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What is the maximum current needed to power (supply current) STM32F401RCT6? I am designing a custom PCB based on an STM32 microcontroller. What is the maximum current needed as a power supply for the STM32F401RCT6 ? I have 3.3 V and 250 mA regulated current to power the STM32F401RCT6 microcontroller. From the datasheet: As per the datasheet, do I need only 160 mA in total to power the microcontroller? Or do I need 100 mA for each Vdd pin? My package has 4 Vdd pins, so do I need 400 mA in total? <Q> Table 12 is in the section 6.2 Absolute Maximum Ratings of the datasheet . <S> These maximum ratings are values that, if you exceed them, damage the chip . <S> Your normal supply currents should be much lower than these values. <S> If that's not the case, your using the chip in the wrong way. <S> Table 21 are the values you need, the \$I_{DD}\$ should be the total current which, according to the table, should not exceed 24.1 mA. 4) <S> the actual current consumption depends on how you're using the chip. <A> The current drawing will depend on how the chip is used. <S> The tables states that if all peripherals are enable you will the chip will draw up to 24.1 mA. <S> This, however, does not include any current your I/ <S> O pins may be forced to supply to external loads. <S> The 160 mA is the absolute maximum current chip can safely pull form the supply. <S> This would only occur if you are heavily loading the I/O pins. <A> Your regulator can provide up to 250mA. <S> But the load will take only as much as current it needs. <S> The MCU itself will never take more than what is listed, about 25 mA. <S> But if you drive loads that need current, like LEDs, they add up. <S> In practice, you are not allowed to exceed 120mA total for non-power pins so you won't get near 160mA ever.
| You are not allowed to exceed 160mA total on the power pins.
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Why do operational amplifiers have an output current limit? From Sedra-Smith: Another limitation on the operation of op-amps is that their output current is limited to a specified maximum. [...] This, of course, has to include both the current in the feedback circuit as well as the current supplied to a load resistor. If the circuit requires a larger current, the op-amp output voltage will saturate at the level corresponding to the maximum allowed output current. The meaning of the paragraph is clear, but I still don't get why op-amps have such a limitation on the output current. How can you deduce that from the internal circuitry of an amplifier? For example, if we consider the two-stage CMOS op-amp below, is there a way to calculate the maximum output current? <Q> The output current that your op amp is able to source is determined by IREF and the size ratio of Q7 and Q8. <S> The maximum output current is sourced when Q4 is in triode and Q6 is in cutoff. <S> The output current that the op amp is able to sink is more complicated, since it will depend on the common mode input voltage, but it happens when the gate of Q6 is pulled high by driving the non inverting input below the inverting input. <S> Usually that is an op amp you will find inside an IC <S> and if it has to source current, that current has to come from the bias current. <S> Usually commercially available op amps have an output stage, usually class B or AB, which buffers the high impedance output of the previous stage and is able to supply mA of current. <A> The meaning of the paragraph is clear, but I still don't get why op-amps have such a limitation on the output current. <S> Because the real-world components have finite physical sizes, their internal connections and the semiconductor materials (and their size) being the most limiting factor (you can only push so much current through them before they started overheating and burning), and their external leads being another limiting factor. <S> How can you deduce that from the internal circuitry of an amplifier? <S> You can't deduce that from the schematic of the internal circuitry (unless there is a resistor specified in the output circuitry which limits the current) because the schematic doesn't tell you the physical dimensions of the internal paths and component areas and the current limit is dependent upon those (besides the limit of the amplification factor of the driving circuitry for the output transistors, in case of BJTs). <S> For example, if we consider the two-stage CMOS op-amp below, is there a way to calculate the maximum output current? <S> Considering the CMOS op-amp schematic presented, there are no numbers included on it which would help us determine <S> the maximum current, specifically the MOSFET channel ON resistance (R DS(on) ) and its maximum power dissipation, so that we can calculate using the formula: \begin{equation}\mathit{I} = <S> \sqrt{\dfrac{P}{R}}\end{equation} <A> Steady-state current to the output has to come from Q6 or Q7 as those are the only DC paths to the output. <S> If we have a drive voltage Vgs on a MOSFET (at most it will be at one of the supply rails), consider what the Ids vs. Vds curve of a MOSFET looks like, and what that implies as to the maximum output current. <A> Bipolar devices, used in the early silicon opamps (UA702 from Fairchild, also UA709), and in SiGe today, often were in emitter-follower output configuration that for slewrate performance have enormous base drive currents which for survival, when the engineer's scope probe tip might slip or when the C_load accidentally becomes 1uf instead of the intended 1nanofarad, needs current limiting. <S> Also FETs at least have a self-protective behavior as temperatures rise,whereas bipolars definitely are not self-protective.
| It is simply because you cannot push an infinite current through a certain area size, and there is a breaking point when you reach a certain current.
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Why does the capacitor start to discharge from peak and not from V(input<0) in a peak detector? My university asked to my do a report about a practice in the lab. One of questions it asks to us to describe the working of a peak detector, it's exactly like this: The output is like this: For me it's strange this output behavior, I expected that the capacitor would only start to discharge when the input was below zero. Can someone explain why this thing is happening? <Q> Once C1 becomes charged to the peak voltage, it can only receive charge again when: - the input voltage rises to a higher peak voltage <S> the capacitor is discharged by the parallel resistor and another input peak comes along <S> So, if your input voltage peaked once and then went to zero volts forever, the capacitor (C1) will receive that peak input voltage and then will be slowly discharged by R1. <A> When the diode is forward biased, you have a voltage follower configuration. <S> The inverting input of the amplifier is connected to the output of the circuit. <S> Thus, the op amp "changes" its output to ensure that the inverting input tracks the non inverting input. <S> The op amp is only able to work as a follower if the diode is forward biased. <S> If you reverse bias the diode, then you break the loop. <S> This happens when the non inverting input voltage is lower than the inverting input. <S> In this case the op amp output slews to the negative rail and the diode is reverse biased. <S> The output capacitor then starts discharging through R1. <S> The output capacitor only starts discharging once you reach the peak voltage of your input signal because until you reach the peak the op amp is configured as a follower and he forward biases the diode. <S> However, if you reach the peak and start reducing the input voltage, the op amp is no longer able to control the output voltage by reducing its output, since it will reverse bias the diode and break the loop. <A> "I expected that the capacitor only start to discharge when the output were bellow of zero." <S> voltage is below the capacitor's own charged voltage and a load is connected across the capacitor; A capacitor is charging whenever the voltage applied to it is larger than its own charged voltage; <S> The diode in this circuit is there to allow charging of the capacitor when the input voltage (Vin) is above the capacitor's voltage, and to prevent the discharge of capacitor back to the input when the Vin is below the capacitor's voltage. <S> Your statement about capacitor discharging "only when the output is below zero" is wrong on at least 2 points. <S> Where did you get "below zero"? <S> Could you explain what do you mean by that? <S> UPDATE: <S> "Sorry, I committed a little mistake, the correct is : "whenever the input were bellow of zero", For me, this phrase means the same than " whenever V(input)<0". <S> Is it correct?" – Lucas Vital <S> Your interpretation of that formula is correct, but the statement is still wrong. <S> The input doesn't have to be below zero, only below the voltage to which the capacitor is charged. <S> This means that as soon as the input voltage goes below the peak voltage to which the capacitor has charged up, the capacitor will start discharging.
| The capacitor and the output ARE the same (at the same voltage, in parallel, on the same leads/rails); A capacitor discharges whenever an input (charging)
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Electric cable for bicycles I am using dynamo on my bicycle to power the lights, as I hate using batteries. This serves me OK, apart from the fact I have constant problems with the end of cables. The end of the copper braid (few cm) becomes whitish and stiff after some time and turns to powder when touching it. Which means I have to constantly cut the wire and repair the problem. I wonder, are there any special electric cables resistant to atmospheric influences, especially water, or I just had a bad luck picking low quality cable? <Q> Use tinned copper wire and high quality terminals. <S> Tinned copper wire is much more resistant to corrosion and is commonly used in environments where moisture and corrosion can be an issue, such as in vehicles. <S> I would not recommend soldering as that has a tendency to break due to vibration and movement at the transition between the rigid (soldered) and flexible parts of the cable. <A> A better plan is to find a way to cover them and keep moisture out. <S> Heat Shrink tubing is probably best, along with periodic spritzing with a water displacement product like WD-40. <A> Spade or ring crimp terminals. <S> These are crimped onto the ends of your wire and may provide a solution if your dynamo (it's really an AC alternator, not a DC dynamo) has suitable terminals.
| If the connections are exposed to the weather, they are going to corrode.
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dsPIC33 ADC: Why is the minimum TAD interval such an oddly-specific number (117.6ns) for 12-bit conversion? I'm writing the firmware for a data-acquisition board using the dsPIC33FJ64GP804 MCU and I noticed something strange reading the electrical characteristics for 12-bit A/D conversion: The ADC clock period (emphasis mine) is listed as 117.6ns, which is an oddly-specific number, especially considering there's no direct hardware obstacle to try running your ADC much faster, e.g. with T AD = F CY , which could be as low as 25ns at the highest officially allowed clock speed. So the limit doesn't come from there. The characteristics for 10-bit conversion seems more like it's been derived from actual characterization testing: So where does this weird value come from? Something related to the settling time of the sample&hold capacitor (esp. considering T SAMP = 3 T AD for 12-bit and T SAMP = 2 T AD for 10-bit)? Edit To clarify, I understand T AD = 25ns would be asking for trouble. My main questions are: Why is T AD different for the 10-bit and the 12-bit case at all? Where does the 12-bit number come from, could IC characterization (which I guess involves statistical methods and uncertainty) really produce a number that precise? <Q> There are some clues in the datasheet. <S> For instance: The AD12B bit (AD1CON1<10>) allows each of the ADC modules to be configured by the user as either a 10-bit, 4-sample/hold ADC (default configuration) or a 12-bit, <S> 1-sample/hold ADC. <S> Then this: <S> • <S> In the 12-bit configuration, conversion speeds of up to 500 ksps are supported <S> • <S> There is only one sample/hold amplifier in the 12-bit configuration, <S> so simultaneous sampling of multiple channels is not supported <S> I agree the unusually precise value is odd, but if you convert that to a sample rate it yields about 607K samples per second (a bit above the maximum stated rate) <S> (14 ADC periods are required for a 12 bit conversion). <S> In the ADC reference manual there is a schematic of the effective ADC input in the two modes; <S> Note the difference in the input capacitance; this is necessary as the sample capacitor has to hold the charge for a longer period of time for the conversion to be accurate and will therefore require a longer time to actually charge up during the sample period. <S> Looking at the values, it appears that the effective capacitance for 12 bit mode is formed of all 4 sample and hold capacitors (4 * 4.4 = 17.6, 18 when rounded) and that makes sense of the statement that there is only one sample and hold in 12 bit mode. <S> Probably achieved by switches isolating the other channel sample and holds from their caps and switching them in to form a single effective device. <S> Hence a longer ADC period (longer charge time and a longer hold time). <A> Although the minimum adc period is 25 ns for 40MHz, due to the hardware limitations (sample and hold), they must specify a minimum time. <S> That does not mean that you can not sample faster, it merely means that if you sample faster they cannot guarantee correct operation. <A> For old parts in this family the Tad wasn't different: the circuit for 10 bits was the same as 12 bits, you just need to wait two more bits for conversion. <S> It does make me wonder :) <S> Assuming it's not an error, I think the Tad increases just so that they can run the conversion slower. <S> A SAR conversion error doesn't just wipe out the last bit: you have to get every bit at 12 bit accuracy to get any bit at 12 bit accuracy. <S> I don't know how the Tad number was calculated: all we know is that they felt comfortable with 117.6, and not 117.5 <S> And I think the sample time increases with Tad just to reduce ADC clock noise. <S> Have I got that wrong? <S> I don't think the sample and hold circuit is connected to the ADC clock at all. <S> With the sample time increase, they've increased both the number of ADC clock cycles (reducing ADC clock noise), and minimum sample time (from 152ns to 353ns). <S> I don't know which was the more important. <S> Sample time depends on the amplifier and capacitor and leakage, so they need to wait until it has settled to with 12 bits, but they could have done that by increasing the multiplier more, not the Tad. <A> I figured out where the odd value came from, but the credits and answer points are to Peter Smith's answer for nudging me in the right direction: if you convert that to a sample rate it yields about 607K samples per second (a bit above the maximum stated rate) <S> (14 ADC periods are required for a 12 bit conversion). <S> Upon closer examination of the datasheet, 14 ADC periods are required for conversion, but 3 more are required for sampling. <S> With only one S&H unit you cannot parallelise sampling and conversion, so it's 17 periods per conversion at minimum. <S> If you do ADC conversions back-to-back at the maximum possible rate of 500 ksps (2µs per conversion), then T AD comes at 2µs/17 = <S> 117.64ns. <S> Which is - TADAAA - the value in the datasheet. <S> So it seems that Microchip had the 500ksps target in mind, characterized it to perform well at that target, and derived the unusually precise value we see, from the not-so-precise planned specification of sample rate. <S> It's entirely possible that the ADC unit works well at 550 ksps too, so lower values of T AD could also work, perhaps even down to the 76 ns bound for 10-bit conversion.
| The value in the datasheet may be from calculations or experimentation (I do not know which).
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Help Identifying a chip from a Schlage lock I'm trying to identify this chip. The info I've been able to gather is the following: Packaging: it looks like WQFN (perhaps 20 pins???) Markings: 8524 C3H OEB22 (the 8 could be a B; the O could be a D; the first 2 could be a Z) The letters X and Y are printed on the PCB next to the chip Thanks for the help. I've added a picture. The lock is Schlage BE469 <Q> <A> The letters X and Y are printed on the PCB next to the chip <S> Considering the way that those letters are oriented, along with that style of IC package, then this IC is probably an accelerometer. <S> Looking at this cropped version of the original photo, I've marked the "Y" and (I think) <S> the "X" which you mentioned: <S> The letter "X" is at the 3 o'clock position, indicating that the X-axis is in an east-west plane. <S> Similarly, the letter "Y" is at the 12 o'clock position, indicating that the Y-axis is in a north-south plane. <S> (The Z-axis would therefore be the plane "in and out of the screen".) <S> In an electronic lock, an accelerometer could be used to detect tampering e.g <S> the lock being removed from the door (change of orientation, while the lock is set), or the shock from external attempts to force the door. <S> Update: I just looked on a product page for your Schlage BE469 lock, which refers to a "built-in alarm" feature. <S> For example, the lock's user guide <S> refers to a "Tamper" setting (page 8), which "Alarms when the lock is disturbed". <S> A classic use for an accelerometer in modern consumer electronics is detecting the mechanical shock when someone attempts to tamper with a device, so that helps to confirm that the lock does likely contain an accelerometer. <S> Therefore I suggest to focus the search for an accelerometer as that IC, if you want to identify a specific device. <S> (On a side note, from the amount of apparent flux that's visible around it, someone might have tried to replace that IC already.) <A> Could it be this one (Renesas Low Skew, 1-to-22 Differential-to-HSTL Fanout Buffer): <S> https://www.idt.com/eu/en/document/dst/8524-datasheet <S> This one actually has 64 pins, but you seem to be uncertain as to the actual number and 22 is not giving a whole number when divided by four. <S> You should have uploaded a photo of the whole board, so that we can see whatever detail is available on the chip. <A> There are a number of LM324 equivalents that include 8524 in the part number. <S> For example NJM8524. <S> Have a look at the pinout and surrounding components on the PCB to see if it looks like a quad op-amp. <S> If it's something really custom, you will find it difficult to get data on anyway, so I would take the approach it's likely to be a jellybean type component, so look at common devices i.e. opamps.
| Its a LIS3DHTR 3-axis Accelerometer from STMicro
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How do I reduce the brightness of an automotive lamp? I would like to know how to reduce the brightness of the running lamp in the tail lights of my tow car. It is a MINI Cooper and each tail light contains one single filament running lamp. These lamps also act as the brake lights in normal(not towed) driving. I believe the car's electronics change the voltage from say 5.5v as a running light to 12v as a brake light although I haven't confirmed this. Since I will tow this car with my RV, the running light feed from the RV will range 12v-14.6v. This high of a voltage will appear that the brake lights are illuminated and also create unnecessary heat within the tail light. I don't know how to calculate the proper size resister to safely reduce the brightness to say maybe half. There are two lamps, 21w each that are fed from one line. I have diodes in the circuit so there won't be any type of back feed into the car's electronics. Thank you for your help. Howard <Q> I don't know how to calculate the proper size resister to safely reduce the brightness to say maybe half. <S> There are two lamps, 21w each that are fed from one line. <S> The relevant formulas are \$R = <S> V <S> / I\$ and <S> \$P = <S> V <S> * I\$ <S> (where V = voltage in volts, I = <S> current in amps, R = resistance in ohms, and P = power in watts). <S> Calculating the required resistance for an incandescent lamp is a little tricky though, because the filament resistance has a positive temperature coefficient so it reduces at lower voltage (causing higher than expected power draw), while the light output spectrum shifts more towards infrared which is less visible. <S> So you might have to experiment with different resistor values to get the exact effect you want. <S> Another option might be to simply wire an identical lamp in series. <S> Each lamp should then drop half the total voltage, and you could use the 'ballast' lamp as an indicator that the tail light is on. <S> So how much current and power would each lamp draw, and what would be the equivalent resistance to do the same job? <S> I tested a 12 V 15 W festoon lamp on a range of voltages from 0 to 12 V, and this is what I got:- <S> At 12 V the lamp drew 1.22 A, so its resistance was 12/1.22 = 9.84 Ω. <S> At 6V it drew 0.86 <S> A, <S> ~70% of the current at 12 V (a fixed resistor would draw 50%). <S> Its resistance at 6 V was 6/0.86 = 6.98 Ω, ~70% of the value at 12 V. <S> We can expect that a 12 V 21 W lamp would act similarly. <S> At 12 V it should draw 21/12 = 1.75 A, and the resistance should be 12/1.75 = 6.86 Ω. <S> 70% of that is 1.23 A and 4.8 Ω. <S> So a 4.7 Ω (standard value) power resistor should drop the voltage from 12 V to ~6 V. <S> It would dissipate ~6 <S> *1.23 = <S> ~7.4 watts, so should be rated for 15 W or higher (derated 50% to keep surface temperature down), and kept away from anything that doesn't like high temperatures. <S> So you could put either a 4.7 Ω 15 W resistor or a 12 V 21 W lamp in series to reduce the tail light voltage to ~6 V. <A> Would it be safer to wire across the Mini's brake light switch, using a relay controlled by your RV's brake light? <S> Then you don't have to break into the Mini's wiring loom (always a bad idea when dealing with a BMW product.) <S> The brake light circuit should be working regardless of the car's key switch position. <A> Each diode would drop about 0.7-0.9V, so you would need about 5-8 pieces. <S> You might need to use 3A diodes, especially if using the same diodes for both bulbs. <S> A better solution would be to get lower wattage bulbs or replace the bulbs with LEDs, which would use a lot less energy (about 10 times less) and would be easier and less wasteful to control.
| Diodes might be more effective in reducing the voltage and current (and therefore power) through your bulbs.
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Turning on an automotive relay when enough voltage is avaialble I'm trying to drive a relay off a dome light in a car, where the dome light fades in smoothly. That is, say, I have a constant ground and the other wire goes from 0 to 12V (up to 14V) over a couple seconds. The dimmer is being run by the car's computer. While the relay does come on, it does so while making a sound of entering a hyperspace. I figured the sound is due to the relay not getting enough voltage during the initial startup. In so far, I've tried a zener diode in series. 5v zener reduces the noise, but not completely. 10v zener is not enough to activate the relay (datasheet claims coil activation is around 8v) due to zener voltage drop. I also tried a capacitor in parallel to the coil 560mu 36V I had laying around seems to work the best. 300mu 24V and smaller did not seem to work. Still, being an automotive application, I'd rather not have a large electrolytic capacitor near the hot roof. Is there a better yet simple solution? A comparator using zener as a reference seems like an overkill. I can also try a much smaller PCB-type relay with a hope of it being more tolerant to the startup conditions. Any suggestions? Bonus points for not frying the car's computer that drives the dimmer! EDIT: I have simplified a bit. In reality, I have 3 wires: 12+, ground, and 'door'. With the door closed, the 'door' wire is at 12+, when the door opens - the 'door' wire goes to close to 0. My goal is to convert the 'door' wire into 'interrupted 12+' (no signal when door closed, positive when door open) required by a dimmable mirror. EDIT2: Would simply comparing ground to 'door' signal make more sense? EDIT3: My ultimate goals are 1) to provide a positive 'door' signal to a Gentex 221 mirror that came from a Ford using a 'door' signal from Toyota that, upon opening the door, goes from +12V to 0V; 2) trigger a Hella timer relay that feeds +12V to this mirror. My original plan was to use the Toyota door signal to trigger a second relay that would send +12V to the mirror's door pin. Upon much struggle with the chatter on the relay, I realized I don't actually need the second relay. All it takes is a PNP transistor like 2N3906. The base goes to the Toyota door signal, Emitter goes to +12V, and Collector goes to Gentex door pin. This setup essentially reverses the +12->0 signal from Toyota into a 0->+12 signal sufficient to drive the Timer relay and the mirror's door function. Now, I'm not 100% positive why the timer relay does not exhibit a chatter; most likely because I'm switching on the full +12V rather than the PWM'd ground. <Q> Here is one approach (this assumes the bulb has one side grounded, if not then just flip everything over and use a P-channel MOSFET). <S> simulate this circuit – <S> Schematic created using CircuitLab <A> I realized I don't actually need the second relay. <S> All it takes is a PNP transistor like 2N3906. <S> The base goes to the Toyota door signal, Emitter goes to +12V, and Collector goes to Gentex door pin. <S> This setup essentially reverses the +12->0 signal from Toyota into a 0->+12 signal sufficient to drive the Timer relay and the mirror's door function. <S> Now, I'm not 100% positive why the timer relay does not exhibit a chatter. <S> One possible reason is because I'm switching on the full +12V rather than the PWM'd ground. <A> Here's one using a transistor to drive the relay. <S> The schematic has been altered to show the relay being fed by +12V instead of the dimmer output.
| The idea is that the PWM signal to the bulb discharges the capacitor C1 (through Lamp1 and D1) during the 'off' part of the PWM cycle until the lamp is almost completely on. Second reason is the timer relay measures 100 kOhms while the normal relay is 10 ohms, perhaps because of a built-in resistor.
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How do I power many devices off of a low voltage microcontroller? I'm new to electronics.I have a GENIE 08 microcontroller, and I want to control multiple LEDs from one pin, and a few buzzers from another, and an LCD from one too.I am worried that the power supplied from the output pins will not be suitable for this.I am going to be running the controller from a 3V supply.Is a relay with another supply going to be the best option for this project?Any help in understanding this would be appriciated <Q> Your concerns are valid. <S> You will need transistors to drive the relay since it takes more current that an MCU pin can supply. <S> You can skip the relay altogether. <S> and only use transistors. <A> The GENIE-08 is essentially a Microchip PIC12F1822 "in disguise". <S> The GENIE app note specifies the pin current limit as 25mA. <S> As a general rule, it is much safer to use GPIOs to trigger external devices like MOSFETs or driver ICs to control high-power loads; it protects the pin from damage and allows you to control much higher power levels than the pin itself can support. <S> Use the GPIO to control the MOSFET gate, the MOSFET will control your LED and piezo loads. <A> You have to look into the Datasheet, there you can see how much current each GPIO is capable of. <S> One LED should generally work... <S> Multiple might be Difficult. <S> Better be safe then sorry and use a MOSFET. <S> If you have many things, you should consider using a Mosfet/transistor array. <S> This is a chip which has all the FET/Transistor integrated. <S> The ULN2800 Series is the first thing which comes to my mind. <S> You definitely shouldn't power a LCD from your MCU. <S> Usually you connect the VCC, GND, LED +, LED - to your Powersupply. <S> The MCU is only connected to the Datalines like D1-D4 (or D8, it depends),R/W and so on. <S> If you want to disable the LCD, you could do it with a Transistor/FET and cut the Power to the LCD.
| The pin would not source enough current to drive a relay coil; your best bet would be to look into logic-level MOSFETs.
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Is the capacity of a capacitor in an RC circuit independent of time? If so why? Is the capacity c of a capacitor independent of time in an RC circuit? In the notes through which I'm working the following step is made: $$\frac{d}{dt}Q=\frac{d}{dt}(uc)=c\frac{d}{dt}u$$ the final equality is implies that c is independent of time but why?(I have a very elementary background in circuits...) <Q> If your ideal circuit uses an ideal capacitor then yes, that capacitance will be constant. <S> Having said that, there are situations in real circuits where the capacitance can change as a function of time or of applied voltage. <A> Basically, for a simple capacitor Q (charge) = capacitance x voltage. <S> So, if there was a voltage across two plates of a capacitor, charge (Q) would be C.V and, if you halved the distance between those plates by bringing them closer AND, with no further introduction of new charge or energy, capacitance would double and the voltage would naturally halve to ensure that charge is conserved. <S> Thus Q = CV is maintained. <S> If you then remembered that the rate of change of Q is current (i) <S> then you could say this: - $$i = C\dfrac{dV}{dt}$$ <S> But equally, you could say this: - $$i = <S> V\dfrac{dC}{dt}$$ <S> Or you could construct whatever mathematical relationship that satisfied the basic charge equation, whether you alter voltage or alter capacitance or a mixture of both. <A> ceramic capacitors, particularly the very compact and cheap varieties, will use mixes of ceramic that VARY A LOT with the voltage---- even 50% or 80%. <S> You don't want to use these high dC/dV capacitors in MUSIC circuits. <A> If it's an electrolytic capacitor, its capacity is a function of time and operating temperature ... on a scale of years or decades it may lose most of its capacity. <S> But otherwise? ... no. <A> The capacitance of a capacitor, its voltage rating, and its frequency characteristics are determined by the kinds of materials used and their physical arrangement. <S> Note that the total energy stored in a capacitor is dependent on both the capacitance and the terminal voltage. <S> A 1 uF capacitor charged up to 10 V stores four times the energy of the same capacitor charged up to only 5 V.
| This is simply by our definition of the ideal capacitor element.
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How to indentify which PCB revision at runtime? I have a board going out into the wild which we know over time will be revised for various reasons including cost, EOL components, reliability etc. We will also be providing firmware update on an infrequent basis. What is a good method to allow the MCU / CPU to indentify which hardware revision the consumers baord is at. <Q> If you have spare pins on a MCU/CPU, you could add pull-up and pull-down resistors on all these spare pins. <S> Let's assume you have 4 of these revision pins. <S> Everytime you update the PCB, update the BOM <S> so the first revision would only have pull-downs yielding revision 0. <S> For the 2nd revision of the PCB, update the BOM <S> so the second revision has one revision pin with a pull-up and <S> all the other revision pins have pull-downs. <S> Etc. <S> , that way you could accommodate up to 16 PCB revisions with 4 pins, this should be enough. <S> Only thing to remember is to update the BOM everytime you change the PCB. <A> Many options. <S> On-chip OTP programmed during production/testing. <S> Eeprom memory and serial number chip programmed during production/testing. <S> Eg: <S> 11AA02UID. <S> These can be done poorly with risk of operator error by having the operator read and enter it. <S> Theoretically you'd want to also store the serial number next to the version. <S> Not sensitive to operator error: <S> Unique Serial Chip number database. <S> (also genuine protection) <S> Hardwired GPIO matrix. <S> Hardwired analog value for the ADC. <S> Pick one that fits into your intended lifecycle program. <S> So apparently the numbered list is broken... <A> The easiest way is to record the version in the MCU's EEPROM at the last address. <S> If your EEPROM is 8bit, you can record 256 revisions. <S> Or you use 2 address to record 65535. <S> Each revision's changes can be documented in your device's documentation. <S> Example: revision 24: <S> In the documentation, for this revision the capacitor values on the PCB were increases, the revision number was replaced in the EEPROM. <A> The only way to absolutely, positively do this is to provide some sort of non-re-writebable value that can be read by the host. <S> EEPROMs can be wiped and reflashed, so that doesn't really work unless you write-protect a section of it. <S> If you can do that you can not only encode the board rev, but a serial number as well. <S> If you have some OTP available, even better. <S> Board straps are a historically common, but not cost-friendly method. <S> Not only does it need extra hardware, it also has the burden of managing a BOM to go with it to configure the straps, so more business units have to get involved on each release. <S> It's painful and error-prone. <S> Some companies keep track by keeping a database of serial numbers and assigning the rev information in a database. <S> This works for a push-update model like a set-top box for example, but not for a standalone device.
| Or you can have your operator scan a sticker or lasered QR code that will be written during programming.
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Do usb and vga controllers consume power if there are not devices connected to them? Let's consider the following situation. There is a server, for example supermicro, that has 4 USB2 ports, 4 USB3 ports, and 2 VGA ports. At the same time we don't use these ports, I mean there are no devices connected to them. Do these ports consume any power in such a situation? I am asking as I need to understand if these ports draw any power from power supply. <Q> USB, no. <S> VGA? <S> Very definitely draws power even if no display is attached - though it may go into a lower-power state. <S> This would be very controller - dependent. <A> The answer depends on the level of implementation of ACPI system states in your particular system. <S> Each device on a computing platform also has special device power states within ACPI: From NCR website: <S> ACPI defines power states for peripherals which are separate from the system power state. <S> The device power states range from D0 (fully-on) to D3 (off) <S> There is also some extension to Sleep States called "Modern Standby" , some improvement over "connected standby". <S> If your system supports Intel-defined so-called "S0ix" mode, and all peripheral controllers (and their drivers!) are compliant, the USB host controller will be forced into D3 state if nothing is connected to it. <S> Even if HID devices are connected, the system will selectively suspend them after some period of inactivity, and put host controller[s] into D3. <S> All these complications are invented mostly to prolong battery life of laptops, tablets, and phones. <S> Implementation of low power saving modes is pretty complicated and requires additional embedded management processors into main computing cores, and it takes some toll. <S> It is quite unlikely that your server system supports them. <A> VGA typically remains active and uses some power, but the amount of power used may depend on the BIOS settings, the OS settings (resolution, activity, power management/savings), the card/adapter itself, and the motherboard. <A> The host controllers will draw a very small amount of power, yes. <S> Probably in the region of milliamps.
| The USB ports themselves don't draw any power when there are no devices attached, but the USB controller for those ports is using a small amount of power to maintain a "presence" in the system and to be ready to communicate when any device is plugged into a USB port. It is the responsibility of the driver developer for each peripheral to define and support the available power states.
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How much current can I draw from a battery? I would like to build a portable bluetooth speaker and I would like to know how much current I can draw from a battery like this? The battery is rated on 3.7V and I want to supply this amplifier with a step up dc dc converter. The amplifier suggests 3A 12V power supply. <Q> From the battery specification that you posted it says that the maximum continuous discharging current is 1000mA. Or 1A if you convert the units. <S> So for safe use of the battery and safety to yourself you would not want to exceed this amount. <S> You were asking about using a boost converter to increase the battery voltage to 12V. A well designed boost converter would be able to achieve an efficiency of say 85%. <S> This means that: Power_Out = 0.85 <S> * Power_In <S> Power_In = <S> Vbat * 1A max = 3.7V <S> * 1A = 3.7W max. <S> Power_Out = <S> 0.85 <S> * 3.7W = <S> 3.14W max. <S> Current_Out = <S> 3.14W <S> max / <S> 12V <S> = 0.26A <S> max. <S> This means that you must not place a load on the boost converter of more than 260mA in order to stay within the safe operational zone of the battery. <A> After commenting about it, I revise my answer. <S> For your battery which is of type LP543450 / 544350, there are different datasheets which state different things. <S> I summurize it to 2 options: Option 1: <S> Specification1 <S> According to this variant:Standard discharge current: 0.2AMax <S> discharging current: <S> 1.9A(2x charge current)Max impulse discharge current <S> : 4AMax charge current: 950mA Option 2: Specification2 <S> Max charge current: 500mAMax discharge current: 1000mA. <S> Result: <A> Try a LiPo battery as used in Drones & RC cars. <S> They're not the same thing as the batteries used in cellphones & computers. <S> Designed to output MUCH higher currents. <S> They also have built-in safety circuits, making it much much harder to blow yourself up <S> ;) <S> They're also available in higher output voltages, making your boost-job easier and less taxing. <S> Example: <S> https://www.ebay.com/itm/Ovonic-50C-7-4V-5200mAh-2S-Lipo-Battery-for-Traxxas-RC-Car-Truck-Boat-Drone-FPV/183819976582?hash=item2acc863f86:g:oEEAAOSwq2heWNGZ <S> This one has 50C rated output - Meaning it can source 5200mA <S> * 50 = 260A ! <S> Notice how thick the wires are.
| According to me its safe to assume 500mA of charging current and 950mA of discharge current.
