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Does AM26C32 leave outputs floating? Does the AM26C32 leave outputs floating, by which I mean, when "off", is the output pulled to GND, or would I need to provide a pull-up/down resistor? Where in the datasheet would this info be specified? <Q> This is specified in section 8.4.1: <S> Device Functional Modes, <S> Enable/Disable: <S> The last row, corresponding to both enables driven with a disable input, shows that the output is not driven ('Z'). <A> It reads on the first datasheet page that it has three-state outputs so they would be floating, not grounded. <A> Quite likely with Schottky diodes, because the output may not exceed V CC + 0.5. <S> Note the upper diode is forward baised <S> when Vcc = 0, so, depending the current injected to the output pin (most likely noise, so, small current), the output will be 0V (for no to small currents) up to max 0.3V - 0.4 V.
| I wouldn't say the outputs are left floating, because they are clamped with diodes.
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Do I need additional power for an IR LED diode 50 feet away from a Raspberry Pi? I'm creating a central remote to control my mini-split HVAC units. Those units are controlled with an IR remote. Each unit is in a different room of my house. I have everything working on my breadboard and the IR LEDs are able to control the HVAC when it's in the same room. However, the Pi will be in my server room and each of the IR LEDs will be wired to a different room in my house. The rooms are anywhere from 30 to 50 feet (10-15m) away from server room. The unit will have multiple LEDs but only one LED will be sending a signal at a given time. I'm using these IR LEDs from Amazon . RX IR LEDWorking Voltage : 2.7 - 5.5VWorking Current : 0.6 - 0.8mACarrier Freq. : 38KHzBMP width : 8KHzReceiving Angle : 70°Receiving Dist. : 15mTX IR LED Working Voltage : 0.9 - 1.3VReverse Voltage : 5VForward DC : 30mALaunch Angle : 30°Emission Dist. : 12 - 13m What gauge wire should I use to connect the LEDs over that distance? Will the 3v power from the Pi be enough to power the LEDs that far away? If I need to use an additional power source to power the LEDs, what would that schematic look like? Note: I'm hard wiring this for reliability and simplicity. I don't want to create multiple wireless WiFi modules that communicate with the Pi. I just want the diode in the other rooms. Here's my schematic Thanks! <Q> TX IR LED Working Voltage : 0.9 - 1.3VReverse Voltage : <S> 5VForward DC : 30mA <S> You can drive an IR LED with higher current pulses (more than 30mA) as long as the maximum average current spec is not exceeded. <S> If you power the LED from 3.3V using a BJT (0.1V Vcesat) and a resistor to limit current, there will be 0.9-1.3V on the LED, and 1.6V minimum on the resistor. <S> For 30mA that would be a 56 ohms resistor. <S> Let's try 25AWG telephone cable. <S> That's about 650 ohms/km. <S> Two 20ft wires make 60ft so about 20m or about 13 ohms. <S> At 30mA current that's 0.39V, it is okay. <S> You can also measure cable resistance with a multimeter. <S> If you want to overdo it, you can substract cable resistance from the resistor value, so yor 56 ohms would end up at say 47 ohms. <S> Use any NPN BJT that will take the current, like BC337 or 2N2904. <S> 1k base resistor gives 2.7mA base current, that'll be plenty to saturate the transistor. <S> R5 value should be 47-56 ohms, and VCC should be 3.3V from your Pi. <S> In that case increase the value of R5 to (5-Vled-Vcesat)/30mA = <S> about 100-120 ohms. <S> Note the resistor is best placed on the transistor side, not at the other end of the cable with the LED. <S> This will protect the transistor if the two wires get shorted. <A> Welcome to the site Michael. <S> If you are only powering 1 LED over <S> ~15 meters, I'm assuming your LEDs are around 20mA max and require maximum of 2V. <S> You should be fine powering them off the 3.3V rail <S> (mind it has a 50mA maximum current limit). <S> Using just about any 2 conductor wire from 18-28 AWG. <S> You should use some sort of relay or transistor instead of driving directly from the IO pin though since the IO pins are only allowed to source 16mA of current. <S> You also could power them off the 5V rail as well, you will just need a larger value resistor. <S> Worst Case Calculation: <S> Driving the LED with 20mA on 15m of 28 AWG wire (round trip 30m) <S> you lose 130mV, which is insignificant. <A> As earlier answer mentioned, yes its doable but I would not recommend it. <S> Raspberry PI <S> I/ <S> Os are pretty delicate and with a few additional components <S> you could create a solid solution. <S> Additionally since the amount of current the LED would consume is unknown, you certainly require a higher voltage power supply and not 3.3V. <S> (Depending on what you can find easily, Your options could be any power supply with more then 5V output, e.g a 12V adapter, 19V laptop adapter etc). <S> The ground between this power source and your Raspberry would have to be common. <S> For each LED, your circuit would look something like below: simulate this circuit – <S> Schematic created using CircuitLab <S> For any given AWG wire, the website below provide a decent table to estimate its resistance per 1000 ft and you could use that to estimate the cable resistance. <S> http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/wirega.html <S> Now addressing the cable selection, as your current is expected to be small, typically between 50mA to 200mA depending on the LED requirement, you could easily use a AWG 28 or so cable. <S> Possibly a simple Ethernet cable as most ethernet cables are AWG24 to AWG28 rated (and usually sturdy, easily available and low cost). <S> by playing around the R1 value, you could actually extend the length even further easily.
| But you don't have to, if the LED is aimed at the receiver, even from the other end of the room, it will probably work even with much less current than 30mA. Schematic: I would suggest you use FET/Transistor to mimize the current you draw from the 3.3V rail. You can use 5V too if you have a 5V supply in your device already, this should be the case if the Pi is powered from a USB "charger". This is enough voltage drop to control LED current properly. The way you are using an NPN transistor doesnt seem to be right.
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Infrared Sensor for drug detection I saw that there has been a push to make a drug detector that can identify cannabis since it has been legalized. I was looking at mouser.com and found this ir sensor: https://www.mouser.com/ProductDetail/Pyreos/PY2435?qs=sGAEpiMZZMvkSJWT1okH7HZM1X5NjtfhYufzstUzGmFpPHtO%2FKJ1pA%3D%3D I also found these IR spectra The ir sensor has filters for 8550nm(1169cm^-1), 8750nm(1143cm^-1), 9650nm (1036cm^-1), 12250nm (816cm^-1). Also your typical methane detector is at 3300nm(3000cm^-1) so alot of these line up with the absorbance peaks, do you think this will work as a detector? I have seen typically for organic compounds that have aromatic rings or pi bonds UV spectroscopy is typically used but i cant seem to find UV detectors online. Is there an easier way to do this? What is a good estimate of ppm that this ir detector could sense? <Q> The ir sensor has filters for 8550nm(1169cm^-1), 8750nm(1143cm^-1), 9650nm (1036cm^-1), 12250nm (816cm^-1). <S> Also your typical methane detector is at 3300nm(3000cm^-1) <S> so alot of these line up with the absorbance peaks <S> , do you think this will work as a detector? <S> That detector is meant for measuring the concentration of a known gas in a purified form. <S> It works because the measurement is of a known, purified substance. <S> You need to define both the substance you want to detect and the additional substances you want to be able to exclude. <S> That determines which and how many wavelengths are needed. <S> I have seen typically for organic compounds that have aromatic rings or pi bonds <S> UV spectroscopy is typically used but i cant seem to find UV detectors online. <S> There are a lot of companies that sell UV spectrometers online. <S> What is a good estimate of ppm that this ir detector could sense? <S> The range of PPM that could be detected depends on the instrument using the detector. <S> With a large volume, ultra-pure sample, and long integration time, you could measure parts per billion or even smaller. <S> With an impure sample, you won't be able to measure at all. <A> <A> I think you will have an issue with false positives because of the overlap peaks with other compounds. <S> Depending on what your tolerance specifications for false positives might be, you may prefer a more accurate method. <S> Why reinvent the wheel, nature has already supplied us with a quite accurate specific compound detection method; canine olfactory detection. <S> Customs and Border authorities have used this method for many years and very reliable. <S> I don't think you can improve on it with these sensors.
| with only four measurements such a sensor will only be able to detect pure THC, there will be 4 innocuous substances that when mixed will confound the measure sufficiently that detection of trace amounts of THC will not be possible.
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How can I make my PNP amplifier circuit to turn off faster? I'm using the following circuit to switch a 24V signal to a PLC. The rising edge is fast and perfect(38 ns) but the falling edge of the signal is not as good(3.5us)!How can I modify this circuit to have faster-falling edges? simulate this circuit – Schematic created using CircuitLab <Q> The basic idea is something like this: simulate this circuit – <S> Schematic created using CircuitLab <S> There are problems with this, though. <S> For example, \$Q_3\$ might oscillate. <S> Some added base resistance is a common fix. <S> But there are other approaches. <S> In this case, I don't think there's much likelihood, though. <S> Just mentioning it, in case it matters. <S> Also, you can always consider adding some emitter resistance for the output BJTs, if you want. <S> But you'd need to know something about what you are driving to figure out those values. <S> So that's also missing. <S> There's also no base protection for either output BJT. <S> I've also not added local power supply capacitors. <S> Again, you may also want those. <S> Or not. <S> I've also avoided the speed-ups. <S> A fuller circuit with all the crap added might look like this: simulate this circuit <S> In the above, I've left off the base protection diodes. <S> But they are pretty obvious, if you want them. <S> With appropriate component values and those fancy BJTs <S> I mentioned above (the BFT93 and BFR93 or BFR91A), the following Spice simulation results (it assumes some source resistance, as well, for what's driving it, and drives a load represented by two \$20\:\text{k}\Omega\$ resistors in series between the \$+24\:\text{V}\$ and ground. <S> (So a \$10\:\text{k}\Omega\$ load, in short.) <S> As you can easily see, it's pretty cut and dried. <S> Nice and sharp edges and very little change in the duty cycle or its delay relative to the input. <S> And I spent exactly zero time trying to calculate resistors or capacitor values when popping that into Spice. <A> Another approach : a Schottky diode from Q1 base to collector; or possibly on both transistors (with the appropriate orientation). <S> As each transistor starts to enter saturation, Vc falls below Vb and the Schottky diode becomes forward biased. <S> This drains further base current preventing the transistor entering full saturation, which is a major cause of increased turnoff time. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> It's also a relatively easy addition to the current design. <S> Details in this Q&A... <A> A "gate driver" integrated circuit may be just the thing here. <S> They're designed to take logic inputs to higher voltage swings. <S> Another possibility is a high voltage analog switch IC. <S> I've used the DG403 as a CCD gate driver and Cockroft-Walton HV exciter in several space missions. <S> There are too many choices for me to venture one for your application ツ <A> The simplest thing to try would be to add a 10k ohm resistor between the IO pin and ground in your original circuit. <S> This should quickly drain the remaining charge out of the IO pin after Q1 turns off.
| You might also consider adding diodes to protect them against short-term reverse-voltage transients.
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Reduce/Remove Sparks in AC and DC Contacts of Relays I see that there are sparks being generated whenever there is a connection established between two contact points in the relays. So is there a way to reduce/remove those spark generated in the relays. <Q> You can even get them in one package. <S> People often use 0.1 uF and 100 ohms, but you may want to use trial and error to get the best results for your application. <S> If the load is at line voltage you might use an X2 for the capacitor for safety. <S> One capacitor company used to loan (sell?) <S> R-C selection boxes to make this convenient. <S> There are also formulas published. <A> Not to sound midevil <S> but... there is another way! <S> For one most relays that take a big amp rating normally have grease inside that helps with arcing. <S> This is a good start to try. <S> Starter car relay for example. <S> If you ever seen like a 80v dc lawnmower for example they solved the problem with 2 relays. <S> You use a small low voltage 2 throw relay. <S> One side will energize your larger relay and the other side will bridge the larger relays gap with a inline resistor, like a 1k ceramic type. <S> This first connect will not power your motor but will pre-complete <S> the circuit so the spark is not present when you engage the bigger relay that does the main work. <S> I know low tech not over complex is never the answer people want on forums, but this method has been used for years. <A> Snubbers as above are useful. <S> One less appreciated source of contact damage is excessively long release times caused by the common "diode across the coil" protection. <S> A series diode and zener across the coil, or at lower cost, a zener across the transistor can significantly speed up relay opening time and reduce arcing especially with inductive loads and DC sources. <S> (see https://www.sciencedirect.com/topics/engineering/resistive-load section 5.1 schematic B)
| There are a number of ways to do this, the most generic is an R-C quench circuit.
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Device runtime with specified battery How long will a 12V 30W device run on 12V 65AH battery, if the battery is first fully charged and the battery is dedicated to this one device? <Q> Details are missing so an accurate calculation cannot be made. <S> A "12V" battery starts at about 13.8V <S> then its voltage drops as it is used. <S> You did not say what is the minimum voltage of the device that you are powering. <S> You also did not say the minimum voltage of the battery when it has provided 65Ah. <A> 26 hours if everything behaves ideally as you described. <S> 12V/30W is 2.5A, <S> so 65AH/2.5A = 26 hours. <A> Ah is Amps x hours, so in theory a 65 Ah battery can produce 65 A for 1 hour, or 32.5 A for 2 hours etc. <S> If we naively also assume that the battery produces 12 V throughout the discharge then the 12 V 30 W device would draw 30/12 = 2.5 A, and the battery would last 65/2.5 = 26 hours. <S> In practice a typical lead-acid battery's rated capacity is for 20 hours or longer. <S> If discharged in 1 hour it would only achieve approximately half the rated capacity. <S> In this case we are expecting over 20 hours so the achieved capacity should be close to the battery's rating. <S> However when a Lead-acid battery is fully discharged the plates are stressed and irreversible chemical reactions occur, so it wears out faster than normal. <S> The usual recommendation for a deep cycle battery is no more than 80% discharge, and for a standard battery 50%. <S> Another factor which can have some effect is the voltage, which drops during discharge. <S> At low current it might start at ~13V and go down to 11V, so although the average is ~12V the load might draw more current at the beginning and less at the end, or the other way around depending on how it operates. <S> A heater or light bulb draws more current at higher voltage, but a switching power supply does the opposite. <S> This may make a small difference to the run time. <A> It would be a bit hard to accurately predict that without knowing a lot about the battery for example:The depth of discharge allowed by battery chemistryBattery internal resistance and its true AH rating which may not he same as claimed by different suppliers. <S> A good way would be to measure this could be by using simple energy meter that measure total energy input and output during a charge and discharge cycle. <S> An answer above gives a best case figure but in reality most battery cannot be discharged to a very low level of charge and neither are claimed AH rating entirely reliable for all brands.
| So depending on the battery type and how well you want to treat it, the practical run time could be 20% to 50% less than the rated capacity indicates.
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Alternating voltages on vintage calculator circuit board I am an electronics novice. I have a vintage calculator with six secondary wires coming out of its transformer and ending on the board, as shown below: Using the AC portion of the multimeter, I measured the voltages across the board terminals marked 1 through 6, with the results as follows: Is it normal to see AC voltages on such a board, or is this because of the age of the product (1979)? Is the AC voltage changed to DC somewhere on the board? Is this a "digital" calculator? (There is an IC chip on the board.) If so, why is there alternating current, unlike say on an Arduino? <Q> Transformers work on AC, not DC. <S> AC voltages can be converted to DC onboard if necessary. <S> Just by looking at the PCB traces it can be guessed that there is a diode bridge for rectifying and the big electrolytic capacitor is the reservoir capacitor. <S> Some AC voltages may be used directly such as the heating filament for the vacuum fluorescent display. <S> But ICs need DC voltages, both digital or analog chips. <A> Yes, it's normal on a PCB with a mains-frequency power transformer to see AC voltages from the transformer. <S> They can be rectified (diodes or bridge rectifier), filtered ( <S> those can electrolytic capacitors) and regulated, as required. <S> That calculator has a vacuum fluorescent display , which uses relatively high voltage for the grids and cathodes (something like 24VDC, often negative with respect to ground, is typical) and also has a filament which is typically a volt or two, and can be driven by AC or DC (possibly directly from a transformer winding in this case). <S> Here is some more information on driving these displays. <S> The calculator IC itself probably runs from a few volts DC. <S> There may be an external chip to drive the display, since the voltage requirements would tend to point to an IC process that would have been costly for the calculator IC. <S> Because of the multiple voltages, it's going to be very easy to damage the ICs by even a momentary short, so be careful probing. <A> it is certainly possible to have AC transformers on old circuits and transformers of this size were common until higher frequency transformers came into play in recent decades. <S> Because almost all logic chips even those before modern microprocessors came in operated with a DC voltage. <S> Some onboard rectifier and capacitors maybe doing the job of this conversion. <S> If there is a chip on board and a digital display then yes it would qualify as a Digital calculator(the definition of digital has quite evolved overtime though). <S> One reason why modern boards like Raspberry Pi or Arduino donot have any AC transformers around is because with advent of modern switching power electronics, transformers and the circuitry required to convert voltages have become really small and can safely be fitted into small adapters which eventually supply standardized 5V, 12V or so voltages. <S> So availability of low cost standard power supply modules mean designers today can focus entirely on core functionality of their circuits and use off the shelf, low cost and standardized devices to manage AC to DC power conversion. <S> If it is really needed and you wish to understand the functionality of this specific transformer in detail it could be worth using a milli Ohm meter/LC meter to measure winding resistances/Inductances which would give some clues on construction of this transformer and whether they are isolated or combined (or by looking at the signals through an oscilloscope) <S> Hope this helps. <A> Looking at this question with your previous question in mind, the information is consistent with the transformer configuration shown below. <S> The measured voltages don't quite add up to 4.5 and 24 volts, bit they are close enough. <S> The continuity test does not show a connection between the two secondaries, but the voltage readings would indicate that there is some connection between the circuits powered by the two secondaries. <S> The voltages between the terminals of one secondary and the terminals of the other are only important if they indicate a fault. <S> They do not necessarily indicate a fault. <S> Since the calculator is working, it seems likely that a common point may have been purposely created for the various DC power supplies. <S> To understand the design, it would be useful to identify the common point for the DC circuits.
| To your question on whether the voltage from this transformer would have been converted to a DC voltage onboard, the most likely answer is yes.
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MCU powered from the same PSU as a LED strip I want to make a simple controller of WS2812B addressable LED strip. I will consist of a MCU (most likely Atmega328P), a couple of buttons (RESET and mode selection) and 7-segment indicator, which will show current illumination mode. Power to the strip will be delivered by a N-channel MOSFET, which gate is also driven by MCU. Max length of the strip is 4 meters, I plan to use 30 leds per meter strip, so maximum current consumption is 6A (1.5A per meter). Since this current is only required if all LEDs are white at full brightness (which I don't plan), i'll use a Mean Well 5V@5A PSU. I think even 5A will never be reached in any application. Main question is, is it OK? Or do I need to use secondary small PSU for microcontroller? This I'd like to avoid, of course. MCU together with indicator would draw no more than 150 mA. Another question is, is it OK to use only two large value capacitors (470 uF), one at the input of power, another at strip output? Do I need a separate small ceramic capacitor close to Atmega pins? It's no problem to add, but I'm not sure if it's required: the board is really small, and distance between input 470 uF capacitor and 5V input pin is just 6-8 mm. Third question is, do I need a fuse on the board? If yes, how to choose it's value properly?Below is my simplified schematic (7-segment indicator and buttons are omitted): simulate this circuit – Schematic created using CircuitLab <Q> Yes, you can do that, however bear in mind that if you make some programming error, and all LEDs are on the adapter might be ruined, so you might want to add some protection (fuse?) <S> , if not already protected. <S> If the PSU does not have a fuse already, I would use a 5A fuse (same as the adapter itself). <S> I also would advise not to use 5A continuosly by LEDs (and MCU), stay at e.g. 4A for have some margin. <A> Yes same powersupply can be used. <S> Yes you need separate 100nF ceramic at AVR supply pins, it is a local bypass for high frequency current pulses the AVR takes every clock cycle. <S> The electrolytics are quite bad at high frequencies. <S> Also, I am slightly worried about using FET to cut ground from the LEDs, as the data pin must then be set high so that it does not pull current when it is low. <S> Think carefully if you even need to cut power from LEDs, and whether high side switching would be better. <A> maximum current consumption is 6A (1.5A per meter). <S> Since this current is only required if all LEDs are white at full brightness (which I don't plan), i'll use a Mean Well 5V@5A PSU. <S> I think even 5A will never be reached in any application. <S> MeanWell usually has proper current limiting so if the LEDs draw too much current it will just shut down. <S> You don't need a fuse. <S> You would need a fuse for fire protection if you used a power supply with a huge current that can actually melt your wires, but that is not the case. <S> Or do I need to use secondary small PSU for microcontroller? <S> No, <S> but the 5V that powers the LEDs is going to be very noisy due to the large PWM currents. <S> If this is some kind of huge vu-meter or anything that uses the micro's ADC then a bit of filtering could be necessary, for example a ferrite bead on the supply line, and the 470µf capacitor you want to put near the micro, that'll work. <S> is it OK to use only two large value capacitors (470 uF), one at the input of power, another at strip output? <S> I think it's recommended with WS2812B <S> but you need to use capacitors that can actually do something useful like decouple the power supply. <S> This means a low-ish ESR like 0.1 ohms and a ripple current rating of an amp or more. <S> Since all the LED PWM won't be synchronized, ripple current shouldn't be the full 5 amps. <S> A general purpose 470µF cap should be rated for about half an amp ripple current with an ESR around 0.7 ohm which is inadequate. <S> Capacitors designed for switching power supplies are a better choice, like Panasonic FC/FM/FR series. <S> These are not expensive. <S> Do I need a separate small ceramic capacitor close to Atmega pins? <S> Usually <S> yes. <S> distance between input 470 <S> uF capacitor and 5V input pin is just 6-8 mm. <S> Are you trying to save a 0805 SMD ceramic cap? <S> That's like 0.1 cent...
| Turning the power on with FET will charge the output capacitor with a high current pulse so it might break. Check the power supply datasheet.
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Can't get this schmitt oscillator to work...did I goof somewhere? Below is my attempt at a schmitt oscillator circuit that uses a 74HC14 and lights up three LEDs in sequence. I built it on some breadboard, but nothing lights up. I'm really at my wits end...can anyone tell me where I went wrong? <Q> You schematic is 100 % accurate. <S> Funny thing is that on a simulator , if all the thresholds and RC values are identical, it won't startup as all LED drivers go high and low at the same time. <S> This circuit works by the voltage difference between stages and can drive about 7mA. <S> So just change any part a bit in value and be sure to have a cap across IC for noise reduction and keep wires as short as possible. <S> Choose any reasonable RC=T product. <S> I used T= <S> RC=0.1s which with Schmitt thresholds can yield 1.6Hz for 3 stages. <A> Please draw your schematics so they flow left to right as much as possible. <S> I have redrawn the left portion so it is easy to follow. <S> At power-on, all the caps will be discharged, and all the three outputs will be high. <S> All the caps will charge and when they have reached the switching threshold, all 3 outputs will switch nearly simultaneously. <S> This is not what you want, I believe that you want the signal to propagate down the line. <S> EDIT: <S> Tony is correct, with a mismatch, it theoretically should startup eventually. <S> But, I dislike any circuit that is not guaranteed to startup. <S> It is difficult to make the 3 circuits completely independent, if you have any coupling through the VCC, they may want to lock on to each other and not startup as you want. <S> I built the circuit on a cheap breadboard and even with 2% difference, it won't always startup. <S> I think you will be better off with a circuit that is guaranteed to oscillate. <S> You can use delays to make the 3 LEDs light in sequence. <S> You will need to fiddle with the values to get each LED to have the same on-time. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> You will have a better chance of success if you connect one of the capacitors to V+ rather than ground. <S> Observe polarity. <S> This will create one low output on power up that will ( hopefully) ripple through the others. <S> The above suggestion will start oscillations, but I am not sure if they will be sustained. <S> Since the goal seems to be to light the LEDs in sequence, I created a circuit that will do this for six LEDs and the same number of parts. <S> After creating this circuit l realized it is a Johnson counter. <S> The LEDs are numbered in the order they will light when power is applied. <A> I can't use all the same resistance values. <S> I swapped out a resistor, and removed one of them for one of the LEDs, and the whole thing works! <S> Thanks!
| Ok...so the problem was that this oscillator doesn't work unless we introduce a small imperfection somewhere.
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Why did Axial Capacitors Fall Out of Use in the Industry? When inspecting the boards from electronics made in 1980s and earlier, one distinct feature is the extensive use of axial, electrolytic capacitors as power supply filter. Axial ceramic decoupling capacitors were used as well, to a lesser extent. For example, this is a C64 motherboard. Source: Wikimedia Common , by Gona.eu, license: CC BY-SA 3.0 This is a Tektronix 1720 vectorscope board. Source: Flickr , by Toby Thain, license: CC BY-NC 2.0 However, although they are still being manufactured, it seems that axial capacitors largely disappeared in most devices since the 90s. Almost none of electronic devices we commonly see have a single axial capacitor. And it's certain that one is going to find something similar to this in a modern device... Source: Wikimedia Common , by Dave Jones from Australia, license: CC BY 2.0 Question Why did Axial Capacitors Fall Out of Use in the Industry? I can imagine that axial capacitors were optimized for point-to-point wiring back in the pre-PCB era, not PCB assembly, and that the introduction of SMT was another shot. But it was just my imagination, backed by nothing. What were the exact sequence of events and/or rationale that led to the disuse of axial capacitors? <Q> PCB area. <S> Axials pre-dated PCBs, their construction was ideal for wiring to tag strips and valve bases, and they were adopted for PCBs because that's what was available. <S> Example of tag strip construction below. <S> (There WERE radial caps in the valve days : they were generally designed for chassis mounting via a ring clamp, and had tags rather than wire leads. <S> The round object bottom centre is the base of one such capacitor) <S> It's actually quite surprising <S> they lasted as long as they did alongside radials, into the 1980s. <A> Single-sided PCBs frequently required the use of wire links to bridge over other tracks on the board. <S> With a suitable circuit layout, the use of axial capacitors (rather than radial) could be used to allow tracks to cross each other, removing the need to use a separate wire link. <S> Axial resistors offer the same capability, of course. <S> With double (and multilayer) PCBs, it's possible to cross tracks with the use of vias between PCB layers instead. <S> This doesn't need the placement and fitting of any through-hole components, so the flexibility offered by axial capacitors was reduced somewhat. <S> Axial capacitors also have a disadvantage of having a large footprint on the PCB. <S> A radial capacitor needs far less space. <S> Take this photo of a PC motherboard as an example; how much more space would be needed for these capacitors if axial ones had been used instead of radial? <A> My recollection of that era was that the selection , size and price of axial leaded electrolytic capacitors was not competitive, so I used radial lead caps in some cases where axial leaded would have been better (production had to lay them down and add a dab of adhesive). <S> You could not find low-leakage caps, for example. <S> Some parts, such as those used in crossover networks, may have been more popular in axial, but I was not involved that area at the time. <S> That was probably a side effect of demand. <S> The radial types just take up significantly less PCB space. <S> Both were available in tape and reel or ammo box <S> so I don't think automation was the issue. <A> Probably the most important reason is difficulties in automated assembly; Wikipedia mentions this .
| Radials use much less PCB space, and standing axials on end is a poor compromise, with a long exposed lead (or the added assembly step of sleeving it) as well as being much less robust.
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Measuring voltage drop across a diode in a switched circuit I have the below circuit where the input voltage is 8V and a PWM pulse is given to the base of the Q101 transistor. From what I have measured, The voltage at the anode of the diodes D102 and D103 is constant like around 8V. But when I measure the voltage at the cathode of the D103 and D103, I get a PWM waveform whose peak value is 7.7V. Voltage measurements are done using oscilloscope. My questions : How to measure the voltage drop across the diodes D102 and D103 in this senario? Why am I getting a PWM voltage waveform only the cathode of the diode and not at the anode? Why is there a 0.3V voltage drop across the diode? Shouldn't it be typ. 0.7V? BOUNTY EDIT : -The 8V is the voltage source. SBATT is connected to a load which draws a minimum current of 80mA and a maximum of 100mA. -The temperature of the circuit , which I have measured is not 150degC. The temperature is around 85degC -The diode is not a schottky diode. <Q> How to measure the voltage drop across the diodes D102 and D103 in this senario? <S> Like you did: with an oscilloscoop. <S> For example by measuring the anode of the diode on one channel and the cathode on the other channel and use the math functionality (if exist) to calculate the difference, or use the cursor to measure the difference. <S> Another option would be to galvanic isolate the oscilloscoop and connect to probe across the diode. <S> Why am I getting a PWM voltage waveform only the cathode of the diode and not at the anode? <S> Because the anode is connected to a voltage source which will regulate its voltage to maintain a constant 8V. <S> Why is there a 0.3V voltage drop across the diode? <S> Shouldn't it be typ. <S> 0.7V? <S> Yes, it should. <S> Only when the ambient temperature would be 150°C (which I doubt it is) <S> the drop would be 0.3V <S> UPDATE with simulation as discussed in comments If the diodes are really 1SR154-400 and noty Schottky's, the only explanation for the \$V_F\$ of 300mV I can think of is that these diodes are hardly conducting. <S> When assuming SBATT is a battery which voltage is about or slightly higher than 7.7V, then that battery provides the main current through R1 . <S> The contribution of the 8V is quite low, say about 2uA, resulting in a VF of only 300 mV. See simulation below. <S> I used 2 shunt resistors of 1 mΩ to obain the current delivered by the battery and the current though the diodes. <S> I made assumptions for the type of Q1 and Q2. <A> How to measure the voltage drop across the diodes D102 and D103 in this senario? <S> You have already done this with scopemeter. <S> This is the correct way. <S> Why am I getting a PWM voltage waveform only the cathode of the diode and not at the anode? <S> Because you are not switching the source. <S> And since the current flows through the diode, you'll see the pwm-like signal at the cathode because of the voltage drop on it. <S> If there were a resistor instead of a diode <S> you'd see the pwm-like signal at resistor's right-side terminal (i.e. the one that does not connected to the voltage source). <S> When the Q100 is off (i.e. the PWM signal is low and Q101 is off) there will be zero current draw and thus zero voltage drop across the diode. <S> When the Q100 is on (i.e. the PWM signal is high and Q101 is on) <S> a current of around 5mA flows through the resistors and Q101 (assuming SBATT-side is unloaded). <S> Thus, the diodes will show a voltage drop across them. <S> Why is there a 0.3V voltage drop across the diode? <S> Shouldn't it be typ. <S> 0.7V? <S> This depends on the diode itself. <S> For silicon diodes, yes, the drop is 0.6~0.7V <S> but under some test conditions (e.g. 5mA forward current) . <S> For schottky diodes, the drop can be as low as 0.2V. <S> The schematic says that the diode is 1SR154-400. <S> The current flowing through the diodes is around 5mA, <S> so 2.5mA per each (assuming they are nearly identical). <S> If you check V F -I F characteristics from the datasheet , <S> V F is around 0.3V for I F = <S> 2.5mA. <S> Assuming the model number of the diode given in the schematic is correct, the only possibility of low voltage drop could be a damaged diode. <S> Or else the diode is a schottky diode. <S> By the way, parallelling two discrete diodes is not a good idea. <A> How to measure the voltage drop across the diodes D102 and D103 in this scenario? <S> On a DSO with 2 probes use CALC Ch A-B <S> Why am I getting a PWM voltage waveform only the cathode of the diode and not at the anode? <S> Anode is connected to a voltage source which in theory is a 0 Ohm to AC <S> but Cathode is high impedance. <S> Why is there a 0.3V voltage drop across the diode? <S> Shouldn't it be typ. <S> 0.7V? <S> When there is no apparent DC load and diode Vf=0.3 V , this indicates there is < 100uA leakage current which is due to Rce across the collector emitter called the Early Effect leakage resistance. <A> When Q100 is on and you have a load connected, it should be down at about 7.3V. <S> In a perfect world, there would be no drop across the diodes when Q100 is off. <S> However, a little bit of extrapolation on the curves indicates that a current of 1-10uA could result in about a 0.3V drop at 25C. <S> This is in the neighborhood of leakage currents and it wouldn't be surprising to get this result. <S> So as you switch off and on, you would see the diode cathodes switch from 7.3V to 7.7V <S> (you did say 7.7V was the peak, not the valley). <S> At the same time, the output would switch from about 0V to 7.1V (due to the 0.2V saturation of the PNP). <S> Does this match what you're seeing on your scope?
| There should be a drop of about 0.7V across those diodes, when there is nonnegligible current through them.
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Same thermistor model (103) with varying response I was using thermistor model 103 to measure temperature of surroundings. To my surprise, when I checked different thermistors of the same model (103) bought from the same shop by changing them in the circuit, each had their own resistive response to the surrounding temperature. Like the first thermistor gave an error of 1.5 degrees celcius and other 2.5, due to varied resistive response. My question is: How can these thermistors of same model differ in the way they measure temperature and what can be a universal solution to all thermistors of this model to be accurate? <Q> The allowable variation in the thermistor characteristics should be defined in the component datasheet. <S> If the parts you purchased are not meeting the datasheet limits then you should return them to the supplier. <S> There is no solution that will make all thermistors of a given type "accurate", because there will always be variation in the components. <S> If you want to limit the variability then you must make this part of your purchasing agreement. <S> The other solution is to perform a final calibration procedure to determine the unique correction needed for each unit, and store the correction parameters in some kind of non-volatile memory. <A> There are (at least) two parameters when you specify a thermistor. <S> Resistance at a reference temperature (in your case 10K at 25°C, most likely) and \$\beta\$ . <S> The latter gives you the nominal resistance change between two specified temperatures. <S> Both of those parameters have an associated tolerance. <S> For example, a TDK B57863S0103F040 thermistor is 10K nominal at 25°C and has a nominal \$\beta\$ of 3988K between 25 and 100°C. <S> This particular part is specified to have a total tolerance including beta and nominal value of +/-0.2 kelvin. <S> If you stick a part with a beta of 3988K into a circuit designed for a different beta, it will read correctly (within tolerance) at 25°C but will deviate from the ideal reading further the further you get from the reference 25°C temperature. <S> If you trim the error out at one temperature it will be inaccurate at other temperatures. <S> So to substitute a part and have it perform accurately you must know at least those two parameters, and the desired tolerances. <S> Typically a 1% tolerance represents a couple tenths of a kelvin. <S> If you want a "universal" solution you can measure the resistance and implement the nonlinear equation for the temperature as a function of resistance. <S> Popular equations include the simplified equation with \$\beta\$ <S> and the Steinhart-Hart equation which requires more parameters. <S> It's a lot easier if you're just measuring near some fixed temperature such as human body temperature. <A> I'll bet if you do a sensitivity analysis of your temperature measuring circuit, and given the initial tolerance and resistance vs temperature characteristic variability of the part you're using, you find that those temperature differences are within expected bounds. <S> By sensitivity analysis, I mean you have know how much the resistance of the thermistor varies with temperature. <S> Then you basically take the derivative of that function (curve) to come up the sensitivity of the circuit, in ohms/deg C. <S> This will probably not be constant, and could vary quite a bit across your temperature range. <S> From that sensitivity, and using the data sheet tolerance and other information, you can estimate what kind temperature error is consistent with your part and circuit. <S> Edit 1: <S> Answer OP's question <S> "Universal solution" - depends on what your measurement accuracy requirements are, which I don't think you've told us. <S> Also, it makes a difference whether you're concerned about absolute accuracy or just repeatability. <S> You can improve the initial uncertainty, at least at 25C, by procuring 1% (or better) tolerance parts. <S> You could do a calibration of your monitoring circuit (excitation method, amplifier chain [if any], ADC error) to help ameliorate those errors.
| Since thermistors are extremely nonlinear you need to have a wide dynamic range in the resistance measurement to get good resolution and accuracy over a wide temperature range.
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Why do my all Atmega328P chips stop responding suddenly in a 12v circuit with Zener diode? I have noticed this strange thing where my atmega328p stops working and when checked through Arduino IDE or AvrdudeSS or ProgISP, they show programmer not in sync, rc=-1 initialization error (while the same circuit recognizes a new chip). EDIT3: Zener removed form schematic as the problem exists even without the zener. My circuit has a 12volt input coming from an adapter, then to 7805 and 7805's 5v output to my atmega chip. The circuit worked all fine until 7805 was moved a bit by jerk and all becomes hell, atmega stops. Is it permanent damage? I am unable to understand what is it and my money is wasted on buying new chips and demoralizing me to continue. EDIT: I must also mention that one of my chips died in the same way while I just had an led tested with the chip's 5volts. All I noticed is a "jerk" to 7805 did it (switched-off the led and then the chip stopped responding to anything) EDIT2 - Is it possible that fuses go wrong if chip resets due to jerk on voltage regulator or anything similar? I know it sounds silly. I always burn bootloader on my new chips using usbasp and Arduino IDE 1.0.1 and then set fuses using AVRDUDESS default fuses for Arduino Uno (L 0xFF H 0xDE E 0xFD, after which blink program blinks at normal rate). <Q> I noticed is a "jerk" to 7805 <S> did it <S> Any interruption in the 7805's ground connection would result in the 12v input being applied to the downstream circuitry, resulting in immediate damage. <S> You really need to arrange things such that a "jerk to the 7805" is not possible, ie, solder the connection or use a good connector in a way that it is not under mechanical stress. <S> Generally speaking, you're better off with a regulated power supply built in some lasting way, not on a breadboard or temporary improvisation. <A> The zener diode is connected the wrong way. <S> Instead of being in series with the atmega it has to be in parallel, with the cathode connected to 5V and the anode connected to ground. <S> In your case you don't really need a zener diode, because the 7805's voltage should be pretty stable (linear regulator) and I would say "paranoia" is not an argument ;). <S> That would be different if you use a switching regulator, where a zener might be usefull to clip the regulator's ripple. <S> Also note that zener diodes are not meant to dissipate much power. <S> Using them to limit a PSU's constant voltage (like in your case) would either have no effect (when the voltage is below the zener voltage) or would lead to a continuous (unlimited) current through the zener, which will probably destroy it. <S> That's why you usually have to apply a current limiting series resistor. <S> This series resistance is undesirable for a power supply and that's why the zener approach is not the way to go. <A> The circuit worked all fine until 7805 was moved a bit by jerk and all becomes hell, atmega stops. <S> Is it permanent damage? <S> I am unable to <S> understand what is it <S> and my money is wasted on buying new chips and demoralizing me to continue. <S> 'Jerking' the regulator should not have any effect, unless one or more pins has a poor connection. <S> If the GND pin becomes disconnected then the regulator will put out ~4V less than the input voltage, in this case ~8V. <S> The ATmega328p is rated for an absolute maximum operating voltage of 6V, so at 8V there is a good chance of permanent damage. <S> Is it possible that fuses go wrong if chip resets due to jerk on voltage regulator <S> Yes. <S> The 'fuses' are actually FETs with floating Gates, which are programmed by applying a high voltage that forces charge onto the Gate. <S> A technique commonly used to break MCU protection is to 'glitch' the IC with a high voltage spike, in the hope that the protection fuses will be reset. <S> Of course this often destroys the whole chip, but if you have enough of them then you may eventually get lucky. <S> The ATmega328p has several fuses controlling oscillator configuration, plus one that disables in circuit serial programming. <S> This makes it quite vulnerable to being 'bricked' by incorrect fuse settings. <S> AVR High Voltage Programming (Fuses Rescue) <A> I don't see how it can even work to begin with. <S> Based on the schematics, the AVR is completely missing the AVCC power supply connection. <S> And lack of proper bypass capacitors at AVR supply pins can cause problems too.
| If the fuses have changed then you may be able to reset them with a 'high voltage' programmer.
