, where any excess can top up the battery, which can be used when solar power is less then what's demanded by the user. Is this possible? Or will the MPPT controller sort this out? Thanks for any help <Q> Is this possible? <S> Or will the MPPT controller sort this out? <S> It is physically possible, but you would need a combined controller for the MPPT, battery charging, and speed control. <S> It would be a fun project for an experienced circuit designer or two, but it's not something that you could likely make from off the shelf boxes. <A> This really difficult to answer without a schematic of your e-bike system. <S> But you almost certainly cannot connect your panels to the motor and expect the battery to charge. <S> You could rig it up to both charge the battery and supply the motor. <S> This would be a more complicated system. <S> Does your MPPT have more than one output? <S> Or does it just have the LiPo charging output? <S> There are some simple ways to do it. <S> They aren't great, but for example you could use a manual switch. <S> In one direction it would connect your battery to the charging port and disconnect it from the motor while connecting the other output of the MPPT to the speed controller. <S> Not exactly elegant, but allows you to charge the battery and power the motor from your panels. <S> Then switch it back when you need battery power. <S> To do it properly you would need to do as TimWescott said, with a combined system. <S> Possible, but you wouldn't be able to just go out and buy the part you need. <A> Thanks for the answers. <S> Apologies for lack of clarification in the question. <S> I will look at the MPPT charger and see if it has two outputs. <S> Otherwise it sounds like a custom job for an electrical expert. <S> The switch idea did occur to me Hekeke. <S> It's probably best to rig it up, and see if there are any performance issues and take it from there.
| I've never seen an off the shelf e-bike that has any sort of regenerative system, I assume because the amount of recovered energy simply wouldn't be worth the bother.
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Is a current varying output or voltage varying output for a 4-20 mA sensor more reliable for a rocket? For the internal plumbing system of a rocket which is expected to reach 100 km above the ground, industrial pressure sensors with a 4-20 mA current output and a 0-10 VDC output are available. What would be a more reliable choice? <Q> As far as signal integrity, 4-20mA is almost always better. <S> Two reasons why: Noise is usually less of an issue. <S> Circuit discontinuity is instantly detected. <S> As for its use in high g applications, that is more up to the sensor itself, both electrical and mechanical construction. <S> Note if it is more then just a sounding rocket, i.e. entering LEO, then you need to start considering radiation from the environment as well <S> and it's effect on the electronics. <A> There are several situations where choice of 4-20mA current is necessary a) <S> Where the receiver already exists and is 4-20mA b) <S> Where you need to transmit over a substantial distance of wiring c) <S> Where there are particularly high levels of EMI Other than those, you have a fairly free choice. <S> Most people find it easier to design around a voltage interface, and you may well use fewer components doing it. <S> In which case you'll least chance of messing it up going the 'easier' route of a voltage interface. <S> Bear in mind that in electronic design, getting an interface to do what you want when everything else is behaving as you expect is the easy bit. <S> Anticipating what can go wrong, and handling it without causing other things to blow up, that's the hard bit. <S> With a rocket, you usually only get one chance to test it live, so figuring out how to test it adequately before flight ought to consume >90% of your research. <A> I assisted very senior engineers in producing 6" BY 6" BY 4" HIGH black telemetry boxes for TRW. <S> In assisting them, I had the opportunity to examine the schematics. <S> The inputs were: ....high-level analog single-ended, .... <S> low-level analog differential-ended into amplifiers ....and digital switch-closures (logic levels) <S> MOSFETS (tho sometimes special-doped bipolars) were used as the analog multiplexors. <S> Analog Digital Converters, using 0.01% red-cased wirewound resistors in R-2R-4R-8R etc DAC structures, using the UA710 single comparator, were the state-of-art and extremely reliable. <S> [I'd written UA711, but was 710] <S> These small telemetry systems had to, after being conformal coated against fungus, sit in storage at Cape Canaveral for 25 years without failing, <S> then be launched in extreme vibration and shock, and then perform for 25 years in orbit --- all sorts of particles impinging on the circuits to degrade current gain and degrade junction leakage. <S> Glass sealed tantalums were the favored large-value capacitors; <S> glass capacitors used in precision timing.
| The reliability of the sensor and the control electronics themselves will be similar, especially if made by the same vendor.
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How to monitor high voltage capacitor bank? I am working with a large ring launcher and a capacitor bank that has the potential to charge up to 3000 volts in a bank of capacitors totaling 12,000 μF. I need to provide a sort of tap to measure the current capacitor bank voltage outside the circuit with a meter. First thought was a 10:1 voltage divider but I'm worried that might leak current and drain the caps too fast. A secondary concern, I'm not sure I could limit the current enough to bring that down to a safer level as well. Is there a more efficient way to monitor the voltage with minimal leakage or would a voltage divider be the more correct solution? Is there way to give a safe voltage/current on a tap in either regard? <Q> A 1000:1 voltage divider night be better. <S> Make sure your series resistor has an adequate voltage specification. <S> If you don't want to buy high voltage resistors, then often what folks do is put several ordinary ones in series. <S> You are unlikely to run your cap bank down too quickly with (say) 10 off 10Meg resistors in series. <A> A simple resistive divider chain using resistors of a very large value plus an opamp with a high impedance input will work. <S> e.g. Use 1G ohm resistor made of 5 x 200 Meg 1206 resistors, and a single 1 Meg 1206 resistor at the bottom of the chain. <S> Feed the voltage across the 1Meg into decent JFET opamp, configured as a buffer. <S> As described above, for 3kV you will need to use a resistor chain to prevent breakdown of the resistors and conformally coat the result to prevent air and surface leakage. <S> This design would create a 3uA discharge of the cap. <S> If you want lower discharge go for larger resistors and spend more time thinking about other forms of leakage. <S> You should be able to get to about 10GOhm before running into issues with other losses. <A> An high input inpedance and low bias current opamp is needed, for example ADA4817-1, ADA4530-1. <S> Then you buffer this signal with cable driver opamp, for example THS3202. <S> This is a Micsig differential probe - new approach and low cost. <S> See resistor string with parallel capacitor string for compensation. <S> The input amplifier is ADA4817 <S> -1 Source EEVBLOG . <S> Older variants of Tektronix, Lecroy,.. <S> probes used a matched differential pair JFET transitor as input pre-amp, but this is very expensive part, newest opamp seems to do the job quite well.
| In addition to Neil_UK and Jason Mason: Have a look to differential probe projects on EEvblog and others, with the exception than you may not need a differential input signal (actually you need just a half of the differential probe) and probably you also don't need a compensation RC circuitry, also you don't need high bandwidth. With any cap bank, it's good practice to have a resistor permanently attached to bleed the charge off when unused. Sounds like you need a 1000:1 potential divider with a very high impedance.
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How Does A USB (non-PD) device modify the current it draws from a charger? I understand at a basic level that a USB DCP simply advertises that it is a DCP, and then basically puts 5V on Vbus, and then the battery can go to town. With that in mind this is really a 2 part question: How does a battery limit it’s current to the 1.5A specified by the USB BC 1.2 spec? Take the example of a charger capable of outputting 2A, advertising as a DCP. Assuming the device being charged is compliant with the spec, how does it limit the current draw to 1.5A given there is no communication between the two devices? Does it modulate some sort of load switch? Does it even limit in the first place? Are USB portable devices responsible for implementing the typical lithium ion battery Constant-Current/Constant-Voltage charge cycle? If so, does it taper the current by some method described in (1)? Or perhaps it relies on the increasing impedance of the battery as it becomes more fully charged? Perhaps it doesn’t implement a CC/CV charge cycle at all? I appreciate any feedback on this. It seems like the answer should be fairly straightforward, but it’s been bugging me for a couple days. Thanks! <Q> Chargers that use the USB VBUS as power source are very sophisticated mixed-signal processors, with extended analog functions and digital controllers. <S> Below is a typical simplified diagram of a charger <S> IC <S> that takes and limits the intake, charges a Li-Ion cell in accord with source capabilities and follow CC-CV charging algorithm, and provides system voltage output to mobile device processors and other circuitry: <S> Brief explanation of entire functionality can be found in the datasheet. <S> The IC takes substantial amount of analog information, digitally processes all thresholds, and generates necessary limits, timeouts, and all other sequencing. <S> In short, modern charging ICs do limit the input current intake by changing switching regulation parameters. <S> IC will reduce the cell's charing current if it is required to maintain the input limit. <A> Even though DCP has "Charging" in the name it does not incorporate any charging functionality inside it. <S> Think of the USB DCP as a 5V power supply with a maximum output current of 1.5A. <S> For Li Ion and Li Polymer batteries the standard way of charging is using constant current (whose value will depend on the battery itself) till the battery voltage reaches 4.2V and then using a constant voltage (4.2V) after that point. <S> The controller takes care of the battery voltage monitoring as well as current and voltage control. <S> It should also make sure that the whole system never draws more than the 1.5A available from the DCP. <S> For small batteries, a linear controller is typically used, where a simple series transistor is responsible for the regulation. <S> For bigger batteries a switched topology (Buck or Buck/Boost depending on the number of series battery cells) is used in order to reduce the internal power dissipation. <A> The simple answer is that often they do not limit the current. <S> If you plug your thing that charges at 2.5A into your dumb wall plug charger that only supplies 1.5A, there is a good chance it just causes the wall plug to burn out (or at least shut down). <S> Apple has its own system where a certain voltage is present on the data pins to indicate what current the source can supply. <S> The other convention is for the data pins to be shorted which indicates a 'high power' source. <S> All charging logic is contained in the device being charged.
| The device on the other side of the cable will have its own charging controller circuitry that will limit the current and voltage that is delivered to the battery.
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What power adapter do I need to charge a 7.2V 4000mAh battery that I made? I made a Ni-CD 7.2V 4000mAh battery. What power adapter do I need to charge it? How many volts and how many amperes should the adapter output? Each battery cell is 1.2 volts.I have a 9V 500mA power adapter.Is this adapter suitable for charging my battery and how long it will take?The adapter will automatically detect if the battery is full to stop charging it? <Q> You made the battery, shouldn't you be telling us what we need in order to charge it? <S> Typically I would say you need a 400mA constant current with a voltage of 1.41V * number of cells. <S> You would then charge it for 16 hours with some sort of timer. <S> At that rate it would be fairly safe regardless of discharge level, but you still don't want to leave it charging for longer than the 16 hours. <S> If you wanted to charge it faster (and you know what the maximum rate of your battery is) you would need to monitor the voltage and/or temperature. <S> But I think this is probably beyond the scope of your question. <S> In regards to your edit: 9V and 500mA are both slightly too high. <S> The main concern is the 9V, since you should be using 8.5V. <S> The 500mA isn't too much higher, but you would need to reduce the charge time. <S> Now if you use that to charge your battery, it will charge. <S> But the over voltage will be damaging it. <S> You could probably solve this problem with a low loss diode. <S> I'm not clear if what you're using is a purpose built battery charger or just a random power pack? <S> In either case I'm going to say it probably wont stop charging, since it seems like it would expect a different battery if it's outputting 9V and wouldn't be calibrated correctly to switch off for yours when it's charged. <A> Depends. <S> If you want to charge it slowly, as stated in @hekete's answer, charge at a rate of C/10 (means 4000mA/10=400mA) for 16 hours and the charger's output voltage should be 1.4V per cell, which equals to (7.2V/1.2V) <S> x 1.4V = 8.4VDC. <S> But with this method, you need a mechanism (i.e. MCU, timer, etc.) to stop charging so that it does not past 16 hours. <S> If you want faster charging, C/3 for 5 hours or C/5 for 8 hours (with the same charging voltage) can be acceptable. <S> But please note that you need to check battery voltage and temperature as well. <S> Also, to ensure any problems not to occur, you need to discharge the battery down to at least 1V per cell then start charging. <A> Back in the days when people used NiCd batteries, you could buy off-the-shelf chargers that would charge them. <S> I still have a 7.2V "racing pack" charger specifically designed for such a job. <S> You would need to limit the current to no more than 400mA (C/10). <S> Without knowing the specification of the adaptor, it's hard to tell how sophisticated that current limiter would need to be. <S> NiCd cells can be trickle charged at rates of up to C/10 for long periods safely. <S> NiCd batteries are largely obsolete now, as: <S> NiMH ones have a larger capacity. <S> Cadmium is toxic.
| A 9V adaptor would over-charge the pack, quickly damaging either the adaptor or the battery.
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Can you actually break an FPGA by programming it wrong? Can you actually break an FPGA by programming it incorrectly? I'm a software guy really. It's no secret that if your software is wrong, you could destroy all sorts of important data, and perhaps even crash the whole machine. But it's really difficult to physically damage a computer just by programming it. (There are endless rumous of a Halt-And-Catch-Fire instruction, or being able to reflash the system firmware to brick the motherboard, or programming incorrect values into the graphics card to fry your monitor. But these all seem to be exactly that: rumours. And all about long-obsolete hardware. It seems really, really hard to break modern computer equipment with bad programming.) With an FPGA, you are (at least nominally) wiring individual circuits together. It seems completely plausible that physical damage might occur in case of a mistake. For example, you could write some VHDL requesting that two outputs get tied together. If they output different logic levels, I imagine that would probably fry something. (I would hope that your synthesis tool would scream at you to not do this... but I don't know if such tools actually implement that level of error checking.) It also seems quite possible to accidentally pick the wrong model of FPGA in the synthesis tool, and thus ending up trying to program your chip with a bitstream intended for some totally different model. I don't know what that would do, but I suspect it would be "bad". For that matter, you could definitely connect the FPGA chip to the rest of the circuit incorrectly. E.g., if you mess up the pin numbers, you might end up with the board trying to drive an I/O pin that the FPGA itself is also trying to drive. Do the I/O pins typically have any "protection" against such a mistake? Or will the chip just fry? <Q> Typically the programming software will query the part being programmed for its part number, and refuse to program in a bitstream meant for a different model of FPGA. <S> The part itself will also generally refuse to start up if programmed with a bitstream that isn't exactly the correct length (and it's very uncommon for bitstreams for different chips to be the same length). <S> you could definitely connect the FPGA chip to the rest of the circuit incorrectly. <S> E.g., if you mess up the pin numbers, you might end up with the board trying to drive an I <S> /O pin that the FPGA itself is also trying to drive. <S> This is the most likely way to damage an FPGA with wrong programming. <S> Another way could be to program a very resource intensive design and run it at a high frequency (so that a high power is consumed), and then run it on an FPGA without an adequate heat sink. <S> Do the I/ <S> O pins typically have any "protection" against such a mistake? <S> Or will the chip just fry? <S> Output pins will "often" survive a short circuit condition for a few seconds or even minutes. <S> But nothing is guaranteed. <A> With a few noted exceptions, tools don't generally give you access to the actual silicon primitives, so it's hard for an end-user engineer to load an electrically invalid design* into an SRAM-based FPGA, except perhaps by inadvertently discovering a tool bug. <S> Flash based FPGAs might conceivably have their re-programmability damaged by certain invalid loads. <S> OTP FPGAs implicitly are "damaged" by even a valid configuration load, as it can never be changed. <S> Ultimately what comes closest to what you are seeming to ask, and to your HCF example, would be a configuration which produced intolerable thermal stress. <S> Power consumption is quite directly driven by clock rate and volume * activity of utilized logic, so if you can trick the tools into uselessly toggling most of the flip flops on the chip at maximum clock (there are ways...) <S> then you can produce a pretty effective heater that would exceed most cooling systems for ordinary usage. <S> Then it's just a question if something protectively shuts it down before it cooks. <S> And of course there are power estimation models in the tools, which are likely reasonably predictive if you're not lying to them about the clock signal provided. <S> (* There is one interesting class of not-a-bug electrical issue you can cause by lying to the tools <S> , that is not necessarily physically destructive, but still surprising. <S> If you feed a clock different than you said you would or simply unstable, you can violate address setup timing on the synchronous block RAM cells, and do something along the lines of shorting them and corrupting their contents - so you can for example see the contents of something designated a ROM in the design actually change at runtime just by trying to read it with a bad clock. <S> But I don't believe this is physically destructive) <A> The most likely thing is violating the current rating on GPIO's by driving a pin that is already being driven. <S> Some FPGA's have settable current limits or changeable output drivers, so this can either help/hurt you if you don't get your port map done right. <S> You should be double checking your port list anyway before programming as mistakes like swapping pins can take hours to resolve <S> , it's best to get ahead of the mistakes and know exactly what the firmware was intended to do. <S> (unless you like the thrill of finding an error) HDL's by themselves usually don't let you connect two output's to the same wire and will stop synthesizing and make you fix your mistake if you do have code that does that. <S> One place that could cause problems is bi-directional ports, but you should have current limiting resistors on those. <A> As with microcontrollers, you can always exceed the maximum total current per IO bank by drawing a maximum current (or more) from each pin. <S> Unless the FPGA has built-in protection against such situation, this may result in damage. <S> Another possibility is making a combinatorial loop which either periodically goes meta-stable, or oscillates at a far higher frequency than the FPGA fabric is designed to handle (several GHz). <S> That will cause a very localised overheating which may do physical damage before the chip-wide thermal protection kicks in. <S> That is, assuming there is such a protection: if over-temperature doesn't lead to shutdown, you can simply come up with a very power-hungry circuit and let it run with insufficient cooling. <S> Dynamic reconfiguration can also work around protections against invalid configuration of internal primitives which may be enforced by the development tools in case of a static configuration. <S> For example, you could configure a PLL in a way that exceeds its maximum internal frequency, or feed the same interconnect line by two sources at once, or force a pin from a high-voltage IO bank to use its low-voltage transceiver like LVDS.
| It also seems quite possible to accidentally pick the wrong model of FPGA in the synthesis tool, and thus ending up trying to program your chip with a bitstream intended for some totally different model.
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For decoupling an IC's power supply pins, is there any reason to use multiple capacitances when all the MLCCs have the same package size? A PCB I'm working with has three decoupling capacitors on an IC's Vdd pin - 0.01uF, 0.1uF, and 1uF. They are all in 0402 packages. I understand that capacitors of multiple sizes are often used because parasitics tend to scale with size, but I was under the impression that this was generally because of different package sizes, not the actual capacitance values themselves. I also understand that putting multiple capacitors in parallel will increase the overall capacitance while decreasing parasitic values, but I don't see why one would use varying capacitances for that rather than just using the largest value possible for each capacitor. Is there any reason to use multiple decoupling capacitors with the same package but different capacitance values? Here's a screenshot of the relevant portion of the circuit schematic (C1, C2, and C3 are all 0402 MLCCs): <Q> Your initial impression is incorrect. <S> Here is what the impedance of various capacitors in the same package looks like. <S> Source <A> Using a Murata's SimSurfing tool, I graphed the impedance vs. frequency curve for a 2.2uF 0402 (1005 metric) MLCC compared to an 0.1uF one in the same package. <S> The 2.2uF cap is shown in blue in and the 0.1uF <S> in green: As you can see, the point of resonance is higher in frequency with the 0.1uF, as would be expected of a larger cap with the same parasitic inductance, but, less expectedly, the smaller MLCC achieves a slightly lower impedance between 10 and 40 MHz, at the expense of higher impedance at lower frequencies, as would be expected given the greater capacity of the 2.2uF cap. <S> So the conclusion is that there's something about the internal structure of the large-valued MLCCs that slightly worsens their high-frequency performance, but below the point of resonance, there seems to be no benefit to the smaller MLCC in a decoupling application. <S> Of course, the larger capacitor will also have worse performance under DC bias, but generally the larger ones will still end up having a larger effective capacitance under DC bias. <A> Indeed there is. <S> The most obvious one is cost. <S> Ceramic capacitors of different values in the same FOOTPRINT (not necessarily package since height may vary) do not cost the same. <S> Beyond that, ceramic capacitors have different impedance curves (due to the different parasitics as you mentioned) and DC bias curves for each combination of capacitance, dielectric, voltage rating, and package size. <S> It's enough to make your head spin. <S> From what I've seen, the tendency is that, all else being equal, larger packages have more inductance and therefore hit resonance at lower frequencies, and that squeezing more more capacitance and/or max voltage rating into a smaller package degrades the DC-bias characteristics. <S> I suggest you go to Murata's SimSurfing website ( https://ds.murata.co.jp/simsurfing/mlcc.html?lcid=en-us ) and filter out their GRM series capacitors and only look at the X7R capacitors (so you don't get overwhelmed and since effects of dielectric are fairly straightforward). <S> Then compare the "Tech-PDF" of different capacitors where all but one of the parameter voltage, capacitance, and package vary. <S> I also understand that putting multiple capacitors in parallel will increase the overall capacitance while decreasing parasitic values, but I don't see why one would use varying capacitances for that rather than just using the largest value possible for each capacitor. <S> Beware... <S> Antiresonance of multiple parallel decoupling capacitors: use same value or multiple values? <A> Impedance of capacitor depends on the frequency of the signal passing through. <S> Our job would have been easier had all signals been ideal. <S> But they aren't! <S> There are different frequency components in any signal. <S> In order to selectively filter out the undesired signals, you need different frequency dependent impedances. <S> Hence, different value capacitors! <S> Check out this video by Dave Jones @EEVBlog <S> https://www.youtube.com/watch?v=BcJ6UdDx1vg <S> To get a visual feel of the concept watch this https://www.youtube.com/watch?v=1xicZF9glH0
| In order to get a sufficiency flat power supply impedance over a wide bandwidth, one needs to use a selection of different capacitors. There may be some small benefit.
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3hp motor contactor size 22A or 40A? So my place has 2 contactor sizes of 22A and 40A. Now i know that both will be more than enough to power up a 3HP motor. To avoid oversizing in this case, would i need to provide additional fusing to help protect the 3HP motor if it has a 40A contactor as opposed to a 22A? This is 480V system that is sized for a 3hp motor ( electrical disconnect that is) There is also a 40A overload to be used with this 40A contactor. It jus means the overload and contactor can handle up to 40A each Basically it's a electrical disconnect like a fuse breaker (sized for a 3hp motor) u have a ol on the starter I dont think there is an ol on the breaker. Normally with < 5hp motor running at 480 your minimum circuit ampacity is <10A so either 20A contactor or 40A contactor is more then enough. <Q> To avoid oversizing in this case, would i need to provide additional fusing to help protect the 3HP motor if it has a 40A contactor as opposed to a 22A? <S> No. <S> Fuses are not for protecting the motor they are for protecting the wiring and other current-carrying components from short circuits and ground faults. <S> They also prevent a fire if the motor or motor protection fails. <S> An overload relay, motor protective circuit-breaker or temperature detecting device in the motor are the possible motor protective devices. <S> There is also a 40A overload to be used with this 40A contactor. <S> It jus means the overload and contactor can handle up to 40A each <S> It does not just mean it can handle 40 amps. <S> The 40A overload may not be appropriate. <S> Some have interchangeable current sensing parts that allow a wide range of motors, and some are adjustable. <S> However there is a limit to the range of motors for which they are appropriate. <S> The documentation for the overload tells what motor voltage & power or current for which they can provide appropriate protection. <A> For more reliable motor protection, use the actual motor current as listed on the motor nameplate which ought to be around 5A. <A> Normally with < 5hp motor running at 480 your minimum circuit ampacity is <10A <S> so either 20A contactor or 40A contactor is more than enough . <S> The pieces protecting the branch are the overload device (motor overheating) and the circuit breaker (short circuit, overload). <S> The contactor provides control, and the disconnect provides a means of locking out the circuit. <S> For the most part a general motor protection branch looks like the following: <S> Assuming that: Disconnect is sized appropriately (if applicable), Circuit breaker/fuses are sized to protect downstream components, Wiring is rated for more than the circuit breaker/fuses, and Overload can be set for somewhere in the 6 A range (3 hp @ <S> 480 VAC = <S> 4.8 <S> A)(4.8 A <S> * 125% = 6 A) <S> The choice between a 22 A and 40 A contactor is moot. <S> A contactor is a glorified switch and the ratings are for its own function. <S> You want to avoid under-sizing a contactor, but you can oversize. <S> Keep the 40 A around for a 30 hp motor. <S> Notes: <S> A 40 <S> A overload may only be adjustable for something like 40 <S> ... 60 A since it is designed for a 30 hp motor. <S> 40 <S> A minimum overload setting is not appropriate for a 3 hp motor.
| More than enough is not the problem when it comes to motor protection. The motor protective devices are always selected and adjusted based on the motor ratings, Hp & voltage, current or winding temperature. We need to be concerned about too much . Go with the 22 A, it will save a couple bucks.