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Choice of buck-boost converter I have a Li-Po battery which has a discharge voltage range of (2.5-4.2)V and I want a regulator which is able to convert this voltage to 3.3V. I was told to use a buck boost converter. I have found one which is a buck boost converter and the other a buck converter. Reading the descriptions of both, I think both will work but I would like to ask for your opinions or if you have any other recommendations of components that'll work. Here below are the datasheets of each component: Buck Buckboost <Q> You will be unable to boost(step up) <S> the voltage with only a buck(step down <S> ) converter when the voltage is lower than the intended output, excluding the conversion losses you will have to have an input voltage of >=3V3 for an output of 3V3 with a buck converter <A> The maximum output voltage of a buck converter is slightly lower than the input voltage. <S> Therefore if you use the buck design, you won't get 3.3 V output when your battery voltage drops below 3.4 V or so. <S> This behavior isn't really explicitly mentioned in the datasheet (except where they call the chip a "step-down" converter), because it's such a fundamental limitation of buck converters that people using them are expected to know about it. <S> There is one section that does at least allude to the limitation: <A> If you use only a buck, you can explore the option of re-specifying your 3.3V rail to be a lower voltage, say, 2.7V. Then choose a buck that has an LDO mode (100% duty with linear voltage regulation) that kicks in when the battery is close to the target voltage. <S> Otherwise, you need to use buck-boost. <S> And, you should not drain the battery all the way down to 2.5V, it will damage it. <S> More here: https://batteryuniversity.com/learn/article/confusion_with_voltages <S> Worth a look.
| Buck-boost integrated with charge / discharge control is available as a single IC from the usual places (TI, Maxim, Analog Devices, etc.)
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Why are these two electrolytic capacitors so hugely different in size? I'm recapping a vintage amp, and that amp has two larger caps at 2200uF @ 50V. I have sourced some possible replacements, and I'd like some insight into why these are SO different in size, and how I should think about that. (These are both new parts. Original cap not pictured. I know they are slightly different in capacitance--the schematic asked for one thing but the original part was another-- but I think the question here still stands since the specs are so close). The big one is a Sprague Atom rated to 85°C. ( data sheet here ). The small one is a JWCO part rated to 105°C. ( Data sheet here )-- it's over an inch long, which makes the Sprague huge at like 2.5"+ and way more diameter. Now, I get that the JWCO is a sort of no-name commodity part that I don't want to use in an audio amplifier-- I get that. But even a Nichicon audio cap at 2200/50 is only going to be a bit over an inch long. But what I don't get here is WHY these are so enormously different in size, and what's up with the Sprague? Small variances wouldn't surprise me at all based on materials and construction but this is not a small variance. Even the 50-year-old cap I'm replacing is much smaller than the Sprague (though not as small as the JWCO or a Nichicon). Thanks for any insight. Piecing together my knowledge of components by example! <Q> Alotta empty air... <S> Those are very old designs (60's), one might use them to restore vintage equipment w/o affecting their appearance. <S> https://www.lespaulforum.com/forum/showthread.php?177151-Sprague-TVA-Atom-vs-Nichicon-VX-Series <S> Guitarists and audiophiles spend stupid money chasing "tone", thinking there's some sort of magic in the parts themselves. <A> Generally speaking, manufacturing has changed significantly since the old Sprague caps were designed, so unsurprisingly there's a lot of blank space in that case. <S> Aluminum electrolytic capacitors can have same specs / be different physically due to a few factors, including lifetime, mechanical robustness, and designed specifically with ESR/ESL in mind. <S> can all be tweaked to make a component for a specific application. <S> Audio capacitors, You would definitely need more of the electrical specs to know why two "equivalent" caps would be different sizes. <A> Be carful,Take a look first to make sure the cap is not in the power rail of the amp. <S> You can't use any old cap due to the fact that if this is a audio amp, the cap is placed across the half bridge if so the type of capacitor is very important. <S> If the cap is blown ,one end I'd blown out are you notice smoke , that means that the cap was used to control what is called pumping you have to make sure the cap has good thermal data, the cap has low ESR, MAKE SURE ITS A POWER CAP WITH WIRE LEADS ARE SCREW TERMINALS. <S> Because when the amp is on you have two power problem PumpingVery fast current coming from the power stage <S> let say 300khz which will be constant, now you have the audio going thru the amp max freq . <S> Are BW band widthThat cap will be under hi stress because you have to keep the switch signal near 85dB <S> now the amp has to control pumping the louder the more pumping <S> ,if you listen with no audio make sure you do no heat any time out of the speakers if you do the cap is going to blow, USE A GOLD SEAL CAPACITOR, USE KEMIT ARE CAPACITOR MADE BY MOLLARY CAPACITORS , USE CAPACITORS <S> THAT HAS A RAPID CHARGE AND DISCHARGE <S> LIKE CAPACITORS <S> FOR ( PHOTO FLASH APPLICATION ) <S> THEN YOU ARE SAFE if you wont take a picture of the board where the cap goes and I will find <S> are locate a better part <S> , now you no the cap is at 2200uF <S> so this is in a part of the bulk dc <S> voltageEmail me at ( wmccain@mccainlab.com ) send picture of the board and the cap and email it to me
| Different electrolytes, layer thickness, encapsulation, electrolyte volume, venting, etc.
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Signal velocity in PCB traces On what does the signal velocity depend for a signal traveling through a PCB trace? According to Wikipedia , it depends exclusively on the relative permittivity (ε_r) of the medium (well, it also depends on c , but that one is a constant), under the approximation that for PCBs, the relative permeability of the medium is 1. My question is: Shouldn't the signal velocity be a function of the characteristic impedance of the transmission line? My thought process is: If I have a trace that has a width change, then there will be a characteristic impedance change, and that would cause partial reflection of the incoming wave. But there is no energy absorption anywhere, so the only cause for a partial reflection can be a change in the wave's propagation speed, right? According to the equation in Wikipedia's signal velocity page, there would be no change in velocity, since the medium is the same, so the relative permittivity remains the same. Can someone shed some light on this? <Q> The rules of thumb given in the previous answer are good enough for many designs. <S> But I want to add one additional thought. <S> The velocity factor is basically going to be the inverse square root of dielectric constant ( \$\varepsilon_R\$ or \$D_k\$ ) of the material <S> the electric field around the transmission line travels through. <S> For stripline, this means it's essentially the dielectric constant of the circuit board material. <S> But for microstrip, it will be an average of the board material's dielectric constant, and the surrounding material (usually air), weighted by the proportion of the electric field that travels in each medium. <S> That means that if your design has both microstrip and stripline, signals in the microstrip will travel at least slightly faster than signals in the stripline. <S> As pointed out in the comments, it's worth mentioning that the dielectric constant of PWB material can vary in service due to factors such as operating frequency, temperature change and moisture absorption. <S> There can also be variation at manufacturing time due to factors such as etch-back of the traces and alignment of the traces relative to the glass fibers in the dielectric. <S> If I have a trace that has a width change, then there will be a characteristic impedance change, and that would cause partial reflection of the incoming wave. <S> But there is no energy absorption anywhere, so the only cause for a partial reflection can be a change in the wave's propagation speed, right? <S> No, that doesn't follow. <S> If there's a discontinuity in the characteristic impedance, then you need a different ratio of current to voltage in the line for the incoming signal and the forward-propagating signal. <S> That means to satisfy KCL and KVL (or, if you want a more mathematical explanation, to satisfy the boundary conditions) at the place the two geometries meet, a reverse travelling wave has to be generated. <S> Then the current to voltage ratio on the incoming line at the junction can be made to match that on the outgoing line, and all is right with the universe. <S> The difference in propagation velocity on the two lines, if there is one, isn't important here. <A> If you consider the PCB trace as a lossless transmission line, the characteristic impedance \$Z_0 = \sqrt{\frac{L}{C}}\$ <S> but the velocity factor is inversely proportional to <S> \$\sqrt{L\cdot C}\$ (where L & C are per unit length). <S> So it should be possible for the velocity to change without the characteristic impedance changing, but it would require two things to change simultaneously. <S> It is true that if you only change the capacitance (or permittivity) then \$Z_0\$ changes proportionally to velocity factor. <S> More in this 1965 paper. <A> Electromagnetic waves travel in a dielectric medium. <S> In theory, the propagation speed depends on the relative permittivity and the relative permeability of the dielectric medium that the wave is traveling in. <S> For all practical materials, the relative permeability is 1 <S> so typically we ignore that, and say that speed depends only on the permittivity of the dielectric. <S> The formula is V = C/sqrt(epsilon) <S> Where V is the propagation speed, C is the speed of light in a vacuum, and epsilon is the relative permittivity. <S> Typical circuit boards are made from a glass fiber epoxy composite called FR4. <S> The relative permittivity of FR4 is around 4, but this can vary with frequency and temperature. <S> However, for a trace on an outer layer, the dielectric is partially air and partially FR4. <S> So for traces on an outer layer, usually an effective permittivity is calculated that tries to average the effects of the two different dielectrics. <S> Since air has a much lower permittivity, outer layer signals are faster than inner layer signals. <S> You can say that they will be roughly 15% faster as a rule of thumb. <S> A detailed calculation can be made using formulas developed specifically for this purpose. <S> But all the details need to be supplied (trace width, thickness, distance to reference plane, exact permittivity of PCB material, etc). <S> For inner layer traces, the speed is approximately half the free space speed. <S> As previously noted, the permittivity in circuit board is a function of temperature and signal frequency. <S> This is something you have to get from the FR4 supplier or the board fabrication house if you want to be extremely accurate. <S> FR4 is not the only choice of material. <S> There are others, some specialized for high heat and some specialized for high frequency, etc. <S> Outer layer traces are typically modeled as microstip transmission lines. <S> And inner traces are modeled as stripline transmission lines. <S> Searching using those terms may be helpful in further research. <A> Consider a 100 ohm trace. <S> Now parallel another 100 ohm trace with it. <S> You have a 50 ohm trace, with exactly the same speed of propagation as in either of the originals. <A> The following is excerpted fro Dr. Eric Bogatin's "Rule of Thumb#3 Signal Speed on an Interconnect", circa 2013. <S> The speed of light (any EM radiation for that matter) is 186,000 miles per second, or 300,000 km/sec in a vacuum, or air. <S> A more useful form of this for our discussion is 12 inches per ns. <S> For example, in FR4, the Dk is 4, so the speed of light in most laminate materials is <S> Keep in mind <S> the dielectric constant Dk is the same as Relative permittivity.
| When an electric field [representing the signal] travels in a dielectric material, like a circuit board laminate, the speed of light slows down with the square root of the dielectric constant, Dk.
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Understanding a simple opto-isolator circuit I have this circuit that is part of a schematic that I am trying to read.I have some idea, but I am not sure that I am correct. All I want is to figure out the D0 and D0_3V3 values. For example assuming that the circuit that is connected to left D0 is set to low (no voltage), will D0_3V3 will be 0 (low) or 1 (high)? (I don't care for real measurements as this is either high or low, as if this is on a micro-controller circuit will get voltage or not.) The bottom part with the R11 and +3V3 gets me confused as it seems to keep the NPN transistor always on, which it cant be. And if the NPN is always on then there's always high signal at D0_3V3? Thank you very much all and sorry for making a such a newbie question! <Q> Both the optocoupler and the NPN invert the signal, so <S> zero <S> in is zero out, high in (with enough current) is high out. <S> When the optocoupler turns on (on a high input), it grounds the base of the NPN, stealing its drive current and turning it off, allowing 3V3 to be pulled high. <A> For example assuming that the circuit that is connected to left D0 is set to low (no voltage), will D0_3V3 will be 0 (low) or 1 (high)? <S> ( D0 LOW <S> If D0 is LOW then the LED will be off and the photo-transistor will be off (high impedance and no current between the collector and the emitter). <S> Pin 10 will be pulled high by R11. <S> Q5 will be turned on. <S> D0 HIGH <S> If D0 is HIGH then the LED will be on and the photo-transistor will be on (low impedance, and current will flow between the collector and the emitter). <S> Pin 10 will pull Q5 base to ground. <S> Q5 will be turned off. <S> D0_3V3 will be pulled HIGH by R13. <A> The bottom part with the R11 and +3V3 gets me confused as it seems to keep the NPN transistor always on, which it cant be. <S> And if the NPN is always on then there's always high signal at D0_3V3? <S> There are a couple of things wrong with these statements. <S> 1) <S> First if the NPN (Q5) is always ON, it pulls DO_3v3 to GND. <S> 2) Q5 is only on if the opto is OFF. <S> If the opto is turned ON then Q5 is OFF also, since the base voltage on Q5 is ~0V.
| D0_3V3 will be pulled LOW.
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Current/power requirements for electric fence? We just ordered some new electric fence chargers for our farm. The new chargers are designed to work off mains or battery. Due to the location, we are considering running off battery with a solar recharger (25w or larger panel) instead of permanently running wire to the nearest fenceline. That would also give us the option to more easily move the charger around and energize different pastures. I'm hoping someone here could help me understand - I looked for a smaller/lighter battery pack (lithium ion, maybe a small bank of 18650 cells?), but the only batteries I'm finding for electric fences are the larger/heavier lead batteries. Assuming everything was weatherproof, is there something fundamental about current draw or 'rechargability' that would make 18650s undesirable in this type of application? I get they might be a little more expensive, but the trade off for easy portability seems worth it. I appreciate any insights you can provide! <Q> After having watched this teardown of an electric fence controller and listening to this podcast which is an interview with a designer of electric fence controllers, I think I might have an answer. <S> The reasons why electric fence controllers mostly use 12 V lead-acid batteries as their power source are: low price of these batteries easy availability can be charged in/from a car can be charged using a cheap charger (of a type many farmers will already have). <S> Li-Ion cells are nowhere near as robust as lead-acid batteries. <S> They require much more delicate charging and handling. <S> A charge controller/balancer is a must-have. <S> A battery protection circuit is also a must-have. <S> I also think that the electric fence controller manufacturers are quite conservative, if there's no pressing reason to change then they prefer not to. <S> If you watch the video and listen to the podcast I linked above you'll notice how safety and reliability is very important. <A> Temperature is probably going to be your biggest consideration. <S> Yes, there are electrical differences between capabilities of SLA and LiIon but your charger should get what it needs out of either one. <A> Li-Ion batteries are not supposed to be charged below 0°C, while lead-acid type can be charged down to -20°C. <S> At the upper temperature range their behavior is similar. <S> However, since you mentioned using solar panels to charge your batteries, I have to say that Li-Ion is a lot better choice for solar because they don't need a long-term, <S> 3-stage charging method like the lead-acid. <S> They are also more receptive and thus faster to charge, unlike the lead-acid that needs relatively lower currents (compared to their capacity). <S> This makes them ideal for solar charging because the power could be available for only a short time due to weather or shorter days during winter. <S> Another plus for them is their charging efficiency which is usually 95% and above, while for lead-acid it is between 80 and 85%. <S> In order to charge Li-ion batteries, all you need to do is limit their charging current and maximum voltage. <S> A plain CC/CV DC-DC converter will do. <S> The charging current limit is usually set to no more than 1C <S> (C - charging rate expressed as a fraction of rated capacity in Ah; charging a 1Ah battery at 1A is 1C rate and should take about 1h to charge, while charging the same 1Ah at 0.2A is 0.2C or C/5 rate and would take about 5h to charge). <S> I would recommend getting higher capacity Li-Ion because it will last you longer and will cover the power needs during very cloudy days when the solar panel output will fall to 3-5% of its rated power. <S> A Li-Ion battery also definitely has to have a built-in BMS (Battery Management System) which keeps all cells charged equally, and which prevents their overcharge and "over-discharge" (they should never go below 2.5-2.8V). <S> If you are expecting freezing temperatures, you could bury the Li-Ion battery deep enough to protect it from going below about 2-4°C. <S> It is possible to charge most Li-Ion batteries below 0°C, but at a VERY low current (0.02C or a 50-hour charging rate). <A> Lead-acid batteries have a long service life, are inexpensive, and have relatively forgiving charging requirements, and can be designed to be basically maintenance-free. <S> They're also very recyclable, much more so than Li-ion. <S> It's very hard to make an argument for Li-ion for a fixed application like this, given its expense and fussy charging profile (not to mention its tendency to explode if charged improperly.) <S> I would hate to be the rancher that set off a conflagration due to a Li-ion 'venting' scenario. <A> While the other answers focus on the technical side, they all miss one obvious point: The farmers / ranchers are knowledgable about the lead acid batteries, <S> They have charging equipment and use solar panels, There is always a suitable battery, or two, on the farm, <S> So having to go to Li Ion just means more expense - which is not a plus for a farmer...
| Otherwise as long as the characteristics of the battery you pick match what the chargers require, there shouldn't be an issue. LiIon doesn't like high / low temps, even if "weatherproof" you'd need to get a cold rated battery.
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Suspending a Nichrome Heated Coil? Scope: The scope of this project is to repeatably light several fireworks simultaneously (Mandalorian Whistling Birds Vambrace build). This requires that the only action required to light more firework rockets would be to load them into the device. I have looked into the ways people have done this in the past but all the options I've seen are consumable or one time use. I've built a proof of concept design and now I'd like to improve it. My initial prototype uses the nichrome ribbon I removed from a car cigarette lighter coil. I chose to start with this since it was a cheap source that uses 12V. The fuses of the fireworks rest on the coil and as I apply 12V to the coil it heats up and they ignite. The issue I am facing is that the coil takes awhile to heat up (no calculations in first design for efficiency) so I would like to design one that heats up faster and has more surface area coverage. I would still like to use my 12V LiPo battery. I can do the equations to figure out voltage, length of wire, amps, and produced heat but what I'm having trouble with is figuring out how to suspend the coil, ensure it maintains its shape, and doesn't short to the metal chamber it's nested in. I plan to use 28-30 gage nichrome wire which is not very rigid. I'm also dealing with a small space of less than a 2" diameter tube. Any ideas? Thank you in advance for any suggestions you may have. Edit: Thanks everyone who answered and provided suggestions! From your responses I have learned that mica, alumina ceramic, or other ceramic compositions will work well as an insulating material. Also, the use of thin stainless steel sheet may work well as the heated element for ignition. <Q> Between the two options of a self-supporting heated element, and a supported heating element, the self-supporting one will have lower mass, so will heat faster. <S> Self-supporting means short and fat, so unfortunately low resistance. <S> However, there are lots of designs, and maybe modules, to get low voltage high current from 12 V sources, think low voltage 100 amp supplies to CPUs and GPUs. <S> I'd go with sourcing some thin stainless steel sheet, which has an astonishingly high resistivity, and cutting some into an inverted 'V' shape. <S> Clamp the two arms between two bolts, the pointy bit would then point up and be able to penetrate into the bottom of the rocket, achieving good contact with the fuel. <S> The bolts would need to pass through ceramic or glass, but maybe you could use GRP and face it with mica or something to survive the short, albeit intense, launch heat. <S> I've been using some fake roofing 'slate' recently as a flame baffle in a propane kiln, and that seems to hold up well. <S> This would be a design for an intermediate life element, robust enough to serve several launches, but cheap and sufficiently easily replaced to not have to be built to survive 100s of them. <A> It has the benefit of providing thermal insulation, so the wire will lose less heat (heating time will be shorter.) <A> Mica sheet might be suitable for this application. <S> In the past it was used for high-temperature fixtures and for electrical insulation. <S> As @SpehroPefhany mentioned it can be sourced from an old toaster or ordered online. <S> These days I believe ceramic is more typical for these applications.
| I would think ceramic (alumina, typically) would work well.
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Identify these potentiometers or variable capacitors I recently came across an old radio that has these odd parts on them. They are labeled "T3" and so on, so maybe they're some kind of capacitor or transformer? They're adjustable at the top and branded/stamped "JET" around the adjustment screw. In the last picture, I carefully dissected one and it's just thin copper wire wrapped around what seems to be a ferrite/iron core. Any ideas? <Q> They're IF transformers (T is for Transformer) used in radios (relatively narrowband amplifiers at the Intermediate Frequency of usually 10.7MHz for FM and 455kHz for AM). <S> The ferrite core adjustment is for trimming the inductance to tune the center frequency, called "aligning" the radio. <S> Below image from this datasheet. <S> There are many variations and you need some equipment to properly align a radio, which does not bode well for your chances at success. <A> <A> These are variable inductors used for trimming specific frequencies. <S> Usually, you also need an oscilloscope for repairing and re-aligning these old receivers. <S> Here's a YouTube channel that specialises in doing exactly that: MrCarlsonsLab <S> Alternatively, you could measure every component in the circuit with a precise multimeter and figure out what value you're supposed to set these at by understanding what particular frequency the component arrangements are filtering out. <S> Here are some online calculators for simple frequency filters: https://electronicbase.net/band-stop-filter-calculator/ http://sim.okawa-denshi.jp/en/Fkeisan.htm <A> They are well known Intermediate Frequency Transformers (IFTs) and not rare. <S> The transformer primary and secondary windings would be on a ferrite bobbin surrounded by a threaded 'cup' core. <S> There would also be a ceramic capacitor across the primary winding. <S> The winding ends and the capacitor leads would be soldered to the pin terminals of the IFT, enclosed in a brass or aluminium housing. <S> By adjusting the cup core the primary winding/capacitor would be made to resonate at the required intermediate frequency. <S> The frequency would generally be 455 kHz for AM and 10.7 MHz for FM. <A> Yes, intermediate frequency transformers used in a superheterodyne receiver. <S> Commonly 455kHz for hf receivers. <S> They cans be set slightly staggered to adjust the bandwidth of the if amplifier. <S> Also 10.7 MHz transformers for FM receivers. <S> The oscillator in the superhet receiver will also use one. <S> The oscillator tracks the input tuner the two frequencies are feed into the mixer and four frequencies can be detected from the mixer. <S> The signal frequency, the oscillator frequency and two heterodyne frequencies one unwanted and one wanted. <S> A double superheterodyne receiver is preferred for the higher frequencies. <S> Crystal filters may be used in the amplifier to give certain bandwidths. <S> The last IF will have an output winding to match into the demodulator. <A> Intermediate frequency 455KHz transformer <A> These are IFT coils used in radio as well as in b/w tv for converting single high to low & low to high for better response.
| They are variable inductors , used for fine-tuning frequencies.
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Sourcing a Relay for AC Power Loss on UPS I have a system outdoors (~200W) that is powered from the house with a battery backup on UPS. When the UPS switches over to battery power, there is no way to currently generate a notification that the main power is lost. The system has a LTE connection and a Raspberry Pi. I'm wondering if there's a relay that could sit between the AC power, and use the GPIO of the Pi to send a notification that the UPS has switched over? What kind of relay should I be looking for? What things do I need to consider? <Q> You need a relay with a 120VAC coil (if in US or Canada, 240VAC elsewhere) and low current SPDT contacts. <S> The relay coil would be connected in parallel with the UPS AC input. <A> You could use an AC-coil relay, but then the wiring would be at mains voltage, and typically the contacts are not appropriate for signal levels. <A> An AC mains-operated relay, with low coil consumption (around 1.2VA) and a set of gold-plated SPDT contacts, would suffice. <S> Here's the schematic.
| Sounds like you just have to detect the loss of AC power, and nothing to do with the UPS, so an AC-adapter plugged into the input power and driving a signal relay with gold contacts (or just driving a transistor) would work.
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How do I decide on a proper PCB ground layout? I'm trying to figure out best practices regarding ground layout, but it seems the more I read the less I understand about the subject as many recommendations seem to contradict themselves. So far I am under the impression that when using a ground plane one should route the specific node through a single via / contact directly to the ground plane. Others seem to recommend that stitching multiple vias to the plane works better since that decreases resistance. From what I've read the latter technique could be prone to introducing ground loops though, which I want to avoid. Right now I'm lost as to how I should layout the following PCB: The first picture shows the top layer ground filled and both caps and casings directly connected to that fill. The second picture shows the same layout as the first, but with an additional ground fill on the bottom and stitching vias connecting both ground layers. The return is for the 3.3V supply only, the two unmarked connectors in the middle transmit the balanced signal to a differential amplifier down the line.I am under the impression that the second way would be overkill for this example, but I'm trying to learn what would be the best practice here.Many thanks in advance! <Q> I think via stitching is only needed if you're doing RF design. <S> If you have high frequency, switching signals, then you will want a uniform ground plane with short via traces attached to your components. <S> If your PCB just uses discrete components with analog signals, then I think what you have in your first picture is okay. <A> For a successful quiet ground, use at least a 2-layer PCB. <S> Have one layer be the ground plane, with NO SLITS cut in the sheet. <S> Only vias are allowed. <S> And the vias should be spaced far enough apart that some grounded-copper always flows between the vias. <S> Once you get experience with ground planes, you can begin thinking about where to place slits so as to steer the return/ground currents, and minimize interactions/crosstalk between your circuits. <S> =============================== <S> = <S> how many vias to use? <S> a square of copper foil of the default thickness (1 ounce per square foot, 35 microns thick, 1.4 mils thick) has 0.000500 ohms per square of foil, measured from opposite edges. <S> A <S> via of 1/16 inch depth and 1/48 inch diameter has 1:1 depth/perimeter ratio, thus also has ONE SQUARE OF FOIL plated inside the drilled hole. <S> That means those vias look like 0.000500 ohms (which varies 0.4% per degree Centigrade in resistance; at 125C, the resistance has increased 40%). <S> The thermal resistance of this default foil thickness is 70 degree Centigrade per square per watt. <S> If you put 1 amp thru such a via (1:1 ratio), the power dissipated inside the hole is P = <S> I^2 <S> * R = <S> 1*1 * 0.0005 = 0.5 milliWatt. <S> Thus the temperature rise is well under 1 degree C, compared to the planes and traces on the PCB surface. <A> Defining the best grounding method is ugly because there is a transition period between DC (path of least resistance) and AC (path of least impedance.) <S> As a quick summary, the higher the frequency (volts per second to be precise,) the more the current in the return path will try to follow the signal trace. <S> Good planning of grounds is not as simple as a rule of thumb, it is usually planning out where the current loops that are either high frequency (in which case you want the distance between signal trace and return path to have the smallest loop possible, this usually means directly underneath the trace, but can also be directly beside it) or high current (in which case if anything is sensitive, you do not want that return path crossing underneath its reference point.) <S> This also means when you jump traces across one another, you should make sure the return path is not broken, usually by providing a path on the other side for it to follow the signal over the break. <S> The vias in the top picture are an inductance. <S> If you can run your signal ground on the same layer it would be slightly better off, but then all the grounds get too hard to route cleanly around all the signal traces on an average layout, so most agree to punch down a via and have the return path underneath the signal trace to make laying out the signal traces as easy as possible, with inductors in parallel reducing the inductance, so multiple vias to ground at these points can reduce how much inductance is added. <S> If a trace has too much inductance (usually hidden behind "loop area,") then it can start acting more like an antenna, either radiating out, or being suseptible to external signals. <S> The second image gets into what is called via fencing, and it is the same principle as why your microwave window is a metal mesh - it is more intended to attenuate high speed signals from coupling to other traces by essentially building a wall that any signals over a certain frequency will be heavily attenuated by, as a down side it technically can also make a PCB a resonant cavity if they are all spaced perfectly in a pattern. <S> Image Source
| That said, the type of ground plane that is needed really depends on what circuit you are designing.
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Why to use signal in this VHDL code? Here is my code for 4 to 1 mux: entity a isPort ( a : in STD_LOGIC; b : in STD_LOGIC; c : in STD_LOGIC; d : in STD_LOGIC; contr : in STD_LOGIC_VECTOR (1 downto 0); z : out STD_LOGIC); end a;end a;architecture Behavioral of a isbegin with contr select z <= a when "00", b when "01", c when "10", d when "11", '0' when others; end Behavioral; But I've seen that this code is also written with a signal declaration: entity a is Port ( a : in STD_LOGIC; b : in STD_LOGIC; c : in STD_LOGIC; d : in STD_LOGIC; contr : in STD_LOGIC_VECTOR (1 downto 0); z : out STD_LOGIC); end a; architecture Behavioral of a issignal control_signal : std_logic_vector (1 downto 0);begincontrol_signal <= contr;with control_signal selectz <= a when "00", b when "01", c when "10", d when "11", '0' when others;end Behavioral; So what's the difference between these two? When and why should I use signals? <Q> I think clock domain synchronization if contr is asynchronous to the FPGA clock. <S> Standard practice to reduce metastability of asynchronous signals is to feed it through a chain of flip flops prior to usage by anything synchronous. <S> But the entire code chunk is not clocked to begin with <S> so I do not see why you would actually need that here since it just adds latency. <S> But this is something you need to be aware of which goes unmentioned when starting out on FPGAs <S> so I will throw it in. <A> There is no practical difference. <S> Synthesis will remove the signal and the final circuit and firmware from the two will be identical. <S> They version without the signal is the preferred design because it is clearer and simpler. <A> This is just creating as signal that is an alias for another signal name. <S> On the other hand, having to know that there is an alias can also lead to confusion.
| Sometimes the signal name in the port list is not very intuitive and makes the code easier to read.
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Can I replace a 3.8V 2900mAh battery with capacitors? I have a 3.8V 2900mAh battery and I was thinking if it's possible to replace it with a bunch of capacitors Here more specs about the battery This is for a mobile phone and was planning to use capacitors instead of batteries since I use the phone as a modem and it always powered by its charger (Not sure how much power is given to the battery by the phone while charging) and I prefer to power the capacitors by what phone gives to the battery if its possible <Q> Yes, you can replace a battery with a capacitor. <S> The energy densities are much lower with capacitors, so the phone will have a very limited power on time, unless you use a lot of capacitors. <S> The voltage of a capacitor falls exponentially also <S> so unless you have a DC DC converter to boost the voltage, you'll get even less time. <S> You need to know how much current the phone is taking and for how long. <S> Just to give you an idea, a 2900maH is about 10000 Coulombs. <S> To store the equivalent energy you'll need at least a 4V 2500F capacitor (or capacitors in parallel). <S> That last calculation is assuming that all the energy in the capacitors can be used, which it can't. <S> And there will also be inefficiencies in the voltage regulator or converter to get a usable voltage to the phone. <S> So you'll need an even greater amount of capacitance. <A> For a short time, a capacitor can behave as battery because it stores energy. <S> This can be useful for keeping a device on when you need to change batteries or if you want to have something turn on temporarily. <S> However, because capacitor voltage and current decrease as capacitor discharges across a load, it wouldn’t be a practical long term replacement for a real battery. <S> This is why batteries can be rechargeable is a useful aspect. <A> So the goal here is to trick the phone into working without a battery? <S> not to provide useful runtime. <S> I suspect you will have to do a few things. <S> First I suspect you will have to pre-charge the capacitor to a voltage that is reasonable for a lithium ion battery. <S> Deeply discharged lithium ion batteries can be dangerous if later recharged, so lithium battery systems will generally have an under-voltage lockout. <S> Secondly the capacitor will have to be big enough to prevent over or under-voltage swings. <S> Just how big may require some experimentation. <S> IIRC the charge circuits in smartphones are often insufficient to cover peak power demand with the battery being expected to take up the slack. <S> Thirdly many phone battery packs will have some kind of temperature sensor in them. <S> I suspect you may need to find a way to fake the response from said sensor (in many cases I suspect this will just be a resistor of appropriate value). <A> The capacity of the capacitors can be back-calculated. <S> Let's do some math. <S> The capacity of the battery is 2900mAh. <S> Let's say that the phone will run for 6 hours on this. <S> Roughly the phone is then consuming 480mA. <S> Also, a typical Li-ion battery goes from 4.2 ish volts down to 3.4 with a hard cutoff at 3.0v. <S> Armed with these facts suppose we want the phone to run for 10 minutes on only capacitors. <S> First, we need the total charge that the phone would consume. <S> 0.480 * 10(60) = 288 Current is coulombs per second. <S> Multiplying current by seconds cancels our the seconds leaving the total coulombs consumed. <S> Next, we use Q = <S> C <S> * V. <S> If the capacitance remains the same we can mess with the charge levels to see what voltage we get. <S> We can then build a system of equations. <S> 4.2 <S> * C = <S> Q <S> 3.0 <S> * C = <S> Q - 288 <S> The first equation calculates our total charge at 4.2 volts. <S> The second equation states that with the capacitance remaining the same, by the time the voltage reaches 3.0 volts the capacitor should have lost 288 coulombs. <S> The final answer is 240 Farads, which is huge. <S> A minute of runtime cuts that value by 10. <S> 24 Farads is more achievable. <S> On Digikey <S> currently, a 5F super cap with a 5v rating runs just over $5. <S> The rating of -0% +100% is terrible but it runs in your favor. <S> You would need 5 of those to get roughly 1 minute of run time. <S> These were very rough calculations, the run time is probably worse. <S> Good luck, you can play with the numbers to see what is acceptable.
| Also to even have a capacitor last for a long time would require very large capacitors and a light load.