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Is it possible to design a dynamical/adaptable functionally/logically complete circuit? Biologist here but I've been reading about functional completeness in digital logic and I am wondering if there is a universally complete circuit that can be modified to express any boolean function for 2 or 3 inputs ? Maybe this could be done by having a big circuit and turning on/off certain logic pathways of the circuit to change its boolean function? Or modifying connections as to skip certain gates altogether? I am sorry if this is a dumb question but couldn't find any information on this. I am thinking this could be interesting for synthetic genetic circuits. ---------------- NEW EDIT Clearly, my question was too vague. In synthetic biology , we are starting to create what we are calling "genetic circuits" (like a genetic flip-flop , circuits with memory , or any other simple circuits), emulating the electrical engineering discipline. But instead of cables that transport electrons, it is genes that produce transcpritors (or RNAP flux) that activate or repress other genes. We are even at the point where we are creating genetic design automation tool s. However, the endeavor of building a genetic circuit and implementing it into a host organism is one that takes a lot of time, money, and effort. So my question is more like: Could a universal genetic circuit be built once, to implement into a host organism once, and then turning one or more gates ON or OFF, have the circuit behave with different behaviors? So I imagine something like this: Where one could cancel/knock out one or two gates or cables and change the function of the circuit to one which the researcher would like. <Q> Any PLD, CPLD, or FPGA (three generations of programmable logic chips) has a programmable gate array that can emulate a wide range of logic functions. <S> In terms of a non-programmable, off-the-shelf chip, the closest thing probably is an AND-OR-Invert gate. <S> https://en.wikipedia.org/wiki/AND-OR-Invert <A> If you are looking for an off the shelf part having two inputs and one output, a multiplexer can be used to do the job. <S> Inputs on control lines and MUX inputs connect to the appropriate logic levels to produce the functionality needed. <A> Parallel PROM and RAM chips are the simplest form of programmable logic. <S> When most people think about these chips, they simply see it as a medium for data storage, not too different from a hard drive. <S> When software programmers think about them, sometimes they visualize it as a table lookup or array indexing process: one sends a n-bit address to the chip, and retrieves the corresponding n-bit word, the RAM is just a huge array. <S> But alternatively, ROM and RAM chips can be seen as a universal logic circuit - <S> Physically, the chip maps an arbitrary n-bit input signals (A0-A11) to an arbitrary n-bit output signal (D0-D7). <S> As long as the control signals and timings are appropriate, they can be used to implement any Boolean function by writing the wanted pattern of data to the memory chip. <S> Speaking of PROM, it's a type of ROM, which one can change its stored data by taking it offline, reprogram it, and put it back online (the most commonly used variant is EEPROM, which can be reprogramed electrically). <S> Thus, it can implement any combinational logic (the type of logic without memory and state, giving the same input, it always returns the same output), and one can change its functionality by reprogramming it. <S> Speaking of RAM, the data on the chip can be modified on-the-fly, thus, it can implement sequential logic (the type of logic with memory and state, the output depends on its current and previous input). <S> Later development of programmable logic, such as PAL , CPLD , and FPGA , all originated from PROM.
| With only RAM and ROM, building a Turing-complete digital computer is not impossible .
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Is plugging a wrist strap in enough to prevent ESD? I've been told by a few people that if you plug in a PC and switch off its power supply that it is enough to prevent ESD. But I was taught differently. I was taught that it's more about difference in potential rather than grounding. The way I was taught was that everything must be unplugged, placed on a matt with a grounding strap for the matt and the installer. If a person follows method 1 will it still cause ESD damage? <Q> I've been told by a few people that if you plug in a PC and switch off its power supply that it is enough to prevent ESD. <S> No, it's not enough to switch off the supply. <S> For most PC's the working voltage has a maximum of 12V, using the Human Body Model (HBM) for Electro Static Discharges (ESD) ranges from 2kV to 8kV, which is hundreds of times more voltage. <S> The point of preventing ESD is not to 'float' a circuit and remove it from ground (which doesn't happen when a PC is turned off, it should always be grounded when proper electrical codes are followed), it is to stop the charge from accumulating in the first place. <S> There are transistors (mosfets) that you can destroy simply by waving your hand over them. <S> most of them are older, and many newer electrical components have ESD protection in their inputs, but that gives you an idea of how easy it is to kill electronic parts with static electric fields. <S> Other ways are, wear an ESD smock, use an ESD mat and to keep humidity high-ish (like 60%) as water vapor increases charge transport through air. <S> Another thing to keep in mind is that some materials like plastic and paper and clothing readily create\buildup charge so only use ESD compatible materials near sensitive parts. <A> No. <S> Method 1 still relies on person also grounded with static bleeding leakage resistance. <S> (1M) <S> But generally if person keeps at least one finger on case, so as at the same potential as board at all times, then there is less risk of generating or discharging a HV charge dump to an ESD sensitive node. <S> But this can restrict your actions to one hand unless you stretch your fingers and don't forget, this is a bandaid solution. <S> Boards should be handled by the edge or supply rail, ground to neutralize. <A> It doesn't matter wether or not the device is powered. <S> Any two objects that are not electrically connected with each other don't happen to be on the same potential, which can lead to an electrostatic discharge. <S> The potential difference (aka voltage) between you and your pc can easily be in the range of several kV, which is out of the specs for most of the components that are involved. <S> From this perspective your pc's 12V don't make much of a difference. <S> So, no, simply shutting down and unplugging is no esd protection. <A> Standard practice at our place is: 1) <S> Always place equipment on an ESD mat, if practical. <S> 2) Wear ESD lab coats. <S> 3) Wear ESD wrist straps. <S> If the equipment is too big to fit onto the mat, it or whatever it is attached to must be grounded. <S> When working on such a configuration the ground clip for the wrist strap is attached to the equipment ground (chassis) before removing or working on any installed components. <A> It will if the source of the ESD is also grounded... <S> in the case of a human, probably by wearing a ground strap connected to the same ground as the computer. <S> However, most people don't bother when using a computer, so they're protected pretty well. <S> If you're talking about servicing a computer, you should do it on a static mat on a bench in a lab with static floor coverings, a wrist strap, and bunny slippers. <S> If you're a private individual without access to all these toys, get a wrist strap and clip it to the computer chassis for best effect. <S> You still need to be careful about anything outside the chassis (spare/new parts, etc) <S> but it's better and safer than working on a unit plugged into the wall.
| The way to prevent ESD is to strap in, with a 1MΩ resistor on the strap to prevent electrocution. Whether a circuit is powered on or off will in most cases have very little effect on where the current from an ESD event flows.
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How are Circuitboards with Multiple CMOS Chips Routed? For my school side project i'm building a digital clock from 40xx series chips. I'm at 11 IC's now, taking up two regular sized breadboards. While I could simply solder this all onto solderable breadboad, I do also have some PCB perfboard with pre-tinned thru holes. I can't help but wonder if I could somehow condense the layout to make the board smaller. While the breadboard size would work, it leaves my clock a little larger than desired. Leaves me wondering... How did people even go about routing boards with multiple, maybe dozens of CMOS chips? For instance: What baffles me the most is those chips stacked vertically above one another. My project uses 6 CD4026 IC's to divide time. All 6 are connected to one another for carry outs. I can't imagine any other way to arrange them other than one after another, like this: And here are my boards.. Possible layout? <Q> The classic layout technique with double-sided boards that mostly contain DIP packages is to run traces (primarily) vertically on one side and primarily horizontally on the other. <S> Power is routed first to keep the traces low inductance, and bypass capacitors are placed at every chip, usually. <S> Typically the power pins are at the corners. <S> That technique is pretty useless in the current year because SMT packages often have leads on each side, and the level of integration is much higher. <S> If you're using perf board you should take care of the power and bypassing first and then run the signal lines. <S> Your proposed layout looks fine to me. <S> Here's the back of an LED display board for an instrument prototype: <A> Wirewrapping - this box was all wirewrapped for connections. <S> LEDs were soldered in place, didn't make sense for sockets for those. <S> And then I wirewrapped right to the LED legs, which are square pins. <S> 8 conductor cables (Dupont-style crimp housing headers with female-female wires) were made up to connect from the back of the card in the middle (what looks like empty sockets are really socket strips) to the back of the LED boards (more socket strips) to drive the 7-segment displays from a MAX7219. <S> I used this for 8 years at my fencing club before we closed up. <A> Tyler mentioned this - don't know why he didn't frame it as an answer. <S> But it's basically multiple layers on the board, at least for commercial applications (not hobbyist). <S> Pretty much everything we do now is 14 layers minimum, up to 40 layers or more. <S> Yeah, they're thick! <S> At least 2 return planes, 6 or 8 routing layers, and no circuitry on the top & bottom layers. <S> Sometimes the layer count is driven by having to break out traces from high density pin count packages.
| I've done many prototype perf board setups using polyester solder-through magnet wire (in the distant past, now it's easier and better to lay out a PCB and have it manufactured).
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What type of capacitor is most stable for use in 555 timer circuit? I am building a 555 timer circuit with a frequency of 15, 30 and 60 Hertz. I will most likely use a 1µF capacitor but I was wondering what kind of capacitor would be best for keeping a stable value even when the temperature changes (goes down most likely)? I will be putting potentiometers so that if the values do change they can be calibrated again, but that will only be done before we deploy the systems. Once deployed, we cannot calibrate again until retrieved days later). Basically, I am asking what kind (ceramic, electrolytic, etc.) of capacitor will keep its capacitance constant at a value of 1µF and with varying temperatures. <Q> However, if your actual goal is to have a stable frequency rather than discuss capacitors, then your original idea of using a 555 timer is not the way to go, as this chip will have worse drift than the C0G capacitor. <S> A much better option would be to use a quartz oscillator, for example a 2.4576 MHz oscillator which will cost about €1 in 50ppm/°C stability, then divide by 4096 using a 74HC4040 ripple counter. <S> You could use a 74HC4060 too, with a crystal instead of an oscillator. <S> Also the quartz oscillator is pre-calibrated and you don't need to adjust it. <S> EDIT <S> I misplaced a decimal point... <S> I mean, a 1.2288 MHz oscillator followed by 74HC4040 to divide by 4096, resulting in 300Hz. <S> Then a 74HC390 or similar which can divide by 5 then <S> 2 then 2 which gives 60Hz, 30Hz, 15Hz. <S> Or a 192kHz oscillator <S> then the same two 74HC chips, but this oscillator is only available on digikey in tiny MEMS flip chip so maybe not the best option. <S> Anyway you get the idea, pick a frequency and a convenient division ratio with 74HC chips... <S> The BOM cost for both solutions will be less than a precision capacitor and a potentiometer, and that doesn't count the salary of the intern who gets to tweak the frequency just right... <S> If you need better stability than what a cheap XO will provide, you can use a TCXO. <S> That will cost a bit more (a few €) and it will be available in less convenient frequencies. <S> So you can use, for example, a 12.288MHz TCXO and divide by 60000 <S> , you'll get a few ppm stability over temperature . <S> Note <S> a cheap microcontroller makes a nice programmable divider if you need one. <S> If you already have a microcontroller in your project, why not use that? <A> Regardless of the caps you use, a 555 is a 0.1-1%-ish stability device over a modest temperature range. <S> Typical <S> (not maximum) drift of the bipolar NE555 is specified at +/-150ppm/ <S> °C, so to get 0.1% typical drift with perfect external components you only need to change the die temperature by 7 <S> °C (either from ambient changes or from the die heating). <S> You can slap 10 C0G 0.1uF <S> 1206 caps in parallel and get 1uF with +/-30ppm/ <S> °C tempco and use some nice expensive 10ppm or better resistors <S> but you're just putting lipstick on a pig . <A> Neither. <S> What you want (or rather, what is practical and most cost effective) is a C0G/NP0 ceramic capacitor. <S> Not any ceramic capacitor, but a C0G/NP0 dielectric. <S> Good luck finding that in a size as large 1uF. <S> Even 100nF is quite rare, and expensive. <S> Other stable alternatives such as mica, glass, or poly max out at even lower capacitances and are much more expensive. <S> Also, temperature isn't the only thing that affects capacitance. <S> The DC-bias the capacitor experiences does too. <A> use a 30.72kHz crystal and a CD4060,digikey is sold out and says it's obsolete, but some other places still have this crystal Q8 will give 60Hz, Q9 30Hz, and Q10 15Hz
| As DKNGuyen said, if you want a stable capacitor, use C0G/NP0 ceramic.
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Can we ground all the nets in a printed circuit board? I am a newbie in electronics and I am designing a printed circuit board. The IC I am using has some pins that I don't use in my circuit. Is it okay to ground those nets or leave them as they are? I am asking this because I have heard that sometimes these nets can act as floating pins and could cause noise and errors. I am not sure whether this is true or not so please clarify my doubt and suggest me the right way. <Q> You will need to check the datasheets for each part with unused pins. <S> In many cases grounding them is a BAD idea, though in many other cases it may be CRUCIAL . <S> And in some cases you SHOULD leave the pins floating. <S> For example, do not connect outputs directly to ground, as this could cause a short. <S> Do not leave inputs floating, unless they have internal pull-ups/pull-downs. <S> The datasheets should tell you what to do with unused pins, but even if they don't you should be able to determine what to do on a pin-by-pin basis by analysis. <S> There is no one-size-fits-all solution for this problem, so you will have to sort through the pins individually to figure out how to connect (or not connect) them. <A> Please refer to Table 4 on p.15 and p.16. <A> Then you can fit 0 ohms for a solid ground, 1K or 10K to stop the pin floating (without damage if something drives it, plus you can pull it high during test), or infinity (no resistor) if you must leave it floating. <S> Final decision can be made for each pin during testing. <A> Unused output pins should just be left disconnected. <S> For unused input or I/ <S> O pins you need to read the datasheet and/or make some discisions. <S> As another answer points out the datasheet for your simcom device says you should leave them floating, so that is what I would do in the first instance. <S> Doing so can significantly increase the current draw of the chip as the transistors in the input buffer start to conduct. <S> One option for IO pins is to configure the pin as an output. <S> Then you can just leave it disconnected. <S> Downside is if you make a mistake as to which pins you should set as outputs you may end up shorting an output driver. <S> Some input pins may have internal pull-ups and/or pull-downs which can be enabled, these will hold the pin in a known state with less risk of mistakes than setting them as an output. <S> In the case of a module there may also be pull-ups/pull-downs on the module PCB. <S> If you aren't hugely constrained for cost or space then an external pull up/down is often the way to go. <S> This can also double up as a conviniant place to solder on a bodge wire when you find out you needed the pin after all. <S> Tieing inputs hard to the power rail or ground is something I preffer to avoid where possible. <S> It may lead to high currents if a pin is inadverantly set as an output, and often makes it harder to modify the board when you realise you need the input after-all. <A> Generally no input should be left floating. <S> Check the data sheet to determine the logic level required for the application you have. <S> If the input should be logic zero, ground it. <S> If it has to be logic one, tie it to logic one. <S> If the status does not affect your operation, check the name of it has a bar on it, meaning an active low input, connect it to positive supply through proper pull up resistor, otherwise connect it ground. <S> Do not live them open because they may pick up noise and change state and these transients may affect the circuit operation and cause undeterministic behaviour.
| If you check the HW Design Manual then you'll see that it suggests the unused pins should be left floating. One thing you CAN do when laying out the PCB is to ground all pins via a resistor on each pin. As a general rule though, a CMOS input should not be just left floating.
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Why we need bandwidth with ASK? Now I know that this question might sound silly, that is because apparently I am missing something here. Basically my question is, if ASK works as on/off keying meaning that provided we have an oscillator with fc carrier frequency and binary input to produce the fc signal (using oscillator) when there is 1 in input we just using that one frequency - the carrier frequency! So why we need a range of frequencies (which as I understand it is what bandwidth basically means)? <Q> The rise time of Carrier Amplitude spreads BW just as the rise time of the signal determines the -3dB BW = 0.35/ <S> t for t= 10 to 90% rise time. <S> So the BW of the signal translates to both sides of the carrier. <S> (unless single sideband carrier is suppressed by design to conserve BW.) <A> You mustn't forget two things: The convolution property of the Fourier transform <S> (i.e. multiplication in time domain is equivalent to convolution in frequency domain) <S> the fact that you're multiplying your carrier with your data signal <S> So, your signal is $$s(t) = <S> c(t) <S> \cdot <S> d(t)$$ <S> where \$c\$ is the carrier (typically, a cosine or a complex sinusoid), and \$d(t)\$ is your amplitude that changes over time. <S> Now, we don't know much about \$d(t)\$ , but assume it's just a rectangular function (so, the thing is "on" for some finite time). <S> That means that your data signal has a spectrum with a bandwidth – the Fourier transform of a rectangle is a (scaled) sinc function, and that has a bandwidth. <S> With the carrier multiplication (the carrier is just one or two dirac impulses in frequency domain), you just shift the spectrum of the data signal from being centered around 0 Hz to being centered around the carrier frequency. <S> The bandwidth stays the same. <S> TL;DR: <S> Your data carrying, i.e. the carrier-modulating, signal has bandwidth. <S> So does the RF. <A> What would happen to the output of a very, very small bandwidth band pass filter when a pure carrier of the matching frequency is suddenly applied at its input. <S> Would the output of the band pass filter instantly spring into action and produce a signal at the carrier frequency or, would you see a very, very sluggish carrier build up in amplitude over several seconds or minutes until finally you saw the full carrier amplitude? <A> You may wish to consider the Shannon Limit from information theory, which states that you need a combination of bandwidth and signal-to-noise ratio to successfully transmit information over a channel. <S> The better SNR you have, the less bandwidth you need, but the latter never actually goes to zero. <S> If you impose a bandwidth limit of 1Hz, for example, then you can expect the signal to take about a full second to respond adequately to each key-on and key-off event. <S> Clearly for a crisp 20WPM transmission, you need significantly better than that.
| The bandwidth of an ASK signal can be very low, allowing many signals to share a limited piece of radio spectrum, but it is closely related to the rate at which the signal is keyed.
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MCP1700 voltage regulator instantly overheats and starts to smoke, no load Newbie question. I have a MCP1700-3302E LDO voltage regulator to power ESP8266 using a 3.7V battery. However, as soon as I connect the negative lead to the ground pin and the positive to Vin the voltage regulator overheats and starts to smoke. I tried 2 different MCP1700s thinking maybe the first one was defective. Tried using a different, stable power supply at 5V, same result. Tried with 1uF input capacitor, same. What am I doing wrong? I just wanted to test the output voltage to be 3.3V. Thank you in advance. EDIT 3: Burned a few more, without seemingly any consistency. Some are stable at 4.6V and live, some go up in smoke right away, some burn out slowly at 2.6-3V. One was in expected range of 3.4V but was very hot. I added a 2K resistor before Vin and input capacitors and it doesn't heat up any more but the output voltage is 0.002V. With a 100 Ohm resistor it's 0.016V, with a 47 Ohm one it's 0.03V (and resistor got hot). So looks like amount of current affects Vout. I'm close to giving up. EDIT 2: In the setup below, it doesn't overheat anymore. Vout is a steady 4.61V, instead of expected 3.3V EDIT: Latest setup, fresh LDO, 0.1uF ceramic + 47uF aluminum caps on input, 1uF ceramic on output. Vin = 5.1VWhen it on, I'm measuring Vout = 4.6V until the problems start. Vout is supposed to be 3.3V. <Q> It appears that the purchased MCP1700 chips were faulty/damaged/fake/mislabeled. <S> They did not function as they were supposed to and there was no rhyme or reason for it. <A> Needs output capacitor. <S> Put it close to the output + gnd pins of the regulator. <S> "A minimum output capacitance of 1.0 μF is required forsmall signal stability in applications that have up <S> to250 mA output current capability. <S> The capacitor typecan be ceramic, tantalum or aluminum electrolytic. <S> TheESR range on the output capacitor can range from 0to 2.0." <A> LDO's with very low Ron and high loop gain have very low output impedance until current limiting. <S> The only way you can burn these out when wired correctly is by turning it into an oscillator charging a battery or large C load. <S> If the protection thresholds are oscillating out of phase with the voltage (OCP , UVP) it can be unprotected. <S> The photo looks like the connections are correct, but the long jumpers are 10nH/cm so with perhaps 100nH of series inductance , with <S> I see 47uF on paper and 100uF+ load on card. <S> This looks like a high Q resonator. <S> Possible Solutions <S> to overdampen the response add <= <S> 1 <S> Ohm between output and load until surge load on startup is reduced. <S> If you want to use a larger Cap, use up to 5 Ohms in series as allowed in datasheet. <S> This may cause low f load regulation errors or static DC drop. <S> Twist the wires to lower the impedance of wire directly to output. <S> use AWG 24 > <S> 5 turns <S> /cm <S> ** for 120 Ohm impedance roughly estimated from incremental sqrt(L/C) ratio. <S> In future consider dynamics of surge currents on startup and LC resonance and if rated , max load capacitance. <S> use current limiting R's as a means to safely turn on experiments and keep an eye on hot spots. <S> Add a good adjustable current limiter to your lab supply without bulk C, until you get more experience. <S> on ESR and reactive impedance with high Q factors. <A> Strange. <S> Try testing the component with a multimeter in diode mode between Vin and GND. <S> If you get ~0.7V, that would suggest you've got the pins the wrong way round (somehow the datasheet could have a mistake? <S> Very unlikely, especially since I have a similar working chip - MCP1703 - with the same pinout). <S> Then try some different pin combinations with a 1k resistor in series with the power supply (to avoid blowing up another one) and see if anything will give you 3.3V output. <S> For each combination, you could also measure Vin to get an indication of the impedance of the input (and hence whether it would blow up the chip). <S> If Vin is close to the power supply voltage, it would imply a high impedance and so it should be safe to short out the resistor. <S> Alternatively, if the multimeter detects a total short (<0.3V) then the chip is probably damaged. <S> Finally, if there is no diode drop detected, try using the chip with a 1k resistor in series with Vin and see if there is a) <S> any voltage on Vin (if not then your multimeter has lied to you) and b) any voltage on Vout. <S> Perhaps you could also check for a short in the breadboard; it always turns out to be something silly like that.
| It could be a bad batch, or there could have been moisture/ESD damage. This design is not stable.
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Does SRAM need PCB track length/impedance matching Well, for educational purposes I've decided to make an FPGA-based board with PCIe, HDMI and other high-speed stuff. I've usually used relatively slow (35-45 ns) SRAM and having traces routed in 'hit or miss' style works just fine. Now I've decided to use a few IS61WV204816BLL-10TLI 10 ns SRAM chips. Also, I want to use AS6C6416-55BIN SRAM as they are relatively compatible in pin-out. I have never routed fast parallel busses myself. In my head, I had a rule that everything above 50 MHz must be routed in some smart way. I've had a read of various articles and got out few things: Length and impedance matching are two very different things Length matching has meaning when you have fast rise/fall times Documentation must somewhere state need of length/impedance matching Each bus (data, address, control) should preferably be routed on its own layer. As I understand it, this is for better impedance. As far as I can see, the documentation of the chips I plan to use doesn't state the need for these techniques. But, out of interest, I did this: (power planes are omitted) All traces are matched within 1 mm.Furthermore, I found out that using 90 deg or 45 deg bends may cause reflections and it is better to use arcs. Does this look okay? So I have a bunch of questions: Have I made things worse? Should I just run straight lines? When exactly (overall frequency or edge time) do I need to start length matching? Is impedance matching necessary in my case? <Q> I don't think you have to length-match the traces. <S> In write mode, the critical timing parameter is from when the last (slowest) address or data line is valid, relative to the WE. <S> In read mode (OE controlled), the critical timing parameter of from when the last (slowest) address line is valid, relative to the OE. <S> By putting the meander line lengths in, you're just making all of the address or data bits have the same prop time as the slowest one. <S> Edit 1: <S> I would use controlled impedance for all the traces. <S> This is something most PWB vendor can do as a matter of course, meaning that it doesn't take much effort on their part and doesn't cost you anything. <A> I've had a read of various articles and got out few things: <S> Length and impedance matching are two very different things Length matching has meaning when you have fast rise/fall times Documentation must somewhere state need of length/impedance matching Each bus (data, address, control) should preferably be routed on its own layer. <S> As I understand it, this is for better impedance. <S> Correct Length matching has meaning when you have fast switching cycles / clock speeds. <S> This is sometimes mentioned in the datasheet, but not always. <S> There might be an application note or something about these requirements though. <S> Not necessarily. <S> It is recommended though to route the whole bus on a single layer, to achieve better matching (both length and impedance), but you can route several buses on the same layer. <S> So I have a bunch of questions: <S> Have I made things worse? <S> Should I just run straight lines? <S> When exactly (overall frequency or edge time) do I need to start length matching? <S> Is impedance matching necessary in my case? <S> Signal integrity point of view, you have not done it worse, it works the same as the straight lines. <S> Your design just takes some more space compared to the straight lines. <S> When the skew in the trace length starts to break setup and hold times. <S> The signal travels roughly 15 cm <S> /ns. <S> Most likely it will work just fine without matching. <S> With short traces the ringing caused by the mismatch is a minor problem. <S> In the datasheet it doesn't mention that there would be any termination or impedance matching in the component. <S> Thus if you want to avoid ringing (and the noise it might cause) you need to use series termination resistors that match the trace impedance. <S> Then you can match the resistors to what ever value the trace impedance happens to be. <S> This can be done by measuring the ringing and adjusting the resistor value until the ringing is minimized. <A> In my opinion, your design looks good with the exception of impedance matching. <S> I'm also assuming you do have ground reference planes stacked-up between signal layers. <S> Assuming you want to achieve full read/write speeds, the datasheet for this SRAM IC states: <S> Test conditions assume signal transition times of 1.5 ns or less <S> (Page 8: http://www.issi.com/WW/pdf/61-64WV204816BLL.pdf ) Taking that 1.5ns value and using a rule of thumb for lumped vs distributed circuit, you find that if all your traces were below 1.5inch (38mm), no length matching is required: <S> http://fullychargd.blogspot.com/2017/02/transmission-line-rules-of-thumb.html <S> I do not know the trace length (you haven't mentioned, they look pretty short) <S> however I do see that your board has the layers and space which makes me believe you did great by using this opportunity to length match. <S> Cornering reflection issues won't be a problem for <10G signals. <S> Take a look to this article (which is based on Dr. Howard Johnson. <S> results) <S> : https://resources.altium.com/pcb-design-blog/pcb-routing-angle-myths-45-degree-angle-versus-90-degree-angle <S> It is however preferred to remove the sharp corners to avoid creating acid traps, which you have done properly in your design. <S> For better signal integrity and system reliability, impedance matching is important, both at design and manufacturing level. <S> However, for a personal project, I would skip the manufacturing part (expensive) and only focus on sizing the trace width following your own stackup calculation (Saturn PCB or equivalent software will do). <S> Impedance matching within + <S> /-10% should give you good results and is commonly used in the industry. <S> As I mentioned above, it does seem like all your SRAM signals on all layers have the same width which probably means you haven't sized them yet. <S> I would highly recommend putting focus on this aspect before sending it out for fabrication.
| Fast rise/fall times alone doen't need length matching, they need impedance matching. If the requirement is roughly 1ns, you can tolerate 15 cm difference in the traces (well, it is not actually quite this simple calculation, but it should give you a ball park figure).
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How does a load decide the amount of current drawn in a circuit? I realise this question might seem similar to other questions asked on the site. But what I'm asking is actually completely different. Suppose I have a 5ohm resistive load hooked up to a 10v battery. The current drawn by the load would be equal to the current through the circuit, right?(assuming the load is connected in series). Using ohm's law, the current through the circuit would hence be 2A. Now,if I connect the same 5ohm resistive load to a 15V battery, the current through the circuit would be 3A, using ohm's law. So doesn't the same load draw different currents when the supply voltage changes? If that's the case, why do we say "current drawn depends on the type of load connected"? Moreover does the above reasoning of using ohm's law hold valid for any type of load? (i.e. L load, C load, RL load or a combination of all 3) Edit: this highly upvoted says that for a constant voltage supply, current drawn depends on load. So how does the load draw two different currents for two different voltages? <Q> why do we say "current drawn depends on the type of load connected <S> We actually say that the current drawn depends on the load AND the power supply voltage. <S> In other words, what we should say matches ohms law. <S> Moreover does the above reasoning of using ohm's law hold valid for any type of load? <S> i.e. L load, C load, RL load or a combination of all 3) <S> No, it becomes more complex with capacitors and inductors because the current drawn by a capacitor depends on the rate of change of applied voltage and not the absolute value of voltage. <S> For an inductor, the current depends on the integral of voltage with respect to time hence, inductor current is proportional to applied voltage x time. <S> That’s for DC circuits. <S> It’s another departure from ohms law for AC circuits and handling complex transients but the same rate of change (or integral) <S> principles apply. <A> If that's the case, why do we say "current drawn depends on the type of load connected"? <S> Usually we're dealing with a fixed / constant voltage supply. <S> Most common examples would be domestic or industrial mains voltages or 12 V automotive systems. <S> In both cases the voltage remains within certain tolerances so that means the other's have an inverse relationship, \$ <S> I = \frac { <S> V}{R} \$ or, since V is constant, \$ <S> I \propto \frac {1}{R} \$ . <S> Figure 1. <S> Each of these various bulbs, connected to the same supply, will present a different effective resistance and will draw a current inversely proportional to their resistance. <S> The higher the power consumption (watts) <S> the lower the effective 'resistance'. <S> (I use the word resistance cautiously here as LEDs are not resistances but <S> it's getting harder to find images of incandescent lamps!) <S> Image source: <S> Banggood. <S> Moreover does the above reasoning of using ohm's law hold valid for any type of load? <S> (i.e. L load, C load, RL load or a combination of all three). <S> No. <S> Ohm's law is specific to resistance. <S> We can extend the law to inductors and capacitors but we have to calculate impedances and use complex numbers. <S> Edit: this highly upvoted says that for a constant voltage supply , current drawn depends on load. <S> So how does the load draw two different currents for two different voltages? <S> Because you've changed the voltage. <S> There's no inconsistency there. <A> It depends both on supplied voltage but also type of load. <S> Then, there are devices that have linear regulators, so they internally convert everything to 5V for example so with a fixed load after regulator, they basically draw constant current regardless of input voltage. <S> Then there are devices with switch mode regulators so with a fixed load after regulator, they always draw constant power, so when voltage is increased, they draw less current. <A> All conductors , semiconductors, capacitors and inductors have resistance even regulated power supplies. <S> Regulated supplies amplify the error between desired output and actual but response time is bandwidth or slew rate limited. <S> Thus load regulation error appears in all regulator specs <S> It can be defined a number of ways. <S> 1st order estimate Zout= ΔV/ΔI[Ω] <S> above some minimal load to avoid other issues. <S> In Batteries and caps, it's called ESR. <S> 2nd order impedance is the storage capacitance which limits slew rate on errors. <S> On Batteries, they recover quickly from a brief short. <S> This memory effect is due to another effect from Double-charge layer effects which are modelled as another RC with bigger C and R that remembers what the brief short circuit that the discharge depleted. <S> Let's look at a simple model. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> I have discussed the models from memory effects of batteries elsewhere, if you are interested. <S> We use many different symbols for the internal resistance, Rs, Ri, ESR, \$r_{be}, <S> R_{ce}\$ and for inductors, DCR and for Zeners, Zzt. <S> Let's look at one of the cheapest LDO's ($0.05653 on 3k reel) <S> recall Zout= ΔV/ΔI[Ω] <S> here <S> Zout= ΔV_{out}/ΔI = 30 mV / 150 mA = <S> 200 <S> mΩ Notice errors include Vout tolerance over temp , Line REG and Load REG. <S> Also note the fast rise time in the spec. <S> This tells you slower loads may be regulated better due to high gain , but this has an internal rise time limit to respond, which is why external Low ESR caps are needed to reduce any source of high freq. <S> ripple.
| Resistive load draws current based on Ohm's law so it depends only on supplied voltage and the load resistance.
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Limiting the inrush current I have a transformer with several secondaries, and this is how I'm planing to control the inrush current: simulate this circuit – Schematic created using CircuitLab I'm using a resistor to limit the inrush current and MCU will energize the relays after a few seconds to bypass the resistors. Can I use a single resistor and relay on the primary to limit the inrush current instead of using a relay for each secondary? How can I activate the relays without an MCU, using only analog circuitry? <Q> Avoiding an MCU means having a small timer circuit that, when timed out (having been initially triggered by the instance of power being applied), activates the relay as you show in your diagram. <S> Or, you could avoid all the complication and use a negative temperature coefficient varistor/thermistor in series with your primary winding. <S> This is a pretty standard solution in many power supplies. <A> This might look like a good idea, but the EMI might be an issue and if the secondary cap fails, the relay will oscillate. <S> EMI is created by the Relay contacts (V=LdI/dt) at some voltage with series transformer inductance and load capacitance can also create very high voltage spikes. <S> There is no evidence or spec that defines that the transformer "caused" this surge. <S> In fact it is more likely that the 10mF Caps caused the surge. <S> If they had a low ESR of 10 mOhm then you can expect a secondary surge of Vout/0.01 ohms. <S> An PTC NTC / <S> aka ICL <S> can reduce the range of peak current, is also a smart choice operating at 85'C. <S> Look for ICL specs rated for your load cap of 10mF. <S> By <S> the way Transformers only cause surge currents in large units with Remanence during a power interruption and re-closure out-of-phase. <S> This causes the peak flux to add to the stored flux then exceed saturation levels. <S> Then XL(f)+DCR impedance drops due to significantly reduced Inductance. <S> Large MVA large cores may hum loudly until stabilized. <S> In this example the only surge is the low ESR caps causing rapid charge currents Vout/ESR that exponentially decay to the load current. <S> This is based on the peak pulse currents because the % ripple voltage also describes the current % duty cycle. <S> Therefore this can be a bad solution for EMI if not done carefully or a good solution for an Audio power Amp (if done right) as done in my old Technics 100W Rx with a 5 second timer. <S> The caps lasted 20 years,, which I considered OK, then I replaced all the e-caps on 1st sign of problems. <S> Typically a series Power R, bulk choke L and preload R were used. <S> But this is ancient technology. <S> Here simulated with low DCR ideal source (secondary) here with improvements <S> The more serious problem or annoyance is when the 10 mF wears out, the relay will be a loud buzzer and burn out in minutes as the ripple turns it on and off. <S> So I added a diode and cap to prevent that. <A> You asked in a different question (which you removed?) <S> about putting an NTC on the primary and bridging it with a relay. <S> If you care enough about light load power consumption since an NTC will waste power, have issues going from light to full load and voltage dip while your NTC is heating up or similar, my recommended solution is NTC + relay on the primary side. <S> As Andy said, inrush into your transformer is zero crossing insertion and transformer design dependent, but your NTC will handle that too if you put it on the primary. <S> The upside of an NTC over a straight resistor is in case your relay fails to open. <S> At tens of watts of load or higher, that resistor will not be possible to have in series during operation whereas an NTC will be, although with some power losses. <S> The downside of this arrangement, apart from the added cost, is that your relay will bridge the primary to secondary isolation provided by your transformer so you need to use an agency approved relay in respect to coil to contact isolation. <S> Plenty of them in the market to choose from. <S> Here is a simplified schematic: <S> simulate this circuit – <S> Schematic created using CircuitLab
| Putting inrush protection on a transformer primary is a pretty standard way of doing things because it also handles primary magnetization current inrush problems too.