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How do batteries connected in parallel give more current than batteries connected in series? Let's say we have three 1V batteries and we're connecting them to a 1Ω resistor. Connected in series, the voltage is 3V and since the resistance is 1Ω, that means that from Ohm's law, the current should be 3A right? And then when we connect them in parallel, the voltage is 1V and the resistance is 1Ω so the current is 1A right? So do I misunderstand the statement that batteries connected in parallel supply more current or do I misunderstand Ohm's law? <Q> As others note "can" and "will" usually differ. <S> Imagine each battery had a chemical to electrical conversion capability such that it COULD deliver up to 0.5A. <S> If you connected a 1 Ohm load, Ohm's law would allow 1A <S> IF the battery was able to supply it. <S> But, as the battery was only able to supply 0.5 <S> A max you'd see V = IR = 0.5 <S> x 1 = 0.5 V across the resistor. <S> ie the battery voltage would sag due to its limitations. <S> Now use 3 similar capability batteries in parallel. <S> Ability is now "up to 1.5 A" and actual with 1 Ohm load will be 1A, as expected. <S> In situations like this which seem to not make sense, work through the problem and see how the resultant "needs" match the capabilities. <S> Here, with one cell, if it CAN provide 1A or more then all is well. <S> If the per cell capability is smaller than the need, work out what is impossible. <S> Here if cell can make 0.5A then. <S> V = IR = 0.5 max x 1 Ohm = 0.5V is false (if battery is at 1V)R <S> = V/I = <S> 1/.5 <S> = 2 <S> Ohms = <S> false I = <S> V/R = <S> 1/1 <S> = 1 <S> A = false <S> if max = <S> 0.5A.Choose <S> any 3 :-) - all false. <S> Now use 3 cells and try again. <S> Ah! <A> You're confusing unrelated concepts. <S> You have correctly calculated with Ohms Law how much current a particular circuit will draw. <S> But this is not related to how much current your supply can deliver. <S> But how can the parallel batteries deliver more current? <S> Well first consider them in series, the current must pass through all of the batteries. <S> So the total current can't exceed what a single battery is capable of providing. <S> This is limited by the resistance of the battery. <S> If a battery can deliver 1A, then no matter how many you put in series only 1A can come out. <A> The parallel-connected batteries are capable of delivering more current than the series-connected batteries but the current actually delivered will depend on the applied voltage and load resistance. <S> You understand Ohm's Law, but the "parallel batteries supply more current" statement <S> should really be "parallel batteries CAN supply more current". <A> This shows some 1.5V battery with some equivalent capacitance and ESR for each. <S> The parallel voltages are matched before putting in parallel. <S> The series batteries are fresh and have same capacity in mAh before loading. <S> Mismatch increases towards end of life so the weakest cell fails 1st. <S> The short circuit test , Isc is momentary. <S> simulate this circuit – <S> Schematic created using CircuitLab Conclusion <S> Yes, parallel batteries "can" supply twice the current when the load is less than the ESR of the battery. <S> ( As shown above, for short circuit current, it is twice.) <S> But otherwise, when the load is equal to battery ESR, the current is the same. <S> With series cells it greater when the load R is higher than ESR, the higher V/R produces a higher current. <S> OHm's Law always works when you can estimate the battery ESR= ratio for some >10% voltage drop for some rising current. <S> Load R should rarely go down to ESR of the batteries but often users underestimate the DCR start current on motors. <S> Some efficient motors can be up to 12X rated current putting the DCR=Vbat/Idc close to the total battery ESR in a string. <A> internal resistor can make a big difference 1- series connection <S> total internal resistorrt= r1+r2+r3 let have an example of 0.5 Ω internal resistorand let the voltage each battery equal to 1v <S> the load is 1 Ω as your exemple rt=0.5+0.5+ <S> 0.5 <S> rt=1.5 Ω total resistor = <S> R load+ <S> rt R= 1 <S> +1.5=2.5 Ω I= <S> Vt/R load+rt <S> Vt=1+ <S> 1+1=3 <S> V <S> I=3/ 1.5+ <S> 1 =3/2.5= 1.2A <S> 2- parallel connection <S> the total resistance is the sume of all resistor 1/Rt=1/r1+1/r2+1/r3+1/Rload 1/Rt=1/0.5+1/0.5+1/0.5 <S> +1/1 <S> Rt=1/7=0.14 <S> Ω <S> I= <S> vt <S> /Rt I= <S> 1/0.14 = <S> 7.14 <S> A !! <S> so what do you think <A> If an amp is flowing through a device with two electrodes, that means that one electrode is losing approximately 6.242×10^18 electrons per second worth of charge while the other is gaining a like amount. <S> If two batteries are connected in parallel to a load, every electron's worth of charge that leaves the negative electrode of either battery will pass through the load before returning to the positive electrode of the same battery. <S> If they are connected in series, each electron's worth of charge that passes through the load must pass through both batteries. <A> Think of garden hoses. <S> Think of water as current. <S> Attaching several hoses in a long line is like connecting in series. <S> You can only force so much water down the single hose. <S> Tying several hoses together in parallel so that there are multiple hose ends sprouting water is analogous to batteries in parallel. <S> Lots more water can come out the end.
| If we put them in parallel however, there is a separate current path through each battery and their total available current can be added together.
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Multiplexing low speed SPI slaves with a high speed SPI master (time-division?) I have 12 SPI slaves, supporting SPI at a maximum clock rate of 2MHz.The SPI master (an ESP32 ), can run at up to 80MHz (and integer divisions thereof). My slaves are all 8-channel ADCs ( MCP3208 ), which I need to poll continously. Therefore, I want to somehow maximize the throughput. I was thinking that it might be possible to do time-division multiplexing on the slaves, by running the master at 40 MHz and create 20 time divisions, each being a 2MHz line for a potential slave. This would effectively combine my slaves into a single SPI device with higher speed (while interleaving all bits for the master). Unfortunately, I cannot find any resources on this topic. Below is a diagram showing the concept, by using an external clock divider to create a 2MHz clock from the 40MHz base clock (which should ideally be offset for each slave..), a multiplexer for MOSI, and a demultiplexer for MISO lines. Unfortunately, I am missing the required background knowledge to judge if this is at all possible, and if the required ICs even exist. <Q> I don't think this is at all practical. <S> First, how are you going to get the SPI interface to interleave data? <S> I doubt very, very much that you can pull this off with a dedicated synchronous serial peripheral, and as @ThePhoton points out, if you have to bit-bang the interface the effective clock rate will fall dramatically. <S> You need a separate 2 MHz clock for each ADC, so that each device sees the appropriate clock edge only when its correct data value is present. <S> You can't just demux the data lines unless you add storage for each data line so that it holds the correct value until the clock edge occurs. <S> For incoming data you need to do the reverse... capture <S> each ADC MISO line on the appropriate clock edge. <A> In addition to what's shown in your diagram, you also need a counter to generate the SEL signals for the [de]multiplexers. <S> And this counter needs to count through 12 states rather than 8 or 16. <S> Before you're done you'll be implementing all this in a CPLD. <S> I'd recommend instead to find a different ADC. <S> Either a multi-channel one with a faster SPI interface. <S> Or a parallel output one that will let you do the readbacks through something like a 74LVC595A serial-parallel buffer (which can then read from your SPI port at well into the 10's of MHz). <A> The ADC chips will not work with a clock frequency higher than the 2MHz specified in the datasheet. <S> Your best bet will be to divide down the MCLK signal generated by the MSP and provide it to all ADCs in parallel. <S> From the datasheet, the ADC chips have a maximum conversion rate of 100k samples/second and this is dependent on a the clock that's driving the chip. <S> If you use common MCLK, MOSI, and /CS lines with each MISO pin tied to a separate GPIO pin, you could bang out a common command to all chips at the same time. <S> Reading the values from all 12 ADCs would then be done by reading the GPIO registers on each cycle of the bit-banged clock. <S> On most micros, with care choosing the GPIO pins, it would be possible to read the state of 12 pins with only a few instructions: a byte- or word-wide read into a register and a bit shift on the register, repeated 12 times for the 12-bit ADC values. <S> The register would then hold the bitstream to be sorted out in software, possibly in a different (not time-critical) section of code. <S> I've never used an ESP32; I don't know how practical/possible it would be to read the GPIO pins in this fashion. <S> If you can't control the GPIO timing of the ESP32 due to interrupts happening (while it manages the wifi and stack, for example) it may be worth considering adding a cheap microcontroller to sit between the ESP32 and the 12 ADC chips and handle all of this for you; presenting instead a single SPI- or UART-connected slave that you could use with the hardware-assisted blocks present in the ESP32.
| One way you could approach this would be to bit-bang the entire thing.
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Thin wall to block LED light from hitting photodiode? I'm designing a small heart rate sensor using green smd LEDs and a photodiode. The photodiode needs to measure the reflected light, and not direct light from the LEDs. The LEDs are near the perimeter of the photodiode, so I'm trying to figure out some sort of thin wall to block their light from hitting the photodiode. The LEDs have a relatively wide viewing angle (110˚) and are ~1mm tall, so ideally something that is about 2mm tall. I'm wondering if anyone has suggestions for some type of wall or component I could place between each LED and photodiode to block light? <Q> My advice is to not bother doing what you're doing, and spend the $3.50US to get the part designed for the task, like https://www.digikey.com/product-detail/en/osram-opto-semiconductors-inc/SFH-7060/475-3174-2-ND/6137022 <A> I have also made little barriers out of thin cardboard stock cut from packaging...easy to cut and form... <S> that worked well. <S> Hotglue, epoxy, and silicone all make decent adhesives. <A> It's not even unheard of to place otherwise useless cheap parts to act as walls: it may be much cheaper than special light-blocking components, especially if you already have a suitable part on your BOM, and ordering it in higher quantities will make it even cheaper. <S> Light pipes can also work wonders, if your PCB / case can support them. <S> They are made of transparent acrylic plastic, but thanks to the internal reflections they can be near-100% opaque from the inside. <S> I'd also take a look at LEDs with smaller view angle and photo-diodes with a smaller field of view. <A> Make some slots in the PCB and solder a small piece of metal there. <S> I've seen it work remarkably well in real-life applications. <A> mylar? <S> The thinnest thing i know of that blocks <S> 100% of light, developed by NASA to block heat and light with minimal addition to the weight of a craft <S> , fuel ain't cheap. <S> Mylar reflects light so if you do not want that then black electrical tape or something black, with a rougher surface than electrical tape, will reflect less. <S> Charcoal finish would be perfect for that.
| Sometimes reworking your PCB a bit allows you to place an existing part between the LED and the sensor which will block the light. I've had luck with drilling holes in lids to microcentrifuge tubes (or other small bottles), cut pieces of copper tubing from the hardware store, or even a piece of black electrical tape in a pinch (my least favorite option). Be careful with silicone, as the RTV stuff can sometimes corrode electronics.
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When you connect two batteries in series, why doesn't the middle short? When you connect two batteries in series, why doesn't the middle short? The connection between the positive terminal on one battery and the negative terminal on the other battery has no load, so shouldn't it be shorted? But I don't notice heat or explosions or anything like that when I connect batteries in series. <Q> When you connect two batteries in series, why doesn't the middle short? <S> The connection between the positive terminal on one battery and the negative terminal on the other battery has no load, so shouldn't it be shorted? <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Here the intended circuit is the loop consisting of the battery and lamp. <S> The short-circuit is bypassing the lamp and providing a "short-cut" between the terminals of the battery. <S> If the resistance is low then high currents will flow. <S> simulate this circuit Figure 2. <S> (a) <S> No short-circuit occurs when the batteries are properly connected in series. <S> (b) <S> BAT3 is short-circuited while BAT4 is not. <S> But I don't notice heat or explosions or anything like that when I connect batteries in series. <S> That's because no current is flowing as in Figure 2a. <S> From the comments: <S> How come in Figure 2a <S> there is no current flowing between the positive terminal of BAT1 and the negative terminal of BAT2. <S> Because there is no closed circuit. <S> I thought BAT1's positive terminal has a deficiency of electrons while BAT2's negative terminal has an excess of electrons. <S> There is no surplus or missing charge at either terminal. <S> So shouldn't the excess electrons in BAT2 flow towards the positive terminal of BAT1, thus creating a current? <S> Current can only flow when there is a circuit. <S> As shown in Figure 1 the circuit is open so no current can flow. <S> I thought a battery separates itself into two sides, one positive and one negative, with a barrier in between. <S> If that were true then batteries suspended by a thread around their middles would all line up due to electrostatic attraction. <S> (They don't.) <A> Your problem is there are two definitions for the phrase a short : A) <S> A short is high current flowing through a wire unintentionally. <S> B) <S> A short is two places connected by a low-resistance interconnection (as a wire). <S> Both are correct <S> but as much as A implies B , this does not mean B implies A . <S> Using the same term a short for both makes you think they are equal, which they are not. <S> When you connect the plus from one battery to the minus of the other, you have a short of the second kind. <S> However, there is no current flowing, as this requires a circuit —a closed loop— <S> so obviously, B does not imply A . <S> As soon you connect the plus from the other battery to the minus of the first also, there is a closed loop, and <S> your short of the second kind grows into a short of the first. <S> EDIT: There was some confusion about the heat production as well, which I explained in the comments before. <S> Where the batteries are connected bumper-to-bumper, you have a very low resistance <S> \$R_b\$ of maybe 0.01 Ω. <S> The outer wire resistance <S> \$R_w\$ <S> be 0.1Ω instead because the wire is a longer and thinner conductor than the bumper-to-bumper contact. <S> Because <S> \$P= <S> I^2\cdot R\$ for each individual connection, and \$I\$ <S> being identical in both "resistors" as they are part of the same loop, this means there is ten times more heat produced in the wire than in the bumper-to-bumper connection. <S> In addition, the batteries are lumps of metal and able to take a lot of heat without a significant temperature rise. <S> Both is why the bumper-to-bumper connection stays considerably cool during an outer short circuit. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> A short (of the type that you mean, that causes high current and thus heat) is caused by a high voltage being directly connected to a low voltage. <S> The positive terminal of a battery is higher voltage with respect to its own negative terminal. <S> But that doesn't give it any particular relationship to a second battery. <S> So, if you take a 5V battery and ground the negative end, the positive end will be at 5V wrt ground. <S> Then you take a second 5V battery and put its negative end on the positive end of the first battery, and its positive end will be at 10V wrt ground. <S> But the positive end of battery1 and the negative end of battery2 are at the same voltage--and there's nothing trying to force different voltages--so no sparks. <A> A battery looks like this: simulate this circuit – <S> Schematic created using CircuitLab <S> A battery is like an ideal voltage source. <S> An ideal voltage source produces the same voltage no matter what. <S> It doesn't matter if you sink 1 million Amps out of it, it will still give you the same voltage. <S> If you put two ideal voltage sources in series, the voltages add. <S> If you revers the polarity of one, it will fight the other one and the voltage will be subtracted. <S> A simmilar case applies to batteries. <S> The ways that a battery is not like an ideal source is: <S> It doesn't have unlimited current sourcing abilities <S> The voltage drops the more current you pull from it (in the short term) it has series resistance <S> The batterys voltage will drop the more energy you pull from it. <S> It has internal resistance that is high (if you leave it there for a long time it will slowly degrade the battery. <S> So if you put two of the same batteries in series, it will double the voltage. <S> If you do this in the real world, it is important to match the batteries (which matches the series resistance, otherwise you can have more voltage running through one battery than the other and it will damage it or explode. <S> The middle doesn't 'short' because the voltage of the battery is pushing against the voltage of the other battery.
| Attempted series-connection of two grounded batteries would result in a short-circuit as the current could flow through the ground connection as indicated by the red arrow. A short-circuit is usually defined as an unintended bypass of the intended circuit. No, the net charge in a battery is zero!
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Characteristic Impedance of a port I thought that the characteristic impedance was a parameter defined only for transmission lines. But now I have this question: is it defined also for a port? And, eventually, what does it represent and which are the physical parameters that determine it? This doubt came to me when I saw that for instance it is possible to evaluate the input reflection coefficient of a transistor, which is not a transmission line. For instance, let's consider this scheme: These blocks are connected by wires, and not by transmission lines. But there are some reflection coefficients, which are defined starting by characteristic impedances. What do they represent? <Q> A port does not have a characteristic impedance. <S> It has an input impedance, or a port impedance, or just an impedance. <S> The characteristic impedance of a line is the ratio of V to I that will produce a travelling wave on the line. <S> It doesn't tell us what the actual ratio of V to I at any point on the line until we also know the magnitudes and phases of the travelling waves in both directions are. <S> The input impedance of a port tells use the actual ratio of V to I at the port terminals due to an AC excitation. <S> We also sometimes talk about the characteristic impedance of a system. <S> This just means we'll design our system with a default characteristic impedance for the transmission lines in the system. <S> (We might also choose to use different characteristic impedances for different parts of our system) <S> If we actually used lines with different characteristic impedance than the system characteristic impedance, we'd have to do some additional math to calculate the actual reflection coefficient where the line connected to a port, since both the line and the port will have non-zero reflection coefficients in that system. <A> A reflection coefficient such as s11, s22 depends only on impedance ratios to measure mismatch, and does not need but may include a transmission line. <S> E.g. a 50 ohm sig-gen puts out twice the expected voltage until a matched load of 50 Ohm load is applied. <S> Why? <S> There is no effective transmission line. <S> Why? <S> Because the propagation delay is almost nothing or at least LESS THAN it's rise time. <S> So matched impedances divides the voltage in half and that gives to excellent Return Loss? <S> meaning minimal signal returned as an echo? <S> or **Maximum Loss of Returned signal <S> ** <S> Yes. <S> RightSo unloaded <S> it is double what you expect if the input or "dial" is calibrated by this method. <S> Yes... <S> So I don't need a transmission line or a delay much longer than the rise time. <S> (10~90%)Right, but you could.ok.. <S> So what does a unloaded step pulse look like on a transmission line with 0 rise time? <S> ( infinite bandwidth) <S> Is Return Loss used with power supplies? <S> Yes, it's used on the grid. <S> but not generally for lab supplies. <S> We expect the load R always to be much greater than the power source. <S> But when huge Caps with low ESR are placed far away, then you need to consider it as driving an AC short circuit transmission line. <S> with a mismatch and echo causing ringing with the damping factor determined by how small the loop resistance was. <A> The port impedance is simply a measure of how the port responds to external sources - regardless of it being a transmission line or not. <S> For an infinite piece of transmission line, the port impedance will equal the characteristic impedance, but for shorter pieces, the two can be different. <S> Think of it this way <S> : A characteristic impedance of 50 ohm just means that when you source a current of 1 mA, 50 mV of voltage will be generated, or, if you apply 50 mV of voltage, 1 mA will flow into the port. <S> A complex impedance means there will be a phase shift in this current-voltage relationship - <S> An impedance of 50*j Ohms will have the same 1 mA and 50 mV but with a 90 degrees phase shift. <S> Now how does that lead to reflections? <S> Imagine the gate of your transistor is a capacitor, and presents a complex impedance. <S> You send in a signal with a 50 ohm source. <S> The capacitor will present a complex impedance, so it has to 'create' a opposite voltage wave to cancel out the incoming one exactly such that the voltage across it matches what it should be.
| Once we have defined a characteristic impedance for our system, we can determine what the reflection coefficient is for a port with a given port impedance if it is used in that system.
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Does it matter if a fuse is connected to the negative or positive terminal of a battery? I am currently going through the conundrum of where to place a fuse in my DC battery-powered circuit to protect the circuit components and the DC battery. I've been googling for a definitive answer, but I came across diverging opinions... Which one is correct and why? Connect the fuse to the negative terminal of the battery since it's where the actual flow of electrons originate which is opposite to the conventional flow of current from the positive terminal. Connect the fuse to the positive terminal. Connect two fuses, one at the positive and one at the negative battery terminals. Also, during my research, I came across a post that advised to connect a fuse at the positive terminal since it would protect both circuit and the battery, but if the fuse is connected to the negative battery terminal, then it only protects the battery. Is this true? It doesn't make sense to me. So, I can't figure out which one is correct and why? I made a simple block diagram to illustrate my question. So, is it position A or B? Or both? And why? <Q> [Should I] connect the fuse to the negative terminal of the battery since it's where the actual flow of electrons originate which is opposite to the conventional flow of current from the positive terminal? <S> Forget about electron flow. <S> It only causes confusion such as in your thinking. <S> Electrical current flows in a circuit in the same way that a bicycle chain flows around. <S> Any mobile charge carriers that leave one terminal of the power source must return on the other. <S> A break anywhere will stop current flow. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Four possible scenarios. <S> Figures 1a and 1b are equivalent. <S> If F1 or F2 blows then current flow will cease. <S> As the power supply has no ground / earth / chassis connection there is no danger of a single fault causing an alternate return path. <S> Figure 1c is the way most vehicles are wired with a negative connection to the chassis. <S> The fuses are placed in the positive lines from the battery and close to the battery. <S> If a fault occurs on the line between the fuse and the load the fuse blows and current flow stops. <S> A ground fault on the return line is unlikely to cause any problems as there is no significant voltage on it. <S> Figure 1d shows a poor arrangement with the fuse in the return wire. <S> It should be clear that a ground fault on the positive wire would be unprotected, F4 would not blow but BAT4 would pass high currents. <S> Don't worry about electrons, just think of it as charge moving from + to -. <A> Since the negative terminal of the battery is normally considered "Ground" or "Zero Volts", a fuse in the negative lead would leave the rest of the circuit "hot" - usually Not a Good Thing. <S> (Of course, if the positive terminal is considered Ground, place the fuse near the negative terminal.) <A> Think about your failure modes <S> You have to ask yourself, "What happens if one of the wires or some of the electronics shorts out to something else? " <S> Suppose I have a + and - pair of wires going out to a left turn signal light. <S> The chassis is bonded to the - side of the battery terminal. <S> What happens if the - wire frays and contacts chassis? <S> nothing <S> What happens if the + wire frays and contacts chassis? <S> Spitzensparkzen . <S> Fuse the + wire. <S> I'm on a Subway Car. <S> Third rail is propulsion +DC volts, rails are propulsion -DC, which means chassis is -DC. <S> My control circuits are 36V isolated, so that if one car's wheels go up on rusty rail, it doesn't try to return 400A of propulsion current to the next car through the tiny 10 AWG control circuits. <S> What happens if control+ shorts to chassis? <S> BADNESS. <S> What happens if control- shorts to chassis? <S> BADNESS. <S> Fuse both . <S> Suppose I'm using LED turn signals, and to save wires <S> , I hook both turn signals to the same pair of wires. <S> Brown wire + means left signal. <S> Blue wire + means right signal. <S> Fuse both, or fuse the source of my reverser circuit. <A> I would add that cars are an exception here: in most low-voltage applicatons the battery is not directly connected to the conductive parts such as casing or heat sinks, and therefore it doesn't matter where you put the fuse. <S> Don't get the impression that connecting the battery the way it is done in cars is a good idea: unless you have to deal with high currents or saving on wires by using the metal case as a conductor is essential, keep your electrical appliance isolated from the case. <S> It will help to prevent electrochemical corrosion should your device get exposed to moisture or water, and a single short to mass will be a non-issue. <S> A blown fuse is better than a fire, but not having to replace the fuse is even better. <S> Also, if your battery has to be electrically connected to the metal frame, don't just put the fuse on the positive terminal <S> : put it on the terminal opposite to the one connected to the frame.
| Recommended practice is to place the fuse near the positive terminal of the battery, so the whole circuit will be dead if the fuse blows. Again, for circuit analysis it is normal to use the conventional current flow from positive to negative.
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Does anyone have problems with micro USB connectors on PCB? In a PCB I've been developing, I included a micro USB connector. The problem is that when I plug in and out cables on it, seems to get weaker and weaker and needs to be resoldered, even the pins on the center seem to give some problems. Is there something I can do on this board that can improve it or another type of connector? Anyone had problems with it? <Q> ( https://gct.co/usb-connector/micro-usb-connector-overview ) Micro USB has its flaws. <S> Phone manufacturers often use mid-mount USB sockets for best mechanical strength. <S> Any vertical version will have the drawback of the length of the plug acting as lever on the joints. <S> You will definitely need additional mechanical support by the enclosure on these. <A> Why don't you take a plug that lies flat on the PCB? <S> That's how I've seen it on every PCB so far. <S> e.g. https://www.mouser.com/datasheet/2/185/CX_catalog-1488784.pdf <A> In addition to Daniel K's suggestion for a horizontal-mount connector, you should also consider the mechanical benefits of through hole (THT) components. <S> I've had issues with SMT-only horizontal USB connectors pulling up from the board and ripping the traces themselves. <S> Source <S> Generally, surface mount solder joints alone shouldn't be relied on for mechanical support of connectors* with significant mating cycles or non-trivial mass. <S> This is especially true for vertical style connectors as shown in your image, where you can easily get a lot of leverage on the connector body and snap the part off the board. <S> *Small embedded connectors that are not user accessible are a different story. <A> You can use hot melt glue or a neutral cure silicone rubber to give mechanical support to the connector. <S> Usually board components are glued like this for enduring long term vibration, but it should help with mechanical support. <S> There are also some special connectors with larger snap tabs on the ground pins, but i'm not sure if they make them like that for micro USB. <S> See here for more info about the different adhesives. <S> Just be careful not to get the adhesive inside of the connector. <S> Another option is to have an adapter cable, like a M to F cable, and always leave it plugged in to the board for development so that it is less wear on the board components and instead wears on the cable's connector instead. <A> Does it have to be micro USB? <S> They are notorious for their fragility and getting dirty really easy. <S> USB type B is much sturdier and if space/height isn't an concern (you have Ethernet on board?) <S> its much better idea than flimsy micro USB.
| A good compromise can be a combination of SMT pins for the USB connections with THT mounting points for the connector body, which I've found provides a lot of help with mechanical strength.
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Testing and safety requirements for selling homemade electricals I am currently researching and designing electrical audio equipment, example devices: Tube and solid state guitar amplifiers Guitar effects pedals Mixing / filtering units Synthesizers Some of these devices will be mains powered and some low voltage. For context I am in the UK and hold a Masters' degree in Electrical and Electronic Engineering. I have had friends, relatives etc approach me to ask if they can buy some of my products. At present I do not plan to make a business out of this but it could be a possible future consideration. The price I would be selling the equipment at would not warrant £1000's of test certification, as I do not have the money for this I would prefer not to sell if test houses are the only option. My primary concerns are safety and liability. What safety standards do I need to comply with and can I self certify? Can I remove / reduce liability by selling as a kit product / prototype? I notice people online selling amplifier kits, presumably they are not liable for incorrect wiring / use. <Q> It's up to you to determine all the requirements that are relevant to your particular application. <S> I'd follow suggestions to use an off-the-shelf power supply that someone else has assumed the responsibility and liability risk for, and maybe just add in some further conditioning and filtering if you're using if for a low noise application. <S> One big reason for this is that demonstrating EMC compliance might be expensive in itself, and an off-the-shelf PSU will likely have the input suppression sorted for compliance. <A> You also have to obey the speed limit and will be held liable when you cause an accident. <S> The same applies for selling products... <S> But if you want to strictly obey the law, read the European site ( <S> well, maybe soon not applicable to UK...) <S> regarding CE marking and seek contact with a testing house and don't rely on (most) amateuristic advises you find on this site. <S> I'm neither an expert, but found and proved using the European site that most advise is only half the truth or partially completely wrong (like this EEVBlog video . <A> Rather than close this for duplication, read my recent answer and search me & FCC. <S> If you use an AC/DC class 2 supply and test it with an AM radio for spurious emissions or better methods of self-certification and document the results. <S> , you are doing some due diligence. <S> Then do some sample testing for HIPOT, and test to failure, with a current limited 25kV from a very long carbon wire pair from your car engine (1M) using mica HV rated caps as a voltage divider with 1M series and NE2 bulb. <S> Or cross your fingers and write to OEM for DVT report on same. <S> OK as long as the PS you choose won’t cause leakage currents when the breakdown voltage is exceeded and arcs a 1 cent capacitor (or designer oversight ) to kill someone from some undocumented distributor on EBay. <S> Lack of surge protection recommended for safety is always a risk. <S> I suggest use with a “surge-protected Common Mode Choke line filter” . <S> I suggest that document all sales and ads with this caveat for best practise, so guitarists won’t get the “wrong” buzz when they plug into their AC powered AMPS in Florida with 100k lightning strikes in one region per year. <S> Then their loved ones can’t sue you successfully for a wrongful death. <S> Expect the worst thing that could happen then offer a solution.
| For compliance to EU requirements and to carry the 'CE' marking you can self-certify without using a testing agency, and to do so you need to create a declaration of conformity, which authorities can request at any time, and maintain a technical file which documents your evaluation of the design and manufacturing process' ability to maintain compliance - so that'd include some quality plan to periodically check production capabilities, and individual item testing if justified.