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How to Prevent CD4017 from Resetting to Random Count on Power On I'm working on a project using a couple of CD4017 decade counter chips sharing a clock signal from a 555 oscillator. The ideal behavior I would expect is that the chips would all start on count 0 when given power. They do this most of the time but sometimes they start on a completely random count and the synchronization between the chips is thrown off. I've tried to get around this by putting a capacitor on the reset pin of the 555 and the enable pins of CD4017s like so. This adds a small lag between the 555 clock being enabled when the device receives power, and a small lag for the CD4017 clock inputs being enabled. However, the CD4017s still start on a random count occasionally when given power. Is there any standard, cleanish method I can use to assure they all power on to count 0? I'm sure I could figure out something hacky with the but I'm trying to keep a low part count. <Q> The best (and most bulletproof) solution is to use a reset chip such as an ADM810 . <S> Suitable voltage ranges are available for 3.3V and 5V rails. <S> If you have a different supply you may have to use a different supervisory chip that has resistor-programmable voltages. <S> The below shows the functionality <S> (the ADM810 has active-high output rather than the more commonly required active low shown in the diagram): <S> When the voltage level drops below the lower threshold the reset signal is asserted for a minimum of 0.24 second. <S> It is not released until the supply voltage is higher than the upper threshold for a minimum of 0.24 second. <S> You may also need to put some resistance across the power supply to cause it to reset reasonably quickly rather than retain the previous state. <S> Sometimes a fancier kind of clamp is used in special situations. <S> The R-C kind of reset is sometimes used in low end consumer products where it's acceptable for it to fail once in a while (the user will just manually cycle the power), but it's a very poor substitute for a proper circuit. <A> The solution I found is this. <S> Turns out it is necessary to use the capacitor reset method for synchronization. <S> Diodes are also necessary to prevent a conflict between the pin reset output levels and the capacitor reset. <S> I've breadboarded this and it's working well, very consistent where nothing else has been: <A> Try this: The capacitor will force the reset for a few hundreds of microseconds after power up. <S> Also: Add a large capacitor (can be electrolytic) at the power terminal of the board, and a 1uF at each Vcc pin on every ic. <S> The reset pin must be connected to ground with a pull down resistor (R1) or set either high or low from an external input (Master Reset). <S> It can't be left floating. <S> simulate this circuit – <S> Schematic created using CircuitLab
| You should pick threshold voltages that guarantee proper operation at the lower threshold meaning it will operate down to the lower threshold and will always be reset properly.
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Which frequency is the best for a 3-speed Motor? Sorry I dont know the appropriate terms that should be used in this subject! I have an inductive motor, it has 3 different speeds (or windings) at 50 Hz: 350 rpm / 700 rpm / 970 rpm I want to drive it with a VFD.the speed I want is around 1000 rpm. When I tested it I found this results (more or less): 350rpm ==> 1000rpm @ 140 Hz ~ 4 amps . 700rpm ==> 1000rpm @ 71 Hz ~ 4 amps . 970rpm ==> 1000rpm @ 51 Hz ~ 7 amps or more. In all cases to motor runs under a normal load without problems.. My questions are: 1. Which one is the best for the **motor , the VFD and the consumption ? 2. Higher frequency doesn't affect the motor? <Q> At or below base motor speed, a motor will operate as a constant torque device. <S> When you use a VFD to operate a motor at above the base motor speed, the motor operates as a "constant power" device, because the amount of torque it can produce is related to the ratio of voltage and frequency, and the VFD cannot create more voltage. <S> So as frequency increases, the ratio goes down and so does the torque . <S> So with a 970 RPM motor forced forced to run at 1000 RPM by increasing only the frequency to 51.5Hz, the operating torque the motor is capable of will only drop to 97% of rated, basically still within "normal" tolerance range (assuming +-10%). <S> But if you use the 700 RPM connection and increase the frequency to 71.4 Hz in order to get 1000 RPM, the motor shaft torque will be decreased to 70% of rated torque and if you use the 350 RPM connection and increase the frequency to 142.9Hz to get to 1000 RPM, the motor will only have 35% of rated torque at best. <S> In reality in both of these scenarios the actual usable torque will be even LESS because losses in the motor will increase and cause the motor to overload faster than normal. <S> Then to make matters worse, the PEAK torque, called the "Break Down Torque" that the motor uses to RE-ACCELERATE a motor after a change in loading, reduces by the SQUARE of the change in the V/Hz ratio. <S> So the likelihood of stalling the motor when a load is applied increases exponentially. <S> Use the highest speed. <A> Your results for the 970 RPM connection seem like there may have been a VFD setup error. <S> I suspect that the 700 RPM connection is best. <S> A frequency that is not so much above 50 Hz and a lower current should be best for the VFD. <S> I neglected to consider that the VFD does not likely have the capability to maintain output voltage proportional to frequency above 50 Hz. <S> That will result in increased slip and motor current when operating at 71 Hz. <S> The question doesn't indicate whether the 4 amp test current is the motor current or VFD input current. <S> If it is the motor current, I see no reason not to operate the motor at 71 Hz with the load conditions that you have. <S> If you did not determine the motor current, you should do that. <S> If the VFD provides that information, that is probably the best way to check it. <S> The 970 RPM connection should provide the most torque from the motor. <S> However if the 4 amp test current is the motor current, the 700 RPM connection seems to results in the lowest percentage of rated winding current. <A> To obtain the 3 speeds, the motor would be configured as a 16 pole, 8 pole or 6 pole motor with synchronous speeds of 375, 750 or 1000 RPM. <S> Run direct on mains, the torque would be 1.5 kgm at 375 RPM, 2.3 kgm at 750 RPM and 2.4 kgm at 1000 RPM. <S> When run through a VFD at 1000 RPM, the low-speed configuration would generate a torque of 0.6 kgm, the mid-speed one 1.8 kgm and the high-speed one 2.4 kgm. <S> Which configuration to choose would be decided by the torque required for the new application. <S> Should a speed of 1000 RPM be not very critical, running direct on mains at 970 RPM would be a better option.
| According to the nameplate, the 700 RPM connection has the most torque per amp. The lowest operating current will be best for the motor.
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Help in choosing correct power supply for LED matrix I'm quite new to electricity and electric components. I'm building an LED matrix from scratch on a small PCB. I will be using an IC for multiplexing, Transistor for controlling external power supply besides Arduino, resistors and of course the LEDs. Now, my general question is - power supply of what type(constant-current vs constant-voltage) and what values(Voltage + Amps) should I buy? My PCB with already soldered LEDs looks like following: 3 parallels of - 5 3mm blue LEDs in series. Forward Voltage 3.1V ; Current: 20mA - each. 3 parallels of - 5 3mm white LEDs in series. Forward Voltage 3.5V ; Current: 35mA - each. Now, after reading tons of articles on Current, Voltage, power sources and so on, there are still questions to which I don't have definite answer. I want to understand the core of all these, so I have following questions: 1) Can I power all 6 parellels, meaning whole PCB with one power supply? If I have understood correctly, I can not. I will need a separate driver for each of the LED color. Correct? 2) How should I calculate the driver values. In my understanding - 5(LEDs in series) x 3(parallels of blue) x 3.1(V each LED) = 46.5V. For Amps, should sum up the values of all 15 LEDs also, or should I calculate by just one LED as the Amps are same across the circuit? To sum-up, I will need a driver of 46.5V / 20mA? 3) Should the same calculation be used for white LEDs, and then two separate power sources connected together somehow? Please help me to understand not just the values of the power source, but also generally the idea behind the topic. Should I be considering something else into my calculations? Any advice for a better design? Thanks a lot. simulate this circuit – Schematic created using CircuitLab <Q> Before building a matrix of LEDs I highly recommend you to build a circuit to drive one single led, then a circuit to drive leds in series, finally a circuit to drive leds in parallel. <S> That way you will know how it works, it will help you a lot because it is an elementary subject in electronics. <S> Answering your questions. <S> It's possible with only one power supply. <S> 46.5v would be excessive, you're calculating it wrong. <S> If all the rows of leds are going to be in parallel, just calculate the voltage of the row that needs more voltage, in this case the row with the white leds. <S> No need of two power supplies or two calculations. <S> Good luck! <A> Using the following info: three parallel branches of five 3mm blue LEDs in series with Forward Voltage 3.1V; Current: 20mA - each. <S> And three parallel branches of five 3mm white LEDs in series with Forward Voltage 3.5V; Current: 35mA - each. <S> To determine voltage of your power supply would require looking at the max voltage for each parallel branch. <S> For the blue LEDs, we have 5x3.1V=15.5V and for the white LEDs we have 5x3.5V=17.5V. <S> So since 17.5 is the highest we’ll have to look at that. <S> Since you’ll be needing current limiting resistors, you need some headroom voltage. <S> I’d say 19V Would be your minimum voltage of your power supply. <S> As for your current specification you’d sum up all currents to determine minimum current. <S> This beingI_total=3x20mA + 3x35mA=165mA. <S> You’ll need a power supply with this current plus any other current drawn from your circuit. <S> If this is the only thing being powered I’d say a supply with 200mA current capability would be fairly sufficient. <A> Post-Edit Answer <S> My PCB with already soldered LEDs looks like following: 3 parallels of - 5 3mm blue LEDs in series. <S> Forward Voltage 3.1V; Current: 20mA - each. <S> 3 parallels of - 5 3mm white LEDs in series. <S> Forward Voltage 3.5V; Current: 35mA - each. <S> No. <S> What you have are 6 parallels of 5 parallel LEDs. <S> Nothing is in series. <S> This will not work well because there is no way to enforce the current through each individual LED. <S> You should never solder anything until you have a completed design and know exactly what you are doing. <S> I'm going to keep and include my pre-edited question answer, because it describes how to drive what you actually said. <S> Pre-Edit Answer <S> You haven't included how you plan to control the LEDs, so I didn't include anything in the schematic. <S> I would recommend you research <S> an N-Channel MOSFET switch which can easily turn your strings off and on. <S> On that note, you don't have to drive and LED at the rated current; I only do for illumination related projects, and then I overdrive them. <S> Blue LED String Voltage: <S> 5 x 3.1V = <S> 15.5V <S> White LED String Voltage: 5 x 3.5V = <S> 17.5V <S> So you need an 18V rail for those White LED strings. <S> I would highly recommend you redesign so you can use a lower voltage power supply, like 12, which is much easier to come by. <S> Use more parallel strings with fewer LEDs in series. <S> Voltage across R1, R2, & R3 <S> = 2.5V. <S> For 20mA, use 125 ohm resistors. <S> Voltage across R4, R5, & R6 = 0.5V. <S> For 35mA, use 15 ohm resistors. <S> Total Current = 3 <S> * 20 + 3 <S> * 35 = 165mA. <S> If you add enough for your microcontroller plus some overhead, you would need an 18V, 250mA power supply . <S> All that said, <S> it seems you need to look up the fundamentals of electricity, specifically Ohm's Law: Electromotive Force (Volts) = <S> Current (Amps) <S> * Resistance (Ohms) -> <S> V = <S> I*R Kirchhoff's Laws: <S> how voltage and current are shared/split in series and parallel connections. <S> Electrical Power (Watts): <S> P = <S> I*V = <S> I^2 <S> * R = <S> V^2 / R
| One power supply can handle it all; you will just use different value current limiting resistors.
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PCB female jack and power supply male melted I've been working in my thesis project which consist on a PID controler to mantain a certain temperature in a case. The heater module is made of two 40W-12V 3D printer resistors and one 15W-10 Ohm wirewound resistor. In addition to these, I have an Arduino Uno, a 20x4 LCD screen, two temperature sensors and two PC coolers (2W each) which all sum up in 5W (rounding up). I'm using a 12V-10A switching power supply. When the PID is at 100% the tension in the resistor (Rparallel= 1.5 Ohm) is supposed to be 12V, but due to the cables and transistor (IRF540N) drop, the maximum tension is 10V. So, all in all, in maximum behaviour I would be drawing 66.6 W + 5W= 71.6W. The PID circuit is as follows: The female jack that is soldered to the PCB is: which is located near decoupling capacitors in the PCB: The problem: I've been doing some test to set up the PID's parameters for over a month (more than 40 hours of tests) without any problems. Suddenly last week while doing a test I smelled like something was melting and found this: The power supply still works well so I wouldn't say there was an overdraw of current. My best guess would be that the PCB jack couldn't handle that much power but then, why did it resist more than 40 hours of testing? Something had to go wrong this time? I need to know the cause of the problem so I can continue with the tests. Obviously I will replace both conectors and try to find a more robust one for the PCB but if the problem comes from the power supply or the PCB circuit itself i will face the same issue again. What would you say the problem was? Could the power supply suffered an overdraw and still be working? Is there a way to determine which component overheated first (the PCB jack or the power supply conector)?Would you recommend to solder the power supply directly to the PCB? <Q> Unfortunately it seems that the female power jack is probably not rated for 10A or anything slightly below it. <S> If you truly plan to use high current like in 5A or up to 10A you should use some power connectors. <S> For example MOLEX power connectors like shown below. <S> You’d split up your positive pin into 3 wires and negative pin into 3 wires so they share the current. <A> You converted the power jack into a melting pot and just had your 1st failure to read max current specs and heat rise calculations. <S> 1st rule is to make a power loss budget. <S> lookup max current rating on power connectors (3A?) <S> Use <S> HDD MOLEX <S> 4 pin ganged ( actual scpecs range from 2.5A min to 3.5 to 4A for best quality) compute temp rise from RthJA = 62 ° <S> C/W (say Pd=0.5W with no sink) <S> determine max RdsOn . <S> R <S> =V²/P= <S> 12²/100W= <S> 1.44Ω and Pd/P * R = <S> 0.5/100W <S> 1.44Ω = <S> 72 <S> mΩ <S> so the IRF540 is OK with Vgs> <S> 10V <S> But your choice of BARREL POWER CONN is a HUGE NO-NO for 8 Amps. <S> Try 3A max. <S> Use this instead. <A> Barrel power connectors like you show cannot handle as much current as you are trying to pass through it. <S> If the center pin of the jack is the 2.1mm style they are typically rated at just 2A current. <S> The voltage of the supply has little to do with the connector capabilities that we are discussing here. <S> The connector contacts have a certain resistance and it is the I 2 R power level that causes the connector temperature to rise. <S> In your case till it rose enough to melt the plastic of the connector. <S> When designing things you should check the data sheets of all parts that you use to assure that your use of all components is within specifications.
| The reason why your power supply is fine is because it can handle the power. My guess is that when you used it over the past month you were using current that was under the maximum specified for that female power jack. Barrel jacks with the 2.5mm style center pin are rated a bit more, sometimes as much as 3A. (This is one reason you see 2.5mm jacks used on powered USB hubs).
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What is the difference between real ground and virtual ground in opamp? I understand what a real ground for a Opamp is.It is a terminal which is physically connected to ground or earth which acts as the reference point for the entire circuit. But what do we mean by a virtual ground? And how do we explain it with the help of opamps? And finally is my understanding of real ground correct or is there something more specific that i can add. <Q> In the context of an op-amp circuit, ‘real’ ground is the common zero-volt referenced to the power supplies and rest of the system. <S> Virtual ground comes up mainly when using a single-ended supply with an op-amp. <S> In order to represent signals that have negative and positive swing, the op-amp’s input and output signals need to be offset so that they fit between ‘real’ ground connected to -V, and positive supply connected to +V. <S> So most often, a low-impedance midpoint reference is used as ‘ground’ to give the desired offset to the op-amp. <S> In fact the virtual ground can be any offset with respect to real ground (including no offset). <S> What makes it virtual is that it’s a driven, low-impedance signal reference and not a direct connection to any ground or supply voltage. <A> Virtual ground means a node that has the same potential as ground but is not passively or inherently equal ground. <S> i.e. a point that is actively being driven to be equal to ground. <A> Ground by definition is a ZERO VOLT REFERENCE. <S> Earth Bonding is a common 0V reference while in electronics it can be floating. <S> Op Amps use negative feedback to force the difference input to become near zero and drive the output in the linear region. <S> So we call the differential null input a virtual ground. <S> Often Vin+ also goes to GND <S> so Vin- must also be near 0V <S> but that's only by design choice.
| ‘Virtual’ ground is a reference used for op-amp signals that represents a signal zero-volt reference.
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How to boost a Power Bank's maximum output voltage? I have an Anker ~27Ah Power Bank which has 3 × (5v , 3A) power output. Is it possible to connect the outputs in series and get 10v or 15v output with a current flow of 3A? <Q> All the 5 V outputs share the same ground hence they cannot be connected in series. <S> To make it easier to understand you can think that the outputs are already in a parallel configuration and hence cannot be changed. <S> What you can do however is use a boost converter, which increases/boost your voltage to a certain extent (depending on the IC used). <S> To clarify, you can connect it to one of the 5 V outputs and increase voltage. <S> The drawback is that the max current at the output will be capped (limited to an extent). <S> Further clarification is required on the use of AH units, AH is a measure of battery capacity. <S> For example 3 AH means that by supplying 3 amp continuously; the battery can survive for an hour. <S> However, this is a theoretical value as voltage decreases as the battery drains out. <S> if the voltage decreases then so does the current when used on a resistive load. <S> Anyways, your battery can supply more current than 3 A but it will last for less time. <A> Not within the same power bank. <S> The outlets all share the same ground, and internally, the same power. <S> If you tried it, all you will do is short two of the outputs to ground which would trigger their USB fault current limiters (so no magic smoke would come out.) <S> Charging would be slightly more complicated though. <S> You'd have to use 3 separate adapters that have floating secondaries (normal configuration for a wall-power charger, but not for a car adapter.) <S> Just something to check on. <A> The outlets are likely not current limiting. <S> USB tends to work on an honor system for current limiting. <S> It's up to the device you plug into the source to limit itself. <S> You could likely draw more than 3A from a single port if you don't care about over loading it. <S> You should be able to use a boost converter to get 10V 3A or 30W (drawing over 5V 7A or 35W due to efficiency)
| That said the ideal solution is to get a power bank that can do QC or USB PD, so it states it can do 5v/9v/12v etc and get a QC trigger module to force it into the voltage you want. If you had 3 separate power banks, then maybe it could work much like connecting 3 batteries in series.
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Got 2 gang light switch, common load, design circuit to know ON/OFF state of switches on each wire Goal is to be able to use standard wall 2 gang light switch to: a) provide line connection with either of two / both hot wires coming from the same 230VAC phase to the ceiling for AC-DC 48V PSU powering DC-DC buck driver, LED array. b) be able to differentiate with MCU which one / both of hot wires is powering the lamp for setting one of 3 brightness levels according to the active combination of switches' states: right switch - left switch - brightness ON - OFF - minimal OFF - ON - medium ON - ON - maximum So how do I design a circuit to know which of two switches is powering my lamp or whether both of them are ON? MCU and other logic could be powered with a separate low power AC-DC module. Delay up to a second for detecting current state is acceptable. Only ideas that I have involve shunt resistors, differential amplifiers, and etc. to measure and/or compare voltage drop that should be present only across resistors on active wire(s). This seems overall complicated and hard to account for 0-1A input current range. UPD1: My case is not a planned installation, but rather conversion of 2x 150W incandescent lamps (thus existing 2 gang switch and 2 wires in the wall) to a single LED array. It is possible to change that wall switch with an alternative that fits existing 68mm cylindrical electrical box, but chances that I will find anything as fancy as DPDT switch are pretty low. I will take a look though. My existing PSU is Meanwell LRS-200-48. UPD2: I tried solution with relays (see finalized schematic below) and it works as intended. I do not have 230VAC relays at hand so I tried it with 3x SRD-05VDC-SL-C general purpose ones and NA05-T2S05-V AC-DC units ( pdf ) as a proof of concept. As expected there was ~1 second load brownout (while turning off SW2), which is acceptable IMO. I could buy something like 230VAC rated RTE24730 to lower the switching time, but I decided against it for 2 reasons. Firstly, it would consume ~1.6W per coil, instead I can operate 12V relay at 9V with 0.23W per coil. Secondly, it means waiting delivery from Mouser and spending 6-9 USD per relay, while I can buy 12VDC RTE24012 locally for 2-3 USD. And I still have to provide my MCU with power, so either this or a dedicated DC-DC converted from 48V that I'll probably would have to make myself. So buying couple of AP09N05-Zero is well worth $8. I tested these PSUs, they have great specs even without input&output filters and manufacturer is pretty ensuring comparing to something like HLK-PM09 / HLK-5M09 which floods the market. D3, D5 are to ensure that MCU absolute maximum requirement for digital input of VCC+0.2V is not exceeded. This was for 5v AC-DC modules, so read a proper voltage divider here.. R2, R3 placed near input pins is to make sure there would be no 50hz noise (somehow I was reading up to 2VAC measured with 10MOhm DMM unless I loaded it with a simple 1.5mA LED). <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> An unusual approach. <S> I offer this solution more as a puzzle answer than as a solution that I would recommend. <S> How it works: <S> Almost certainly the 48 V supply you are using for your LEDs is an SMPS (switched-mode power supply) has a full-wave rectifier on the input so it can be run on full-wave rectified mains. <S> While the current through the internal bridge will be the same only two of the four diodes will be employed <S> so there is a risk that they may be stressed. <S> You're not interested in the zero-cross but rather that the mains turns on. <S> There are plenty of examples on this and other sites but bear in mind that a peak detector will suffice for you so you may be able to design for minimum power dissipation in the zero-cross circuits. <S> BR1 and BR2 rectify the AC coming from SW1 and SW2 and prevent backfeed from one circuit to the other. <S> It gets around the relay back-feed problems and avoids current-sensing solutions which might not work if the current isn't shared equally when SW1 and SW2 are both on. <A> gangable douple pole double throw 230V switches are available such as Clipsal 30MD2this greatly simplifies the signalling. <S> simulate this circuit – <S> Schematic created using CircuitLab here <S> X2 gets the power to run the lamps and X1 carries the signal - either open-circuit (SW1 only) with resistance (SW2 only) or with no resistance (both) <S> the resistor could be instead a diode or a capacitor whatever simplifies the analogue end most. <S> assuming 10V signalling the other end of the wire could something like: <S> simulate this circuit <S> I'm running the 10V signalling at mains potential as the signalling is galvanically isolated from ground and from the mains will be fine, but all the resistors used will need to be treated as live and need good insulation. <A> simulate this circuit – Schematic created using CircuitLab <S> This solution requires two mains voltage relays, one with at least two contacts <S> NO contacts and one with two NO and one NC contact. <S> How it works: <S> With SW1 and SW2 off LED PSU has no power. <S> When SW2 is on RLY2 is energised and power is passed to LED PSU. <S> At the same time RLY2b is energised disconnecting the supply from RLY1 and preventing a backfeed from holding on RLY1 if SW1 is switched off. <S> The tricky part is ensuring that going from both on to only one on always unlatches the other relay. <S> We've already seen in 3 above that switching off SW1 will work because RLY2b has broken the backfeed. <S> If instead we switch off SW2 then RLY2 will drop out and RLY2a <S> NO contact will open before RLY2b's NC contact will close. <S> RLY1 will then supply power to LED PSU. <S> Note that there will be a very brief interruption in power but the LED PSU's internal capacitance should bridge that without noticeable flicker. <S> You need to be very choosy about the relay as you require guaranteed isolation between the mains contacts and the low-voltage control circuits.
| Zero-cross detection circuits provide a simple and opto-isolated means of monitoring the switches' on/off status. With SW1 on RLY1 is energised and power is passed to LED PSU. RLY1c and RLY2c contacts can be fed to your logic controller to determine the four switch combinations.
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Mosfet controlled LED strip not bright enough So, I'm trying to control a LED strip with a NodeMCU using N-MOSFETs FQP30N06L , one for each channel. However, the strip is not nearly as bright as with the original IR controller. (I saw the similar question, however my transistors are definitely logic-level) The main problem SEEMS to be the mosfets, as if I connect the LED strip chanels directly to ground i get a much much brighter light. Also, there seems so be a quiet high pitch sound when using the transistors. The circuit is very simple, strip (5m, 30 leds/m) runs on 12V and should take 1.5A, so 0.5A for each channel, and that's what the power supplies. Looking at the datasheet for the MOSFETs, it looks like at 3V G_gs is should let through 10A, way more than the 0.5 I require. (However it is my first time looking at a datasheet so that's probably where my mistake is, though I have no clue what). I have a multimeter, just wouldn't know where to even start measuring things. Also I'm not connecting the MCU to 12V as I know that's too much for the voltage regulator, I just couldn't find a good enough cirtcuit drawing tool, so any suggestion for that is welcome as well. <Q> Fortunately, the graph you want is actually next to the one you used <S> and I have marked red lines on to explain where you problem is: - <S> So, on the graph on the right (the one you used) I have marked a vertical red line corresponding with a gate source voltage of 3 volts. <S> Now, at the point where the red line hits the traces if you looked across at figure 1 you can see what that means. <S> Note - the current is about 13 amps. <S> Can you see that the 2nd line I drew (the horizontal red line) crosses a bunch of traces in figure 1? <S> It doesn't cross the lower trace (not quite) and that is the trace when \$V_{GS}\$ is 3 volts - this means that driving it at 3 volts isn't going to deliver the goods (13 amps) typically. <S> Take the next trace up where \$V_{GS}\$ is 3.5 volts - you can see that my horizontal red line crosses it at a drain source voltage of about 0.7 volts (vertical blue line). <S> This typically means that if you wanted to deliver 13 amps to your load, the MOSFET would drop about 0.7 volts from drain to source. <S> If you drove \$V_{GS}\$ at (say) 10 volts, can you see that to deliver 13 amps, the MOSFET would drop about 0.35 volts (vertical green line). <S> Given that you only need 0.5 amps <S> I would say this MOSFET would drop significantly less than 0.1 volts across drain and source even when fed with a \$V_{GS}\$ voltage of only 3 volts. <S> So, my conclusion is that it can't be the MOSFET that's giving you the problem. <S> Maybe your MOSFET is "fake"? <S> However it is my first time looking at a datasheet so that's probably where my mistake is, though I have no clue what <S> Hopefully this clears up where the problem doesn't appear to be - use fig 1 in future <S> - it delivers what you need to know in this respect. <A> Based on you mentioning noise, I'm assuming you are using a PWM control scheme of some sort? <S> Are you sure you are driving the devices on 100% duty cycle. <S> Also if you are switching with PWM, you may need a dedicated mosfet driver, <S> the IO pin on your micro controller is not ideal for driving mosfets with a PWM like frequency. <S> The gate acts like a capacitor that is charged and discharged. <S> Doing this quickly requires more then the 20mA drive of most microcontrollers. <A> The datasheet says that the gate-source threshold voltage could be as high as 2.5 Volts. <S> Maybe you have a batch that came out at the high end, and your 3-volt gate-source voltage is not quite enough. <S> Do you have a way to temporarily apply a higher voltage to the gate as a test?
| Maybe it's the interconnecting wires or your power supply isn't able to deliver the current you need.
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Multiplying 3 bit number with 3 bit number using 4 bit full adder ive been trying to do this for hours to multiply 3 bit number with a 3 bit number ,can some here tell me whats wrong here ? <Q> 7 * 7 = 49, which needs 6 bits <A> There are lots of ways of arranging it. <S> Since you have full 4-bit adders to work with, this is pretty easy to understand: simulate this circuit – <S> Schematic created using CircuitLab <S> Each group of three AND gates represents a product term . <S> And it helps to lay it out so that this is clear. <S> (You need to improve your layout skills.) <S> Clearly laid out, it helps reduce the chances of making mistakes when you hook things up in the schematic. <S> All that the last adder does is add one part of one product term as the carry-in. <S> But it's necessary. <S> (It could be hooked to the A0 input of the last adder, instead.) <A> It looks like you have all of the adders wired backwards. <S> For instance, a0 AND b1 should feed input 0 instead of input 2. <S> Also, a0 AND b0 should directly feed the output LSB, bypassing the adders. <A> Bit number 0 is the least significant bit. <S> Your carry with this logic goes to the wrong direction. <S> Maybe it's useful to learn to do the binary multiplication manually on the paper at first.
| You have totally reversed the significance numbering.