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Question about Vbe and operation regions of BJT Updated/Corrected question: Realizing what confuses me the most, I will make this more straight-forward. Say that a transistor works in the forward active region and it receives an ac input signal at its base. Is there a chance that the change in the base voltage will decrease Vbe and force the transistor into the cut-off region? Is there any chance this might happen before first entering the saturation region? Original post: I'm soon taking an exam on transistors and I've just started my revision. After having a look at the conditions for each region of operation and some example problems, there is something I don't understand or that maybe I don't remember very well. The condition for an npn bjt to be on is that the base emitter voltage is above 0.7 volts. From what I remember this is not an accurate value but let's say it is around there. When we analyse the small signal model of a circuit , the values we calculate are superimposed on the dc values, right? So, when we apply an ac signal to the base of the bjt, the base emitter voltage changes. So, my questions are : Is there a chance that the ac input signal will take the bjt in the cut off region? If yes, how do we check it? If yes, when deciding on the swing of the output voltage are we more worried about entering saturation or cut-off region? I mean, which comes first as the swing increases? Does it depend on the operation point? <Q> Is there a chance that the change in the base voltage will decrease Vbe and force the transistor into the cut-off region? <S> Yes this is possible, and it would result in non-linear operation of your amplifier. <S> Is there any chance this might happen before first entering the saturation region? <S> Yes it's possible to (mis-)design an amplifier <S> so this happens. <S> when deciding on the swing of the output voltage are we more worried about entering saturation or cut-off region? <S> You need to consider both when designing your amplifier. <S> Assuming we're talking about a common emitter amplifier, you need to be sure the amplifier can handle the complete range of input amplitudes desired without going into cut-off, and you need to be sure it can handle the complete range of output amplitudes without entering saturation. <A> I guess this question is more about vocabulary than about BJTs. <S> "AC small signal analysis" or "small signal conditions" is a useful approximation to simplify the analysis of a circuit. <S> This approximation assumes the AC signal voltage wiggles around the DC operating point in a "small enough" interval that the operating point and all characteristics of circuit devices can be assumed to be constant. <S> This allows to approximate all non-linear devices (like transistors) with linear equations. <S> For example, in small signal analysis, a diode is approximated by a resistor, and the value of this resistor depends on the DC current flowing in the diode. <S> So, my questions are : Is there a chance that the ac input signal will take the bjt in the cut off region? <S> No... <S> and yes <S> In a real circuit, yes of course. <S> In small signal analysis, no. <S> If the "small signal" is large enough to cause a significant change in device operating parameters, like your example of turning the transistor on and off, then the small signal approximation is no longer valid ... and this is no longer small signal analysis, but large signal analysis. <S> Small signal analysis replaces the system under study with a linear approximation. <S> This is 100% correct if the AC signal is infinitely small, and will give correct results for gain, transfer function, impedance, etc. <S> The accuracy of the approximation breaks down progressively as signal level increases, which manifest as distortion (for example the \$ <S> g_m \$ of your transistor will vary a bit during the voltage swing). <S> When the signal is strong enough to make devices switch on/off, the approximation breaks down completely. <S> If yes, how do we check it? <S> First do the small signal analysis. <S> Then calculate the actual voltage swings and check by how much the operating point of your transistor varies or if will turn off or saturate. <A> Some amplifiers are even designed to have the signal always do this, these are called class B or class C amplifiers. <S> Amplifiers that have the singal always be in the linear region are called class A <S> To know whether your amplifier is operating as a class A, B or C amplifier, you need to calculate whether the input signal causes the base-emitter voltage to fall into the cutoff region, to do this you need to know the base-emitter bias current, the input impedance, and the source current (or voltage) and impedance. <S> You always design your amplifier so that it does not enter the saturation region, as long as the signal amplitude is within the operating range Going into the saturation region means having a forward biased base-collector junction, and in this region the transistor stops amplifying the signal and just conducts large currents from base to emitter and collector.
| Yes the AC signal superimposed on the DC bias voltage can take the base-emitter voltage into the cutoff region
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Difference between the following two very basic circuits What is the difference between the below two circuits? I know the bottom circuit has 'ground', but what does that mean exactly? Why would someone want to construct the second circuit over the first? And, what would be an example of doing the bottom on a breadboard -- where does ground 'go'? Finally, when doing the above in EveryCircuit, the first example doesn't even work, or allow current to flow through it. Why is this? (The same circuit works on my breadboard). <Q> There is no functional difference in reality. <S> However, your simulation software is stupid and needs to do calculations, so you need to tell it which node you want to represent zero and will become the reference to which every other node is measured (remember that voltage is a potential DIFFERENCE). <S> The circuit with a GND connection has designated that node as zero volts, he node with which all other nodes are measured with respect to. <S> Note that physically making that GND connection to something else in reality doesn't make your circuit operate any differently, but does make it so your circuit doesn't "float" if you are using a battery source or some other isolated source. <S> It anchors your circuit to a known potential and not doing so can become an issue for things like noise, shielding, EMI, and safety (arcing for high voltage things). <A> It does not usually imply a connection to the earth - we can talk about Ground in portable equipment, or in battery-powered circuits on a plastic breadboard. <S> The Ground symbol has no effect on the operation of the circuit, but is usually required by circuit simulator programs so that they know what point to use as a reference when calculating voltages elsewhere. <S> In AC power wiring and some radio antenna systems "Ground" (particularly the green "Safety Ground") does indicate an actual connection to the earth. <A> But the position of ground has significant role in the circuits. <S> The ground in a electrical circuit is the node where we have our "zero volts". <S> Pre-Information: A battery is something that maintains a certain potential difference across its terminals depending upon its rated capacity. <S> A 1.5V battery keeps its +ve end at 1.5V higher potential than its -ve end. <S> If I have 0V at its -ve end, I have 1.5V at its positive end. <S> If I have 10V at its -ve end, I have 11.5V at its +ve end <S> and so on. <S> Notice how the +ve end is always higher than -ve end. <S> Keeping this in mind consider two batteries connected to each others as shown; simulate this circuit – <S> Schematic created using CircuitLab <S> In first figure, the ground is connected to -ve terminal of 1V battery and +ve terminal of 2V battery. <S> Since we refer to ground as our 0V, the -ve end of V1 and +ve end of V2 is 0V. <S> So, you get the + <S> ve end of V1 to be 1V greater than -ve end of V1. <S> So, you have 1V at node A. <S> In case of V2, the +ve end is at 0V. <S> So the -ve side must be 2V lower than the +ve end. <S> So, you get -2V <S> at node B. <S> In case of second figure, we have the -ve side of V2 grounded(0 volts). <S> So the +ve side of V2 is 2V higher than the -ve side. <S> Meaning 2V at node <S> C. Also, the -ve end of V1 is connected to + <S> ve end of V2(which is at 2V). <S> Since V1 is 1V, it should maintain its +ve end at 1V higher potential than -ve end. <S> This implies we have 3V at node A. <S> This is how you can see the difference between your two circuits. <S> The ground and -ve end of battery may or may not be at same node. <S> The ground can be placed at +ve end of battery as well. <S> But this gives -ve voltage at your -ve end. <S> Hope this helps!! <A> Many simulators compute all voltages relative to some "ground" node, and require that some node have a voltage which is set relative to ground. <S> To accommodate schematics where some groups of nodes are "floating" relative to ground, however, simulators will can often identify islands that don't have any ground connection and, if any art found, select some arbitrary node in each island and set its voltage to 0 relative to ground. <S> In the upper schematic, if you were to attach scope probes to the ends of the resistor, they might report readings of +1.0V and 0.0V, or 0.0V and -1.0V, or possibly (though I've never seen such behavior) <S> +0.5V and -0.5V. <S> In the second schematic, however, the voltages would definitely report +1.0V and 0.0V. <S> In the third schematic, the upper portion's voltages are all measured relative to ground, but the lower island's voltages are not. <S> Thus, a simulator would treat the lower island like it treated the second schematic, adding a "ground" node connection. <A> There is no functional difference inside the circuit , the same 1V potential will take effect on the specified resistor, but there can be a big difference from an external point of view . <S> In the 1st case, your 1V difference can be generated by multiple methods:Having a +1V and 0V, having a 0V and -1V, having a +0.5V and -0.5V or any values that give the difference of 1V. <S> Although not largely in use, some circuit boards today still use +12 and -12V to obtain a 24V potential. <S> In the 2nd case, you can only have a +1V and 0V (the specified grounding).
| Here, as in most electronics, "Ground" simply marks the point in the circuit that we want to call "Zero Volts", and use as a reference when measuring voltages elsewhere in the circuit. There is no any difference between these two circuits in this case.
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How does the magnetic field in a servo motor acts according to the PWM signal? I have read about servo motors but have problem grasping the main idea of obtaining fixed position by the magnetic field. The servo motor position is said to be controlled by PWM signal. And a toy DC motor speed can be controlled by PWM. I'm mostly wondering the electro-mechanical part rather than the feedback. So in toy DC motor case the PWM controls the speed which I can make sense. But in servo motor case how come the PWM signal causes a non moving magnetic field? I guess servo motor works like a compass where the magnetic field is fixed(non-moving), is that correct? If so, is there a way to grasp the how the PWM causes such a magnetic field? <Q> Most small servos don't use magnetic fields for position sensing, they use potentiometers. <S> Here's a good beginner's level explanation in text with a video link. <S> They all seem to have the same basic components – a potentiometer hooked to a voltage regulator and one shot generator, which converts position of the armature to a PWM pulse. <S> This PWM pulse is compared to the original one sent by the microcontroller. <S> This logic board finds the difference between the pulses, which is called the error. <S> The magnitude of the error is sent to a pulse stretcher and the direction of the error is send to a flip flop to be stored as a high or low. <S> The magnitude of the error is stretched out by the pulse stretcher. <S> Then both parts of the error are sent to the output driver, which is probably an h-bridge. <S> Most servos are 50HZ, which means this control loop is happening 50 times every second until the error is “zero”. <S> The dead band for the pulse stretcher sets a minimum pulse length that it will stretch. <S> Anything below this limit is considered zero error. <S> (Not to be confused with: stepper motors, or the use of EMF to sense position in BLDC motors, or situations where Hall sensors are used to measure shaft position. <S> All of those are not usually called "servo" motors.) <A> You might need to make a distinction between 'power' PWM, and 'signal' PWM. <S> To command a typical servo, we use a PWM signal which has a high pulse of 1mS full left, 1.5mS centre, 2mS full right, which most servoes will handle being repeated at somewhere between 20Hz and 100Hz. <S> These times are interpretted by the control electronics as a position, and the motor is driven to the commanded position. <S> To control the effective voltage, and therefore speed, of a DC motor, we might use power PWM, which chops the input voltage between supply and 0, to give an effective voltage at the motor of the time weighted mean. <A> The PWM you send the servo motor isn't the same as the PWM you send the motor itself. <S> The one you send is just a command signal. <S> It could just as easily be an analog signal voltage or serial message. <S> The actual drive signal containing power which is sent to the internal motor can do whatever it wants. <S> Electronics in between can change it to reverse direction as required. <S> The motor in most servos is also brushed so the magnetic field rotates, but in discrete steps, not continuously. <S> So the motor could easily be in a situation where the winding is energized but between commutation points, so the magnetic field isn't rotating and is just applying a force to the shaft and the shaft is static because it is being opposed by an equal load torque. <S> Conversely, if it is near a commutation point, combined with gearing, the servo motor output shaft could remain relatively still while the inner motor shaft dances around a position (being commanded by the electronics internal to constantly reverse direction to do so) to hold that position.
| The PWM signal is not aimed at the motor: it's aimed at the control circuit inside the servo, which then chooses whether to turn the motor clockwise or anticlockwise if the PWM signal differs from the position measured by the potentiometer.
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How can I make a mod for VGA cable to display only grayscale or Black & White? My computer connects to the monitor using a VGA cable. I want to turn the display grayscale or black & white. I don't want to use any software solutions as I fear it will not be as effective as I want it to be. So I researched the Internet for RFC of VGA pin and cable and I couldn't find such an RFC and pin layout configuration was not helpful. Is there a way I can build a mod which I can attach to either the output of the computer or to the input of the monitor, which discards the color information and display either a grayscale image or black & white image? If possible I would also like the mod to throttle the refresh rate to either 24Hz or 30Hz. It has to display a resolution of 1920 x 1080. <Q> You do not need an FPGA. <S> As you can see there are three separate signals. <S> You could combine them to a single "white" signal and then re-connect that to the R/G/B pins, following this example: VGA to composite circuit. <S> Does it work? <S> Changing the refresh rate is more complicated since that's entirely under the control of the sender. <S> It's based on EDID . <S> What you could do is disconnect the I2C lines from the target monitor and route them to an EEPROM or microcontroller of your own, which would then send back only the data for those video modes that are in 30Hz. <S> 24Hz seems rather too low unless it's interlaced? <A> Simply connect R, G and B lines on cable together (pins 1, 2 and 3). <S> The matching resistors on VGA card output will do the mixing. <S> Any FPGAs, opamps etc. are overengineering. <A> Simple analog summing circuit with video op-amps should work fine for this, when frame rate conversion is not needed. <S> If frame rate conversion is needed then this must be done in digital domain with DRAM frame buffer, but most likely no display will eat 1080p24 or 1080p30 signal from analog input. <S> But for a one-off hobby project, a single FPGA can be easier to get.
| The pinout of the VGA cable can be found in lots of places. If doing this in digital domain, video processor ASICs are cheaper when mass producing these and easier to get it working, and you also need video ADCs and DACs.
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What does it mean when my car inverter's labeling says "Neutral Floating"? Is it safe to use? I recently came across a fairly old car inverter. It plugs into a 12v DC lighter socket, and outputs 115 Vac at 60 Hz, and has a normal American outlet on the side of it (with a ground pin, which seems questionable) and is rated for 100W continuous or 150W peak. This tool could be very useful to me, mainly to run a laptop in the car (when someone else is driving, of course). The output is also non-sinusoidal, but, as far as I understand, that does not matter for switch mode power supplies, and all of the things that I want to power in the car run on switch mode power supplies. More concerningly, however, it says "NEUTRAL FLOATING." My understanding of electrical engineering tells me that this means that the neutral wire isn't attached to anything, except maybe a capacitor. This may still work with lower current loads (perhaps up to 100W based on the labeling, although that seems high) due to the capacitance on the neutral wire, but it would also cause the device that I am using to start floating above 0V on its neutral or ground side, a dangerous situation. It seems, however, that such a dangerous device would not be legal to sell in the United States, where I am pretty sure that this was bought (it is very old, so I don't know exactly). As such, it seems like I am misunderstanding what "NEUTRAL FLOATING" means. What does it actually mean? Is it safe to use my inverter, or do I need to buy another one? <Q> Many smaller inverters (and my Honda EU2000 generator) have "floating neutrals". <S> I don't think "floating" is quite the right term here, as the Neutral will alternately be connected to +170 V or Ground. <S> What happens is that the inverter has a single high voltage supply (+170 V or so), and connects the Line to that supply on one half-cycle, and Neutral to that supply on the other half-cycle of the AC waveform. <S> A load connected between Line and Neutral will see an alternating voltage. <S> As neither Line or Neutral should be exposed to the user, this technique should not present a hazard, and simplifies the inverter as it does not need to generate a negative high voltage supply. <A> It means that the output 115VAC is not connected to the input power in any way (not common with the minus, which is connected to the car chassis, nor to the +12V input). <S> Similar to adding an isolation transformer to your mains that comes out of the wall (well, without the pesky design, safety testing and certification ... ). <S> That's a fairly subtle difference with an inverter or a generator.. <S> it's generally likely safer because getting your body between the "hot" and car chassis ought not to give you a shock ( <S> though I would not suggest testing this given the typical slapdash construction of cheap inverters), however it might have some (non-destructive) ill effect on the operation of devices that use capacitive touch sensors for operation such as phones and computers with touch screens. <S> Possibly they have incorporated a capacitor to AC-ground the output, or maybe not. <S> If one is added it should be a Y-rated safety capacitor. <S> The receptacle ground pin is likely connected to the (-) lead (car chassis). <S> It is not connected to the neutral. <A> On a car, ground would be the chassis. <S> The ground pin might be connected to the car chassis, or it might be totally disconnected. <S> If the appliance is double insulated, with a 2-pin plug, it's largely irrelevant anyway. <S> Even if the appliance has a 3-pin plug, then the lack of a connection between the 115V neutral and the vehicle ground doesn't mean that you will be electrocuted if you touch the appliance and the car bodywork at the same time - because there's no connection between them.
| What it means is that the neutral output of the inverter is not connected to ground. If you are just plugging one appliance in, it isn't necessarily unsafe.
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Grounding for a battery Given a 9V battery as the only voltage source for a circuit, where is the Ground usually put? Is there a standard, for example, that puts the negative terminal at 0 and the positive at +9? Does it every vary, or are there usually hard-set rules for grounding? <Q> You can put ground wherever you want; the grounded points in your circuit will then all be at 0V (by definition/convention), and, theoretically, they will all look as the same node (ie, no impedance whatsoever between them).Look at it this way: <S> Batteries, and other voltage sources, only set a voltage difference between their terminals, which means that ungrounded circuits have an indeterminate voltage at every point, with only voltage differences being defined. <S> So hooking some point to ground is simply a matter of fixing an origin for the scale. <A> In most circuits these days, "Ground/Reference" is the negative terminal of the power supply, but in some cases it might be the positive terminal. <S> (Old 6 Volt cars were often "positive Ground", but negative ground is common now.) <S> In many analog circuits, "Ground/Reference" is the center point of the power supply, so you can have both positive and negative voltages in the circuit. <A> Is there a standard? <S> Not a universal one writing... <S> but by popularity positive grounds and coaxial center negative have faded fast in the last 2 decades, due to user error. <S> Since ground only means a 0V reference, it does not matter locally, only when interacing. <S> Zero Volts, 0V is always floating , unless connected to anything that is connected grid earth or PE, Protective Earth Gnd, which, coincidentally is also floating ;) in the Universe, but locally , it is our biggest common-mode 0V reference. <S> Yet with trans-atlantic cables there can be a thousand volts difference. <S> E.g. VGA monitors are PE grounded. <S> Like all voltage sources, even 0V has impedance so ground shift voltage depends in V(f)=I(f)*Z(f). <S> It has both differential mode impedance (Zdm) and common mode (Zcm). <S> So floating means at DC high R but at some frequency that Z can drop low by conductor length/gap ratio due to inherent capacitance. <S> Grid tower grounds need to be < 100 Ohms for lightning protection. <S> But grid to home much less, so it is distributed.
| "Ground" in most circuits is just the point we want to call "Zero Volts" and use as a reference when measuring voltages elsewhere in the circuit.
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Driving a 74HC245 in one direction with 3.3V when Vcc is 5V I have a 5V device that accepts parallel data in. That data is buffered by a 74HC245 IC ( TI datasheet , NXP datasheet ) where the Dir pin is hard wired to drive in one direction. I can't change/modify this device hardware-wise. To date I have been driving this device using an ATMega328P, running at 5V, with no problems. I now have to switch to a 3.3V microcontroller for other reasons out of my control. I am hoping that I don't have to add logic level adjustment circuitry in-between the microcontroller PCB and the device being controlled. Reading the datasheet of the 74HC245, the input high threshold is 2.4V(typ.) to 3.15V when Vcc is 4.5V. Being that the 74HC245 IC is supplied by 5V, I am concerned that 3.3V logic out of the microcontroller is going to be right on the limit. Will I be able to get away without any logic level adjustment circuitry? <Q> There are 74HC devices and there are 74HCT devices. <S> The HCT device works with TTL levels at the input: <S> - Guess what the "T" might stand for. <S> If you can swap-out the HC for a HCT you'll be good to run. <A> This will not work reliably. <S> If you look at the TI datasheet you will see that \$V_{IH}\$ is specified at \$3.15\$ V when \$V_{CC} = 4.5\$ V, and it is clearly a strong function of \$V_{CC}\$ . <S> If we assume that \$V_{IH}\$ scales linearly then it would be \$3.675\$ V when your power supply voltage is \$5.25\$ V, which could be the case if your voltage regulator tolerance is 5%. <S> In fact, this arrangement won't work is the supply voltage is above about \$4.7\$ V. <S> Furthermore, if the supply voltage for the microcontroller is \$3.3\$ V <S> that doesn't mean that the logic signals will be right at \$3.3\$ <S> V. Check <S> \$V_{OH}\$ for the microcontroller. <A> That particular one has a 3.6V limit on the B side so you'd have to flip it and invert the DIR line, but there are others such as 74LVC8T245 that will work from 1.2V to 5.5V on either side. <S> Many voltage translating types can avoid certain problems with power domain sequencing. <S> With the HCT option you have to ensure the 3.3V supply is not present when the 5V supply is less than about 3V in order to stay within maximum limits. <S> From the 74LVC8T245 datasheet : <S> The devices are fully specified for partial power-down applications using IOFF. <S> The IOFF circuitry disables the output, preventing any damaging backflow current through the device when it is powered down. <S> In suspend mode when either VCC(A) or VCC(B) are at GND level, both A port and B port are in the high-impedance OFF-state. <S> The performance is generally higher as well, being a newer logic family. <A> Maybe the following is the solution for you. <S> I just found this when I searched for a level shifter. <S> And after I found that I found your question. <S> That is a commercial product from a reputable Thai electronic component supplier. <S> I didn't try it myself. <S> It seems you just have connect DIR to 3.3V. Problem solved. <S> This is just one circuit. <S> In the download below you will find similar with 5V and 3.3V. <S> https://www.es.co.th/detail.asp?prod=005904425 Sorry, it's all in Thai but the circuit works in any language. <S> You can download the data sheet with more info on that page. <S> I just looked a little on the internet. <S> Here is the ready build circuit with English description. <S> https://www.futurlec.com/Mini_Logic.shtml
| As an alternative to using the HCT chip, you could consider something like the 74LVC4245A, which is an octal transceiver with voltage translation.
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Connect a Switchboard to Arduino Sorry if this is a simple question, but I‘m quite new to Arduino. What I‘m looking for is a way to connect 100+ strain gauges to one Arduino board. Now, those each need a separate ADC (HX711) to convert the input from snalog to digital (best would be if only 1 was needed)(Also, I use 3 resistors to make it a quarter-bridge, so only 1 strain gauge per HX711 is needed.) What I thought would be a good way, is to connect the HX711 to the Arduino like normal, but also connect the Arduino and the HX711 to something like a switchboard, in which the Arduino tells it to which setup to connect to, and the switchboard then changes the connections (pretty much like a transitor.) Not sure if I‘m clear here, so something like this: (This - means connected to) Arduino - HX711 Arduino - Switch HX711 - Switch Switchboard - StrainGaugeQB [1,2,3,4...100] (All 100+ quarter bridges = 400 wires.) Something like this: Basically, the Arduino switches between which strain gauge is to be read currently. Although they should at the end still each be read at least 4-5 times a second. I read something about multiplexers, but they don‘t seem to be exactly what I‘m looking for. They seem to be only serial, or did I get that wrong? They only have one in/out connected, but if it‘s not serial, could I maybe use 4 or those each getting one of the wires of the HX711? Something like multiplexer 1 being connected to wire e+ of multiple quarterbridges, 2 being e- of those, 3/4 being connected to their a+/a-? <Q> If I understand correctly: You would want 4 on/off switches in parallel in front of each signal line from the strain guage. <S> Each of the switches would be wired to one of the four quadrants of the wheatstone bridge. <S> Only one (or none) of the 4 switches would ever be on. <S> Repeat this for all strain gauges. <S> Then you would just have to be careful what to only turn on 4 load cells (in general) for each quadrant of the load cell. <S> You would also have to calibrate every combination. <S> Something along the lines of this: https://www.analog.com/media/en/technical-documentation/data-sheets/ADG1633_1634.pdf or similar to do the switching. <S> I have no experience with these chips, so maybe someone else can answer better. <A> i use 3 resistors to make it a quarter-bridge <S> I don't see your wiring in the schematics, but if what you are doing is what I think, then you basically killing the performance of the strain gauge. <S> Not a good idea. <S> You need to switch all four lines. <S> Also, i read something about multiplexers, but they don‘t seem to be exactly what i‘m looking for You were reading about wrong multiplexers. <S> What you are looking for is called analog switch , and it is designed exactly for switching analog signals. <S> Some of them controlled by binary address lines (e.g. ADG726 ), others have internal shift register (e.g. ADG725 ). <S> In first case you need more free control lines but can randomly access any specific gauge. <S> In the second case you save control lines but an access is sequential. <S> The problem, however, is that analog muxes will affect the signals somewhat, reducing precision and linearity of your measurements. <S> So, it all depends on your requirements. <S> I think the best solution would be to use multiple HX711 permanently connected to their gauges . <S> Since you are talking about hundreds of gauges, they cannot possibly be located in close proximity to the control board. <S> Running analog signals all the way from the gauges to your "switchboard" will be another way to look for troubles. <S> By placing HX711 next to each gauge you will be shifting wiring into digital domain, greatly improving resolution and simplifying switching. <S> In this case you can still use analog muxes mentioned above but since you only be switching 2 lines you need only half of chips and wiring is easier. <S> UPDATE: <S> You've commented that you are going to look into ADG725. <S> Unfortunately, ADG725 does not have carry output, so you'd need additional lines to select groups of MUXes. <S> Using directly addressable ADG726 MUX might be more convenient. <S> Connect clock pin to 7-bit counter (74HC4024), connect first 5 bits of counter to address inputs of all muxes, then use decoder chip (74HC138D) to convert next two bits into 4 enable lines for groups of muxes. <S> This will enable you to switch up to 128 gauges with only 2 pins (3 if you count optional "clear" signal). <A> You could use 100+ sets of 2 DPDT relays, with 3 signals+power+Gnd from the Arduino to shift data into TPIC6B595 shift registers to sink the relay's coil current, one pair at a time. <S> The ADC is then connected to one H-bridge at a time. <S> You can use 5V relays, telecom relays need 31mA or less to energize them from 5V.Examples: <S> https://www.digikey.com/products/en/relays/signal-relays-up-to-2-amps/189?k=relay&k=&pkeyword=relay&sv=0&pv1989=0&pv72=208767&pv1409=340465&sf=0&FV=-8%7C189&quantity=&ColumnSort=0&page=1&stock=1&pageSize=25 <S> Each TPIC6B595 can control 8 relay sets, so you'd need a string of 100/8 = 13 shift registers, with SPI.transfer() used to send in 13 bytes to the daisy chained shift registers. <S> I've daisy chained 45 of them and updated them at a 20 KHz rate, so 13 will not be an issue. <S> At the worst, maybe a buffer for SCK and the latch signal, both of which go to all chips. <S> digitalWrite <S> (ssPin, LOW);for (x = 0; x < 14; x = x+1){SPI.transfer (dataArray[x]);}digitalWrite (ssPin, HIGH); dataArray[] is 13 bytes long, start with it filled with 0s, = no relays energized. <S> Then march a 1 across the array to energize one relay at a time: 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x01 <S> 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x02 <S> 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x04 <S> 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x08 <S> 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x10 <S> 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x20 <S> 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x40 <S> 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, <S> 0x80 <S> etc. <S> up to 0x40, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00 0x80, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00 for the last two relays (assuming 104 relays, 8 x 13).
| For better precision you could use a hundred 4PST relays, but those are more expensive than HX711 modules and need additional switching circuits.
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What’s the difference between led current draw with and without PWM? I have a LED that’s rated for 3W (aprox 3V and 1A max current) Do I get the same result in terms of current draw and led’s durability if I supply the led with a 1A Led Driver (at full current supply, or full PWM signal) or if I supply the led with a 2A led driver, but only allow to the led half the PWM max current? Attached is the datasheet for the Cree XP-E neutral white led I’m using. XPE-Datasheet I found this document about over-current pulsed driving leds of this manufacturer: Cree over-current pulsed driving leds I’m also using a constant led driver, but it only provides 500mA. I have another led driver that is set for more than 1A. That’s why I thought of using this one but in a lower PWM setting. Thanks for reading! <Q> A led had usually two kind of max current specified in its datasheet: Pulsed and DC operation. <S> Usually, max pulsed current can be much (e.g. 5-10 times) higher than max. <S> DC current. <S> If you consider only the current, you will have the same result if you operate 1A at 100% or 2A at 50% (Both have 1A mean current). <S> But because the forward voltage of the LED is higher at 2A, you will generate more heat losses at 2A/50%. <S> It could be that you have more brightness at 2A/50%, because the current to brightness curve of a led is not linear. <A> LED current draw with PWM has ripple that averages out to a value but instantaneously might have higher and lower voltage/current values. <S> For example, would you rather have the temperature be 25C? <S> Or alternate between 0C and 50C? <S> They both average to be 25C after all. <S> But if your LED can only handle 1A and gets 2A at 100% and try to PWM at 50% to stay at 1A average, then you are hammering the LED at 200% what it can take half the time. <S> Not to say you "overdriving" will necessarily destroy the LED. <S> LEDs can handle high power bursts for short periods of time as long as it kept under control, but you have to know what you're doing. <S> It does shorten their life though. <S> You might do this with something like a pulsed laser diode where you need really high power pulses and it's too expensive to just use a bigger diode, but really not worth it to overdrive an LED that is just a visible indicator. <S> This is all if your driver has no filtering (you're just PWMing the LED to control brightness, not to run something like a constant current driver which has filtering to smooth things out <S> so it's less peak and the actual currents more closely resemble the average current. <A> The specific LED you have mentioned does not provide any indication in its datasheet that it can be pulsed at a current higher than 1000mA. <S> So, if you choose to operate the LED using PWM at 2A you are using the part beyond the manufacturer's specified maximum ratings. <S> It might work, it might fail immediately, or it might fail next week.
| If your LED handles 1A and gets 1A at 100% PWM, then PWMing lets you reduce the visual brightness (to the human eye anyways) more efficiently than using a resistor because you are dissipating less heat.
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Spikes on a square wave of a oscilloscope reading This question is not really a problem i am having but a thought that want to get off my mind. Usually on a reading on a oscilloscope especially on a square wave and sometime triangle waves, there are these spikes or ringing. See sample images from google below I too have sometimes observed this when i am using an oscilloscope but i pay it not mind. From what i read it would seem that this is sometimes caused by bad probing. It got me thinking is this a oscilloscope specific phenomenon? if its caused by probing that would mean that whats actually happening on those lines are very clean signal (maybe not as steep voltage change but there is no ringing) <Q> Picture it this way <S> ...the wire from the driver to the receiver has intrinsic inductance, and most digital inputs have [gate] capacitance. <S> Once the output starts driving the line, current builds up in the (inductive) trace until the capacitance fills up to the same voltage as the driver. <S> There's still a current flowing through that inductance, though, so it continues pumping the capacitor higher until the current stops... <S> now the voltage drives the current back to the driver. <S> That's the first peak, and the next time the current stops is the first valley. <S> Back and forth until resistance dissipates the stored energy. <S> Of course, the trace is actually a transmission line with distributed inductance AND capacitance, as well as a speed of light limitation, but that's about half a semester worth. <S> This can affect the signal integrity of the circuit when the ringing is large enough, as well as causing permanent damage to ICs over months or years of repetitive overvoltage/undervoltage. <S> Of course, the scope probe becomes a part of the circuit when applied, and can change the behaviour of the node. <A> It is certainly not specific to an oscilloscope measurement. <S> Fast edges relative to transmission line length and any unterminated or poorly terminated line results in (real) ringing at a frequency related to the length of the line. <S> Rule of thumb is that the unterminated length in FR4 is about 1.5 * period of oscillation (in ns), so your relatively long ~13ns period represents about 20" of trace or somewhat more length in a badly-terminated coax cable. <S> A bit long for a PCB trace, but not far off a scope probe cable length. <A> This is classic Ground wire inductance MEASUREMENT ERROR. <S> (0.5nH/mm and probe custom low C coax capacitance ~ <S> 60pF/m) <S> (LC resonance > 20MHz) Solution . <S> Search this site for proper 10:1 scope probe RF measurement methods with tip & ring only or terminated Coax with R divider to 50 Ohms. <S> repeat experiment using above methods until you get textbook quality results with calibration square wave on DSO. <S> you do the search when that is perfect and your experiment rings. <S> Fix ALL the signal paths using coax or twisted pairs .. Or Ignore error and use Scope 20 MHz filter. <S> I see no one was motivated to appreciate this. <S> http://www.cbtricks.com/miscellaneous/tech_publications/scope/abcs_of_probes.pdf <S> What you expect .... <S> But what you actually have that causes a 2nd order overshoot problem. <S> One way to improve response, either use an other R divider on board or direct to 10:1 probe with tip & ground clip removed. <S> Or more convenient with nearby signal and short ground track. <S> You may accept 3% overshoot within 30ns as a good measurement.
| That's called ringing, and it's an effect of inductance and capacitance in the circuit (often just parasitic).
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Best way to drop 10volts at 0.25A with linear regulator? What would be a minimalistic circuit design to convert 14V to 3.3V at 0.25A with a linear regulator? I'm thinking about two linear regulators. I've tried LM7805(14V-5V) to LD1117(5V-3.3V.) LM7805 gets hot even at 0.1A. What would be best solution without switching power supply? I'm trying to power efficiently ESP32 module from a car's battery. <Q> you're planning to burn up 2.5 watts <S> a TO220 part is going to need a heatsink at that power level, but you can put a resistor in series with the input and make the heat there where it's not damanging anything simulate this circuit – <S> Schematic created using CircuitLab at low current the reguilator sees near full voltage, but the current is low <S> so there's not much heat made. <S> at 250mA R1 is dropping about 8V so the regulator is only working with 6V in and so making less that 1W of heat <A> Best? <S> no , Simple Yes Get a junkyard Festoon lamp holder but instead of a 12V bulb, buy a 24V 10W bulb designed for 250mA constant current with just enough voltage drop to run any 3.3V LDO and only dump < 0.5W in the LDO and 2.2W out of 10W in the Lamp, so it will be dim. <S> If the bulb is 10W rated at 24V it will be 3 Ohms cold and if rated at 28V, it will be 4 Ohms cold and you may need to bypass it with two 470 Ohm resistors in parallel across bulb. <A> In a word: don't The best way to drop 10 volts at 0.25A is already answered by other people, so I will not cover that. <S> Your question, however, aims at powering an ESP32 module specifically, and you don't disclose more details about the intended usage. <S> However, if you are targeting an IoT scenario with WiFi communication, you need to consider that the ESP32 does <S> NOT consume 0.25A. <S> It's power usage varies with the distance to the AP, and the required increase of transmit power when RSSI falls down due to distance. <S> ESP32 supplies are typically rated for 500 or 600mA, for this reason. <S> Personally the highest I've seen is 450mA, but your mileage may vary, and you may have other things on board, so... do the math. <S> You can go with 12 to 3.3V supply as well, but the 5V supply gives you the option to add a Li-Ion battery later, and the 5V supplies are more common. <A> I'd use a circuit similar to one given in Tony's answer . <S> simulate this circuit – <S> Schematic created using CircuitLab <S> \$D1\$ <S> (ES2J - but can be anything suitable ) protects the circuit against reverse polarity. <S> Power rating of both \$R2\$ (470R) and \$D1\$ <S> (3V9 zener) should be at least 1W because their power dissipations increase under no- or light-load situations. <S> \$R3\$ <S> (1k) ensures the collector current of \$Q1\$ to be non-zero and thus the unloaded voltage to be 3.3VDC. <S> \$DROPPER\$ can be a resistor with enough wattage, or a 12V car lamp or even an ATX chassis fan .
| If I were you, I'd use a cheap car USB supply (12-to-5V) with 600mA rating or more (which likely features a switching regulator inside), and use a LDO to drop the remaining voltage to 3.3V.
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pulsed current with 200 kHz squarewave signal I use this circuit to obtain pulses as close as possible to the input signal pulses on the 1.5 Ohms resistor. The MOSFET is FPQ50N06 and the transistor is BC239. The circuit gives good results on 25 kHz, 4.8 Vpp and 1.5 V offset. However, I need 200 kHz, 4.8 Vpp and 1.5 V offset. How can I make it work when the signal is 200kHz? (I connect oscilloscope probes shown on the scheme and I exactly need on channel2 (CH2) the pulses shown on the other image which is an oscilloscope screen. ) <Q> You have a massive power MOSFET with a lot of capacitance. <S> If you reduce the size of the MOSFET you can go faster, or change the type to a different technology than silicon. <S> A cheap one that should show a lot (like 5 or 10:1) improvement is the FDD1600N10ALZ . <S> They're cheap enough, but you can run some simulations to see what the predicted performance is. <S> You can also try reducing the 600 \$\Omega\$ if that's not enough improvement. <A> It seems to me like you could just use an emitter follower circuit. <S> If you do this with the NPN transistor then the output voltage would just be Vout= <S> Vd-0.6 <S> If you need a current then just scale the output resistor accordingly. <S> If this answer isn't enough then you need to restate the problem in a more complete way. <A> There are a few problems with your layout , I can imagine from your resonance. <S> The 10:1 probe 10pF to 15pF causes Drain resonance with your wire jumper inductance from Vcc low ESR 5V Cap to Drain @ <S> 10nH <S> /cm. <S> Solution add 50 OHm series near drain for a test pin <S> The inductance in the power to drain increases the resonant gain which can be reduced by using at least hal the length of a few cm in the Source to Rs resistor to ground. <S> Eliminate 600R gate resistance to drive with 50 Ohm gen. <S> ~ 0 to 10V . <S> Summary: to decrease Q of resonance; <S> Add series 50R to Drain for probe capacitance and add wire inductance between source and Rs to gnd. <S> Also increase base load to 1k with 1k pullup to gate to Drain for negative feedback. <S> With these loads it draws 14W peak from the 5V supply and >5W peak in the FET when driven properly as below draws <S> 0.25W <S> Avg or < 2%. <A> Basically, in a FET, lower RdsON means a bigger chip which means higher Qg, higher capacitance, ie a slower FET and higher gate drive current if required to switch fast. <S> RdsOn <S> i sonly relevant <S> if you use the FET as a switch, ie fully turned on. <S> Since your circuit uses the FET in linear mode, RdsON is not a relevant selection criteria for your FET. <S> There will be about 3V across the FET when it is ON. <S> So using a very low RdsON FET results in high capacitance which only makes it slower, but it does not lower dissipation since the FET is never used as a switch. <S> You need a FET with a lower capacitance, and a driver capable of pushing more current into the gate. <S> I simply added a push-pull driver and the simulator thinks it will work: <S> Note I picked a random one from the simulator library, so don't use that reference, instead please carefully select the FET, something like: Vds > 10V <S> I <S> d <S> > 5A (depends on your current) <S> RdsOn <S> about 100-200 mOhms at Vgs=4.5V -- This is mostly to make sure it will be well into linear mode at Vds=3V and 0.6A. <S> Check the Id vs Vgs graph in the datasheet (graph below), green operating point is OK, red is not OK as the FET behaves as a resistor so it will need much larger gate voltage swing to control the load, which means it will take more time to charge/discharge the gate. <S> Qg as low as possible ; be aware that Qg spec depends a lot on the Vds at which is is specified... <S> SOA should not be violated when the FET is ON in linear mode, so you need a FET that is usable in linear mode, not just switching, so probably not a TrenchFET. <S> You could try FQP13N06 or FQP10N20 (above graphs are from FQP10N20). <S> Spehro's FDD1600N10 looks even better. <S> You can also use a BJT, for example D44H11 which should be a good fit. <S> However you will need to adjust your signal generator to add a bit of negative offset voltage, the low level part of the waveform should be about -0.5V to make sure Q2 really brings the power transistor's base down to almost 0V to fully turn it off when it needs to. <S> Or use a more complicated driver circuit. <A> Ignoring the ringing (Which might just be poor scope probe technique) <S> you see basically straight line rise and fall, which says to me that the mosfet capacitance is introducing a slew rate limit. <S> Or go with a BJT in place of the mosfet, will need a little base current but should be faster.
| Pick a mosfet with much smaller gate capacitance's or lower the 600R resistor (or both).