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Why do some PCBs have exposed plated perimeters? I've seen a number of PCBs, largely high speed and RF boards, that have exposed copper, either at the perimeter of the entire board, or in various sections, often with stitching vias. I've never fully understood the purpose of these. Some explanations I've heard called them "ESD Rings" used for handling the board, but that makes less sense to me when there are a lot of individual perimeters, specifically ones more inboard like in the image below. Are these just the top ground plane exposed? If so, what's the point of exposing it? I don't see how it would make a difference from an EMI perspective whether said ground pours are exposed or not. I've also heard, and more or less accept that for an outer perimeter plated ring of this kind, it's often connected to GND and then used to connect via mounting hardware to an enclosure. Thanks! <Q> At very high frequencies, the area between planes can function as a waveguide/antenna and high frequencies can move between planes and out of the edge of the PCB. <S> In addition to this the planes on the top layer can be plated to accept EMI gaskets/shields. <S> Source: <S> https://sc01.alicdn.com/kf/HTB1uZSNRXXXXXahXpXXq6xXFXXXd/Photo-chemical-etching-RFI-EMI-shielding-box.jpg <S> Are these just the top ground plane exposed? <S> If so, what's the point of exposing it? <S> The vias most likely connect to the ground plane and the trace/plane on the top layer, but doesn't have to be. <S> The point of exposing it is to make it conductive and continuous. <S> The layer is then plated with a surface finish that is low impedance/resistance metal like ENIG (with gold). <S> This allows for the high frequency currents to be shorted to the ground plane with an EMI gasket (conductive foam or deform-able metal mesh) and return back to the source. <S> Without the conductive layer on the top of the PCB, RF could potentially leak underneath the RF shield. <S> Many of the chips are generating RF on the board pictured above, to prevent cross talk and leakage, the EMI shield prevents RF from moving to other areas of the design, or outside of the board (entities like the FCC regulate how much devices can radiate radio frequencies). <S> This is why the shield also partitions different sections of the PCB design. <S> Here is the distance charge for via fencing, <S> if you wanted to see what frequencies they were trying to shield on the board above, you could measure between vias to find the cutoff frequency. <S> Source: https://www.edn.com/Pdf/ViewPdf?contentItemId=4406491 <A> There will be shielding added from above and attached to the PCB by screws (therefore the holes). <S> The shielding will not only shield the circuit from outside world but also subcircuits from each other. <S> Here is an example of such a shielding: image from http://tennvac.com/custom-shielding-solutions <A> Answer: <S> The mounting holes are a clear contact area for braided contact to a cover. <S> https://images.app.goo.gl/usAPvRQmVPCfHtQu9 <S> You won’t find any online because these are all custom designs. <S> The above is just a simple rectangular shape. <S> When you mix logic speeds and RF frequencies which both overlap here in this design up to 6 GHz range for these types, you need a good common ground with many layers yet isolate the logic impulse currents from conducting through the RF grounds. <S> So you will see microvias <S> every \$\lambda/20\$ for the highest frequency of interest, to reduce the loop crossectional area of these logic current spikes (CMOS FETs have capacitance when switched). <S> The surface is likely immersion gold-plated copper to buried layers to reduce oxidation and prevent irregular thickness solder level plating that affects impedance of transmission lines. <S> You will not see microvias for all the linear RF stuff because their ground plane is isolated from logic ground plane. <S> and they are only connected near the RF ports. <S> This minimizes the crosstalk of conducted and radiate ground currents between logic and RF. <S> The wide boundary around each zone is like the Mexican-US border. <S> It sinks stray radiation fields, reduces crosstalk, but it doesn't stop migration of current or voltage fields it all together. <S> It is coplanar after all and stray coupling is always reduced with a ground track in between. <S> But the digital side is also analog with edge jitter and internal processes that are still sensitive to adjacent module crosstalk. <S> It is common for Faraday shields to be soldered on top when needed to further reduce crosstalk using reflow. <S> If you've seen a number of these boards without shields, then they did a pretty darn good layout design. <S> Nortel and others also did some of these designs without shields up to 1 Gbit/s with very balanced differential microstrips (also bankrupt). <S> I have some pre-Y2K designs we did for 1 GHz ISM band for the AMR market with in-house tin plated brass boxes etched by local PCB shops. <S> Unfortunately, this company went bankrupt. <S> It had over 130 patents and many roots like HP microwave and a dozen others all experts in mobile wireless technology. <S> Intel bought all the patents.
| They are called via fences, they are placed on the outside of the board to "fence in RF", they do this by creating a barrier smaller than the wavelength that needs to be shielded.
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Securing mod wires to PCB I have done a few mods to a PCB with 36 gauge mod wire. Obviously this wire is so thin that it only takes a few bends back and forth to break it. What is the best way to secure this in place on the PCB? I have tried super glue but that makes a mess because it flows too easily. <Q> Kapton tape. <S> Enamel paint. <S> (Nail polish might even work in a pinch.) <S> Hot glue for larger wires. <A> At work I use Dow Corning 3145 RTV silicone adhesive for this very purpose. <S> We also use it to secure flex cables in their connectors, secure right-angle parts such as capacitors to the board surface, fill small gaps to make the product waterproof, electrical component insulation, and so on. <S> It is chemical-resistant, water-resistant, corrosion-resistant, non-conductive, and sticks to just about anything. <S> There are many different uses for it, and in my opinion it is absolutely indispensable. <S> And to prevent this from being a specific product plug (I am not affiliated with the company in any way, I just like the product) <S> , I will say that just about any silicone adhesive designed for printed circuit board use will probably work great in this sort of application. <A> Use a needle which you dip into the super glue to make tiny glue dot and (right angled) tweezers to keep the modification wire in place. <S> Or use hot melt; same procedure. <S> Hot melt has the benefit you can melt it again and you can kind of control the cool down using your hot air station. <A> 1 glue dot per 5 cm. <S> I prefer annealed magnet wire. <S> AWG30 or smaller. <S> I was Eng Mgr for C-MAC WINNIPEG and global Design Services Mgr with over 100 factories <S> now all sold twice and owned by the largest EMS provider overseas.
| Our PCB EMS facility used wire similar to WW AWG30 or smaller , soldered and Loctite bonded with 1 part instant adhesive for strain relief from vibration.
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Help designing a voltage divider I said voltage divider on the title, but could be anything. I have a DC source of 19.7V. I need an output that varies from 0V to 18V. I have: Resistors (1/8W); 1k potentiometer; 5k trimpot; 2N3904; LM358; Capacitors. My only requirement, besides part count, is that both potentiometers be used: the 1k being the main, and the 5k acting for fine tuning. And the adjust... not too sensitive. The more linear, the merrier, and taking advantage of the whole excursion of the potentiometer. In the example below I could not obey the requirement of sensitivity and excursion. The transistor I'm trying to use has a cutoff around 680mV and saturation around 820mV.Very small window. Hence, my trouble. More: The output current can be small. 5mA will work; This whole system will control the GND pin of a LM7805, and later on, of a 7905; LM317 unavailable; The 1k potentiometer can be turned by hand and is what I have available. The 5k trimpot needs a screwdriver. History: At first I thought a voltage divider could do it, but in my attempts, the voltage hardly approaches 15V. Still open for this type of design, though (just bear the requirements if you'd like to help me this way) After giving up, I tried using OpAmp. It worked. But later I will need a negative version of this structure I'm trying to build, but 19.7V-(-19.7V) my opamps won't support. Maybe I could reduce the supply voltage of this OpAmp, but then... part count. Third attempt, more promising but difficult, was with a transistor. <Q> simulate this circuit – Schematic created using CircuitLab DONT USE ABOVE poor simulator part of EE.SE Don't burn it out by over loading the pot with 1/4W type or excess current. <S> Use this simulator instead Every part serves a purpose. <S> If you understand this design and tried it out, say OK (+1) <A> Is this an assignment or similar. <S> If so, we will still help but in a more 'tutorial' manner. <S> Here is a fairly complete outline. <S> Purposefully with no circuit and somewhat skeleton detail assuming an assignment. <S> Ask more if wanted. <S> The opamp is your great friend. <S> An UNLOADED pot will give you the whole voltage range. <S> Add a trimpot at the top and you can reduce the range. <S> If the pot is linear (and not log or semilog) <S> you get linear Vout with pot position. <S> Use the opamp to drive the transistor as en emitter follower. <S> Compare pot voltage and vout (scaled into common mode range of opamp) and drive transistor (drive scaled to allow for opamp not being a rail to rail one). <S> Divide Vout by a factor to bring it into opamp common mode range and compare it to the pot voltage. <S> Result should be reasonably superb. <S> Instability is possible. <S> Judiciously placed small cap <S> MAY be needed somewhere if it oscillates. <A> I'll share my own solution, which loses on two aspects: <S> Voltage will never reach my desired minimum of 0V; The part count is a little higher. <S> Luckly, it's all resistors. <S> The only advantage here is the low current through the components. <S> I encourage those interested to enhance this model. <S> My next goal is a negative counterpart: 0V to -18V from a -19.7V supply
| Pot Vout or pot total Voltage range can be scaled down to lower voltage to match opamp Vout related input.
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MT3608 gets very hot only handle 1A In the datasheet they mention the MT3608 can handle up to 2A however in some stress tests with a load tester it gets already very hot at 1A. It is so hot, it is impossible to touch the board, all components around it get very hot because it spread all of the heat via paths. The chip itself is that tiny it is almost impossible to attach a heatsink. Because I don't have a second one I have to ask this, is this normal or abnormal? Board (arrow point at chip and I have the one with protection diode): Test config: Battery 18650 2600mah boost 4.10V to 5.6V (drops down to 4.87 when heated up) Load 1A Datasheet: https://www.olimex.com/Products/Breadboarding/BB-PWR-3608/resources/MT3608.pdf <Q> I have the identical board. <S> I connected it to a power supply set to 4.1V, adjusted the booster output to 5.60V and connected a 5.6Ω 10W resistor. <S> The output stayed solid at 5.60V while the input current rose to 1.56A. <S> After 6 minutes the chip temperature was stable at 39°C according to my infrared thermometer, 23°C above ambient. <S> The inductor was 41°C, the diode was 42°C, and the load resistor was 112°C! <S> 4.1V <S> * 1.56A = 6.4W input power, so the conversion efficiency was 5.6W / 6.4W = 87.5%. <S> So this board should do what you are asking of it no problem. <S> However if the input voltage drops the current will have to increase to get the required power, and efficiency may suffer. <S> At 3.7V my unit drew 1.8A and its efficiency dropped to 84%. <S> At these relatively low voltages and high currents, any losses in wiring or connectors can make things worse. <S> I used thick wires and heavy duty connectors, and measured voltages directly at the board terminals. <S> I also put 22uF capacitors across the input and output to help reduce voltage ripple. <S> Before putting the blame on your board you should consider two things:- Boosting to higher voltage requires more power for the same output current, and therefore higher input current. <S> Higher current causes greater voltage drop in wiring and connectors (as well as inside the booster itself) making its job harder. <S> When input voltage drops the chip draws more current to compensate, but this could cause the input voltage to drop even more. <S> If the power source is too weak to deliver the required power then the booster will continue trying to draw more current until maxed out. <S> 18650 Li-ion cells have typical 'knee' voltage at end of discharge of ~3.4V (it might start at 4.1V, but it won't stay there for long). <S> To get maximum usable capacity out of your battery the booster must be able to work down to this voltage, and you need to keep input losses down with short wiring and low-loss connections. <S> The MT3608 datasheet only shows output current up to 1A at 5V, and you may need a similar boost ratio as the 3V to 5V curve shown (where efficiency is heading downhill fast at 1A). <S> Therefore, despite being rated for 2A, under these conditions I think 1A is a more realistic maximum. <A> Your load is 5.6V @ <S> 1 <S> A = <S> 5.6W. <S> If overall efficiency is 80%, that means an additional Power lost = <S> (20%/80%) <S> x 5.6W <S> = 1.4 Watts is dissipated. <S> That's more than an IC of that size can handle at sensible temperatures with PCB copper (if any) heatsinking plus a small amount of direct radiation and convection. <S> For better insulation (and somewhat worse heat transfer) use an insulating washer as sold for use with larger ICs. <S> This could be thermally conductive rubber or mica or ... . <S> The thermal compound needs to be in good contact with the IC <S> **and* spread out around it to allow good contact over a reasonable area with the heatsink materials. <S> For retention you could make larger holes in or near two of the Vin and Vout pads and fasten with nonconductng screws. <S> Or sandwich a suitably cunning metal piece and the PCB and wrap a cable tie around the whole "assembly". <S> Prettiness may suffer but it should be doable. <A> It's the experience you gain from these failures about thermal resistance, power calculations, test methods with accurate measurements and thermal sensing <S> , that's far worth more than the cost of this inexpensive board. <S> Next time consider a thermometer and record hot spots and use a voltage controlled power transistor on a heatsink to control load current. <S> The datasheet for this IC design it gives examples for only step-up to 12V and 18V for a reason. <S> The lower the input voltage, <S> the lower the efficiency <S> It drops from 90% to 80% as Vin drops to 3V. the higher the step-up voltage before limits, the higher the efficiency. <S> expect a 200'C rise per 1W loss on this tiny chip unless there is a huge 4x4cm copper heat spreader
| Heatsinking can be obtained by adding a blob of thermal transfer compound and positioning a metal heatsink against the IC and compound. Yes, this is a really tiny power supply, yet very efficient if used as intended.
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Odd PCB Layout for Voltage Regulator I am reverse-engineering a board which has a Xilinx Spartan 3E FPGA, with VCCAUX powered by a 2.5 volt regulator. Below is the PCB layout for the regulator part of the circuit, and something seems very fishy to me. My apologies for the horrible pixelation, this was the highest resolution I could get with the equipment I had available. Anyway, the SOT23-5 component labeled "LFSB" is a Texas Instruments LP3988IMF-2.5 linear voltage regulator . I have traced out the schematic below from the board layout: You may already have noticed the source of my confusion: I have no idea why they would have placed a 316 ohm resistor directly across the output of a 2.5 volt regulator. All that does is waste 7.9 milliamps. I cannot seem to find any reason for doing this. I wonder if it is a design flaw, and that resistor is actually supposed to be connected to the PG pin instead of to ground. I have triple-checked the original PCB, though, and it definitely connects to ground and the PG pin is not connected to anything. If this is an error, however, it would explain why they used a separate trace on the low side of the resistor instead of connecting it to the copper ground pour that's right there. I also wondered if the regulator may require a minimum load in order to maintain a stable output, but that is not the case for this regulator. There are no minimum load requirements. I also considered the possibility that it was intended to bring up VCCAUX more slowly for sequencing purposes for the FPGA, but reading the datasheet this also does not seem to fit - there are no strict sequencing rules for powering up the Spartan 3E. Can anyone think of a reason why someone would intentionally place a 316 ohm resistor directly across the output of a 2.5V regulator? I considered it might be a bleeder resistor for the output capacitor, but it seems like too low of a value for that. EDIT: Perhaps this additional information will help. The datasheet for the Spartan 3E specifies what the VCCAUX supply is used for: VCCAUX: Auxiliary supply voltage. Supplies Digital Clock Managers (DCMs), differential drivers, dedicated configuration pins, JTAG interface. Input to Power-On Reset (POR) circuit. <Q> I would have done the same design, in order to reduce dynamic and static load regulation error. <S> The details for the reasons are evident in the datasheet. <S> look at dynamic load regulation error and input step regulation error. <S> I can only guess what error budget the designer had in mind, but it common for every LDO to have the above responses , although this FET LDO is exceptional low power and dropout voltage. <S> 5mV error {input step=0.6V} with 1mA step load, 200mV error with 150mA step load <S> * <S> the static load regulation error is only rated above 1mA as 0.007%/mA . <S> This implies it is worse below 1 mA and improves with a dummy load of 7.6mA to the designers satisfaction. <S> It also improves dynamic step load regulation error above. <S> * <S> This 1mA ensures the rise fall time of Gate drive to speed up response. <S> 7.6mA is even better with diminishing returns above this. <S> static load regulation error is only due to RdsOn of the PFET used in the LDO divided by its internal Loop gain. <S> This true for any voltage regulator whether it is FET or BJT. <S> But infinite loop gain can increase stability errors or more ringing, under certain load , (ESR, C ) conditions so it is finite. <S> Fishy? <S> No way <A> I am not convinced that the resistor is grounded. <S> I have labeled the parts and the copper pours as per your "reversed engineered" circuit. <S> If R14 was grounded, why would a via be wasted when there is GND pour right next door to it. <S> How did you test it was ground? <S> did you just buzz between lines? <S> There is a very high chance there is an LED to ground hanging off that via. <S> This would provide a visual indication 2.5V is powered and a resistor around 316R would be ok for a RED/YELLOW/GREEN LED ( 4mA). <S> This would aos give the "indication" of a short if you mis-read a DMM or depending on specifics of the DMM. <S> https://reference.digilentinc.com/_media/s3e:spartan-3e_sch.pdf <S> This is a reference design for a Spartan 3E. <S> There is a 2k2 loading on the 2.5V regulator but also an LED off the 3v3. <S> This could be to provide some damping to the circuit downstream <A> That leakage would typically cause the regulator output to shut off and to rise up and go to a higher voltage. <S> A designer makes a design tradeoff between how much sink capability to allow for versus the amount of extra load the resistor places on the voltage regulator. <S> Leakage conditions can exist during power on and power off sequencing of complex semiconductor devices and the sink capability can be important to keep things in check. <S> In some cases the voltage regulator may have a feature called over voltage lock out that shuts down the regulator if the output rises up too much. <S> This can be detrimental to system operation, especially if the power good (PG) indicator pin is monitored to control a voltage regulator chain on a complex board. <S> The current sink resistor can play a role of preventing an unexpected shutdown due to a small amount of leakage into a particular rail.
| As already suggested by some other comments that 316 ohm resistor is placed there to allow the voltage regulator circuit some ability to sink some current in the case that the 2.5V rail gets some leakage from a higher voltage rail.
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How many transistors are there in a logic gate? How many transistors are there in a logic gate? If anybody asks me, I tell them: A NOT gate is 1 transistor. A NAND gate is 1 transistor per input. A NOR gate is 1 transistor per input. An AND gate is basically a NAND gate + a NOT gate, so it takes 1 transistor more than a NAND gate. Same for OR vs NOR. An XOR gate is built from multiple other gates, typically about ~4. Sounds pretty reasonable, right? Thing is, I just realised... I have no idea why I think these are the correct numbers. I don't remember reading that somewhere or anything. I'm beginning to think maybe I just made it up. Sure, it sounds convincing; but that doesn't make it correct ! So what is the actual number of transistors per gate? I imagine it's different depending on which logic family we're discussing. (My brain is telling me that the numbers above are for TTL, and CMOS is exactly 2× the that, but again I don't know if there's actually a shred of truth to that.) If it does make a difference, I'm most interested in TTL and CMOS. <Q> See the first answer to this question . <S> It shows a TTL inverter circuit with a totem-pole output; there are four transistors. <S> A TTL NAND gate would also have four transistors, but the input side would have a dual-emitter transistor. <S> An unbuffered CMOS inverter has just two transistors, yes, but a buffered inverter will have more ( <S> either four or six, I can't remember which, or perhaps it varies). <A> If you are making gates out of discrete transistors, diodes and resistors, you can make an inverter with one transistor, a NAND with two transistors, or with diodes. <S> If it is in a integrated circuit, where a resistor is more complex to make than a transistor, more transistors are used either for polarisation (for example ni old NMOS), or now as part of more complex circuits with complementary logic such as CMOS. <S> A basic CMOS inverter uses 2 transistors. <S> Inputs can be added by using transistors with several gate contacts. <S> It works when that gate is one among many others, driving a few similar gates. <S> For large fan-out gates, or when these gates are sold as components, like a 7400 or a HC4000, there are additional transistors, with different geometries for conditioning the input, multiply the output current rating. <A> For a basic unbuffered CMOS NAND/NOR/NOT you have two transistors per input. <S> For an unbuffered NMOS or PMOS NAND/ <S> NOR/NOT you have one transister per input, but you also have a pullup or pulldown, this pullup or pulldown can be a resistor, but it's more commonly a weak transistor. <S> Many practical discrete gates will be buffered, that is they will have additional inverter stages added to improve the output drive strength and prevent the drive strength depending on which combination of inputs are active. <S> TTL is more complex, a basic TTL NAND gate has two transistors, plus some resistors, but one of those transistors is a special one with two emmitters, furthermore many TTL gates have a "totem pole" output which adds an extra transistor or two. <S> In TTL NOR is substantially more complicated than NAND, so it's often avoided.
| It depends on the logic family, it also depends on what exactly you count as a transistor.
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What is this wire to board connector called? I'm looking for the correct terminology of the connector in the pictures below, which, as far as I understand, serves to connect unstriped wires pretty easily to some form of socket. I encountered it in a washing machine. I tried googling and going through a distributor's online catalog, but I'm not getting far without the correct jargon. The first two pictures show the connector opened up. The metal pieces serve as both connection pin for the socket and to pierce trough the protective layer of the wires which have to be pushed trough the top holes. I suspect that first the wires are pushed trough the holes, then the metal pins are placed, and finally the whole is closed. The last two pictures show the connector when assembled. Edit:To further clarify the use of the connector. It is used on the thermostat / thermistor of the the dryer in the machine. <Q> An insulation-displacement contact (IDC) , also known as insulation-piercing contact (IPC) , is an electrical connector designed to be connected to the conductor(s) of an insulated cable by a connection process which forces a selectively sharpened blade or blades through the insulation, bypassing the need to strip the conductors of insulation before connecting. <S> Source: <S> https://en.wikipedia.org/wiki/Insulation-displacement_connector <S> See also: Molex about IDC <S> To further narrow down, measure the pitch, estimate AWG (it may be printed on the red cables), try looking for markings on the connector (some markings are molded with the connector having the same material/colour: changing the incidence of light may make it more clear) <A> You are looking for stackable edge connectors . <A> I finally found the exact name after a long search. <S> The exact terminology is a RAST 2.5 plug / wire housing (e.g. here from molex ). <S> "2.5" stands for 2.5 mm pitch (there is a 5 mm variant). <S> This is indeed an IDC connector as Huisman suggested, but that didn't narrow the vast assortment of connectors enough to give a result. <S> Especially different to other IDC connectors is the form of the pins. <S> It are no simple metal bars, but a kind of clamps. <S> Note also the ribs on one side of the connector. <S> I eventually discovered the exact naming by googling on the device it was used, a washing machine dryer thermostat. <S> After noticing that many such sensors had this kind of connector, I found through the datasheet of a similar temperature probe assembly the specification of the connector type: RAST 2.5. <S> For people wondering wherefore this connector servers. <S> According to this press release from Lumberg (another connector manufacturer), the RAST connector has "a higher mechanical retention (...) without taking up additional casing space, made possible by the reinforced locking toes inside the reliable geometry of the casing". <S> This makes sense given the application. <S> It was mounted on the tumbler/drum of the washing machine, which of course exposes the connector to considerable & prolonged vibrations. <S> (Unfortunately in my machine, not the connector, but the wire attached to it failed due to fatigue...)
| The technology is called Insulation Displacing Connector Technology
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Are these two equivalent Are these two equivalent? Z is characteristic impedance of the transmission line and theta is electrical length of it. <Q> No. <S> As you can see, the top one has only one terminal, and the bottom one has two terminals; thus they cannot be equivalent as any two equivalent subcircuits must have the same number of terminals. <A> In the top diagram, the return path is implied by the ground symbol, while in the bottom diagram, it is explicit. <S> They both represent the same physical system. <A> Yes they are the same if your parallel lines in the lower once represent and transmission line with the same effect as the symbolic mode above it with a lumped characteristic impedance with phase at some f. <S> But if you expected everyone to. <S> Assume your lower diagram was a ‘logic” or symbolic schematic then you are confusing physical parallel tracks with a dielectric constant between them or just showing a mathematical model of two lines with Z(f) <S> then they are equivalent. <S> Schematics are usually logical like the 1st and not physical with a math function in shunt with a load. <S> A series transmission line also has a shunt input characteristic impedance and a transfer function at some f such that the transmission line can convert or “transform” the load to a different equivalent input impedance. <S> Somewhat but not same as a Norton or Thevenin equivalent cct. <S> As it is frequency sensitive. <S> The 1st shows a 0V Gnd reference while the second is a floating differential input which is never used for transmission lines but may be grounded on the lower side AT SOURCE. <S> So technically it could be argued that the 2nd diagram is ambiguous as there is no 0V reference and only arrows showing 2 lines with opposite current flow. <S> So logically the same but bad way of representing the circuit without a ground to define the 0V reference to a transmission line. <S> So this is only done in academia to show you the “theoretical” equivalent circuit.
| Yes, they're equivalent.
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Resistor & capacitor are getting too hot I made a transformarless power supply with capacitor. When i connect it on ac, it light the leds. But after some second the 100ohm resistor and 470uf capacitor get too hot. Its too hot that i could not touch it then. What's the problem . Pls help me. <Q> THIS CIRCUIT WILL KILL YOU Or your family Or your friends <S> The capacitor MUST be X or Y rated for AC mains use. <S> If it is getting hot you are using the wrong cap. <S> The same applies to the zener diode. <S> This circuit is suitable (if ever) only for applications where nobody can EVER touch ANY part of the circuit when operating. <S> The cost savings are so small compared to using a commercial supply that the only application that may make sense is in a high volume product where it is properly designed for the task and the limitations are properly understood and allowed for. <S> THIS CIRCUIT WILL KILL YOU <S> ALL parts of the circuit MUST be considered always "live" - ie at full mains voltage. <S> Some parts are, others can be at any moment if something fails. <A> Make sure that your cap is a safety cap ,like say X rated 275VAC <S> .Not <S> 4oo <S> VDC.Your Zener will get very hot when the supply is not loaded .If <S> your hot zener goes open circuit the cap will get too much volts and get hot and fail. <S> These non isolated supplies can present a shock hazard .Seek <S> local advice and do not get electrocuted. <A> With 12Vdc drawn from 230Vac, you have a current-limit by 100Ω and 2.2uF @ <S> 100Hz <S> = -j800 <S> Ohms for an apparent current of ~170 mA . with 230mA rms thru 100 Ohms <S> the power dissipation is I^2R-Pd=5.3W exceeding its rating and operating a finger burning max temps of 150'C+ Conclusion <S> You must use an e-Cap rated for > .4A ripple current with low ESR for safety or reasonable life. <S> Don't use Chinese Caps with no specs <S> ( Japanese are far better) <S> example <S> Rubycon Manufacturer Part Number <S> 25ZLH470MEFC10X12.5 <S> 730 mA ripple @ <S> 120Hz <S> Proof with SIM <S> - Always derate power resistors at least 33% and pre 50% - so use a 10W resistor or add another 100 OHm 5W in series. <S> which only reduces overall current by 5mA. <A> Your circuit does not show LEDs or any other load. <S> If you have LEDs connected to the DC output then maybe they are overloading the resistor and capacitor. <A> simulate this circuit – Schematic created using CircuitLab <S> Simulating that shows you get a 15V peak drop across the 100 Ohm resistor. <S> 15V*15V/100Ohms <S> = 2.25W. <S> Whether the resistor can survive that and is rated for it, it's still burning a lot of energy <S> and it's <S> going to get hot. <S> Also, what everyone else said. <S> Be safe.
| Regardless of what formulae and calculations say, if the resistor is getting too hot you must use a larger wattage one or cool it better. You should never touch any part that is in a transformerless circuit to avoid a lethal shock.