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Free-as-in-beer power from neutral and ground? I live in a house where each socket has 3 connections,: hot (phase) neutral ground Where: The voltage between hot and neutral is about 220V AC. The voltage between ground and neutral is about 2-3V AC. Can I use those 2-3V as a 'free' (not paying for it) power source?Do you know any project using them? <Q> Short answer: <S> Kirchhoff current law says... no. <S> That ‘free’ current makes its way back to the utility feed, through the meter, and back to the pole, along with all the other currents in your house. <S> Long answer: in the video, they’re using the voltage difference between neutral and ground at the plug. <S> This is caused by I-R drop on the neutral wire carrying the return current back to the panel. <S> There’s a corresponding I-R drop on the hot wire, which you could see if you were to measure the hot voltage directly from the panel to the receptacle. <S> You will only see this if there is a load on the circuit feeding the receptacle . <S> Otherwise it won’t be there. <S> Here’s the thing: neutral and ground are tied together at the panel. <S> All that’s happening in the video is that some of the neutral current is being shunted back to the panel via the ground wire. <S> But make no mistake: it still makes its way back to the panel, through the meter, and to the utility. <S> And the meter will see it and tally it. <S> So it’s not ‘free’ as in beer at all. <S> (Don’t ever connect neutral to ground at the socket. <S> It defeats the safety function of the ground wire, which normally never carries current.) <S> Now, about that I-R drop. <S> 2-3V is an acceptable neutral drop per electrical codes, so <S> there’s no need to ‘fix’ it. <S> If it were more than that, it would be time to call an electrician to fix the faulty neutral, which he or she would do by making the run shorter, installing a new run, using thicker wires, or converting the run to use higher voltage. <S> The latter ‘fix’ is why some runs in a US home are 240V (dryer, oven, A/C) <S> while normal plugs are 120V. Using higher voltage reduces wire loss for these high-power devices, so in that sense it’s ‘free’. <A> Of course not. <S> The connection are called Live or phase, Neutral or return, and earth or ground. <S> The neutral is connected to earth at some point in your home, so ideally neutral and earth should have no voltage difference. <S> However depending on how the wires are routed, there will be return currents on the neutral wire. <S> Since all wires have resistance, the current on neutral wire will cause voltage drop over it, so neutral will only have same potential as earth only at the place where they are connected. <A> What you're measuring is probably a combination of neutral shift, due to the neutral conductor not being a perfect superconductor, but having a resistance and, unlike the protective earth conductor (what you called "ground"), carrying significant current high-frequency crosstalk and radio pickup being misinterpreted by your multimeter as useful AC voltage other measurement artifacts. <S> If it's actually a ground shift, it's already energy you pay for (and that voltage difference will instantly break down as soon as you stop having a load that leads to current in the neutral wire) and your devices suddenly using less payable energy at same power as soon as you add another current return path. <S> Any energy that you have here HAS already passed your meter. <S> And thus, you're paying for it – in this case, the energy is just used to ever so slightly warm up a neutral wire, but you're still paying for it. <S> The energy doesn't come out of nowhere – it's actually missing from the useful power of your intentional loads. <S> If it's radio pickup: it's much more high-frequency than you think, and it will instantly break down as soon as you try to attach a load. <S> Radio receivers deal with femto- to microamperes; that's the amount of current you could theoretically, after somehow magically converting that to DC, use. <S> Even then, it's not free – the radio transmitter used electricity to induce it. <S> Most of it is probably broadcast FM radio and TV; and in most countries it's forbidden by law to abuse broadcast radio to power anything but a receiver circuit. <A> It's not free. <S> It's dangerous (that 2-3V could momentarily become hundreds of volts when an appliance is turned on or off). <S> Unless you have faulty wiring you won't get much power out of it. <S> This is not electrical engineering , it's a disgusting hack.
| No, you cannot use it, and it's not free:
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Amplifier speaker outputs: is the black always the chassis? Any amplifier or A/V receiver that I have seen has a black and a red terminal for each speaker. From 2 pairs for a simple stereo system up to 7 or more for an A/V receiver. It is quite a while since I have looked inside an amplifier but, for at least some, the blacks are common and connected to the chassis. Can I safely assume this or are there any common or likely amplifier designs in which the black terminals might not all be at the same potential? Edit: Chassis might not have been the best word; as people seem to have picked up, the significance is whether they are all necessarily at the same potential. My outlook is still biased by early experience with valve devices: radios, TVs, amplifiers, in which a metal chassis (not an external casing) was usually a common 0V reference. <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Two common amplifier arrangements. <S> Can I safely assume this or are there any common or likely amplifier designs in which the black terminals might not all be at the same potential? <S> Bridge mode amplifiers use two amplifiers - one an inverted version of the other. <S> The advantage is that on the same supply voltage you can now get double the peak to peak voltage. <S> This is common in car audio systems where the supply voltage is fixed at 12 V nominal, but it is increasingly common in domestic amplifiers also. <S> The result is that you can't assume that black is common ground. <S> Investigation is required. <A> Can I safely assume this or are there any common or likely amplifier designs in which the black terminals might not all be at the same potential? <S> Especially in stereo applications that means that left and right "black" can't be at the same potential for every point in time. <A> I don't know if there is some sort of home-entertainment industry rule, but it certainly is very common practice. <S> NOTE that the black terminal is the reference potential ("system ground" or "signal ground"), but not necessarily connected directly to the chassis. <S> The system ground might be floating with respect to earth ground. <S> Separate from that, be wary of class D amplifiers or linear amplifiers with a BTL (bridge-tied load) topology. <S> In these cases, the black terminal is not connected to the system ground - it has 1/2 of the output audio signal on it, and the signals are different for each channel, so the black terminals are not interchangeable. <S> Red and black still indicate correct phasing. <A> You can check this with an ohmmeter. <S> Why? <S> Typical A/V gear will use a line-derived high voltage supply with a half-bridge driver, capacitively coupled to the output. <S> Again, why? <S> A line-powered amp has lots of voltage to work with. <S> There is plenty of headroom to avoid voltage clipping, even at high power levels using just a single-ended drive. <S> So it's just not worth the extra expense of using a full-bridge driver - even a switchmode one - for normal consumer <S> A/V gear like a receiver. <S> Car stereos - not so much. <S> Higher power in-dash ones will use a full-bridge drive to maximize the use of the 12V available. <S> Example: single-ended 12V power amp into 4 ohms: <S> Largest AC RMS output: 12V / 2 <S> * 0.707 = 4.24V Power into 4 ohms = <S> 4.24 <S> ^ <S> 2 / 4 <S> = 4.5W <S> Bridged 12V power amp into 4 ohms: <S> Largest AC RMS output: 12V / 2 <S> * 0.707 = 4.24V <S> Bridged: 4.24V <S> * 2 <S> = 8.48V Power into 4 ohms = <S> 8.28 <S> ^2 / 4 <S> = 18W <S> That's a big win for a car stereo, so it's worth the extra expense. <S> Now, car power amps are a different matter. <S> These will use DC-DC step-up make higher rails to achieve higher voltage swings at the output. <S> Some may use bridge, some will not. <A> In many older vacuum tube-based radios with inductively-coupled output stages (not utilizing negative feedback), it was common to have the speaker connected directly to the secondary of the output transformer (T15 in this example) floating it completely. <S> This obviously wouldn't be used in a modern consumer-designed amplifier, but an example of where the speaker output wouldn't be tied to chassis or even circuit ground. <S> EDIT: <S> Inspired by @hacktastical's comment I looked up another schematic with multiple secondary taps. <S> I don't have one for a McIntosh 240 <S> so this is for a Fisher 500C, also a home/consumer integrated amplifier. <S> I expected this to be more straightforward, but looks like the 4 ohm tap is tied to circuit ground, and the "common" <S> (I assume this is why historically it isn't called "ground") <S> floated below circuit ground via an inductor in series (or I'm not familiar with the construction of a transformer symbol where a winding extends past the core). <S> Either way, another example of where "common" or "ground" may not be circuit or chassis ground. <S> Fisher generally had (back then) good reasons for why they did what they did. <S> Would be curious if anyone has any thoughts on the advantages of this design? <S> (Mod note: if this is veering off topic of the OP I can re-post as new. <S> Now I'm genuinely curious.)
| In many amplifier designs (especially the power-efficient and well-integrateable and hence popular class D driving a H-bridge), both speaker conductors might alternate between any voltages within the range of the internal power supply. For most home A/V gear, yes, the black is ground (chassis ground).
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When does the transformer equation hold? The "transformer equation" is as follows, $$\frac{N_s}{N_p}=\frac{V_s}{V_p}$$ What is assumed when this equation is used? (Primary coil on the left, secondary on the right, in both cases) Is transformer 1 an ideal transformer, or transformer 2 (I have seen both in books and I'm not sure whether in the first picture, the other field lines are just not shown because they are irrelevant). It would make sense that transformer 1 is ideal and is 100% efficient because no work is done by the magnetic field in the air, but wouldn't a magnetic field through "ideal air" do no work regardless? How does this relate to flux linkage? Surely the flux linkage in both coils must be the same for the equation to hold, so in transformer 2, conceptually, less magnetic field lines go through the secondary coil than the primary coil, therefore the equation would not hold. <Q> For this assumption to be true, generally a magnetic core is needed to trap and guide the flux from one coil to the other. <S> Without the core, much of the magnetic flux generated by one coil does not pass through the other coil. <S> After that, there are many issues that lead to deviation from the ideal behavior, such as various forms of losses. <A> The transformer equation is for an ideal transformer. <S> A real transformer behaves like an ideal transformer with added components as shown in the equivalent circuit shown below. <S> The added components account for the flux linkage between the primary and secondary being less than 100% as depicted in your transformer 2. <S> It also accounts for the winding resistance and core losses. <S> It is the winding resistance and core losses that make the transformer less than 100% efficient, not the incomplete flux linkage. <S> Incomplete flux linkage results in leakage reactance, Xp and Xs in the equivalent circuit. <S> There is no energy lost in those components, but there is a voltage drop across them that is proportional to the current. <S> That means that the expected secondary voltage from the transformer equation could be reduced significantly as the load current increases. <S> Some leakage reactance is unavoidable, but good designs minimize that to the extent required for the type of use intended. <S> For the equivalent circuit presented below, the transformer equation holds for the ideal transformer part of the circuit: $$\frac{N_s}{N_p}=\frac{E_s}{E_p}$$ <S> For additional information about the above figure, see my answer to: Equivalent ciruit parameters of a 1kVA transformer . <A> What is assumed when this equation is used? <S> All the magnetic flux produced by the primary couples to all other primary turns i.e. there is no leakage flux that can give rise to leakage inductance that gives rise to formula errors when there is a load or magnetization inductance (which there always will be). <S> All the flux generated by the primary couples to all the turns on the secondary - this ensures that the voltage in the secondary is truly dependent on the turns ratio and the voltage ratio <S> However, meeting the above (and satisfying the equation) can be done theoretically with what we sometimes refer to as a non-ideal transformer. <S> But only if that non-ideal transformer has zero leakage components (see below for the listings in the equivalent circuit). <S> It can have magnetization current and, it can have core losses and <S> the basic equation still works - this is because core losses and magnetization current are parallel to the ideal "inner transformer". <S> Transformer equivalent circuit showing leakages and magnetization inductance and losses: - Picture from here <S> \$L_P\$ is the primary leakage inductance \$R_P\$ <S> is the primary copper loss <S> \$R_C\$ <S> is the core losses due to eddy currents and hysteresis <S> \$L_M\$ <S> is the magnetization inductance <S> \$L_S\$ <S> is the secondary leakage inductance <S> \$R_S\$ <S> is the secondary copper loss <S> If there is no loading on the secondary, \$R_S\$ and \$L_S\$ won't affect the output voltage. <S> If the magnetization inductance and core losses ( \$L_M\$ and \$R_C\$ ) are both very high impedances, then some leakage components on the primary ( \$L_P\$ and \$R_P\$ ) can be tolerated without much detriment to the equation in question (providing there is still no load on the secondary). <S> Is transformer 1 an ideal transformer, or transformer <S> 2 T1 might be an ideal transformer, T2 is not an ideal transformer <S> so in transformer 2, conceptually, less magnetic field lines go through the secondary coil than the primary coil, therefore the equation would not hold. <S> Correct.
| The most basic and important assumption required for the equation to hold is that all of the flux generated in one coil (primary or secondary) passes through the other coil.
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Connecting two DC converter's inputs in series I know it won't work, but I want to make sure I have the right explanation why it won't work. Given are two devices which have a build in converter (I have this digital voltmeter in mind). They need 12 - 24 V AC or DC supply. And apparently the supply is galvanically isolated from the measurement inputs. Also given is a let's say 40 VDC power supply. One could falsely come up with the idea of just connecting the two voltmeters power inputs in series with the 40 VDC supply in order to save two (e.g.) 24V Converters. My explanation why this won't work is as follows: Since those digital voltmeters while have internally some sort of buck converter with a feedback-loop (due to the variable input voltage), the voltage will not divide 50:50. The ratio will either tilt to one or the other side, depending on which device has a slightly higher load when turning everything on. I would say the measurement input connectors have no influence on this and could theoretically be connected since there is a galvanically isolation anyway. The only problem is that these two feedback loops behave unstable and in the end tend to apply almost all the voltage to one of the two devices. This will most likely kill one of them (or at least the fuse). Does this explanation sound valid, or is there a more reasonable explanation that I have overlooked? <Q> The problem is what happens if your two devices want to draw different amounts of current. <S> The feedback mechanism in their internal power regulators makes things more complicated, but it very likely leads to things not working well at all. <S> One device might end up getting the whole 40 V and the other getting nothing, the system might oscillate, etc. <S> A simpler solution is get just one 24 V AC-DC converter, and connect the two devices to it in parallel. <A> It’s possible. <S> Use a 21V shunt Zener regulator across the input of each device to ensure the input voltage doesn’t go too high if the loads are unequal. <A> The Photon's answer agrees with my explanation attempt. <S> However, it does not state clearly a reason for this. <S> Doing some research I found a paper which mentions the following: It is important to note that stable operation does not result when two standard, independent dc-dc converters, without any special control for sharing, are connected in the input- series and output-series mode. <S> For example, if a small disturbance causes a converter input voltage to increase, its control loop reduces its duty ratio, thereby reducing the average input current. <S> This results in further increase in the input voltage of the corresponding converter, ultimately leading to the entire input voltage appearing across this converter...
| In the absence of input voltage sharing control, any small disturbance causes a runaway of the converter input voltages due to the negative input resistance property of dc-dc converters.
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How do computers output a signal without an input? I'm curious about how an output pin gets turned on at a low level when there is no input like pressing a button to complete a circuit. If you program a raspberry pi to turn on and off an LED how does the output pin get switched? If its transistors doing the switching with a small current at the base allowing a larger current to flow through, what turns on and off the current at the base? <Q> Software and circuitry is just a bunch of switches (transistors) that trigger each other in a big complex loop so they continue to trigger each other. <S> You manually crank an engine to turn it over and start it, but what makes the engine continue to spin after that when you stop cranking? <S> The engine does, of course. <S> It keeps itself spinning by hitting all the right valves and switches while spinning to keep spinning. <S> A processor is the same. <S> Also see: What happens at hardware level when we feed a code? <A> You may be missing the concept of a clock: a circuit which produces a string of pulses at a high frequency. <A> They do get input - via the software that’s loaded into them. <S> At some point, someone or something put that software in the machine. <S> The software not only defines the algorithm it runs, but also the initial data used to run it. <S> Further, as the program progresses it can get more data that was loaded along with the program. <S> There’s another input too: power . <S> Turn on the power, and the machine runs its software. <A> The output is managed by a transistor. <A> Program is just a sequence of pulses/switches. <S> Switch on your fan, bulb and motor. <S> Name the buttons F, B, M. <S> And you have made a language. <S> Bring a microcontroller or rasberry or what ever. <S> Feed the squence in their memory, registers or flip flops. <S> Play it whenever you want and you will turn on the fan,bulb,motor with the same sequence. <S> Basically its your clicking on the keyboard that turn on the switches of transistors etc. <S> Transistor is i guess the smallest unit of computers. <S> These transistors make many things. <A> It's not clear if your question is about how computers can do things behind the scenes (e.g, clock plus a set of instructions and some memory), OR, how computers bridge digital logic with analog circuits. <S> I'll answer the latter. <S> The line between digital and analog circuits is actually really blurred. <S> A computer can have a state, or some memory, that it can set to some value (true or false in the simplest case). <S> So it may be simple to understand the following computer code. <S> if our calculation is finished, turn on that light. <S> if (calculationFinished <S> ) lightOn = <S> true <S> this sets some memory (referenced by 'lightOn') to some value (true, possibly a 1) <S> But when you set some variable or some computer state to a value (such as true, or a number, etc..), it's still just a circuit state. <S> Somewhere, that memory is stored inside a circuit. <S> There's no special computerized ether outside the circuits - it's all analog under the hood. <S> In digital circuits, you typically represent true and false with some voltage like 0v (false) and 5v (true), to turn that analog world into digital. <S> So if you have a value in software memory, somewhere there is actually a pin or a wire (or silicon or...) <S> that either has 0v (against ground) or 5v (against ground) in our case. <S> In the simplest circuit you connect that pin to the LED <S> and it turns on. <S> Of course, that pin may be too high resistance (not enough current), so you feed it into a transistor as a signal that can let lots of current flow to power your light. <S> Computer World <S> | Analog Circuit world<amazing stuff>--some wire connected to lightOn -|- transistor -> <S> LED<here be dragons> | <S> <here be other dragons <S> > <S> They're in the same world. <S> If you're interested in how you can have memory in a circuit, e.g., that remembers that true <S> even after the computer moves on, check out flip flops : If you're wondering how a computer could possibly have dedicated wires for EACH BIT , when your computer can have giga BYTES of memory, it comes down to addressing and multiplexers: you set some memory to tell it /which/ memory you want to look at, and it acts like an old fashioned telephone operator to connect your wire to the right memory location. <S> Check out multiplexers
| The current at the base of this transistor is turned on and off by an output pin at the raspberry core processor.(Between the final transistor and the core processor there are intermediary components, but to make it simple, that's how it works)
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Power dissipation of a current limiting resistor (connected to an input pin of uC.) Below I have attached a circuit which divides the voltage and gives the output to an input pin of a microcontroller. Input voltage = 24V The resistor R0012 is for current limiting. The Zener diode breakdown voltage is 5.1V. The microcontroller which I am using is - Microcontroller Datasheet I could not see an GPIO input pin internal architecture of diagram inside the microcontroller datasheet. I tried to simulate this portion of the circuit with the MCU end floating. I found that the voltage before R0012 is 2.9486561V and after R0012 is 2.9486144V. The current through it was 2.953uA.I couldn't understand how this current is obtained (even though I left the other end of the diode floating - nothing connected at the MCU end.) My objective : I want to calculate the maximum power dissipation of the R0012 current limiting resistor. I know that it will not dissipate a huge amount. Just want to understand. My questions: The microcontroller datasheet attached mentions a GPIO sink current of +3mA max. So, what would be the current flowing into the microcontroller pin in my case and what would be the voltage detected at the MCU pin? How to go about calculating the power dissipation of the resistor? How did my simulation show a current of 2.953uA? <Q> First question where is that trickle of current going? <S> It's leakage current through the diode. <S> Got a datasheet? <S> Its probably specified Second question ( actually labeled #1). <S> The max 3mA is when the port is an OUTPUT You seem to be wanting an INPUT <S> yes? <S> The current would be close to zero. <S> Except if you have am internal pull resistor active. <S> In other words... "What is the input impedance of your I/O port? <A> Even capacitors have a certain leakage current. <S> See this picture which shows a reference model of a non-ideal capacitor. <S> The maximum power dissipated by the resistor can be calculated by getting the maximum voltage on the resistor \$(24V/87.75kOhm*5.75kOhm)^2/11.5kOhm = <S> 215uW \$ <A> R0012 will never see any appreciable current. <S> Assume that D0005 shorts out completely. <S> That effectively puts R0011 and R0012 in parallel. <S> The total resistance is (R0011||R0012)+R0010. <S> That works out to 87.75 kohms. <S> At 24V going in, that's a current of 0.27 milliamperes through R0010. <S> Half of that goes through R0012, so 0.13 milliamperes. <S> The power dissipated in R0012 would be: <S> P= RI^2= 11.5k <S> *0.13mA^2 = 0.215 milliwatts. <S> Any resistor can handle that power. <S> The dissipated power does go up as the input voltage goes up. <S> I'd be more concerned about R0010 and R0011, though. <S> They are always exposed to the full voltage and current. <S> The calculation above applies to R0011 as well.
| It'll take a very high voltage to exceed the power rated on even a 1/4 watt resistor.
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DC Relay overrated for max. 100ms In my case I need a relay to switch continous 0.5A at 24V DC , but in time of closing the contacts there will occour an inrush current from the further circuit of 6A, but only for max. 100ms because there is a contactor (DC) in line of the switch. I dont know the exact amount of damage of the contacts in that case, because its only for milliseconds. Do I need a Relay with max. switching current rating of 5A or 8A rating?Could there be a significant amount of damage in general? I already checked the datasheets according to that, but the graphs are ending at the rated current. I personally think I schould just take the overrated big one, but in my case a smaller component size would be better and it will be fully underrated wit 0.5A for the continous time At a 5A type I would take the omron G6B-2214P-US or the Panasonic DSP2a-DC5V.They are DPST relays, the second pole will be used for a signal. Thanks for your help! <Q> In-rush current is a killer for contacts in relay (and elsewhere).A short current spike during operation, when the relay is properly closed might be tolerable (ask the manufacturer). <S> In-rush current on the other hand is much worse: It starts flowing at the very first instant the contacts barely close. <S> Usually contacts bounce for a short moment, causing a current arc and extensive heating of the contacts. <S> The same holds true for high currents when switching off the relay (i.e. when shutting down an inductive load). <S> As soon as the load gets capacitive or inductive, the allowable limits are much lower. <A> Contact rating is almost always limited by the switching current, not the steady state. <S> It is at switching that the arcing and damage will occur. <S> Photo 1. <S> (Left) <S> Pristine contacts from a relay(Right) <S> The nearly destroyed contacts from a relay operated under power for nearly 100,000 cycles. <S> Image source: <S> Arc <S> suppression on Wikipedia. <S> While the damage in Photo 1 may be as a result of normal wear over a large number of cycles, the results due to over-current may be similar in a very low number of cycles and could lead to failure of the contacts to open. <S> i.e., They may weld closed. <S> Don't think you know more than the guys who make the relays. <A> This will end badly (probably with welded contacts, if not immediately, after some period of time). <S> If you can find a tungsten rated relay they use an appropriate contact metallurgy to handle surges of the order of 10x the continuous current. <S> Compressor (motor) rated relays may also be appropriate <S> but I doubt you'll find them in a relatively small size. <S> Another thing to look for is "TV" rated relays, such as TV-3 or TV-5 that are rated for generous amounts of inrush current. <S> Cadmium is no longer acceptable in the contacts for RoHS reasons so they tend to use silver oxide and alloys. <S> Unfortunately, pretty much all such relays will be larger and more power-hungry than "signal" relays designed with contact forces and contact metallurgy to switch small currents.
| Expect the contact life (number of operations before it wears out) to be significantly lower than if you are switching the rated resistive load. As a result, the rating of relays are typically for resistive loads only.
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Cost reduction by eliminating vias from PCB design I am designing a small (30mmX20mm) and not complicated two layer PCB. It is possible to do the design without using vias. Is it possible to determine how much can I save approximately by eliminating vias from design? Is it worth it? <Q> Centuries ago, when I first started designing PC boards, I believe we were charged for the number of vias/holes (or number of different hole sizes) on a board, but that charge was dropped long ago. <S> There may be an extra charge if you have an excessive number of holes - like for a board matching the hole pattern of a plastic breadboard - but there is not usually a per-hole charge for "normal" PC boards. <S> (However, this may depend on the board shop..) <A> Basically, the cost breakdown for doing the drilling of vias and other through-holes is: (1) <S> setup time (putting the board on the machine) <S> (2) time on the machine (3) removing the board from the machine (4) (later) plate-through holes <S> Costs #1, 3, and 4 are the same regardless of the number of vias. <S> Cost #2 does vary, not just with the number of vias, but how long it takes to drill them, and how often the bit needs changing. <S> So there is some benefit to reducing vias if you have to have them, and also keeping the hole sizes large enough that the bits don't wear out as quickly. <S> But in the big picture, for a very simple board this won't matter much. <S> Also consider that if your board has a special shape or uses 'mouse bites' or other breakaways for panelization, that takes routing time on the same machine that does the drilling. <S> This in fact will dominate time spent on the drilling table. <S> Using v-scribe singulation avoids this. <A> It is not worth it, unless you make million or more units per year. <S> The cost per via on a 2 layer board, assuming no HDI, laser drilling, back filling or similar high cost requirement is negligible (less than 0.001USD). <A> We've been involved in high volume, consumer electronics, and produced mostly overseas in 50k quantities of product runs. <S> And the answer is that we've never seen it presented as a cost driver. <S> I wouldn't be shy with the vias from a design perspective vs. cost analysis. <S> As others have said, the layers of the board and size of finished board are the biggest drivers. <A>
| For small orders, it is likely that your board design will be put on a panel with other customers' designs, and if those people use vias, the panel will be put into the via drilling machine nonetheless, so there is no cost difference other than machine time, which is really cheap.
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How does this voltage adjustment circuit work? This is a real-world application. My circuit 101 knowledge is not up to the task. I cannot understand the purpose of the portion of the circuit highlighted in yellow. With the presence of diode D3 and D4, I can't see that R_10K ever gets utilized. The only scenario R_10K will have some current passing through it is when V@pointA < 0.7Vdc and V@pointB > -0.7Vdc, but that's an impossibility, you can't reconcile the currents flowing in R14 and R15. The circuit just seems impossible to operate. I must be missing something here. In retrospect, judging from all the answers to this question , the place I was stuck upon was the notion that: Once D3 and D4 are conducting, they will short circuit R_10K. In a real-world situation, a forward-biased diode is not a short circuit. <Q> R15 and R15 provide enough current to bias D3 and D4 "on" so that each has roughly 0.7V across it. <S> So the 10K pot has about 1.4V across it and 140uA of current through it. <S> DR14 will have about (15V - 0.7V)/1.5K or about 9.5mA through it. <S> That current, minus the 140uA in the pot flows through the diode D3. <S> Similar reasoning for the negative branch of the circuit. <S> So the pot can set a voltage of anywhere from ~ <S> +0.7V to -0.7V depending on the wiper position. <A> A is 0.7V, B is -0.7V, approximately. <S> 10K pot has about 1.4V over it. <S> Pot can select between -0.7 and 0.7V by turning the pot. <S> 1k5 has slightly less than 10mA flowing. <A> If you remove the diodes, there would be about 23 volts across the 10 K pot. <S> However, with the diodes, D3 will be forward biased <S> so will hold point A at about +0.7 volts. <S> Similarly, D4 will hold point B at -0.7 volts. <S> Therefore, the 10K pot will have about 1.4 volts across it, and will pass 0.14 mA. <S> The voltage on the pot wiper can vary between -0.7 volts and +0.7 volts. <S> You could have much higher value resistors as R14 and R15, and still have this work. <A> While John, Peter and Justme have answered your 'how' question exactly, you may still be wondering 'why'. <S> These diodes are being used as a cheap voltage regulator. <S> If the diodes were not present, and if the resistors were adjusted to provide +/- <S> 0.7 V by themselves, then if the rails changed to (say) +/- <S> 10 V, the bias voltage would drop in the same ratio. <S> Even worse, if just one rail moved to 10 V, then the bias would change from being centred around zero, to being offset well away from zero. <S> The 'diodes have a 0.7 V drop' model ... <S> The diodes clamp the pot voltages to +/- <S> 0.7 V to allow the rails considerable freedom in how stable they have to be to get a stable adjustable bias voltage. <S> The 'diodes have an exponential voltage/current relation' model ... <S> The diodes reduce the sensitivity of the pot voltages to the rail voltages to allow the rails considerable freedom in how stable they have to be to get a stable adjustable bias voltage. <S> Let's say we get a change of rail from 15 V to 10 V, and the current drops from 10 mA to 6 mA. <S> At the average current of 8 mA, the diodes will have a dynamic resistance (that's <S> dV/dI, not V/I) of about 3 ohms (approximately 25 mV / diode I, for silicon diodes), so the 5 V 33% rail change would give an expected voltage variation of about 12 mV or less than 2%. <S> As you can see, a change of 10 mV in the roughly 700 mV across them, is 'clamped at 0.7 V', for most purposes.
| This circuit will give a fairly stable range of voltages from the pot wiper.
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When is a wire considered high-Z? I need to make sure I understand perfectly the high-Z state and its relation to voltage.I intend to explain what I have understood based on three example, is there something wrong ? Let's consider the three following examples : If the LED is red, the point 1 will approximatively be at 1.3 V because the voltage drop across a RED LED is approximatively 2 V. Point 4 is at high-Z so we do not know its voltage. Can we assume it is near 1.3 V, or can it really be anything ? If I measure it with a voltmeter I would get around 1.3 V, but it is because the voltmeter is injecting a small current resulting in measuring the LED's resistor value ? Point 3 is at high-Z as well. But is point 2 considered at high-Z ? It is not really an unconnected wire as it is connected to a resistor. More generally, if we replace the resistor with any other passive component, would 2 be at high-Z or no ? <Q> Think at an Hi-Z wire in this way: can I connect its lead to any (reasonable) point in the circuit without having current flowing in the wire? <S> If yes, then the other lead of the wire is connected to a Hi-Z point of the circuit. <S> so no current can flow. <S> In your schematic NO wire (neither of 1,2,3,4) is Hi-Z because if you connect one of them to GND, current starts flowing and the LED lights up. <S> UPDATE : <S> Wire number 3 can actually be Hi-Z, provided that the resistor attached to it has a very high ohmic value. <S> Hi-Z is used for example in microcontrollers when they start-up. <S> Given that any I/O pin can be connected to anything, and at the start time the microcontroller does not even know whether a certain own I/ <S> O pin is (will be configured as) an input or an output, those I/ <S> O are Hi-Z: <S> you can pull them low or high, you can attach them to an input of some other IC, but nothing happens, they seem to be connected nowhere and no current can flow. <S> Another example where Hi-Z is desirable is in scope probes: a scope probe should be Hi-Z (as much as possible) in order to not pollute (or worse) <S> the signal you are trying to measure. <A> Hi <S> -Z just means high impedant. <S> So, when connecting such a wire to any voltage with any frequency, there will flow hardly no current through that wire, because $$ \frac{V}{Z}=I \text{ }\text{ }\text{ or }\text{ }\text{ } \frac{V}{\text{high impedant value}}=\text{very small current} $$ <S> In your applications, this only applies to wire 3 when the resistor above it is high ohmic (and so, high impendant). <S> UPDATE Reading the wikipedia page about High Impedance , I see there is a distinction made between analog and digital electronics. <S> In digital circuits, a high impedance (also known as hi-Z, tri-stated, or floating) output is not being driven to any defined logic level by the output circuit. <S> So, in that case, it depends on your application how these logical levels are defined... <S> In analog circuits <S> a high impedance node is one that does not have any low impedance paths to any other nodes in the frequency range being considered. <S> This is what I wrote in the first part of my answer. <S> In your schematic the load connected to the nodes 1,2,3 or 4 determined whether the LED starts to (seriously) conduct. <S> So, it is load dependent. <S> When you want to explain digital Hi-Z, you'd better use the following schematic: simulate this circuit – <S> Schematic created using CircuitLab <S> Here, the output pin can be configured as output high (by driving M1), low (by driving M2) and Hi-Z (by neither driving M1 or M2). <A> If I measure it with a voltmeter I would get around 1.3 V, but it is because the voltmeter is injecting a small current resulting in measuring the LED's resistor value ? <S> Kind of. <S> The voltmeter is not "injecting" a current. <S> A voltmeter behaves (almost) like a very high resistor, so putting the voltmeter in there effectively turns the circuit into your first example, and you see 1.3V. <S> The voltmeter's impedance is enormous, so in almost every other case you wouldn't notice a change in circuit behaviour with the voltmeter attached. <S> But here, because the only thing completing the circuit is the voltmeter's impedance, you do. <S> If you didn't have the voltmeter there, you have no current through the LED, and hence no voltage drop on the LED. <S> So point 4 will be at 3.3V. <S> Unfortunately you can't prove that with a voltmeter though! :)
| A Hi-Z point is one connected to a very high resistance
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How to digitally control a MOSFET? I'm trying to make a temperature controller for peltier module using MOSFET. The peltier uses 3.3 to 12 volts and 15A. So far I've only come across a way in which the circuit uses a potentiometer at the gate & drain to control the MOSFET source output. I'm looking for a way in which the microcontroller can automatically control the voltage from a feedback loop. I've come across some digital potentiometers but they can't handle 15A. Any help? By the way, I'm not an electronics engineer. Thank you. <Q> A good way to mechanically kill a Peltier is to switch it on and off sufficiently infrequently that it changes temperature, anything longer than a few seconds for instance. <S> Thermal cycling degrades them. <S> A Peltier shifts heat proportional to the average current, but generates waste heat proportional to the current squared, so even fast PWM gives you a much lower efficiency than steady DC of the correct value. <S> Use a proper buck converter with an output filter for controlling the current rather than simple PWM. <S> If you compare a linear FET control with buck converter, the overall efficiency will be very low for the linear regulator. <S> In addition, a FET does not make a good linear regulator. <S> Most BJTs, but few FETs, are rated for steady power dissipation. <S> A good way to destroy a typical power FET is to let it dissipate steady power, even when well heatsunk, and well within its headline maximum power. <S> So, get a design for a buck converter, and build one of those round your controller. <S> Or buy one. <A> First, establish a linear relationship between the temperature and PWM. <S> I have plotted temperature in the x-axis and OCR register value (OCR controls the PWM) in the y-axis. <S> For example, I want that when the temperature is 0-degree <S> Celsius, the pulse width will be 100% (i.e. OCR value 255). <S> And when the temperature is 70-degree Celsius, the pulse width will be 0% (i.e. OCR value 0). <S> So, the relationship between the temperature and pulse width will be: y=255-3.642857x <S> Here, x is the current temperature and y is the OCR value corresponding to it. <S> If your sensor is analog then you have to use ADC. <S> Then the x-coordinate value will be the ADC value. <S> Use the pulses to drive a MOSFET or IGBT. <A> One other way could be to use an ADC to an OP-amp to buffer the signal to the gate.
| I agree with Neil_UK, but to answer your question, you could use PWM-modulation where you filter the output with an RC-link in such a away that the ripple will be negligible.
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Is there a method to know the temperature value of a microcontroller? I want to know if there is a method to know the temperature of the microcontroller to avoid its over heat. <Q> The vast majority of microcontrollers on the market today have some kind of built-in die temperature sensor. <S> Often this is implemented as an extra channel on the ADC multiplexer. <S> It may not be especially accurate, but should be fine for detecting an overheat condition. <A> There are million digital temperature sense IC. <S> You can glue one to your CPU and connect to your MCU like any other IC. <S> Just read the temperature directly. <S> It is important to know alot about the IC if you want to do this correctly. <S> You need to know the thermal impedances of the IC to case, and case to ambient (or case to heatsink, then heatsink to ambient). <S> THese values are found in the IC datasheet. <S> The math is simple. <A> If you already have an I2C bus in your project you can hang a digital temperature sensor that has a unique address off the bus without consuming any additional I/O pins. <S> If there is no conflict between polling the temperature and other usage of the bus (which only you can decide) then you have a potential solution. <S> An example of such a chip is the MCP9808 , but you should evaluate the various offerings should you decide to go this way. <S> Accuracy: <S> ±0.25°C (typical) from -40°C to +125° <S> C ±0.5°C (maximum) from -20°C to 100° <S> C <S> ±1 <S> °C (maximum) from -40°C to +125°C
| If you can calculate the thermal impedances, you can indirectly approximate the temperature of your IC die.