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emitter follower oscillation cured by adding base input resistor? I made a simple emitter follower headphone amp using fairly high speed BJT- 2SC5994. Vcc 16V R1- 10K R2- 40k RE- 100 load - 300 ohm After powering up there it was clear there was serious issue, the amp would pop in and out oscillation randomly. This manifested as buzzing sounds through headphones and after checking with cheap 2mHz oscilloscope there were HF oscillations clearly visible. They are high speed transistors so this was anticipated, Firstly I added some resistance at the output to isolate from headphone's capacitive load. This seemed to make it less prone to popping into oscillation but it would push up the output impedance too much so wasnt a suitable solution. Surprisingly adding resistance to the base input ('base stopper') also fixed the problem. With 40 Ohm there hasnt been any oscillation when loaded with HPs and playing test tones of various frequency. So if it wasnt capactive load on the output causing the issue what else could it be? <Q> Yes, adding a base resistor (usually a 50 - 1000 \$\Omega\$ ) is one solution to RF oscillation of an emitter follower caused by capacitive loading. <S> It doesn't take much capacitance, tens or hundreds of pF can be enough, which is not much length of cable. <S> Often a ferrite bead is used, either in the base or emitter lead because it doesn't degrade the high frequency performance as much as a resistor. <S> You don't need it for audio. <S> See also Brian Drummond's answer here <S> Your actual question is "what else could it be?". <S> I think it's the most likely cause, and I can't think of anything else that would be improved by the 'base stopper' resistor. <A> Any common collector ( emitter follower) or FET source follower with a reactive load where the slightest resonant gain is >1 can result in emitter oscillation with any wire L and C, <S> capacitance loads. <S> This can be avoided by attenuating the loop gain, improving phase margin or matching emitter towards load impedance. <S> This may be done by; <S> adding either lowering the input shunt impedance to hundreds of Ohms or adding at least 22 Ohms series R to the emitter to closer match the emitter to the transmission line or add a ferrite R which adds similar values at unity gain BW freq. <S> by minimizing C(Vbe) <S> copper coupling capacitance in the layout <S> e.g. Design by impedance <A> It is probably layout related. <S> If you have inductance on the collector it tends to turn the circuit into a Colpitts oscillator. <S> By adding the resistor you are damping the loop gain below 1. <A> Your schematic is missing important power supply decoupling capacitors close to the transistor. <S> 0.1uF <S> ceramic (marked 104) for VHF plus 10uF electrolytic for audio. <A> So if it wasnt capactive load on the output causing the issue what else could it be? <S> This is serious oscillation, if it can be seen on a 2Mhz-bandwidth 'scope. <S> My numerous encounters with similar circuits oscillate at very-high or ultra-high frequency. <S> The audible "pops" and "buzzes" suggest oscillating amplitude is large enough to seriously modulate audio. <S> Less-serious oscillations can lurk under the audible-radar. <S> Check carefully that oscillations have really been snubbed by your efforts, and not simply moved to a much-higher frequency. <S> Oscillation is often the result of some inductance resonating with small transistor capacitances. <S> But oscillation can also result from a ground loop feeding current back into earlier gain stages that drive this circuit. <S> Lower-frequency oscillation (rather than VHF or UHF oscillations) make ground-loop a serious possibility. <S> Re-wiring the headphone return current to a different circuit ground can reduce or eliminate such oscillations. <S> Substituting a lower-frequency \$ F_T \$ <S> transistor won't likely cure a ground-loop oscillator, just lower its frequency. <S> In search of inductance <S> If a small series base resistor cures oscillation, this is more likely a local LC oscillator rather than a ground-loop oscillator. <S> Where could the "L" of an LC oscillator be hiding? <S> An oscillator in the MHz region requires significant L. <S> Possibly the headphone itself. <S> It is a coil of wire. <S> Or possibly the headphone cable. <S> It is a long transmission line that appears inductive at some frequencies. <S> A bypass capacitor from transistor collector to ground might bypass this inductance. <S> Where \$ R_E \$ is grounded might be the right spot. <S> BTW: a series base resistor less <S> than about 100 ohms seems a decent solution.
| But be careful where the capacitor ground is grounded - you can add a ground-loop if it intercepts headphone current with the wrong phase. Possibly the inductance of the wire back to the DC supply.
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Could 48V arc across this small gap? I have soldered two wires onto this Cree XHP 70.2 LED, tested it in a circuit, and everything works fine. I intend to wire four of these LEDs in series and drive them together at about 48V and 2.4A (12V each). My only concern is how close the solder is to the copper LED backplate (see images below). I'm worried that it might arc to the backplate and create a short circuit, bypassing the LED. Is this a legitimate concern? <Q> Even a very small gap is sufficient to stop quite a bit of voltage. <S> IPC-2221B would specify 0.6mm for 48V for an exposed gap on a PCB. <S> Breakdown voltage of an air gap is much higher than that-- 48V <S> will hardly start an arc over any non-microscopic gap. <S> Of course if you get conductive dirt or liquid on it, all bets are off. <S> For example, by using plumber's solder with acid flux. <A> No problem, Air gap is >1kV <S> /mm with humidity and a bit less with dust which would just burn if you had an inductive transient exceed this level. <S> But then you need reverse LED protection if you had inductive switching. <S> The wire looks well tinned and bonded. <S> You can always add a 1cm stretch of Polyurethane rigid plastic (subfloor adhesive) for wire strain relief as well for the solder pad protection. <S> This also acts as short circuit protection. <S> The copper bottom surface should be greased & screw clamped to a heatsink. <A> You never know what's gonna happen. <A> If you're worried, you could shallac your worry spot. <S> But with that voltage and amps it's as someone answered and won't arc that air gap.
| It won't be a problem but just for the safety I would use some shrink tubing/electrical tape.
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How can I power a circuit while charging its Li-ion battery pack? I'm creating a portable device that can be powered by a battery pack, or plugged in and used while charging the battery. At 4:18 in this video (shown below), it shows that the circuit can be powered by the battery charger while it is charging the battery pack through the BMS. I feel like the current draw from the device would interfere with the charger's CC/CV modes and battery state monitoring, causing incorrect battery charging, or insufficient power for the device. Am I missing something, or is the video wrong? <Q> For applications like this better choice is charger chip with power path, something like BQ2403x , <S> BQ2407x etc. <S> Or as simple as device not working if charger cannot supply enough power. <S> Check this application note form Microchip for many good ideas. <A> It will just take longer during CC mode but may never finish in CV mode as the load current may exceed 5 or 10% of CC current used for cutoff. <S> If this is the case, there should be a timer with CV time limit and cutoff minimum time in order to reduce aging effects at 4.2V. <S> I would just leave it float at 3.8V if you do not have a separate battery current sensor. <S> Normally the device senses battery current <S> but here it is the charger. <A> You are correct, but the question is how big a deal it’s going to be. <S> With the device powered, the charge will take longer. <S> The charger switches modes based on the battery voltage at the charging current, but in this case, the charging current will be off. <S> I suppose if the charging circuit were super optimized for a particular battery, you might feel bad that your design was no longer optimum. <S> Then again, what if the battery life were reduced by 10%? <S> Would it be worth the extra design trouble to get this life back? <S> As the designer, you can decide. <S> (By the way, the 10% number is <S> just a gut feel guess. <S> If you are designing something really important or critical, I guess you would have to experiment to find the actual number)
| If you want to use these components somewhat better solution would be to create additional path from voltage regulator to the load using diodes or mosfets. You are correct that this setup will change charging profile, potentially causing charger overheating, undercharged battery or reduced battery lifetime.
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How to extend a microcontroller GPIO to ~300 inputs? I need to find out the address of buttons pushed. Say I need about 300 Input port required dedicated for individual buttons. Then I would need 300 I/O pins to read the address of a pushed button which is not feasible for Microcontroller without using any sort of Integrated Circuit. My question is is there any IC that would dedicate an individual address for push button so whenever I push button it would send/give me an address like 0xA ? Or is there work around? <Q> If 256 buttons will be enough, you can use two <S> MCP23017 <S> (16 port I2C GPIO): the first one as 16 outputs, the second one as 16 inputs, of course with 16x16 matrix of buttons. <S> If not enough, you can use three of them, or "borrow" additional three lines form MCU (so, it will make 16x19 = 304 buttons possible). <S> As @jonk mentioned, if you want to detect multiple buttons pressed at the same time, you need to add diodes ( 1N4148 , 1N914 ) to each button, like below: <S> If you don't need to handle multiple pressed buttons, you need to protect only the output lines by diodes, like below: <S> MCP23017 <S> is pretty cheap and works very well. <A> I would also choose multiplexing or Charlieplexing for this role. <S> /SN74HC165. <S> You could also combine the two, use 595s to shift out a pattern and 166s to input the button states. <S> It's easy to use these chips with a microcontroller's SPI interface, and they're very cheap if you buy more of them at the same time. <A> Yep. <S> The very simplest solution is called a shift register. <S> You can use them for both input and output -- though generally each chip is dedicated to one or the other. <S> Well, that is to say, they're either "serial to parallel" or "parallel to serial". <S> Say you have an 8-bit parallel-to-serial shift register, it'll have 8 parallel input lines, and when you read the serial line, it'll give you the value of each of its 8 parallel input pins in sequence. <S> Or for output, the other way around (writing 8 bits to one line sets the output on all 8 parallel output pins). <S> Usually you can chain them together, so that reading/writing beyond those 8 bits shifts in the values from the next register you've connected. <S> But if you have a LOT of inputs, like building in a keyboard, then it makes sense to do the multiplexing arrangement others have mentioned. <S> That way with N GPIOs you can read N 2 inputs. <S> In addition to the advice here, you should find plenty of tutorials online since DIY keyboards are a thing. <S> Still, since shift registers are so cheap (I've seen them at 50 for $1), sometimes it's just all a matter of managing complexity in whatever way makes the most sense for you at the time.
| If you insist on using dedicated I/O lines for every single button, a handful of shift registers could also do the job, e.g. SN74HC166
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5V LDO blowing up only when testing current draw I have a circuit to power some 5-volt logic from a +24VDC source. It has some basic reverse-polarity protection and uses a TI 5VDC LDO. It all works great during normal operation however, when I use my FLUKE multimeter in amp mode to test the current draw like so: The LDO blows up immediately. Smoke, sparks, a bright light, all of it.Anyone know what's going on? Thanks <Q> I think people are distracted by the PMOS, but the real reason is a voltage spike when the 24V is connected, due to inductance of the wiring that is even higher when the multimeter is connected, vastly exceeding the max voltage of the LDO. <S> I assume you are using a ceramic capacitor at the LDO input, which is a reason someone may be experiencing this problem for the first time -- it wouldn't happen with an electrolytic cap, as I'll explain. <S> Forget the transistor for now and model the circuit like this, with a source that has some resistance and inductance in the wiring (R1, L1). <S> Also model the capacitor as having a very low ESR (R2): <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Simulate this <S> and you'll see that the RLC circuit rings, peaking at about 42V when the 24V turns on: Now try the same thing after changing R2 from 10 <S> mΩ <S> (which could be a typical ESR for a multilayer ceramic capacitor) to <S> 1 Ω (which could be a typical ESR for an electrolytic capacitor) and the peak will be much lower, about 27V: <S> The added resistance in series with the cap significantly dampens the oscillation, keeping the input voltage well with range of the LDO. <S> Yay! <S> So the simple solution is to add an additional 1 Ω resistor between your 1 uF capacitor and GND to account for the low ESR of the cap. <S> You should be able to observe this behavior in the real circuit pretty easily with an oscilloscope. <S> Remove the LDO (to avoid blowing it up) and watch the voltage across the capacitor when you first attach the 24V. <S> You can tweak the value of the added resistor as necessary to ensure the peak never goes above the LDO's limit. <S> Alternately, the ringing will never exceed twice the input voltage regardless of any of the parasitic values, so you could switch to a LDO that has a >48V input limit <S> and you'd always be safe. <A> I agree with VillageTech in incriminating the PMOS. <S> It has a capacitive load in the source and an inductive load in the drain (your mutlimeter wires), add a bit of parasitic inductive/capacitive coupling and you get a colpitts oscillator . <S> However, since your supply is 24V and you're using a LDO to make 5V at 80mA, I wonder why you need to use a PMOS to make a quasi-ideal low dropout diode... <S> If you implement your reverse polarity protection with a dumb old diode, dropping 0.6V, the LDO will still have 18.4V headroom instead of <S> 19V! <S> If the PMOS provides reverse polarity protection for other stuff on the 24V rail that you didn't mention in your question, then of course don't remove the PMOS, but instead add a cap on the 24VDC input. <S> Preferably an electrolytic with a bit of ESR, like a 10-100µF general purpose cap, to damp transients or resonances due to cable inductance and low-ESR caps in the power supply. <S> A low-ESR ceramic cap may produce ringing due to this. <A> I'm suspicious that you may have insufficient capacitance on the output of the LDO. <S> Personally, I think it unlikely the MOSFET is involved in this, the body diode severely limits what mischief it can get up to. <S> If the LDO is hovering on the edge of instability, a small change to the input could push it over the edge. <S> If you're using a very small (physically) capacitor, the capacitance under bias may be much less than even the nominal 1.5uF <S> you've shown. <S> And you're under the recommended minimum 2.2uF even with the nominal value. <S> But that may not be the main thing wrong. <S> Consider the massive power dissipation for such a tiny device, as Andy points out. <S> If I read the suffix code correctly you have an SOT-23-5 device, which has a datasheet (possibly optimistic, depending on the footprint) junction-to-ambient thermal resistance of 212°C/W meaning at room temperature the junction is heading for 350°C. <S> FAR too high for a silicon chip. <S> Now, it's supposed to shut down gracefully under thermal overload (at a very high die temperature), but that's not a good thing to provoke and with such a high input voltage it may well be dying as a result. <A> It seems that your PMOS protection circuit starts oscillating because of adding long wires (inductance) in series to power source. <S> Try to add 0.1uF capacitor between drain of PMOS (pin 3) and ground. <A> The 50v input capacitor will survive a spike above 30v, which I is fatal to the regulator. <S> You may have an over voltage spike at power up. <S> So as soon as the inrush current falls back to stable value, the ammeter reacts by raising the voltage, effectively acting as a booster. <S> It has little energy but enough voltage over 30v to fry the regulator while the input cap is protected up to 50v, which apparently is never achieved. <S> Sometimes the industrial 24vdc power supplies are adjustable up to 28 or even 32vdc. <S> Make sure yours is set to 24vdc. <S> If you disconnected the load, then powered down to insert the ammeter, you kept the capacitor charged, thus prone to carry a spike should one occur. <S> This not an easy troubleshooting. <S> The starting point should have been the 24vdc supplied, so close to max 30v supported by the regulator.
| It is possible that at power up, an inductor present in the ammeter tries to overrun the stabilizing inrush current, thus raising the voltage beyound the capability of the regulator. So I'd say just get rid of the PMOS and replace with a diode. The input cap (50v) needs discharged at power down.
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How does a microwave oven modulate its output power? My microwave has multiple power settings, ranging from 500 watts to 1500 watts. My question is how is the power reduced/modulated? I can think of a few options (but I don't know which one is correct/most common): Is the microwave constantly switched on and off with different 'duty cycles' (similar to PWM) (different on/off times with longer total 'off time' when the microwave is set to a lower 'power level') in order to create lower power output? Is the electromagnetic frequency of the photons being emitted by the microwave altered resulting in more or less energy per photon? * Is the amplitude of the waveform (in other words the amount of photons per second) altered resulting in more total energy per second? * Have I made any wrong assumptions in the three options listed above? Are all three options above theoretically possible in order to achieve modulation of the microwave's power output? Are any of these options used in a microwave oven in order to handle the modulation, if so which is most common? According to this forum post photon energy is determined by a photon's electrical frequency and a waveform with same frequency/wavelength and different amplitude has more photons being emitted per second. <Q> It's really simple. <S> They have one fixed power level, and turn it on and off. <S> You can hear it when it is on low power. <S> It'll hum louder for a few seconds, then have a longer period when it is a little quieter. <S> On is louder, off is quieter. <S> On full power, it is on all the time. <S> Lower power switches it off for a short pause with a longer on period. <S> Low power is a short on period followed by a longer off period. <S> It is sort of pulse width modulation, but real slow. <S> The frequency of the microwaves is fixed. <S> More or less - it isn't strictly controlled because it isn't critical. <S> The construction of the magnetron sets the frequency - you can't easily vary it. <S> The frequency is more of a range of frequencies. <S> Modulating the power of a 1000 watt transmitter isn't easy, so the manufacturers don't try. <S> They don't even closely regulate the power all that closely. <S> The transformer is made such that it limits the current to the magnetron, and that's about all the regulation there is. <A> Most microwave ovens use the on-off method of control described in other answers. <S> There are microwave ovens advertised as "inverter" products that rectify the incoming power to DC and then invert it to AC at a high frequency. <S> That allows the use of a smaller transformer to boost the voltage to the 2000 volts or so required by the magnetron. <S> The inverter can also modulate the voltage using PWM to control the voltage and thus control the power. <A> Most, if not all microwave ovens, use the method 1. <S> At full power it is always on, and at lower levels, it is just turned off for a period of time. <S> The starting and stopping can usually be heard when observing the device operation.
| A fixed power level is enough, if you can turn it on and off.
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What does 'ground' look like on a breadboard? Given the following circuit: What would this look like on a breadboard? For example, what is 'ground' on the right? Is there a particular 'component' that can be used in a breadboard to act as ground (I'm usually using a 9V battery as the voltage source). <Q> I remember having similar confusion when first introduced to circuits. <S> There is really not much to it.. <S> The grounds on the left and right need to be connected together. <S> In schematics, they are often split, for convenience and clarity. <S> For the 9V battery, the negative terminal would typically be used as ground. <S> However, you could also connect two batteries in series and use the mid-point as ground to get -9V to +9V supply. <S> Image source <A> The GND symbol is not a component as much as it is a label or net name. <S> Ground is just a conductor, an electrical node we designate as zero or the reference potential for our circuit. <S> Those two ground symbols are the same point. <S> You could just replace them with wire that all connects to each other. <S> It's just messy and difficult <S> actually draw wires everywhere if you have 50 connections to ground in your circuit. <S> Also difficult to read if you have to follow a long trace that seems to run everywhere just to find that everything else ends up there too without shedding additional light onto the circuit section you were following since you already reached its end bit <S> didn't realize it because it wasn't labelled. <A> On a breadboard, I tend to use the bottom one of the long common rails as the ground or common terminal. <S> This is because ground tends to be the terminal with most components connected to it. <S> If you connect meters, or 'scopes, or signal generators, you usually need one connection to the ground terminal. <S> Using a long trace gives plenty of points at which to make the connection, and I like to be able to identify quickly where it is. <S> Often, if a circuit doesn't include a long line at the bottom labelled 'ground', it's trivial to add one in, lining to all the ground symbols. <S> Then the breadboard can be built to look like a copy of the schematic, as far as that's possible, to help with component identification. <S> One the one hand <S> , ground is 'just' a named node, and doesn't need any particular treatment. <S> It certainly does not need to be connected to an external ground like an earth spike or the safety ground in the mains wiring. <S> One the other hand, ground does tend to be a special node. <S> It tends to be the largest, so stray capacitances from other nodes will be larger to the ground node than to others, which might make a difference in RF circuits, and even stability in lower frequency circuits built with fast components.
| In other words, ground can be anything you choose, it is just a "reference point".
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BLDC motor current sensing for load torque I would like to control 14V BLDC motor of battery powered drill. What I want to do remake the drill to nut runner (torque wrench), so I need to sense a motor current which is proportional to motor torque. My controller will include microcontroller. Here are some questions: How battery voltage discharge can affect motor current measurement dependingon the same load (for example I would like to wrench the same nutant the same torque Nm)? Does it will affect my current reading andthe current will increase? Could I sense the BLDC motor current directly on the battery high side or do I need to sense it on all three the motor phases? Is it enough to sense only current to create closed loop torque control, or do I need some additional parameter for torque control based on the different battery discharge levels? UPDATE: Okay, for now this is more clear. Please correct me if I am wrong: I do not need to sense voltage of the battery for any calculations or compensation. The only thing I have to sense is motor current, because the current is proportional to the torque. The battery discharge will not have any impact for the current measurement, depending on the same load torque. So shortly, I will get the same current readings when the battery is fully charged and the same current readings when the battery is discharged let say to 60% of full voltage. What I plan to do is to create relationship table between current readings and torque (Nm) of a load, and save these values in a microcontroller memory. So when the torque wrencher reaches knowing current value, controller will know that the bolt or nut is wrenched with a fixed known torque and motor stops. UPDATE 2: The things with current and voltage sensing seems ok. But I forgot another thing - temperature of motor. What impact to the torque (current) will have temperature changes in motor? What is the relationship between temperaute and current of BLDC motor? Thank you for your answers. <Q> How battery voltage discharge can affect motor current measurement depending on the same load... <S> Does it will affect my current reading and the current will increase? <S> Torque is directly proportional to current, independent of battery voltage. <S> Could I sense the BLDC motor current directly on the battery high side or do I need to sense it on all three the motor phases? <S> Yes, you can sense battery current on the high (or low) side. <S> In a battery powered application sensing the low side may be easier. <S> Sensing phase current could be more accurate, but is also more difficult. <S> Is it enough to sense only current to create closed loop torque control, or do I need some additional parameter for torque control based on the different battery discharge levels? <S> To control torque you must use PWM. <S> If you sense battery current then the motor current will be higher at lower PWM ratio, by the same proportion eg. <S> if PWM ratio is 50% then motor current will be double battery current. <S> This occurs because during PWM 'off' time the controller recirculates back-emf current back into the motor. <S> This current never reaches the battery leads, so to get the true motor current you must multiply your battery current measurement by the inverse of PWM ratio. <A> I guess I can sort of answer your questions. <S> battery discharge usually relates to voltage drop, so with a lower voltage the current will have to increase to deliver exactly the same torque. <S> Before discussing what you can measure I think you need to decide a control scheme for the motor. <S> This is an example of a scheme, to answer your question you need to be able to affect all 3 phase currents to control the motor, thing is how do you plan to do that? <S> you got a micro controller, do you have MOSFETs or some sort of switch to change the current on each phase. <S> The equation for the torque looks like <S> this E is the voltage per phase, I is each current, w is the angular speed which is 2pifrequency. <S> $$T=\frac{EuIu+EvIv+EwIw}{w}$$ <S> https://www.digikey.jp/en/articles/techzone/2016/dec/how-to-power-and-control-brushless-dc-motors <S> you sorta need to make the control scheme more clear to advance any further in this project. <S> In theory with just the current, considering the voltage is constant in all phases you could do a closed loop. <S> Sadly, with a battery this is not true <S> so you should also monitor the voltage <S> so it doesn't drop below a certain level to ensure proper functioning. <A> A couple of background notes to hopefully help your thinking. <S> In six step commutation with PWM speed control, the effective applied voltage to the motor, V_APPLIED is: V_APPLIED = <S> VBATT <S> * D <S> where D is the duty cycle. <S> The resulting phase current, I_PHASE will be: I_PHASE = <S> (V_APPLIED - V_BACK <S> ) / R_PHASE <S> where R_PHASE is the phase-to-phase resistance and V_BACK is the back EMF voltage from the motor. <S> In theory, you could solve for D and do current control (which is torque control) without sensing current. <S> But in practice, you can't use this equation to do open-loop torque control because very small errors in back emf calculation or battery voltage sensing will result in large errors in torque output. <S> How battery voltage discharge can affect motor current measurement depending on the same load (for example I would like to wrench the same nut ant the same torque Nm)? <S> Does it will affect my current reading and the current will increase? <S> This question is not super clear. <S> Motor current will be directly proportional to torque. <S> Could I sense the BLDC motor current directly on the battery high side or do I need to sense it on all three the motor phases? <S> No and no. <S> The motor current is not the same as the battery current. <S> But you only need to sense one of the three phases, not all three. <S> Is it enough to sense only current to create closed loop torque control, or do I need some additional parameter for torque control based on the different battery discharge levels? <S> If you sense phase current, it won't be necessary to sense battery voltage. <S> You may be able to fine tune your loop to close more quickly if it knows RPM and battery voltage. <S> But it should not be strictly needed. <S> There is one more thing I should add. <S> You can implement closed loop control by placing the current sense resistor on the low side of one of the switching FET's. <S> It doesn't have to be on the phase conductor. <S> Your sampling needs to be designed to read the current sense value at the right time when the FET is on. <S> Ideally, the sampling should also be synchronous to the PWM frequency, but not all processors support this.
| If you use phase current sensing plus a closed loop to control torque, you don't necessarily need to know battery voltage.
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How to protect a battery terminal after removing corrosion? I have repaired a corroded battery terminal by neutralizing it in vinegar and then cleaning it with a wire brush. The result is a terminal lead which now has bare steel visible where the protective chrome layer has been destroyed. What kind of treatment can be done to this surface to protect it from oxidation now that the steel is exposed? Even if it doesn't corrode again in the future from a battery, I believe oxidation of the steel can still lead to further damage to the surface and it's electrical conductivity. I am interested in both inexpensive solutions such as an aerosol spray as well as other more expensive options for an electronics hobbyist that relate to this type of repair. <Q> WD40 displaces moisture but allows contacts to penetrate the thin film for a reasonable resistance ~ 100 mOhms, if there is sufficient force. <S> It will eventually dry out and leave a thin residue so it is not permanent. <S> If you had liquid Tin, after cleaning off oxide, dip then rinse in water. <S> That is better but you have to buy bulk (2L?) <S> so this is not practical. <S> Solder works but will oxidize rapidly in humid weather >90% RH. <A> Any spray is likely to be an insulator, negating the function of a contact. <S> Most solders, whether pure tin, older (non-RoHS) Pb/Sn, or newer Bi-RoHS types, are more corrosion-resistant than plain steel. <S> Having had the same issue, I sand the actual contact area to remove any oxidation and cover just that area with solder. <S> Making solder adhere to steel or iron is more difficult than to copper, though, and may require acid flux and higher temperature. <S> BTW, you've already done a good job of surface preparation. <A> And don't use any vegetable or animal fat, or vegetable oil - these tends to hardening because of oxidation, or to decomposing, and can cause corrosion.
| You can use technical Vaseline (best solution), or - if you haven't - any grease or machine/engine oil.
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Why doesn't a short-circuit immediate consume all voltage? If we assume a short circuit, with zero resistance (not even within the battery), would this 'infinite current' deplete all voltage within the battery? For example, if I "short" a 9V alkaline battery, it will run for about an hour (though the wire gets pretty hot, if not melts) until it depletes all its power. Is this only because the battery has internal resistance, or why does the battery 'keep running'? Additionally, does a battery ever give its Amp-hours or how long it can run? <Q> Because in the real world there is nothing such as a perfect wire or a perfect battery. <S> There are parasitic resistances, inductances, and capacitances everywhere. <S> Plus there is the issue of the chemical reactions in the battery that is taking place to produce that electricity. <S> That can only happen so fast even if you had a battery with zero parasitics. <S> Also, I find it pretty hard to believe that your shorted 9V runs for an hour. <S> What is your definition of depleted? <S> Batteries are considered dead long before they reach 0V in the same way <S> you are too tired to do useful work long before you drop dead from exhaustion. <S> I can only assume you are talking about reaching 0V across the terminals for purely academic purposes. <A> Just so this actually has an answer: <S> Information about the capacity of a given battery can be found in the battery's data sheet, which should be available from the manufacturer. <S> However, batteries are generally not designed to used with an extremely high load current like this so the actual capacity with a (near) short circuit is probably much less than what the manufacturer has specified for a more reasonable load scenario. <A> The bigger the battery in a single cell, the lower it's internal resistance. <S> Of course battery chemistry like electrolytic caps has a wide range in ESR (effective Series Resistance) <S> So shorting its voltage across it's internal ESR will produce the short circuit current. <S> This is dangerous on Lithium cells and lead acid cells due to the energy stored. <S> Voc/ESR= <S> Isc <S> But for Alkaline which has almost the same energy of Li Ion cell <S> it is safe since Alkaline has a higher ESR. <S> We call the voltage drop with load resistance, Load Regulation Error, measured @ <S> Amp with % V error also being a ratio of ESR to Load R. <S> For bigger battery short current pulsses Isc must be safely. <S> It is not necessary to go to 0V to measure ESR, since you now understand where Load regulation comes from with ESR=ΔV/ΔI. <S> Although this image has slightly more energy than a Lithium cell both have a thermal runaway effect, once the activation energy threshold is reached. <S> It accelerates to fuel a lower resistance with exothermic reactions of heat to ever increasing currents. <S> In the case of a nuclear bomb, it takes a hydrogen bomb to Trigger it. <S> Once triggered it has a negative incremental ESR , so the rise time is so short <S> , it's equivalent bandwidth BW=0.35/ <S> Tr reaches gamma wave frequencies.
| yes, it is because the battery has internal resistance and because your wire is not an ideal short circuit.
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LED lamp - connecting LEDs in parallel I have a question about LED lamp: Quantum Board It has 16 strings of LEDs connected in parallel. Each string consist of 18 LEDs connected in series. It's supplied with constant current source (with max current 2800mA, what is much more than safe current for single LED/led string.) I know that connecting LEDs in parallel is not the best solution because of imperfection of LEDs while manufacturing and as a result inequality of currents through every string. But I see that is quite common solution for such lamps. I have several assumption how it is possible: using LEDs with very similar parameters (Vf), using LEDs with same binning. But actually, according to datasheet Samsung LM301B leds have 1V of Vf variation even inside every binning group. Possible in practice this variation is much lower. very good thermal dissipation due to metal-core PCB Could anyone explain how does it works? Are my suggestions the keys for the answer? Is putting a series current limiting resistor for every string and using CV source a more robust solution and why not do it in this way? <Q> Look again: - I see R1 to R10. <A> There are plenty of LED lamps that don't have current limiting resistors and rely on the LEDs being equal. <S> This often is not a problem or not seen as a problem. <S> The LEDs might be "equal enough" for the current to divide itself properly. <S> Maybe the current isn't exactly divided equally but as long as the difference in the current through the LEDs isn't too small or too large, the difference in currents is not a problem. <S> The LEDs themselves have some series resistance. <S> Is putting a series current limiting resistor for every string and using CV source a more robust solution and why not do it in this way? <S> Indeed this would be a more robust solution to add current limiting resistors. <S> The reasons why this isn't always done might be cost (a resistor cost money and might take up valuable space where there could be an LED). <S> Also: it works without resistors <S> so why add them? <A> The LM301B has the spec, Vf=2.8V and If=200mA. <S> The voltage on 18 LEDs in series is roughly 54V. If each branch has 200mA current, 16 of them will have 3200mA in total. <S> The current source you have sources 2800mA. <S> So each branch will get 175mA (2800mA/16) on average, which is less then the spec of <S> 200mA. Let's say one branch is broken open, the other branches will get 187mA (2800mA/15) each on average. <S> Even two strings are broken open, the current on a average on each string is 200mA. <S> So, it seems OK or normal operation as long as the LEDs are functioning and cooling is in place. <S> If you ask what would happen when three or more strings break open, or two strings break but variation of each component can cause some components operate outside of the spec, you are right, the design is not completely safe. <A> using LEDs with very similar parameters (Vf), using LEDs with same binning. <S> But actually, according to datasheet Samsung LM301B leds have 1V of Vf variation even inside every binning group. <S> I think you missed a decimal place there? <S> The AY bin LM301B LEDs is actually speced with a min/max Vf of 2.6 to 2.7V. <S> That is a very tight spec, so <S> the difference between individual diodes will be very small. <S> Could anyone explain how does it works? <S> The absolute maximum difference is very small, the average difference smaller still, and 18 diodes are put in series. <S> If you take a small variation and average it 18-fold, you get an absolutely tiny variation between parallel strings. <S> For example, if you assume the average variation is 50mV, and averaging that 18-fold will reduce that to about 12mV, or about 0.02%. <S> That is insignificant. <S> Is putting a series current limiting resistor for every string and using CV source a more robust solution and why not do it in this way? <S> If you bought a quantum board, you paid many times what a normal LED costs for a high efficiency diode with the lowest possible forward voltage so that you can save electrical power. <S> Putting a series resistor to drop voltage would defeat the purpose of having spent that money. <A> As the others explained, if the LEDs are binned and everything is at the same temperature, it'll work. <S> However, with a constant current power supply, if one of the LEDs fails open the entire string will turn off and the current will be shared between the remaining strings, which will take more current. <S> There are a lot of LED strings here, so this is much less of a problem than if you had only two LED strings in parallel (in which case one failed string would double the current in the other). <S> Your Meanwell PSU has a current setting, and also a voltage limit setting. <S> So it is a good idea to measure the output voltage with the LEDs on, then unplug the LEDs. <S> Without a load, the power supply will output its maximum voltage. <S> Then adjust the "output voltage limit" potentiometer so that the no-load voltage is maybe a volt higher than the voltage when loaded by the LEDs. <S> Then check again with the LEDs. <S> This ensures even if a LED string blows open, and the PSU tries to output constant current, it will hit its voltage limit.
| With many LEDs in series, differences in forward voltage will "average out" reducing the difference between LED strings.
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Is there a difference between measuring the resistance using a multimeter ohm function and using the voltage / current relation? I need to measure the electrical resistivity of a concrete sample following the procedure described in ASTM C-1760. The procedure requires a 60V DC potential to be applied across the concrete sample (cylinder with 10cm diameter and 10cm length) and the current measured to obtain the bulk resistivity of the sample. The problem is that I don't have a power supply capable of delivering 60V. I was wondering if a regular multimeter can produce the same result using the ohm function. The expected sample resistivity should range 50 - 500 Ohm.m <Q> The Ohm function on your multimeter will use the voltage-current relation to determine the resistance of a sample. <S> It will either apply a voltage and measure the current delivered, or force a current and measure the voltage required to deliver it. <S> But it will almost certainly not use a voltage as high as 60 V, so it won't meet the needs of your test. <S> Probably the ASTM spec requires a 60 V source to provide a repeatable test method that allows for breakdown of some of the thin layers of oxide material within the concrete, but not others. <S> The higher voltage used may also be needed to accurately measure currents due to ionic transport within the never completely cured concrete. <S> Therefore I'd expect you need to use an actual 60 V source to achieve a compliant measurement. <A> Probably not. <S> Most DVMs use low voltage and are designed around typical electrical components and wire. <S> Items made from other materials may not conduct until a certain voltage threshold is reached. <S> Also, without enough voltage the material may not conduct uniformly, which will also skew your results. <A> If on the other hand what you want to measure the resistance of is something complicated, with perhaps non-linear resistance, then the way you measure it is indeed relevant. <S> Something damp like concrete will react to current flow with internal chemical reactions which will modify the current flow, this is called polarisation. <S> The grains may be coated with thin insulating films that break down at particular voltages. <S> If you want to measure a sample to ASTM C-1760, then you need to use the procedure that ASTM C-1760 defines.
| If you have a perfect, aka linear, resistor, and components called 'resistors' that you can buy are more or less 'perfect', then there will be no significant difference between what you measure with a DVM on the ohms range, and what you compute having measured voltage and current.
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What's wrong with this PWM-controlled constant current sink? T10 gets very hot and starts melting within 30s, even with a heat sink and values way below its rated 15A. Edit: I know it is not a true PWM controlled constant current, but that what I can do with the parts that I have access to. As for the current being used at, it does not exceed 4A at full duty of the 65kHz signal. Edit 2: So, basically, what's wrong with my circuit is my inability to correctly use the information in the datasheet. Thanks for your answers. <Q> I would not call this a PWM controlled constant current sink. <S> The analog voltage is just being generated by a filtered PWM. <S> The 15A rating of T1 isn't for operating as a linear device. <S> It's for operating as a switch. <S> That 15A rating is meaningless here. <S> If you're expecting the efficiency of a switching circuit, don't because it isn't one. <S> It is inefficient (and hot) in the same way a linear regulator is hot. <S> So it's operating as expected. <S> You've basically built a linear regulator but have sized it according to the efficiency expected of a switching regulator. <S> For something like this to produce 15A, you need parallel transistors with balancing measures, mounted to a heatsink and possibly a fan. <A> With the shown 2 ohm resistor the dissipation in T10 is maximum at 2.4A (60% duty cycle on your PWM) <S> when the transistor will be dissipating 14.4W, which is a lot for a TO-220 package. <S> It will be more if R5 is lower. <S> It's also possible <S> it's oscillating, which you can check with an oscilloscope, but I suspect the problem is simply power dissipation. <S> The BDW93C is only capable of 80W, and for that you have to hold the case <S> temperature to 25° <S> C, which is rather difficult (and for that, the transistor die will be at 150°C, very very hot indeed). <S> BTW, if your PWM frequency is low, that little 10nF capacitor won't do a good job filtering. <S> Cutoff is about 200Hz as shown. <S> The 12A ( <S> not 15A) maximum continuous collector collector current is only one limitation. <S> The limitations stack up like a Venn diagram and you have to obey all of them. <S> Another limitation is the Safe Operating Area as shown in the datasheet figure. <S> With 10V across the transistor the maximum DC current is 8A, at 12V 4A is the maximum permissible. <A> When the load (2 ohm) is driven at 3 amps, it develops 6 volts across it and, conversely there is 6 volts from collector to ground because the supply is 12 volts. <S> At this current the emitter resistor drops 1.5 volts therefore, the collector to emitter voltage has to equal 4.5 volts. <S> So, the power dissipation in the transistor is 3 amps x <S> 4.5 volts = <S> 13.5 watts. <S> The thermal resistance of the transistor is 1.5 degC per watt <S> and if this is on a Heatsink of (say) <S> 10 degC per watt, the temperature rise of the transistor’s junction is 13.5 x 11.5 degC. <S> So, do you see the problem here? <A> Your output transistor isn't operating as on-off switch, it's total dissipation is = <S> I(12V <S> - I * 2,5Ohm) where I is the output current and R5+R6 is assumed to be 2,5Ohm. <S> The max dissipation in the transistor occurs when I*2,5 Ohm = 6 volts; that means I=2,4A. <S> Then the dissipation is 14,4 watts. <S> I guess that's much more than you have assumed. <S> The design details are beyond the scope of this answer.
| It is a linear current source that is being controlled by an analog voltage. It's heating up because it's linear current source, not a switching current source. If you want to dissipate less consider to make a version where the output transistor is switched on-off and the current is smoothed with a series inductor like in constant current led drivers.