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Why do we need a bootloader separate from our application program in microcontrollers? Why do we need a separate program in the same flash program memory of a microcontroller, specifically STM32F103, which is called a bootloader? What is special about it to keep it separate from the main application program? Generally speaking, does a bootloader of a microprocessor-based system (say PowerPC MPC8270) do the same job as that of a microcontroller (say ARM STM32F103) or are they doing fundamentally different jobs from each other and yet both are called a 'bootloader'? <Q> A bootloader on a microcontroller is responsible for updating the main firmware over a communication channel other than the programming header. <S> This is useful for updating firmware in the field over BLE, UART, I2C, SD cards, USB, etc. <S> It would be extremely inconvenient to require customers to purchase programmers just to update the firmware on their devices. <S> The reason why the bootloader is kept separate is for reliability. <S> If the bootloader and application were kept together, then the bootloader code would need to be copied to RAM before it could run, since any firmware update would erase the bootloader code in flash. <S> If power were cut with the bootloader code in RAM and the flash erased, the device would be bricked. <A> So that the loading process can recover from errors. <S> Suppose there is a communication error or power disconnects during an upgrade. <S> If the boot loader were part of the application you were upgrading then the user wouldn't be able to try again without using special hardware to reflash to boot loader. <S> Some microcontrollers can't execute code from RAM. <S> If the boot loader was mixed in with the rest of the software then you wouldn't actually be able to upgrade your software because you can't erase pages of flash that you are currently executing out of. <S> The work around is to first burn the new code to the second half of flash, then jump to it. <S> The new code then copies itself to the first half of flash. <S> Of course the downside is that burning flash is usually slow and now that you have to do it twice the loading process might take up to twice as long. <S> Also this work-around limits your application size to be no larger than half of your total flash. <S> Well written boot loaders try to verify that valid code exists on the device before trying to execute it. <S> If the boot loader and other code were mixed together then how could you be sure that your validation routine would work if all the code didn't load? <S> Authentication. <S> Secure boot loaders try to verify that the loaded application matches a digital signature before executing. <S> But if the boot loader and other code were mixed together <S> then you can't control what runs on the device because once the user loads new code you can't control what happens at startup. <A> You need some code which knows how to erase and reprogram some of the internal flash, that can't be the main program as when it's erased itself it wouldn't be able to reprogram. <A> The bootloader allows the MCU to communicate with something else to accept a new program, store it, and run it after a reset. <S> If you didn't have a bootloader, then a Programmer is needed to access the memory and put the program in place. <A> In addition to the other correct answers about allowing reprogramming of the main firmware from the bootloader, another benefit of having the bootloader be separate is that you can logically separate the "do once on boot" tasks from the code you need during runtime. <S> Then, after the bootloader finishes its initial configuration tasks, the main firmware can evict the bootloader with all its no-longer-needed code from memory, saving significant RAM space. <S> It's possible to achieve this in other ways, but the bootloader/firmware split makes it much easier on many architectures. <A> The short answer, is because software is awesome. <S> You could have everything the bootloader does be "pure hardware". <S> But it is far, far, far easier to have the tasks the bootloader does be written as software, then interpreted by hardware. <S> These tasks can involve setting up the hardware for the "real" software to run (for example, on a Raspberry Pi (via @ErikF)), having a protocol to replace the "real" program before it runs (check a pin, if that pin is set then reflash the real program), or even setting up the software environment for the "real" program. <S> On less micro-scale software, when you run an executable the application loader moves does stuff like loading parts of your data into memory, sometimes fixes up addresses, sets up arguments to main or other global stuff, spins up your OS <S> provided libraries, and then jumps to the start of the _main code. <S> Some of these things can be done by a bootloader. <S> In a microcontroller, some of the tasks that a bootloader does could be split off into the program. <S> The compiler for your platform could automatically inject the "setup" code into every executable. <S> But, having it in the bootloader means that the same compiler might work on different hardware, as the bootloader can "hide" the difference between the platforms. <S> Top that off with the fact that a flash of the main program doesn't risk the bootloader (and the ability to reflash the main program), and having a non-trivial bootloader is a pretty great thing.
| The bootloader and application code are placed in separate sections of flash, so that the application code can be erased and re-written by the bootloader without changing anything related to the bootloader code. They're generally there to allow you to update your main application program.
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Voltage loss on DC circuits I want to connect 4 bulbs in parallel. The bulbs need 12V DC and rated 18 watts. P=V*I, that means I=1.5 A on each bulb. From that the resistance of the bulbs are 8 Ohm (R=12/1.5). The circuit will look like this: simulate this circuit – Schematic created using CircuitLab But in real life wires have resistance too. The resistance of my wire is 12 mOhm/m. I need around 12 meters of this wire. The bulbs will be 3 meters from each other. If I calculate this, on the last bulb only 11V will fall and 1.375A will flow through. Which is only around 15Watts. simulate this circuit If I want to connect more bulbs (lets say 8 or even more) there won't be enough Voltage across the bulb and it wont work. Did i calculate it wrong? How is this problem problem solved in real life? <Q> You have come across the reason why high voltages are used in the home (230V, 110V), even higher are used to transmit power over longer distances (10kV and higher). <S> This is also the reason many cars are moving away from 12V power to 48V, and lorries/trucks use 24V rather than 12V. <S> As you mention, Power is the current times the voltage, P=IV. <S> If you remember Ohm's law: <S> V=IR, you can put the two equations together: P=Ix(IxR), P=I^2R. <S> Looking at the wires, they have a fixed resistance, other than increasing the diameter of the wirse(expensive and additional weight) or using some very fancy high end superconductor (very very expensive and most are still just not practical for a whole plethora of reasons), the only thing we can change is the current. <S> To get the same power, if you half the current, you need to double the voltage, which isn't too hard to do. <S> But with half the current, the power lost due to resistance is reduced by a factor of 4. <S> So, to answer your question: how is it solved in real life, we increase the voltage. <S> Other ideas you may want to consider for your application:Reduce wire lengthIncrease wire diameter (reduce wire resistance)Check any connectors, which often have higher resistance than the wires <S> As mentioned in the comments, a "star" configuration could also improve matters. <S> The reason for this is pretty obvious, as using a "star" (each load being powered by its own wire) decreases the load per wire. <S> But you are having to spend more money and space on wire, which may not be suitable for your application. <S> If those aren't useful for whatever reason, increase input voltage. <S> Increasing input voltage may mean you need to drop the voltage back down at the point of load to make things work. <S> But that can be done with a simple buck converter. <S> There are lots of things to consider when designing the power connection system, but that is beyond what I can do in a quick answer on here. <A> So there’s two ways of solving that problem. <S> One is to wire the bulbs in series and use a boost LED DC-DC driver, and let Kirchhoff’s Law do the rest. <S> This is arguably the cheapest and most efficient way, with the downside that when one bulb fails the whole string goes down. <S> The other is to regulate current locally at each bulb using a current sense regulator. <A> What is the application? <S> - it probably matters. <S> Do you care about volts. <S> If so why? <S> You may. <S> Or not. <S> Brightness of about 2:1 can just be discerned by eye for most people. <S> Tungsten bulb resistance rises with voltage so 18V:15V difference will be less than 1.2:1 brightness - but IF incandescent, colour change may be unacceptable. <S> Use LEDs, assuming "bulb" <> LED. <S> FAR less current than incandescent per lux out. <S> If these ARE LEDs then a current regulator per LED leg may suit. <S> A bjt jellybean and a FET and a few resistors maketh a current source with about 1V min headroom. <S> At the cost of more wire, feed them at the midpoint. <S> You then have 54 mOhm from feed common point to outside bulbs and half the current in each half so Vdrop difference is substantially lower. <S> For "perfect" results you could use a number of feeds optimised for R and Ls (and more wire). <S> If you don't care about power and only about relative brightness, an eg LM317 regulator at each node will allow you more even V but need a higher feed voltage. <S> (LM317 Imax varies - select to suit). <S> |A <S> TL431 and MOSFET / lamp allows customisable results. <S> (Probably need a jellybean bjt and few Rs as well per lamp.) <S> Circuits can be drawn for any of the above if of interest.
| If you care about matching brightness (which I assume you do based on the nature of your question) then what you want to shoot for is equal current through each bulb.
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Clock Dividers with Clock Domain Crossing I am doing a design in FPGA that looks like this: 100 MHz is the clock available in my FPGA board. It feeds Module 2. Module 1 is needs a slow clock of 10 MHz clock. So I used a clock divider with flip flops. I thought of not using PLL to make my code purely RTL. Module 1 is transferring some data along with control signals to Module 2. My current assumption is that: The main clock and the divided clock are synchronous, and their phase relations are known to the synthesiser. And hence there is no Clock Domain Crossing here. So that I can simply constraint both the clocks and put multi-path SDC constraints between Module 1 and Module 2. But I am not sure whether I am right. Is this a case of asynchronous CDC ? Will I need any synchroniser between Module 1 and Module 2 ? Any feedback is welcome. <Q> Depends on your vendor On most modern FPGA's of the two biggest vendors, your first assumption is correct and you won't need synchronization registers for the CDC if your clock divider is written correctly. <S> The synthesis tools will take care of everything. <S> Synthesis tools by smaller vendors don't always take into account the known phase relationship between the clocks, which might mean you do need a proper CDC. <S> From experience I can say that the old Libero tools for at least the older RTAX and ProASIC lines allow you to specify synchronous clocks, but the specification just gets ignored and proper CDCs need to be added at the RTL level. <A> Yes, it's all one clock domain as far as synthesis is concerned. <S> You can draw a circle around all three modules, and there's only on clock entering that circle. <S> The fact that some of the logic inside that circle runs at 10 MHz is largely irrelevant. <S> Depending on how your implemented your ÷10 logic, you'll make things easier or harder for the synthesis to meet timing, because it's going to be doing everything based on the 10 ns period of the master clock. <S> Using a PLL would be good in this sense, because it can effectively eliminate any delay between the two clocks. <S> If you used a ripple counter, that would be the worst, because it creates the largest amount of skew between the clocks. <S> A synchronous counter would create a low, but controlled amount of skew. <A> The design will synthesize and meet timing easier if you source the 10mhz clock from the same MMCM / PLL (Xilinx terms for their clock tiles) for the 100 MHz. <S> You would treat the domain cross as a multicycle path and constrain it accordingly. <S> An app note from Xilinx that touches on this. <S> https://www.xilinx.com/support/answers/62488.html
| Generally not recommended to use logic to drive clock in an FPGA: the clock skew is harder to control as you’re routing logic onto a BUFG (a Xilinx term for an entry point onto a low-skew clock network; other FPGAs have similar concepts.)
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Why is the audio input from the microphone connected to the inverting input on the TPA6111A2? As far as I know, when we connect a signal to the inverting terminal of an op amp, we get an amplified output which is 180 deg out of phase from the input. Going through the datasheet of TPA611A2 , the input from the microphone goes to the inverting input. However, the LM386 has the input given to the non-inverting input. Can someone help me out regarding this? The way I'm interpreting this is that giving the microphone input to the inverting terminal will shift the phase of the sound we want to hear by 180 deg, making it a different waveform from the original signal, changing the sound we'll hear altogether. <Q> Any sound picked up by a microphone can be assumed to come from some point distant to the microphone (maybe mm to several metres or more). <S> Take a sound at 1 kHz at 10 metres distant to the microphone. <S> 1 kHz has a wavelength in air of around 0.34 metres and this means that between the source of the sound and the microphone diaphragm, the signal has undergone 2*10/0.34 (= 59 <S> ) phase inversions. <S> If you moved the source 0.17 metres closer it will have undergone 58 phase inversions. <S> I'll leave it to the reader to decide if it's at all important to worry about the phase change effects of a mono source on the listener or any amplification stage after a microphone. <A> In some cases it will make no difference, as the microphone signal is an AC signal, and in many cases the inversion does very little, if anything, to the audio <S> (there are cases, like sending differently phased signals to different speakers, can do bad things to sound). <S> In other cases, it can make a big difference, often depending on the type of microphone. <S> Piezo-based microphones, for example, require an amplifier with a very high input impedance. <S> Generally, non-inverting amps can be build with higher input impedance than inverting amps. <A> (EDIT: Complete rewrite) <S> The TPA6111A2 is a stereo speaker driver with an inverting input. <S> To use it correctly, the source would need to have negative polarity. <S> The datasheet shows that, labeling the inputs IN1- and IN2-. <S> The LM386 is a single channel with internal biasing that accepts a positive input. <S> Neither chip is really a mic amplifier: they don't have enough gain, and a mic input should have some AGC function. <S> That all said, by convention a mic input should not invert: positive voltage from the mic results in positive voltage at the final driver. <A> An LM386 power amplifier works fine with an electret mic since its inputs are a higher impedance than the other inverting amplifier and its frequency response is fine for all audio frequencies. <S> But it produces hiss because it does not have a low noise preamp. <A> To solve my question, I did a simple experiment on Audacity. <S> I took an audio sample and loaded it on Audacity <S> and I did a phase inversion by 180degrees. <S> Sure enough, as per the answers, there was absolutely no difference in audio quality, proving that phase of an audio sample doesn't change the sound when we're listening to it.
| The TPA6111A2 headphones amplifier is not designed for electret microphones as inputs because with enough gain then its input impedance is too low and the external compensation capacitor required will cut high audio frequencies.
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Weatherproofed Connector For Outdoor Valve To connect an electronic outdoor DC solenoid valve to a timer, I am getting recommendations to use a silicone-filled wire nut OR solder and heat-shrink. I understand that wire-nuts are simpler for those without a soldering iron, but will heat-shrink technically form a more weatherproof and reliable connection? https://www.homedepot.com/p/DryConn-Small-Waterproof-Wire-Connectors-Aqua-Orange-20-Pack-62114/202889871 <Q> Heat-shrink tubing does not always hermetically seal a connection. <S> The advantage of the wire nuts is that they are filled with silicone ( not silicon ) grease that prevents water from corroding the connection. <S> What I have done is to make my solder connection and then cover it liberally with a high-quality silicone or urethane caulk. <S> Put shrink tubing over that. <S> When you shrink the tubing it forces the caulk into all of the little gaps and gives a better seal than shrink tubing alone. <A> Knot and loop the wire to strain relieve the joint. <A> I just repaired my rusty wet ground low voltage AWG16 solder joints with Wire Nuts and filled with Polyurethane(PU) , an excellent moisture inhibitor. <S> Many plastics slowly allow Moisture absorption but in my experience, Polycarbonate (PC) is best but not liquid and PU is available in low VOC now which means it takes a day to cure exposed to air. <S> This cable drives my fence LEDS with 60W 14V silicone <S> is good when applied to dry air voids ( to prevent condensation) on cleaned surfaces, but the seal has low adhesion strength , so interface moisture creepage is more likely or at least possible along the silicone interface. <A> I have had good luck with 3M 314 Series connectors. <S> They are not hermetically sealed but Gel filled(EG-3 Grease). <S> They say no underground or submersion. <S> Can use with Solid or Stranded wires. <S> Might be worth taking a look. <S> Here is one of them 314-Box <A> You can get heat shrinkable tubing with hot-melt glue in it. <S> That has worked well for me.
| In the marine environment I use butt format crimp connectors then adhesive lined heat shrink this forms a reliable connection proof against the environment.
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Measure 10kV with 600V rated multimeter My multimeter is rated for 600V but I need to measure a 10kV AC source. I was thinking of building a voltage divider but I wouldn't know how many resistors it would take to safely measure this high of voltage. <Q> Figure 1. <S> Source: gfuve.com . <S> The Model GFJDZX0978-10BG voltage transformer is single-phase multi-winding whole sealing epoxy pouring product. <S> It is used both indoor and outdoor for measurement of voltage and electric energy as well as relay protection in the electric system of rated frequency 50/60Hz and rated voltage 12kV or below. <S> 10 <A> You can easily make a 50kV probe for free (surplus/scrap junk yard) or specs=? <S> Typ. <S> 1000:1 probes are designed for 100 M, which you can make with a string of 500V or 1kV cheap resistors inside a polycarbonate tube with polycarbonate hand shield. <S> Using rubber wire tip >10cm from hand position <S> Alligator clamp for ground to Blk wire. <S> But all sources of grid power at >=600Vac have Arc Flash minimum distances with fire protection gear , so <S> ,I trust this is not your goal. <S> I have measured 200kVac using a very large size low pF <S> Cap transform to low voltage. <S> This works better than a voltage transformer. <A> You can buy a HV probe for your DMM. <S> Fluke 80K-40 High Voltage Probe. <S> https://www.amazon.com/Fluke-80K-40-High-Voltage-Probe/dp/B000LDQ672/ <S> If you are a hobbyist this might be more than you want to spend, but how much is your life worth? <S> B&K Precision PR 28A High Voltage DMM Probe, 40kV, <S> X1000 Attenuation: https://www.amazon.com/Precision-PR-28A-Voltage-Attenuation/dp/B004PA02Q8/
| Depends on your Probe Load R specs. 10 kV AC is measured using a voltage transformer. kV measurement is not something you do with a few 1/4 W resistors and a hobby multimeter. Here is a less expensive model. Please don't fool around with building your own voltage dividers at this voltage unless you really know what you are doing.
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Is kapton suitable for use as high voltage insulation? I need to solder together some wires which will eventually run at about five kV, and I'm trying to figure out the best way to insulate the exposed leads. Heat shrink (my usual go-to) is apparently not rated above about 600 V. Kapton seems to be rated much higher (100's of kV/mm), so I am thinking of wrapping the leads in several layers of this. Given the high voltage hazard, I'm looking for advice: Would this be a good solution, or is there a better way? <Q> Kapton is not appropriate to wrap cables: it's relatively stiff, and thus won't be conformal to the cable, which leaves air gaps, which have lower insulation per mm. <S> So, even multiple layers of heat shrink would be better. <S> At 5 kV, I'd start by trying to keep cables mechanically separated far enough that the air distance ensures sufficient isolation, even in the absence of any dedicated isolating material. <S> Use high-voltage cables, which come with the necessary isolation. <S> Don't connect them at the same distance – cut one conductor shorter than the other, so that you don't break isolation in close vicinity. <S> There's isolating potting that you can use to fixate a solder joint. <A> KAPTON (tm) is excellent (Polyamide) But must be sealed from humidity to prevent creepage. <S> Rubber insulated wire is best. <S> Exposed terminations can be somewhat sealed if dry when applied with RTV Silicone to 25kV/cm safely from high impedance sources. <S> Ignition wiring uses Rubber insulation but now is all carbon, but you can still get copper wire. <S> Rated for spark plugs. <S> You MUST be aware of ARC FLASH protection gear if high energy source is being tested. <A> However... Kapton is susceptible to embrittlement over time https://www.mitrecaasd.org/atsrac/intrusive/Chapter_6.pdf https://www.faa.gov/documentLibrary/media/Advisory_Circular/ac25-16.pdf https://www.dfrsolutions.com/hubfs/DfR_Solutions_Website/Resources-Archived/Publications/2002-2004/2004_AgingPolyimide_Hillman-Murray.pdf <S> If it is to encapsulate the solder connections then yes you could rely on Kapton or go for some form of potting to encapsulate the joint
| Kapton is a really good insulative medium and aircraft cable use to be wrapped in it.
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Wouldn't putting an electronic key inside a small Faraday cage render it completely useless? Am I missing something? The Engadget article Kia made a tiny Faraday cage to protect your wireless key from thieves says: Many existing keyless entry systems aren't secure, but few people are likely to replace their cars just to reduce the chances of a determined thief making off with their ride. Kia UK has an official stopgap solution, though. It's taking a cue from third parties and releasing KiaSafe, a case that serves as a minuscule Faraday cage to block the key's wireless signals. There's nothing particularly special to it -- it's ultimately a metal-lined pouch -- but that's all might you need to prevent someone from swiping your car while you're asleep. I'm confused in more than one way. I'd thought that the point of a Faraday cage for RF signals is to block RF inside from getting out and RF outside from getting in. So then you'd have to take it out of the Faraday cage to use it and then of course the usual intercept mechanisms can still take place. If your radiating source is positioned flat up against or even a fraction of a wavelength away from (at least a wire mesh) Faraday cage, don't you then get significant leakage anyway? <Q> The idea of that case is to protect the keys while you're sleeping. <S> You need to take the keys out when you want to use them to drive the car. <S> The main problem is trying to cope with key-relay car theft. <S> This tends to happen at night, with thieves making use of the fact most people keep their keys near the front door. <S> So the key signal just needs to be relayed from just the other side of the wall (where the key's signal just about reaches), to the car. <S> Then the car can then be unlocked, started and driven away. <S> Then you'll need to take the key out of the case in order to get in your car. <S> So your key-less entry and start isn't quite as easy as it would otherwise be. <S> So yes, the keys are out of the case then, but you are in a position to see/use the car, so that should stop anyone stealing your car. <S> Alternative solution to this is my prefer ed one: keep keys further away from the door. <A> Wouldn't putting an electronic key inside a small Faraday cage render it completely useless? <S> That is entirely the point of putting it in a small Faraday cage. <S> You'd take it out of the cage to access your car or drive. <S> Some new-fangled keys chat with the car. <S> The car asks 'are you there?', and the key replies 'yup!'. <S> Which is all fine and dandy when you are near the car and want it to work. <S> Unfortunately, when you are asleep, and your keys are in your jacket pocket hanging in the hall, some well-equipped car thieves might put one end of an RF relay link near your car, and the other end next to your front door. <S> When the car asks 'are you there?' <S> , thanks to the link the key can hear, and the car hear its answer. <S> Next morning, there your car isn't. <S> The Faraday cage blocks the unwanted RF access. <A> Addressing the 2nd part of your question <S> If your radiating source is positioned flat up against or even a fraction of a wavelength away from (at least a wire mesh) <S> Faraday cage, don't you then get significant leakage anyway? <S> Richard Feynmann showed the attenuation of a parallel-metal-wires Faraday cage, with wire spacing of D distance, with L spacing between the wires and the circuit to be shielded, to be AT LEAST <S> $$2 \pi \frac{D}{L <S> } $$ in NEPERS. <S> Thus 1mm wire spacing, and 1mm distance from wires to circuit, provides 6.28 nepers which with 8.6 dB/neper == <S> 54 <S> dB. <S> If the circuit is 2mm inside the wires, at least 108dB. <S> Why are many IR receivers in metal cages?
| The idea of the case, is you put the key in the case, a lot less RF gets out, so the signal is so low that it cannot be relayed to steal your car.
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Why are core losses mechanical and not electrical in an AC power flow diagram? The power flow diagram of a generator (a) and a motor (b) are shown. But why are core losses mechanical losses? <Q> The synchronous motor/generator has two power ports, a mechanical power port at the shaft and electrical power port at the motor/generator terminals. <S> Rotational losses are mechanical in nature and are thus subtracted from the mechanical power. <S> That's why the core and stray losses are shown before the conversion from the mechanical domain to the electrical domain. <S> So, stray losses and core losses are not mechanical power losses, but they are subtracted from the mechanical input power (port). <S> The mechanical power losses themselves are subtracted as well. <S> This subtraction is shown by the down pointing wide arrows. <S> Literature: <S> http://electricalacademia.com/synchronous-machines/synchronous-motor-generator-efficiency-losses/ <S> https://www.motioncontroltips.com/faq-what-are-rotational-losses-in-dc-motors/ <A> Core or iron losses are not mechanical. <S> They are mainly due to frequency and voltage, so they are constant. <S> Eddy currents (batteries set up by impurities in the iron) and hysteresis (heat created by reversing magnetic field) are the main iron losses. <S> Eddy currents are the main reason the core is made of laminated segments. <A> Eddy current losses are the result of Faraday’s Law <S> http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html <S> which states that, “ Any change in the environment of a coil of wire will cause a voltage to be induced in the coil, regardless of how the magnetic change is produced.” <S> Thus, when a motor core is rotated in a magnetic field, a voltage, or EMF, is induced in the coils. <S> This induced EMF causes circulating currents to flow, referred to as eddy currents. <S> The power loss caused by these currents is known as eddy current loss. <S> Motors armature cores use many, thin pieces of iron (referred to as “laminations”), rather than a single piece, because the resistance of individual pieces is higher than the resistance of one, solid piece. <S> This higher resistance (due to smaller area per piece) reduces eddy currents, and in turn, eddy current losses. <S> The laminations are insulated from each other with a lacquer coating to prevent the eddy currents from “jumping” from one lamination to another. <S> Each layer also has capacitance (uF) but in series and burrs on the edges from cut Transformer shears can/may/will shunt laminations.
| Because mechanical losses and magnetic losses (core and stray losses) are (approximately) proportional to the motor’s rotational speed , they are often taken together and referred to as rotational losses. Core losses are resistive due to eddy currents that rise with f^2 and iron thickness, t^2 and and hysteresis losses due to nearing the L-10% saturation threshold.