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Matched biasing transistor vs. emitter resistor in current mirror If adding emitter resistors can reduce the temperature variation in \$V_{BE}\$ , why do we still need to find a matched pair of transistors to construct a current mirror? <Q> You are right about the emitter resistor reducing the temperature variation of the current mirror. <S> However, not the collector current is mirrored but the base current / voltage is mirrored, meaning that you still have some dependency on the gain of the second pair of your current mirror. <S> In the following circuit, the current through \$Q_1\$ <S> will be approximately given by: $$I_{C,Q_1} \propto <S> I_B\cdot \beta_{Q_1}$$ <S> $$I_{C,Q_1} \propto I_{C, Q_2}\dfrac{\beta_{Q_2}+1}{\beta_{Q_1}+1} \dfrac{\beta_{Q_1}}{\beta_{Q_2}}$$ simulate this circuit – Schematic created using CircuitLab EDIT #1 Consider the following circuit: <S> The base voltage which is common for both transistor is given approximately by (for the sake of simplicity the internal emitter resistance is neglected ): <S> $$V_B = V_{BE,1}+ R_{E,1} <S> (I_{C,1}(1+\dfrac{1}{\beta_1})) = <S> V_{BE,2}+ R_{E,2} <S> (I_{C,2}(1+\dfrac{1}{\beta_2}))$$ <S> For a matched transistor and resistor, the above formula yields: <S> $$I_{C,1}=I_{C,2}$$ Assumming now that <S> , \$\beta\$ <S> is not matched, the collector current of \$Q_2\$ is given by: $$I_{C,2}=I_{C,1}\dfrac{\beta_1 +1}{\beta_2 +1}\dfrac{\beta_2}{\beta_1}$$ <S> As already explained in this question , \$\beta\$ , which is defined by collector to base current ratio, has some temperature dependence given by: $$\beta <S> = \dfrac{I_C}{I_B}=\dfrac{I_C}{I_E - I_C}$$ where \$I_E\$ can be written in terms of thermal voltage \$V_T\$ , which is dependent on the temperature according to: $$V_T=\dfrac{k_BT}{q}$$ <S> The following simulation, can show that although not very significant, the beta of the transistor still play a role on the mirrored current despite the introduction of matched emitter degeneration resistors. <S> This simulation, modifies the beta of the second transistor, by adding a <S> \$\pm 50%\$ tolerance to its nominal beta value. <S> All other parameters are left untouched. <S> Furthermore, the simulation is run for 3 different temperatures in order to accout their variations on the final collector curent. <S> As you can see in the above plot, the output current ( \$I_{Q,2}\$ ) has a dependence on the temperature, and consequently on beta. <A> The tilt of the collector-current versus collector-voltage (Early Effect) also causes errors. <S> Some current mirror topologies address that effect. <A> In a designed current mirror IC <S> the Vbes would be matched but in say a power amplifier's input stage, which would make use of discrete transistors, the emitter resistors should be included, reducing the effect of Vbe tolerances and thereby matching the collector currents much more closely which reduces distortion in the amplifier.
| The reason for adding the emitter resistors, a technique known as emitter degeneration, is not for temperature reasons but to reduce the effect of any mismatch in the transistors' Vbes.
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Different readings of voltmeter on simple circuit Hi I am trying to learn the basics of electronics and puzzled by the following qucs simulation. Why the readings of voltmeters are different? I would expect, by the way zero readings on both voltmeters as they are basically breaking the circuit one for another. At least the readings should be symmetric but they are not. <Q> I suspect the impedance of the virtual meters is infinite - breaking the circuit. <S> You may get better results by putting a 10 megaohm resistor in parallel with the meter. <S> Edit: <S> this modification of your circuit clears up what's going on. <S> Falstad makes three simplifying assumptions: (1) It defines an absolute voltage for purposes of simulation, rather than real voltages which are defined between two points. <S> (2) <S> Real voltmeters do not do this. <S> (3) Voltmeters have infinite impedance. <S> (4) <S> In the original circuit at the top of the page, it looks like it arbitrarily decided that the positive terminal of the battery was to be used for "absolute zero voltage". <S> That explains all the rest of the readings - the resistor and the right-hand terminals get assigned "absolute zero". <S> Not entirely broken, but all you're doing is showing how the simulation is a simplification of reality. <A> In theory, both voltmeter should be reading 0V. <S> So you're right there. <S> But in real life, if you did this experiment, both meters would not equal 0V. <S> This is because wires have resistance that varies depending on the size wire and its length. <S> A small voltage is being divided along the wires when the circuit is closed, so that's why your meter would not equal 0V. <A> In real life, I would expect the sum of the two meters' readings to equal 1 V, and that's what you've got there. <S> So in a sense, it's not inconsistent with real life. <S> If I had two identical meters, then I would expect each to read 0.5 V. <S> The simulation looks suspect. <S> If the meters are identical, then they have resistance, or they do not. <S> If they do, then they should each read 0.5 V. <S> If they don't have resistance, then the simulation should fail, with a 'floating node' error (or something more obscure that means the same thing, like 'zero determinant matrix'). <S> There's no reason to think the meters should be different in the simulator. <S> It appears therefore that the first meter it meets somehow breaks the circuit, and it does something different for the other one. <S> Try the simulation again with a 10 Mohm resistor across each meter.
| A voltmeter with nothing on the other end therefore produces a voltage as if the other end was connected to zero volts.
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What is purpose of Balun transformer in RF front end In following RF receiver circuit (30 MHz), two wires of a dipole antenna are connected to SMA connectors and the signal goes to an amplifier. I don't understand what is purpose of this balun transformer in circuit. If anybody can explain? <Q> Without much information about the signal connected to the input it is hard to say. <S> However, one of the main applications of baluns is the rejection of common mode signals. <S> The common mode signals which are in phase in both lines, will cancel each other out through the opposite magnetic field generated by the balun. <A> The purpose of the transformer is as follows: <S> To create a ground reference for the balanced signal being fed to the input of the differential amplifier. <S> This would also improve the balance in the system. <S> To create a DC path to ground for static charges on the dipole elements. <S> To nullify common mode currents in the feeder. <S> This offers redundancy to a prime function of the differential amplifier. <S> The schematic has been redrawn to make it easier to comprehend. <A> Possibility: - If you consider that with the balun in place, either input individually can now properly drive the input amplifier. <S> This might mean: - The system might be alternatively used with a monopole antenna and then the balun will provide an anti-phase signal to the other amplifier input and might help improve performance. <S> The implication of the above is that this might also be used as a test fixture allowing a single ended input (such as from a signal generator) to be used on either input.
| In you schematic, since both connectors are expecting to receive the "same" RF signal, it seems that the balun is being used to attenuate common mode signals , e.g. noise.
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Mount a capacitor on a connector without any solder I am looking for a specific connector that I can solder / press-fit in my PCB and mount a capacitor on the connector without any solder such that I can easily remove or replace the capacitor without desoldering. I have one option to use a press-fit capacitor. But I am open to any other suggestion for the connectors. AEC 100uF, with 2 leads, cylindrical, axial mouser.ca/datasheet/2/212/KEM_A4011_PEG124-1104316.pdf <Q> You might consider using screw terminals. <S> Such as shown below. <S> No solder required. <S> Just a small screw driver. <S> It's reliable and durable. <S> And you can try different parts if you need to. <A> For a tool-free solution, use a common female header. <S> (Image from sparkfun.com ) <S> The male leads of the header go in your PCB. <S> Insert the capacitor leads in the holes of the header and they are pinched in place. <S> You can choose one that spans enough width for your capacitor. <S> (0.1"/2.54mm pitch shown above). <A> The Wago 733 or 734 series might also be interesting to you. <S> Cable and capacitor can easily released by inserting a small screwdriver into the upper slot. <S> Note: <S> The connector shown in the picture is the one attached on the cable side. <S> If you plug out the cable, the capacitor comes out with the connector and loses contact to the PCB. <A> These would be physically larger than most, but if you go to an electronics supplier that carries hundreds of thousands of different capacitors, they will have both "snap in" and "screw terminal" capacitors. <A> Turned pin Sil sockets. <S> Cut to length required. <S> An alternative is the crystal sockets although their pin connections can be larger in diameter leading to intermittent contact with the capacitor leads. <S> Some force is required to connect up and on some capacitors the leads require trimming to equal length. <A> Similar to Robert Neill's suggestion, there are single-contact sockets similar to turned pins available in a variety of sizes to accommodate different pin diameters. <S> They can be had in both SMT and PTH varieties. <S> Some even allow the mating pin to pass straight through the socket and the board, although these are only available in through-hole for obvious reasons. <S> There are also a variety of spring-loaded tabs that are usually intended to make electrical contact with a housing for EMC purposes, but some of them are shaped such that they can accept a pin. <S> Finally, you haven't mentioned if this is meant to be easily removable or not, but for a permanent connection there are single wire-to-board IDC terminals, that are designed to be soldered into the board and have an insulated wire pressed into them. <S> The terminal displaces the insulation and makes contact with the wire inside. <A> Maybe you can try with one of this. <S> It's easy to solder on the board. <S> After that insert or remove a cap is really practical. <S> I think you should consider also if you can, get a radial cap, this connector suits best on that kind of capacitor. <S> Hope it helps! <S> https://www.digikey.in/product-detail/en/preci-dip/323-87-102-41-001101/323-87-102-41-001101-ND/4099478
| You might be able to locate a part that is designed for solid conductor wire and matches the lead diameter of your capacitor well enough to work. You may want a header with 3+ pins so as to be wide enough.
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What do these numbers represent in this schematic diagram? What are those 1 4 and 2 3 on both sides of the button? They look like having connected to same places at both ends, it looks like redundant cable to me. So, what are they representing? <Q> Tactile switches usually have four pins, with pin pairs already connected to each other. <S> Very often you'll see that the switch closes contact between the pins on a particular side. <S> Here is an excerpt from a common tactile switch datasheet (C&K PTS645) : <S> The schematic (top right corner) shows a slightly different arrangement where the pairs are [1,2] and [3,4]. <S> You can use a switch with either arrangement, just be sure you know <S> which pins are electrically "normally closed" and which are "normally open" by examining the datasheet before you solder. <A> I was induced to post this answer here because another question I started to answer was closed as a duplicate of this one. <S> There are two other things to consider regarding the two pins on each side connected together that are not mentioned in other answers and or comments. <S> The internal construction of these switches often has the lead frame made in two pieces each of which extends to the two connection pins on either side. <S> These lead frames extend internally to periphery contacts and a center contact that sit under a small click dome switch. <S> The disk, when clicked down, makes contact with the mentioned contacts closing the switch. <S> Here is a series of pictures showing construction of the typical TACT switch all the way down to the lead frame. <S> Another valuable feature of the pairs of pins connected together on the switch is that it allows for the layout of matrix of rows and columns of these switches on a one sided circuit board (or one layer section of a multi-layer board). <S> The following picture shows a layout that takes advantage of this to allow making the matrix without jumper wires. <A> Those are pin numbers. <S> It's useful to have them on the schematic to verify that the correct pins of the physical part will be connected to the correct places in the layout.
| Since pin numbering is typically done on alternating sides, but the contact is done between adjacent pins, you can have pairs [1,4] and [2,3] electrically connected.
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effects of hFE in emitter follower? There are multiple hFE options for BD139s, with the higher hFE being slightly more expensive. What advantages or disadvantages, if any, would the higher hFE parts have in emitter follower circuit? <Q> Advantages of higher hFE: <S> Higher input impedance to the circuit Lower output impedance <S> This way the emitter follower load effects to previous and next stages are reduced. <S> Disadvantages: <S> References: <S> Emitter Follower, Ruye Wang , Electronic Circuits <S> - I. pp. <S> 1-21, A.P.Godse, U.A.Bakshi <A> given the beta results from carriers injected by the emitter, attempting to impact carriers arriving from the base pin and yet missing thus continue on into the collector, with THINNER BASE regions producing HIGHER BETA, we have some interesting changes in behavior with high beta. <S> Higher beta means thinner base regions. <S> Thinner base regions mean FASTER TRANSISTORS. <A> For a given bias current, the relationship between input impedance and hFE is one of proportionality.
| Depending on the polarization of the transistor, high hFE can affect negatively the stability due to variations on temperature.
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Why would one mount components above the PCB? Inside of an automotive ECU I noticed that all three crystals are mounted about 1/4" above the PCB. What is the advantage of doing this versus mounting the case flush with the board? This seems like it could lead to vibration cracking the solder joints over time. Is there a thermal issue - crystals don't really dissipate much heat? Manufacturing constraint? Some of the capacitors are mounted in a similar way, although this looks more normal to me. <Q> The bent leads set the component height during wave solder. <S> It does four things: During wave soldering: retains the component in the board with spring tension <S> limits the solder wicking to just the pad <S> provides thermal relief to improve soldering <S> After soldering: provides extra lead length to reduce mechanical stress on the component. <S> The extra lead helps prevent cracking at the lead/component junction in high-vibration environments by spreading the flex over the lead, rather than concentrating flex at the component itself. <A> One reason for mounting a 'through-hole' quartz crystal away from the PCB could be to avoid solder shorts between the pads and the can. <S> Likewise the reason for raised mounting of capacitors could be to avoid damage due to heat, during soldering. <A> Look at all other TH components, they are all above the PCB. <S> This way solder flows through the hole. <S> Otherwise, it could only stick to the pad outside the hole on the bottom side, which would eventually lead to cracks and disconnection.
| Another reason could be to prevent heat damaging the glass seals as well as the crystal itself, during soldering.
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How to check SMD capacitor for being shorted, without disconnecting it from the electric board This YouTube video shows that you can check SMD capacitors for being shorted using buzzer mode, by touching the ground of the electric board with the negative terminal while touching each side of the SMD capacitors with the positive terminal, the one that has both of its sides making a buzz is identified as shorted. "SMD bad capacitor test / laptop - desktop computer & electronics troubleshooting" My problem is that I couldn't identify the "ground" on my electric board. It's a 2 circuits (AC & DC) board. P.S: The 2 golden squares near the microprocessor don't exist on my board, the rest is identical to mine. Also when I checked some of my SMD capacitors using resistance mode, I got the following values: 11K Ohms 12.4K Ohms 11.6M Ohms 11.3M Ohms 0.L Ohms (after moving upwards to reach 40MOhms, the maximum of my DMM) What ohm value should be used to consider an SMD capacitor good or bad? <Q> Without a schematic, you will need to engage in some sleuthing to find the ground plane. <S> A good place to start checking would be the negative (-) pins of the large electrolytic capacitors, which are clearly indicated on the plastic sleevings of the capacitors (the lighter-coloured section). <S> Using the plated-through-hole mounting pad may or may not work, because they are not always connected to ground. <S> You can check this by measuring between the electrolytic capacitor negative pin and the mounting pad. <S> You can still take direct continuity measurements across the capacitors. <S> A short is a short. <S> One final point to remember - these capacitors are almost always connected in parallel with other components in the circuit. <S> A short indicates that one or more of the devices on the circuit have failed short - not necessarily the capacitor. <S> The most common failure mechanism for ceramic capacitors to fail short is mechanical stress causing the ceramic layers to crack and internally short out. <S> Unless you dropped the assembly, I doubt the caps are bad. <S> If they were exposed to excessive electrical or thermal stress, you would see burning, discoloration, etc. <S> The measurements you took already - all with some resistance or beyond the DMM measurement capability - imply that none of the circuits connected to the capacitors you checked are shorted. <S> Your problem is likely elsewhere, or different in nature (i.e. not a short). <S> DO NOT PROBE THE POWER SUPPLY. <S> If you touch the wrong area you could get electrocuted. <S> Also be very careful to not short your probes together when measuring, or you may get a really big spark. <A> How to check SMD capacitor for being shorted, without disconnecting it from the electric board <S> And then, what if your buzzer tells you unambiguously that a certain SMD capacitor is short - these capacitors have no marking on them <S> so what do you do next? <S> I also note that the video you linked has a title with these words: - SMD bad capacitor test <S> I reckon the word "bad" implies it's a bad test. <A> Power the board and check the power rails. <S> If one of them is way too low then a short could be the cause. <S> If there is indeed a short, it could be a capacitor, or other component like a chip or transistor. <S> This short circuit will usually not be a very good conductor, instead it will have enough resistance to heat up. <S> This means you can find it by looking for a hot component. <S> If there are dangerous voltages on the board (which may happen on a washing machine board as JRE noted in the comments) then you'll have to use an IR thermometer or a FLIR camera if you can get one. <S> You can make sure there are no dangerous voltages by disconnecting the power supply and powering the board from a bench supply. <S> This will also allow you to measure current. <S> Then perhaps you can simply use your finger to look for a hot component.
| Where I would start is putting your DMM in DC voltage mode, plug the AC power in and check for DC on the logic board by carefully measuring voltages across the ceramic caps and e-caps on the logic board. Basically you can't because you are reliant on other components connected to it being non-zero (or even close to zero) impedance and this won't always be the case. You can also squirt some 95% isopropyl alcohol on the board, and look where it evaporates fastest.
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Cascade amplifiers with different corner frequency In these 2 cases I have the output of a non inversor circuit connected to the input of an inversor circuit. The only difference is the resistors. What I have noticed is that the cutoff frequency is different for both cases, but I dont know why, since both separate circuits have the same cutoff frequency and gain. <Q> What I have noticed is that the cutoff frequency is different for both cases, but I dont know why, since both separate circuits have the same cutoff frequency and gain. <S> They certainly don't have the same gain. <S> The non-inverting stage has a gain of: - $$1 + \dfrac{1400}{100} = <S> 15$$ <S> The inverting stage has a gain magnitude of: - $$\dfrac{2000}{100} = <S> 20$$ <S> Based on this, they are bound to have different cut-off frequencies. <S> Here's why (using a typical bode plot from an op-amp) <S> : - At a gain of 20, the 3 dB point will be 430 kHz <S> At a gain of 15, the 3 dB point will be 630 kHz <S> OK it's a different op-amp to the LM324 <S> but the principle is the same; the lower gain setup will have a cut-off frequency that is circa 63/43 higher than the higher gain setup. <A> Each amplifier should have some gain bandwidth product. <S> The first amplifier stage has a gain of 14 and the second stage has a gain of 20. <S> Therefore I would expect that the cutoff frequency for the second amplifier would be lower. <S> The value should be 14/20 <S> * the_cutoff_of_the_first_amplifier. <S> Real op-amps have parasitic capacitance's at their terminals (usually in the pF range). <S> The combination of the feedback resistors and that parasitic capacitance can create a low pass filter with some cutoff. <S> On the other hand, the ability of an op-amp output-stage to drive current usually decreases with frequency. <S> So if the feedback resistors are too small then it will also lead to a reduction in cutoff frequency. <S> It looks like that is the case here. <S> Even though the resistor ratios are similar in the two designs, the resistor values are different by a factor of 10 <S> so I would expect some change due to that. <A> The results you should get are: <S> Non inverting: fc = 66.7kHz (from 1MHz/15) <S> Inverting: fc = 47.6kHz (from 1MHz/21) <S> Cascade: fc <S> = 37kHz. <S> 1MHz is the GBW of the LM324. <S> Your two simulations show that you are getting fc = <S> 37kHz <S> (nominal) for both cascades, although the two results are slightly different due to the different loading on the op amps' outputs.
| If the resistors are large enough it can reduce your cutoff frequency or lead to instability in the feedback.
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How much current for a 5V 60 LED/m WS2812B 5m LED strip? I'm planning on controlling a 5m 60 LED/m WS2812B chip 5V strip with a Raspberry Pi Zero W. According to the website it draws 18 watts per meter (90 watts total). What current would this strip draw, and what would you recommend on powering both the Pi and the strip from one PSU? Any link to some decent, budget ones would be appreciated. <Q> <A> 90W at 5V comes out to 18A. <S> You can find several 5V 20A power supplies around. <S> P (power) = <S> V (volts) <S> *I (amps) <S> therefore I= <S> P/V <S> 90W/5V=18A <S> I have built several similar projects using WS2812b with Arduino and LPD8806 with Raspberry Pi. <S> I used this 5V 60A power supply and it has been running in my attic for years with no issues. <S> Its a bit large for what you need but at $20, you're not going to find much cheaper. <S> Unrelated to the power supply, its a lot easier to use the LPD8806 or some similar 4 pin cable with RPi because you can control the clock. <S> The three pin addressable strips need to have their data output at a very specific frequency and it can be troublesome to run long strips. <S> It can be done, especially with the short run you have, and the 3 pin version are usually cheaper. <S> P.S. <S> Hardware Recommendations is a good site for these kinds of questions. <A> I'm surprised that existing answers didn't explain how to calculate this. <S> If you don't know the relationship between voltage, current, resistance, and power, you'll forever be asking the same sorts of questions. <S> Ohm's Law and electric power (sometimes called Watt's Law) are fundamentals that might well be considered Electronics 101. <S> Ohm's Law: <S> \$I = <S> \frac{V}{R}\$ <S> Electrical Power: <S> \$P <S> = VI\$ <S> Where: \$I\$ is current measured in amperes (A), \$V\$ is voltage measured in volts (V), <S> \$R\$ <S> is resistance measured in ohms (Ω), and \$P\$ <S> is power measured in watts (W) <S> The two relate to each other, so you can rearrange the equations to find missing terms. <S> There are countless helpful "cheat sheets" like this one: <S> (Sometimes you will see E for 'electromotive force' instead of V for 'voltage'. <S> They're the same thing.) <S> Now that you have this information, you can rearrange to use what you know <S> (power = 90W, voltage = 5V) to find out what you want (current = ? <S> A). <S> If \$P = VI\$ , then <S> \$I = <S> \frac{P}{V}\$ . <S> 90/5 is 18. <S> That's how the current was determined from the other information you gave. <S> I also suggest that you read this excellent Q&A by Olin about selecting power supplies, which will explain more about how current is drawn by the load and may shine some light on why you want a power supply that can meet your requirement of 18A as a minimum.
| 90 Watts at 5 volts is 18 Amps, so you need a 5 volt power supply that can supply at least 20 Amps - perhaps 25 Amps to allow a safety margin.
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What are these four pin components labeled LB-4 on a motor control circuit from 1979? This board performs the speed control logic for a 3 HP shunt-wound DC motor driver in a CNC mill built in 1979. The spindle turned on intermittently last week and then stopped working, so I'm reverse engineering the circuit to debug it. I've ID'd everything except these three black components with the blue dots near the card edge connector. They each have four leads, and the only markings are "LB-4", a little chat bubble like logo, and +, - markings for two of the leads. There's no continuity between any of the leads. Can anyone help me identify these? I'm also not certain about the three big green bricks. They seem to just be power resistors but the markings on them say E M L2.2K 100(logo) 0.60for the large two and 1K 100 (logo) E.776for the small one, where (logo) is a little downward arrow in a circle. What has me skeptical about the 'power resistor' guess is that they measure more than 2.2 kohms. Edit: Back image: <Q> I agree, they look like bridge rectifiers (+1). <S> They probably go to individual transformer windings (the center pairs of connections). <S> You can check the inputs and outputs. <S> They are unlikely to be bad, but it's possible. <S> At least two rails appear to be zener shunt-regulated with the power dropping resistors (discolored PCB under them) and top-hat diodes. <S> Design looks older than 1979 but with some newer parts. <S> Weird hand-made Japanese-looking sort of, but no plated-through holes and no gold plating on the connector pins, very cruddy. <S> If I were to guess I'd say ca. <S> 1970 Taiwanese design. <S> Check the ESR or just replace them. <S> The green "bricks" are film capacitors. <A> The resistor in the bottom left looks burned out. <S> Try replacing that first. <S> Excellent job overlaying the mask you added to the latter image. <A> I agree on the electrolytics (test them or replace them) & that the green components are caps. <S> I might be missing something but the black components with blue dots only have two pins and there are only three of them <S> so they can't be bridge rectifier either together or separately. <S> I would think they are just diodes that are big enough to carry some current. <A> First of all why don't t you try to fix those corroded trace lines. <S> Use the soldering iron and some solder to plate them and measure continuity.. <S> Then measures those resistors where you see the overheating areas. <S> Use a pencil eraser to clean both sides of the edge connector. <S> Measure all active and passive components. <S> Retest it <A> First guess is that they are MOV or TVS devices for suppression of noise and spikes on long cables to the motor; once you figure out the schematic it may become obvious. <A> Black components are bridge rectifier.
| Those black components look like bridge rectifiers to me. The green components are capacitors 2.2 uF and 1uF 100v. If the electrolytic capacitors are that old, they're all suspect, particularly the two larger ones. All three of them seem to just be connected to the edge connector pins. You problem is the cooper traces, if you check continuity you will find some open.
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What is the purpose of this polygon arrays on PCB? I'm interested what is the purpose these polygon array on below picture. Thermal management? Thanks for the help! <Q> It's called "copper thieving". <S> The copper is added to areas with low density of copper to create a more even distribution across the board. <S> Even distribution helps ensure that the plating thickness is fairly even, and you don't get areas with little copper and very thick plating, and areas with heavy copper density and very thin or even missing plating. <S> This can also affect hole plating for critical areas such as press-fit connectors. <S> The added copper is typically added by the PCB manufacturer, in collaboration with the customer, obviously, since adding copper in otherwise blank areas may violate some other intent of the designer. <S> Altium comments on it. <S> Here is a useful blog entry with design tips for avoiding problems when the fabricator adds thieving to outer layers. <S> Edit: Typical note to fabricator: <A> Most likely thermal management, to average the thermal mass so that on the production line the board heats uniformly, to allow solder to melt at the same time, and to reduce issues of thermal expansion. <A> Those copper circles are on the inner layers and are called flow dots. <S> In large areas with no copper the epoxy doesn't flow and fill evenly and you end up with not quite flat areas of surfaces across the PCB.
| Copper thieving is generally done on the outside edges of the panel where the current density is the highest to promote more even plating of interior area of panel.
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Charge indicator with comparator - very sensitive to noise I need an indicator when the current from my charger is larger than 100mA. I tried the below circuit but this seems very sensitivity to noise. Probably I am getting some ripple in the mV range from the charger. Is there a more robust, less sensitive way to set this up? <Q> Yes, there is, add hysteresis to the comparator so the ripple is less likely to trigger the comparator from one state to the next. <S> Source: <S> https://www.electronics-tutorials.ws/opamp/op-amp-comparator.html <A> Add a tiny bit of hysteresis. <S> [ edit: And increase the R_sense to ONE OHM. <S> Plenty of headroom in that 29 volts .] <S> Insert 10 kΩ below that 100 kΩ resistor. <S> That will have 2.9 V across it. <S> Now decide how much you want the voltage across the 15 Ω to vary..... <S> 1mV, 5mV? <S> If 1mV, you need 100,000/15 <S> = 6,000:1 more variation, requiring 6 V variation at the top of the 10 kΩ you just added. <S> Now add a SECOND resistor from comparator output to that 10 kΩ. <A> You can try something along these lines: simulate this circuit – Schematic created using CircuitLab R7 and C1 form a low-pass filter with a cutoff frequency of 1.6Hz to filter out load spikes and SMPS noise. <S> R6 and R9 provide some positive feedback amounting to about 5mA hysteresis. <S> Increase R9 if you want to increase the hysteresis. <S> R10/C2 balance out any bias current, and can be eliminated if the bias current is small enough. <S> The op-amp needs to be a low Vos type (your whole signal is ~3mV), high voltage power supply capable and with an input common mode range that includes the positive supply rail. <S> There are some such op-amps available, but only a few. <A> You can also add low pass filtering to the input of the opamp as in this answer: Using low pass filter for a differential ended receiver system?
| I used a circuit like this when I had a lot of ripple on the inputs. If your prototyping this, make sure you have good grounds and Vcc to the amp with bypass capacitors
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How can I find the current draw of an SN74HC595 when OE is disabled? I am trying to daisy chain 4 SN74HC595's together in an "always on" battery powered system. When the system is "off" I am disabling a few components via their EN pin to lower current into the 10's of uA. I can't seem to determine how much current these SN74HC595 chips will draw when I set the !OE pin to disable the output. Any help would be wonderfully appreciated! I could always add a load switch for this, but I'd rather do it elegantly rather than add more components. Specifically, this chip: SN74HC595BRWNR . Here's a snapshot of (I think?) the pertinent information from its datasheet. I can't tell from this if the current would be up to 1000 nA (maka 1uA) by disabling via !OE, or if I just need to be sure to set all logic inputs to 0 to obtain that level of minimal current draw. Anyway, again any help is greatly appreciated. Thank you!! Edit: actually after reading through my post again, is it fair to assume that Icc is the current draw? and based on its Test Conditions , if Io (Iout?) is 0, then the device should only draw 8uA at normal temperatures. and I can be sure Iout is 0 by disabling the device with !OE?.. I think? <Q> Maximum is 8uA at room temperature, 80uA at up to 85°C. <S> But those are very loose specifications. <S> The typical Icc is about 27nA with a 5V supply at 25°C. <S> As a rule of thumb, it will approximately double for every 10°C above room temperature. <S> That's assuming all the inputs are near the rails and static. <S> If inputs are not near the rails the current can go up <S> considerably due to conduction of the input transistors, and similarly if they are switching. <S> Cpd is specified as 400pF (typical) and there is an explanation of how to use that here . <S> If the input strays too far from the rails, then shoot-through current can greatly increase the tiny leakage current broadly similar to this effect: (from this document): <A> The input current or I_i won't change, it will remain at 1000nA. <S> The output current Ioz will go to 5uA Max when the Output is disabled. <S> The output enable won't change the power consumption of the part much (the power consumed by quiescent current and inputs), but it will change the power delivered to the loads. <A>
| Yes, it is shown on the datasheet as Icc, it is either 8uA, 160uA, or 80uA, depending on your temperature range.
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Disabling IC by removing VCC I've made level shifting circuit to shift pulsed ~10 kHz 5 V signal to 15 V by using gate driver and level shifter IC: TC4427 and CD4504 . There is a condition that the circuit output should be 0 when "enable" bit is zero. I have done that by placing an optocoupler to disconnect main 15 V supply from the ICs, but only after assembling and testing I have found out that ICs have these ESD protection diodes from input to VCC which leads to condition that input has to be always < VCC + 0.3 V. Now when enable is 0, the +15V VCC floats. Now when I connect input i get some voltage at the VCC rail and outputs of IC:s because the voltage is "leaking" through the esd protection diodes. Is there any workaround that could be done with minimal changes to the design so that the output will be as close to zero as possible (while there is still pulsed 5V input)? Of course by still keeping the lifetime expectancy of ICs as high as possible. Condition where enable is 0 while there is some input voltage is quite rare and not normal operating condition. <Q> If your intention is to make sure that the voltage rails are nearly zero when enable is low, you'd have to drain the voltage store at these capacitive nodes, either passively or actively. <S> Passive Draining <S> Draining it passively is accomplished by connecting a resistor in parallel to the voltage rail of interest. <S> A low resistor will lead to a continuous power loss in the system during normal operation, but will make sure to drain the node very rapidely once you disable the optocopler. <S> For a highly resistive resistor, you would have to wait for some time until the supply voltage rails would go down to an acceptable level. <S> Since you are using \$15V\$ and \$5V\$ in your system, bleeder resistors would have to be connected to both rails. <S> This approach is the easiest one and require small modifications, however, it has the drawback of adding additional power loss. <S> Active Draining <S> Implementing an active draining circuit, reduces the continuous power loss but it comes with the burden of adding more complexity. <S> The idea here is that the nodes are only drained once you disable the optocopler. <S> For the the \$15V\$ rail it would look like something like: <S> Here, <S> the \$V(out)\$ which represents your \$+15V\$ , is drained through a \$100\Omega\$ resistor and a pair of MOSFETs. <S> The draining "speed" can be configured through the active bleeder resistor \$R_3\$ . <A> I once inherited a design that had this issue. <S> I solved it by clamping the switched voltage to ground when not enabled. <S> You could do this with a second opto-MOSFET, but connect its diode anode to +5V <S> so the logic is reversed. <S> You probably want a small value resistor in series with the second opto, maybe 10-30 ohms, so you don't stress your power source during the transitions. <S> Now you will stress the ESD diodes more, since they will be driving a lower impedance. <S> You will need to find a value that still allows the circuit to work without stressing the input diodes too much, maybe a few hundred ohms. <A> Pulling the Vcc pin to GND could be a solution: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The totem-pole drive stage consisting of Q2 and Q3 is a buffer with a very low output impedance. <S> I've marked Q2 and Q3 as general-purpose BJTs but they can be MOSFETs. <S> Rb prevents the input to be floating, which can be 10k or so. <S> When the EN signal is high, a non-zero current through the opto's transistor will trigger the totem pole <S> so the Vcc will be +15V. <S> When the en signal is low, the totem pole's input will be pulled down via Rb and so the Vcc pin will be effectively shorted to GND. <S> IMPORTANT NOTE <S> The circuit above shorts the IC's Vcc pin to GND <S> but it's still dangerous if the inputs are non-zero when the Vcc is grounded. <S> Probably the inputs have clamping diodes, so when you apply non-zero inputs those diodes may fry out -- this is also dangerous for the pins outputting those signals.
| A small resistor in series with the inputs can solve that.