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Can I somehow control a PSU-powered fan using a 4-pin fan output on the motherboard? I have a 12V 29.4W 3-wire Delta Electronics fan that I want to power using an EVGA 750 G3 PSU and control via one of the 4-pin fan terminals on an MSI MSI Z390-A PRO ATX LGA1151 motherboard . Not considering the wattage of the fan, I originally plugged it into a (known working) fan terminal but it of course would not turn (and the BIOS correctly reported the RPM as 0). Connecting it directly to the PSU is simple enough, but this would result in it turning at full speed at all times. I am curious if there is a "DIY" way that I can supply the fan with power from the PSU and control how much actually reaches the fan using the output of the motherboard (using something like an optoisolator to keep the motherboard safe). The motherboard is able to control 4-wire fans using PWM, and 3-wire fans using DC voltage level. Is there a product or combination of parts that will do this? I originally asked this on Super User, but it was deemed more appropriate here due to the involvement of DIY electronics. Thanks in advance! EDIT:The 4-pin connectors on the motherboard are indeed keyed to force correct connection of 3-pin plugs. The BIOS includes a Fan interface that allows you to plot desired RPM vs temperature. Each fan output can individually be set to either PWM or DC voltage level control. The motherboard auto-detects what is plugged in (4 or 3 wire) and selects the appropriate setting, but it can be changed manually if desired. The fan was not purchased through DigiKey, and came terminated with a keyed 3-hole connector (F). It also shipped with a 2-wire (but 4-wide) molex connector (M), unterminated on the opposite end. <Q> fans using DC voltage level . <S> I did not see anything about 3-wire fan support on that page, but if user manual says so then you don't need anything else but correct wiring on 3-pin plug. <S> The PWM control pin will not be used, and the taho output from 3-pin fan is compatible with tach input on 4-pin fan connector. <S> Furthermore, 4-pin connector is usually keyed in a way that forces correct connection of 3-pin plug. <S> Note, that if motherboard does indeed support both types of fans then most likely you have to configure fan type somewhere in BIOS. <S> I originally plugged it into a (known working) fan terminal but it of course would not turn <S> There is nothing "of course" about this. <S> The fan should have been working at full speed all the time and motherboard should have been able to read its RPM. <S> That's, of course, if it is configured for 4-pin. <S> From the link to a fan it seems it comes without a connector. <S> So, check your wiring first. <S> UPDATE: <S> If your problem turns out to be insufficient motherboard output wattage, then one DIY solution would be to add OpAmp in voltage follower configuration, supplied directly from PCU and controlled by voltage output from motherboard connector. <S> Depending on motherboard driver's design you might need to add a load resistor on the control line to simulate fan motor. <A> Not with a 3-wire fan with a tachometer (TACH) output like the delta fan, the FG signal is for speed monitoring and it outputs a signal. <S> Source: <S> https://www.delta-fan.com/Download/Spec/AFB1212GHE-CF00.pdf <S> And with 24W you would have to be able to dissipate Watts of power in a resistor, which is not good. <S> A variable DC DC converter or voltage regulator might also do the job, the fan has a working voltage of 6-12V. <S> The best thing to do would be to get a fan with PWM control. <A> What you have. <S> What you need to make it work. <S> A 20A Pch FET.
| The motherboard is able to control 4-wire fans using PWM, and 3-wire One way would be to change the voltage which could be done with a variable resistor in series, but one problem with this is approach is the remaining power will be dissipated in the resistor.
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Hypothetically, 6 or more phase line with same mechanical energy? So I don't understand a concept of mechanical to electrical energy conversion. Let's assume I have a 3-phase generator and need X amount of mechanical power in order to get 200 Volts Phase-Neutral. Now suppose I add some windings etc. so that I have a hypothetical 6-phase system. Would I need more mechanical energy in order to get 200 V for each phase, or can I get it by same amount X? So basically, if I have more phases do I need more mechanical power? <Q> Let's assume I have a 3-phase generator and need X amount of mechanical power in order to get 200 Volts Phase-Neutral. <S> You are confusing the relationship between power and voltage. <S> For a given winding arrangement the voltage will depend on the speed. <S> The mechanical power input required when unloaded will be that only to overcome the mechanical resistance. <S> Now suppose I add some windings etc. <S> so that I have a hypothetical 6-phase system. <S> Would I need more mechanical energy in order to get 200 V for each phase, or can I get it by same amount <S> X? <S> You don't need much power to generate voltage. <S> You do need power to generate current at that voltage. <S> So basically, if I have more phases do I need more mechanical power? <S> No. <S> In general: $$ P_{OUT} = P_{IN} \times efficiency $$ <A> If it's a hypothetical generator, then if you put no load on the generator, it will take no effort to spin. <S> Adding more windings then makes no difference. <S> The voltage output also makes no difference, as there is no current being drawn. <S> Add an electrical load to the generator, and the mechanical load on whatever is driving increases by the same amount. <S> Again, that's true regardless of how many windings you put on. <S> If you double the windings and attach a load to each, then you double the mechanical effort. <S> Mechanical power in = electrical power out <S> Of course real generators aren't perfect. <S> They have friction in the bearings and the windings get warm, all wasting power. <A> The 0-120-240° three phase system is used because it utilizes the three wires as good as DC would. <S> 100% of possible current and 100% of possible voltage without overloading either. <S> A single-phase AC system in contrast only has an utilisation of 70% because of the zero crossings. <S> You cannot do better than 100%. <S> That's why more phases are pointless. <S> There are systems with more phases for other reasons, for example for smoothing the output of a connected rectifier. <S> But this is within a machine or building at most.
| You need more mechanical power if you draw more electrical power from the generator no matter how many phases you have.
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AC to DC power supply questions I have built the circuit in the diagram below with a couple of minor changes. I desire 3 DC outputs - +12, -12, and +5 volts. So, the changes I made is to use an MC7812BTG for the +12 and an MC7912BTG for the -12. Then I simply added an LM7805CT after the 7812 to get my +5. I made 2 heatsinks from a couple of L brackets I had laying around. I also added LEDs to the 3 "rails" so that I could verify they are operating. Everything else is the same as the diagram. This is for a very small modular synthesizer case. I tested the power supply with no load and it checked out fine. Then I installed 4 modules with approximately 350 mA draw on +12 and 100 mA on -12. The +5 volt rail was not used at all. I powered the case on and it all worked as expected for about 35 or 40 minutes. Then the modules glitched and the audio "froze". These are digital modules. Anyway, turning the power switch off and back on restored the function of the synth, but just a couple minutes later it glitched again. That's when I realized how hot the case was underneath where the voltage regulators are at. My main questions are: Am I pushing the regulators too hard by supplying 25 volts and dropping it to 12? Should I consider a 24V transformer? What would a "proper" heat sink consist of for these regulators? Is my "L" bracket piece of metal (3/4" wide, 1/8" thick, and 4" long, bent to L shape at center) just not enough of a heatsink? If the 7812 is overheating, what is occurring to make the modules "glitch"? (overvoltage, undervoltage, etc.)? Thanks in advance for any advice, comments, or stern reprimands. <Q> Yes, the 7812 is overheating and should be shutting down. <S> A 24V center-tapped transformer will produce a rectified and filtered 16v which will be fine if your electricity does not have brownouts. <S> The datasheet for the 7812 shows its thermal resistance and finned aluminum heatsink manufacturers show how much heat they remove. <A> As others have said - a lower voltage input to the regulators will decrease power losses. <S> Ideally Vin is about Vout + Vdropout + Vheadroom <S> Where "Vheadroom" is a volt or two. <S> Where you have excessive Vin, dissipation in the regulator can be moved into a series resistor that can be dimensioned to withstand the dissipation. <S> An LM340 datasheet here has a dropout voltage of 2V typical at 1A at 25C, and somewhat less according to fig 11. <S> Allow 2V. <S> At present the LM340 dissipate P = <S> V <S> x I = (25-12) <S> x 0.35 <S> = 3.9W <S> Adding a series resistor that drops Vin to 14V min <S> gives: Vin min = <S> 12V <S> + 2V at 1A = <S> 14V. <S> Rseries = <S> V <S> /I = <S> (25-14)/1A = 11 Ohms. <S> At 350 mA that drops only V=IR = 0.35 x 11 = 3.9V and dissipates only P = <S> I^2R = <S> 0.35 <S> ^2 <S> x 11 <S> = 1.35 W <S> That's a help - but not as much as you'd wish. <S> If you NEVER draw more than say 400 mA continuous and have enough capacitance at the regulator input to handle any higher surges then. <S> P_LM340 no resistor = <S> V <S> x I = (24-12) <S> x 0.4 <S> = 4.4W <S> With series resistor: Vsupply = <S> 25V. <S> Vin = 14V min. <S> Imax = 0.4A <S> Rseries = <S> V <S> / I = <S> (25-14) / 0.4 <S> A = <S> 27 <S> Ohms. <S> Power in resistor max = <S> I^2.R = 4.3 W Power in LM340 = <S> V <S> x <S> I = 2 <S> x 0.4 <S> = 0.8W. = <S> 0.8/4.4 = 18% of initial !!! <S> Use a 10W resistor to dissipate the <= 4.3 W (A 5W resistor would notionally work, but running it at < 50% power rating extends its life.) <S> Use an aircooled resistor - readily available, so no resistor heatsinking is needed. <S> Keep the resistor away from the regulator AND away from any electrolytic capacitors. <S> This SE <S> Q&A <S> My linear voltage regulator is overheating very fast <S> covers the subject in (great) detail. <A> There are some couple of solution to bring down the power dissipation <S> Transformer secondary voltage can be bring down to somewhere near 15V <S> Use Switching regulator instead of LDO's
| If 5V is used, connecting the 5V regulator to Vin via a resistor will take the large thermal load off the 12V regulator. The +28V input is way too high and creating too much heat.
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Impedance matching in common collector The last stage of a power amplifier to an 8 ohm speaker is usually a common collector. Now many textbooks mention that it is used as a buffer as it has a low output impedance compared to the speaker (Zout << Z speaker). To me this is what people call voltage impedance matching as the full voltage of the amplifier is applied to the speaker. But I always thought that what is of interest is the maximum power transfer. So what matters is power impedance matching which implies that Zout =Z speaker.So which one is true : voltage or power impedance matching in the last stage of a power AMPLIFIER? <Q> Think about it this way. <S> If you had a 12 volt light bulb that has a resistance of 12 ohms <S> and you applied it to a 12 volt power supply, it would produce a certain amount of brightness based on the power consumed. <S> Then, you insert a 12 ohms resistor in series with the bulb and looked at the bulb’s brightness. <S> In which scenario is the bulb the brightest? <S> In your question you are putting the cart before the horse and assuming maximum power is obtained when the source impedance matches the load impedance. <S> In fact it’s when the load impedance matches the source impedance that you have the maximum power scenario. <A> In a power amplifier the designer is typically trying to present a perfect, undistorted voltage signal to the speaker. <S> The output voltage should simply be the input voltage multiplied by the gain of the amplifier. <S> You get this in part by having your source impedance be much lower than the load impedance. <S> Without this buffer there would be some voltage dropped across the power transistor and some across the speaker. <S> The amount of drop across the transistor would be dependent on the instantaneous current through the transistor. <S> Since this current varies from one instant to another, the voltage across the transistor would also vary from moment to moment, causing distortion in the output voltage. <S> As a side note: Zout << Zload results in high efficiency as opposed to maximum power transfer. <S> As Zout drops, more power is dissipated in Zload, less in Zout. <A> You're confusing two slightly different things. <S> Example: a signal generator (like this one , they're used in electronics labs), <S> these have "50 ohm" outputs and <S> only when you load them with 50 ohms will you get the maximum power. <S> A much higher or lower load will also just work but the power will always be less. <S> Audio amplifiers generally have a very low output impedance, like in the order of less than 1 ohm. <S> If you impedance matched that with a very low impedance speaker you might think that you would get maximum power. <S> Theoretically, yes you would <S> BUT I have yet to see an amplifier that behaves that way <S> (theoretically ideal I mean). <S> Real world amplifiers cannot deliver more than a certain amount of current and a certain amount of voltage. <S> Since power = voltage x current, for maximum power you want voltage and current to be as large as possible. <S> That's where the recommended impedance (4 ohms, 8 ohms) comes in <S> , that is the load impedance that will make the amplifier operate at it's maximum voltage and current that it can handle. <S> Use a lower impedance and the current will become too large. <S> Use a higher impedance and the current will be lower than the amplifier can deliver (you'd be wasting that current capability). <S> I'm deliberately not writing: " <S> Use a higher impedance and the voltage will become too high" as that doesn't happen. <S> Audio amplifiers are practically all voltage output amplifiers which means that they amplify the input signal (also a voltage) to an output voltage. <S> The current will depend on the load. <S> Without a load, you still get the voltage. <S> With a load that is too low, too much current will flow and the amplifier will clip (distort) and/or switch off to protect itself and/or can be damaged.
| When you're using an amplifier with a certain output impedance then you can extract the maximum power from it when the load impedance has the same value.
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DCDC converter in series I need to convert 5V (main power supply) to +/- 30V. I have two +/-15V DCDC converters and was wondering if I can connect them in series to generate the +30V and -30V rail? My simple schematic I tried simulating this circuit using Mplab MINDI, but there are only DC power sources and I am not sure it is valid to connect them the way I use them. The result show +/-30V. (When I connect the DC source on both ends the Amplitude is split to +15V and -15V therefore I set the amplitude to 30V). The datasheet for the converter is here (pdf, 224kB) UPDATE:After the discussion in the comments I updated my circuit. This would mean I have the separate the supply GND from my circuit GND and also adding two more converters for -30V <Q> Per the datasheet these supplies have isolated outputs, so you can offset them from the input. <A> I suspect leaving 0V outputs floating will result in unstable operation or oscillation. <S> The datasheet seems to imply that minimum load for regulation is 10% of max current. <S> Therefore I'd suggest adding small loads to the outputs. <S> For example 1.5k for 10mA current. <A> It depends on whether your 15v power supplies are isolated . <S> Some are, most aren't. <S> If it's not isolated, then the -15v outputs of the PSUs are both at -15V in reference to the input ground, not just 30V below their own respective high rail.
| All you need to do is avoid creating ground loops, and let the (nominal) 0V outputs float to the midpoint of the 30V rails each one produces.
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Is it possible to wire up a simple camera module with a separate wifi module to create a live UDP stream without full fledged OS? Could be any simple camera module, like for example that: https://www.arducam.com/product/arducam-5mp-plus-spi-cam-arduino-ov5642/ Wifi module could be also any available. I imagine I would also need some kind of H.264 encoding module, is there even anything like that? Would I need something else to produce an RTP stream? Is there anything for that? I fully expect the solution to be very difficult (if it exists) but could it even be wired up at home? Currently I have setup with RPi, and I just wonder if this can be achieved without Linux or any full fledged OS, just with combining multiple modules, some electronic work and some code (I imagine it would be difficult, but I would willingly accept that challenge). <Q> One such example I've found is the " AI-Thinker ESP32-CAM ". <S> Image source: <S> AliExpress @ <S> https://www.aliexpress.com/i/4000000845941.html <S> When you go for a known module you get the benefit of having plenty of resources available to help you get it up and running. <S> For example, you could follow a tutorial and more easily end up with a working system you could use as a starting point for your final project. <A> Yes, it is possible. <S> But, you don't want to use an RPi, everything about it assumes Linux. <S> All the hardware drivers assume Linux. <S> There are very simple microcontrollers that are usually programmed without an OS (AKA bare metal ). <S> And at the other extreme, there are complex systems that are usually programmed with an OS. <S> Somewhere in-between is a gray area, where either method could be used, which is used is often decided by the programmer's background, they will often use the method that they are more comfortable with. <S> I grew up using bare metal, so I will tend to use that method if possible. <S> You should start with a MCU that people often program with bare metal. <S> I use the TI MSP430 series, there are many others. <S> Then pick peripherals that have bare metal drivers available. <A> Yes, specifically that module you linked has an example arduino sketch "ArduCAM_Mini_5MP_Plus_Video_Streaming" and you can see a discussion about reaching the limits of framerate for streaming here: <S> https://github.com/ArduCAM/Arduino/issues/173 <S> Basically instead of writing to your SD card you'd just be writing the data out to a UDP socket. <S> I've done similar streaming with small amounts of data from 8x8 thermal imaging arrays so that I could view the current camera image on my laptop.
| There are products based on well known modules like the ESP32 that provide camera and network interfaces.
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Can I use a 2.4 GHz rated parabolic MIMO antenna for 700 MHz Verizon 4G LTE data? Can I use a 2.4 GHz rated parabolic MIMO antenna for 700 MHz Verizon 4G LTE data? It seems 700 MHz are a long ways from the range of a 2.4 GHz antenna, but I have seen people marketing similar looking parabolic antennas that claim they are wide spectrum and go down to 600 Mhz and up to 5GHz... I'm just not sure at what attenuation though? Will the same Verizon tower switch me from 700 MHz to one of their higher frequencies with better reception? Verizon Wireless appears to utilize multiple 4G LTE frequencies like 2.1 GHz, 1.9 GHz, 1.7 GHz, 850 MHz, and 700 MHz. I don't know if they would they would responsively switch me from 700 MHz to a higher frequency if my signal improves though as some people report ??? This would be important if I need an antenna that capable of working from 700 MHz to 2.1 GHz. How I know the band I'm using: I used my iPhone to figure out I'm using band 13 for communication with my local Verizon tower, which is 700 Mhz by following these direction . I have LTE data miles away from the tower but it's not a good connection so I'm looking to get a highly directional antenna for my JetPack 7730L . Specific use case Here's the 900mm-wide (about 3 feet) parabolic antenna I'm looking at so you can have a specific example to pick on. Added 2020-01-28: I did my own test to see if Verizon switches bands based on signal quality. You can see the video here . <Q> Can I use a 2.4 GHz rated parabolic MIMO antenna for 700 MHz <S> Verizon 4G LTE data? <S> The gain of an antenna, that is, how much it focuses in one direction, is pretty much proportional to its size divided by the wavelength. <S> Generally, antennas don't work (well) for frequencies they're not designed for, but you can consider a parabolic dish more of a reflector than an actual antenna – the feed (which is mounted in the focal point) <S> is what actually converts the electromagnetic wave from free space to a coax waveguide (or vice versa). <S> So, replace the feed with something that works for 700 MHz and you got a directive antenna that's <S> roughly only 700/2400 = 7/24 ~= <S> 1/3 <S> as directive as it would've been for a signal at 2.4 GHz. <S> Now comes the problem: you don't seem to have such a feed (and you're right <S> , you need a multi-band antenna as feed). <S> In this situation, you'd just buy a cellular-optimized multi-band antenna that has some gain in itself: it typically will have less gain than what you'd get from a well-matched dish, but considering this is all a bit impossible to calibrate and do right <S> , it'll probably work much better than trying to attach a low-gain multi-band feed to your dish. <S> Generally, what you want to do is quite possible illegal: in many legislations the power times the gain is limited, and if your phone is legal with the little gain its antenna has, it's probably illegal with the gain of a highly directive antenna. <S> Will the same Verizon tower switch me from 700 MHz to one of their higher frequencies with better reception? <S> Only the base station configuration of Verizon's cellular infrastructure can answer that. <S> But, generally, the lower the frequency, the better your reception is over a long-distance link: <S> so, 700 MHz is pretty much the best you can get. <S> "as some people report": Um. " <S> some people" are a terrible source. <A> The gain and directivity of a parabolic dish is inversely related to the wavelength of the signal and directly related to the size (diameter) of the dish. <S> The wavelength at 700 MHz is approximately 3.5 times greater than 2.4 GHz. <S> Hence keeping the dish size constant and lowering the frequency (increasing the wavelength) by 3.5 will result in a drop in gain and directivity (that is, the beamwidth increases.) <S> There are formulas for computing the gain of a parabolic dish as a function of it's size and the operating frequency out there on-line. <S> Here's one I've used: GAIN <S> \$=10\log[0.5(\frac{\pi D}\lambda)^2]\$ <A> I am going to simplify this for you. <S> No, you cannot use the 2.4 GHz antenna for 700 MHz. <S> The parabolic part of the antenna may still provide a little bit of gain, but it won't work nearly as well as it did at 2.4 GHz. <S> More importantly, the antenna is probably not passive. <S> I base this on the fact that the band specified <S> is very narrow if it were a purely passive antenna. <S> Even if there is not an amplifier, there is no guarantee that the radiating structure at the focus will perform well at 700 MHz.
| There is (or may be) a low-noise amplifier (LNA) at the focus of the reflector that won't work at 700 MHz.
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Why are there so many grounds? A bit of context: I'm working on a smart tracker watch for an 8 year old. A hobby project, and I'm looking for parts to use. Anyways, that's not what I'm confused by. This is my first project, so please don't judge me if I've done something wrong. But I can't for the life of me figure out why these parts have more than one GND. One of them actually has upwards of 30. What the heck is going on here? The part in question is from a WL1835MOD WiLink 8 from DigiKey. I've linked the datasheet here: http://www.ti.com/lit/ds/symlink/wl1835mod.pdf It's a Wifi/Bluetooth combo, and the part shows up as two separate parts on Upverter. That's why sometimes I refer to it as 2 parts and sometimes just 1. <Q> Because wires and traces are not perfect. <S> Needless to say: Radio frequency = really high frequency. <S> The bad effects are increased noise, voltage spikes when the chip's current demand decreases and voltage dips when the chip's current demand increases. <S> So what do you do if are trying to move lots of water and all you have are tiny pipes? <S> You use a bunch of tiny pipes in parallel. <S> Parallel inductances results in an overall lower inductance. <S> Also, because IC packaging is standardized and if you don't need all those pins, you might as well connect them to ground because it reduces noise due to the aformentioned high frequency currents trying to flow through trace inductance. <A> Grounds <S> like you're seeing on that package can serve several functions. <S> 1) <S> Multiple ground pins <S> /pads reduce the inductance of the ground paths, which is important to reduce ground bounce. <S> Ground bounce is a shift in the ground voltage within the packege (on the die) caused by having multiple outputs switch at the same time (from dV=L(di/dt)). <S> This sometimes goes by the term SSO, for Simultaneous Switched Outputs. <S> 2) <S> Multiple ground pins <S> /pads also help to isolate one part of a circuit from another. <S> This is especially important in mixed signal (analog or RF + digital) devices, where you don't want digital noise to corrupt the RF, or you don't want one RF path leaking over into another. <S> EDIT1: <S> Example of ground use on a new chip <S> We are currently designing a 400 bump-pads (IO) mixed signal (RF+digital+power) chip. <S> Of those 400 pads, 325 are some form of ground (return). <S> Even though the device's primary function is RF, only 10 of the IO are actually RF. <A> The electrical reasons for the numerous GND connections are already well described by others. <S> But they are also used to conduct heat out of the module. <S> Find it from the application notes in the datasheet. <A> I once used 74HC00 logic biased linearly, as a 200MHz fet probe to capture enough signal from an NE602 oscillator/mixer's tank circuit that the production line couldmonitor a spectrum analyzer and set the 3 LOs to 101/107/113MHz. <S> When first built (my design) I used 2 Inverters in DIP package.... <S> the fet-probe oscillated. <S> cause --- the two inverters *shared" gnd pins (and vdd pins). <S> And the 2nd inverter had to drive a AC_coupled 50 ohm input on the Spectrum Analyzer (was 50ohm in the gain_chain, plus the 50ohm in SA, so 100 ohms, thus 10 milliAmps per volt of load current) <S> problem---using Vgnd = <S> Lgnd <S> * dI/dT, at approximately 50 MHz Foscillation, assuming 2 volt output <S> thus 1 volt peak <S> , the dI/dT was <S> 50Mhz * 6.28/100 ohms or 300,000,000/100 = 3,000,000 amps per second. <S> The Lgnd was 10nanoHenries. <S> TheThe Vgnd was 10nH * 3Million = 30 milliVolts. <S> Assuming a gain on only 10X in each of the 2 NAND_biased_linearly gain stages <S> , thus 10*10 = 100X <S> , our Vground becomes 0..03 <S> * 100 = 3 volts. <S> And we'd assumed 2 volts (peak-peak). <S> Hence the oscillation. <S> cure ---- use two 74hc00, use only one inverter in each package. <S> An alternate cure would be using 150 or 240 ohm output resistor, for 14dB reduction in the Vground. <S> what was the problem? <S> not enough gnd pins. <S> ( and vdd pins) <S> [so the down-voter could not provide an actual EXAMPLE, and cast aspersionsinstead?] <A> Multiple grounds is common in RF applications explained by others. <S> Remember, those almost chip-scale packages like QFN are tiny , in a 3 mm x 3 mm package, having a single connection is far from insufficient to handle the large current output, and inadequate to dissipate the heat generated internally in converters. <S> For SOIC and QFN chips, a common solution is using many pins in parallel, and employing a thermal pad at the center of a chip, usually electrically connected to ground to achieve both high electrical conductance and thermal conductance to ground. <S> Leaving the pad unconnected may cause the chip to malfunction. <S> If you leave the thermal pad on a power converter unconnected, the chip may appear to work under low current, but when you attach a load, it immediately turns off due to thermal protection. <S> And in BGA chips, there is no thermal pad, the only path for heat dissipation is the solder balls. <S> The number of grounds in such chips is even larger, 20-30 ground connections are typical. <S> All the gray pads are connections to ground. <S> The power and ground and output pins are distributed alternatively for thermal balancing.
| They all contain some inductance which impede high frequency currents trying to flow through them. And it's worth noting that using multiple grounds is also an extremely common practice for power converters and power transistors.
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Choosing a digital oscilloscope for a self learner I'm about to start a course about electronic interfaces. My intention is to learn to debug hardware by doing measurements etc. with oscilloscopes and multimeters. According to the course, the recommended device to buy (digital oscilloscope / multimeter) is the following: National Instruments myDAQ The main problem with the above is that it needs LabView to work (which is not for free). I want to buy a tool that I can use without paying exorbitant software licenses. Looking on the internet, I have found this one: PicoScope 2000 series . I see that it is an oscilloscope, but I'm not sure if it will work as a digital multimeter. Is the PicoScope a valid option to replace the NI myDAQ ? If not, do you know anyone I can buy that doesn't need non-free software to run? <Q> National Instruments makes great stuff, I used NI products for 20+ years at work, but is a poor choice for a hobbyist. <S> IMO, a serious hobbyist should have a stand-alone scope and DMM, not something that attaches to a computer. <S> I have always used Tektronix at work, so I wanted the same for home. <S> I got the TBS1052B for $100 off from Amazon during their holiday sale last year. <S> If you aren't tied to one brand, there are many other choices that will give you more features for the same price. <A> NI equipment is usually very expensive for what it is. <S> I'd go with the Picoscope. <S> They have many serial decoding schemes which will be useful for learning "electronic interfaces". <S> I'm not quite sure what the NI DAQ offers. <S> While you can measure voltages on a picoscope, you've got to be careful with it being referenced to ground (Outer shell / common point is tied to ground). <S> So be careful with what you put the common alligator clip on, as you could be shorting something to ground that shouldn't be. <S> The picoscope also comes with an arbitrary function generator, which can be handy. <S> As for multimeter, I'd pick up a cheap fluke on Ebay. <S> An 87V or 187/189. <S> You can usually get on for ~$100 <S> or so. <S> This multimeter will allow you to measure currents, resistance, capacitance, and higher voltages than the oscilloscope <S> can (unless you have a high voltage probe, which they don't normally come with). <A> There is no need to have a multi-meter function on the scope. <S> In fact, it is better to buy a separate multi-meter which is more handy than having to wait till the scope powers on let alone it is more portable. <S> Uni-T have very high quality and affordable ones. <S> Having a bench power supply and function generator also helps with your course. <A> <A> I'd like to frame-challenge this. <S> I don't think you should use a digital 'scope at all, just starting out. <S> Find a good analog one instead. <S> (as deep as your arm, heavy, green CRT display, etc.) <S> Same for the other gear. <S> The reason is to connect you closer to what's actually going on, instead of abstracting things away in an attempt to be helpful. <S> (the auto-set button on a digital 'scope, for example) <S> Once you know what's going on, how to set everything manually to see what you want to see, and found a few oddities in the process that can't possibly be caused by smarts gone awry because there aren't any, then you can look for smart tools. <S> They are nice, and very helpful IN CERTAIN SITUATIONS! <S> But they can also outsmart themselves and end up lying to you. <S> If you can't tell the difference, then you're going to have to unlearn <S> some things later.
| I would look into the Digilent Analog Discovery 2 for anything upwards of a few megahertz it is excellent. Hantek have very affordable and high quality scopes.
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Will a 950 nm IR emitter work with a 870 nm phototransistor? I'm planning to build a simple IR light barrier across a distance of 15 mm. I plan to use this receiver with this emitter . The reason is to have some SMD PNP capable components. The emitter and sensor have a slightly different IR wavelength. Will it work anyway? <Q> Side by side: - The receiver will be slightly off centre compared with the transmitter (purple lines I added are sitting at 950 nm) but the receiver will still be above 80% of its optimum sensitivity. <S> You should also consider that as the emitter warms up, the peak centre emission can move by 0.3 nm per degree C <S> so, if it warmed 40 degC, the peak wavelength could move 12 nm. <S> This could lower the overall receiver sensitivity into the 75 % area. <S> Bear also in mind that the graph of receiver sensitivity has a centre point that is defined as “typical” in the data sheet <S> so, there could be some temperature dependency or drift associated with the receiver too. <S> The data sheet does not appear to give details on this. <A> Yes, your system will work. <S> Consider the system spectrum (red line below), which is each sensitivity at each wavelength times the emission at that wavelength. <S> You can see that the transmitter is operating on a 0.84 efficiency portion of the receiver, which ought to be well within any error margin you might have, considering losses for distance, misalignment, focus, etc. <S> Compare with an imagined transmitter whose peak is at 875 nm to match the receiver (dashed blue line) and <S> whose corresponding matching system spectrum is also shown (thin red). <S> This is engineering: how good is the fit in numbers? <S> The area under the 950 nm pairing is 90% of the area under the 875 nm pairing, and this considered a small loss. <S> Of course, the receiver is also dropping compared to a perfect "white" receiver, and it's about 5% (see for example difference between dotted blue and thin red at 800 nm.) <S> The spectrum for the receiver is scraped from a graph on the datasheet. <S> For the transmitter, Osram publishes it in "optical simulation" section, ASCII file download. <S> (The staircasing on the graph is just an artifact from the graph drawing.) <A> Look at the sensitivity/emission vs wavelength curves. <S> It is a spread and there is a point of overlap where both devices have reasonably high values.
| Yes, it will work.
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Why output signal is not symmetrical? What to do? I am trying to design a 2-stage amplifiers with first stage C.E with PNP transistor and second C.C with NPN transistor.The amplifier is supposed to have a gain more that 40. After designing the problem is my output voltage wave form is not symmetrical. Anyone has an idea? <Q> Your problem is that the base-emitter junction in the common-emitter amplifier does not have a stable voltage. <S> Hence the gain is variable dependent on where you are on your output wave. <S> Your quiescent emitter current is around \$1.5\ \mathrm{mA}\$ . <S> Hence \$r_\mathrm{e} \approx 16.6\ \mathrm{\Omega}\$ . <S> If the output swings by half a volt, the collector current swings by $$\frac{0.5}{5000} = 0.1\ \mathrm{mA}.$$ <S> The current swings in both directions, hence \$i \in [1.4, 1.6]\ \mathrm{mA}\$ . <S> Thus \$r_\mathrm{e} \in [15.6, 17.9]\$ . <S> Therefore the gain is in the range <S> \$A \in -[320, 279]\$ . <A> Look at the current in Q7. <S> What limits its range? <S> For 500mVpk into a 100 Ohm load, you're trying to supply +/- <S> 5mA. <S> Can that stage do that? <S> The emitter follower is a class-A stage. <A> Why (the) output signal is not symmetrical? <S> The asymmetry or non-linear gain causes more collector current and negative Vce swing to look asymmetrical and also cause total harmonic distortion, (THD). <S> To reduce this distortion, the Vbe AC input swing needs to be reduced. <S> This will also reduce the AC gain. <S> One way is by emitter degeneration with a series R added to emitter bypass cap. <S> The better way is to use negative feedback and choose R ratios to allow the feedback to reduce the Vbe variation using some of the forward gains. <S> It depends on hFE and also your (missing) SPECS for THD, GAIN and MAX SWING, but it is possible to go almost rail to rail using 20~25% of the open-loop gain and reduce the THD by the amount of feedback from 10% to 1% to 0.1%. <S> It also depends on your (missing) specs for Rin, Rout, Fmin, fmax (-3dB) What to do? <S> I prefer to only use negative feedback for significantly reduced THD. <S> (and non-visible but measurable asymmetry THD <S> ~ %(Vp+-Vp-) / Vpp . <S> Design Specs: THD=1%, Rin=TBD, Rout=TBD <S> , V Gain =>20, hFE=250 for large output swing, f_min = <S> TBD, <S> Idc <=1mA @10V Results THD=0.9%, Gain=21, Idc=1mA <S> OK <S> You must choose the tradeoffs between gain, THD and output swing.
| There are various ways to fix this problem: decrease the emitter capacitor so its impedance is larger than the dynamic resistance; increase the biasing current so that the dynamic resistance is smaller; put a resistor in series with the emitter capacitor such that the voltage drop across said resistor is larger than the base-emitter voltage.
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Is keeping batteries rolling loose inside a drawer a fire hazard? Simple question. At some point I began keeping batteries (AA, AAA, 9V, etc) inside a random drawer in my wardrobe that I use for other electronics (like cables, pen drives, etc.). And I've started to worry that they might randomly arrange themselves (they roll a lot when I open and close the drawer) into a short circuit inside the drawer and start a fire. I imagine a lot of people do that, and I've never heard of this causing fires. Is it a fire hazard? Extra: If that is indeed a fire hazard, what would be the ideal way of storing batteries at home? <Q> like the other answered it, the battery didn't enough current source but definitely can SPARK if your battery somewhat miraculously (in bad term) contact with metal body of another battery. <S> but Spark thing is not likely will be happens, the most occurence is leak by battery. <S> the smell is awful, it's greasy like oil spill, and didn't remove easily with single wipes not even with hair dryer. <S> that way you will be fine except the leak <A> The 9V maybe <S> but tbh I sincerely doubt the batteries alone would short. <S> If you had random metal scraps in the drawer than I would be worried. <S> No fire fortunately. <A> Most batteries cannot source much current (meaning it has internal resistance), in most cases this will not be a problem. <S> This will probably be true for all house hold batteries. <S> However, if all of the following prerequisites are true, a fire hazard could occur: <S> Battery can source lots of current (a higher voltage helps too) <S> There is a shortcut (wires, components, metal in general form a short cut from the plus to minus pole) <S> The heat through a metal part is higher than its temperature to get into fire. <S> Note also, that in case one component probably will not directly go to fire, but because of heat deform, and stops the circuit to be a shortcut circuit (see it as a fuse). <S> Most thinner metals (or wires) will be bent for example, breaking the short circuit. <S> I keep my batteries in a box like below, so you can easily sort your batteries to 9V, AA, AAA, <S> rechargeable/non rechargeable, low mAH, high mAH. <A> It depends on the recipe. <S> If you want we can definitely create a fire with only batteries and no other components involved. <S> One sure way is to have a few NiMH 9V batteries, fully charged dumped together. <S> 9V are easy to short among each other. <S> A disciplined approach(I too seldom follow) would be to have a holder or atleast keep them in a place with a proper orientation and less movements when you open and close the drawer.
| The batteries might get overheated or even blast. if you must to drop all of your battery in drawer, i suggest you put them out all in one direction or place it straight vertical in boxes. When I took circuits in college I had a 9V and a ton of scrap wires/parts in my backpack and the battery got bloated from a short circuit.
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What circuit simulators can also simulate ESP8266 and/or ATmega328? I'm going to start soon some ESP8266 and Arduino Nano projects. I've used some simulators in the past, but none of them simulated these ICs. Dou you know if there is any simulator, that would also include basic passive components? Clarification: ideally, the simulator will take and run the Arduino/ESP code. Can be online or desktop (Win/Linux/Mac), free or paid. <Q> Microcontrollers do not have comprehensive simulation models. <S> You can maaaybe (if you try incredibly hard) get a vendor model for a GPIO block but even that would be an engineer's fantasy come true. <S> Typically, reading the datasheet and doing some characterization on your own is enough to verify the Chip's analog performance. <S> If you're talking about a CPU emulator, that's something else altogether... <A> Labcenter's Proteus VSM is the only simulator I know of which can read code for several different micro-controllers and emulate them in near-real-time. <S> I'm not sure if they have ESP8266 capability <S> but it is not listed. <S> The price tag for micro emulation and other "VSM" features is significant however. <S> Personally I would use/design a debug/serial port connection to the micro and simulate any other circuitry as needed ( LTspice , etc.) <A> and it also has the posibility of load the page in the web explorer of you computer i.e., firefox, chrome, edge, etc.. <S> For linux users I've ran it using playonlinux whose use wine and there the loadind of the web page doesn't work because a bug where the http protocol is not supported (message shown by the depurator of playonlinux). <S> NOTE: <S> The version of proteus in was ran it that function is the 8.8 SP1
| the Proteus version which has the IoT capability can simulate the ESP8266, it even shows the web page coded into the module (inside the simulator) whith the visual designer window
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Feeling a tingling sensation by touching stuff, Is this safe? I'm putting on my desktop computer on steel rack as usual. A while ago, when I touch the steel rack for seconds, practically I feel a tingling sensation, sort of electrical shock in my finders. After that I measured how voltage the steel rack has, the multi-tester meter indicated almost 50V by AC mode. Is this safe to touch or keep the PC on the position, or I should get away the PC somewhere electricity-safe? I'm not sure this site is suit for asking this kind of question, but I'm very worrying whether this is safe or not since this experience is my very first time. Thank you. <Q> A comment by the OP: <S> In Japan almost all outlets have no third pole for grounding, only two poles they have. <S> Because modern SMPS wall warts (or equivalent) don't tend to use the ground wire in the socket because many times it's just not available, they have to reliably reduce emissions on the DC output wires by connecting a capacitor from output to rectified line voltage as per this answer : <S> - Now, the impact of adding the capacitor in the red box is that it can pass a small (safe) amount of AC current through to the secondary and, if the secondary isn't earthed via the laptop or PC then you will feel a (harmless) tingling sensation if you make contact with it. <A> And with a supply cord that connects the grounds. <S> A standard ATX power supply connects mains ground to metal case. <S> If it is not grounded, the EMI filter capacitors from Live to ground and Neutral to ground will act as a capacitive voltage divider and thus PC metal case measures half of mains AC voltage. <S> The slight shock can also surprise people and their reflexes can cause movements that might hit something. <A> The case of the computer should be grounded. <S> If in doubt, the steel rack can be grounded as well. <S> A stray 50V suggests that there is no ground, and you are feeling some small current leaking to the rack, perhaps via a power supply capacitor (see the answer by Andy aka). <S> Check your wiring - it may be in a dangerous state.
| Not hazardous itself, but connecting other mains powered devices should be made while mains cords are unplugged to prevent discharge currents from destroying devices. Any device such as a computer that has a mains input with ground pole really should be connected to a mains socket that provides ground.