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What does the multimeter dial do internally? Everyone who has ever handled a multimeter is familiar with these dials. The position of the dial indicates the maximum range of the quantity that can be measured. But why are we required to adjust the maximum range ourselves? What happens internally in the multimeter when the dial is adjusted, say, from 20V to 200V? If we have the dial on 20V, and the voltage measured is 50V, why can't the meter provide a measurement? I don't have much knowledge on the internal workings of a multimeter, but I understand that voltage is measured by letting an infinitesimal amount of current through the meter and measuring the magnetic field (something along these lines). But why can't the meters adjust their range themselves? EDIT: I know there are autoranging meters, but I'm interested in knowing why others have to be adjusted manually. <Q> This image ( source ) ought to tell you all you need to know about how it works. <S> There are wiper contacts on the dial, shown at the bottom, that mate with pads on the meter's PCB. <S> These pads are connected to different taps of a voltage divider to divide the voltage, or to pass current through a current shunt. <S> Internally, the meter can only measure voltages from, say, -0.2V to +0.2V. <S> The range switch changes the voltage divider to prescale the input voltage to be within that range, and on most meters will also send a signal to the LCD to tell it where to put the decimal point. <S> As for why you have to do it yourself instead of the meter doing it for you: Nothing more and nothing less than price. <S> A meter that auto-ranges is more expensive than one that doesn't due to the need for additional hardware to detect when it's over-range and perform the switching. <A> An old analog meter may be easier to understand (source http://fourier.eng.hmc.edu/e84/labs/lab1/node1.html <S> ): <S> Since the gauge deflection depends on the current passing it, and the deflection/current ratio does not change, different contact positions will form different voltage or current dividers to adapt the scale. <S> Note: the coil resistance is also fixed. <S> It's not that different with digital meters, except that instead of a coil that needs a certain current it has a Analog to Digital Converter (ADC) which needs a certain voltage for full scale indication. <A> If it is a digital multimeter, it works differently than an analogue one. <S> The schematics for an analogue one is shown in Vangelo's answer. <S> The schematics for a digital one looks like this: simulate this circuit – <S> Schematic created using CircuitLab <A> A voltmeter is not measuring any currents. <S> There is an ADC (analog to digital converter, that is the real measuring device) inside, which compares the input voltage with a reference voltage and gives the factor of the input to the reference as a digital output. <S> Let's say the reference voltage is 2V. <S> Then the maximum input voltage is also 2V, because this would be 100%. <S> If you turn the dial to 20V input range you are switching in a voltage divider to the input, so that you reduce the input voltage by the factor of 10. <S> This results in only 2V at the ADC with 20V input voltage. <S> The microcontroller knows that and shows you the real value on the display. <S> Autoranging meters can choose the different voltage dividers by switching relays. <S> This requires additional parts and therefore some meters just have the user set it manually and by that reducing cost. <A> In simplest form the dial switches suitable resistors between probes and meter coil so that multiple ranges of voltages and currents cam be measured with a single meter. <S> If you try to measure 40V at 20V range then the meter would have to go to 200% reading because there is two times the current flowing it needs to display 100% reading. <S> The coil cannot move the meter physically and the coil might burn because of excess current flowing unless there are protection components to limit it. <S> The meter does not know how much voltage you are about to measure and has no active components to detect or switch the range automatically.
| For the voltage measurements, the dial will select the ratio of the voltage divider (as shown in the schematics above) and it will change the position of the decimal dot on the display.
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Solutions to reduce start-up currents from multiple DC motors I have some DC motors (8) which are connected to the same power supply (Murata, 12V, 20A), each motor needs 1A and is driven with a MOSFET. When I start these motors at the same time (a need for my application), I have a big start-up current. The voltage falls and the automatic over-current control of the power supply is triggered. How can I properly start these motors at the same time? Does it need a bulk capacitor? Which size? <Q> There are a couple simple ways of doing this. <S> The most straightforward is to slowly ramp up the voltage from zero instead of applying 12V all at once; this "spreads out" your inrush pulse and makes it a lower current for a longer time. <S> If your power supply is fixed, however, one simple option is shown below: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> There's a resistor to limit the current, along with a relay to short the resistor out when the motor is up to speed. <S> This is similar to ramping up the voltage, as the resistor's resistance will reduce the voltage across the motor by an amount proportional to the motor current, and then the relay pulls the voltage the rest of the way up. <S> The relay can also be any other type of switch that works for the currents and voltages needed; a MOSFET would be a perfectly acceptable choice as the application is DC. <S> You can also get what's called an inrush current limiter, which is simply a thermistor with a negative temperature coefficient. <S> As current passes through the ICL, it heats up and its resistance drops, eventually balancing out with the motor current. <S> You can also use an ICL in conjunction with a relay as above, which helps reduce the power lost in the ICL. <S> Note that relays used in this way do not necessarily need to be rated to break the full motor current. <S> If you can ensure that the relays do not open while the motors are still spinning, you can use cheaper relays. <A> You don't say what motion you need out of the motor. <S> DC motors driven by constant voltage draw more current <S> the further away they are from their no-load speed; this means they'll draw a lot if they're stopped and you just switch on the voltage. <S> In fact, in that interval they draw stall current. <S> There's a few things you can do: <S> Limit the motor current individually. <S> Current-output amplifiers suitable for this sort of thing abound; if you don't know how to do that ask a separate question. <S> You mention that you're wrapping the motors with PID loops -- you could also ramp up the target motor velocity. <S> If the motors must be synchronized, give them all the maximum current command but then limit all the faster motors to match the slowest. <S> If you really want a control systems challenge, do the above, but reapportion more current to the slow motor -- but keep in mind that this is likely going to be a very difficult system to keep stable; I could see academic papers being written about it. <A> At the moment of starting the motors, it will prevent the high current. <S> Soon it will work as a short circuit path. <S> So, during running condition, it will not resist the flow of the current. <A> You would ideally measure the inrush current of one motor. <S> Use a very-low value resistor in series with the (-) motor lead and use an oscilloscope. <S> What you are looking for is a current vs time curve. <S> Now decide how much voltage drop you can tolerate when the motors are starting. <S> You haven't provided any data about the power supply that you are using, so <S> I'm going to make some assumptions. <S> You can change the numbers as required. <S> Let's say that your scope curve shows that the peak inrush current is 17 Amps and decreases to 2 Amps in 100 ms. <S> The change in current is 15 Amps. <S> Again: this is for a single motor only. <S> Now let's assume that the power supply will accommodate a drop in output voltage of 1 Volt before shutting down. <S> I may be way off here <S> but, again, you can adjust the numbers to suit your specific situation. <S> Now choose an approximate capacitor value. <S> Delta <S> I is 15 Amps. <S> Delta V is 1 Volt. <S> Delta T is 100 ms. <S> Now we are going to make some approximations. <S> A 1 Farad capacitor will have a voltage change of 1 Volt if a current of 1 Amp is drawn from the capacitor for 1 second. <S> Really simple relationship. <S> In your case, the current is not constant. <S> What we need to do is choose a capacitor value that results in a voltage change of 1V in 100ms. <S> I'm not going to do the math for you - holler if you need help with that. <S> Multiply that capacitor value by the total number of motors that you have. <S> This should get you in the ballpark. <S> Again: I'm pulling numbers out of the air because you did not provide any data about your specific devices. <S> But the above should help you make this work.
| I think the simplest way to limit the high starting current is to add an Inductor in series with the motors. The inductor will have to be of a high value (1 Henry or more).
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Need a non-volatile memory IC with near unlimited read/write operations capability I need a memory solution which is going to be used to keep track of an accumulated count on a micro-controller based project. By accumulated count, I mean to say that the micro-controller uses this memory location to keep count of the occurrence of an event. The count needs to be preserved during power outages, hence the need for NON-VOLATILE memory. Also the occurrence of the count increment event is frequent hence there will be a lot of writes to the memory hence my hesitation to use EEPROM. The preferred communication interface will be I2C, but other alternatives are welcome. Off the top of my head, I envision an SRAM low-power volatile memory IC with the option of being powered by a backup battery like a coin cell on power-downs. <Q> Three non-volatile memory types match your needs, in order of available size: <S> Wear leveled EEPROM/FLASH. <S> Battery backup SRAM. <S> FRAM. <S> In terms of cost, FRAM is best. <S> All you need is inside the chip, including backup capacitors to complete writing. <S> However available sizes are low. <S> Battery backup SRAM is large and costly in materials. <S> Wear leveled EEPROM requires firmware to handle the wear leveling. <A> Here is what I did on a product that's still in mass production. <S> Keep all the parameters and counters in RAM Hook up an interrupt line to a power supply voltage threshold detector <S> When the interrupt triggers, shut off everything that consumes power (most peripherals, LEDs, etc) and back up all the RAM to flash. <S> Turns out there was about 10-20ms of time between the low voltage trigger and the time when the power management IC kicked in and shut everything down (in an orderly fashion). <S> Whether this works or not depends on the energy storage in your power supply, but even a small-ish supply can slow this down enough so that you can write a small data set reliably. <A> Toggle MRAM (magnetoresistive RAM) is claimed to have an effectively infinite write endurance (they're not aware of any mechanism that would cause writing to wear it out). <S> I'm not aware of any such chips that speak I2C, though, so you'd have to settle for SPI. <S> Here's one such part: https://www.digikey.com/product-detail/en/everspin-technologies-inc/MR25H256ACDF/819-1064-ND/8286370 <A> These have battery backup, extra SRAM for user data and come with I2C interface. <S> Or just use a MCU with battery <S> backed SRAM to begin with, so no external components needed. <A> Cypress makes what they call Nonvolatile SRAM . <S> It is standard SRAM that automatically backs up when the power fails. <S> Since it only writes to the non-volatile memory on power failure, it has potentially much greater durability. <S> It comes in serial and parallel versions. <S> It might be a bit overkill, since the smallest one is 64Kb. <S> Under normal operation, nvSRAM behaves like a conventional asynchronous SRAM using standard signals and timing. <S> nvSRAM performs parallel random access reads and writes as fast as 20 ns. <S> On a power failure, nvSRAM automatically saves a copy of the SRAM data into nonvolatile memory, where the data is protected for over 20 years. <S> The transfer between SRAM and nonvolatile memory is completely parallel, allowing the operation to complete in 8 ms or less, without any user intervention. <S> On power-up, nvSRAM returns the data back to the SRAM and system operation <S> continues from where it left off. <S> nvSRAM also provides user controlled software STORE and RECALL initiation commands, as well as a user controlled hardware STORE command in most versions. <A> For a single 4 byte variable, EEPROM would be totally fine. <S> Let's say you are writing to it once per second <S> and you have a typical 32Kb EEPROM and we go with a conservative endurance of 100,000 write cycles. <S> You can write your 4 bytes 8000 times before you need to do a clear. <S> So that should be 800 million times that you can write it even using a conservative estimate. <S> Now there are only 31.5 million seconds in a year, so at one write a second it would take 25 years to reach the low end estimate of EEPROM endurance. <A> There are plenty of options here, but the real issue is stopping the data from getting corrupted. <S> Power loss during a write could corrupt the data. <S> I2C is a good option for avoiding this, because e.g. with SPI you could find that a write appears (from the memory's point of view) to complete half way through updating say 4 bytes of a 32 bit word. <S> I2C is a little more robust, but only a little. <S> My advice would be to store 4 copies of the value. <S> That way even if writing is interrupted, two will always match.
| FRAM or similar is probably the best option. Sounds like you can just use a RTC clock chip or module.
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Does "1KV" on this disc capacitor mean 1 kiloVolt? One of my power supplies died. Investigation turned out that this capacitor is broken. "102" means 1nF, but does "1 KV" mean 1 kiloVolt or is that some sort of tolerance code? I've tried several "capacitor code calculators" but none of them mentions "KV" (only K, which is indeed a tolerance code). <Q> Yes, it means 1kV (DC) rating. <A> Tolerance code is usually on the same line as capacitance, for example 102K would be 1nF +-10%. <S> But to be on the safe side, I would also do an educated guess about the requirements for that capacitor in that application when replacing it. <S> Maybe it has to be of a specific type <S> (desired failure mode in particular) or maybe it is over-specified and operates at much lower voltage. <A> Yes, it’s rated for 1000V (1KV), a high voltage rating. <S> If this cap is part of a safety-critical system (and at that voltage, very likely) review your circuit closely. <S> If there’s line voltage involved, consider using a safety rated X or Y-cap which is designed to fail in a way that will render your circuit safe. <S> X caps go ‘across the line’ and are designed to fail as a short, which should blow the input fuse. <S> Y caps go ‘line to ground’ and are designed to fail open, preventing a shock hazard. <S> More here: https://www.allaboutcircuits.com/technical-articles/safety-capacitor-class-x-and-class-y-capacitors/
| Looks like 1 nF, 1kV capacitor.
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How to do this USB transfer? I have a scenario that, When receiving data through UART (STM) ,same time the data is displayed in the notepad(opened in PC) through USB. How can I implement this? I am using stm32 micro controller Figure(as given) For example, The data received by STM is employee id,at the same time data should displayed in PC. Note: I too have LPC1768 <Q> If you want data to Notepad, then the STM32 must be a keyboard HID to the PC, and it must convert serial data to keyboard presses. <S> STM32 must have USB. <S> But this may not be what you want. <S> If however the point is to watch serial data on screen and log it to a file, you would use a terminal program like putty to open a serial port. <S> No microcontroller involved, just get a suitable serial cable or USB to serial adapter <S> and it is done. <A> You can use Notepad++ to view data in a text file. <S> From within Notepad++: <S> Settings > Preferences > MISC. <S> From there, you can check "Update silently" and "Scroll to the last line after update" in the group "File Status Auto-Detection" <S> Or you can use the command prompt as well, simply use Tail -n 0 -f pathtotextfile <A> You can build that on a breadboard. <S> You can used FTDI IC. <S> SOme microcontrollers are programmed through FTDI chips which converts RX/TX signals to USB's D+/D- signals. <S> As for a software to view received data from UART, you can use Terra Term. <A> I'm not sure if the STM32 is a requirement or just happens to be the hardware you saw and began trying with. <S> However, there is much easier hardware (than stm32) to use to do this. <S> If you get an arduino Leonardo you can do this with a few lines of code. <S> You can pick up three Leonardos for $20USD . <S> The Leonardo is small and will be seen as an HID device as soon as you connect it to your computer. <S> It uses the ATMega <S> 32u4 <S> which implements HID USB <S> so you don't have to do all that work. <S> Once you get the Leonardo the code is as simple as this example shows: https://www.arduino.cc/en/Tutorial/KeyboardSerial <S> #include "Keyboard.h"void setup() { // open the serial port: Serial.begin(9600); // initialize control over the keyboard: Keyboard.begin();}void loop() { // check for incoming serial data: if (Serial.available() <S> > 0) { // read incoming serial data: <S> char inChar = <S> Serial.read <S> (); // Type the next ASCII value from what you received: // <S> this sends the data to your computer as if a user typed it on a keyboard. <S> Keyboard.write(inChar + 1); }}
| It's simple, build a UART to USB converter.
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How to remove solder mask in Eagle I want to remove solder mask in particular area for mount a TO220 package. As shown in below image I want to mount the 7805 using a M3 nut and need to remove solder mask only in covering area. (like a thermal pad) <Q> Create a rectangle on the solder mask layer, which is a "negative" layer - anything you create there will not have soldermask. <S> Because the layer is negative and because the program stacks features, you can do this on the board, if it is application specific, rather than in the library part's footprint. <S> The advantage of doing it in the footprint (first copy or export the part to a custom library) is that you may have more ready access to the part geometry when defining it, and something that is a common need can stay with the part as a footprint variation. <A> Put a polygon named GND under the tab <S> so it is grounded, it will also help with cooling. <A> Draw a polygon in the tStop layer(21). <S> It stops the solder mask in side the polygon.
| Conversely, the advantage of doing it on the board is that what exactly you need may depend on the particular board you are designing.
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Why does this potentiometer in an op-amp feedback path cause noise when adjusted? I have placed a dual-gang 100k potentiometer in the feedback path of two op-amps that work as a Sallen-Key filters. This controls gain but also the Q factor of the filter ( somewhat demonstrated on wikipedia . However, when I move the knob, it makes some rubbish noise. The noise is demonstrated in this youtube video . The schematic of the filters is shown below. The Q factor pot is RV3A and RV3B. The minimum Q is set by R7 and R10. The power supply for the op-amps is from a boosted 9V battery up to 18 V with an LM27313 regulator. I'm really not sure what is causing this noise or how I could go about preventing it. My only thought is having the pot in the feedback loop is not a good idea, but there is no other way of controlling the Q factor. How could I alleviate this noise? Edit : I accepted @Catalyst's answer since it was the most technically correct. I appreciate all of the suggestions on how to fix it. I simulated some tests on placing capacitors in parallel with the potentiometer but they really messed with the frequency response. What did fix the circuit was by placing it in an enclosure. Using an aluminium enclosure linked all of the pots strongly to ground which seems to have generally improved noise performance hugely. Here is a link to the a new video demonstrating a lack of noise . The audio is recorded in exactly the same manner as before. I don't fully understand why a better ground fixed the travel noise but I'm certainly glad it did. <Q> The noise is caused by minute mechanical vibrations of the pot wipers on the rings (the latter are the resistive material.) <S> Since neither the wiper nor the ring material are atomically smooth where they touch, rubbing the wiper on the ring produces slight vibrations. <S> Some of that vibration is perpendicular to the contact patch between wiper and ring. <S> The resistance of the wiper/ring contact varies with the normal force. <S> This transient/AC variation of the contact resistance is what you're hearing. <S> How to fix it is (IMHO should be) a separate question. <S> And non-trivial because adding caps across any pot terminals will change the filter characteristics. <A> How could I alleviate this noise? <S> Probably not what you want to read now, but: don't use pots in audio circuits where they are subject to DC currents. <S> This is the case for the 2 pots you mentioned, but not for the other 4. <S> Due to the mechanical nature of the device, the resistance variation is not "clean" and continuous. <S> To state with other words: <S> quick small changes to the DC operating point cause the same effect as an AC signal being injected in the circuit. <S> These small resistance variations do not result in large voltage changes if the the potentiometers are only subject to small DC currents (due to capacitor leakages or op. <S> amp. <S> inputs, for example). <S> Update: this is a crude attempt to document what I suggested in the comments: <A> simulate this circuit – Schematic created using CircuitLab Datasheets for audio amplifier ICs strongly recommend using a DC-blocking capacitor between the input potentiometer and the IC's audio input in order to prevent the slider noise caused by a very small DC current (on the order of microamps or less) between the IC input and the ground leg of the potentiometer. <S> I apologize for taking this long to get back, I had been busy and forgot about this. <S> HERE IS A BETTER ANSWER <S> (refer to the schematic above!) <S> : <S> The suggested additions to your circuit are boxed in dashed red. <S> CY1 and CY2 could be up to 1uF, see if there is any difference. <S> CY3 can be up to 47uF. CY4 and CY5 are suggested if you have no capacitors before and after this circuit in the audio chain. <S> As for your glitch problem, I would say it is because you have a low resistance (RV2a and R9) going from the positive supply rail (+8V) to the non-inverting input of the U2a, and that causes clipping and distortion. <S> Op-amp inputs should never be driven to either positive or negative rail with DC, as that will reduce their positive or negative swing (headroom) and cause clipping and distortion. <S> What is usually done is that the resistors connecting DC to op-amp inputs are either at real ground (when you have a dual voltage supply and the ground "sits" in the middle) or you create a virtual ground by using a simple voltage divider like 2 resistors (of equal values, 10k to 100k each) in series between positive and negative rail, use their middle point as ground but make sure to use a 1-2.2uF decoupling capacitor from that point to the actual ground (usually the negative rail). <S> So, your R9 would go to this (virtual) ground instead of the positive rail of <S> +8V. <S> In fact, I would suggest that you connect all 3 signal points that are going to +8V to this virtual ground instead, and see if that solves both of your problems. <S> If it doesn't solve the noise when adjusting the Q factor, you could place 1uF capacitors at either end of both of the 68k resistors, and see if that helps. <S> Try all these changes in the schematic one by one and let me know which ones have worked for you. <S> I would like to know if these suggestions have helped you. <A> If the pot is not sealed, you can apply a small amount of Deoxit Green , to both wafers, cycle the pot several times, and see what happens. <S> Amp techs and hobbyists have eliminated scratchy pots on many an amplifier this way. <A> I have had some Moog synthesizers with noisy pots. <S> After replacing them with new, clean pots the noise remained. <S> Turned out the electrolytic capacitors had gone bad and were leaking DC on to the pots causing it to sound just like dirty pots. <S> I replaced the ecaps and the noise was gone. <A> For voltage controlled gain you may use either JFET, which behaves like linear resistor, or a transconductance amplifier. <S> The latter has voltage gain input. <S> Equally, you may use another opamp instead of your resistor to shift the DC point and regulate gain.
| On a more practical note, it's possible the pot is simply dirty. If the potentiometer is subject to DC current, the quick variations on resistance result in voltage changes which are handled by the circuit as signal, depending where the potentiometer is.
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Arduino nano resetting after solenoid valve fires I'm trying to control small, plastic 12V solenoid valves with an Arduino nano. I have an Arduino nano hooked up to a Sainsmart 4-Channel 5V Relay Module, and finally connected to the solenoid valves, which are powered externally with an AC to DC converter. The valves are controlling gas flow to tanks of seawater for a marine biology experiment. As such, I need the solenoid valves to open and close very quickly (on the order of tenths to hundredths of a second). The problem is that after the solenoid valves fire, my Arduino resets and reinitializes, and this messes with the timing of gas flow. I've isolated the problem to the powered solenoid valves (the code, arduino, relays, and solenoids all work perfectly when the solenoids are unpowered. The second I plug them in, the problem resumes). I think that reverse voltage spikes are causing my arduino to reset, but am not entirely sure. I've looked into adding flyback diodes, but heard that those slow down the relays and again, I need the timing to be very fast. Does anyone know if adding a diode will solve my problem, and if so, how much will it slow down the relays? My knowledge in electrical physics is limited, so if you could keep your explanations to more basic language, I'd appreciate it. But ANY and all help would be welcome! Thank you! <Q> Using a zener diode should solve your issue while keeping the speed fast enough. <S> You should definetly use a flyback diode when working with induction load. <S> Zener should do the trick well. <A> You're on the right track. <S> Inductive loads, like solenoids, usually need a flyback diode to prevent voltage spikes when they are switched from the on state to off state. <S> Tenths to hundredths of a second isn't all that fast for electronic devices. <S> I suspect you'd be fine with nearly any diode type. <S> Even relatively slow diodes will still have a reverse recovery time measured in microseconds, and your switching speed is considerably slower than that. <S> Getting an oscilloscope on the solenoid power line should help confirm that this is your root cause. <S> It could also help you determine a specific diode to install based on the size and shape of the overshoot. <A> Consider that you most likely already have flyback diodes in the relay module, and you don't seem to notice that your relays are "slow". <S> Your valves won't appear any slower with flyback diodes unless the EM field decay is the limiting factor, as opposed to mechanical movement, and I doubt that will be the case.
| Since solenoids have inductivity fast switching causes the solenoid do demagnetize into the circuitry.
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How does one position control a voice coil actuator? I am looking at trying to position control a voice coil actuator (something like these ). On the end of the voice coil actuator will be a small load (let's just imagine a 10 oz cube), screwed onto the end of the shaft. Here is where I come from. When position controlling a spring-return solenoid with a feedback sensor, it's relatively simple: to move the solenoid out to position X, you control the power based on the feedback signal's value. The more power you input, the further out the shaft moves. I am wondering, how does one actually conduct position control on a voice coil actuator? In other words, to hold the shaft static at position X, does it require a constant voltage/current? Or, can you remove the input voltage/current, and have the shaft not move? <Q> I am wondering, how does one actually conduct position control on a voice coil actuator? <S> In other words, to hold the shaft static at position X, does it require a constant voltage/current? <S> A constant voltage or current will hold the shaft at a constant position provided the load, angle, and temperature don't change. <S> To control in that fashion is generally called "open loop" and can work well enough for some applications. <S> There is no feedback built into the units in your link. <S> Or, can you remove the input voltage/current, and have the shaft not move? <S> There is no mention of a shaft brake on the datasheet either. <S> The brake, if supplied, would require additional wiring as you couldn't expect the brake to release on low voltage operation (for a small movement). <S> Response to comment: If a brake were to be included, what would be the brake's technical name? <S> I'd be looking for "holding brake" or similar. <S> Would it be a linear magnetic brake? <S> No. <S> Linear magnetic brakes apply a decelerating force proportional to the velocity of a conductive plate through a slot in a magnet. <S> At zero velocity the have no holding force. <S> You'll need a mechanical holding brake. <S> Also, to interpret what you're saying in the 2nd half of your post... <S> If the brake were added, it would prevent motion at low voltages because the power input to the VCA would have to be enough to generate a magnetic field that could overcome the magnetic field from a brake <S> Let's say we had a 12 V actuator with a 12 V brake. <S> The brake requires some voltage - say 8 V - to release it fully. <S> If the brake is sharing the actuator wires then the actuator won't move well or at all until the voltage reaches 8 V. <A> There are many ways to control a voice coil positioner, here are some of the analog systems: <S> You can connect the shaft to a linear differential transformer (LVDT). <S> This is probably the simplest high accuracy analog position control you can get at reasonable cost. <S> Neither is easy to interface. <S> Potentiometric linear sensor. <S> I've seen many DIY implementations of this, and they are quite easy to implement. <S> Probably the cheapest implementation you can do. <S> You can get magnetic and capacitive linear position sensors, but they are problematic to design with. <S> This article explains some of the choices. <S> All of the analog systems are easily included in an analog feedback loop for closed loop position control. <S> For digital sensors the choices are more difficult and typically more costly, except for AMS magnetic sensors I've used. <S> Probably for your application you could use the AS5306 with Quad outputs that would easily accommodate a closed loop position system with an MCU maintaining the position counts. <A> A voice coil actuator alone simply produces a force that is proportional to the current passing through the coil (with a bit of friction in the bearings). <S> Ideally that force is independent of the position of the shaft, but in practice it will vary a bit throughout the stroke. <S> To move it to a given position you need to measure the position by some means and use a controller and output amplifier to provide an appropriate current, resulting in a force, to move and maintain the position against whatever external forces and friction that may be present.
| For precise position control some form of position feedback is required to adjust the output so that the position error is reduced. You can connect the shaft to a torque bar with a strain gauge or piezoelectric sensor attached.