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Looking for tools or technique to design circuits on PCB prototype boards Once I test my prototypes on a breadboard, I have a hard time to migrate the breadboard circuit to a double-sided PCB prototype board like this: I am looking for a design tool (similar to thinkercad circuits) that helps me to place (draw) my components on a double-sided PCB prototype board and add connector wires. The drawing helps me to print and implement my circuit easier on such PCB prototype boards. I know there are many free online tools to draw and design PCB board. Is there a similar drawing tool for double-sided PCB prototype boards as well? <Q> I do these types of layout using Eagle. <S> But any PCB layout software should work. <S> I set the grid size to 0.1” to match the pitch of the protoboard through holes. <S> At each point the grid lines cross is where a through hole would be on the protoboard. <S> You can then align your components’ pins to match those points. <S> You then run traces making sure to stick to 90 degree bends. <S> If your protoboard is two layer, you can use both TOP and BOTTOM layers in Eagle, which makes it clear which traces go where. <S> For single-sided protoboards <S> I use the TOP layer to draw jumper wires. <S> Again, makes it very clear. <S> This also works for strip/veroboard <S> but you’re obviously limited to horizontal traces. <S> I layout the components and then draw in the horizontal traces where they need to go. <S> This helps show where you need to break the tracks. <S> It can be a little bit of a faff to setup, but works great once you get going. <S> Using Eagle or similar means you can take advantage of the rats nest tool to make sure you’re not missing any connections. <S> Update Pictures always make things better, so here we go: <S> This image shows the layout. <S> You can see the grid, with the intersections being the through holes. <S> All the parts are aligned to this. <S> Blue traces are the Bottom layer. <S> Red traces are the jumpers on the Top layer. <S> In the past I have tried to minimise jumpers etc, but you could spend hours and hours trying to get everything on one layer. <S> Plenty of jumpers give you more places to use as test points! <S> What is a good idea is to space things out as much as you can. <S> It makes things much easier when soldering, and gives some wiggle room for on the fly adjustments - or for bodging down the line! <S> This second image is the final result. <S> I don't have a picture of the bottom, but it more or less matches the board layout in the first image. <S> The LED arrangement is slightly different, as I adjusted this on the fly. <A> <A> To add up on the previous comments, I'll share this resource that has saved me lots of time when designing PCBs: SnapEDA. <S> It’s like Google for components where you can find free symbols, footprints and 3D models, and export them to your PCB design software. bit.ly/EngineerSnapEDA
| There is a software called VeeCAD that allows you to design circuit on protoboard.
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What's the difference between these plugs? I'm not an electrical engineer, but I play one on TV. Can you help me be more realistic? Our facility has three items that run on three phase 208 and plug in at a temporary building using CS (California Style) connectors. We are in the process of speccing a new building. To give the users maximum flexibility in where to put their stuff, I want to scatter five of these receptacles around the new building. I just realized that the three equipments do not have the same CS connectors, oops. Two of them have a CS8365, and the other one has a CS6365. UH-OH! We have the competence to change out the connectors to match each other, but before I recommend that, I want to better understand the differences to make sure I don't create a safety problem. Here are some data on both of these connectors: https://www.hubbell.com/hubbell/en/Products/Electrical-Electronic/Wiring-Devices/Locking-Devices/50A-Twist-Lock/CS8365/p/1634245 https://www.hubbell.com/hubbell/en/Products/Electrical-Electronic/Wiring-Devices/Locking-Devices/50A-Twist-Lock/CS6365/p/1634242 I can see in the description that one of them claims to be for 250 "delta" power while the other claims to be for 125/250 power. That's all good, and I think I understand it, until the next sentence--they both claim to be "3-Pole 4-Wire Grounding". Now that does not make sense....the 125/250 one seems to be claiming to have a neutral, so how does that add up with still only four conductors? Now, my bottom line questions: Would it be safe to remove a CS6365 from a Harrington Hoist VFD246 series crane and replace it with a CS8365? Alternatively, would it be safe to remove two CS8365 from a Toyota forklift battery charger and a Grainger Speedaire 13G709 air compressor and replace them with CS6365? As far as I can tell, all three of these devices are straight 208 three phase, with no need for a neutral or 115 power, but I give the model numbers to let you all educate me if needed. Thank you for your time! <Q> Be careful -- receptacle type usually reflects voltage and phasing First, the "250V" rating relates to socket insulation limits. <S> It has nothing to do with expected/conventional voltage on them. <S> (You can't exceed it obviously). <S> The CS8365 is for 3-phase delta 208V, its NEMA equivalent is NEMA 15. <S> That is plainly stated in their catalog <S> (if you know what to look for). <S> I leave it to you to look up the bolded phrases. <S> By the way... all 208V installations are actually "wye" (120V line to neutral), and can be used either wye or delta. <S> It is quite common for 120/240V split-phase appliances to be used in 120/208V 3-phase territory. <S> They simply attach 2 phases and 1 neutral, so the 120V sections work normally, and the 240V sections get sqrt(0.75) less voltage than they are used to. <S> Resistive loads can live with this, they simply work at 3/4 power. <S> Motor loads, you'll have to check the nameplate to see what they can accept. <S> However, the important part is these 120/208V 2-of-3-phase appliances are wired up NEMA 14 - same as 120/240V split-phase . <S> But why on earth do some have a different connector? <S> The worrisome thing is that some of your machines have the different connector. <S> Usually, that means this machine requires a different voltage. <S> It's like if you go to plug in your RV <S> and it has a TT30 type connector, that always means it is a 120V/30A RV. <S> An RV with a NEMA 14-50 means it wants 120/240V split-phase. <S> So on these machines that have CS6365 -- look very, very closely at their labeling, nameplates and the like. <S> In fact, post them here . <S> You need to assume the CS6365 is correct and the machine is 120/240V split-phase (or 120/208V) requiring neutral, until clear evidence on the unit proves otherwise. <S> Don't fall to wishful thinking. <S> I suspect that converting these to 208V delta would be a mistake. <S> You must assume they have sections which require 120V line to neutral, won't like 208V there , expect 240V line to line, and may be able to tolerate 208V line to line. <S> If that's the case, then you need to provision sockets with this H-H-N-G wiring. <A> Change the CS6365 from a Harrington Hoist VFD246 series crane and replace it with a CS8365. <S> Reasoning behind that is the CS8365 will only be expected to work with 250V lines, so you are restricting the equipment a little bit, but it's perfectly safe. <S> Changing the CS8365 to CS6365 opens the possibility that the equipment could be plugged into a supply it's not rated for, potentially causing issues. <A> I’m going to accept Harper’s answer because it is the “correct” answer for most future readers. <S> I’m also writing my own answer for the small portion of future readers whose situation is the same as my own. <S> The difference is Harper’s answer is correct in the case where previous electrical work was done correctly, but that is not the case for me. <S> The electrician that installed the CS6365 several years ago made either a careless mistake or a poor choice. <S> The Harrington crane I verified 100% in its schematic that it requires three phase power with no neutral, which is exactly what the CS6365 is putting out. <S> This is actually a potentially dangerous situation, so we are going to replace the 6365 receptacle and plug with 8365. <S> Dmitri thanks to you also, <S> and I am going follow your recommendation, but for different reasons than you suggested.
| The CS6365 is for split-phase 120/240V power, and its NEMA equivalent is NEMA 14.
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Simple circuit to protect 2.7V supercapacitor from exploding I need advice and help. I am newbie and kind of interested in building things. Please bear with me. Situation: I want to power my wireless modem/router using a solar panel so when there is no electricity I can still watch videos and surf internet as there are frequent brownouts in my area. Buying a lead battery is expensive and not nature friendly, secondly I want to explore things or building things. So I started reading about solar panels, capacitors, diodes and transistors, etc. I bought soldering (irons,leads) and digital multi-tester. I bought a super-capacitor rated as 100F 2.7V I read that when the capacitor is charged or exposed to more than the values it is design it will explode. I tested the solar panel using multi-tester and it shows 7.2V. QUESTION : If I connect my solar panel (7V) supply to my super-capacitor which is only 2.7V then my super-capacitor will explode, right? Is there a simple way (circuitry) you can suggest to avoid my capacitor from exploding? The input is 7V and I read that capacitor explode when they are exposed to a higher voltage source. I cannot buy resistors in my area so I am salvaging some resistors from older radios and TVs that are still working. I can order online but it will take weeks to come. How to make sure that the input will never be above 2.7 volts since super-capacitor is rated only 2.7V 100F. [ EDITS ] Sorry I like to correct the capacitor value: 2.7V 10F I would like to add more about my situation. My location is always sunny and in-case there is a power cut and it's daytime I want to power my router/modem using the solar panel that we have. The solar panel supply is not fixed depends on the sunlight. It sometimes produces 5V up to 9V. I read that capacitors function like a temporary battery. The goal is to provide a steady flow of 9V for the router. In case the solar power source drops then the super-capacitor will still have the power to supply the 9V. Please note I have 6 super-capacitors (2.7V, 10F.) I bought them online. See the image below. Bear with me. Thank you. <Q> Few underlying issues with your idea, Lead acid is actually one of the few battery technologies that can be almost completely recycled. <S> (That's why most scrap metal places separate them - just melt down the lead and purify the acid and you have a new battery.) <S> Super capacitors hold usually less than 1% of the same charge of an equivalent size battery, assuming you could use a boost converter to drain every last bit of charge in your capacitor. <S> It only holds 0.1 Watt hours of energy, so say your router draws 5W, it would only store about 1 minute worth of power. <S> On to the actual questions you would be after a regulator, e.g. an LM317 with the correct resistors on the output to keep the output set to something like 2.65V. I am not clear what modem/routers have an input voltage as low as around 2.7V, and suspect there is another issue. <S> Edit: <S> Based on your update of having 6 of the capacitors with a lower capacity, The bast configuration you could have would be 4 of them in series, with some balancing resistors in parallel with each capacitor, this would make your capacitor breakdown voltage 10.8V <S> , you would still need a regulator or similar to make sure this is never exceeded, but would work better to your end goal, <S> So we have a 10.8V capacitor arrangement, assuming your solar panel + regulator can keep it held at say 10.6V, and we take a guess that the modem / router will not cut out until about 6.5V (assuming internal 5V rail) <S> your left with about 21 Joules of energy, so about 35 seconds of power. <S> It will get you something usable, but not much power to hold you over. <S> For a lead acid based approach you would use a panel that could charge the battery to say 13.2V, and use a regulator to drop that to the routers 9V, assuming a scooter battery which is 7Ah, and about 80 Watt hours, a 5 Watt load could be powered for at least half a day. <A> I am sorry, but a super capacitor will not be able to power your wireless modem for very long. <S> A few minutes perhaps. <S> To answer your question, you could use a buck regulator to step down from 7V to 2.5V. <S> But then you need a boot regulator to step up that voltage again to power your modem. <S> If you really want to use a capacitor for this purpose, then you should buy a capacitor with higher voltage rating. <A> Event your voltage is 7 voltage is 7 voltage <S> but it have limited current flow there for v(t) = <S> (1/C) i(t) <S> dt so your voltage to increase from zero overtime. <S> To limit voltage to below 2.7 voltage you need voltage regulator (recommended switching type) and set it below 2.7V or if your capacitor bug enough <S> you can do voltage cut off with voltage comparator and mosfet.
| With older TVs, you may be able to find a zener diode for something easier to find, a 2.7V one "may" be suitable, but it will need a resistor between it and the capacitor to limit the current, and will get quite hot.
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Source voltage and voltage across inductor waveform I did an RL circuit experiment. It is a series circuit with a potentiometer, inductor, and a DC source. Here is the resulting waveform at 2 different resistances (ch 1 is source voltage and ch 2 is voltage across the inductor.) Why does the voltage across inductor on the first picture have 'jumps' and why does it curve like that in the second picture? I am also confused as to why the source voltage is affected on the second picture. <Q> Regarding the RL circuit: Notice the spikes in the inductor voltage are aligned with the rise/fall times of the source voltage. <S> With inductor voltage equal to L(di/dt), we can conclude there is a substantial di/dt from the rapid dv/dt of the input source voltage during the brief rise/fall intervals, causing these briefs periods of substantial positive/negative voltage across the inductor. <S> Also note that the voltage remains slightly positive after the spike for positive input voltage (likewise, slightly negative after the spike for negative input voltage). <S> Regarding the RC circuit: This unfortunately appears to defy our expectation, given what we know of the circuit from your question. <S> We expect an exponential charge of the capacitor voltage of the form: vc(t) = <S> Vsource*(1-exp(t/RC)) <S> which occurs after each inversion of the source voltage polarity (see image below). <S> This time to charge ought to increase when you increase the resistance of the pot. <S> You may consider making a second attempt at this test and check for any errors. <S> Best of luck! <A> It might help to think of a voltage divider plus the nature of a coil charging and discharging to help understand this. <S> In the first image the resistor is a very high value. <S> This means that there is a large voltage drop across it <S> so V2 is very low when V1 is high. <S> The spikes you see on the edges of the pulse are due to the small amount of magnetic field collapsing and generating a spike of current on both rising and falling edges as the inductor stores current on BOTH rising and falling edges (it just flows the opposite direction as the field collapses, hence the negative and positive voltage increases). <S> The large voltage drop is also why V2 looks almost zero: it is effectively at ground during steady state when the inductor is a short when charged with magnetic field (whereas capictors act as an open circuit when charged). <S> In the second image, the resistance is lower so more current is entering the inductor. <S> When V1 switches, you are seeing the current flow out of the inductor and decay back down to Vhigh and Vlow just like in the first picture, but the current is so much higher that the exponential decay curve lasts longer and makes that nice curve. <S> Since the resistor is so low, V2 is close V1 when high, however the current across R1 is much higher with the boost from the inductor, and that is read by BOTH voltage probes. <S> It is likely that the voltage scales are different in the first image which is why you do not see spikes on the V1 probe. <A> You can intuitively understand this inductive circuit if you think of the inductor as of a "recheargable current source". <S> Its operation is considered in five steps in the picture below which is a snapshot of a Flash movie (I cannot put a link to it because StackExchange policy does not allow it).
| We can credit this to the ESR of the inductor, which has a voltage drop from the current in the circuit.
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Two types of LEDs, serial - parallel combination I am trying to design small DIY grow light for Chilli Peppers. I found ideal to use a combination of Samsung LM301H( https://www.samsung.com/led/lighting/mid-power-leds/3030-leds/lm301h/ ) and LM351H V2 ( https://www.samsung.com/led/lighting/high-power-leds/3535-leds/lh351h-deep-red/ ). Now I am not sure what is the best serial-parallel combination of these two types of LEDs together. I want to use a Constant Current source. For LM301H(Uf=2,75) I want to use 65 mA and 350 mA for LM351H(Uf=2.2). Is it a great idea to put 6 LM301H together in parallel, series to 1 LM351H? Then I will have a block that will have a voltage drop of 5V and driven with a constant current of 350 mA. Is this correct design, or am I missing something? Thanks! :) <Q> I found ideal to use a combination of Samsung LM301H <S> ( https://www.samsung.com/led/lighting/mid-power-leds/3030-leds/lm301h/ ) and LM351H V2 ( https://www.samsung.com/led/lighting/high-power-leds/3535-leds/lh351h-deep-red/ ). <S> The efficiency of the red diodes is no better than the whites. <S> You could get similar results just using white LEDs of the color temperature you want, at least assuming those exist. <S> Is it a great idea to put 6 LM301H together in parallel, series to 1 LM351H? <S> You'd have to look at the spec sheet to get an idea of the variation in forward voltage, but in general, no I wouldn't prefer that. <S> Typical grow lights put 10-30 lights in series, and then build parallel strings from that. <S> This way the current is more reasonable. <S> Unless you have a very small grow light, usually you want to be working at somewhat higher voltages to reduce wiring losses. <A> This looks like a reasonable shared current balance for up to 350 mA if you have adequate heat transfer in the design for each part with shared temperature for each parallel LED ( most critical without a series R equivalent for each White LED equal to its ESR of 2 Ohms. ) <S> However, this is a fixed design of < 2 watts. <S> Always start from the system specs design for radiated power at each wavelength in PPF or PE <S> then design according to the range of system application requirements rather than a potential array and see what it can do or if it works. <S> Also learn about thermal runaway and thermal balancing and ESR of each part = <S> incremental R= <S> V/I and tolerances. <S> Samsung will have lower tolerances than no-name brands. <S> https://www.samsung.com/led/about-us/news-events/news/news-detail-49/ <A> I would be concerned with manufacturing variances in the devices causing 1 or 2 of the 6 LEDs to flow more current than the others, heating up more, flowing more current still, and creating a vicious cycle. <S> That could be at least contained by putting a resistor in series with each LED.
| Instead I would try to have a few in series so that the variation in individual units averages out and the current is more uniform. Plus the voltage is probably too low to make sense for something like a growlight where you want many watts of power.
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Fade in and out 12v LED strip Is there any easy way to continuously, slowly fade (dim) in and out a couple (5 at most) 12v monochrome LEDs (5050)? A wild guess would involve a large capacitor (for the fades to be slow) and a MOSFET, but I have no idea where to start designing such a circuit. Since my project won't need any smarter logic, I wanted to try to avoid using a µC for it. <Q> Any circuit that could generate PWM of varying duty cycles would suffice. <S> You don't need to drive the LEDs ; your circuit only needs to source a few milliamps. <S> You will send its output into the input of any common readily available $9 amplifier , and that will drive the lights. <S> You want the PWM to be up in at least the multiple hundreds of Hz to avoid visible flicker, and it's OK to change the frequency. <S> The amplifier will simply follow the input. <S> Also keep in mind LED strip tech is all wired "positive ground". <A> A bipolar transistor seems best for this option. <S> Examples . <A> If you allow me to blow my own trumpet, I made something similar: <S> This circuit generates a sine triangle wave around a Vbias voltage. <S> The frequency is fixed, but you can vary it by changing the values of the different components. <S> The circuit itself has been ripped from "The Art of Electronics" so if you have a copy, you can find it explained there, in the OpAmps section (don't have it in front of me, sorry, can't reference the page). <S> You feed the Vfade voltage to a transistor <S> And job's a good'n. <S> You can find more info at my repo ( shameless plug ). <S> There's an spreadsheet in the support folder to calculate the passive values. <S> In my case, as I was driving a blue LED with a 3V battery, I wanted to keep the voltage from 3V (or whatever the battery was outputting) to just around 2V (so the LED is not on), otherwise the LEDs would be off for most of the time, so instead of a 1.5 Vpp sine <S> wave over a 1.5V bias voltage, I ended up generating a .5Vpp <S> sine <S> wave over a 2.5V vias voltage. <S> EDIT: I originally said that it generated a sine wave, but that's not true. <S> U2 generates a square wave, which is fed to a integrator, so you get a triangle wave. <S> EDIT2: <S> I used this circuit because it has the side-effect that if you rise the bias voltage, you get uneven positive and negative cycles <S> (the triangle wave has different ramp up and down, so you get a nice breathe-like effect). <S> I will modify the excel on my repo to better represent what's going on and what effect tweaking each component has.
| A PWM as suggested in the previous comment is possible, you can use a 555 timer or an MCU.
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Connecting 10AWG wires to PCB I have 4 wires 10AWG stranded, that I need to connect to a PCB that I'm designing. They will be in a high vibration environment, so direct through hole soldering may not be the best solution. I've thought about ferrules, but those are usually used with screw terminals, which are also not ideal for high vibration env.I've also thought about crimping the ferrule onto the wire and then soldering the ferrules to the PCB, but I can't find any information on that idea, good or bad. What other options are there? <Q> I would consider adding holes in your PCB specifically for wire retention, like in the image below. <S> You can use superglue or a potting compound to adhere the wire to the PCB to further prevent the solder joint from vibrating and failing. <A> Ferrules on each conductor, then into spring cage connectors on the PC board. <S> I do a lot of work on very large earth-moving machinery and the field techs tell me that connections done that don't ever fail. <S> I've since started using those spring-cage terminals on some of my products and have had zero failures since going to them. <S> The following link shows Phoenix rail-mount terminal blocks but will allow you to find PC mount variants. <S> https://www.phoenixcontact.com/online/portal/ca?1dmy&urile=wcm:path:/caen/web/main/products/subcategory_pages/Spring-cage_connection_P-15-03/c252a030-3764-4828-a692-d0ba1dd45a27 <S> Sorry for not enclosing the link in the proper hyperlink format <S> but I can't find how to do that on my new phone. <A> Can't you just use a Faston? <S> They'll do AWG 10. <A> I used to work in a lab that tested high speed cables. <S> These cables were much thinner than what you're dealing with, but you may be able to use the same technique for strain relief. <S> We would zip-tie the cable to the PCB an inch or two away from where it needed to terminate and then solder the wire to surface-mount pads on the PCB. <S> The zip-tie passed through holes in the PCB and wrapped around the cable <S> Here is a blurry picture of how it was done: <A> This really depends on the vibration strength. <A> depending on vibration strength, you might probably want to consider pcb mounted circular connector. <S> might be expensive depending on size and number <A> A terminal block is a pretty normal way to get big wires onto boards. <S> They come in a dizzying array of shapes, sizes, styles, etc. <S> If you don't like the spring clamp or screw terminal types, there are also ring terminals of a variety of types, etc. <S> Whether you choose something like this depends on whether or not the wire ever needs to be disconnected again (service, replace parts, user connection point, etc). <A> How high is high vibration? <S> My default answer would be to use an appropriate connector. <S> There are many good, high current options. <S> Spring loaded contacts from Wago, Phoenix Contact, and others are available in this wire size. <S> Watch the current ratings carefully. <S> Many of these have surprisingly low current ratings for larger wire sizes. <S> You might also consider a locking Faston - these have a locking tab that goes onto the hole on the tab. <S> You could even reflow it after it's riveted. <S> It all depends on your application. <S> 10AWG implies pretty high current. <S> How are you getting the current capacity in a board as small as you imply?
| You could also consider riveting a ring terminal to the board. Spring loaded terminals are one option worth considering. My general strategy for dealing with vibrating wires is to provide strain relief.
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Is the maximum rated current of the cable the same for any voltage? Cat5e cable has a rated current of 0.577A ( wiki page ). I am wondering if this is voltage dependent? I am almost certain it is not, but if it is, something might blow up. Say I need to get 2A at 5V across the cable. I boost it to 24V / 0.5A, send it across the wire, and step it down to back to 5V and a bit under 2A. Would that work? If the cable is proper Cat5e could I do this trick with up to 100V as long as I don't exceed 0.577A? I am worried if the cable heats up at near maximum rated current, for instance, the safe voltage could drop as the insulation softens up or something. I will certainly test this, but I wonder if there is some rule that I don't know about. <Q> The heating you get in a wire is only dependent on the current you pass through it. <S> The voltage makes no difference. <S> But you need to check the voltage rating of the cable you are using. <S> For Ethernet cables, it can be very low. <S> A random cable <S> I just looked at is only marked 30V (which was rather surprising since power-over-Ethernet is nominally 48V). <S> If you are pushing a cable to its limits, then where it is run can also make a difference. <S> A cable enclosed in insulation can get rather hot, while one in free air cools a lot better. <S> Don't exceed the temperature rating of the cable. <S> Again, the random cable I have to hand is only rated for 60°C. <A> You shouldn't mix up rated and maximum , the are not the same. <S> The wiki page says "Maximum current per conductor". <S> Moreover, I don't believe the wiki covers all cables and also doubt the reliability of that number (1) . <S> In all cases you'd better check the datasheet for the rated and maximum current. <S> Stepping up the voltage to reduce the current through the cable may help reduce voltage drop and losses in the cable (but note stepping up and down also gives power losses). <S> (1) <S> The wiki refers to a AWG table with a (in my eyes) <S> a very conservative current rating. <S> They quite likely used the Maximum amps for power transmission for AWG 24. <S> Considering 10 mm of a 24 AWG cable carrying 0.5 A, using the resistance per unit length of 84 Ω/km, this piece of 10 mm has a resistance of 0.84 mΩ. <S> Assuming all 8 cables carrying each 0.5 A, gives 8*(1 A)^2 <S> *0.84 <S> m&Omega = <S> 1.7 mW per piece of 10 mm CAT5e cable. <S> I wouldn't worry about it... <A> Most wiring regulations specify the max temperature for cables. <S> So, check the regulations for your country as getting this wrong can cause a fire.
| This depends on the environment where the cable is, such as open air or in a wall, in a conduit, in a conduit that is buried in insulation etc etc When the rated current would be 0.5A and the rated voltage would be 125V, then, indeed the voltage doesn't matter as long it doesn't exceed this rated voltage.
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PCB Layout of a circuit So, I drew a two sided PCB of a circuit represented in the picture below. I used the program DipTrace to draw it. The first picture represents the top side. And the next one represents the bottom side. I wanted to ask is it preferred to change anything and does the space at the bottom right corner that is left there cause any problem? This picture represents the schematics: <Q> Im no expert on the subject <S> but I've done some simple PCBs <S> so here's my two cents 1) <S> Don't route 90 degree angles <S> Why is there such a strong preference for 45 degree angles in PCB routing? <S> 2) Make sure your silkscreens are going to be visible when the components are populated. <S> Once R5 is populated, R2's silkscreen is covered. <S> 3) Also make sure there is enough useful information on the silkscreen as well. <S> You'll probably be able to make out the line on the diode part outline but may be a good idea to add a "+" to the positive side of the diode, offset from the bottom of it. <S> This may be opinionated since the square pad does the same <S> but I think it's a good practice in case <S> it were an SMD diode. <S> Same with the connectors. <S> Add a positive indicator <S> so you know which pin is positive. <S> Sure you can flip the board, but say you end up mounting this to something and cant access the bottom, it wont be so easy to tell. <S> Switches with more than 2 pins, add a silkscreen showing where pin 1 is. <S> And then on the second row show where pin 4 is since it could be either on the right or left. <S> 4) Change "SW2" to "SW1". <S> When I see "SW2" I will assume that's the second switch on the board. <S> Also U1.B, again makes me think there should be a U1.A. 5) <S> Make the bottom layer a ground plane. <S> What are the advantages of having two ground pours? <S> 6) <S> I don't know what U1.B is, but maybe read the datasheet again. <S> I suspect it might suggest adding a bypass cap across the power input http://www.learningaboutelectronics.com/Articles/What-is-a-bypass-capacitor.html <S> EDIT: <S> Also, why use a DPDT switch? <S> Overkill for your application <A> This is always better to use 45° angle and not 90°. <S> I don't now what is U1 but often, IC need a decoupling capacitor. <S> Also, this is maybe what you want, but you don't have mounting holes. <S> No problem for the space. <S> After that, without schema, I can't say nothing more. <A> You've got a label off the edge of the board (s2) and labels covering pads. <S> Press f10 and you can drag these around to fix that. <S> They will most likely be removed by the fab if you leave them that way. <S> You've got a couple of right angle traces for no reason. <S> Why not make them shorter? <S> You should add a date to the board so you can keep track of which board is which once you have several revisions. <S> Double check that the through holes are sufficient diameter (looking at D5 and 6 specifically).
| The empty spaces on the PCB will not cause a problem.
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I want to reprograme my wirelss mouse so that it shall generate a nudge if it is on and inactive for 30s, is this possible? My problem is that my work computer locks up if there is no activity for only a few minutes. One simple solution is to buy a wireless mouse (or even a wired mouse) that automatically sends a small nudge signal the PC every 30s if the mouse has not been moved for that much time. However, I can't find mouse with such capability built in. I don't know if I can reverse engineer my mouse and reprogram it to do this. Please let me know if a mouse of this nature (not an expensive USB dongle) exists with this function. And, if it is possible to reverse engineer the code and reprogram the mouse. I have an ET X-08 gaming mouse. Thanks. EDIT:I should have used the term "locked" instead of "locks up". Basically the computer goes back to the log-in screen and requires me the enter the very long password again. <Q> That's madness. <S> If you're using Windows then try this: set wsc = <S> CreateObject("WScript. <S> Shell")Do WScript. <S> Sleep(5 <S> *60 <S> *1000) ' <S> Every 5 minutes ... <S> wsc. <S> SendKeys("{F13}") '... press the F13 key. <S> Loop <S> Save it in <S> nosleep.vbs <S> on your Desktop and run it when required. <S> F13 is chosen so that it doesn't interfere with anything else. <S> To check if it's running: <S> Open Task Manager (Ctrl-Shift-Esc) and go to Details tab. <S> If a VBScript or JScript is running, the process wscript.exe or cscript.exe would appear in the list. <S> - Right-click on the column header and enable "Command Line". <S> This should tell you which script file is being executed. <A> Seems the type of device / program your looking for is called a "mouse jiggler", intended to shift the cursor some small amount in known time steps, From a quick glace some only move the mouse about one pixel to prevent how much it effects the end user, as there is no easy way for that mouse to know to move. <S> As for modifying an actual mouse, they are very tolerant to most external distractions so you would likely have the most luck with a small bodge circuit inside. <S> Most wireless mice dim their LED a few seconds after motion or clicking has stopped. <S> When it dims you have something that times out 30 seconds, then close say the middle mouse button contacts with a mosfet, the mouse lights up, the circuit waits another 30 seconds and so on and so on. <A> You should first try the script proposed by Transistor. <S> However if you do not have permissions to change screensaver settings, perhaps you can't execute the script either. <S> Windows will allow multiple mice <S> and they all control the cursor simultaneously, so this will work. <S> Here are some answers <S> but you can simply google "usb hid microcontroller", you'll find example for pretty much any microcontroller that has a USB port since making a HID device is a common code example. <A> Simpler to change the lockout time, no? <S> If you can't change it yourself, most likely it is set via company policy and you might be violating the policy if you circumvent it.
| Simplest solution I can think about would be a microcontroller development module acting as an USB HID mouse, and sending mouse commands every once in a while.