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Convention to label components in a circuit Given an un-labeled schematic like the below, is there a convention on how to label the various components? For example, would the 'top-left' resistor be R1? Would the one to the right of it be R2, or the one branching downward from it, etc. Are there conventions for this, or it just comes up to 'label it however you want' ? <Q> There is no standard convention, or even a widely-used ad hoc approach. <S> These labels are called reference designators. <S> Do not reuse reference designations... <S> if you delete R1 from the schematic you should never have another R1. <S> You should also maintain a list, somewhere on the drawing, of reference designators that are not used because their corresponding components have been removed. <A> On a schematic, I would generally label components left-to-right and top-to-bottom, while trying to keep parts in a logical block together. <S> For example, in a multi-stage amplifier, stage 1 might have R1 - R5, and stage 2 has R6 - R10, even if that violates the left to right and top down rule. <S> Often, once a PC board is designed, the components are re-labeled left to right and top down on the PC board, then that numbering is back-annotated to the schematic. <S> This may mess up the order on the schematic, but makes it easier to find parts on the PC board. <A> Not that I know of. <S> And I think the RefDes being relatively ordered and close in number matters more on the layout than the schematic. <S> On the schematic all components associated with that part of the circuit are already right there in front of me. <S> On the PCB, if I am looking for C15, and I see C14, C17, or C18 first <S> then I expect C15 to be nearby and not across the board. <S> It doesn't matter to me if all those caps are far apart from each other on the schematic and associated with completely different parts of the circuit.
| You must make sure that the first letter is correct for the type of component being labeled...that much is standardized.
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Considerations for transmitting the signal from a MEMS microphone over a short cable? In my application I have the signals from 6 analog MEMS microphones (CUI Devices CMM-3729AT-42108-TR ), each amplified by a pair of op-amps in a voltage follower -> inverting amp configuration. The microphones are on tiny PCBs buried deep within a chemical analysis cavity, and the only way to get the signals out is through a short cable (150mm, conductors are +3V3, SIGNAL, and GND). After amplification, it's off to an ADC to be digitized, and then I'm demodulating the signal to sniff out the amplitude of a ~1500 Hz signal. Due to the stupidly small space I have to deal with (can't change the geometry of the analysis cavity), I'm prototyping with the Molex PicoBlade system. DigiKey sells pre-fabricated cable assemblies, but they're all unshielded. Many of the cables are touching each other. Should I be considering a way to shield the cables? The DC blocking capacitor is near the op-amp end of things, so the signal on the SIGNAL wire will be biased around +550 mV. This can be changed, I can move the DC blocking cap closer to the microphone if recommended. If I should be shielding, what're some recommendations for that? Did I miss anything else in the circuit? The microphones each have power supply bypass caps very close to them (though not shown in the circuit I've included). The output impedance of the microphones is 300 ohms at 1 kHz. How does that interact with the input impedance of the op-amp? <Q> Several factors affect noise pickup. <S> First is the actual level involved. <S> The higher the desired level sent down the conductor, the less effect external noise has. <S> It's a matter of ratios. <S> 2) Output impedance of the source determines just how much external noise will affect the signal. <S> In an ideal world, zero output impedance is immune to external EMI fields. <S> We don't live in an ideal world but in general, the lower the source impedance, the less effect external EMI has on the signal. <A> As others say, shielding might not be necessary... <S> But watch out for voltage drop on the ground wires, if the current in them changes. <S> For example if something else on your board turns on, that current step will transfer to all microphone outputs. <S> Large power supply capacitors on the board will help. <S> Separating power ground and signal ground, and using differential amplifiers before the ADC will solve this. <S> Check out the U.Fl RF connector <S> - it's much smaller than a picoblade and is well shielded. <S> It doesn't solve the common current problem though. <A> The unshielded short cable is 300 ohms <S> so it probably will not pickup much interference. <S> The voltage follower opamp does nothing and should be removed. <S> The input resistor of the inverting opamp can use the 1.6k resistor then the 0.1uF capacitor feeding it cuts frequencies below 1kHz. <S> The 2.2uF output capacitor will cause the inverting opamp to oscillate. <S> To cut high frequencies a capacitor can be parallel with the negative feedback resistor. <A> Your input highpass filter and feedback lowpass filter have simple gradual slopes <S> so they actually reduce the level of your 1500Hz signal and also do not make a good bandpass filter. <S> A multiple-order Sallen-Key highpass and lowpass filters make a good bandpass filter. <S> A Wien Bridge, Multiple Feedback or Twin-T bandpass filter has a high Q at its peak with sharp cutoffs but reduces its slopes when away from the peak frequency. <A> You don't say what level of control you have over the design, but you might consider MEMS mics with a digital (PDM) output. <S> Many audio ADCs will take this directly; if not, you can easily resample and LPF the signal to recover the analog version. <S> If you have particular noise concerns (crosstalk, motor noise, line noise, etc.) <S> edit them into your question and we could probably address them specifically.
| My opinion only: so long as the output impedance of the preamp is fairly low, you should NOT need to provide shielding between the output of the preamp on the card containing the microphones and the other card containing the ADC.
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Why does current change when adding in an LED? In the following diagram: The current is 1 mA from: V = IRR = 1 / 1000 = 1 mA However, when I add in an LED with forward voltage 1V: Why does adding in the LED drop the current as well? Does an LED have resistance as well? (I thought an LED only causes a voltage drop but doesn't have resistance). How does that work? <Q> A bog-standard red LED has a forward drop of about 2 volts at 20 mA <S> so <S> yes, it has a significant impact in what current might flow when the supply voltage is only a few volts: - <S> So, with 2 volts across it and 20 mA flowing, the perceived resistance is 100 ohms. <S> With 1.9 volts across it, the current is only 5 mA and the perceived resistance has risen to 380 ohms. <S> At only 1.8 volts, the current might be 1 mA <S> and this means it looks like 1800 ohms. <S> Do you see the issue? <S> For instance, if your supply voltage was only 1 volt then forget it because it won’t overcome the innate forward voltage of the LED and, barely nano amps will flow. <S> Other LED colours can have significantly higher forward voltages <S> so, what you are trying to achieve depends on the LED type. <S> So, like the game rock, paper and scissors, the LED dictates current flow at low voltages but, at higher voltages, the series resistor wins. <A> Please do following things. <S> Change the supply voltage to 5V. <S> Calculate the current through the resistor considering the LED as a voltage source of voltage equals to froward vorlage drop of the LED (polarity will be reverse). <S> The currents wil then match. <S> The reduction in current is due to reduction in voltage across the resistor with and without the LED. <S> Note that, the LEDs will have very small currents until the applied voltage across the LED is less than their forward Voltage. <S> Once the applied voltage crosses the LED forward Voltage, the current will be very high (the series resistor will define the current) <A> A first-order approximation of a red/yellow LED is V f <S> = 1.85 V <S> + I f <S> *R s for <S> I f > 15% rated current, while R s has a wide tolerance deviation that results in V f tolerances at rated I f . <S> Thus <S> if V f = 2 V @20 mA, R s = 0.15 V/20 mA = 7.5 ohms. <S> Below this threshold the diode voltage had a logarithmic relation with I f . <A> Why does adding in the LED drop the current as well? <S> Does an LED have resistance as well? <S> (I thought an LED only causes a voltage drop but doesn't have resistance). <S> An LED does have resistance in that it resists current flow. <S> If it didn't there would be no voltage drop across it and, since P = VI, with V = <S> 0 <S> it could not consume any power or produce any light. <S> We don't treat it the same as a resistor though because the relationship between voltage across an LED and current through it is not linear. <S> (Ohm's brilliance was showing that there is a linear relationship for a true resistor.) <S> Hopefully the following will help. <S> We can, however, make a simplification and model them over a range of currents as a combination of a resistor and a voltage source. <S> Figure 1. <S> An LED can be approximated as a resistor with a fixed voltage source. <S> If we look at a typical LED IV curve we can see that it is approximately linear over much of its useful range. <S> This allows us to model the LED as a resistor and voltage source. <S> Figure 2. <S> LED equivalent circuit model. <S> In Figure 1 the grey line is reasonably close to the LED curve from 20 mA to 100 mA. <S> We can work out the resistance that this represents from Ohm <S> ’s <S> law V=IR <S> but in this case we will look at the change in voltage and current in the area of interes. <S> $$R = \frac {ΔV}{ΔI} = \frac {3.5–2.0}{100m–0 <S> } = \frac {1.5}{100m} <S> = 15 <S> \ <S> Ω <S> $$ <S> We can also see that the line crosses the X-axis at V f <S> = 2.0 V. <S> Our equivalent circuit for this region of interest is (referring to Figure 2) R1 = 15 Ω and V1 = <S> 2.0 V. <S> All images and text from my article Resistance of an LED . <S> There is a little more in the article.
| LEDs do not have a linear relationship between current and voltage so they cannot be modeled as simply as a resistor using Ohm’s Law, V=IR.
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Capacitors in Series - common node I'm a bit confused on the sign of charges in circuits like these. The bottom (negative) plate of C1 has -Q charge and the top plate (positive) of C2 has +Q charge. However, these two plates are at the same node, so they are the same plate essentially. How can they have a different charge each??? <Q> The +/- represents the voltage relative only to the capacitor's other plate only. <S> If I am standing on your head, our feet are always below our heads, but that doesn't mean my feet and your head can't be at the same height. <S> The voltage potential at the bottom plate of C1 is not different than the potential in the top plate of C2. <S> This is different than the charge on each plate. <S> If I have a large and small air tank and connect them together, their pressure (voltage) will be the same but their air volume (charge) won't be. <S> In the same way, the larger plate of a larger cap needs more charge to produce the same voltage than a smaller plate. <A> It is an electrostatic behavior as shown below, Does this make sense? <S> Comparing to <A> The C1 and C2 definitely share a common node. <S> The same plate can be treated as positive when the lower plate of C2 is taken as reference. <A> Just remember that in electronics, when you see this type of notation, it means "with respect to the other side of the device". <S> So the pluses on C1 are in comparison to the minuses on C1, not necessarily anything else. <S> By looking at the design of the circuit, however, it is implied that the plus side of C1 is positive in comparison to the plus or the minus side of C2, for example. <A> The bottom (negative) plate of C1 has -Q charge and the top plate (positive) of C2 has +Q charge. <S> However, these two plates are at the same node, so they are the same plate essentially. <S> How can they have a different charge each??? <S> In electrostatics, a conductive material will be an equipotential (the potential will be the same throughout its volume). <S> That doesn't mean that the charge in the material will be uniformly distributed. <S> If (like in this case) there's an external electric field, the charge will distribute unevenly. <S> In fact, the charge has to re-distribute for the object to remain an equipotential. <S> But some negative charge has moved to the lower plate part of the object, leaving a net positive charge on the upper plate part of the object. <S> This is necessary or else the whole object wouldn't be at the same potential.
| The charge on lower plate of C1 will be negative with respect to the top plate of the C1. In this case you have a metallic object (two plates joined by a wire), so the potential is the same throughout the object.
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For electrical outlets that have on/off switches, is energy sent to the socket when nothing is connected to it but the switch is on? In several countries, electrical outlets have physical on/off switches used to control the energy flow sent to the connected appliance. If the switch is set to the on position but nothing is connected to the socket, will any energy be used/wasted? <Q> A switched outlet with nothing plugged into the outlet will generally not consume any power when the switch is turned on. <S> There are a few cases where this may not be totally true. <S> Some outlets may have an indicator to show that the outlet has a live power connection and the indicator circuit will consume a small amount of power. <S> Some indicators would be a neon bulb, others LED and some may be small incandescent bulbs. <S> Each type will consume a small but different amount of power. <A> If the switch is set to the on position but nothing is connected to the socket, will any energy be used/wasted? <S> simulate this circuit – <S> Schematic created using CircuitLab <S> For electrical energy to be consumed current must flow. <S> Air is a very good insulator. <S> We use air as an insulator on the high-voltage lines you see criss-crossing the country. <S> When we open a switch a millimeter or so the gap is sufficient to prevent current flow. <S> The gap between the contacts of your socket is much more than a millimeter. <S> No current will flow from one contact to the other. <S> No energy can be consumed. <S> If the switch is set to the on position but nothing is connected to the socket, no energy be used/wasted. <A> Power outlets which are switched on with nothing connected will use no power. <S> However, if devices are turned off but still on standby and still connected to the on power outlet then they do still waste/use a little electricity such as TV’s , phone chargers etc. <A> From the title: is energy sent to the socket when nothing is connected to it but the switch is on? <S> Yes it is, in preparation for something to be plugged in and use that energy. <S> From the body of the question: If the switch is set to the on position but nothing is connected to the socket, will any energy be used/wasted? <S> It’s very small but nonetheless energy is burned. <S> The cable has capacitance and series resistance and a small reactive current flows that passes through the copper resistance. <S> If the cable is very long then more energy is wasted in transporting voltage to the socket.
| Energy cannot be used by the socket because nothing is plugged in but, energy is slightly wasted in the transportation of voltage to the socket by the cable.
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Latching Relay That Follows +12VDC Trigger and Uses No Power Unless Switching I have an application where I have a +12VDC circuit that I need to use as a trigger for a latching relay. So, when +12VDC is present, the relay would SET.When +12VDC is not present, the relay would RESET. However, I need to have no power consumption when the system is quiescent (i.e. power draw during switching is fine, but there can't be a power draw during non-switching times). I've tried a few things, including wiring the latching relay coil through its own output contacts (so that the latching side only received a ground when the relay was in the unlatched state), however this doesn't provide enough time to latch the relay. Any ideas would be greatly appreciated. P.S. I'm using an Omron MY2K DC12, but happy to change to something different. Thanks. <Q> Minimum pulse width is stated as 30ms. <S> Assuming a 1V drop would be okay and allowing for say 45ms time, you'd need a capacitor C = <S> \$\frac{t <S> \cdot I}{\Delta V}\$ or 0.092A <S> * 0.045s/1V <S> = 4,000uF. <S> So maybe <S> 4700uF/16V. <S> Then you need a control circuit to detect the power loss and pulse the set and reset coils. <S> It may be possible to do an adequate job in your case with a few passive parts and a 2N6028 <S> but if you need reliable operation in more pathological situations such as slow brown-out of the 12V line you might want to consider more sophisticated circuitry or even a microcontroller. <A> The circuit below does what you want. <S> It Provides an ~= 100 ms latching pulse when ignition is turned on And similar delatching pulse when ignition is turned off. <S> At all other times the relay coils draw no current. <S> This is a "cut down" version of my circuit from this answer to "Power Latching Relay off/ <S> on in response to 12V ignition and 1 second ground pulse". <S> If there is a single ignition feed input then an overall power supply capacitor - charged via a diode (not shown) needs to provide enough energy to hold the IC on and to provide relay delatching energy. <S> If an unswitched 12V feed is available then the circuit can be powered from that. <S> After the turn-off action current drain is around 1 microamp. <S> I have left an additional a 1 second input pulse option to show how the relay may if desired follow a shorter pulse. <S> The two inputs are essentially identical. <S> A "better" solution may be to use an Arduino or similar as more flexibility is possible if the spec "evolves", as happens. <S> But, this should do the specified task. <S> Inverters are a package of hex Schmitt trigger inverters. <S> MUST be Schmitt trigger. <S> (CD40106, 74c14, ...) <S> D1 accepts 12V positive on ignition D2 (optional input) accepts a 12V positive going alarm pulse. <S> R1C1 and R3C3 plus following inverters provide 1 second delays. <S> A negative going edge pulls the cap low and the resistor then charges it high. <S> When the alarm pulse activates C1R1 the output pulse lasts the lesser of the alarm pulse or C1Ra timeout. <S> As the set/reset relay only needs a 10 ms pulse 'this is not a problem". <S> R1C1 & R3C3 <S> both provide ~~= <S> 100 ms time constant but values vary by a factor of 10 as examples of how different values can be used. <A> I had the same idea as Spehro Pefhany use a capacitor to store energy and a thyristor to dump it through the relay simulate this circuit – <S> Schematic created using CircuitLab <S> I came up with this, but I'm a bit worried that it might trigger from a supply ripple and short-circuit the supply. <S> but this isn't going to work with the Omron MK12K because this relay seems to use a mechanical latching mechanism and <S> the relay coils probably have the same action with reverse polarity. <S> simulate this circuit
| You need a capacitor large enough to hold enough energy to reset the relay when the +12 disappears. There are two spare inverters which can be used in various ways if desired.
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Why is my program stored in flash memory instead of EEPROM in ATmega328? As I was going through the differences (basically they are same, but they still have some differences) between flash memory and EEPROM here , I figured out that the flash memory can handle lesser amount of write cycles than the EEPROM. But, the program I upload to my ATmega328 is written in flash memory. This means I write the flash memory every time I upload a program into the microcontroller. But I think I have never written on the EEPROM. So I wonder why flash instead of EEPROM? Going through the datasheet, I figured out that the flash memory has the storage of 32 KB while EEPROM has 1 KB. If this was the reason why my programs are uploaded to flash, why didn't the designers make 32 KB of EEPROM and 1 KB of flash? The only reason I found to program on flash was the speed. Is this the only reason? <Q> EEPROM is normally used to save data between power downs/ups or for backup (in case power goes down). <S> So if you save some data (e.g. acquired by your sketch) to EEPROM, and the Arduino is powered on again (or not, that does not matter), the data is still in EEPROM and your sketch can continue with the data without losing anything. <S> And it's a good thing that EEPROM can be written more often than Flash, because once your sketch is finished and you build it in your 'product', you don't need to change it anymore. <S> Only during development you will write to Flash. <S> Maybe it happens a lot, but I doubt you do that many thousands of times (if you do, you need to change your workflow). <S> But the data on EEPROM might be written daily, hourly or even more (depending on the requirements). <A> EEPROM is more expensive per bit than flash, so it's only used where it is needed. <S> Sometimes the EEPROM is left out entirely, and applications that require EEPROM have to use an external chip or use some trickery to make flash pretend to be EEPROM. <S> It doesn't make much sense to compromise 2048 Kbytes of flash for a few bytes of non-volatile memory. <S> If I recall correctly, there were a few early MCUs that had a small amount of EEPROM for program memory (and no flash), but they could not compete. <S> The datasheet states 10,000 times minimum endurance for the flash on the ATmega328p, <S> so that's enough for most practical situations, even development where it may be reprogrammed many times. <S> ● Write/erase cycles: 10,000 flash/100,000 EEPROM <S> However, note that the data retention time may be reduced by many erase/write cycles, which is a reason for not shipping development units into the field. <S> From the ATSAME70Q21 datasheet: <S> Earlier MCUs often used OTP (one-time programmable) <S> EPROM (but with no physical window for the 'E' on production chips) for the program memory ( <S> and they may have had a bit of EEPROM). <S> There are a few MCUs that use a different technology for the program memory and can be reprogrammed more times- <S> for example TI's Ferroelectric FRAM technology MCUs. <S> Market disadvantages of FeRAM are much lower storage densities than flash devices, storage capacity limitations and higher cost. <S> Like DRAM, FeRAM's read process is destructive, necessitating a write-after-read architecture. <S> FRAM memory has a limited number of lifetime read cycles so even if the higher cost is okay <S> , it's not a panacea. <S> There certainly disadvantages of flash- <S> it's relatively slow compared to RAM, it's erased in large blocks, so <S> a lot of information has to be re-written to change a single bit, limited number of write cycles, requirement for an on-chip charge pump etc. <S> but the high density and low cost is what makes it the dominant technology. <A> The EEPROM is used for things that can change a lot, like storing some user configuration settings so they survive during reset or powerdown. <S> It is not directly sitting on memory bus, but it is accessed via address and data register. <S> Therefore code cannot be executed from it. <S> It can be erased, read and written one byte at a time, and because it is EEPROM it can handle the many erase and write cycles. <S> The page size is only 4 bytes and the hardware handles a single byte write operation as read-modify-write of 4 bytes for you. <S> It is merely a peripheral. <S> The Flash on the other hand, sits on the address bus <S> so code can be executed from it. <S> Erase can happen on 128 byte blocks. <S> If you want to change one byte, you must read 128 bytes to temporary buffer, change the byte you want, erase the 128 byte page, and write 128 byte page back. <S> And it can only work for less erase/program cycles. <S> So while it can be used for configuration storage and updates, it juat has some downsides. <S> But code execution is fast from Flash. <A> Flash and EEPROM technologies use different physics (Wikipedia explains this in some detail). <S> That explains why flash is denser (and thus cheaper) but has less write cycles. <S> You can on some chips use Flash from your programs to store information that fits this requirement (like implementing a music player). <S> (And on some ARM-based microcontrollers, there is no EEPROM <S> so you have to do this).
| In most cases flash will only be programmed once or twice, and having to erase it in large blocks rather than byte-by-byte is not an issue, so the designers will use the least expensive memory that meets their high-volume end-user requirements.
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How to raise voltage output of charge controller IC due to cable loss? I am using a CN3703 IC to charge a 3 cell li-ion pack (layout as below.) The charge current is programmed using a resistor between its CSP and BAT pins which develops a voltage drop of 200mV. The output voltage on the BAT pin is preset to 12.6V (+/-1%.) Due to cable and connector voltage loss, I want to raise the output voltage so that 12.6V (instead of 12.4V currently in my previous layout) reaches the battery. I am not sure exactly how to do this as there is no mention in the datasheet, but I came up with the following circuit (circled in red.) The idea is is that 99% of the voltage is fed back to the IC so that it raises its output by 1% to around 12.72V. Would this modification likely work? How large can I make these resistor values without them introducing significant error? EXTRA INFO: I should mention that I have already modified (U6) INA219 breakout by swapping the 0.1ohm resistor with a 0.01ohm resistor. Full Schematic from working prototype and picture of PCB UPDATE: Some very helpful comments. I have redone the layout to put the shunt before the battery sense so that the chip sees the voltage of the battery directly. I have also switched to a XT30 connector for the battery so that i am not using the same connector for the input power and the output charging cable. I have also moved the battery connector closer to the IC and used a short wide plane to connect them. Finally I have avoided using vias for the high current path by placing the low current signals on the other side. <Q> I believe your problem is in those absolutely unnecessary voltage dividers R7/R8, <S> R11/R12.You basically adding 500 Ohm load to the charger's output and also shifting feedback voltages, messing up all the charging profiles the chip is designed to follow. <S> Remove R8, R12, C10 and put jumpers instead of R7, R11. <S> Also, what on Earth is R2 doing there? <S> If you want a discharge path (again, absolutely unnecessary) for that tiny 10uF capacitor, put 1M there, not 22k <S> Finally, don't use VIAs in high current path . <S> You have through-hole pad on VIN+. <S> Why would you take a trace on one side only to bring it to the other with VIA if you already have same pad on the other side? <S> UPDATE: <S> One more thing. <S> The whole premise of this post is an attempt to tweak output of the charger to accommodate the voltage loss in some (supposedly) too long or too thin charging cable. <S> This is like attaching RV trailer to Fiat 500 and then tweaking the carburetor hoping it will go. <S> What if you change the cable tomorrow? <S> What if somebody uses different gauge cable?IMHO, you are fighting self-created problem here. <S> UPDATE 2: <S> is there a way to compensate for the lower than expected output voltage? <S> There might be, but IMHO <S> it is not worth doing. <S> First, the standard way to detect end of charge of lithium battery is to monitor <S> current while in constant voltage stage. <S> If you use correct shunt resistor and use short appropriately sized feedback traces then battery will probably be charged OK. <S> Second, you should focus on minimizing voltage losses in your wiring. <S> Heavy and as short as possible PCB traces, no VIAs, good connector, heavy and short cable. <S> See if you can find alternative to INA219 with lower than 20uA input bias. <S> Then you might be able to share shunt, further reducing voltage drop. <S> Also, I'd definitely try another chip or two to see if this lower than expected voltage is a fluke or a norm for this IC. <S> Regarding discharge resistor - similar ICs from TI and AD with built-in power path technology often use ideal diodes internally, to minimize both forward drop and leakage current. <S> Maybe it worth checking those out. <S> In any case, I would definitely use larger value. <S> Keep in mind that it essentially drains the battery when it left connected with no power applied. <A> The idea is is that 99% of the voltage is fed back to the IC so that it raises its output by 1% to around 12.72V. <S> And the battery will get charged to 12.72 V. <S> Not a good idea! <S> Due to cable and connector voltage loss i want to raise the output voltage so that 12.6v (instead of 12.4v currently in my previous layout) reaches the battery. <S> Cable and connector losses are resistive, so their voltage drop is proportional to charging current. <S> Once the charger detects peak battery voltage it will wind the current down, so the cable/connector voltage drop reduces and the battery still reaches full charge voltage. <S> This the same reason an old battery (with higher internal resistance) takes longer to charge. <A> The charge current is programmed using a resistor between its CSP and BAT pins which develops a voltage drop of 200mV. By adding this INA219 breakout, you add another 0.1Ω and thus another 200 mV voltage drop. <S> This is the cause of the additional voltage drop. <S> You should swap the positions of both sense resistors: <S> remove the INA219 breakout board and connect R10 directly to connector J2. <S> Better not to share one 100 mΩ sense resistor for both current sensing ICs because cabling introduces noise/additional resistance.
| The only bad effect of resistance in the charging path is that the current will start to taper off sooner than it would without the extra resistance, so the charge time will be a bit longer. That way the battery voltage is correctly measured and maintained at 12.6V. insert the INA219 breakout board between L1 and R10 (and have CSP and BAT still connected to R10).
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How would an alarm button work in a bathroom scale? I am given a bathroom scale that I have to reverse engineer and beneath the bathroom scale, there is a button that sets an alarm for 24 hours from when you pressed it so that it can remind you the next day at that same time to check your weight. On the circuit board, there is the button that sets off the alarm and there is a soundbox where the alarm noise comes from. How are the seconds measured? Is it in cycles, what units would they be in? How does the scale know when 24 hours is up and how does it display that information onto the LCD screen? And when the alarm, rings, it can be turned off as soon as you step on the scale and if not the alarm will sound for 20 seconds; how does this function work? All I know is that when you press the button, the LCD icon displays an alarm icon which indicates the alarm has been programmed to sound at the exact time every 24 hours. <Q> The metal cylindrical component is the lower left is a crystal. <S> The crystal oscillations are used to create an accurate timebase. <S> The black epoxy blob in the center covers the fragile bond wires of an integrated circuit (IC). <S> The IC is probably a microcontroller. <S> Most microcontrollers have internal timers to create delays, but the delays are relatively short. <S> To get long delays, software takes over. <S> Reverse engineering an IC is extraordinarily difficult. <A> "How are the seconds measured?" <S> - The exact details of this would be difficult to determine without further information about the design. <S> These are either driven by an internal oscillator or an external clock source. <S> See the component just south of C6 and C7. <S> That is a crystal that oscillates at a very specific frequency. <S> This could be the known time base. <S> The specifics of this product would be known to the designer but could potentially be determined with some poking around. <S> "Is it in cycles, what units would they be in?" <S> - This again depends on the design, but with a known clock source as described above, they would be in ticks. <S> Each tick would be the length of the clock source period, potentially divided by some prescaler. <S> If the clock oscillates at fc = <S> 32.768 kHz, the period would be 1/fc ~= 30.5 us. <S> The button is probably an input to the microcontroller, so it knows when the button is pressed. <S> Therefore, can clear/start timers when necessary. <A> The little metal cylinder is almost certainly a timing crystal. <S> Most likely with a frequency of 32768Hz. <S> At least the people I worked with referred to these small low frequency crystals as "watch crystals". <S> Inside the microcontroller there will be a crystal driver that drives said crystal and produces a clock signal. <S> This signal can be divided down and fed to timers inside the microcontroller. <S> Many microcontrollers only have generic timers, designed to track relatively short periods, it is possible to track long periods of time based on these, but it requires waking up the main CPU frequently to count the number of timer periods that have elapsed. <S> However some microcontrollers have a specific "real time clock" peripheral that can measure long periods of time* in hardware and can set an "alarm" to wake up the main CPU core later. <S> For example the PIC18F45J11 has a real time clock, intended to represent dates in the range 2000 to 2099 and can generate either a single one-shot alarm or a repeating alarm that repeats every half-second, every second, every ten seconds, every minute, every 10 minutes, every hour, every day, every week, every month or every year. <S> Unfortunately I don't think you are going to get very far with reverse engineering your bathroom scale. <S> You can figure out what the items connected to the microcontroller do, but figuring out what is going on inside the microcontroller is going to be an uphill struggle, you would likely have to. <S> Remove the epoxy blob, I believe this is possible chemically but the chemicals involved are fairly nasty. <S> Try and figure out what the micro-controller is, at best this is likely to require a powerful microscope, and even when you do get an image it may or may not have any useful identifying markings. <S> Try and figure out how to get the code out of the microcontroller, while many microcontrollers support readback of the code they often have "code protect" options to disable this, there can be possible bypasses though glitching the clock, but the exact details will depend heavily on the particular microcontroller.
| In generic terms, many microcontrollers (my guess that would be U1 in this case) have internal timers that can be programmed. With a lot of money and motivation it can be done, but you need specialized equipment and very experienced people.
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Is there a reason that dryers aren't grounded One minute ago I cried out in pain, after being electrocuted again by my clothes dryer; probably the thousandth time in my life that this has happened. I was taking out clothes after they're dry, made the mistake of touching the metal parts of the dryer, and a jolt of static electricity went through my hand. I've had this happen in different brands of clothes dryers over the years. The one I'm using right now is a Bosch, which is supposed to be a good brand. I'm guessing that the static electricity is generated by the clothes continually rubbing against each other, robbing each other of electrons. That's understandable, electrons will be electrons. What I'm wondering is, why wouldn't the metal body of the dryer be connected to the mains grounding? <Q> The metal body of the appliance is connected to ground. <S> The problem is that the clothes are charging you with respect to ground as you unload them. <S> When you touch something grounded, then a large current flows to ground through your finger. <S> You have a relatively large capacitance to ground, many 10s of pF, which stores quite enough energy at a few kV or 10s of kV for the discharge to be felt through your fingertips. <S> There are several things you can do about this. <S> As this is an electronics engineering site, the first thing to mention is what airline companies do to prevent the same problem, which is a discharge between the airplane and the fuel bowser, two large conductive objects on insulating tyres. <S> You obviously do not want a spark between the fuel nozzle and the plane fuel filler! <S> Before refuelling is started, a grounding lead is attached between them, to equalise their potentials. <S> Use an ESD strap, grounded to the drier, when you unload your clothes. <S> Although each item of clothing will carry a small charge, the energy that's released by each small zap of the clothes to you will be too small to be felt. <S> As the clothes are not conductive, you only get the charge from the tiny area in contact with your hand. <S> The problem with the drier shock you describe is that you are conductive, and the shock you feel is the entire charge that was transferred to you, going all at once. <S> ESD straps are useful as well when you're working with static sensitive components. <S> Remember you can kill a component with voltages and energies far too low to be noticed, even by your fingertips. <S> This takes some anticipation, but you will learn fast. <S> Knuckles have an order of magnitude or two less sensitivity than fingertips to an electric shock <A> The metal body of the dryer has to be grounded . <S> If this is not the case the machine could be lethal in case of malfunction - specially the heating elements tend to break and the live potential becomes shorted to the metal casing. <S> Another story is whether is about static discharge. <S> The clothes are insulators and static charge is produced by rubbing, these charges can't discharge since the cloth is not conducting. <S> When you touch it, the charge is discharged trough your body into the ground while you are touching the metal part of the machine, which is supposedly grounded. <A> What they said. <S> May work: <S> If there is some time since the dryer was running, try turning it back on on cold for maybe 20 seconds. <S> Distributing the clothes so they brush against the metal body MAY help. <S> Will work with due care: <S> Either obtain an antistatic earth strap or (much cheaper) <S> obtain a 1 megohm resistor with a wire connected to each end. <S> Connect one wire (clip OK) to a part of the dryer's metal body. <S> Then: Open the dryer door but DO NOT touch the dryer body or the clothes. <S> Wearing insulating shoes or stand on an insulating mat (eg rubber sheet). <S> Hold the other wire end and reach into the open dryer and "stir" the clothes by hand being sure not to touch any metal dryer body parts. <S> The static charge should pass from the clothes to you then via the 1 megohm resistor to ground. <S> This should ensure charge discharge at a slow enough rate to not cause pain - but still very quickly. <A> You should consider that the body of the dryer is grounded but the rotating drum might not be. <S> Perhaps the drum is supported on rubber mounts and is driven by a rubber drive belt. <S> And it would be easy to discharge that static by touching the drum and body at the same time. <S> The purpose of grounding the body is to protect you in the event that a live mains wire gets shorted to the body. <S> If you get static shocks when you touch the body <S> that is a good sign.
| So the grounding is an essential safety feature. Another method is to anticipate that you will have been charged by the clothes, and discharge yourself by the knuckles to the drier before touching the door with your fingertips. It would be easy for the drum to develop a substantial static charge, which will not kill you.
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Number of errors that can be detected with a single-bit even parity code word system? How many errors can a transmission system which uses code words with 7 data bits and a single parity bit (Checking for Even parity) detect at the receiving end. While I am aware of the fact that any odd number of errors can be detected with this kind of a system, what exactly would be the answer? Or is the answer just one? Ps. I'm not sure if this is the right Stack site to post this question. Do let me know if it isn't. <Q> I'm not sure exactly what you're asking. <S> If you fail a parity check, you are assured that there is at least one bit of the 8 in error (it may be the parity bit). <S> There may be 3, but you have no way of distinguishing the two results. <S> Further, if you have an even number of errors, they will result in no detected error at all. <S> No matter what, you have no indication which bit is in error, so parity is useless for any error correction. <S> Parity is generally used in systems which are reliable enough that the probability of having more than one error in N bits (N=8 in this example) is considered negligible. <A> A single parity bit within an 8-bit field allows 50 % of field bit errors to be detected. <S> Of its 256 possible values, 128 (50 %) of them have valid parity and 128 have invalid parity. <S> Parity on your 7-bit value does not let you specifically detect errors in the actual value, only that the overall field's bit pattern is invalid. <A> We can see the parity bit as a simple checksum. <S> No checksum system, included the parity, can assure the received datum is correct. <S> These system can however reduce the possibility of taking bad data as good, which is totally different. <S> Let's start talking about a one byte checksum. <S> For every datum (a packet) you receive, you have 256 possible values for the checksum, and of those 256, only one is correct. <S> So, 255 among 256 possible errors are detected. <S> That's no bad. <S> If now we go to the parity bit, we have only two possible values for the checksum: 0 and 1. <S> So we can detect 1 error among 2. <S> The bottom line is this, I think: we can not speak about "how many errors can we detect? <S> ", instead we can speak about "how much we improve the error detection?". <S> Depending on which side we look at the question, either we doubled the quality (because we halved the probability of undetected errors), or we have made an infinite step ahead. <S> Last thought: the parity bit protects every single byte transmitted on a serial line. <S> Often a meaningful message is made of several bytes; if we invalidate the whole message as soon as any single byte fails the parity, then every byte in the message doubles the quality of the error detection.
| Using no parity bit, there is no error detection; using a single bit of checksum (a parity) we [can potentially] detect 50% of the errors.
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Stabilize voltage input for Arduino nano on a drone I am doing a small drone project, and the 8V battery provides power to both the motors and the Arduino Nano. I expect a lot of noise from the motors, however I assume that the Arduino Nano power supply should be quite stable. Is it recommanded to use a RC circuit to low pass filter the input voltage? Or is it possible to directly connect the Arduino Nano to the battery? EDIT : Given all the comments, it seems that the circuit should look like the one below. No resistor, smaller common wires between motors and Arduino power supply, and at least 2 capacitors in parallel <Q> I would not use this arrangement, except in some rare scenarios. <S> The lipo batteries are a very low impedance voltage source, so if you use good routing, and decoupling you shouldn't need any filter. <S> Route <S> the power from as near the battery as possible. <S> You don't want to share wires that are supplying current to the motors if possible. <S> Add a large electrolytic cap, 100-1000uF <S> (depending on the size of the drone), right near the battery in the circuit. <S> Almost all drone builders do this, and it seems to reduce overall system noise. <S> Add 1 or more 0.1uF decoupling caps near the arduino, and maybe an additional smaller electrolytic. <S> Make sure your voltage regulators have enough input voltage to keep working even when the battery is near dead and heavily loaded. <S> The voltage may sag 15% or more under load, depending on the battery quality. <S> If you're regulator is 5v and your battery is 2s <S> , you're pretty much at the limit. <S> Consider switching to a LDO regulator, or switching the system to 3.3v if this is the case. <S> LC filters are sometimes used on sensitive electronics in drones, but I can say from first had experience that they can easily make things worse instead of better. <S> Explore those as a last resort, after you've tried all this. <A> If I needed to filter the power supply to the nano, I would use a ferrite bead instead of the resistor. <S> The thing about linear regulators, such as that in the Arduino Nano, is that they don't react fast enough (have a high enough bandwidth) at higher frequencies so they can't filter out high frequency noise from things that switch (like motor drivers). <S> They react fast enough to slower changes (lower frequencies) appearing on their input so that the output remains constant, essentially filtering out the noise, but not high frequencies. <S> You normally want to get short out the noise <S> (give it a low impedance path to ground) before it gets to the sensitive components with parallel capacitors if you can. <S> They also have less loss than than series filtering components <S> But capacitors have a parasitic series inductance that will limit how well they pass really high frequencies which directly influences how well they appear as a short-circuit to GND to noise. <S> It is usually at this point when you start throwing in ferrite beads. <S> You tend not to want to throw in ferrite beads or inductors because they resonate with the capacitor (and other parasitics) and if they resonate at your more prominent noise frequencies, they will actually amplify that noise rather than supress it. <A> Lose the resistor. <S> You can add extra bypass caps if you have problems, although they may or may not be necessary. <S> I do this often for robots with a large number of actuators (such as a hexapod). <S> Obviously the extra weight is less than ideal for a drone, but it's a good easy trick.
| In addition to the suggestions in other answers, a quick and dirty fix for high frequency/motor noise getting passed through a linear regulator is to simply use a separate battery (for instance a small 1S lipo) to power your sensitive control electronics. Capacitors are more ideal than like inductors, ferrite beads (a specialized lossy inductor).