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LDO vs Switching Voltage Regulator for Drone Application I have been designing a drone flight controller and currently have been using LDOs to regulate the voltage to 5V. As I have started to use higher voltage batteries the LDOs have started to overheat. I know that switching regulators are so much more efficient which is definitely a plus with battery-operated devices, but heat is not my only concern when considering which power supply to use. When the drone's motors are on, the voltage supply becomes very noisy, and I have RF ICs on the board that are incredibly sensitive to power supply noise. Would a switching regulator be better for this application, or would it generally perform worse? TLDR: When comparing LDOs and switching voltage regulators, which are generally the better performing option in terms of output voltage noise when considering an already noisy power supply? <Q> As you have probably found searching internet, a LDO has generally less ripple/noise/EMI than a SMPS. <S> heat is not my only concern when considering which power supply You could distribute the heat using resistors in front of the LDO. <S> Adding capacitors to form RC filters help to filter "an already noisy power supply" at the same time. <S> For example: You can also preceed the circuit above with a SMPS to gain better efficiency. <S> You probably don't need the RC filters. <S> When the drone's motors are on, the voltage supply becomes very noisy But addressing the cause of the problem rather than patching the effect of the problem would be a better approach. <A> You can use both. <S> Use the DC-DC to drop the rail down to a voltage the LDO can use. <S> You get the best of both worlds: DC-DC efficiency, and low noise for your RF stuff. <S> This is a really common design approach, that is, sub-regulating a local supply from a bulk DC-DC supply to create a rail with special requirements. <S> The TI NMOS ‘cap-free’ LDOs are particularly good for RF applications (TPS731xx, 2xx, etc.) <S> As far as motor noise corrupting the supply, a way to mitigate that is to make separate runs to the battery for the motor drives and to your radio board. <S> That is, isolate each power on its own loop so that ground/power noise from the motors doesn’t find its way into the rest of your system. <S> You can also apply some common-mode filtering to the individual power runs to stop the noise from coupling between them. <S> More on that idea here: Noisy Voltage Rails (Vcc, Gnd) - Noise Isolation for Automotive Circuit <A> A switch mode regulator by its nature of 'switching' generates noise. <S> You can kind of think of it like this: An LDO produces waste heat to drop the voltage while a SMPS produces waste noise . <S> Regardless of which you use, you generally want to do something about minimising the noise into them. <S> While your LDO will produce less noise for a given input, it may not do all that <S> much when it comes to reducing existing noise. <S> In a drone I would really be looking strongly at the SMPS, simply because you're looking at an extra 50% or more efficiency over the LDO. <S> Noise can be filtered out and you already have noise problems that need addressing regardless of which you use.
| When comparing LDO's and Switching Voltage Regulators, which are generally the better performing option in terms of output voltage noise when considering an already noisy power supply.
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Electronic load, Problem turning MOSFET on I am trying to make an adjustable dummy load to test a 12V supply, I’m having trouble with the op-amp feedback circuit.Attached is a simplified schematic. I have tried lots of things and stripped it down to this. Basically when +In of the op-amp is 0V, the output of the op-amp goes to the negative rail. In this case the negative rail is 0V, but when powered with a +/-5V supply, the output goes to -5V. In this state the MOSFET is fully off When I add just a tiny bit of positive voltage to the +In of the op-amp, The output shoots straight into positive saturation and goes as close to the positive rail as it can get (about 3.7V) and stays there no matter how much voltage I give to the +In. At this point the MOSFET is on, and allows approximately 250mA to flow through though. Any ideas on what this could be? Things I have tried:- removing R6 and Q1 and connecting R8 to -In. At this stage I get a normal voltage follower.- Changing the FET- Raising the resistance of R6 On a side note, there is one way that I can get the load to be variable, and that is by removing the heat sink from the MOSFET. I have only done this for short periods of time. <Q> The LM324 is old. <S> Without too much analysis... <S> It could be that the voltage offset (which for the LM324 would be high at 3mV) is causing problems with the feedback loop as there would be unertanty when the voltage across R6 goes lower than roughly 3mV (the input offset voltage will vary from amp to amp and temperature). <S> Another problem could be the common mode range. <S> Anyway this is an old opamp, my suggestion would be to find a new rail to rail opamp with a low input voltage offset. <A> What you have built is an oscillator. <S> There is nothing wrong with the LM324, it has been updated through the years, but I do agree with @laptop2D <S> you could make a better choice of devices. <S> You don't need a low offset opamp in this application. <S> The problem you have is that the FET you chose has a huge input capacitance (>5000pf). <S> With the 180 Ohm series resistor you guarantee that the circuit will oscillate around your setpoint. <S> You can try simply removing the 180 Ohm, which allows that LM358 to current limit into the gate, that might just work though it isn't pretty. <S> The circuit has another problem though, the VGS(th) range for the IR1404 is 2-4V. <S> You may have lucked out and got one that is closer to 2V ….but <S> you will run out of drive voltage and not be able to use one that comes in at close to 4V. <S> You should at least double your supply to 10V (perhaps you could use the 12V supply you are testing to generate this). <S> To charge your gate capacitance at a reasonable slew rate I'd suggest you need to allow for up to +/-500mA drive. <S> Perhaps something like this: simulate this circuit – Schematic created using CircuitLab <A> Yes. <S> The loop gain is to high, so the circuit will oscillate even if you replace the opamp with better one. <S> You would need at least a low pass filter on feedback to damp the high frequencies. <S> Since it is not oscillating yet, the problem could be a poor opamp as laptop said.
| However to fix the problem you really need a power driver stage for the FET gate, or chose a FET with much lower Gate capacitance. You could do this by using an NPN/PNP pair on the output of your opamp.
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Problems with dc motor driver design: Mosfet fail I made a brushed dc motor driver. The motor speed is controlled with PWM (from an Arduino). The motor that I'm using draws 3 amps without load. This is the schematic: As you can see there are two N channel mosfet in a parallel configuration. The first problem that I had was that when I tested the PCB for the first time R2 burned and Q2 was damaged (all three pin where internally connected). I though that Q2 was defective so I change it and the PCB worked perfectly.Then I tested the same circuit, different PCB, and happened the exact same thing, R3 burned and Q2 damaged, that can't be coincidence. Right? Then I ran the motor at max speed for around 2 minutes, and the circuit worked fine, no overheating at all. When I test the circuit with load (in a rc tank) the Heatsink heat a little bit (as expected), but then one of the drivers failed and the motor keep spinning. A mosfet failed. Any suggestion on how to make the circuit work reliably, without the mosfet failing? <Q> You need to add a high current diode from the mosfet drains to battery positive. <S> This will give that energy somewhere to go. <S> You'll want to connect the diode physically close to the mosfets (the motor wires will produce their own small inductive kick). <S> This is probably the reason your mosfets are dying. <S> You'll want to read up on how to pick this diode, but the key parameters are current (a good rule of thumb is to handle as much as your motor draws), and reverse recovery time (I'd go with a schottkey). <S> A second problem is that you're driving the mosfets at only 5v Vgs (slightly less actually due to the resistors). <S> Looking at the current/Vgs curve in the datasheet. <S> The Rds will be approximately 0.25 ohms. <S> In parallel that will be 0.125, so at 10A you're looking at 12.5W of dissipation. <S> That's quite a bit. <S> The third problem is that you're driving the mosfets from the arduino directly. <S> Since the arduino can only supply 20ma or so to the gate, the switching times will be slow. <S> This means more power dissipation, especially as the frequency increases. <S> This will drastically improve the power handling of your motor driver circuit. <A> Freewheel diode is needed as Drew has stated +1 .Also <S> check C2 ,if it is too large the charging currents will stress out the mosfets .Also check the stall current of the motor <S> .The lithium battery can supply large currents .What <S> if the current was 30 Amp? <S> This 30 Amp will be drawn when the motor starts .Can <S> your mosfets on the little heatsink handle this ? <A> The hexfets you are using look like they have Zener diodes inside the package, but there is no reverse voltage spec. <S> Voltage pulses may be killing the fets. <S> When I looked at the schematic I expected a reversable driver as some kind of half-bridge, but I see it is one way only. <S> I wonder if current spikes in the ground side of the fets may cause an over-voltage condition on the insulated gates. <S> Since the driven current is one-way, the fly back voltage will also be one way. <S> When the fets turn off, current will continue to flow into the fets. <S> The diodes in the fet package are not zeners. <S> They will not conduct current at any reverse voltage. <S> Diodes to the M+ rail are needed. <S> Also, keep the grould lines short back to the bypass caps on M-/M+. <S> That leakage inductance could fry the gate, and even if not, the voltage bump will work against your $V_{gs}$ and increase the turn on time, further heating the fets.
| One huge issue I'm seeing is a lack of any flyback diode for the motor. You can solve problems 2 and 3, by using a gate driver IC .
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Does a potentiometer need a resistor in series between power and ground? I have a potentiometer I use with a circuit to get data readings. This circuit is connected to an arduino mega 2560. I was wondering if I need a resistor in series with the potentiometer? Attached are two picture of what I mean, sorry if the circuits are a bit rough first time I have used an online schematic maker. I would also like to add that the potentiometer values are likely wrong, it was a generic one a friend gave me. All I know is that it goes from 5 ohm to 10k ohm. <Q> First, a couple of CircuitLab tips. <S> Double-click a component to edit its properties. ' <S> R' = rotate, 'H' = <S> horizontal flip. ' <S> V' = vertical flip. <S> Note that when you use the CircuitLab button on the editor toolbar an editable schematic is saved in your post. <S> That makes it easy for us to copy and edit in our answers. <S> You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. <S> The important thing is that your potentiometer circuit shares the same ground as your microcontroller. <S> If you leave out the ground connection then you have an open circuit and no current can flow from the potentiometer to the micro. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Both circuits share a common ground with the microcontroller. <S> There is a difference between the two circuits. <S> The wiper on Figure 1a can go from 0 V at the bottom to 5 V at the top. <S> The wiper on Figure 1b can go from 0 V at the bottom to 5/4 V (1.25 V) at the top because R1 and R3 form a potential divider with the maximum voltage give by \$ \frac {R3}{R1 + R3 <S> } V_2 \$ . <S> If your micro's analog input is 0 to 5 V then (a) uses the full scale (typically 1024 counts on a 10-bit ADC). <S> (b) would give a maximum of 1024/4 = 256 counts for 1.25 V. <A> No, the wiper pin is the part of the pot which "sees" a different resistance based on its position. <A> No, you don't need a resistor between the positive side of the potentiometer and the power, assuming that it's the same voltage level as the arduino runs on. <S> But, taking your circuits litterally, note that you do need a ground connection between the battery/potentiometer and the arduino.
| There will always be the max resistance between power and ground pins on the potentiometer.
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Serial communication between STM32F4 and PC -USART- The question I will ask now is not about how to solve something, I am currently able to make the connection as it will be mentioned below, but my problem is I really did not understand WHY ? Thanks in advance. Hey, I am trying to connect my STM32F4-DISC board to PC via USART. It is a very common and basic example like sending a string to PC COM port which is covered by lots of tutorials on the web. But my question is, I first used this product and directly connected to my board's RX TX pins and I was able to get data but all the data was a mess.(Not random, always same but some characters I have never seen before.) I think there was nothing wrong with connections because when I disconnect the jumper which connects the Rx of converter and Tx of DISC board, there was nothing ongoing even that messed up characters. After lots of thinking about "what is wrong? is it code?( It could not be because only 3 lines of code and all examples writes the same thing) ", I found a little device which has a chip on it( I can't read what is the model or manufacturer) and one side is D sub 9 male and other side has 4 jumpers come out of that named as Rx,Tx,GND and 3.3V. Then I also used it and created a connection like this: PC --USB ENTRANCE-- DIGITUS USB to serial adaptor -- D SUB 9 ENTRANCE -- THE IC I MENTIONED ABOVE -- JUMPERS -- STM32F4 After this setup I was able to see the correct text on the screen and everything worked well. I researched about why this could be but did not understand. The digitus product itself looks like the only necessity to connect my board to PC but It does not work alone. I wonder what is the real case? <Q> USB to Serial adapters with a DE9 connector use RS-232 Voltage levels , where logical 0 is a positive voltage between 3 and 15 volts, and logical 1 is a negative voltage between -3 and -15 volts. <S> STM32 Microcontrollers use TTL-like signaling, where logical 0 is between approximately 0 and 0.7 volts, logical 1 is between 2 and 5 volts. <S> Voltages below 0V would be registered as logical 0, as long as the MCU works, but they can likely damage it permanently. <S> Apart from being extremely lucky that your board didn't blow up, you've got the bits inverted, that's why strange characters have appeared. <S> The other device would be a level converter, which translates the signals between the two standards. <S> TTL 3.3V adapter. <S> I've disassembled an USB-RS232 adapter similar to that of the question. <S> There is the <S> FT232RL <S> doing the USB to serial-TTL conversion, and an RS-232 transceiver , doing the voltage level shifting. <S> You could theoretically remove the SP213E part, and connect the microcontroller UART port directly to the <S> FT232RL . <S> Top side IC markings on top side Bottom side IC markings on bottom side <A> The first product is a USB to RS232 interface converter, the standard RS232 interface sends UART data between devices using RS232 voltage levels, Logic 1 is negative and Logic 0 is positive, for voltages about +/- <S> 5V to <S> +/- <S> 15V. The STM32 Discovery board does not have a RS232 interface, <S> the UART pins that use 3.3V MCU voltage logic levels are at the connector. <S> Therefore connecting this to a RS232 adapter does not work as the data is inverted, and the high voltage and current driving ability can break the MCU. <S> So this is why you need a RS232 to logic level interface chip between STM32 and RS232 port, or the other solution would of course be to just get a 3.3V level USB adapter for direct connection. <A> Your STM32 board most likely has a logic-level USART (also known as TTL-level) while the Digitus device uses RS232 signaling. <S> The difference, other than potentially different voltage levels, is that logic-level serial signal idle at 0 volts and a one bit is indicated by 3.3 or 5 volts while RS232 signals are inverted, idling high and going low for a one’s bit. <S> The intermediate device probably has an RS232 level converter chip, such as the MAX232 (Wikipedia) . <S> This both inverts the signals and contains a charge pump circuit to generate the higher voltages specified for RS232. <S> As a rule of thumb, if you have a DB connector or a 9 or 10 pin connector, you are probably working with RS232. <S> If you have a simple row of 3, 4 or 5 pins, you probably have logic-level signaling.
| There are products that can directly connect an USB ports to a 3.3V microcontroller UART port, look for an USB RS232
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When to use slots instead of barriers in high voltage design? I have read a lot of information about the clearance and creepage strategies, distances, materials, etc.. I am designing my first HV PCB, main LV circuit will be on TOP and HV (6KV) on the bottom. But I can' t avoid some low voltage points on the same side. Here I will be using insulation barriers to increase creepage and clearance at the same time. But I would like to use slots in order to separate and give some air gap between "hot" points. But when you put slots, the only factor that increase will be creepage. Clearance stills being the same with or without the slot. So, here are my questions: When to use slots instead of barriers? When can be the clearence shortest than creepage? Is it the same to put a 3cm slot between two points and to simply put a distance of 3cm without slot between 2 points? How can it helps to avoid breakdown if clearance is the same at two both cases? <Q> One of the risks with attaching barriers to a pcb is that they aren't consistently attached ("cemented" is the UL term) to the board with variation in effectiveness both along their length, and from part to part in production, leading to the possibility that there's a small gap or wormhole under the barrier - so you then have to look at the creepage along that path. <S> For very high voltage systems, it's common to move to a molded insulator with integral ribs, and the conductors are stampings that can be insert molded as part of the molding - or attached by screws or rivets. <S> Obviously, adding slots to a pcb is more cost effective than (carefully) adding a barrier, since it's just part of the milling of the board that is undergoes regardless. <S> The standards, like UL840, are often insufficient when it comes to very high voltage design, as other factors become important. <S> There is a risk of corona discharge that is especially bad if there are sharp points e.g. through hole soldered lead ends or angles in the tracks, and the presence of the high voltages tend to attract dust that will over time compromise the tracking resistance of the surface, so air gaps tend to be better than barriers for this. <S> If the device is used in locations where air pressure is reduced Paschen's law applies, this is relevant in aerospace applicaitons. <S> The HV circuits I worked on (10-15kV) were always encapsulated to avoid the risk of reduction of contamination reducing the tracking resistance, once this is done you then only have to worry about the breakdown of the materials involved. <S> Asphalts were used for the large assemblies, since this was more cost-effective than the siicones or epoxies used on smaller units. <S> TLDR - airgaps are practically better than barriers, but once you get to multi-kV applications, encapsulation is preferred. <A> When to use slots instead of barriers? <S> Easier to produce. <S> When can be the clearence shortest than creepage? <S> Clearance is always at most as long as creepage path. <S> That's the simple geometric triangle inequality . <S> Is it the same to put a 3cm slot between two points and to simply put a distance of 3cm without slot between 2 points? <S> No, as your pictures clearly illustrate. <S> How can it helps to avoid breakdown if clearance is the same at two both cases? <S> Not quite sure what you mean. <A> Not an answer, still an important thing to show (and comments don't show images): <S> When the bottom layer has the high voltage, there is another creepage way, e.g. the one like shown in purple in image below. <A> Please ignore people giving you fixed numbers. <S> I work with high voltage design and it does things you dont expect, so make sure you get a creepage and clearance calculator which includes your material group, pollution degree and altitude your working at (the higher you go the more distance you need). <S> The calculator will tell you the clearance you need. <S> There are three distances, functional (bare minimum), basic and reinforced. <S> If it is safety related then you must work to reinforced, however if none is going to handle this thing in operation use basic clearance. <S> I would link you one but my company has developed it's own internal one. <S> The standards you are working with will tell you what voltage to apply. <S> If the material withstands then it means you dont have to worry about creepage. <S> Some standards say you have to get a single layer to pass <S> then double it. <S> Physical barriers are usually more expensive as someone has to manually put it in <S> but I've seen cases where a 'box's has slotted over things to reduce clearance <S> so in places like these it's not too much more expensive.
| You typically use slots where you can however if you have things like pcbs stacked or younphypically cant meet the clearance, you have to use barriers and these effectively reduce your clearance (still check the creapage).
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Detecting if 120v AC wire is live with a 2 or 3 VDC line For a background, I have 3 DPST bistable (latching flip-flop) relays that I have 120VAC input and 2 outputs. On one side, I have a 120VAC appliance. On the other side I need to find a way to feed it back into a Raspberry Pi's GPIO pins to detect whether it's live. What is the simplest way to accomplish this? I thought about step-down transformers and induction, but the transformer would get expensive, since I can't seem to find any other way to do it besides chaining 120vac to 5vdc and 5vdc to 3vdc and I have no idea how to use induction. UPDATE: Thanks for all of your answers. The Bistable relay requires a 0.2s pulse to trigger to switch, so as long as the response time is less than or equal to that, it will work just fine. As such, using a cheap USB charger should work if @Transistor is correct in his assessment about ~100ms response time. However, it seems a bit hackish and I'd rather find something ready to be used in a DIY circuit rather than tearing apart the casing to extract the guts of a product. <Q> @Finbar's suggestion is the easiest and safest if you just want to plug it in and don't care about response time. <S> Use a genuine safety-agency approved charger , not some dollar store item made of Shenzhen street sweepings, some of them are very bad (and unsafe) indeed. <S> You should add a bleeder resistor to drain the output votlage in a reasonable length of time and you'll also need to reduce the 5V USB output to 3.3V or so for the Raspberry Pi. <S> A single pair of resistors can accomplish both requirements, say 1.8K and 3.3K. simulate this circuit – Schematic created using CircuitLab <S> Test it to make sure it responds fast enough, some chargers can take close to 1 second to start up and there will also be a noticeable turn off time. <S> Because the turn-off is "soft" you should apply software debouncing if you're looking for edges. <S> They are multiple-sourced and made to be safe and reliable. <A> If you know how to make compute an impedance voltage transformer with caps and resistors like a scope probe, you can divide 100:1 and rectify with some slow rise or fast risetime specs and threshold to logic and shunt Neutral to 0V with another cap. <S> A suitable detection window for time and frequency must be specified by you for noise immunity, depending on your noise environment. <S> If you define your input , output and range / tolerances & thresholds, then a solution can easily be defined. <S> Vin min(on), Vin max(off), Logic “0” Vil(max), Logic “1” Vih(min,max) <S> , Time detect, min, max) <S> Noise reject at 1kHz to 1MHz= __ dB, <S> ( nearest interference <S> dI/dt, dV/dt and distance to input.) <A> If your neutral is (or can safely be) connected to GND, you can use a diode, a capacitor and a voltage divider. <A> Since you haven't specified a response time I guess that it is not important for your application <S> so 5 V USB power supplies which are available for a couple of €/$/£ would be suitable. <S> You will find that they take maybe 100 ms or so to turn on and the turn-off time will depend on how long it takes to discharge the internal capacitors. <S> You can speed this up by adding a load such as a resistor and LED. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> PSU with indicator / discharge LED. <A> I was looking at a mechanical relays internals and had an epiphany - relays are just electromagnetic versions of an opto isolator. <S> So I started looking for relays <S> and I found a relay that has 120vac control and a 28vdc contact. <S> By the answer provided in my other question , I found it was feasible to use the relay for my purpose. <S> I feel this is the simplest way to do this. <S> If anyone has any better ideas, please feel free to add another answer because I'm always open to suggestions.
| Rectify the AC to DC (keep in mind that this will be peak voltage!), buffer in the capacitor and pull down to the voltage you want. Alternatively, you could use a standard "AC input" module made for industrial use. The difficulty there is that when you do that, your circuit ground will no longer be floating, which brings you into "you need a suitable enclosure" territory.
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Crowbar circuit causes unexpected behavior for op amp circuit I am using an op amp to amplify an input signal from a microcontroller, which in general is working fine. For over voltage protection, I added the crowbar circuit taken directly from figure 32, page 27 of the TL431 data sheet and that added some undesirable behavior to the circuit I don't quite understand. With the TL431 triggering at a voltage of 2.5 V and the voltage divider \$R_3\$ / \$R_4\$ the crowbar should trigger at an op amp output voltage of 4.8 V and blow the fuse. But what I see is, that as soon as the output voltage reaches 3 V, the output drops to 0.75 V and stays at that level until the input voltage drops far enough, that the output should be below 0.75 V in normal operation. After that, it works as expected again, until 3 V or more output is reached. I found in this discussion of this crowbar circuit , that the placement and size of the capacitor as depicted in the data sheet might not be ideal. Could that somehow cause my problem? If not, what else might be responsible for this behavior? EDIT: For proper context for the added crowbar, I regulate the power of a laser with the op amp output. I have to make sure that the laser is not permanently turned on by a short circuit of the output to the 5V that is used as +Vcc for the op amp and for other parts on the pcb. Since I don't need more than 4.2V output and shouldn't get more that that during regular operation, blowing a fuse with the crowbar was the best I could come up with to protect against this case. Datasheets: Fuse: https://www.mouser.de/datasheet/2/358/typ_MGA-A-1388649.pdf Op amp: https://www.mouser.de/datasheet/2/609/AD8605_8606_8608-877839.pdf Triac: http://www.ween-semi.com/sites/default/files/2018-11/BT137S-600D.pdf simulate this circuit – Schematic created using CircuitLab Update: Removing C1 completely does not eliminate the described behavior, but increases the voltage at which it occurs to 3.3V <Q> The opamp does not provide sufficient current to blow the fuse. <S> The fuse is rated for 200mA (for the lowest current rated fuse in the family), the opamp can only supply 80mA (if running at 5V at 2.7V it's only 30mA) , or less then half the current to blow the fuse. <S> Lets suppose that ground was attached to the other end of the fuse, instead of the crowbar circuit, only 80mA would flow through the fuse, and it still would not blow, even if you raised the voltage as high as the output of the AD8605 would allow for it's rating (6V). <S> Crowbar circuits are for voltage circuits that have low source impedances/high currents like a power supply. <S> EDIT: <S> There are a few options, one way would be to limit the opamp's output by changing Vcc of the opamp to 3.4V. <S> The other method would be to use a zener diode on the input, however this would sacrifice some linearity and the load impedance as seen from the Vin. <S> The resistor could be raised to a higher value, but would also change the slope of the limiting curve and make the upper range of the Vin/Vout curve where the diode clamps inaccurate/less linear. <S> This is not a good option, I prefer clamping the output or limiting the Vcc ( which would probably be the simplest and only add a regulator to the circuit). <S> The last option would be to use a series resistor and diodes on the output, also with some loss due to the series resistor and the leakage current from the diode. <S> ANOTHER EDIT: <S> If current limiting is what you want there are plenty of circuits that can accomplish this task. <S> (there are also many IC's suited to this task). <S> Most involve detecting the current with a current sense amplifier as the one shown below (the amplifier IC1 switches the optocoupler which in turn switchs a pmos high side switch): <S> Source: <S> https://www.electronicdesign.com/power/current-limiter-offers-circuit-protection-low-voltage-drop <S> OR many circuits that are listed here <A> In the circuit as you show it you require no C1. <S> As pointed out in the prior discussion, the capacitor may turn on the Triac on sudden rises in opamp output. <S> The TL431 is not really suitable for what you are trying to do since it requires a minimum Ik to set the reference (0.4mA). <S> However assuming you want to blow a fuse (and as already pointed out the fuse you selected is not suitable) <S> I'd suggest the following change may solve your problems: <S> simulate this circuit – <S> Schematic created using CircuitLab R3 ensures the TL431 internal ref is always adequately serviced and is not dependent on the signal level. <S> M2 shorts the output of the opamp ….but <S> here things are hazy. <S> The opamp is only capable of 80mA, so I assume you are trying to blow the fuse when the opamp is dead (and the current is uncontrolled). <S> However if the opamp is ok and the signal just too high, then this circuit would clamp the output sinking the 80ma without problems. <S> Getting a fuse to blow is hard work. <S> Update: <S> What is the reason you want to limit the output swing to 4.8V when the rail-rail operation limits it to 5V already? <S> Explain your needs more fully for a better hope of a viable answer. <S> Looking at the problem from a pure opamp perspective, is your specification as follows: <S> Output of opamp must NEVER go above 4.8V with a 5V supply <S> Input must be high impedance <S> The opamp is not broken (so the output current limits work) <S> Clamp the input rather than the output <S> This might be a viable approach to simply clamp the input signal: <S> simulate this circuit <S> The <S> TLV3011 provides a very accurate reference voltage and R4/5/6 <S> provides an adjustment for the output threshold. <A> I found the original reason for the unexpected behavior of the TL431. <S> It turned out that the Eagle library I was using had the SOT23 package for the TL432 under the TL431. <S> Since the two have switched cathode and reference pins, my circuit didn't work properly with the TL431. <S> The original circuit still had some instabilities, which is why I accepted Jack Creasy's answer, since his alternative circuit works very well.
| The strange conduction you are seeing is in all probability due to the impact of the internal reference generator.