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How to add an emergency stop switch in a circuit? I am into a project, where I have made a a bot, which has a 6 degree of freedom robotic arm with four wheels. I need to add a kill switch that is connected to a relay, which stops all the 10 motors. I don't know the circuit connection between the switch and the relay which then goes to the motors. It is powered by a 24V battery. <Q> Sounds like you need to begin with a conventional latching stop-start circuit: - <S> So, the circuit above will have another relay contact that feeds 24 volts to all your motors and, when the circuit is first powered, the relay will not be energized. <S> To energize it press "start". <S> A relay contact latches the start button contact. <S> To de-energize press "stop". <S> Stop switches can be a series of emergency stop switches wired in series with the stop button above. <A> you can get to no matter the situation, and with this method, you can have more than 1 of these in series. <S> If you already have a relay that controls power to the motors, you would fit it in series provided you want the control circuits to remain powered. <A> emergency stop may require a BUTTON on the robot, so if the Radio Link or the cable is broken, you can (try to ) just slap that big red button.
| If you do not have much power storage other than the battery, the easiest way would be a normally closed latching switch, like the big red e-stop buttons, in series with the battery, you hit that and when what small amount of charge is left in the system is used, all the motors loose power, Just remember to place it somewhere
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I thought I understood the basics of DC circuits and Ohm's law but now I'm confused So I was taught in school and also read a lot of articles of the basics of DC circuits. I know the formula R=V/I, and therefore that resistance is directly proportional to voltage and inversely proportional to current. So if I increase the resistance the voltage increases too right? Now I'm working on an old RC car I used to play with. The battery pack consists of 6 AA batteries which provide 9V, but I noticed that there's another output that outputs 4.5V. I guess it's being used to power the driver board.I tried reading the voltage but I also put a 10 megaohm resistor in the circuit. I was surprised when I saw that the multimeter was reading around 2.2V. Isn't the voltage supposed to be higher because of the resistor? I am completely confused now and I'm pretty sure that I have a lack of understanding or just confusing some things. Please explain why did this happen. <Q> I know the formula R=V/I, and therefore that resistance is directly proportional to voltage and inversely proportional to current. <S> I would say that, "The resistance defines the relationship of voltage across a resistor to the current through the resistor". <S> So if I higher the resistance the voltage increases too right? <S> Only if the current remains constant. <S> In very many cases the voltage is fixed so increasing the resistance <S> decreases the current as you would expect from \$ <S> I = \frac V R \$ . <S> Now I'm working on an old RC car I used to play with. <S> The battery pack consists of 6 AA batteries which provide 9 V, but I noticed that there's another output that outputs 4.5 V. <S> It sounds as though they are tapping off at the battery pack mid-point or they have a voltage regulator stepping down from 9 V to 4.5 V. <S> I guess it's being used to power the driver board. <S> It could be. <S> Many types of logic will run will on about 5 V. <S> I tried reading the voltage <S> but I also put a 10 megaohm resistor in the circuit. <S> I was surprised when I saw that the multimeter was reading around 2.2V. <S> Isn't the voltage supposed to be higher because of the resistor? <S> Bring this to its logical conclusion: with an infinite resistance (no resistor present) you should get infinite voltage? <S> No, you have connected 1 MΩ in series with a multimeter with what appears to be a 1 MΩ input impedance. <S> This has divided the available voltage in two. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> (a) Direct measurement. <S> (b) Voltage measurement through a series resistor. <S> You can use Figure 1b as a way of measuring the internal resistance of the multimeter. <A> So if I higher the resistance the voltage increases too right? <S> You're not thinking about it quite right. <S> If a voltage source is being used (i.e. the voltage is being kept constant) then increasing the resistance will reduce the current. <S> Do you understand F = ma? <S> It is the same thing. <S> Increasing the force being applied doesn't increase the mass of whatever you are pushing. <S> If the mass of whatever you are pushing is held constant though (as is usually the case) and you increase the force then its acceleration will increase. <S> But if the acceleration was somehow kept constant instead of the mass then increasing the force must also be accompanied by an increase in mass, otherwise the acceleration cannot remain the same. <S> This is a lot more difficult to do since we apply forces which result in accelerations rather than the other way around. <S> There is one exception to this: gravity. <S> Gravity automatically applies more force to objects with more mass but they have more mass to accelerate so the acceleration due to gravity stays constant regardless of the object's mass. <S> In a sense, gravity applies the same acceleration to everything and a force results from this which is the reverse of most cases we see. <A> what you made was a voltage_divider. <S> the math on that is Vout = <S> Vin <S> * R1 / (R1 + R2) <S> and your DVM is measuring the Vout. <S> With both your resistors (the external 10Megohm and the DVMs internal 10MegOhm, the math becomes Vout = <S> 4.5v <S> * x / <S> (x <S> + x) = <S> 4.5 <S> * 0.5 <S> = 2.25 <S> Try other resistors instead of your 10Meg OHm. <S> Imeg ohm, 100Kohm, 330K ohm, 10Kohm <S> If you go below 10Kohm, your DVM will just read the battery voltage (within 0.1%). <S> Have fun. <S> ====================================== <S> If you remove the DVM, then there is no current thru that 10Meg OHm resistor you had connect to the 4.5v node. <S> If there is no current, then there is no voltage drop across the resistor. <S> Both ends of the resistor are at 4.5 volts. <S> Not a very useful thing, because you cannot connect anything to it ...... ahhh a gold-leaf-electrometer is acceptable.
| Increasing the resistance doesn't increase the voltage UNLESS the current is kept constant (i.e. a current source) is being used).
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What is meant by 'On-chip 2-cycle Multiplier' for AVR microcontroller? While checking the datasheet of ATMEGA 32 I have found a feature called 'On-chip 2-cycle Multiplier'. Can anyone explain to me what's that and what's the advantage of it? <Q> This is faster than doing multiplication using software (e.g. adding multiple times in a for loop). <A> It means it has a hardware multiplier that takes can complete a multiply operation in two instruction cycles. <A> Some processors can't multiply numbers, they only have addition and bit boolean logic. <S> This website discusses how to implement multiplication on a 6502 (The old Apple ][ computer, as well as other computers of that era). <S> https://www.lysator.liu.se/~nisse/misc/6502-mul.html <S> These 8bit x 8bit operations take about 100 cycles per multiplication. <S> You can see that a 2 cycle multiplication (done in hardware) is far superior. <S> You can do simple DSP type calculations if you can do multiplies at half the clock rate. <S> The best CPUs can still only do 1 multiplication per cycle, because that is the definition of a clock cycle (although they can do many in parallel). <S> (Now division is a lot harder. <S> Most CPUs take at least N clock cycles to do a division, where N is the number of bits. <S> For some reason, dealing with the "carry" bit in multiplication can be done quickly, while the "carry" bit in division is much more difficult.) <A> It's a feature that would perhaps have been impressive a decade or so ago. <S> Traditionally basic 8-bit processor cores did not have multiply instructions, silicon area was considered more important than performance. <S> Higher end <S> 16 and 32 bit processor cores on the other hand typically did have hardware multiply instructions and furthermore worked with larger data word sizes. <S> Some of the high-end 8 bit (data word size) micro-controller architectures such as the "atmega32" and the "pic18" decided to add a multiply instruction. <S> Since this was a new feature for 8-bit microcontrollers the manufacturers saw reason to shout about it. <S> Both the "pic18" and "atmega32" implementations took a pair of 8 bit inputs and produced a 16 bit output, but the atmega32 version gives more options regarding the exact type of multiply that is desired. <S> but the elephant in the room is the arm cortex m0 <S> +. <S> You can now get micro-controllers with a 32-bit data path and a single-cycle 32 <S> *32->32 multiplier for prices comparable to the high-end 8 bit micro-controllers.
| It means that the ALU in the microcontroller has a hardware multiplier, which takes two instruction clock cycles to perform a calculation.
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Generating 5V from 37V DC with no EMI I'm making a PCB for an audio amplifier + power supply (circuit here ). The power supply voltage rails with my transformer end up being about +/- 37V, and I want to generate a 5V rail from this for powering some cooling fans on the same PCB. This is pushing the limits of a typical voltage regulator (e.g. the max. input of a L7805 is 35V). I tried a DC/DC converter (a R-78C5.0-1.0) but this produces audible noise on a speaker at the amplifier output (I put 10 uF decoupling caps at the input and output). One thing I didn't try yet is the EMC filter suggested in the datasheet here . I found "high-voltage regulators" (e.g. an LR8 ), but the max. output current is 30 mA, and my fans need about 100 mA. Is there a better way for generating 5V from 37V without producing EMC that I can hear as noise on the amplifier? UPDATE: I'm an idiot. It turns out the noise was coming from having some signal wires too close to the transformer. I rearranged everything and now I can use the DC/DC converter without producing any audible noise at the speakers. <Q> Look at it this way - you've found a weakness in your amplifier <S> so, if you solve that you'll be in a much better position to cope with all sorts of EMI issues that can occur from time to time in pretty much all designs. <S> EMI is always produced by a switching regulator so your options are: - Choose a linear regulator (and a big heat-sink with significant power loss) or, <S> Solve the basic problem with your amplifier or, Pick fans that run on a higher voltage <A> Custom regulator Build your own linear regulator to get the flexibility you want, you won't need anything advanced here. <S> simulate this circuit – <S> Schematic created using CircuitLab Q1 is an emitter follower and R2+D1 sets the output voltage, which has to be about 0.6 to 1 V higher than what you want, because of the voltage drop through Q1's base. <S> R1 is optional and is just there to take some heat off of the main transistor. <S> It may be easier to keep a power resistor cool or within limits, than it is to keep the transistor cooled. <S> Without R1, Q1 will have to dissipate 3 watts, which requires a heat sink. <S> If R1 is installed, Q1 should only dissipate 1 watt at most. <S> R1 also acts as a crude current limiter, because the voltage will drop too much at high currents. <S> I just picked Q1 from something that seemed reasonable and existed in circuitlab. <S> Which one you can get varies depending on where you live and where you order your components. <S> It must be able to handle the voltage, current, and a couple of watts. <S> It's worth noting that no matter what you do, using a linear regulator such as this, a 7805, or an LM317, will necessarily create at least 3.2 watts of heat as losses, while delivering 0.5 watts to your fan. <S> Better solution <S> I noticed this: […] powering some cooling fans […] <S> If you have more than one fan, you should connect as many of them as you want in series , the optimal being 3 x 12 volt fans which you can connect directly to your unregulated supply. <A> If the transformer is of a toroid type, it is often easy to add an extra winding on top of the existing ones. <S> Just wrap wire around the core until output voltage is about 7 volts AC, then rectify it and use 7805 to regulate it to 5V. <S> If you cannot add an extra winding to the transformer, you could use a small secondary transformer. <A> A note on that pi filter — if you just decouple the input of the switching regulator with a capacitor, that's fine for making the regulator happy, but on its own, it's not a filter. <S> The lower the inductance of the wires bringing that power supply to the regulator, the less work the capacitor has to do, and the more EMI you'll see on those wires. <S> You can view the inductor in the pi filter as deliberately raising that inductance, which makes the capacitor at the regulator input work harder, and "traps" the high-frequency component from flowing back to the power supply. <S> Then we provide an extra capacitor (the other leg of the pi) on the power-supply end for good measure.
| As you already have a transformer as the main power supply and have AC voltage available in the circuit, you could use a transformer to generate the +5V supply also. You can feed it either from the mains AC supply, or from the ~37 VAC supply from your primary transformer. It does not scale well with multiple fans in parallel either, seeing how you would have to remove a lot of heat somehow.
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In which region should a MOSFET be operated as a switch? I have a question about MOSFET switching operation. According to an article: In order to operate a MOSFET as a switch, it must be operated in cut-off and linear (or triode) region. According to another article: MOSFET in saturation region is preferred to make it work as a switch. I am very much confused about the operating region of MOSFET to be used as a switch . Should I operate the MOSFET to "Turn ON" in a (linear/ohmic/triode) or saturation region? <Q> When your article says this (wrongly): - MOSFET in saturation region is preferred to make it work as a switch. <S> It's because it's written by someone who thinks that the name of the equivalent section of the BJT's characteristic is 100% transferable to MOSFETs. <S> To clear this up: - When a MOSFET is operated as an on -switch <S> it works in the triode or ohmic region <S> When a MOSFET is operated as an off -switch <S> it works in the cut-off region <S> conducting <S> Should I operate the MOSFET to "Turn ON" in a (Linear/Ohmic/Triode) or Saturation region? <S> Answer: the linear/ohmic/triode region <A> If you want to use a MOSFET as a switch, you probably want to have a low V DS , as an ideal switch has no voltage drop across terminals, but it can have an infinite current through it. <S> As you can see, if you need 10 A (supposing the scale is in ampere), the V DS will be lower if you increase V GS . <S> Your confusion may come from the fact that for a bipolar transitor, the name of the different regions is not the same: <S> So if you use a bipolar transistor as a switch, you should use it in the saturation region. <A> A MOSFET transistor can be used as an on switch in both triode and saturation regions, but it gives us different advantages. <S> In saturation, a higher current can be obtained, but in triode, because of its lower resistance, lower losses can be achieved. <S> Normally in the digital circuit design, the triode region is more common. <A> Be it mosfet, fet or bipolar, for switching purpose, it is only in one definition: it is in saturation or cut off. <S> Saturation is min voltage across device, and Max current. <S> In cut off or open state, all the voltage across the device, but negligible current. <S> In this switching mode, in both the above cases, the dissipation in the device is minimum, and Max power is controlled. <S> To ensure saturation, you may have to overdrive, within limits, and similar under-drive, or even negatively drive the devices to ensure full cutoff. <S> Go by the datasheets and don't overdo it. <S> The only time these devices are used in the active region, or linier region, is in amplifiers (not even all). <S> Bringing "triode mode" in to this is meaningless. <S> NOT to mention, who said triodes don't have cutoff and saturation?It is advantageous to use 'English' sometimes when you want to describe the physical picture, but there are no cmos language, triode language etc. <S> Common sense does not depend on language. <S> Engineering does not preclude common sense.
| When a MOSFET is operated as a controlled current device it works in the saturation region "Saturation" refers to the channel being saturated When a BJT is operated as a switch it works in the saturation region and cut-off regions "Saturation" in the case of a BJT refers to the saturation of the base in that both PN or NP junctions are (somewhat) So according to the characteristic curves of a MOSFET, you have to be operating in the linear region.
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DC/DC system has voltage between grounds My system is powered by 60 V battery. First converter is 60 V / 12 V isolated, second converter is 60 V / 12 V non-isolated. Two converters power own system component. When I measure the voltage between converters' grounds, the voltage is -6.90 V. Why is the voltage like this? And How can I find the problem? editing: I measure the voltage between converters seconds(outputs) grounds. <Q> You should be able to connect the two grounds together, which will solve your 'problem'. <A> If I correctly understand your problem, your converters are connected in parallel for powering their own load and you are measuring the differential potential between the ground of your non isolated converter and the primary ground of the isolated converter ? <S> How do you do your measurement ? <S> With an oscilloscope or with a multimeter ? <S> --------------------------E <S> D <S> I <S> T <S> -------------------------- <S> As you can see you are trying to measure the voltage across your transformer isolation, ie the potential difference between the primary ground and the secondary ground. <S> As @hacktastical said, the isolation is not perfect. <S> Nevertheless I think your measure is bad as you are paralleling a high impedance (the one of your multimeter) and the high isolation of your transformer. <S> Your measure modify the system. <S> It is an "intrusive" measurement. <S> And the voltage value given by the multimeter is not correct as you need to measure an impedance negligeable with respect to the internal impedance of your multimeter for having correct measurement. <S> So it is complicated to say (at least to me) <S> what is the meaning of your voltage measurement. <A> I think the question you’re asking is, how does an isolated power supply develop a measurable voltage difference between the primary (grounded) and secondary (floating) side? <S> The answer is, the isolation isn’t perfect. <S> There’s some leakage across the galvanic isolation (transformer in this case). <S> Its measureable using a high impedance meter from floating secondary to primary ground. <S> You can measure leakage current too <S> - this should be quite low, less than a mA <S> probably for this kind of supply. <S> Is there a specific reason one part of your load is isolated? <S> A medical device or special instrumentation for example? <S> Then you may need to find or design a supply that has less leakage than the one you’re using.
| If one converter is isolated, there's no reason there should be any specific voltage between its output ground and any other point in the system. I think you are trying to measure the average voltage across the capacitor between the primary and the secondary of your transformer of your isolated converter.
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Is R2 needed on this circuit? I am working on a circuit to drive some water pumps. I intend to achieve a double galvanic isolation between the pumps and the CPU to ensure nothing bad will happen to the CPU. Here is my schematic: I am using a 4N35 optocoupler to achieve the first stage galvanic isolation between the 5V control signal and the relay driver circuit. The relay driver circuit consists of a BC337 NPN BJT(Q1) a flyback diode (D1) and a resistor to the base of Q1 (R2). As Q2 is intended to be used in saturation mode, I learned that Ic = VCC/RC, in this case VCC is the relay supply which is 5V and the coil resistance is 70Ω with 10% error, hence the worst case resistance should be 63Ω, knowing this Ic = 5/63 = 79.3mA. Knowing Ic is now posible to solve for Ib, Ib = Ic/Beta, I am using the lowest HFE found on the BC337 datasheet, this value is 60. Hence Ib >= 1.32 mA. So now I know that I need at least 1.32 mA feeding through the base of Q1 for it to be properly saturated. The thing is that I don't know if R2 is needed on this circuit, because the optocoupler is already sourcing current. If R2 is needed, how can I solve for its proper value? <Q> Your beta is actually closer to 100 for this use case (Ic = 100mA) <S> The optocouplers CE junction voltage drop should be small, but lets say it drops 0.2V, So 5V - 0.2V - 0.7V (transistor base) <S> = <S> 4.1V <S> 79.3mA <S> / Beta of 100 <S> = 0.793mA <S> 4.1V <S> / 0.793mA <S> = 1265 <S> Ohm <S> So R2 should be about 1.2K <S> You do want to include R2, otherwise the transistor base looks like a diode to the 5V rail, and may damage both the transistor, the optocoupler and possibly your power supply by drawing too much current. <A> If it weren't there you would risk destroying one or the other of those components. <S> To calculate the needed value take the saturation voltage of your optocoupler output at your LED current, or look at the curves to see where Vce should be, then add the Vbe voltage of the BJT at your desired base current. <S> Subtract that value from the relay supply voltage <S> and that's the voltage across R2. <S> Size R2 to give your desired base current plus whatever safety margin <S> you feel comfortable with, as long as it's below the max ratings of your parts. <A> R2 stops the base emitter of Q1 from simply shorting via diode to ground. <S> It need only prevent the current from being so large something fries while stil being large enough to saturate Q1. <S> You can assume the opto output NPN is also saturated in your calculation. <S> You know your minimum Ib required. <S> You can look up the max current through opto or Q2 and pick a reasonable Ib in between then calculate how much excess voltage appears across R2 after \$V_{be. <S> Q1}\$ is removed from the supply voltage.
| Yes, you need R2 to limit the base current, and the current through the output transistor of the optocoupler.
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Can I use a mosfet to discharge quickly a capacitor without resistor? I have to monitor a PV cell with a capacitor and to discharge I was thinking about using a mosfet whithout resistor to do it, whithout time constant. This, produce a short circuit when the discharge is on. Could I do that? What happen with the energy ? <Q> This is not a good idea. <S> It may work OK if the capacitor is very small. <S> But you will be operating the transistor in a pulse mode and you will have to refer to the peak pulse power allowable from the datasheet. <S> It is always better to dissipate heat with a resistor rather than with silicon. <S> Even a small resistor (a few ohms) will likely completely shift the power dissipation from the transistor to the resistor. <S> Try to choose a resistor with a pulse power rating listed in the datasheet. <S> Also, you may want to put a fuse in series so that if/when the mosfet fails, nothing catastrophic occurs. <S> MOSFET's usually fail short. <A> From offline computations, you'll have about 0.05 joules of heat storage in the silicon and the copper flag directly under the silicon, per degree C heating. <S> If you want only 100 degree C temperature rise, then the capacitor can only store 5 joules of energy. <S> ================================== <S> here is the math <S> Assume <S> the FET is 5mm by 5mm, atop a copper flag; assume the total thickness is 2mm. <S> This is volume of 50 cubic milliMeters or (each mm^3 being 1,000,000,000 cubic microns) <S> 50 Billion cubic microns. <S> Or that is 0.05 Trillion cubic microns. <S> Now assume the thermal capacity of copper and silicon are the same, using the silicon value of 1.6 picoJoules per cubic micron per degree <S> C. Lets multiply our 2 numbers <S> 0.05 Trillion cubic microns * 1.6 picoJoules per cubic micron per degree <S> C <S> and we have 0.08 joules per degree C temperature rise. <S> Again, this is for the silicon and the copper immediately under the silicon. <S> Why is this important? <S> Because the thermal time constant to spread heat into the rest of the flag (the copper slab) is very slow: 1cm is 1.14 seconds, 4cm is 16X slower. <S> What is the time constant to dump heat from the silicon (probably 100 micron thick, atop 1.9? <S> millimeter of copper? <S> Well, 1 cubic meter has time constant of 11,400 seconds. <S> 1mm is 1000 * 1000 faster at 11.4 milliseconds (0.0114 seconds). <S> A 2mm thick slab is 4X slower at about 0.05 seconds. <S> Thus you MUST slow down the heat generation, to spread out over 0.05 seconds. <S> or you will merely be heating the silicon so fast the heat cannot move down into the copper mounting slab (the TO-220 base_plate). <S> ====================== <S> answering the commentor: yes these are 63% τ just so the idea gets out there that total punt-and-hope is not needed. <A> Sound like a sample and hold circuit. <S> Just be aware the heat will be split between the mosfet rds on and the capacitor esr. <S> But things will get warm depending on how often. <S> P = <S> I 2 <S> x R, so for an accurate wattage if your conscerned you would need to find the rms current over the discharge time.
| The amount of energy in a small capacitor should not be an issue. You can discharge the capacitor with a mosfet.
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Need to keep re-soldering SMD resistors on PCB I made a PCB to scale a +-100V signal into a 0v to 3.3V signal that I could read using an ADC on a microcontroller(see link for idea .) I used this solder paste flux to help with the soldering. After soldering 16 of these, a bunch resistors were shorted. I got some tips from a friend that is good at soldering (use a microscope and go slowly) and got them all working. Now though every few days a voltage divider will either get stuck at 3.3V or ground or somewhere in the middle. If I re-solder the resistors it'll work again. Even if I just reflow the solder on the resistors it will start working again. Any idea why they keep breaking after a few days? Any ideas how I can get them permanently working? The resistors are all 0603 and 0805 footprint. FYI: I used two 150k ohm resistors instead of one 300k resistor because I was worried about a 100V drop across just one. <Q> This seems to be plumbing flux, which often is higher in corrosive components than electronics fluxes. <S> I've never head of that happening within weeks, however. <S> So, my hypothesis is: you got a "cold" solder join, meaning that there's no wetting of the contacts on both sides. <S> (you can google image search for "cold solder joint") <S> You have a large capacitance on the OutCh1 side that you don't show. <S> After having charged that to a specific voltage, it takes long to discharge it (depends on how shortly after changing InCH1 you look) <S> Whatever you have on the output actually latches up. <A> As others have pointed out, your issue will be one of using the wrong type of flux. <S> Plumbing flux is highly corrosive and will damage your board and components. <S> It should not be used for soldering. <S> Below is an example of why you should not use plumbing flux. <S> The board was a DIY PCB without a solder mask that was tin plated. <S> Because of the lack of solder mask I required a large amount of flux to solder the board to keep shorts from forming. <S> This was before I knew any better about fluxes, so I used the plumbing flux I had to hand. <S> The picture shows the board after about a month. <S> It had corroded to the point that several areas had actually been stripped entirely of the tin plating, and the terminals of the resistors and capacitors have corroded. <S> Fortunately it was a temporary test board that lasted long enough to serve its purpose. <A> Your flux is probably eating away everything. <S> Reading on the page you linked for the flux properties: cleans and fluxes copper piping Compatible with all common plumbing solder alloys Good for large diameter copper piping <S> This is a plumber's flux. <S> Reading the Wikipedia page on flux , we find that… Plumbing and automotive applications, among others, typically use an acid-based (hydrochloric acid) flux which provides rather aggressive cleaning of the joint. <S> These fluxes cannot be used in electronics because their residues are conductive leading to unintended electrical connections, and because they will eventually dissolve small diameter wires.
| Considering solder wire comes with a rosin core, you'll probably be better off without that flux, as such fluxes have the tendency to corrode electronic contacts over time.
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What is failed closed switch? What is the meaning of switch failed open or failed closed in electronics?I did a search but without results. <Q> As I understand your question the answer would be this. <S> A failed open switch is a switching device (be it mechanical, relay, semiconductor) that has lost it's switching capabilities (it's defective) and <S> the switched contacts are now constantly open (disconnected) regardless of the control signal to the switch. <S> Similarly a failed closed <S> switch is the same just with the contacts connected through. <A> As you usually do not get a say in how physical devices like a toggle switch will fail, I am going to assume this is for active circuits, such as relays or transistors, <S> In these cases, I would consider it the "Fail-safe" state, e.g. if there is an error in the system, or even if it looses power, what state do the outputs change to, <S> So a fail closed contact pair might be a relays Normally closed contact pair, if the circuit looses power, the relay closes, and keeps something that should be powered in a failure state to keep things safe under all reasonable conditions. <S> e.g. a cooling fan going to 100%, <A> You typically encounter terms like this when doing safety calculations. <S> For example, a relay supplying some actuator etc could be dangerous if the relay contacts get stuck in the closed (active) position, but safe if they get stuck in the open (inactive) position. <S> So when you do theoretical calculations of how long the relay supposedly lasts before it fails in a dangerous way, you only need to consider the cases where it gets stuck in the dangerous position. <A> The terms 'switch failed open' and 'switch failed closed' may be applicable to any type of switch whether mechanical, magnetic, electromagnetic, mercury, vacuum tube or semiconductor. <S> A switch is said to have 'failed closed' when continuity between its terminals cannot be broken when actuated.
| A switch is said to have 'failed open' when continuity between its terminals cannot be established when actuated.
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How much 12V DC current can I safely draw from an AC 230V 13A socket and why? In my cabin I have a sollar cell and a 12V battery that I use to power my electronics. However I'm dissatisfied with the standard CL sockets used in cars (apparently I'm not the only one ) and I decided to reuse a foreign AC socket instead, specifically the United Kingdom's “13 amp socket” (BS 1363) . I chose the UK socket because it's well designed and it's very rare that anyone has UK plugs in my country. I will take actions to ensure no-one will thoughtlessly plug an UK electronic device to these sockets! The socket is designed for 230 V and up to 13 A of alternating current. That means it can supply up to ~3 kW of power. However with the 12 V I'll be using , thirteen amps can barely reach 150 W. Since some popular water pumps need ~700 W, this seems like a serious limitation. In a different question here on Electrical Engineering SE, someone asked whether they can use a 10A@250V-rated socket to draw 13A@230v and the most popular answer says that “the only important parameter is the current rating, [...] 10 amps through the socket will heat it up just as much at 1 volt as it will at 5,000 volts.” Sadly, they don't provide an explanation why. Does this apply to my case and why / why not? Another thing that concerns me is I'll be using DC instead of AC. There are several things that work differently with direct and alternating current. Is it safe to use the socket this way? EDIT: There is a question about the current rating of fuses where they explain why fuses blow at a certain current , not power . However I failed to understand the reasoning in the answers, so I can't tell whether it also extends to my case, or not. <Q> Power is current times voltage (P = IE). <S> You don't mention if you're converting from one voltage to another. <S> Are you using a step-down converter? <S> Are you merely charging the battery with the solar system and want to know how to achieve equivalent power from a charged battery bank? <S> 230V AC 13A is 2990 watts. <S> 2990 watts at 12V DC would be ~249 amperes. <S> Because you said "13A at 12V can barely reach 150W" <S> it seems like you're already aware of the relationship between voltage, current, and power. <S> It also sounds like you're looking for a better socket to use for your system, and chose the UK mains style because it won't likely be confused with the real thing. <S> So here's what you are maybe missing, related to my question from 8 years ago about fuses. <S> Voltage is what "motivates" electrons to go through a particular thing, whether it's a fuse, a wire, or a connector. <S> The current is "how many" of them. <S> The larger the current, the greater the friction, and thus heat. <S> A connector or wire rated for 13A is not going to handle more power unless the voltage is also higher. <S> In the case of a connector, the voltage rating will be mostly applicable to the distance between conductors (to avoid arcing) while the current rating will be applicable to the robustness of the conductors. <S> Put another way, 13A can have wildly different power values based on voltage, but it's always going to be the same quantity of electric current flowing. <S> You will likely want to use something capable of much higher current. <S> Automotive applications that use 12V systems often have fuses and wire rated for 50 or more amps. <S> But be absolutely sure to look at the specifications for your battery (or battery bank). <S> There's no point in installing 60A-capable wiring and connectors if your source can only supply 40A. <S> Also be sure to install fuses to limit current to less than the weakest link in the system. <A> From your comment: But why is it I 2 R and not eg. <S> IU? <S> It's because the plug contacts know nothing about the voltage of the circuit. <S> All the contacts know about is the current through them and the contact resistance. <S> The voltage on the load is irrelevant as far as the contacts are concerned. <S> Figure 1. <S> A Schurter C20, 16 <S> A plug. <S> Image source: RS <S> Online . <S> I couldn't find a specification for a 13 A plug but <S> the C20 plug has a contact resistance of <S> < 10 mΩ according to RS. <S> At 16 A the power (heat) dissipated in each of the live and neutral pins would be given by \$ <S> P = I^2R = 16 <S> ^2 <S> \times 0.01 <S> = 2.56 \ \text <S> W \$ or 5 W for the feed and return. <S> This is not insignificant. <S> Current is what matters. <S> Figure 2. <S> Something like Anderson Power Products Powerpole connector series may be more suitable. <A> Current is the pseudo-actual electron flow ( https://www.chegg.com/homework-help/definitions/current-flow-2 ) and the reason power lines use very high voltage and little current (to minimize losses due to heat). <S> http://www.betaengineering.com/high-voltage-industry-blog/transmitting-electricity-at-high-voltages <S> NOTE: <S> the electrons actually move relatively slowly, it's the energy itself that moves at near the speed of light. <S> However, many devices have specifications lower than their theoretical maximum due to variability in manufacturing. <S> I am in no way recommending to find out their max specs though. <S> I would see if there is a way to use a power transformer to step up the voltage to the rating of the outlet to deliver more power at approximately the same current. <S> With your current 12vDC, there is no way to deliver ~700W at only 13A.
| This means your battery(ies) (and all the connectors and cabling) would have to be capable of safely delivering 250A in order to have roughly equivalent power to your UK mains example. A main point regarding current and voltage: more current produces more heat. If the voltage is 12V, you're right, it isn't a lot of power, but it still requires thick wire and connectors to safely handle that current without having an unsafe temperature rise. As for the ratings of the device, it is never a good idea to exceed either limitation. Note that the voltage across the load doesn't enter into the calculations.
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Solar DC circuit: Preventing Buck Converter from Dropping out Due to Motor switching on/off (in Parallel) I have an off-grid orchard which has a well pump running directly off of a 24v PV array. (27-32v when at full power) Parallel to this direct circuit, I have a buck converter to power a Raspberry pi (via GPIO), which has a switching solid-state relay and a camera to watch the pump and its gauge. When the Pi triggers the SSR, the voltage coming into the buck drops from 27v to 24v on my multimeter (I don't have an oscilloscope, so perhaps its more of a severe drop), and this causes the output voltage to briefly go to zero, causing the Pi to reboot. While it's entirely possible this is the fault of the cheap 12v/24v to 5v buck I bought on Amazon , I'm wondering if this is expected, and if so, can I do something like add a capacitor, either on the input or output to buffer this, and keep the pi running. <Q> To keep the input voltage to the buck converter from dropping down to some really low value during the inrush current period of the 150 watt motor, use a diode and capacitor. <S> This will "hold-up" the voltage at the input to the buck circuit despite it dropping down across the motor. <S> The diode gets reverse biased during this short period and therefore the motor load doesn't discharge the capacitor. <S> But, you have to choose a capacitor that is sufficient. <S> For instance, if the buck regulator will work down to (say) <S> 8 volts without a problem, then you can start to calculate the hold-up capacitor. <S> Based on the pi taking (or needing) say 1 amp we can say: - $$ <S> I = <S> C\dfrac{dv}{dt}$$ <S> Where dv is the permissible drop from 24 volts to 8 volts <S> i.e. 16 volts and dt is the the inrush time for the motor (say) <S> 2 seconds hence: - $$1\text{ amp} = <S> C\dfrac{16 \text <S> { volts}}{2\text{ seconds}}$$ C therefore equals 125,000 uF or above. <S> The numbers used above are my guesstimates and may not bear much resemblance to the real values needed by the buck regulator or the pi current. <A> If the power converters & supplies you have are well-made, they should already have capacitors in parallel between the +5V and ground lines to smooth any jumps in voltage. <S> However, you can definitely add a cap as close to the RPi as possible to smooth out the voltage even more. <S> An alternate (and pricier) option is to use an un-interrupted power supply for the RPi power, or connect a battery pack that is both continuously 'charging' the pi, and being charged by the buck you already have. <S> The battery method isn't recommended due to the current draw from the RPi but is a possible short-term fix depending on the type of battery. <A> The buck converter apparently can't keep up with the abrupt change in input voltage. <S> A cap probably won't be enough to smooth the voltage drop, but if you are willing to build a circuit, you could put in a capacitance multiplier to smooth it much more effectively. <S> This solution could work if the voltage drop is not in fact much more severe than a few Volts. <S> This circuit is a common ripple rejection circuit. <S> Q1 is a Darlington transistor. <S> R1 and R2 form a low-pass filter with C1, which prevents abrupt changes at the collector of Q1 from passing through to the emitter. <S> If the PV array output steps abruptly by 3V, it should take 10 seconds or so for the Q1 emitter output to make that transition with the component values shown below. <S> That should give the buck converter plenty of time to adjust. <S> The RPi and camera probably draw somewhere in the ballpark of 200 mA at the input of the buck. <S> With these component values there is about five Volts across the Q1 collector-emitter terminals, so Q1 will dissipate about a Watt. <S> It should be able to handle without a heat sink, but it might be a good idea to put one on in case the Pi & camera draws more than that. <S> The disadvantage of this approach, apart from the complexity, is that you give up a Watt or so to the ripple rejection circuit. <S> simulate this circuit – <S> Schematic created using CircuitLab
| The problem is also coming from the RPi triggering a power hungry device, so you may be able to add an inductor in series with the device to slow down power surges.