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What are these types of logic diagrams called? They are not quite "logic circuits" because when I look that up I get mostly the ones with logic gates drawn, not the transistors and such like these. Also if I wanted to find a book that would teach me specifically about transistors and understand these types of circuits specifically, not digital systems in general, what should I look for? Any recommendations? <Q> Figure 1. <S> Simple logic gates made of discrete NPN transistors and resistors. <S> This are schematics of simplified logic gates built using discrete components. <S> From left to right they are NOT, NAND and NOR. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> Logic gate symbols NOT, NAND and NOR. <S> The symbolic representations allow the designer to concentrate on the logic operation rather than the electronic operation of the circuit. <S> Table 1. <S> Truth table for NOT. <S> a NOT--------0 <S> 11 <S> 0 <S> Table 2. <S> Truth table for NAND and NOR. <S> a b <S> NAND <S> NOR---------------------0 0 <S> 1 <S> 11 0 <S> 1 <S> 00 <S> 1 <S> 1 <S> 01 <S> 1 <S> 0 <S> 0 <S> (edited truth table) <A> This looks like a circuit schematic diagram. <S> What would you like to know about transistors exactly? <S> The ones shown in your diagram are called NPN transistors. <S> They consist of three pins: base, collector and emitter. <S> A small current applied to the base of the transistor <S> creates a much larger current between the collector and the emitter. <S> Therefore they can be used as both switches and amplifiers. <S> You could always just Google transistors and see the different types. <A> As others have said, those in the picture are in fact schematics, or schematic diagrams. <S> They represent with symbols actual, physical components with their connections. <S> I'd like to add one more detail regarding the specific type of circuits represented in the pictures. <S> Compared with complementary logic (or TTL by transistor-transistor logic), the RTL logic only uses pull-down or pull-up transistors with the other branch made by a resistor. <S> This simplifies the circuitry, at the expense of performance (speed, area, power efficiency). <A> I think you are looking for Schematic Diagrams - they show the detailed connections between individual components - transistors, resistors, capacitors, etc. <S> If you are using digital integrated circuits, a schematic will show the gated symbols. <S> In schematics, more complex ICs will be shown as blocks with many connections. <S> The sketches you show are simple schematic diagrams. <A> I read the original requests as the OP wanting some reasonable search terms (for Googling, for instance) to find schematic diagrams of circuits that implement logic gates. <S> His problem is that unfortunately "logic gate circuits", would generally yield circuits that use logic gates rather than circuits that implement logic gates. <S> This would appear to be analogous to "audio amplifier circuits", but this search would more likely actually provide circuits that implement amplifiers. <S> This sort of ambiguity can turn up any time you use a noun to modify another noun with no preposition or other phrasing to connect them. <S> It is an ambiguous form of the distinction between a "cardboard box" (a box made of cardboard) and a "shoe box" (a box made to contain shoes). <S> This example is unambiguous because one does not generally store or pack cardboard in a box, nor does one generally makes boxes out of shoes. <S> But because circuits can be used in a modular manner within more complex circuits terms like "logic gate circuit" are much more ambiguous. <S> Having said all this, no I don't have a solution to this problem. <S> Even appropriate use of prepositions and such likely would not help because these tend to be ignored by search engines.
| They are a form of logic gates (it depends on the input signals, but that's the meaning of the captions), built with the resistor-transistor logic (RTL).
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Does lifetime of tungsten incandescent lamp shorten by every turning on, directly connecting to electric grid, and how much? Does lifetime of tungsten incandescent lamp shorten by every turning on, directly connecting to electric grid, ie not using special device to turn it on smoothly, and how much? It was said to us in school that it shortens its lifetime. How much it (lifetime shortening) is compared to lifetime shortening by stable ligthting? If a lamp user goes away from a lamp, planning for some time of absence, how much time of absence worth turning the lamp off and then on, for the time of absence? When it is too long, it worths, because saved electricity is more expensive than the increase of price by having to buy new lamps more frequently (because of such behaviour). When the time is very short, it also worth, because lamp does not get cooled and do not get shock. Do not take in account that frequent flickering causes decreased efficiency of getting light per amount of used power. (Ie if user for some reason makes frequent flickering, calculate only price of electicity and price of lamp, do not calculate amount of light produced; and it anyway produces heat efficiently and heat+light may be goal of user, anyway, consider it is ok for him since he does so). I have found also confirmation of that the lifetime shortens in an answer on this site: https://electronics.stackexchange.com/a/214397 : During rapid switching the filament would not cool completely before the next switch-on so it would not be stressed so much. The longer the 'off' periods the harder it is for the lamp. My domestic experience was that lamps on dimmer switches lasted 'forever' relative to the other lamps in the house. This was due to the rotary dimmer being turned up over, say, half a second and gradually increasing the current while the element warmed up. There was no sudden inrush current. And this question is similar to that question, but in content part of the question it is asked about 120 hz flickering. And an answer there is already accepted. And the 2 answers there do not answer about how much the lifetime shortens, ie there is no formula nor links to investigation result papers are given. <Q> The tungsten filament thins with use and it will tend to fail when switched on because the current surge causes mechanical stress. <S> This gives the user the impression that the lamp would have lasted much longer were it not for the turn-on surge. <S> The life of an incandescent bulb is strongly (exponentially) related to RMS voltage so even a small decrease in voltage results in much longer life. <S> Unfortunately, the efficiency also drops with the voltage so you have a choice of getting more light per watt that is whiter or getting much less light that is redder and having longer bulb life. <S> Consumer bulbs were designed to have a certain trade-off like 1500 hours average life. <S> It's no problem making a bulb that will last 5000 hours, it just will be very inefficient and will use a lot more $$ in electricity to get the same amount of redder light. <S> I don't think there is much evidence that soft-start materially prolongs the life of an incandescent bulb. <S> Many years ago I did a feasibility study on prolonging lamp life in a high-reliability application and we concluded it was not worthwhile. <S> If you still have any incandescent bulbs, I suggest turning them off the moment they're no longer needed, even if you're just leaving the room for a few minutes. <A> There are actually two mechanisms involved. <S> An incandescent tungtsten filament in a vacuum will not melt - but it does lose surface material due to evaporation. <S> As a practical matter, the thickness of the filament will vary slightly from place to place on the filament, and because the filament is thinner (less cross-sectional area) <S> its local resistance will be slightly higher, and its temperature will be higher. <S> This means that the evaporation rate at that point will be slightly greater than the rest of the filament, which means that, while the filament as a whole will grow thinner, the thin spot will lose mass even faster. <S> Eventually, the thin spot will get very thin, and usually a turn-on event will provide enough mechanical stress to break the wire and cause the bulb to fail. <S> This is not necessary, and sooner or later a bulb which is run continuously will fail due to the hot spot getting so thin that it can no longer maintain structural integrity. <S> But in general failure will occur when the bulb is turned on. <S> The reason that halogen bulbs have longer lifetimes than normal bulbs is that the halogen scavenges tungsten from the inside of the bulb, then dissociates when it contacts the high-temperature filament and deposits the tungsten back on the filament. <S> You might think that you'd get more tungsten plated on hot spots (because they're hotter, naturally), but the difference is not significant. <S> So halogen bulbs are just as likely to fail on turn-on as regular bulbs, rather than failing during steady-state operation. <A> Arrhenius Law prevails for predicting reliability due to thermal stress alone. <S> The way to accelerate life testing is to raise the ambient temperature around the bulb (glass bowl). <S> Old porch light globes can reduce the life span to 3~6 months. <S> The other factor is the surge force of 10x cold start current when in sync with the 141% peak of an RMS voltage during switch On. <S> Perhaps you recall the times when the bulbs go poof, yet worked perfectly the fine last time. <S> (Yes we know this is a joke like saying the match was working last time I tried it.) <S> A bigger joke is that Trump announced the cancelling of the ban on incandecent bulbs yet said nothing about climate change issues in last nite's state of the union. <S> So you want to reduce the ambient temp. <S> A good way to extend bulb life span is to use use a dimmer or a soft start or get warm white dimmable LED ceiling spot lamps or 4500'K ~5000'K <S> 4ft FL tubes with high efficacy in tri-phosphor and 50kh life span.
| In this respect, turning on a bulb does indeed shorten the lifetime of the bulb, but especially in the early stages of the bulb's life it has no noticeable effect.
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Maximum safe gate current for IRFZ44N MOSFET I'm designing a spot welder using a high-current car starter battery, and a handful of N-channel mosfets of the given type (IRFZ44N, datasheet1 , datasheet2 ). A microcontroller (through some drive logic, probably another mosfets) sends a short pulse to open and close the gates of the power mosfets in parallel (I mean all the gates, sources and drains are connected, making one cheap "supermosfet" from 10 mosfets), sending a high-current pulse to the spot welder's pins. Since the fets dissipate most of the power in transition (either its resistance or the current is negligible otherwise), I want to open and close them very fast. For that I need to charge and discharge the gates, and for that I need to know what is a reasonable current to do that. Fast transition also helps with balancing out small differences among the mosfets. (10 pcs in parallel, with 160 or 200A pulse current depending on the datasheet per piece, and the battery supposedly can provide around 3-400A) The datasheets (linked above) didn't help me. Any idea how much current can I use? <Q> You can pretty much use as much gate current as you can pump through the parasitic inductance and gate resistance. <S> A typical strong FET driver has a peak current capability of a few amps, but layout is critical to getting good gate driver performance. <S> If you look at the Vgs vs gate charge curve you can see how much charge you have to transfer to the gate to get the Vgs and therefore RDSon you're looking for. <S> So work back from your desired switching transition time to give you needed charge in needed time which will give you the required current. <S> For example, as an approximation if you want a Vgs of 12V, (around 45nC on the chart) <S> and you want to switch in 100ns <S> , you need 45nC/100ns which is 450mA. <S> You can switch faster, but as noted above the parasitics will ultimately limit your switching speed. <S> Another consideration is EMI. <S> The faster the edge <S> the more radiated and conducted EMI you will have due to the harmonics generated. <S> You may also get voltage overshoot on the drain with extremely fast switching, again due to layout and parasitics. <S> If the overshoot exceeds the VDS rating it can degrade or destroy the device. <A> Since the fets dissipate most of the power in transition (either its resistance or the current is negligible otherwise), I want to open and close them very fast. <S> 0.35 <S> V <S> * 20 <S> A = 7 W, which is not 'negligible'. <S> However it should be OK for a single pulse of a few seconds. <S> For safe power dissipation in the transition region you could look at the Maximum Safe Operating Area graph. <S> This says the IRFZ44N can handle 20 A at 12 V for 1 ms. <S> So the 'opening' and 'closing' times don't have to be very fast. <S> These resistors also limit maximum Gate current, making the driver's job easier and evening out the current distribution to each FET. <A> With FETS rated at 17.5 milliohm max at 25A with a 400 uS pulse width, the assumption that the main losses are during turn-on "seems suspect". <S> Based on datasheet Figs 1 & 2: <S> At even 100 A/FET you need under R= <S> V <S> / <S> I = <S> 12/100 <S> = 120 milliohm total circuit resistance to get the 100A. <S> If the battery is good for say 400A <S> then you'd need significant turn on imbalance, no series inductance and very good contacts and low lead loss and ... to achieve the sort of currents that matter. <S> At 100 A at Tj=25 °C a single FET drops over 1V and at 125C, over 3V. That's far larger than the overall circuit resistance so having one FET hog most of the current <S> is 'very unlikely'. <S> Even at 400A/10 = 40A/FET <S> you have 0.5V drop (fig1) <S> = <S> 20W (but not for overly long) <S> - so both imbalance and switching losses compared to actual conduction losses seem not too much of an issue. <S> A date drive rise time of even 10 uS seems liable to be very adequate. <S> It is usual to add a series gate resistor - typically 1 to 10 ohms, to reduce gate ringing and to LIMIT gate current peaks. <S> If you have those FETs available they are usable as you propose, but if buying to suit the task, FETs with lower Rdson and higher current ratings may be no dearer and make life easier. <S> ______________________ <S> I'd always add a gate to source zener of somewhat higher than Vgsmax "just because". <S> While a spot welder is unlikely to look too much like an inductive load, Millar coupling from drain spikes can cause gate destruction if they occur - and a gs reverse zener is cheap protection.
| When using several FETs in parallel with common Gate drive you should include an individual resistor in series with each Gate, to prevent ringing between them.
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How to design an inductor which consists of four separate windings in the same core? I need to design an inductor such that there will be four separate windings at the input side (each input is coming from a different DC-DC converter) then these four windings are combined at the output. All these actions will occur in the same magnetic core. "The different converters are interleaved but they have their own power source so they can be thought as independent at each other until the inductor. The inductor will be employed in the LC filter of the DC-DC converters. I have attached the schema of the configuration and the voltages of each winding. I have modeled it in Simulink by using the magnetic domain blocks (electro-magnetic converters and reluctance blocks) but can't be sure if it theoretically makes sense or not because the simulation gives different flux (for each winding flux) results at each run, although other currents or voltages on the simulation are same at each time. My research on the web did not end up well so far. I would highly appreciate it if anyone can give an idea or show me a reference. Figure 1: Inductor configuration Figure 2: Winding voltages <Q> If the intent is simply to provide smoothing on the incoming pulses, then you do NOT want coupling (transformer action) between each winding. <S> There are screened inductors available to further reduce mutual coupling if necessary. <A> if you connect all the windings together at the non-dot end then you may as well have just used a single inductor because all the windings are coupled magnetically. <A> Sounds like you are implementing a multi phase coupled inductor buck converter. <S> This search term could set you back months or years, researching for the many ways it can be done, optimised and tuned. <S> I have seen research papers dated back to 2001 regarding them, and they are still an active research area within converters. <S> They seem to be 99% concentrated on the opposing flux version. <S> Main features of this setup are: No change in capacitor ripples compared to your non coupled version Individual inductor current ripples can be highly reduced, depending on duty cycle and coupling factor. <S> Above could be traded for smaller inductance = <S> > <S> Less turns = <S> > <S> High efficiency = <S> > <S> Faster step response. <S> If fast step response is required, like in a cpu supply, output capacitance could be decreased. <S> This as you get an effective transient ripple = steady state ripple/N-phases. <S> Opposing flux will much reduce core flux, resulting in less air gap, which could result in less turns again for the same inductance. <S> A downside is that if duty cycle goes above 1/N-phases, you will need a lot leakage inductance, as delta flux (in other words inductance) is otherwise completely cancelled. <S> Another downside is the core material is effectively handled with N-phases times the switching frequency. <S> So peak B field must decrease to keep the same core losses. <S> Below are some examples of implementations And a link to a maxim article to get you started.
| Simply use 4 separate inductors, and common the output terminals. There is nothing to be gained from coupling the windings on a common core, and quite a lot to be lost (e.g. edges from one input creating spikes doubling the voltage on another; possibly destroying its driver).
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Are screened transistors an endangered species? In the germanium transistor era, the RF transistors were PNP small signal devices. Some of these small metal cased BJTs had 4 leads instead of the usual 3. The metal can was isolated and the 4th lead was called the screen and was connected to an internal screen. Common european example part numbers of such devices are: AF116 AF117 AF126 I have even seen such transistors in early solid state MW AM car radios, and the popular Phillips 1960s electronics set had an AF116. Sure there were some reliability issues with the screens shorting due to tin whiskers but they were not that bad. The long leads of the day would undermine the effectiveness of the screen at HF but the lead could be cut shorter. Why don't they make SMD screened Si or GaAs or GaN devices today? Is a screen expensive to manufacture? Would there still be reliability issues? <Q> The normal approach for small SMT components is to put a single big shield or "can" over a number of components: (image from this link ) <A> They did make it into the silicon era ... <S> just. <S> Look for any transistor in the TO-72 package for example : it's a 4-lead version of the TO-18, with the extra lead connecting to the case. <S> A few examples : BF115 (perhaps following the AF116?) <S> but also BF173, and 2N4416 JFET. <S> And the 2N5179 - still <S> (just?) <S> available from Farnell. <S> I believe what killed them was the move away from metal cans (TO-18, TO-72, TO-5 etc) to epoxy (TO-92 etc) packaging, where screening would cost extra. <S> Inside those shiny cans in pjc50's answer you will often find further internal partitions, between stages, so the idea hasn't gone away altogether. <A> For SMD components, that would only happen in the millimeter wave regions. <S> For millimeter wave circuits, you basically need to consider your whole circuit including the circuit board as RF element that you'd want to simulate to know what's happening: you'd not expect any "stray" radiation coming from the outside, since you'd typically RF-encase the whole circuit, and no stray radiation from the circuitry itself, since everything needs to be designed as transmission lines rather than simple current-carrying traces, anyway.
| RF screening makes sense if any of the elements come even close to a length where RF of relevant frequencies can couple in.
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Why do some SMPS power supplies require an input voltage select switch? I have 2 SMPS power supplies near me. One is an ATX power supply, the other is for a 3d printer. They both are rated for 110-240v in, but the ATX supply can handle the full range automatically, while the other supply has an input voltage select switch One quirk I noticed with the supply with the switch, if I flip that switch to 230v, with my 115v mains, the output is half the rated voltage. I assume that is not a good idea to do that. What does that switch do? What is so different about these powersupplies that one needs a switch and one does not? What are the risks of having the switch in the wrong position? <Q> This was common on older ATX power supplies as well, to use single design that can be used globally. <S> Today the PFC stage at the input can handle larger input range automatically. <S> In the supply, there might be two 200V bulk capacitors for storing rectified mains. <S> In the 230V position, the bridge rectifier is connected as a full bridge to the two series connected capacitors to effectively use them as a single 400V capacitor. <S> Stored voltage is about 325V peak, or about 162V per capacitor. <S> In the 115V position, the bridge rectifier is connected as a half bridge to a single capacitor, where positive mains cycle rectifies the 162V peak into one capacitor, and negative mains cycle rectifies the 162V peak into other capacitor. <S> The total voltage over both capacitors would then be again 325V. <S> So if you use the switch in 230V position and feed in 115V, the input voltage is half what is needed and due to undervoltage it may not work at all <S> or it might be so simple design that it just blindly tries to work but output halved as well. <S> As this is not intended operation it may damage the parts. <S> Using it the other way, switch in 115V position but feeding in 230V will be much more dangerous and spectacular. <S> The fuse might blow immediately, and if it does not, there will be a lot of overvoltage at the capacitors so they can heat up and vent up electrolytes as vapour after a while (seen this) or <S> the capacitors may just explode outright. <A> That switch could do anything. <S> More often than not, especially on Chinesium power supplies, having a switch like that indicates a lack of a PFC stage. <S> 1) <S> That switch most likely switches which taps on their transformer are used. <S> Say their supply needs 60V on the secondary, for 120 you need a 2:1 ratio, for 240 you need a 4:1 ratio. <S> They may have 2 2:1 primaries that are either wired in series or parallel based on that switch for both voltage ranges. <S> This may be why your output voltage is half of what it should be. <S> Different cheapo supplies use different methods. <S> 120 uses twice the current as 240 for the same amount of power so things need to be sized/done differently. <S> 2) <S> Like I said earlier, your ATX supply most likely has a PFC stage which takes the AC input and boosts it to ~400VDC <S> before the DC/DC converters turn that back into 5, 12, etc. <S> Because the boost can work on anything less than 400Vpk, the AC input can be pretty much anything (the "universal range" is 90-264VAC). <S> 3) <S> The risks? <S> Fire, death, bodily injury, you know, the usual things. <S> If everything in the supply is sized for the 120V currents, then it probably just won't work well, but copper is expensive and corners can be cut. <S> If you see an agency label on it then they test all these things and you'll be fine either way. <A> Figure 1. <S> The AC input section of STEVAL-ISA018V1: 150W, 24V @ <S> 6A, 85 <S> ~ 185/185 ~ 265VAC PSU. <S> Source: Digikey . <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> (a) and (b) 240 V configuration. <S> (c) and (d) 120 V configuration. <S> The peak voltage of the AC mains is \$ \sqrt 2 V_{RMS} \$ . <S> 240 V AC peaks at 340 V and 120 V AC peaks at 170 V. <S> With the voltage selector switch open as shown in Figure 1a the circuit becomes a simple bridge rectifier as shown in ! <S> b. <S> At 240 V AC there will be 340 V DC (approx.) between DC+ and DC-. <S> With the voltage selector at 110 V the circuit becomes two simple half-wave rectifiers with 170 V DC across each capacitor resulting in the same 340 V DC between DC+ and DC-. <S> The linked schematic is worth studying. <S> Note in particular their use of three different ground symbols. <S> simulate this circuit Figure 3. <S> Earth / ground symbols. <S> They've used (a) as the chassis ground. <S> They've used (c) as DC GND. <S> My guess is that they only had one earth/ground symbol left so they used it, the earth symbol, to indicate the internal ground for the mains parts of the circuit. <S> It is not connected to Earth. <S> I would have tried to avoid using it in that way. <S> For more on ground symbols you can see what I've written on Ground, earth and chassis explained .
| Most likely the switch controls how the mains input is rectified into the input bulk capacitors.
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Is this lab power supply linear or switch mode type? I'm using a lab power supply which has a manual here . It has the following block diagram: From the manual or the block diagram can we infer whether the supply is made up of switch mode or linear type of supply? If it is impossible from the manual or the diagram, can one perform a test to verify this? <Q> Its a linear supply. <S> The manual you linked has as image of the unit displaying a 3303s badge. <S> As others have mentioned the transformer and series regulator also points to this. <S> For bonus points, feel the weight of the unit. <S> Typically linear supplies with output current ratings like this have pretty hefty transformers <S> so if your arm aches its most likely linear Datahseet here: Datasheet <A> From the sheer size of that unit, it must be linear. <S> A switching supply of that current capacity would be very small. <A> The block labeled "Series Regulator" is a pretty good clue that it is a linear supply. <S> These words imply that a hefty transistor in series with the current flow is used to control the output voltage, and the fact that it is driven by a "Voltage Comparator" suggests linear control to me.
| Looking up the datasheet for that model confirms its linear.
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Powering LEDs, dim LED in parallel I want to build an light fixture for my aquarium and decided to use LEDs, almost everywhere it's suggested to wire the LEDs in series. but that will need a high voltage for many high power LEDs. also if one of the LEDs in series for whatever reason burnt, all LED's will turn off. The main reason to wire LEDs in series (if I understand correctly) is because of the manufacturing variation in LEDs, that may cause one LED pull more current than the others if wired in parallel. Let's assume that we've 5 LEDs, each needs 1 V / 10 mA and it's connected to a power source which provides 1 V and 60 mA constant current: simulate this circuit – Schematic created using CircuitLab In above diagram each LED can't pull more than 10 mA and resistors will not get hot either because the power source have a current limit. But what if the little variation in LEDs cause one LED to pull less current than the others? what's the solution for that? <Q> The solution for this is to use a higher supply voltage. <S> If there is a large voltage drop across the resistor, then small variations in the forward voltage of the LED will cause relatively small variation in the voltage across the resistor, and therefore small variations in LED current. <S> With a supply voltage close to the LED forward voltage <S> there must be a very small voltage drop across the resistor, so that small changes in LED forward voltage cause large changes in resistor voltage, and large changes in current. <S> So, LEDs that have a bit higher forward voltage get significantly less current and appear dim. <A> But what if the little variation in LEDs cause one LED to pull less current than the others? <S> what's the solution for that? <S> For the combination you have provided, the current will be less than 10mA per LED resistor combination because of voltage drop of the LED. <S> This will not have any influence on the other LEDs as such because all of them are in parallel. <S> Even if the LEDs are in series, the little variation in the LEDs will be there as a variation in the luminous intensity but they will be barely palpable. <S> Choosing LEDs from a reputed supplier from a same batch and bin will yield the best results. <S> You are wasting about 40% of the power considering forward voltage drop of 3V. With a 5 V supply. <S> It is my assumption but in any case, it is a huge waste in power and there are better ways to drive the LED. <S> Opt for a constant current LED driver if possible <A> LEDs cannot be made the same even if they have the same part number. <S> The datasheet shows the range of forward voltage. <S> You can buy many LEDs, test them all and group them into piles that have the same forward voltage then throw away the ones that cannot be in any group. <A> I want to build an light fixture for my aquarium and decided to use LEDs, almost everywhere it's suggested to wire the LEDs in series. <S> The reason you see this suggestion is that you should run them in series. <S> Parallel <S> like you are doing is more difficult, less efficient, produces worse results and is more dangerous. <S> There is no reason to do it. <S> also if one of the LEDs in series for whatever reason burnt, all LED's will turn off. <S> Being fail safe is a very good thing. <S> If you circuit is burning, you want the power disconnected. <S> Dumping power into an overheating circuit is a fire hazard.
| You can buy some LEDs that are already grouped into what they say are binned. If one of the LED pulls lesser current than the others, it will be simply less brighter due to a higher forward voltlage drop compared to the rest of the LEDs.
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What is the purpose of the ferrite bead on DC Adaptors? According to WikiPedia : Ferrite bead suppress high frequency noise. In the case of wall power adaptors, are they trying to protect the device being powered or are they protecting the PSU from high frequency noise? Why is it placed on the plug end rather than on the adaptor end? What would be the consequence were the ferrite bead not installed? Thank you! <Q> These are similar to all the ones found on VGA cables. <S> The DC power also conducts broadband spectrum on unbalanced impedance wires from the switched current transients. <S> The purpose of this Ferrite, Folding clamp, split clamshell, ungapped component is to act as a BALUN or to BALance UNbalanced lines at RF frequencies. <S> The other functional name is a Common Mode (CM) Choke. <S> Since the AC-DC converters carry lots of harmonics up to 10MHz this ferrite raises the impedance of both lines DC+/- such that they become balanced radiators and thus cancel each other out for far-field EMI emissions. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Two main bands exist: MnZn ≤ 10MHz(AM band range) <S> (most common for DC power cords) <S> NiZn ≤ <S> 300MHz(FM band range) <S> options available •Split construction , precision mating surfaces ref <A> It removes noise, specifically high frequencies. <S> See first paragraph from Ferrite Beads <S> Demystified : <S> Also see the related question Ferrite Bead: How much impedance do I need? <A> Firstly, it’s not a ferrite bead; it is much bigger than what is generally regarded as a ferrite bead and, as such, it will have a noise reducing effect in the low hundreds of kHz to probably over 10 MHz. <S> Generally, ferrite beads are only useful above 10 MHz. <S> I suspect that the ferrite core is placed on the wire to ensure that the wall wart complies with emitted noise regulations. <S> Which regulation? <S> There are many but the device markings might give a clue. <S> However, given that it is on the barrel plug end of the cable, one can be suspicious that it is conducted emission regulations that were the problem. <S> If the non compliance were radiated emissions then it’s more likely that the ferrite core would be much closer to the body of the wall wart. <S> But who can really say? <S> You can’t rule out that someone in the manufacturing side of the product was leant on by someone from sales (to move the core) after many devices were returned due to the core being broken by people smashing their vacuum cleaners into it when plugged into a low wall socket. <A> The permeability of the ferrite bead increases the inductance of the wire going through it. <S> This increased inductance increases the impedance of the wire at higher frequencies. <S> The increased impedance reduces the amount of RF common mode power that can flow in either direction: reducing harmonics from a power supply's DC-DC converter switching frequency from entering the device as RF noise, and reducing any digital RF noise created by the device from using the power supply cord (and building's AC wiring) as an antenna that can radiate EMI into the environment (potentially exceeding EMC regulations, such as CFR 47 Part 15 in the U.S.) <S> Both of the above are especially important with any radio equipment in the room, where one want to reduce the RF noise from interfering with signals of interest. <S> (radiated and conducted EMI, etc.) <S> To hear this RF noise, take an old analog AM radio, tune it between stations, and hold it next a cheap LED light or USB power supply. <S> You often will hear a very nasty buzzing. <S> You might want to reduce the amount of this power supply noise that gets conducted into a radio receiver or other circuits or equipment that might be sensitive to RF noise. <S> A poorly filtered power supply can increase a radio's RF noise floor and completely obliterate amateur radio weak DX signals. <S> Often a single ferrite bead is not enough, so one might add multiple clamp-on ferrite beads to the cord(s), or wind multiple turns of the cord(s) through a ferrite toroid of a proper mix to increase the impedance at an RF frequency range of interest. <S> Amateur radio operators are known to do this, not only for power supply cords, but for every network, USB, and video cable in the room. <A> As you stated, the ferrite is installed to suppress high frequency noise. <S> As you have also pointed out, the location of it is close to the plug end that goes into the device it is powering. <S> This may indicate that its purpose is to suppress noise generated by the end device from getting onto the cable. <S> Cables can act as antennas, and any high frequency noise on the cable can radiate, potentially beyond limits imposed by governing agencies. <S> If the end device requires EMC compliance testing, it may be certified with a specific power supply PN, and they are certified together. <S> The ferrite this present to provide the best chance of passing emissions requirements. <S> If it is removed, the emissions may change rendering the certification invalid.
| An effective method for filtering high frequency power supply noise and cleanly sharing similar voltage supply rails (that is, analog and digital rails for mixed-signal ICs) while preserving high frequency isolation between the shared rails is the use of ferrite beads.
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Supporting SMT electrolytic capacitors in high vibration environments I recently experienced a situation where an engine ECU failed. The post-failure analysis showed that two capacitors had broken loose from the board due to stress from vibration. I have replaced the capacitors, but without any changes, I assume that this will happen again. How can I provide proper mechanical support for these? Edit: the capacitors are Nichicon UUT1V220MCL1GS <Q> RTV silicone is used for this (and to adhere components to PCBs permanently in general). <S> The neutral cure kind that doesn't produce acetic acid and smell like vinegar when it cures. <S> So not most of the stuff you will find in hardware stores. <S> RTV-162 is one of the purpose-designed silicones for this <S> but it's almost more than double the cost of some other neutral cure alternatives like RTV-108 or RTV-6708. <S> From comparing the datasheets, the main difference appears to be that RTV-162 is twice the hardness and strength. <A> <A> We use some sort of structural adhesive for these kinds of applications, where staking is required. <S> Since this is for a high-volume ECU, I assume the proper structural/vibration analysis was done? <S> If so, the first thing you need to do is go back to that analysis analysis and figure out where the disconnect is between the model and the real world. <S> Then you will be in better position to ascertain whether staking is a way out of your problem. <S> EDIT 1 - Added typical vibe profile <S> The plot below is what comes out of a properly done vibration analysis. <S> The colors represent the amount of deflection, with blue being the smallest (the board being constrained along its edges), and red the highest. <S> A structural engineer then takes the mass properties of the components, the amount of movement and frequency (which are used to compute the G forces) and figures out what the stresses are, and where. <S> Lacking such an analysis, you're left with just using guesses as to what kind, where, and how much adhesive to use. <A> Interesting failure, and not surprising when you feel how loose the bases often are. <S> I think you'd be more than fine with a couple drips of epoxy or non-acidic RTV silicone between the can and the plastic base before the cap is mounted. <S> Personally, I'd use 5-minute cure dollar store epoxy.
| Perhaps looking at the case, you can add some thick tape or something else that you come up with to the top part, which would push down on the capacitor (gently), since the case will be vibrating the same direction as the PCB it should be fine.
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Why we need two resistors to control the gate voltage of a PUT transistor? Looking the follow schematics you can see two resistors 15k and 27k that are controlling the gate voltage of the PUT (programmable unijunction transistor) transistor. My question is why we need two resistors to control the PUT's Gate, while we can do our job with only one carefully selected resistor? <Q> When the 15K resistor is left out, the gate of the SCR will be pulled down to the negative terminal (which is probably ground, but isn't shown in the schematic). <S> When it is pulled down, the SCR will never turn on. <S> When the 27K resistor is left out, the gate of the SCR is pulled up. <S> This will (almost) always turn on the SCR (provided that the terminal voltage (6V in case of this schematic) is high enough such that the gate trigger voltage is exceeded and <S> the resistor (here 27K) is such that the gate trigger current is exceeded). <S> Using a voltage divider only makes sense if the terminal voltage can differ. <S> Otherwise, for a fixed voltage, one can leave out the 27K resistor. <S> For varying terminal voltage, tweaking the voltage divider determines at which terminal voltage the SRC is triggered. <A> The voltage source needs to be 3.86V and the resistor would be 9.6K to duplicate the behavior of the circuit shown. <S> The voltage determines the trigger point of the PUT and the resistance determines the valley current. <S> Below from the Onsemi datasheet. <S> With the normal way PUTs are used, we pick R1/(R1+R2) to program the trigger point (usually something like 2/3 of the supply voltage) and R1*R2/(R1+R2) to program the valley current. <S> ~2/3 <S> or about (1-1/e) of the supply voltage <S> gives you about 1 time constant timing, without getting into the part of the exponential rise where the capacitor voltage rises too slowly and thus becomes much more sensitive to PUT temperature and noise on the supply. <A> Two resistor form a voltage divider that set the reference voltage. <S> One resistor dont doo dat.
| You can use one resistor if you have an appropriate voltage source available.
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What consumes the most power in a computer I have a quite basic question: As far as I understand (correct me if I am wrong), we can summarize a computer by saying it is composed of wiring, transistors and some electronic components like capacitors for example. The electric consumption of a computer mainly come from heat dissipation if I am not wrong. Does this heat dissipation mainly comes from Joule effect in the wiring between the components, or from the components themselves (transistors) ? <Q> A computer is usually defined by what it does rather than how it is built. <S> Computers can (and have been) built from vacuum tubes, relays, discrete transistors, and integrated circuits (not a complete list). <S> A simplified definition of a computer is that it is a device that takes inputs, operates on them in accordance with a set of instructions (which can be changed by the operator or even during operation) and then provides outputs which can be used to perform operations such as printing, running machinery, playing games, etc. <S> All of the electrical energy consumed by a computer is turned into heat except for the relatively small amount that is contained in its outputs. <S> This energy is especially significant when the transistors change their states by switching between logic levels. <S> This is why microprocessors dissipate more energy as their clock frequency increases. <A> A desktop computer produces very little in the way of energy output relative to what it consumes. <S> A bit of RF deliberately for Bluetooth and WiFi, a bit of current in Ethernet cables, some RF that is accidental, and some light. <S> Just guessing perhaps a watt or less, if we draw the 'box' around all the peripherals that it is powering such as USB keyboards and such like. <S> So yes, most of the power input ends up as heat. <S> Aside from the losses in the various power supplies, the chips themselves have a leakage current that is always there (in simple circuits, anyway) and a dynamic current that is caused by the charging and discharging of millions or billions of tiny capacitances within the chips. <S> The faster the nodes change, the higher the voltage change, the larger the capacitances, and the more nodes are changing, the more current and thus the more power. <S> Both of them act as loads on the power supply rails. <S> Most of the power would be dissipated in the transistors (and for the dynamic power during the actual switching), not in on-chip or off-chip wiring. <S> The dynamic power is proportional to the square of the supply voltage. <S> Each year, the chip features tend to get smaller (so less capacitance per node), the internal core supply voltage lower, and thus the power consumed per calculation tends to drop. <A> The consumption in a computer is mostly due to transistor power consumption in the IC's. <S> The display (whatever type) also consumes a lot of power. <S> In a laptop, the LED backlight might be roughly 20% to 50% of the total. <S> That is just kind of an estimate. <S> The joule heating in wires and such is not very significant compared to semiconductor power consumption.
| The heat is partially from energy consumed in running current through the computer wiring but is primarily from the energy consumed by the transistors in the logic circuits that simultaneously have voltage across them and current through them.