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How Can I connect My Solar Power System? I brought this 50A Solar Controller now I want to connect an inverter to it and they said to connect the inverter directly to the battery. I also brought a 100W Solar Panel and MC4 Wires, etc What is these "60A/50A/40A/30A/20A/10A "? I choose the 50A one and wanted to know the Amps are used for DC Load or what's the use of selecting "60A/50A/40A/30A/20A/10A" if it is to be connected to the battery eventually? The reason for not wanting to connect it to the battery is: The Protections (Load, Over Charging, Deep Charging, etc) go away if I connect the inverter to the battery directly. <Q> It is designed to control the charging of the battery bank only, not to control the inverter / battery connection. <A> There's a diagram on the page you link showing that the inverter should be connected to the battery direct. <S> These controllers have a DC out that is intended to allow it to disconnect the loads at a particular terminal voltage, so that the battery doesn't over-discharge, but in this case the capacity of that output is limited (it's probably done with a MOSFET), and the inverter can potentially take more than that - unless it's a only a small one. <S> Given that limitation, you need to check that the inverter has its own undervoltage trip if it's going to be left unattended, or you risk pulling the battery state of charge low enough that it'd be permanently damaged. <A> The "60A/50A/40A/30A/20A/10A" is the maximum amperage the controller can handle. <S> You would want to size the controller based on the total amperage output of your solar panels. <S> You have a 100 watt solar panel and if you hook it up to a 12 volt battery: 100watts <S> / 12 volts = <S> 8.33 amps, you would need a 10A solar controller. <S> If you had a 1200 watt solar array and were going to use a 24 volt battery: <S> 1200W <S> / 24V = 50A, you would need a 50A controller. <S> Hope this helps..... <S> Good luck <A> If you want whatever battery protection the charge controller provides, feel free to attach it to the DC Load terminals of the charge controller. <S> That is what it is for . <S> The inverter is a DC load. <S> However make sure the charge controller can handle the DC ampacity the inverter can draw. <S> If the inverter instructions are saying to hook it up to the battery, those may be generic instructions because they don't know your system.
| The solar controller you bought goes between the solar panels and the battery bank.
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Am I testing diodes properly? I decided to test some diodes that I have, using a multimeter to measure their forward bias voltage. I set my multimeter to voltage test and connected the diode across: I tested a bunch of different diodes, both germanium and silicone type, so I expected to see voltages around 0.2 - 0.7V. However, all diodes showed 0V! I tried reversing the polarity of the diodes, as well switching multimeters. Always my measurement came out as 0V. So am I making some kind of mistake in measuring diodes, or are both of my multimeters/all diodes broken? <Q> In voltage mode, a multimeter just measures what voltage is present between its leads. <S> On your meter, it's the option one to the left of your voltage mode--set the dial to that and press the mode button a few times to put it in diode mode; it'll say on the LCD. <S> In diode mode, the meter applies a known current to the diode -- you can check the meter's datasheet or instruction manual to know what current it uses, and fancier meters might even let you select a current -- and then measures the voltage across the diode. <S> Note that this usually won't work for LEDs, as most meters limit their diode test voltage to only one or two volts, which is too low to turn on any LEDs except maybe some red or yellow ones. <S> But for a conventional diode like the one you show in the picture it will work fine. <A> The guys in forensics had a difficult job enhancing the dodgy photos but the problem is clear. <S> The meter is set to measure DC. <S> Volts. <S> (Barely visible in this rendering) <S> the range selected is 'V'. <S> The correct switch setting. <S> Since you selected DC V and a diode doesn't generate any voltage <S> the reading is zero. <S> This is correct. <A> Nope. <S> You need to set your multimeter to diode test mode. <S> One click anticlockwise, and then press the "mode" button until the LCD shows a diode symbol. <A> Perhaps you are being misled by the fact that circuit simulators model a diode as a voltage source in series with a resistor. <S> That is a fiction that is required to make the circuit simulate correctly. <S> A Real diode does not produce a voltage across itself, but will produce a voltage drop when you pass current through it in the forward direction.
| What you want is diode test mode, which is usually indicated on the dial with a diode symbol.
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What connectors should I use with 5V DC current? In a personal project, I need to power several 5V appliances which, together, use up to 12A. I want to use a 60 inch cable between a case to which those appliances are connected, and the transformer with 230V AC in and 5V DC max. 30A out. When I look at the specification for the cables and the connectors, they are all rated pretty low in terms of maximum amperage allowed. For instance, standard C13/C14 connectors used for personal computers are rated 10A. I suppose that those ratings are likely to be for 250V AC. Is it safe to simply scale the ratings down to 5V, so 10A for 250V would mean a power of 2500W, and so a maximum of 500A at 5V? If not, what calculations should I use to determine the required AWG size of the cables and the type of connectors that I am allowed to use? <Q> These include the ring, spade/forked connectors that get screwed down. <S> The hardware store also has similarly crimped Quick-Connect connectors that have a male and female end that you can plug into each other wire-to-wire. <S> The female terminal is often just sitting inthe hardware store, however the male end is less common to find sitting in a hardware store, so you may have to order online. <S> The easiest thing is just to google an ampacity table and look up the AWG for current unless you want to do super-physics calculations and simulations for wire heating. <S> There's little point to making a connector that can fit a 12AWG wire but can't carry the current of a 12AWG wire. <S> Usually the limiting factor is the insulation or plastic on the connector or wire since that will melt/burn long before the metal conductor does. <S> Voltage and current ratings are independent. <S> You can't trade one for the other. <A> Is it safe to simply scale the ratings down to 5V, so 10A for 250V would mean a power of 2500W, and so a maximum of 500A at 5V? <S> No, one limiting factor in connectors is the contact and conductor resistance. <S> A ballpark number that I usually use is max 0.1Ω which works for most connectors. <S> Many connector datasheets also have numbers for connector resistance. <S> P <S> =I^2*R <S> If we have 30A, this means we will dissipate roughly 90W in a connector with 0.1Ω of resistance and 9W in a cable with 0.01Ω's of resistance. <S> Most connectors will fall in that range. <S> That is a lot of heat to dissipate in a small area. <S> The connector can have a temperature rise and actually melt with enough current. <S> If not, what calculations should I use to determine the required AWG size of the cables and the type of connectors that I am allowed to use? <S> Always use the ratings on the connectors, they are tested and also tested in a variety of conditions including temperature. <S> Dust\Dirt can also create resistance. <S> So use the ratings found in the datasheet. <S> If you must go down to low voltages (there is a reason we use high voltages) then parallel connectors. <A> Your best budget solution is to use 2 pairs of Molex 4 pin HDD contacts rated for 10A ea. <S> for your 5V distributed loads. <S> These are used in PC towers to interconnect 5V/12V peripheral power with the centre pair joined at source and load, as ground return. <S> Voltage drop on each wire is I*R <S> /m <S> * m length. <S> [ or ft] LV contacts are always for current as temp rise increases with \$I^2Rs= <S> Pd\$ for contact Rs. <S> Hot insertion is not recommended to reduce pitting of contact surfaces and rise in Rs. <S> Use the largest AWG that you can to minimize the voltage drop or according to your load error budget, based on Ohms/ft or /m Vdrop= <S> IR X2 for R length of cable. <S> If you care to use AWG 14 stranded or welding cable with jumpers to crimp or soldered pins that would be ideal, but if you can get away with xxx mV cable loss, lighter cable is O.K. Use AWG tables <S> Ohm/ft to compute drop and <S> (x) for each wire length x Amps to achieve less drop than your voltage error tolerance. <S> It’s not rocket science. <S> The contacts may be 5 to 30 mOhms max initially then may increase with aging abuse. <S> So again from Ohm’s <S> Law V=IR , each contact has a small,voltage drop.
| Crimped electrical terminals that you find at the hardware store are the simplest and most available high-current connectors that requires the cheapest and least amount of tools for most people. Rule of thumb: if the wire fits in to the connector then the connector can probably support at least as much current as the wire can. Use 1 connector pair for each peripheral and use wiki AWG tables for cable resistance per unit length of each cable.
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Can I use a current control buck converter to charge a battery? As shown below, could I use a microcontroller to charge the Li-ion battery? I'm unsure if I'll run into issues with the voltages on the ADC. I've done this exact setup before with a generic load, I'm unfamiliar with batteries though. simulate this circuit – Schematic created using CircuitLab <Q> Generally, yes, you can build a switch-mode power supply using a microcontroller. <S> It'll certainly be more work than just buying a dedicated controller IC, though, and your digital control scheme must take software reliability risks into account. <S> Here, these risks are increased: if your MCU software fails for some reason, or the MCU doesn't properly power on, there's a realistic chance the MOSFET will stay "on", and that'll quickly render the inductor effectively a short, delivering as much current as your 12V can source into the battery. <S> Lithium batteries, under these circumstances, will heat up, produce gas, bulge and tend to explode. <S> No fun. <S> Considering that, I'd very much recommend using dedicated third-party LiIon charge controllers that integrate all the switch driving (often even the switching mosfet), sensing and overtemperature and -current protection you need. <S> Your ADC just checks the voltage across the battery, not the current flowing into it. <S> Thus, this can't be a safe charger. <S> Again, a dedicated LiIon charger IC has an "understanding" of the state the Lithium cell is in. <S> It will charge at a limitable rate, and it will slow down and stop when the cell has reached a certain state. <S> Such chips are typically not very expensive. <A> For Li-ion batteries you need to guarantee three things: <S> don’t discharge below 2.8 <S> ~ 3.0V don’t charge above 4.2V don’t exceed the battery 1C charge rate <S> Your proposed solution really doesn’t address these items. <S> It could damage the battery, or worse, cause the cell to catch fire. <S> Please consider a purpose-made controller IC that’s designed for managing Li-ion safely. <S> A pretty good write-up here: https://www.digikey.com/en/articles/techzone/2016/sep/a-designer-guide-fast-lithium-ion-battery-charging <S> A good resource here: https://batteryuniversity.com/learn/article/charging_lithium_ion_batteries <A> Assuming this is a single cell Li-Ion due to the buck converter circuit topology <S> so you'd stop charging at a voltage within the ADC range (higher voltage would need accurate resistor divider). <S> Charging Li-Ion batteries is a two stage process and you can find lots of info on the charging particulars online. <S> Generally the first stage is Constant Current (CC) until the measured battery cell voltage hits the desired "fully charged voltage" <S> (depends on your exact battery chemistry) and then it switches over to Constant Voltage (CV) mode. <S> Therefore the current vs time profile is a flat line at your desired charging current <S> (manufacture should specify safe charge rate as a fraction of the battery capacity) and then the charging current will drop off and asymptotically approach zero <S> once you switch over to CV charging. <S> From Battery University charging lithium battery post: <S> So to answer your question, your circuit will work for the CV portion but you don't have a way to check what the charging current is into your battery. <S> Unfortunately there's no accurate way to calculate this from the circuit you already have because even if we know the source voltage (Vs) and the voltage drop over the FET (Vfet) <S> we could calculate the voltage over the inductor using your ADC input, but the integral to calculate current would quickly diverge and you'd lose control of your CC charging. <S> \begin{equation}I = \frac{1}{L} <S> \int_0^T V \it{dt} + I_0\end{equation} <S> I'd recommend using a shunt resistor directly after your inductor with current sensor amplifier (should be details elsewhere in this forum) to increase the measured voltage to something you can more accurately detect with a separate ADC pin. <S> It'll be a variation of this sort of setup. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Be aware that you may want an external entity to verify the battery never goes above <S> it's max voltage in case the FET fails in conduction mode or the microcontroller freezes.
| That being said: No, this is not an appropriate LiIon charger circuit; Lithium batteries want to be charged by a controlled current, not using a controlled voltage (the voltage is usually used to check the charging state).
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Where does return current flow for a differential signal? As far as I understand, return current flows the "negative" channel for a differential signal connection. Only if it is not balanced well, there might be unwanted return current in reference GND or power plane. But some say the return path for both positive and negative are always the reference plane, which get me confused... <Q> It depends on the physical geometry of the transmission line. <S> I can route two tracks on a PCB, not too close together but matched in length, and use them as a differential pair. <S> In this case, each of the individual lines will have its own return current on a nearby ground plane. <S> Or I can make a closely coupled pair of tracks on the PCB, with the line-to-line spacing less than the line-to-ground spacing. <S> In this case most of the return current will be in the complementary line, but there will likely be some return current in the ground plane near each line. <S> Or I can make a purely differential line, like unshielded twisted pair, and pretty much the only return current for one line will be in the other and vice versa. <A> I'll take a slightly different approach to the other answers and use a differential transformer as an example. <S> These are common on balanced microphone circuits. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> A balanced microphone circuit showing differential (green) and common mode (red) signals. <S> Note that the differential signals are in-phase on the transformer primary and so generate an output on the transformer secondary. <S> Meanwhile the common-mode signals are out of phase and cancel out giving no output on the secondary. <S> The transformer rejects the common mode signal. <S> As far as I understand, return current flows the "negative" channel for a differential signal connection. <S> This depends purely on which you take as the reference. <S> Only if it is not balanced well, there might be unwanted return current in reference GND or power plane. <S> That would be true. <S> In Figure 1 there is little possibility of this as the microphone capsule is floating with respect to ground. <S> In practice there will be some capacitance to ground and in other applications it may be using a differential amplifier output with a direct ground reference. <S> If, say, there are an uneven number of turns on the two halves of the transformer primary <S> then the common mode rejection will be reduced and the CMRR will decrease. <S> But some say the return path for both positive and negative are always the reference plane, which get me confused... <S> If you think about this a bit, what happens if two out of phase signals are transmitted through the same path? <S> They cancel out. <S> Only the difference between the two signals will go through the ground path. <A> A properly balanced differential line has equal amounts of current flowing in both directions. <S> A properly balanced differential transmission line does not need a reference plane (there are many transmission lines that don't have reference planes,anything that uses UTP or STP being some examples). <S> Of course if the lines are not balanced, current will flow along the reference plane (or sheild) <S> but this will be minimal, because an unbalanced differential pair not longer functions like a differential pair. <S> The reference plane does change the characteristic impedance of a differential pair so reference planes must be taken into account when routing. <A> Any imbalance in parasitic capacitance will limit the CMRR for AC signals. <S> The return currents will flow thru whatever bits of metal form the 2nd plate of theparasitic capacitor. <S> Its your job to identify all the parasitic Efields, and bring those fields under your control.
| If you start with the negative or inverting input then the "return current" is on the "positive" or non-inverting input.
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Digital electronics - how does a circuit determine when some element (like adder) is done? (delay handling) Take for example a ripple carry adder, how does a circuit handle delays of some elements?If ripple carry adder starts working from some clock cycle, we can't expect that it will have correct results on the next clock cycle. <Q> The circuit doesn't, the designer has to do. <S> Maximum delays are typically fixed, you have to plan the circuit appropriately, which potentially means that you design it to wait one clock cycle. <S> If needed, the designer can route out a "ready/conversion done etc" signal. <A> You imply in your question that delays are being "introduced" or "implemented". <S> They are neither implemented nor introduced Delays happen because gates take time to pass signals. <S> They are intrinsic . <S> If you have a long adder with a lot of gates, it will have a long delay, and you have two options: 1) <S> Lower the clockspeed so that each cycle lasts longer than the delay. <S> 2) Implement a multi-cycle adder where part of the addition happens on the first cycle, then part on the second, etc. <S> Then set your clockspeed so that the slowest stage of your adder has less delay than the duration of your clock cycle. <S> This can be done using common logic elements such as D flip flops that take a value on clock cycle transitions. <A> No. <S> The ripple effect takes as long as it takes. <S> It is up to the designer to make sure that the result is stable when it is used. <S> On purpose I don't write <S> "the next clock edge" as the designer can invoke a so called multi-cycle path. <S> In that case the result is not used until two or three clock cycles later. <S> If this is supposed to be the next clock edge, the designer must make sure the clock frequency is low enough. <S> How are delays introduced?Are counters used for that purpose? <S> Sometimes but in general you try to avoid adding delays. <S> From what I gather from your questions is that you are looking for the answer how to do what we call in the profession timing closure . <S> Timing closure <S> is a very complex process which involves various solutions depending on if you have set-up, hold time or maximum frequency problems. <S> It would take many pages and is thus beyond the scope of stack exchange to tell you even the beginning principles and techniques, but I hope with that as search term you learn a lot from the internet. <S> Just for completeness: delays are only added when a register has hold-time problems but has slack in its set-up time. <A> The designer must ensure that the operational speed of the function can meet the overall requirements. <S> Take your ripple counter example - an asynchronous design. <S> If I were to implement such a circuit using separate flip flops such as this one , I would need to know the worst case propagation delay which is 14 nsec. <S> In addition, if using D flip flops (where the D input is fed from the #Q output), the setup time would also need to be evaluated which in this case is 7 nsec. <S> So the maximum rate at which such a counter could be operated at for a single device is 47MHz for guaranteed operation (provided the maximum toggle rate of the device is rated for it, which in this case is true). <S> For more bits, all the delays need to be analysed if there must be a particular duration where the output result is correct. <A> Delays, rise times and setup times are defined by design due to CMOS voltages, RdsOn, Ciss, Ciss etc and given in no times for each discrete part. <S> You can choose a multiphase clock to decide when the output is guaranteed to be settled to improve clock rates by a state diagram pipeline so you get what you compute for worst case delays to guarantee the logic is valid within these delays for the next clock edge and state values. <S> All FET including those used in CMOS have a 50% production tolerance for Vt = <S> Vgs(th) and a temperature with Vdd <S> (=Vgs) dependency for all these delay times.
| Ensuring various types of delay are properly evaluated so a circuit operates properly is the responsibility of the designer of the circuit.
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How to monitor switch state? Background I am building a simple web application using raspberry as a server. I am using this raspberry also to control an arduino. A user can send commands through the browser to the raspberry, and it can control some arduino functions! What I want I want to use an arduino to turn a lamp on and off, parallel to a normal manual switch that I can use to turn on the lamp on and off, like this: 1: If the lamp is off and the switch is also on OFF position (open switch), I can turn the lamp on by using the switch, putting it on the ON position (closing the switch). Then, the monitoring system is gonna tell arduino (and the web server) that the switch is on/closed and the light is on. 2: I can use arduino to turn off the light, without necessarily puting the switch on the OFF position. If the light is OFF (because I turned it off using an arduino) and the switch is on the ON position (closed), I should need to switch it to OFF position and then to ON position again, in a manner that I can sense these changes and send it to arduino (and consequently, to the web server). 3 (Possible improvement?): Whenever I change the state of the switch, the light would change its state to whichever is the opposite of the current one: if the light is on, it goes off and vice-versa, regardless of the switch state. The problem How can I actually know if the switch is on and off? There is anyway to use a sensor, some electronic circuit or component that can if tell the switch is on or off? I was thinking about using a relay to control the light, depending of the state of the light that is stored on arduino/web-server. Is this a reliable way (I mean, using a relay) to do that? I appreciate any possible help on this, I have been interested on IOT and I want to do some simple applications for home automation; this is my first one! Thanks in advance for any help, again. <Q> This is just like having two switches that control the lights from either end of a hallway, except that one of them is controlled by your MCU. <S> Sensing the state of the light is a separate matter. <S> There are many circuits on this site and others for safely detecting the presence of line voltage at the light socket. <A> If you're not comfortable working with 120VAC (could be disastrous if you don't know what you're doing) <S> then a mechanical solution is much safer. <S> You're going to hate this, but for proof of concept you could hot glue <S> a limit switch onto the faceplate underneath your light switch like the picture below (painting it white might make it more tolerable). <S> I'd use a pulldown resistor to give you an active HIGH signal which is usually more intuitive. <S> Check this switch tutorial for guidance. <S> For a more advanced solution, I'd reduce and rectify the voltage across the switch to see if there's the same thing on both sides of the switch (CLOSED) or different things (OPEN). <S> You'd ideally have a small isolation step-down transformer between any mains voltage and your sensitive microcontroller. <S> For the sake of a cheap solution though, you can use "X rated" capacitors for isolation. <S> Either solution will need to be rectified to DC to be readable by your microcontroller operating between 0 and 5 volts. <S> simulate this circuit – <S> Schematic created using CircuitLab Beautiful output I'd say, and it only sips about 25mW according to the simulator <S> (blue is the isolated, rectified voltage and orange is the zener regulated voltage). <S> Read that output with one of your microcontroller inputs and you've got an indicator! <S> (FYI 0V means no difference across switch = CLOSED) <A> As you would need to control mains anyway, you can connect only relay and lights in series. <S> Existing switch you wire to GPIO of Arduino. <S> This way Arduino know state of switch and can control lights without extra connections and sensors. <S> There is a problem that your Arduino have to function correctly to control lights even if you only want to use physical button. <S> So this solution is only appropriate if you can live with inability to switch state of lights until you repair Arduino. <S> Regarding reliability. <S> I have exactly same solution as in this answer <S> and I have experience some minor problems with relay not disconnecting (contacts welded), so I would advise to buy quality relay and not some unknown part from ebay even if its "parameters" looks OK.
| To get the functionality you describe, you need to have "3-way" (SPDT) switches — a manual one and a relay that the Arduino controls.
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What is a near IR blackout material for Lidar testing? This question seems not a best best question for this place but still related. I am doing a project regarding to lidar that involves in testing lidar behaviour when beams get absorbed.So, what is a good off the shelf material that is good on absorbing 850nm-1000nm bandwidth? <Q> which is very black from UV down to >10um wavelength. <S> From the manufacturer's website : <A> You can try thermographic paint. <S> It's made to make the surface non-reflective at IR wavelengths. <S> It's needed to let IR radiate effectively for temperature measurement, but as well it can absorb external radiation. <S> I haven't tested it. <A> IMO you'd be better to find a planar material that is reflective to IR so that you reflect the beam away from the target and away from Rx. <S> For example here is the response of ZnSe as an absorption example: There were some great Lidar shots from Google <S> (I can't find them) that showed just the effect you want. <S> The had a bus in view and the windows were black (no return at all) due to the IR reflectance of the glass. <S> Yet the camera shot showed all the people in the bus. <S> It was a great example. <S> So almost any heat reflectig glass may do a reasonable job.
| You could use Vantablack
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FPGA CPUs, how to find the max speed? I'm just getting into FPGAs, and if I understand correctly, you are connecting logic gates together using code. So if I design a CPU in Verilog, it should connect some logic gates together and work, but how do I know how fast my DIY CPU can run? What does it depend on? <Q> The speed of a design is limited by several things. <S> The biggest will most likely be the propagation delay through the combinatorial logic in your design, called the critical path . <S> On a lower end FPGA, for example a Spartan 6, a reasonable figure is probably more like 250 MHz. <S> This requires pipelining everywhere <S> so you have the absolute minimum amount of combinatorial logic between stateful components (minimize levels of logic), low fan-outs (minimize loading on logic elements), and no congested rats-nests (efficient routing paths). <S> The fabric logic of different FPGAs will have different timing parameters. <S> Faster, more expensive FPGAs will have smaller delays and as a result can achieve higher clock frequencies with the same design, or run a more complex design or design with less pipelining at the same frequency. <S> Performance within a particular process can be similar - for example, Kintex Ultrascale and Virtex Ultrascale are made on the same process and have similar cell and routing delays. <S> It is impossible to say how fast a given design will be without running it through the tool chain and looking at the timing reports from the static timing analysis. <S> When doing toolchain runs to determine maximum clock speed, bear in mind that the tools are timing-driven: they will try to meet the specified timing constraints. <S> If no timing constraints are specified, the result can be very poor as the tools will not try to optimize the design for speed. <S> Generally, the tools will have to be run several times with different clock period constraints to find what the max achievable clock frequency. <S> If you can optimize your design so that the critical path is not the limit, then you'll run in to limitations in the clock generation and distribution (PLLs, DCMs, clock buffers, and global clock nets). <S> These limits can be found in part datasheets, but getting near them with a non-trivial design is difficult. <S> I have run stuff on a Virtex Ultrascale at 500 MHz, but this was only a handful of counters to provide triggering signals to other components. <A> You synthesize your design in the target technology (a particular FPGA) and let the static timing analysis tools tell you what the minimum clock period is. <S> Or, you add constraints to the design in the first place, and then the tools will let you know whether they're met or not. <A> The flop-to-flop delay will include clock-to-Q, routing, logic/LUT, and flop setup time. <S> These added together form the critical path of your timing, which you can inspect in the timing report output by the place-and-route tool. <S> There are entire design disciplines devoted to making architectures that minimize this delay to get the most out of a given process - pipelining, parallel execution, speculative execution, and so forth. <S> It's a fascinating, involving task, wringing that last ounce of performance out of an FPGA (or for that matter, an ASIC.) <S> That said, FPGA vendors will give different speed grades for their parts, which correspond to a max MHz rate. <S> For example a -2 Xilinx Artix is a '250 MHz' part roughly speaking although it's capable of higher clock rates for highly-pipelined designs. <S> When you interact with the FPGA synthesis and place-and-route tools, you will need to give constraints for your design. <S> These tell the tool flow the target flop-to-flop delay you're trying to achieve. <S> In Quartus (Altera) and Vivado (Xilinx) these constraints use a syntax called SDC, which stands for Synopsys Design Constraints. <S> SDC came initially from the ASIC world and has been adopted by the FPGA industry as well. <S> Get to know SDC - it will help you get the results you want. <S> Altera and Xilinx have online communities for help with how to use SDC syntax and many other topics. <S> That all said, if you care about speed you should consider an FPGA that has a CPU hard macro in it, such as Zynq.
| The speed that your CPU will run will be based on your longest flop-to-flop delay in your synthesized design. If you use a fast FPGA and write your HDL very carefully, you could probably hit 700 MHz on something like a Virtex Ultrascale+.
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Understanding this comparator circuit for AC voltage protection The circuit shown below is taken from this whole design . I understand the main idea. They want to measure some voltage condition and trigger something else with the comparator. R39 and C38 form a low pass filter. R37 is the pull-up resistor as the comparator's output is open collector. D19 and D20 are protection diodes as the input goes positive and negative. What I don't understand is the arrangement of the resistors R36, R38, R35, and R40. Why are R36 and R38 referenced to ground in the midpoint? How is the comparison process achieved? According to the reference design, the input AC voltage is 230 VAC 50/60 Hz. <Q> Note that the whole shown circuit (so, +3.3V, +12V, ZC_IN, etc) should be considered live and thus dangerous). <S> The shown picture has cut off the name of the output that gives an important hint: ZC_IN , which stands for Zero Crossing in. <S> D19 and D20 clamp <S> the floating line voltage to a floating 0.7V volt, where R35 and R40 limit the current through D19 or D20. <S> I mean floating with respect to the circuit's ground. <S> R36 and R38 'align' <S> this floating 0.7V with respect to ground. <S> So, the comparator will see a sine wave which matches the line voltage at the zero crossing , but will be clamped between -0.35V and 0.35V. <S> This zero crossing is required to make a good PFC. <S> Below a picture of the (clamped) <S> voltages the comparator sees at its input terminals. <A> The design appears to assume that L and N can be swapped. <S> This is a common problem in countries with reversible mains plugs. <S> The circuit, therefore, needs to accommodate live on either pin. <S> Let's ignore D19, 20 and C37 for a now. <S> With the junction of R36 and R38 earthed one of the comparator inputs will be grounded. <S> Remember that the neutral is so called because one of the supply wires has been earthed or 'neutralised' at the supply transformer <S> so its voltage will be close to zero. <S> As drawn, 'AC_N' will be at zero volts and <S> since the top of R38 is at zero volts the junction between them will be at zero volts. <S> The comparator '+' input will now also be at zero volts. <S> Any time AC_L goes negative the comparator output will switch high. <S> The result is a squarewave which goes high when AC_L is negative and low when positive. <S> Reversing the input polarity inverts the result. <S> It causes the voltage on IN+ to rise above zero somewhat. <S> This is probably a good thing as it gets the inputs away from operating at 0 V (although the chosen device will work down to zero on the inputs.) <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Equivalent circuit while AC_L is positive. <S> You can work out what happens on the negative half cycle or when the live and neutral are reversed. <A> The depicted ground is not a chassis ground, it is a floating ground. <S> R36 and R38 are bleeding resistors and they set this floating ground potential at mean voltage between N and L. <S> The chassis ground, which is also conencted to PE wire is depicted differently:
| D19 does two things when AC_L goes positive: It limits the voltage on IN- to one diode drop above that on IN+.
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Identification of an AC transformer I have a broken Philips alarm clock FM/AM radio broken, so I stripped some parts out of it. One of it seems like an AC converter, but I cannot find any reference/datasheet about it. Do you have any information about this? (or what the meaning of the yellow/green wires are, I assume the black is GND and Red is VCC). The dimensions of the yellow 'block' are 35.3mm x 29.7mm x 11.6mm. The text is: SIL35T00000953004 So all wires going out of this transformer and I would like to know the meaning of the yellow and green wires: Blue wire (on the top back), going into the black cable you can see on the right page, it's going to the AC mains power plug (220V). Brown wire (on the top back), going into the black cable you can see on the right page, it's going to the AC mains power plug (220V). Green wire: What is the meaning of this wire? Yellow wire: What is the meaning of this wire? Red wire: I assume this is output? (AC/DC?) Black wire: I assume this is ground <Q> Usually Yellow\Green lines are for AC mains (according to international standards, but it also depends on the age of the device). <S> Your best bet is to hook up a signal generator, and see what the step down ratio is if you want to use it. <S> The other thing that will be of use is any markings on the outside of the clock that specify the current and voltage (you need the current) because that will give you a good idea of the saturation point of the transformer and how much power can be run through it. <S> EDIT: <S> I didn't see the blue black line in the picture which is not clearly indicated:The primary is most likely blue brown and there are two secondaries, in this case red/black are one secondary and the other secondary is yellow/green Source: https://airlinktransformers.com/post/chassis-mounting-toroidal-transformers-technical-notes <A> It's a transformer. <S> It outputs lower-voltage AC on the secondary wires. <S> As you surmised, the red/yellow/green/black are secondary windings and the blue/brown are primary going to the mains voltage. <S> You have that figured out so far. <S> Now the next step is to figure out how the secondary windings are connected and what their outputs are to see if they're useful (they very probably are.) <S> The 'safe' way <S> : Use an ohmmeter to determine which coils connect to which, and how they relate to each other. <S> Determine this <S> by seeing the relative series resistance between each wire. <S> You may find that there are two separate coils, or that they're a single coil with multiple taps. <S> More about this in a bit. <S> The 'fun' way: test it. <S> Connect AC up to the primary and use the voltmeter to measure the secondary AC voltages, wire-to-wire, to figure this out. <S> Do this with care obviously, though the secondaries are low enough voltage that you're not at significant risk for shock. <S> "More about this". <S> If this unit came from a VF (vacuum fluorescent) type clock, my guess is that it has two separate windings: one for 3VAC for VF tube filament, and the other 7~9VAC for logic power which would have fed to a full-wave bridge on the board (maybe you can check this if you still have the board.) <S> If I had to wager further, the logic power pair would be the black and red pair. <A> You probably won't find a datasheet as the transformer itself quite likely is not a consumer part but a Philips internal design, tailered to this specific product (alarm clock FM/AM radio) and to mass production. <S> They're not that expensive...
| It's an AC/AC transformer, the rectifier was probably on the red/black lines. I wouldn't bother investigating it, but just buy a transformer that fits your need, made by a reliable manufacturer with a well defined datasheet.
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Would a 9V alkaline power a 12V DC motor? I am thinking about buying a 12V DC motor but want to keep everything, including the motor, battery, and power switch somewhat compact. I am not very knowledgeable about these kind of things and am curious whether or not my 12V DC motor would run at all on a 9V battery. This is the motor <Q> Not for long! <S> (And the speed would decrease as the battery wears out and the voltage falls) <S> 9V batteries are pretty "weak" because the other half of the story is how much current they can supply at that voltage <S> (Power = Voltage * Current). <S> If you want to spin a big motor, the battery voltage will sag because of the battery's internal resistance. <S> The lower the supply voltage, the slower your motor will spin. <S> Ultimately the battery and motor will reach an equilibrium speed for the load you have on it. <S> Check out this page to see how the 9V batteries hold up under load (hint not well). <S> For reference, Adafruit's 130 size hobby motors consume only 70mA with no load (battery will last ~5 <S> hours but won't be very useful) and up to 500mA when stalled <S> (worst case scenario and if the motor doesn't burn up, the battery will last ~20 minutes). <S> As you can see, the alkaline battery chemistry is not designed for rapid discharge. <S> A NiMH, PbA or LiPo battery would be a better bet if you need more power (and you could recharge the battery for more fun in the future!) <A> If you reduce the voltage to 9 volts, the speed will be reduced proportionally. <S> The torque capability will not be reduced. <S> If the battery is too small, it will be discharged quickly and the internal resistance may reduce the voltage significantly below the nominal battery voltage. <S> The rated motor current is 1.1 amps and the rated torque is 14 kg-cm. <S> That means that the motor will require 1.1 amps to drive a load that require 14 kg-cm to turn. <S> The motor should be capable of operating at the rated speed and torque continuously without failure for the normal lifetime. <S> Unfortunately, the rated speed of the motor is not given. <S> It is likely considerably less than the 30 RPM no-load speed. <S> A small motor with a gear reducer that provides a lot of speed reduction is not going to be very efficient. <S> A lot of power will be lost in the winding resistance and in the gear friction. <S> A search of battery specifications should allow estimation of motor operating time and the minimum battery size. <S> Battery life is given in ampere hours or milliamp hours. <S> See also: Calculate battery life <A> According to your link the motor specs are: <S> Rated Voltage: <S> DC 12V <S> Reduction Ratio: 1: 172 <S> No-Load Speed: <S> 30RPM <S> Rated Torque: <S> 14Kg.cm <S> Rated Current: 1.1Amp <S> With a permanent magnet brushed DC motor, speed is proportional to voltage and torque is proportional to current. <S> Your motor is specified to produce 30rpm at 12V with no load, and 14kg.cm (presumably) at 1.1A. No-load current draw and rpm at rated torque are not specified. <S> A typical PP9 Alkaline battery produces ~7V at 1A when fresh, dropping to 5~6V after a few minutes. <S> At 1.1A you might only get 4 minutes run time before the battery goes flat, at about half normal speed due to the lower voltage. <S> If the torque load is less the motor will draw less current, so you should get longer run time with a light load. <S> However without a no-load current spec we have no way of knowing what the maximum run time could be. <S> We also don't know how accurate the specs are. <S> The only way to find out what it can really do may be to test it. <S> Discharge tests of 9 Volt transistor radio style batteries <A> In principle, it's probably possible to run a 12V motor on 9V -- it'll just run slower and with much less torque. <S> But if by "a 9V battery" you mean a typical PP3 size battery (i.e, the rectangular ones with a snap connector on top), the answer is almost certainly "no".
| These batteries are designed for low-current applications like handheld radios, remote controls, or smoke detectors; they cannot supply much current, and will be unable to drive a motor for long, especially a motor that's designed for a higher voltage anyway.
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Do I need a 50/60Hz notch filter for battery powered devices? Given that battery powered devices are not plugged in the mains power, can I assume that there will be no 50/60 Hz frequency noise? This question came up while thinking about what the circuit for a portable ECG monitor would look like and it seemed obvious that the 50 Hz notch filter could be eliminated. But is it so? Thank you <Q> An ECG circuit picks up 60Hz "common mode" signal from both probes then inverts it and feeds it to the patient's foot to cancel it. <S> The heart does not produce a common mode signal so its output is not affected by the common mode cancellation. <S> Here is the circuit: (Image source: Analog Devices AD620A datasheet (old version) ) <A> Assuming you're talking about the input signal, you may still want the filter because the long leads running to the chest pads (or wherever your detection location is) could pick up a 50/60Hz signal like a transformer's secondary from a nearby loaded mains cable. <S> If you're only referring to the power supply, you're correct, a battery produces very stable DC voltage, any instantaneous fluctuation in battery voltage would predominantly come from sudden loading/unloading of the battery. <S> Therefore no filter would be needed. <A> If you are talking about power inputs. <S> There will be no ripples in a battery powered system to be filtered. <S> But a parallel capacitor can be useful for reducing instant voltage change during power off/on. <A> Unlike AC, where − and + do alternate between poles. <S> Depending on where you live, the mains will be 50 Hz or 60 Hz. <S> The UK is <S> 50 Hz and the USA is 60 Hz for example. <S> So, you don't need a 50 Hz filter.
| Batteries are DC, the − and + poles do not alternate, therefore there is no frequency in batteries. You don't need to use a filter.
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What is this small circuit board used for in a 12v light? Repairing a home-made disco light that was used in a nightclub during the 80's/90's and it consisted of 16 12v 50w lamps arranged on 8 legs. I have discovered a small circuit board within each lamp which has an LED on the back. What was this small board used for and can it be removed? The lights were powered in serial, so would this board of allowed for the rest of the lights to be powered if one light blew? <Q> As it is in parallel with the lamp, it can be removed. <A> Others have noted that this is a blown-bulb detector. <S> When a bulb blows you get AC mains across the resistor-diode-LED combination. <S> The 15K resistor limits current to about 20 mA peak hald cycle DC on 230 VAC. <S> LEDS: <S> With incandescent bulbs, WHEN the light blows you get AC mains across [(the diode resistor LED combination) + (all other bulbs in series)] . <S> The other bulbs more or less look like resistors of 10% or maybe less of on resistance so for 50W 12V lamps <S> Ron ~= <S> 0.25 <S> Ohms <S> so Roff is maybe 2-10 Ohm range. <S> SO the series 15k dominates. <S> | With LEDS Voltage <S> MUST BE DC. <S> A full bridge is in order. <S> If you run this on AC with LEDs you will probably destroy ALL of them nearly instantaneously. <S> LEDs do not like high reverse voltage and draw little reverse current. <S> You may be lucky. <S> You may not. <S> Use a bridge rectifier. <S> You MAY get 50 Hz flicker. <S> Better still, see below, abandon this method. <S> LEDS <S> MUST be 12V (or more with DC) if 16 used, or the number must be adjusted to run OK on mains if voltage per LED is other than 12V, as you have mains voltage across the string. <S> They do not have to be 50W as long as all are of equal power and voltage. <S> With a blown LED you get 15 x LEDs in series with the blow-detect-circuit. <S> It will PROBABLY still work. <S> I'd consider testing it with a variac and turning it up slowly to see if there are any unexpected gotchas - but, probably OK. <S> This is a nasty method of operation. <S> It was done to allow 800 Watts of lamps to be run off mains directly with no transformer. <S> Each socket potentially has mains on it at all times. <S> V at a given bulb depends on its place in the string. <S> When a bulb blows ALL bulbs on the phase / <S> live side of the dead bulb are at full mains voltage. <S> It's lethal. <S> The same light (or more) can be obtained with perhaps 5% - 10% of the power with modern LEDs. <S> 12V supplies are not overly expensive. <A> it looks like it only lights the LED and is a place to join brown wires to blue wires. <S> if you don't need the LED you could find some other way to join the brown and blue wires.
| It looks like the LED will light up if the 12V lamp blows, so as all are in series and all lamps turn off if one blows, it is easier to locate the broken lamp by just looking at the LED. Abandoning this nasty kluge is a good idea - and you can easily have an arrangement where a single blown LED (which will be very rare) does not extinguish all the others.
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Removing output voltage ripple in high voltage outputs I am looking to reduce the voltage ripple on a DC/DC converter operating at 5kV. I have looked extensively on the internet and have found multiple ways to reduce output voltage ripple, an interesting reference is by EEVBlog which uses capacitance multipliers with BJT/MOSFET/Darlington transistors. Multiple other sources state to use multi-stage LC filters, and high value capacitors, etc. However, when the output voltage enters the kV range, the use of inductors is avoided as much as possible, if possible. Furthermore the size of capacitors with large values rated for kV values are very bulky and again, would be avoided if possible. This brings limitations in the amount of voltage ripple that can be attenuated in this range - which makes designing low output voltage ripple DC/DC converters in ripple sensitive applications (such as travelling wave tubes) very difficult. Splitting a transformer secondary into multiple windings reduces the ratings of inductors/capacitors but to what extent does this beat having a single inductor? Does anyone have any tips on how the voltage ripple can be minimzed in such cases? I am planning on using HV rated silicon carbide devices in the isolated power supply so possibly something that utilizes a MOSFET in the rectifier can reduce the ripple. <Q> You should consider higher frequency switching operations and a means to minimize filtering needs. <S> Look into interleaving options (running n converters shifted by 360/n electrical degrees for their PWM periods) as well - this will have a similar effect as a single converter running at n*f frequency and can provide a modular/scalable solution. <S> Rob B.'s suggestion was for a 2x interleaved system, but it's extendable to many stages/systems. <S> Depending on the application... an important point to consider for high-voltage supplies is their performance under arcing conditions. <S> You may particularly want to minimize filter energy for this purpose as much as size/cost/weight reasons. <A> There are 100 V 5 W Zener diodes on market. <S> You can construct a voltage limiter circuit. <S> All you need to do is building the limiter circuit with less voltage level. <S> For example consider a 5 kV DC with +250 V peak to peak ripples, a voltage limiter with 4.9 kV will kill all the ripples. <S> This solution is likely cost you more because larger diodes that on market is around 15$. <S> But it is a trivial solution. <S> Because this method will likely increase the power loss. <A> You are concerned about bulky capacitors but can trade a little loss for the benefit of lower ripple, so have you considered using a MOSFET gate follower after the RC filter? <S> A capacitance multiplier? <S> The capacitance at output will be amplifier gain <S> x C4.You'll need to have a steady load current and good heat dissipation, but you already said you wanted a MOSFET solution. <A> You did not state the power or current output nor the frequency so it is difficult to understand best approach. <S> However, one approach is to have 2 converter stages operating 180 degrees out of phase to help cancel the ripple. <S> If you can't use an inductor in your output how did generate the 5kV - without an inductor? <S> A common output filter is an LCL filter with a resistor in series with the capacitor. <S> The capacitors at that voltage are larger but you should not need a very large value. <S> The resistor is for damping. <S> Another approach is to consider a feed forward control.
| Another solution is dividing the output voltage and filtering them seperately with simple RC networks.
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How can I determine if this computer fan is dead? Desktop computer fan started making noise then stopped. We noticed it requires 12V and will run at 200mA.I wasn't sure about that at first. Does that mean it has an internal resistance of :12V / 200mA = 60 Ohms? If so that means I could supply it with 12V and it would consume 200mA and spin. Question : Is that what I should assume from reading those details on the fan? I decided to try it out with a bench power supply, but first I decided to measure how much resistance I could read. Check Fan Resistance When I hooked up my meter to read resistance I read 20 KOhms of resistance. Hmmm...?? Well, wasn't sure about that because if it were true then 12V would only supply it with only 0.6mA (600uA) and I knew that wouldn't work properly. Applied 12V I applied 12V to it figuring nothing lost here since it wasn't working. The the fan just barely moved then stopped. I'm assuming the fan is dead, but I have a couple of questions. What are the "normal" reasons these fans die?Dirt? What generally makes these fans die? There really aren't any parts touching anything to wear out are there? Is there any way to bring it back? (extreme cleaning of some sort? - blowing out dust or is it carbon build-up or something?) Does the high resistance (20 KOhms) indicate that the fan is dead or would a working fan read that amount also? One Last Interesting Point After I was done testing it I put it back in the desktop case and hooked the computer's power to it again and it started spinning slowly 20-30 rpm or so (guessing). Why would it do that? <Q> This sounds like a bearing is seizing or a winding has gone open circuit. <S> It needs replacing. <A> The fan is not a resistor, so you cannot use Ohm's law to find out its resistance. <S> This is brushless motor, so there is switching electronics between the power wires and the motor windings. <S> One very common cause of the fan failure is dirt in the bushing, and excessive noise is usually associated with this. <S> If you peel off that green sticker, there should be an access to the rotor's shaft with either plastic or metal snap ring on it. <S> Carefully remove the ring, pull out the rotor, clean everything, lubricate with a drop of machine oil and reassemble. <S> While you at it, the PCB inside should be visible with rotor off. <S> Inspect it for any signs of burned-out parts. <A> We noticed it requires 12 V and will run at 200 mA. <S> I wasn't sure about that at first. <S> Does that mean it has an internal resistance of : 12 V / 200 mA = 60 ohms? <S> No, <S> but when running it behaves like one. <S> A motor isn't a resistor. <S> If so that means I could supply it with 12 V <S> and it would consume 200 mA and spin. <S> They may have a "soft start" to prevent hogging the power supply on startup. <S> When I hooked up my meter to read resistance I read 20 k-ohms of resistance. <S> Hmmm...?? <S> Well, wasn't sure about that because if it were true then 12 V would only supply it with only 0.6 mA (600 uA) <S> and I knew that wouldn't work properly. <S> Again, a fan motor is not a resistor. <S> In this case there is an electronic circuit running a three-phase motor. <S> The electronics will present a high resistance until powered up and operating properly. <S> I applied 12 V to it figuring nothing lost here since it wasn't working. <S> The the fan just barely moved then stopped. <S> Is there any way to bring it back? <S> (extreme cleaning of some sort? - blowing out dust or is it carbon build-up or something?) <S> If you can find the fault and fix it then maybe. <S> Does the high resistance (20 KOhms) indicate that the fan is dead or would a working fan read that amount also? <S> No. <S> See above. <S> After I was done testing it <S> I put it back in the desktop case and hooked the computer's power to it again and <S> it started spinning slowly 20-30 rpm or so (guessing). <S> Why would it do that? <S> Maybe your bench PSU hadn't enough power to drive it when it is nearly stalled. <S> Check if the voltage is collapsing when powered on the bench. <S> The computer PSU may be more powerful.
| It may draw a higher current when starting but that depends on the internal electronics. It could be a mechanical failure, an electronics failure or a coil failure. Furthermore, even if you do know the winding's resistance you cannot measure it on the wires. If that is the case, don't bother cleaning, it is most likely dead anyway.
|
I want to know the name of this below component Its a turn Knob switch for setting timer and push start in Samsung microwave ovenI also want to desolder this component. How to do that? https://www.buyspares.co.uk/microwave/mw87l-mw87l-bxeu/assy-key-module-dkm-mw87l-key-module/product.pl <Q> It is probably a rotary encoder with a switch. <A> It might not be soldered on. <S> If it is, you might need a hot air gun to desolder it. <A> I'd guess that there are no solder points, there's a body with the rotor in it, riveted to the pcb, and that rotor has spring contacts that bear on pads directly printed on the board i.e. the board forms part of the assembly <S> and it won't be any use by itself - <S> just like the snap action switches on that board, they will bear down onto pads printed in the board itself. <S> Here's a transparent rotary switch that shows what is printed on the pcb that's part of this assembly. <S> Typically these small encoders are quadrature encoders, so there's two traces with sectors offset from each other, and a center pad for the push action. <S> As for repairing this board - check the connections at the end of the ribbon cable, looks like there are some corroded joints there. <A> It is a surface mount component. <S> You can apply heat to the rear of the board behind the part with a heat gun to remove it. <A> The trace in the top right corner of the red circle could be broken. <S> Maybe it's enough to fix the trace.
| It looks like it is a rotary encoder riveted to the copper side of a single sided board. You could drill/grind the rivets from the back of the board and see if the encoder detaches.
|
Should I use a resistor between the gate driver and MOSFET (gate pin)? I'm making a PWM driver for a DC motor. After some questions were answered in this post , I ended up with this schematic: And I have some new questions: Do I need a resistor between the gate and the output of the gate driver (U?)? Do I need a pull-down resistor at the gate of the MOSFET? MC34152 gate driver datasheet <Q> Maybe. <S> The MC34152 datasheet pp.8 on shows a series Rg to damp oscillations, and reverse-bias Schottky diodes for catching negative ringing spikes at the driver. <S> Wouldn't hurt to have these in your layout. <S> You could stuff zero-ohm for Rg if you don't need damping, and no-stuff the diodes if you find the ringing isn't too bad. <S> Place them near the gate. <S> But you will want a pull-down on the driver input to make the default state 'off'. <S> Also, while we're discussing the inputs - tie them together and use both of the separate outputs, one for each FET. <S> The way you have it - driving 2 FETS together <S> - kind of defeats the purpose of the buffer. <S> Finally, if your motor is a normal brush type you will want to use a freewheel diode across it to catch the flyback spike when the switches turn off. <S> For BLDC this isn't an issue. <S> Yes, C2 does this too, but the diode is better. <A> You might not need a gate resistor but if this is a custom PCB, you should make space for one and just jumper it if it is not needed. <S> Gate resistors slow down the rise time which can result in EMI, ringing, and damaging spikes. <S> Ferrite beads can also be used instead of resistors. <S> You want one resistor per gate and you want the resistor close to the gate. <S> This is for ringing issues should they arise. <S> You don't want to share gate resistors because parallel MOSFETs can ring with each other. <S> You should have a pull-down if your drive circuit doesn't sink it when powered is off since your MOSFET gate capacitance will just remain charged and keep the MOSFET on. <S> It also helps keep the MOSFET in a default state if your driver is busy powering up if your driver doesn't handle that well either. <S> I don't think you need one with the gate driver you are using. <S> You should have a flyback diode anti-parallel to your motor. <S> C2 does help though. <S> Your gate driver IC also needs a decoupling capacitor. <A> Some FET circuits, with various parasitic capacitances (including inside the FET) and inductances (including the FET packaging) and lumped loads, WILL OSCILLATE. <A> No, you do NOT requires a pulldown resistor for the gates. <S> This is already built into the driver and the driver when powered on will immediately have a defined state. <S> You MAY require a series resistor since the the driver is rated for only 1.5A maximum. <S> The series resistor is used to limit maximum current and reduce parasitic ringing which may increase dissipation in the output FETs. <S> As shown in the datasheet : <S> As mentioned in another answer you could put a diode across the motor to control back EMF instead of the capacitor, but neither of these are a good solution. <S> The best way to get rid of the stored energy as fast as possible is to use a Zener across your drive devices. <S> If you do this you will definitively require a series resistor from the driver to the gates since the Zener will hold the drive FETs on to dissipate the back EMF.. <S> Update: <S> One of the comments exposed (@DKNguyen) exposed the potential for gate oscillations between the paralleled FETs, and linked to this paper. <S> From this it would appear prudent to include a separate series resistor for each FET gate to reduce the Q of any parasitic elements. <S> Now however you definately will only use one of the FETs (the one with the Zener Drain-Gate) to limit the back EMF, even if the are well matched. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> A resistor in series between the MOSFET gate and driver give you the chance to adjust the speed of your MOSFET switching. <S> A pull down at the gate of the MOSFET, may or may not needed. <S> You can refer to here .
| Having gate resistances (as an option) provides a means to dampen the circulating energy, if the C_gate_drain or C_gate_source is in the resonant loop. No pull-down is needed at the gate drive. Have one resistor/diode per FET, don't share them.
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Is it a good practice to use resistors for MCU inputs for this case? I want to use Arduino Uno or a Nano which uses ATMEGA328P as micro-controller. D2 is now set as INPUT. During the start up I want the MCU to detect whether pin D2 is HIGH or LOW depending on the setting set by the user. The setting will be done by an SPDT switch as shown below: So if the switch pole is coupled to the +5V the state of D2 will be HIGH and if coupled to the GND the state will be LOW. Depending on this state in void main(), the program will perform a specific task. I'm just not 100% sure whether it is very fine to connect pins that way. Is there any need for a pull-down resistor between the switch and the GND or between the switch and the +5V pin? (What if the worst case D2 is accidentally set as OUTPUT pin?) <Q> Assuming the MCU pin will be damaged/destroyed if configured as Output, when externally forced to some fixed level, yes, you should insert 10Kohm resistor in the pin's PCB trace. <S> Do not configure the pin with any on-chip pullup or pulldown current. <A> As a quick-and-cheerful experiment there isn’t anything inherently wrong with this really, but it could be improved. <S> Out of reset the MCU will set the pin as an input. <S> The worst that can happen is if you make a programming error and change the port function or direction such that it's an output. <S> The resulting current is limited by the Rds(on) of the port pin driver so nothing bad will happen, no cratered IC, no white smoke. <S> If you’re paranoid, add a series resistor. <S> It is questionable practice to have a floating signal. <S> With no pull-up/pull down, the signal will float while the switch moves from one position to another. <S> It’s vulnerable to noise pickup during that time, for example, from your finger as it touches the switch. <S> So you should program the MCU to have a pull-up or pull down. <S> Hacktastical idea: <S> Add a small capacitor on it (100pF, say) to the input also. <S> Then you get a debounce action too as well as reducing the impedance of the input to improve noise immunity. <S> Finally, in a real product there is one other issue: ESD immunity. <S> Switches are an entry point for static discharges. <S> If the switch has a housing, ground that too. <A> Does your switch break-before-make or make-before-break? <S> If the latter, a resistor is needed to prevent a short from 5V to ground. <S> The resistor should be between the switch and 5V, or between the switch and ground. <S> Setting D2 as an output can cause that output to source/sink very large currents. <S> This may cause the processor to behave incorrectly, or in extreme cases, to be permanently damaged. <S> In general, you should avoid this possibility either in software or by adding a resistor in series with D2 (between the switch and D2).
| A TVS diode on the line next to the switch would be prudent.
|
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