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Estimating current when voltage drops > voltage input Let's say I have a circuit with a 5V battery and a 2V LED and a 100 ohm resistor. Ignoring certain properties of an LED, I can estimate the current by doing: 3V / 100 ohms ==> ~30 mA However, what happens when I have 3 LEDs in the circuit? Using the above approach, I would get a current of zero (or an open circuit or whatever it would be called). But this isn't really the case, as shown in a simulator, where I get ~1 mA: Does using the voltage-drop approach not work where there may be more drops than the input can support? If so, what would be an example of calculating the below circuit to get something 'close to' 1mA . <Q> In practice its voltage will vary depending on the current. <S> In your simulated circuit, you put 3 LEDs in series across a 5V source. <S> Assuming the LEDs are identical, then each one will drop 1/3 of 5V or 1.67 volts which is what your simulator shows. <S> However the current depends on the actual current/voltage characteristics of your LED. <S> The simulator probably assumed some generic model and came up with a current of 823 uA. <S> That is not unreasonable for an LED operating well below its rated voltage but <S> the exact value, as already mentioned, depends on the particular characteristics of your LED. <S> This is true in general when operating devices above or below their ratings. <A> You can solve this graphically, if you have a current vs. voltage curve for the LED. <S> This is an actual measured curve for a red LED. <S> I can't use your exact example because the voltage is outside of the curve. <S> But, here is a similar analysis. <S> If your voltage was 6V, and you had 3 of these LEDs in series, what would the current be? <S> 6V / 3 LEDs = 2V per LED. <S> Find the intersection of the curve and read the current, about 4.4 mA. <A> Again, I don't know the actual parameter values that your simulator is using. <S> But it is probably using something like \$\eta=4\$ <S> (the emission co-efficient.) <S> You know about \$V_T=\frac{k\,T}{q}\$ as the thermal temperature voltage and again I'm going to assume a saturation current of <S> \$100\:\text{pA}\$ for the LED and thereby move directly to a closed solution. <S> The diode voltage is: $$ <S> V_\text{D}=\eta\cdot V_T\cdot\operatorname{ln}\left(\frac{I_\text{D}}{I_\text{SAT}}+1\right)$$ <S> Using KVL, we find the following ( \$V_\text{CC}\$ is the power supply voltage): $$\begin{align*}V_\text{CC}-3\cdot\eta\cdot <S> V_T\cdot\operatorname{ln}\left(\frac{I_\text{D}}{I_\text{SAT}}+1\right)&=0\:\text{V}\\\\\frac{V_\text{CC}}{3\cdot\eta\cdot V_T}&=\operatorname{ln}\left(\frac{I_\text{D}}{I_\text{SAT}}+1\right)\\\\e^{^\frac{V_\text{CC}}{3\,\cdot\,\eta\,\cdot\, <S> V_T}}&=\frac{I_\text{D}}{I_\text{SAT}}+1\\\\e^{^\frac{V_\text{CC}}{3\,\cdot\,\eta\,\cdot\, V_T}}-1&=\frac{I_\text{D}}{I_\text{SAT}}\\\\&\therefore\\\\I_\text{D}&=I_\text{SAT}\cdot\left(e^{^\frac{\left[\frac{V_\text{CC}}3\right]}{\eta\,\cdot\, V_T}}-1\right)\end{align*}$$ <S> Which is what you'd expect. <S> This is just the Shockley diode equation, except that \$V_\text{CC}\$ has been divided into three parts before computing the result. <S> Using \$V_T\approx 26\:\text{mV}\$ , I find \$I_\text{D}\approx <S> 912\:\mu\text{A}\$ . <S> Which isn't far from what you got. <S> (For others not yet following along with all the "David" questions recently, there was another one where I used \$\eta\approx 4\$ and <S> \$I_\text{SAT}=100\:\text{pA}\$ <S> as approximations for the diode he is simulating. <S> I kept these values here, too, for consistency.) <A> You can't force voltage drop. <S> Suppose you have three identical things which drop 5V each. <S> And you connect them across a 9V supply. <S> What actually happens? <S> They drop 3 volts each . <S> Why? <S> Because they have to . <S> So what happened? <S> Clearly you got that 5V voltage drop number from somewhere : a data sheet, or calculations. <S> Both would have made certain assumptions about the default or nominal conditions. <S> The data sheets explicitly say that. <S> For instance a lot of lighting LEDs that advertise 1400ma, you read the data sheet and all the data on it is based on testing at 350ma, and the data sheet typically shows you how performance varies up to 1400ma, expressing it as a percentage of nominal at 350ma. <S> So you need to figure out the drop you're going to get, either by guesswork (the 3V above) or measurement, and go see what the data sheet says about that.
| First of all, in your initial calculation, you assumed that the LED will drop 2V no matter what current is flowing through it. If you have a data sheet for the LED, it should specify the current vs. voltage for the LED over some reasonable range.
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Determine the voltage across the 5.0-Ω resistor in the drawing I have tried to solve this problem but I haven't seen a dc circuit with a battery on the shared portion so I don't know how to include the 2.0 volt battery in the circuit rules. Please help <Q> You will get a resistor in series with a voltage source. <S> Then combine that with the circuit on the left. <S> It will be simple to solve since everything is in series. <A> If you are studying circuits you look in you text book or online how to solve circuits like this with loop mesh equations. <S> Very easy with just two loops like this circuit. <A> You can use superposition. <S> Replace each pair of voltage sources in turn, leaving one, with a short and calculate the voltage at the T-junction for each, then add all three voltages. <S> So you'd be adding the voltage due to the 10V battery, the voltage due to the 2V battery and the voltage due to the 15V battery to get the total. <S> The difference between that and the other side of the 5 \$\Omega\$ resistor is your answer.
| One approach is to find the Thevinin equivalent for the circuit to the right of the 5 ohm resistor.
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Can I use a low dropout voltage regulator to drop a 5V 100 nSec pulse to 3.3V? I am using an analog switch IC with 3.6V as a maximum operating voltage. The input to the analog switch is the output pulse from a multi-vibrator which is 5V and 100nSec width. I need to drop the vibrator output to less than 3.6V without affecting the pulse width. In this case, can I use an LDO to drop the voltage? <Q> A regulator is not a good option. <S> Too slow and possibly not stable without capacitors. <S> You could use a resistive divider, but another option is to use a purpose-designed level translator chip such as the 74LVC1T45 , which can work from 1.65V to 5.5V in either direction. <S> When converting 5V to 3.3 it has a propagation delay of less than 5.4ns over the whole temperature range. <S> It also draws hardly any current normally when not switching. <A> In this case, can I use LDO to drop the voltage? <S> No because virtually all (LDO or non LDO) <S> regulators require input and output capacitors and, all voltage regulators are far slower in operation than that needed to main the signal's shape and edge integrity. <S> Use a resistive potential divider. <A> An LDO would probably not turn on in 100 ns (note standard SI units). <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> A voltage divider to convert a 5 V signal to 3.6 V. <S> The resistor values were chosen for easy calculation. <S> You can scale up or down but watch out for loading effect of following circuit. <A> The resistive divider solution so the simplest one. <S> Go ahead with those. <S> If you see loading effects (sagging of signal transition) you can also opt for active clamping using zener diode or BJTs (consider the inverted logic, if it going to MCU it can. <S> be handled, else invert it again..or go with Zener clamping or dedicated level shifter IC) <S> depending on what you have handy. <S> Image source: Fig. <S> 7, Resistor and Zener diode from "How to interface a 5V output to a 3.3V input" on next-hack.com Image source: Fig. <S> 10, BJT/MOSFET as inverter can be used as level shifter from "How to interface a 5V output to a 3.3V input" on next-hack.com <A> No. <S> LDOs are there to reduce any noise, disturbance on their input and provide a clean and stable output voltage. <S> They are voltage regulators. <S> Their stability also implies that they are slow. <S> You can use a resistor divider as @transistor suggested, but then you might have other troubles depending on the capacitive load of the switch and on the node on its output. <S> Basically you introduce a series resistance; Your signal source is not an - approximately - ideal voltage source any more, but it will have a series resistance (look up 'Thevenin equivalent circuit'). <S> In the mentioned example it is 1kOhm, which will form a 10MHz low-pass filter with only 16pF, which you can get easily on a PCB if you are not careful. <S> The period of 10MHz is 100ns, which is your pulse width. <S> To compensate the filtering you might add a capacitor in parallel with the upper resistance to cancel the pole. <S> Moreover your signal will be driven with 1kOhm, which means it is more susceptible to capacitive coupling. <S> You can add a unity gain buffer amp after it to alleviate this if it is really a trouble, but it is an extra component with additional power consumption. <S> In general I would use components which all can withstand the supply voltage of the system, or the other way around: choose a supply voltage where all of your components are safe. <S> There are way too much trouble in supply domain crossing, and very often it can be avoided. <S> Most of the analog ICs even nowadays works with 5V. <S> The exceptions are really high speed circuits, but the 100ns pulse is - arguable though - not extremely fast. <S> Why analog switch? <S> Isn't it a digital signal? <S> A signal pulse seems to me a digital signal. <S> If this is the case, just pick a digital level translator. <S> They are failsafe, do not burn any static current <S> and they drive their outputs properly. <S> I am really confused by the description of using an analog switch for a digital pulse. <S> If the signal source is digital, then you do not need an analog switch.
| Instead you can probably use a voltage divider. There are digital level translators.
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Why is the metal tab of a BJT power transistor is always connected to its collector pin? Until now I thought the metal tab of a BJT power transistor is connected to ground but it seems like it is connected to its own collector pin´as shown in the below illustration: Is there a particular reason for this? <Q> The electrical connection at the bottom of the diagram also makes a good thermal connection. <S> Very important for a power transistor. <S> So, why can't you just reverse the Emitter and Collector in the diagram. <S> You could, but it would be a lousy transistor. <S> https://en.wikipedia.org/wiki/Bipolar_junction_transistor#Structure <S> The collector surrounds the emitter region, making it almost impossible for the electrons injected into the base region to escape without being collected, thus making the resulting value of α very close to unity, and so, giving the transistor a large β. <A> If you can access the website Wikipedia, see the article titled 2N3055 . <S> Scroll down the page to the figure titled 2N3055 transistor internals ; it shows how this particular power NPN transistor is mounted inside a TO-3 case. <S> The collector is the gray colored substrate material (look at the edges of the square shape, and in the "trough" that snakes back-and-forth between the base and emitter, which are on top). <S> The collector is mounted/bonded directly onto the metal can, and this is why the collector voltage is present on the metal can. <S> Bonding the semiconductor die directly to the metal can in this manner optimizes the flow of heat from the semiconductor die—which dissipates power and produces heat—to "the outside world" through the metal can. <S> For what it's worth, many electronic component case styles bond the silicon die to a metal pad to facilitate heat flow out of the die—e.g., TO-220, TO-262, leadless chip carrier, just to name a few—and quite often that metal tab has voltage on it. <S> (NB: When attaching a metal heat sink to a component, quite often an electrical insulator of some sort is installed between the component's metal pad and the metal heat sink, and special fastening hardware is used, to ensure the heat sink is galvanically isolated from the voltage on the component's metal pad. <S> For example, try an Internet search using "TO-3 insulator kit". <S> In this way a single block of metal can be used as the heat sink for multiple electronic components, and if necessary (for operator / equipment safety reasons) <S> the metal heat sink can be electrically bonded to ground potential.) <S> (NB: A specially-engineered material called thermal grease (a.k.a., thermal compound, thermal paste, heat sink compound) is used in some heat sink applications when attaching a heat sink to an electronic component. <S> If air is trapped between a component and its heat sink, the air acts as a thermal insulator and limits the flow of heat from the component into the heat sink. <S> If this happens, heat gets trapped and accumulates within the component, and the component can be damaged or destroyed as a consequence of excessive heat buildup. <S> Therefore, thermal grease is applied to the various mating surfaces to remove/replace as much air as possible between the mating surfaces and thereby optimize the flow of heat out of the component and into the heat sink.) <A> Collector receives both majority charge carriers from base and emitter. <S> The collector is also the widest section. <S> Thus, connecting it to the external heatsink, metal conjunction facilitates better heat transfer to the environment. <A> Note that not all power transistors have the case or tab connected to the collector, some RF power transistors have the case connected to the emitter.
| This provides a lower inductance path to RF ground, which in an RF amplifier is critical to maximize gain.
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Leads of a particular electrical component Is lead the most common term used in electronics to indicate the number of connectible input/outputs that a component has? Are there other terms that are used interchangeably with leads? Are there different qualifications for the type of lead it is? For example, ground would have something like one input lead but zero output leads, whereas a resistor might have two leads that could be almost interchangeable/symmetric in that they both accept incoming/outgoing directions, whereas a transistor/diode has specific leads with specific properties. In other words, is there something like a glossary or something of all the different lead types or how 'connectable areas' of a component are defined? <Q> Electrodes (tend to be for larger, higher voltage things, especially older equipment like vacuum tubes) are all distinct things. <S> But no one cares. <S> It's all used interchangeably. <S> "Terminal" is also another term that would cover all of the above. <S> I tend to use "pin" or "terminal" across the board. <S> The specific terminal itself on a particular type of component has different names depending on the component. <S> For a diodes, and generally two-terminal polarized components (such as polarized capacitors or cells and batteries), it is the anode and cathode. <S> For BJTs and other three-terminal components of "similar NPN arrangement" such as thyristors <S> , it's base, collector emitter. <S> For MOSFETs and JFETs, it's gate, source, drain. <S> IGBTs have gates, collectors, and emitters since they steal construction and operating traits of MOSFETs and BJTs for those respective connections. <S> For op amps, it's inverting input, non-inverting input, and output. <S> There's no glossary. <S> You just run into it when you use the component <S> and it's obvious what it is if you know how you're supposed to use the component. <S> Also, the datasheet. <A> To make a distinction between the electrical type of leads/pins/pads/balls there could be on an electronic device, there are: Inputs : <S> Signal input, usually small voltages/currents, analog or digital. <S> Outputs : <S> Signal output, ditto. <S> Can be further refined as "push/pull" (driven high or low, never floating), or "open drain/collector" (floating or low) and "open source/emitter" (floating or high.) <S> Bi-Directional : <S> Signal goes either way through this pin type, usually digital but could be analog also. <S> Tri-State : Three states (digital only) - usually high/active, low/inactive, and a high-impedance state (off, not connected.) <S> The state is selectable via other pin(s). <S> Passive : Pin doesn't have a noteworthy function. <S> Power Input : <S> Designed for power input, sinking current from a supply. <S> Power Output : <S> Sourcing current for a load. <S> Not Connected : <S> Internally not connected to anything; mechanical only. <S> The pins on a resistor could be thought of as passive , as it is well-known what each does. <S> Things get a little more complicated with something like an op-amp, as there are definite inputs and outputs, plus some power and possibly some offset pins, trim, null, etc. <A> Leads, legs, pins, pads, and whatever the holes/sockets are called that BGA parts have.
| Technically: Leads (thru-hole and SMD packages that have metallic protrusions suchas DIP, SOIC, SOT, and TQFP), Pads (SMD like QFN without metallic protrusions), Balls (BGA)
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Most simple way to detect small voltage changes(~150mV) I have one unit of museum cabinet detector that works on 18v. I want make simple security circuit with detector. I've figured out that two output ports on detector are normally contact(NC), but it'll shows ~150mV difference charge when it sensed something. So main alarm system get signal. I have only units and I need to control it with my circuit with battery. I assume It is quite simple logic so I sorted out with arduino at the moment. (Analog input from the detector, digital output to latch up alarm circuit. Alarm rings whenever it get signals for 3 seconds.) But I believe there must be much simpler way to build it by only electronic components. I tried to sense voltage changing of the detector with FET, Opto-coupler, relay but has failed. Probably my component's specification doesn't match well with the situation but also I am not sure if I approach proper way. When activated voltage changes 0~150mv what would be best and cost efficient way to detect the signal so that alarm circuit could get it as input? simulate this circuit – Schematic created using CircuitLab <Q> simple transistor or opamp will do to give a gain of say 10 .This <S> nominal 1.5 Volt peak signal could go into an analog input for processing . <A> If the signal is relatively clean, 150mV might be fine to work with. <S> Sample it with Analog.read on A0, spit the number to the serial port, and view it in the Serial monitor in the Arduino IDE. <S> This might be exactly what the Analog.read example in the Arduino environment does. <S> In the Serial monitor, sample with the detector signal off for a while. <S> You will see integers nearish to zero on your screen. <S> Then set the detector output to it's high state and watch the numbers. <S> They should be around 31 or so. <S> If there is no overlap between the two steps, you can use an if statement to set D1, using a threshold of maybe 20. <A> You can do something like this: simulate this circuit – <S> Schematic created using CircuitLab <S> I have included both a relay output and an opto output, depending on what you want to use. <S> R3 is for hysteresis for clean switching. <S> Because R3 must connect from output to non-inverting input for hysteresis, the reference divider must go on the non-inverting input which makes the output signal reversed from what is required to switch M1 and M2. <S> M3 and R6 exist to reverse the logic output. <S> D1 is a flyback diode because when you interrupt current through aninductor, it makes voltage spikes that destroys things if you don'tgive the current in the inductor a path to decay. <S> R1 lets you tune the switching point. <S> R2 is just <S> so you have moreresolution in your tuning since you are only interested in adjustingaround 150mV. <S> R4 is only required if the relay coil voltage is lower than your supply voltage. <S> If they match you don't need it. <S> Do not power your relay off the regulator if you use one. <S> This circuit won't wait 3 seconds.
| You may or may not power the opto off the regulator if you want. You are wanting to detect Changes in Voltage ,not absolute voltage .This means that you can capacitively couple to remove the nominal 18VDC component .You can easily amplify the AC signal that represents changes in voltage .A You will still need a regulator of some kind since the accuracy of the switching threshold of R1 is dependent on V1.
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Why does voltage sign matter for electrical devices and components If you look at the specifications for certain electrical devices, computer hard drives for instance, the voltage requirements would be something like +6.6v or +12v. Why does the sign matter? Why couldn't -6.6v or -12v work? I thought when it came to voltage, only the magnitude of the difference of voltages matter? <Q> Why does the sign matter? <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Polarity sensitive components. <S> D1, a diode, has a symbol which shows the direction of current flow (in the direction of the arrow). <S> If the polarity is reversed current cannot flow. <S> Q1 and Q2 are transistors. <S> Again, the arrows in the symbols show the direction of current flow, top to bottom as drawn. <S> A circuit's design will take this into consideration and it will not function if polarity is reversed. <S> Integrated circuits such as op-amps and microcontrollers contain several or thousands of transistors of one type or another and the whole circuit is polarity sensitive. <S> Circuits using passive components only, resistors, capacitors and inductors, are not polarity sensitive. <S> simulate this circuit Figure 2. <S> A bridge rectifier accepts input of either polarity or alternating and converts it to DC of the required polarity. <S> Circuits can be designed in certain applications which are not polarity sensitive and can even work on AC power. <S> Typically these use a rectifier to "rectify" the input voltage to the correct polarity. <A> Using your example of a hard drive, the ground reference is typically connected to the chassis of the computer, which is connected to the mains earth, which finds its way to earth potential in a building. <S> So the chassis is at 0V. <S> The supply voltages are something like +3.3, +5 and/or +12V. <S> If you put a reference voltage somewhere else, perhaps the +3.3V terminal, the voltages you measure might be -3.3, 0, +1.7 etc. <S> but the main thing from the hard drive's point of view is that the terminal +3.3 on the hard drive sees 3.3V higher than the terminal for 0V or ground. <S> If you connect -3.3V to the +3.3V terminal relative to the ground terminal the drive will be destroyed. <S> That's because virtually all electronics requires a certain polarity of electricity to work, and is often damaged with reverse polarity. <S> Sometimes it is protected (for example, when the user can insert batteries backwards it might be wise) sometimes not. <S> In some cases, I've designed products with a bridge rectifier so that if the user reverses the power it will still function perfectly, but that's fairly rare, and not always practical. <S> Even a circuit as simple as an LED + resistor will light up if the polarity is correct, and remain dark if the polarity is reversed (usually without damage in this case). <A> That is because there are devices in a computer that need both +12V and -12V, referenced to a GND in the middle. <S> RS232 transceivers are the classic example. <A> Why couldn't -6.6v or -12v work? <S> I thought when it came to voltage, only the magnitude of the difference of voltages matter? <S> The square of the magnitude of the difference is proportional to power, essentially you are asking why isn't it enough to just make sure each device gets the right amount of power. <S> For something like a heating element that just takes electrical power and produces heat, that is all that matters. <S> However for most interesting things the direction of current flows is important. <S> Consider a permanent magnet DC motor that spins clockwise for positive voltage and counterclockwise for negative voltage. <S> If your hard drive used one of those, and you gave it the reverse voltage, clearly it is not going to work right because the disk would be spinning the wrong way (among many, many other reasons). <S> Direction of current flows is very important in this case. <A> Yeah, don't think so much about the actual numbers of the voltage (magnitude), but more so how the voltage is referenced to ground and how current flows in loops. <S> Your power supply is looking for a high voltage at 6.6V and a lower voltage at the GND pin, and current flows from high to low potential. <S> If you switch the two potentials then current would flow from the ground into the positive terminal, which isn't good if your component has polarity, and most things beyond resistors do and care about current flow. <S> That's just how I think of it. <A> The simplest answer is that some components are polarized, that is to say they are only designed to work in one direction. <S> If the flow of current is moving in the opposite direction, they will not work as expected. <S> For instance if you were to add a diode ( a component that only allows current flow in one direction) to a simple circuit that connects the two poles of a battery in reverse, it will prevent current flow in the wrong direction. <S> Though given enough current it can heat up, causing the diode to burn out. <S> These are frequently used to provide reverse polarity protection, and can be quite useful when protecting integrated circuits (ICs) which are susceptible to reverse current.
| It matters because many electronic components only work with current in one direction.
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Can I still use this power supply if I clean off the rust from the transformer? I have power supply for a vintage Timex Sinclair. The unit was sealed but the cable felt loose so I pried it open to inspect the connections. There was a knot on the other side so it was just the cable slipping between that and the outside and not on any solder joints themselves. But I also found the transformer to have a lot of rust on it. Can this be cleaned off with vinegar and a wire brush or will I need to replace the whole thing? <Q> No, do not get vinegar on this part or you could cause it to fail. <S> Vinegar and things like Naval Jelly (a thixotropic concoction of dilute phosphoric and sulphuric acids) are highly conductive electrically. <S> The rust causes no harm, just ignore it. <A> Do not put any liquids on the transformer, whether water based, oil based, or alcohol based. <A> Yes. <S> Rust does not alter the performance of the transformator as long as it is only on the surface. <S> Strictly speaking the iron core of the transformer is not needed for its functioning principle. <S> It "just" increases its efficiently, but it increases it so well, that it is worth to add anywhere where power loss and thus thermal dissipation is an issue.
| Leave the rust where it is, it's not doing any harm.
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Is there a simple circuit that can change the AC output voltage from a transformer or AC-AC power supply? I have a device that requires 9V AC and rated for 2.1 amperes. I thought I could modify the output of either of the following voltage supplies to meet those values: A power transformer that outputs 24V, 50VA A power supply wall-wart - output 15VAC, 1.5 amperes Is there a simple way to do this - any schematics/component values possible? <Q> The component you're looking for is.. another transformer. <S> Are you looking for "9v a.c" RMS? <S> Peak-to-peak? <S> What does the input to the device you're trying to power look like (electrically)? <S> It's not trivial to adjust the voltage of an AC signal after transformation; <S> but it's generally pretty easy to simply get a new transformer. <S> You can try disassembling the power supplies and checking if they have an additional tap that happens to provide the voltage you're looking for (which is rare in low-cost parts; <S> but you can often salvage multi-tap transformers out of other equipment!), or simply replace the transformer inside with one that has the correct turns ratio. <S> (This is NOT efficient; but for testing and temporary use; would also allow you to convert the incoming power to the output you're looking for.) <S> If you're feeling very adventurous (I'd HIGHLY recommend against this unless you really know what you're doing!) <S> you can remove some turns from the secondary side of the transformer to get closer to the voltage you expect ( <S> though you WILL de-rate the total power capacity quickly!) <S> Depending on the rest of the power supply; any AC supply should run at a lower voltage without a problem. <A> Use the secondary output from the 15 volt AC supply in antiphase series with the secondary output from the 24 volt AC supply and, you should get the difference. <S> That difference is 9 volts AC. <S> Well, it’s worth a try anyway. <S> Make sure you test with a meter first and leave it to soak on load in a safe area for a while. <S> If you measure 39 volts just reverse the secondary connections on one of the transformers. <A> If you have a split-bobbin transformer (primary and secondary on separate segments of the bobbin) you could consider removing the existing secondary and winding a new one. <S> You would probably need something of the order of 35 turns, but you would want to measure the volts per turn and base the number on an actual measurement. <S> Gauge would probably be something like AWG 18 or equivalent parallel strands of thinner wire. <S> Ideally you would try to mostly fill the winding window <S> but if you're going way down in VA rating from the original that's not essential. <S> If it can't be done without removing the E-I laminations, it's probably not worth the hassle. <S> Of course it's important to maintain proper clearance between primary and secondary for safety and it would best be lacquered afterwards to prevent hum, and taped for protection. <A> Best is to buy a new 9v transformer. <S> If you are adventures and spoiling one of the available transformer is no problem then use the 24v transformer, because its wire gauge will give you the required current rating, and start removing few turns at a time of the secondary winding and measuring the voltage every time till you get the required 9v. <S> Even if it is not split bobbin or side by side winding, because normally the secondary winding is always on the top.
| Alternatively; depending on how long you need such a solution; you could also simply power either existing power supply with an Autotransformer ; (A device that simply allows you to "tap" the output anywhere along the coil, giving you a variable turn ratio transformer) to step down the voltage going into the power supply before it steps the voltage down again.
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Does coiling and straightening a wire change its resistance? I’m looking to make my own kiln coils and I wanted to know why are the coils coiled? When the coils are stretched too far apart they run cold and when they are close together they run hot. What is the reason for this if resistance doesn’t change whether a wire is coiled or straight? <Q> I wanted to know first off why the coils are coiled? <S> Suppose the wire is 10 m long. <S> If you don't coil it, some of the heat it produces is "here" and some of the heat is 10 m away. <S> Coiling it means you can heat a small area instead a long skinny area 10 m long. <S> when the coils are stretched too far apart thy run cold and when they are close together the run hot. <S> What is the reason for this if resistance doesn’t change whether a wire is could or straight? <S> The temperature of the coils depends non only on how much heat they produce ( \$I^2R\$ ) but also how much heat they lose to the environment. <S> If you stretch the coil, it has an overall larger surface area over which heat is carried away by conduction and convection. <A> In addition to the accepted answer, coils also offer physical advantages in taking up the change in length when heated without sagging. <S> The wire becomes brittle after use so <S> the spring in the coil makes it easier to reroute into the channel in the firebrick if a coil pops out (heat the wire up when you do this). <S> I think sharp bends are subject to more strain with heating/cooling cycles <S> so coils avoid those failure spots. <A> In addition to all of the correct answers: coiling a wire does change its inductance , which is something like resistance except it only affects the flow of AC current, not DC. <S> This isn't the reason for your coils (which are fed from DC or 50/60 Hz AC — at those frequencies, the inductance isn't enough to matter much, and doesn't contribute to heating). <S> However, it is a reason why you will see coils in other kinds of electronics, including radios, motors, and power supplies. <S> They're not trying to keep heat in (usually they want to get rid of as much heat as possible), but they are trying to regulate the flow of current by storing energy in magnetic fields. <A> It is definitely negligible for a heating coil, but can be an important source of error in a current measurement shunt. <A> As the other answers point out, resistance doesn't change with shape. <S> In a kiln though, like a light bulb, much of the heat is lost radiatively. <S> in a coil, the infra-red radiation emitted towards the center of the coil will simply be reabsorbed the wire the other side, hence as the wire is coiled, the effective surface area for IR emission goes down, hence temperature goes up. <S> Resistance does increase as temperature goes up, for most if not all metals. <S> This gives something of a regulatory effect limiting the temperature.
| If you compress the coil, it loses heat over a smaller area, and much of the heat produced by one turn of the coil actually heats the neighboring turns, rather than being lost to the environment. The resistance generally goes UP for almost any deformation and the effect is negligible in most cases. One more addition: most wires DO change their resistance when subject to mechanical stress and deformation.
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What purpose are the transistors serving in this CRT oscilloscope circuit? So this is part of a CRT oscilloscope diagram, and I don't understand the function of the transistors on the bottom right, Q18 and Q19. Their outputs (41 and 42, crossing through the dotted line) go directly to the horizontal deflection plates. Furthermore, this diagram says the base voltage is 1.2V, but all three points on each are measuring 20-30VDC for me. Keep in mind this scope is broken, something's wrong with the horizontal sweep circuit. I just need to understand the purpose of these transistors and what kind of voltage they're supposed to be carrying to even attempt to diagnose this thing. Thanks a heap in advance. <Q> You have a differential amplifier that adds two inputs: one input is the horizontal position <S> the other input is the saw-tooth sweep <S> Gain is perhaps near 50. <S> Q18 has about 5mA flowing from +180, through R57, through Q18, through R54 to -10V DC. <S> Q19 is similarly biased. <S> VR7 sets horizontal gain. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Q18 and Q19 are the X-output amplifier. <S> A CRT may need hundreds of volts peak-peak to deflect the beam over the width of the screen. <S> The signals generated by the sweep generator (From Q14 and Q15 via S103) will only generate only a few volts peak to peak. <S> Q18 and Q19 amplify the voltage to the levels required with a few other requirements: <S> The voltage to the X1 plate needs to sweep with the opposite polarity to the voltage on plate X2 (ie when one goes up the other goes down). <S> The mean voltage needs to be constant at 105V to avoid defocusing th eCRZT or causing astigmatism. <S> The bandwidth and pulse response of the amplifier needs to be appropriate to amplify the sawtooth waveform from the sweep generator with acceptable accuracy. <S> Most oscilloscopes use a similar arrangement with a differential output stage. <A> What you have shown us is a classic oscilloscope schematic. <S> I designed 'scopes in the 1960 and 70s. <S> Judging by the load resistors (15k) of Q18 & 19 this must be a very low bandwidth, low performance scope with low sweep speeds. <S> CRTs require around 5 -10 volts of signal per division. <S> So to deflect the spot across the screen (assuming it has 10cm wide screen) requires somewhere between 50 and 100 volts pk-pk. <S> This signal appears at the collectors of 'push-pull' stage Q18 & 19, so between 25 and 50 volts per side in antiphase. <S> The gain of the output stage is roughly (15k +15k)/500 ohms or 60 X when the gain potentiometer is mid-position. <S> This is necessary to drive the reflected capacitance from the output stage. <S> Q18 & 19 will have a collector to base capacitance (Cob) of somewhere between 5 & 10 pF. <S> This is multiplied by the stage gain. <S> This is called the Miller capacitance. <S> Hence Q16 & 17 have to drive around 300 to 600 pF <S> You will see the Y amplifier uses a different technique. <S> It uses a cascode stage consisting of Q11 & 13 and Q12 & 14. <S> Again it is a push-pull output stage <S> but because the output transistors are used in a grounded base circuit there is no Miller capacitance, hence this kind of technique lends itself to higher bandwidths. <S> Any more questions, just ask! <A> It's not really a Darlington pair as the two transistors are common emitter and the output is inverted. <S> It is a simple amplifier section which provides gain and probably uses a higher voltage transistor for the deflection voltage. <S> If you have 20 volts everywhere I'd check vr6 r54 r55. <A> I think that if one of the transistor i.e. either Q18 or Q19 is short you will get around 20-30 volts on all the three terminals of Q18 & Q19. <S> Just check the transistors.
| These transistors form the output stage of the X amplifier which drive the plates of the CRT. Q16 & 17 are emitter followers.
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benefits of removing reset in an FPGA design The benefits of removing reset from your design are: fewer timing paths, performance and the ability to infer more dedicated hardware Is this true or false? I think that is false because the performance has nothing to do with fact of having/not having reset. What do you think? <Q> It is true in general. <S> Most obvious reason is that the reset net is a very large fanout net, and removing loads from the reset net will make it easier to close timing. <S> Removing reset connections also means less nets need to be routed, alleviating some amount of routing congestion enabling the router to close timing more easily. <S> However, resets are important. <S> The thing to keep in mind is that not every register needs to be reset. <S> When you have control signals such as "valid" or "enable" alongside a wide datapath, generally you only need to worry about resetting the control signals; the much larger datapath is usually fine without any reset connections. <A> The answer to your question depends a great deal on how the reset is actually implemented in a given FPGA's architecture. <S> It the reset logic is hardwired into the FPGA's logic <S> then it would probably not make any difference whether the reset was used or not. <S> On the other hand, if a synchronous reset is used and the reset signal is merged into the other logic then it is possible that the presence of the reset would have a negative effect on the path delay and on the required logic resources. <S> It's always dangerous to try to make a blanket statement about such things. <S> It's a complex world out there. <A> All the above answers are correct <S> and i think i have some more things to add about correct usage of resets in context of FPGA/ASIC as resets architectures are complex involving reset sequencing and Reset domain crossings <S> once cannot give direct suggestions <S> https://www.eetimes.com/how-do-i-reset-my-fpga/# <S> I think that is false because the performance has nothing to do with fact of having/not having reset. <S> *** please go through this 2 part article ,Atleast read 2 article for the above assumption https://forums.xilinx.com/t5/Adaptable-Advantage-Blog/Demystifying-Resets-Synchronous-Asynchronous-other-Design/ba-p/882252 <S> https://forums.xilinx.com/t5/Adaptable-Advantage-Blog/Demystifying-Resets-Synchronous-Asynchronous-and-other-Design/ba-p/887366 <S> http://www.sunburst-design.com/papers/CummingsSNUG2003Boston_Resets.pdf <S> http://www.sunburst-design.com/papers/CummingsSNUG2002SJ_Resets.pdf
| Additionally, freeing up the reset pin on FPGA flip-flops means that the synthesis tools can connect other signals to that pin, which could mean a reduction in the number of levels of logic, which can improve timing performance.
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measurement errors: add up or cancel out? Take an accelerometer, or a hall sensor, or a thermometer, or any sensor. It will have an accuracy, which means the readings will contain some error of this sort. My guess is that by taking a big number of readings of the measured parameter will cancel out the sensor inaccuracy to some extent. Or am I wrong ? And if I am not wrong, how do I compute the accuracy of the averaged number of readings that I take over a period of time ? <Q> You always have two kinds or errors: Random error and systematic error <S> Systematic errors are there always with the same amplitude, they don't change (over long periods of time and extanded temperature range they do have drift as well, this will be defined e.g. in ppm /°C or ppm /y). <S> This means these errors won't cancel out with averaging, because all measurements will have the same systematic error. <S> But if you can determine the systematic error (e.g. by some calibration routine) you can compensate for this error by subtracting a calibration value from your measurement result. <S> Random errors behaves like you descripe, every measurement varies by some degree due to noise from all kind of sources. <S> You can not correct this error from a single measurement, because there is no way you could possibly know how this exact measurement was altered. <S> BUT you can average this random error down. <A> My guess is that by taking a big number of readings of the measured parameter will cancel out the sensor inaccuracy to some extent. <S> Or am I wrong ? <S> The law of large numbers (LLN) says that when averaging a large number of random variable realizations (here: measurements), the result will converge to the expectation of the underlying random distribution. <S> You assume that's the "real" value. <S> That's only true if you can model your measurement to be a random variable whose expectation is the actual physical entity <S> your measuring; if you model it as correct value + noise, then the noise must be "zero-mean". <S> That's generally not the case; usually, you have some systematic measurement error that you don't know. <S> So, yes, you're wrong. <S> However, repeated measurement will reduce the variance of the overall measurement, and hence make it more reliable (but can't erase the systematic error). <S> Attention: Your measurements should be stochastically independent , i.e. no use to measure things that are very correlated multiple times. <S> This poses a problem. <S> how do I compute the accuracy of the averaged number of readings that I take over a period of time ? <S> "Accuracy" is not the right term here. <S> You want to read up on what variance is. <S> If your measurements are independent in zero-mean noise, but identical in the actual value thing, then averaging \$K\$ of them will reduce the variance by a factor of \$K\DeclareMathOperator{\Var}{Var}\$ of your measurement \$Y\$ : $$\Var(Y):=\Var\left(\frac1K\sum\limits_{i=1}^K X_\text{actual, const.}+N_i\right) = <S> \frac <S> 1K \Var(X_\text{actual, const.}+N) = <S> \frac1K\Var(N)$$ For general estimation theory, you'll have to look into the Fisher Information of your measurement-based estimator, but hint: no fun at this stage. <A> The easy way is averaging a large amount of samples. <S> The better way is to evaluate the normal distribution and set a confidence interval say 95% .. <S> Then discard the samples that aren't within that confidence interval of the normal distribution. <S> I guess it is also worth taking the precision into account. <S> Say if the accelerometer at rest measures 0.9g you would have to offset that precision by 0.1g to get 1g as expected. <S> Now it may measure 0.9, 0,91 0,89 and so on... <S> That would be the accuracy of your sensor. <S> So if you test the sensor in a controlled environment you would be able to handle this.
| If you take an infinite amount of samples your average random error will approach 0 (you can still have a systematic error, meaning the mean of your measurements is not representing the real mean value).
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Change state hardware interrupt handled by micro-controllers How are external interrupts handled by microcontrollers? My question is general to micro-controllers, but if it is architecture specific, leans to ARM- based micro-controllers. Assuming we have 2 interrupt pins PIN1 and PIN2 set to interrupt when the pin state change. At start both of them are in low state, PIN1 is then interrupted as the signal pulls high. In software the uC is still processing data although the data is very small due to coincidence PIN2 pulls HIGH which in most will be missed since there is still an interrupt being processed. Now when the interrupt on PIN1 exits will PIN2 immediately trigger another interrupt since its state changed? EDIT: since the question is much broader than expected i have chosen a micro - controller for the sake of the question. The micro - controller is: AT91SAM3X8E the one used in arduino due. And will be programmed using the ARDUINO IDE. I have project that uses this chip and knowledge on its interrupt might come in handy. I have always thought there will be a standardized way. <Q> The behavior of these external interrupts is not part of the ARM architecture itself. <S> This is determined by the individual chip manufacturers. <S> The answer depends on whether the interrupts have the same or different priority, whether they are triggered by edges on the pins or levels on the pins, and what happens if a level-sensitive interrupt is serviced <S> but the pin remains in the interrupting state. <A> In ARM cortex M3, there is a concept same, where the stack pushed data will not be retrieved back because the system already knows that it has to serve interrupt 2. <S> It is called tail chaining. <S> https://community.arm.com/developer/ip-products/processors/f/cortex-m-forum/10147/the-means-of-tail-chaining-of-interrupts <S> You have to narrow down your question to your MCU atleast, else this answer has to be a book and a page more. <S> Why we can't tell exactly what will Happen <S> Interrupts can have same priority <S> Interrupts can be in same group priority but different level Interrupt 1 has higher priority than interrupt 2 <S> Interrupt 2 has higher priority than interrupt 1 <S> Second interrupt was already serving Interrupt was already pending Interrupt was disabled <S> And so on.. <S> For every case above there is a defined way which depends on the architecture. <S> ARM Vs SHARC vs many others.. <A> When an active interrupt is detected, the processor will generally start processing instructions in a different location. <S> This is usually a predetermined location or a location whose address is stored in a predetermined location. <S> If an interrupt is being processed and another interrupt is detected: -- <S> If the second interrupt is at the same or lower priority, it will be processed when the first interrupt routine is complete. <S> -- If the second interrupt is at a higher priority, the first interrupt routine will itself be interrupted and will continue only after the second interrupt routine is done. <S> Note that some of these processes, such as manipulation of the interrupt flags, may not be automatic <S> so the interrupt routine may have to handle them explicitly. <S> This will depend on the architecture. <A> Multiple external interrupts will not chain together if the IO share interrupt flags which get cleared when handling one or the other. <S> Read a specific manufacturer datasheet to see how multiple GPIO might share interrupt lines since it os overkill to have a dedicated interrupt for each GPIO. <S> You can even enable things so higher priority interrupts interrupt a lower priority interrupt currently running. <S> So-called nested interrupts <S> but it makes programming very tricky.
| In general pending interrupts (not just external) should run when the current one is done, in order of priority. ARM Cortex has a feature called tail-chaining to speed this up.
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