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Max watt bulb for an old "660w 250v" table lamp? I have an old table lamp that's imprinted "LEVITON / MADE IN USA / 660W 250V / LAMP HOLDER". I'm used to lamps with stickers that specifically tell me what max-watt bulbs I can use, usually 60W, 75W, or 100W. Why is this lamp printed "660W", and what's its actually safe, maximum-wattage bulb? Photos below. There's a UL sticker, but otherwise nothing else is printed or stuck on the lamp sides or bottom. <Q> Without any other information, you are left with trial and error. <S> With an incanddescent bulb you need to be concerned about overheating the shade. <S> With an LED bulb the concern would be the temperature of the control circuit in the bulb. <S> The 660 watt rating of the lamp holder would indicate that should never be a concern, other parts of the lamp will always be the limiting factors. <S> Unless the shade is ridiculously small, I would think a 60 W incandescent would be fine and 100 W probably ok. <S> For 150 W, you should probably have a larger clearance between the bulb and shade than would be usual for a single-bulb lamp. <A> If the socket faces up like that (bulb goes in base down), then the 660W limit has only to do with heat that might transmit down to the socket. <S> Anything less than 660W is fine. <A> The 660W rating you see is the rating of the socket component of the lamp. <S> Note that actual 660W lamps with Edison bases are basically unheard of. <S> The lamp fixture (luminaire) will have its own rating. <S> For that you must consult its labeling and/or instructions, which are part of its UL Listing. <S> Search the luminaire, including the base, for anything about that. <S> Otherwise, 60W is a very safe assumption. <S> And that is more than you'll ever want in the 21st century since CFLs came and went and LED is king... <S> 60 actual watts of LED would be an insane amount of light. <S> The watt ratings of luminaires are thermal; they are the max that the lamp is certified for to not start a fire. <S> As such, it is relevant to the heat output, not the light output.
| The safety issue comes down to the possibility of the shade catching fire, but you might also want to avoid discoloration. You can use anything UP TO 660W, which for a table lamp would be ridiculous.
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Converting a digital incremental encoder to an analogue voltage Is there a way to convert the output of an incremental encoder (50% duty cycle) to an analogue voltage using analogue components? This is because the interface that I'm using doesn't handle a large enough sample rate to accurately determine the shaft angle with just using interrupts directly from the encoder. (Any other possible solutions are greatly appreciated) P.S. using a potentiometer attached to the shaft is not a solution. <Q> so you have a bit stream with 2 bit flows (quadrature, right?) <S> use a binary up-down counter 74160/161 <S> should be useful <S> be able to ZERO_SET <S> the counter then convert the counter's binary number to an analog voltage, using a digital-voltage convert <S> (ADI should have some) <S> ======================== <S> You need to prepare a error analysis. <S> You have THREE quantizers here. <A> There is no way to do this with analog[ue] components. <S> Easiest way is to use a microcontroller, scan the two quadrature inputs and store a digital voltage. <S> Most MCUs have some EEPROM so you could store the last position over a power interruption provided you take care to not write too often (for example, you could write the position after a delay of 30 seconds). <S> That's because most EEPROMs may wear out after 100K to 1M write operations. <S> There are also wear-leveling techniques. <S> You could also consider a battery backup, which might be attractive if you need that for some other reason. <A> Any solution that uses a simple counter for incremental encoding suffers a problem: glitching at counter transitions. <S> So, for example, wiring up some '163s to an R-2R DAC will yield a noisy ramp signal. <S> For this reason, encoders that output absolute values (like what's being proposed here) use Gray code. <S> With Gray code only one bit changes at a time. <S> More about that here: <S> https://www.quantumdev.com/understanding-gray-code-for-rotary-encoders/ <S> You also need to detect the direction. <S> There's some logic there, even specialized (and now obsolete, like the HCTL-2000 <S> ) chips too. <S> If, however, the sensor was processed by something that masks the glitches and detects the direction, then outputs these to some kind of a DAC or PWM output... <S> And one other thing: index. <S> Your actuator will need a sensor to establish a 'home' position. <S> Again, if you just had something that could see that too, and reset the count... <S> All these things point to using a small microcontroller. <S> Or, buy it. https://www.genesisautomationonline.com/site/pdf/products/GAO/Motrona/UpdatedPDFs2017/Zu252_CountModule.pdf <A> It will be a little convoluted, but possible, First start out with some circuitry to break out a step and direction signal, below is the direction, for the step from memory it is an XOR gate between the 2 signals (you will need to filter the encoder outputs to prevent multiple transitions from noisy switching) <S> Image Source <S> From that you feed it in to a binary up/down counter e.g. SN74ALS867A to give you a binary output from those counts (with some other logic to prevent it overflowing if you desire it to behave that way), Then finally run it into a parrellel dac, or use the output of the binary counter as a crude R2R dac. <A> It depends what you want to do. <S> For a crude measure of speed, you could use a differentiator if only pure analog technology is allowed (or monostable if you're allowed a 555 or 74123) to generate a fixed pulse width from every positive going edge. <S> This is followed by a low pass filter to convert that pulse density into an analog voltage representing speed (or an integrator to get a crude indication of position). <S> Things get a bit more complicated (but still feasible) <S> if you have to handle rotation in both directions; I'll let you imagine the details. <S> For practical purposes or any real accuracy, Spehro's answer is correct.
| You can then use a DAC (or the MCU PWM) to convert to an analog voltage.
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Physical size of the registers for x86 We have an 8-bit register called CL on x86 CPU architecture. Does it mean that physically it contain 32 bits, but we have access only to lower 8 bits? Or does this register physically have only 8 bits? So in general my question is: Are all the registers physically have the same size (say 32 bits) or not? <Q> Not really greatly researched, a quick search on x86 registers led me here , but <S> yes: <S> The <S> L in CL is for Lower, its the lower byte of the 16-bit counter register <S> CX , which itself is the lower half of the 32-bit ECX <S> register. <S> This logic applies to most registers in IA32 (not all). <S> Are all the registers physically have the same size (say 32 bits) no. <A> Back in the olden days, the 8088 and 8086 had four general-purpose registers, named AX, BX, CX, and DX. <S> Those registers were all 16 bits wide. <S> Each register could also be accessed as two 8-bit chunks, the high byte and the low byte. <S> Those 8-bit chunks were AH/AL, BH/BL, CH/CL, and DH/DL. <S> Assigning a value to AH changes the contents of the upper byte of the AX register; assigning a value to AL changes the contents of the lower byte of the AX register. <S> Then the 80386 came along. <S> It used 32-bit wide registers, but preserved the 16-bit and 8-bit features of the 8086. <S> That way, assembly programs written for the 8088/8086 could be re-assembled to target the 80386 without having to rewrite the code. <S> The 32-bit registers were named EAX, EBX, ECX, and EDX. <S> The lower 16 bits of each of those registers could be addressed by the old names, AX, BX, CX, and DX. <S> The low 8 bits of each register could be addressed by their old names, AL, BL, CL, and DL. <S> Further, the upper 8 bits of each of the lower (16-bit) registers could still be addressed by their old names, AH, BH, CH, and DH. <S> I'm sure you can guess what happened when Intel introduced 64-bit processors. <S> Yup, all the old names were still there with the same meanings, but those 8-bit, 16-bit, and 32-bit registers were each part of a 64-bit general register. <S> The names of the 64-bit registers are RAX, RBX, RCX, and RDX. <S> RAX is 64 bits wide; EAX is the lower 32 bits of RAX; AX is the lower 16 bits of RAX; AL is the lower 8 bits of RAX; and AH is bits 9-16 of RAX. <S> That's the story of the general registers in the 8086 family. <S> There are other registers with various specialized roles, and their sizes are appropriate for what they do. <S> For example, on the 80386, as mentioned earlier, the general registers are 32 bits wide. <S> The segment registers (which are used to calculate the physical address of the memory being accessed) are 16 bits wide. <A> I'm a digital IC designer, but I have no knowledge on x86 CPUs. <S> I think CL / CX / ECX are just software terms which are used to refer to different portion of a hardware entity <S> (let's name it 'REG'). <S> The 'REG' with certain bit width is the real memory device (probably flip-flops) inside the CPU circuit which is addressible. <S> Why is CL for narrower 'REG' introduced? <S> Perhaps shorter instruction is generated, hence optimized machine code can be achieved. <S> Or there're other reasons, I don't know. <S> Different machine code for CL / CX / ECX can lead to assertion of different control signals after the decode block of CPU decodes it. <S> Those control signals plus the address of the entire 'REG' allows you to access specific portion of 'REG'.
| So, no, not all registers are the same size.
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Regulatory considerations for 500W, 400V DC power tether I'm working on a robotics project where we intend to deliver ~500W of power over a 1000 ft tether. The current plan is to run ~400VDC down the tether on a pair of 16 AWG wires. These power wires will be bundled with other control wires as part of a cable assembly. We are intending to include basic safety features like fuses and overcurrent/overvoltage/temperature sensors. We're considering including some type of DC Ground Fault Interrupter. Additionally, this is not a consumer product, it's an industrial tool for infrastructure maintenance work. I have seen in this post that many of these standards are handled by UL standards, but I am not sure if complying with UL standards is a legal requirement, or only necessary if we are pursuing a UL certification. What sort of regulatory standards do we need to comply with to distribute power to our robot in this manner? <Q> That means the state, county and perhaps city electrical code. <S> Those codes are based on the National Electrical Code (NEC) but contain some additions and may contain some exceptions. <S> You need to study the codes carefully for industrial use exemptions. <S> NEC requires equipment to be acceptable to the authority having jurisdiction(AHJ). <S> The NEC has provisions calling for listing and labeling by a nationally recognized testing laboratory (NRTL). <S> UL and ETL are the most well known electrical NRTL's on that list. <S> Although there may be several NRTL'S to choose from, most of the applicable standards are published by UL. <S> Regardless of which NRTL you may want to use, you probably need to get advice from someone who is familiar with UL standards. <S> You can probably find a consultant to advise you. <S> It is best to get advice during design rather than after the design is substantially complete. <S> For an industrial establishment, OSHA requirements may be the primary concern. <S> The local AHJ may defer to OSHA. <S> You may also want to talk to your insurance underwriter. <A> Your proposed Power Cable is DC not AC .Sure <S> this has some advantages <S> But a disadvantage is Arcing .DC <S> Arcs much worse than AC generally. <A> I would as a minimum add a portable earth leakage to the AC part of the circuit - before the 300 V rectifier. <S> This would protect against any sensitive earth fault on the tether. <S> This can also be done by electronics. <S> I would also not include any storage or ripple at the source side - use a diode to block back-feed of the stored voltage into a fault on the tether.
| You need to comply with the safety codes where the equipment will be used. You should incorperate Arc Protection. The Occupational Safety and Health Administration (OSHA) maintains a list of NRTL's.
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How to avoid shorting the boards while using Probe or Multimeter? I have been working on multi layer boards at my work. Occasionally I short the boards because of probe hits a wrong spot. I measure various power rails and most of the output capacitors are 0201 and 1005 size. Can people here provide some tips on avoiding shorts while probing? <Q> Use a microprobe. <S> The smaller the width of the probe. <S> The less likely you are to hit multiple pads from its radius alone <S> A sharp one is possible. <S> If it can stab into the solder. <S> It means its less likely to slip. <S> In the past I have used sharp single point pogo pins for this end <S> Set up the measurement <S> so you do not need to look away from where your probing. <S> The moment it leaves your vision your hand positioning gets much worse. <S> If you can hold a probe to a 0.1mm spot when seen. <S> Your probably closer to 2mm unseen if you cannot stab the board for some feedback. <S> Use something similar to varnish or heatshrink to make sure only the very end of the probe is conductive. <S> This will reduce the sides from touching other parts of the circuit. <S> In complex boards. <S> I prefer rigid ones like resistor legs. <S> But for 01005 and 0201 size it would be more likely to be enamelled magnet wire to reduce the risk of tearing stuff off the board. <S> Then you measure those wires. <S> And finally if you get any say in how the devices are made. <S> Yell at the designers to include some uncoated vias for measuring the power rails with a small exclusion area of say 0.5mm. <S> Makes things much easier for barely any change to layout. <S> And as the probe can rest in the hole your less likely to drift. <A> There's not really much to say, beyond: <S> Use magnification, unless you have excellent eyesight! <S> Be careful <S> The first is probably the most important; You want a probe that will dig into the solder pad. <S> The purpose if this is twofold: You're less likely to slip <S> It will cut through the oxide layer / flux and make a good contact with less pressure, which also means you are less likely to slip. <S> And less likely to damage those little 0201's. <A> I often solder a wire to the point I want to measure (while the board is not powered). <S> Then I attach the probe to the wire; <S> Sometimes I solder a pin from a pin header to the board that I can attach the probe to; <S> The scope very ground is likely connected to mains earth unless your scope is not connected to the mains earth! <S> A battery powered oscilloscope can help. <S> I've modded a low cost (low performance) DSO120 scope to be powered from a USB Power Bank by integrating a DC-DC booster in it. <S> Add some isolation tape to cover the conducting parts that you do not want to touch.
| I anticipate debugging by adding test pin locations to the PCB while designing it; If you have test pads, you can even solder a wire on to those; Make sure that the "ground" connection of my probe can't easily loose grip and wander around - the tip rarely makes a short alone; Make sure that the board is "floating" with regards to the scope's ground. While the power is off tack on some measurement wires. Make sure you have a set of good, sharp probes (needle probes). You can also just isolate your circuit under test (while considering your safety of course).
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Does data rate depends on Frequency not on BW? If we have two RF signals of 900Mhz and 1800Mhz both of equal Bandwidth 20Mhz which one has higher data rate? <Q> The value 900MHz and 1800MHz refer only to the carrier frequency. <S> A pure perfect sine wave doesn't transmit / carry any information. <S> Only when some parameter of the carrier changes, is data transmitted. <S> This process is called modulation. <S> The parameter could be something like frequency, phase, amplitude etc. <S> More frequently you change the parameter, more data gets transmitted. <S> The frequency with which you modulate the chosen parameter determines the bandwidth. <S> So a signal with higher bandwidth transmits data at a higher rate. <A> The Shannon–Hartley theorem . <S> If the signal to noise ratio is identical for both systems and they are both presumed to use the same modulation method, the data rate (C) is proportional to B, bandwidth and nothing to do with the carrier frequency. <A> If, as you say, everything about the two systems is the same (modulation, etc.), and the only difference is in the carrier frequency (one at 900 MHz, the other at 1800 MHz), then the data rate would be the same, ideally. <S> In practice, if there's more interference in one band than other, if the performance of RF components varies from one band to the other, etc., you may observe differences. <S> But I don't think that's what you asking about in this question? <S> So, then the answer would be that they are the same.
| Two signals with same bandwidth and modulation scheme at two different carrier frequencies transport data at the same rate.
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To which voltage rail should I connect pull-up resistors? I designed a circuit which have some components with different voltage ratings. There is an ESP8266 which should connect to 3.3V. There are more components such as PCF8574 (Multiplexer which is reading active-low buttons), DS1307 (RTC), 24L08 (eeprom), ... which should connect to 5V. There are some pull-up resistors needed in this circuit. Pullup for I2C bus, PCF8574's interrupt, buttons, reset and program pins of ESP8266, ... My Question is: To which voltage rail should I connect these pull-ups? For example, PCF8574 is connected to 5V and its I2C bus is connected to ESP8266 which is 3.3V tolerant. How should I determine its rail? I2C bus is not my problem. I just want to know: Is it important which rail to pull up to? Another example to clarify: Can I use 5V to pull up ESP8266's reset pin and calculate a suitable value for its resistor? Is this really important which rail? <Q> For unidirectional lines, connect the pull-up to match what the input requires. <S> Just make sure that when the output is high impedance that it can tolerate the rail voltage because that is what it will experience. <S> If you're talking I2C busses where the signal goes both ways then you need to take additional measures because now you're doing bidirectional open collector voltage translation <S> so you probably need to get an IC to do the job. <S> There's probably a circuit that can do it too with clamp diodes and resistors <S> but I've not given it much thought <A> The pullup voltage is extremely important. <S> You need to check if a 3.3V device is 5V tolerant, and if it isn't then pullup to 5V can't be used. <S> The same must be done to 5V devices, they need to be checked if they can work with 3.3V logic levels. <S> If not then level translation is needed between 3.3V and 5V buses. <S> The PCF8574 interrupt pin is easy, it's open drain. <S> So because ESP32 only works with 3.3V IO voltages then pullups of ESP32 inputs including the reset must be to 3.3V <A> If the 5V devices can accept TTL levels (Vih > 2.0V) their inputs can be pulled up to 3.3V. <S> That’s likely the case with all the I2C pins other than the ESP8266, but check their data sheets. <S> What you don’t want to do is pull up a 3.3V <S> I/ <S> O pin to 5V. <S> That will cause a leak path though the pin’s protection diode unless the pin is designed to be 5V tolerant. <S> For translation between voltage domains there’s a number of techniques - diodes, FETs and so forth. <S> I like to use the LSF0204 chip as it’s compact and can do bidirectional signals, including I2C. <S> That said, why make your life more difficult than it needs to be?
| Every component you list has a 3.3V functional equivalent.
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Soldering a thin spring steel wire to a PCB Pretty noobish at electronics here. That said... I want to attach a very thin 1080 spring steel/music wire (30 AWG or ~0.25mm) to a pcb with a thru hole, hopefully in an automated way. The choice of spring steel is for the mechanical properties. It seems like soldering stainless steel is doable, see the below links. I think that applies to spring steel as well. https://www.rcgroups.com/forums/showthread.php?35563-Soldering-Music-Wire-!-!- ! Stainless steel soldering Soldering Nichrome wire to Stainless steel and steel to PCB Though it looks like I'd need special acid (HCl) flux for this. A few questions: Does this seem doable? Is acid flux necessary here? If yes, can that be done in a reflow oven or would this need to be done totally by hand? I think I want the thru hole diameter to be as small as possible to roughly match the wire (Say 0.5mm), though I'm not super sure how small of a hole diameter I can get. Thoughts? Is there any good way to automate this? I don't think I could do a pick and place type approach here, but is there a machine that could apply the right flux, put the wires in place, and then put the thing in a reflow oven? If this is crazy and spring steel won't work, is there a better springy material? <Q> We solder stainless steel from time to time. <S> It needs more heat than I'm really happy bringing to a PCB, and the flux (we use a HCl flux, as you say) will quickly eat the tracks off the pcb if you leave any residue. <S> Our process is to tin the steel with the acid flux, then clean very thoroughly, then solder to the PCB in a separate operation. <S> If you can buy the wire pre-tinned, all the better. <S> We can't use conductive epoxy for this for various reasons. <S> If we could, we might consider it. <S> But in my experience it comes a poor third place for mechanical strength, and good epoxies (i.e. Epotek H20E, not "silver paint for PCB repair") can be expensive and a little faffy to work with. <S> Another approach we've used is to crimp a ferrule to the end of the steel wire, then solder that to the PCB. <S> That avoids the nasty flux, and gives a mechanically very satisfactory join. <S> But it does take up a little more space. <A> Since you seem to be thinking about possible quantity production, consider sourcing plated wire which is solderable. <S> This kind of material is used for such applications as battery holder springs. <S> You'll have to consider how the high temperatures in reflow soldering will affect the spring metallurgy. <S> Soldering temperature is within the normal range of tempering for carbon steel (and much briefer), so I'm guessing it won't have a huge effect. <S> I don't have a lot of optimism about how sturdy a soldered carbon steel wire would be stuck through a PCB hole. <S> I think the joint will tend to fracture near or at the wire-soft-solder interface if stressed. <S> For small quantities consider silver soldering the wire to a ferrule (can be done easily with a propane or MAPP torch) and <S> staking/soldering the ferrule into the board. <A> Rather than soldering, consider using a conductive epoxy. <S> Example. <S> The amount of heat this needs to work is higher than most PCBs can withstand without de-laminating. <A> If you have aspirin tablet at home <S> then put tip of the still wire (spring) on the tablet of aspirin and heat it up with soldering iron. <S> Be careful as evaporated aspirin is harmful for health. <S> Once the tip of the wire is tinned, solder spring into the PCB. <S> This approach was used by radio amateurs for decades.
| Steel can be soldered using tin/lead solder and the correct flux.
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Powering a very low current application In my application I want to give power to an ATmega328p and an MMA8451 accelerometer. When active, these devices require 5.2mA and 24uA respectively, and when in power saving mode, they require 4.20uA and 1.8uA respectively. I only need to be taking a measurement once every 100ms, so as a result, both components will be in power saving mode most of the time. Also, my main requirement is to make the system as energy efficient as possible. My question is, how can I provide power to these components efficiently? What kind of power supply should I use? From what I have seen, dc/dc converters' efficiency, even the ultra-low power ones, give an efficiency of 20-40% at 100uA, and give no info about my system's sleeping current level (around 10uA). Due to application specific reasons, powering the devices directly from a battery is not an option. EDIT: The project will be powered by a LiPo (3V-4.2V). As a result the accelerometer, which takes voltages up to 3.6V, cannot be connected directly to the battery. Furthermore, the reason behind my idea of using the LiPo and not another kind of battery, is because the application also has an SD card. I will be required to write to the SD card once every few minutes, requiring bursts of 100mA or so, at 3.3V. The reason I did not mention it initially was that I planned to use a different DC converter just for the SD card, because its current requirements are well above the rest of the circuit, and because obviously there is no converter which is efficient from 10uA until 200mA. <Q> Probably your best bet is going to be to use a linear regulator of the low drop-out variety. <S> These are usually called "LDO's". <S> The parameter you need to scrutinize is the quiescent current, also sometimes given as "Iq". <S> This is the current consumed by the regulator when the output current is zero. <S> Some LDO's have very low Iq of less than 1uA. <S> Some switching regulators also have low Iq, but the best LDO's will have lower Iq than the best switching regulator. <S> Because of your usage profile, the low Iq will win over the higher efficiency during the brief instant when your devices are not sleeping. <A> Just insert a forward-biased silicon diode in series with VDD to the sensor, and run that directly from the battery. <S> Regarding your 100mA bursts <S> : using I = <S> C <S> * dV/dT <S> With I = <S> 0.1 <S> amp, <S> dV = 0.5volt <S> (sag in the provided voltage) <S> , dt = 10milliSec,we find the necessary LARGE capacitor on the 3.3 volt VDD is <S> C = <S> I <S> * dt <S> /dV = 0.1 amp * <S> 0.01 second/0.5volt = 0.002 = 2,000 uF <A> Something like S-1313A33 will only consume 0.1uA in shutdown and with 0.32V drop <S> it will let you discharge down to 3.6V without over-stressing the battery. <S> This way you don't have to bother with sleeping modes of devices other than MCU, you simply cut off their power. <S> Keep in mind that you need to 3-state MCU pins connected to other components before powering them down, as many devices do not tolerate inputs higher than VCC.
| For your specific application I'd recommend powering MCU directly from the battery and the rest of the components via LDO with "enable" pin controlled by MCU.
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switching 1 phase 230Vrms with a relay I have a 230Vrms 1 phase output and i need to switch this with another circuit with a relay.The relays are all given in ...VAC. Is VAC= VRMS or do i need to take a relay with 330VAC(Vrms*sqrt(2))? <Q> Unless otherwise specified, yes, VAC is assumed to be the RMS value. <A> This came from the DC-AC switchover (after Edison lost the war of the currents). <S> There was a huge installed base of 110VDC light bulbs. <S> "110" V AC was chosen to be the voltage at which those light bulbs would be happy, i.e. produce the same amount of net power, thermal rise, light, and therefore not burn out prematurely. <S> It would be bad press for AC if light bulbs were dimmer or burned out prematurely. <S> You know with resistive loads, power (watts) is proportional to voltage squared. <S> As the math works out, the difference is the square root of 2. <S> But anything rated in AC should be accounting for this, yes. <S> If you find a relay that is only rated in DC, you might need to deal with that. <S> But I doubt anyone would ever fail to rate a relay for AC, because AC voltage is much easier to interrupt (since it crosses zero frequently). <S> Here's the thing. <S> If you see an AC relay, don't multiply its AC rating by 1.414 and say "Oh, it can interrupt that high a DC voltage". <S> It really can't . <S> It does not ever cross zero, and so nothing snuffs a DC arc . <S> DC voltage above about 50V is a very different animal. <S> It does not know pity or remorse or fear and it absolutely will not stop, ever, until all conductive paths are utterly consumed. <S> (wait until 2:30 in this video for the second round of mad arcing, and that's only 600V). <S> Relays rely heavily on this characteristic of AC, and typical relay DC ratings are 1/4 to 1/10 the voltage of their AC rating. <S> I saw a circa-1995 design on a light rail vehicle. <S> For a 600VDC main contactor, they chose a Siemens 960VAC 3-phase contactor, and wired all 3 phases in series . <S> Even this was off-label; the manufacturer did not approve. <A> Is VAC= VRMS <S> If the relay is sourced from a reputable supplier or sourced from a reputable manufacturer and it has a proper believable data sheet then the default values for AC voltages and currents is RMS. <S> Wikipedia quote
| Yes, whenever we talk about AC, we really mean the RMS value.
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Daisy chaining ground to buttons I am building a small controller box based on an Arduino or similar MCU board. There are upto 6 buttons on the controller. In regards to ground, can I daisy chain the ground round all switches? Fundamentally it seems to work ok.For example:Arduino GND--> Btn1 --> Btn2 --> Btn3 etc. Or is it better to have a single wire to each button from the Arduino GND pin. This seems an easy way in terms of less wiring. The wire is 7/0.2 (24AWG). What are the pro's and cons here? Additionally, there are 3 GND pins on the Ardunio, why would you use all 3? <Q> You can use single ground wire that connects to all buttons in a chain. <A> You could mitigate this by adding another GND wire in parallel of the first one (the probabilities of failure happening twice at a given location are low, except if you have those wires through a mechanism). <S> With AWG24 over short distances <S> (you say in a box), you don't even have to take into account the resistance of the wire - and at the frequencies you certainly deal with (since those are mechanical buttons), propagation times of signals do not matter either. <S> You're fine. <S> Note <S> that the above ONLY applies to buttons, biased using an extremely low voltage (<50VDC according to BS 7671 18th ed.) <S> using a pull-up sufficiently strong to force next to no requirement on signal quality. <S> This is the case when driving directly from IO pins with internal pull ups enabled on microcontrollers such as Arduinos. <S> If the number of wires is important to you, you could even chose to use Charlieplexing. <A> Daisy-chaining +ve / -ve lines to inputs / outputs of a controller is standard. <S> Daisy-chaining safety ground wires between equipment is a no-no <S> as disconnection at one point could render all of them a shock hazard. <S> Single point grounding is a must. <S> Daisy-chaining ground wires also causes noise and interference in signal paths between systems.
| Daisy chaining the buttons will introduce a single point failure in your system: if the wire fails (cut or otherwise) anywhere along the buttons chain, none of those beyond that point will work.
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Smallest 5V 2.5mA supply from 110VAC I need the smallest power source possible to get 5V to power ATTINY402 and one 100K pullup and turn ON a MOSFET with a 2K resistor for 5 seconds once a day (to switch ON 1W LEDs run by a dedicated LED driver like AL5890 ) So the continuous power consumption will be around 0.5mA with peak of 3mA for 5 seconds. The power supply does not needs to be an isolated one as it will be enclosed in a sealed container and requires no user interactions. The power supply needs to be inductor less and should not require an EMI filtering cap. Since the current consumption is so low the Power Factor and Efficiency should not matter that much right? So its OK if the they are low. The preferred size is less then or equal to SOP8 package.If there are no external components then that is preferable. If this power supply requires a DC source like a bridge rectifier then it is already in our circuit so that should not be calculated in the size consideration. simulate this circuit – Schematic created using CircuitLab What I have considered so far: Capacitor dropper circuit: The reason I did not choose this is because the capacitor needs to mains rated and they are pretty big so I have discarded that thought. Resistor dropper Circuit: Placed as the last resort if nothing else is available. SR086 Based power Supply: Requires a lot of external components and pretty big 470uF electrolytic capacitor at the output. Small Transformer:I could not find a very small mains rated 60Hz transformer that could fulfill my requirements. Suggestion: 5. MP100L : This looks promising but it requires an X capacitor at the mains to keep the EMI under Class B (Which is the requirements). SMJR-N-1-24 :I used this LED driver in my previous circuit and it provides 5V for my ATTINY as it has an internal 5V auxilary supply inside. So I was thinking maybe there is a chip that has only the 5V power supply circuitry inside. Ideally I need a small chip that can give me 5V 2.5mA power from 120VAC. With minimal external components. <Q> There are LDOs available for that purpose. <S> Some examples, I know of:- NCP785A, NCP786A from On Semiconductor- TPS7A78 from Texas Instruments- LR8 from Microchip <S> The devices from On Semi for example only need a diode and two caps to deliver 10mA of stabilized power at the rated voltage, eg. 3.3V, 5V, 12V, 15V, directly from the mains with 85VAC to 260VAC or from a DC source with 25V to 450V. <S> So, they should work everywhere. <S> The LDO is 4.1mm x 2.5mm in size. <A> With a capacitive dropper you would be using a 47-86nF X2 capacitor <S> (depending on how much margin you want for aging), (peak voltage / current = impedance), thermal dissipation would be about 0.5W, which is plenty cool for this size capacitor, looks to be as small as 10x6x12mm from most suppliers for this size capacitor For a resistive dropper it would be something around 47K, again about 0.5W, so probably use a 2 or 3W resistor for both the wattage and the voltage rating. <S> For both of these, afterwards you need to rectify the output, and use say a small linear regulator / LDO to bring it to exactly 5V, If you drop the requirement of no external components, to just as compact as reasonable, then there are other options. <S> but they likely won't be smaller than either of these. <A> Why not just use another LED driver? <S> simulate this circuit – <S> Schematic created using CircuitLab R5 provides a 5.6 V drop and D1/C2 protects U3 from dropouts (at, e.g., zero crossings). <S> Increase C2 as needed to support peak loads.
| For what your asking, your perfectly in the use cases for a capacitive or resistive dropper, There may be bucket brigade chips that could work, but they will need external capacitors and some other passives.
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Checking amps with manual-ranging meter: Jack/dial consistency required? Multimeter: UNI-T 89XD manual ranging I received a comment or two that said I will not get an accurate amp reading if I use the 20 amp jack and set the dial to anything other than the 20 amp setting, such as 60 or 600mA. I have used the 20 amp jack and set the dial to 60mA and did not get an OL on the screen . I got mA readings. What is dumber, me or the meter? (Don't answer that!!!) How true is the admonition that I will not get accurate readings if the 20 amp jack is used with other amp dial settings? <Q> The 20A socket has its own shunt resistor to common terminal. <S> It measures the current by measuring the voltage over the shunt resistor. <A> The 20A current path is not even switched internally (you do that when you move the probe), so you can't expect a reasonable reading. <S> Likely you'd get something like the expected count with a decimal point position that depends on which incorrect position the switch is in. <S> Of course if you leave the probe in the 20A setting and attempt to measure voltage bad, or very bad, things can happen depending on the source of the voltage, since the 20A current path is pretty much a direct short between the probes. <S> Some meters have an alarm if you set the knob to volts with a probe in the 20A socket. <A> If you're trying to measure 15 amperes and need accuracy to within 10mA, that meter likely won't do it <S> (I didn't go look at its specifications). <S> The ranges available affect the accuracy because they use different resistors internally. <S> The basic way that current measurement works is by putting a low but known resistance in series with the load, and then measuring the voltage difference across that resistance. <S> The dial is selecting different resistors as well as selecting which port is used (20A uses its own port, bottom left of the unit; all other current measurements use the <S> mA µA <S> port above it). <S> Therefore, measuring current with the probe in the 20A port but the dial on something else will probably result in random values (noise). <S> When you select 20A (and use the correct port), the meter is using a higher-current fuse and resistor, which provides less resolution for low currents. <S> In other words, measuring 50mA with the 20A setting may result in a reading of 0.0 A (it's too little to register) or potentially something like 0.1 <S> A (if it rounds up or something weird). <S> If you do the opposite, measuring a current that's higher than the dial setting and port, you'll blow the fuse. <S> Meters like this have two fuses for current, a 20A fuse for the 20A setting, and a 600mA fuse for the others. <S> As @Spehro said, don't measure voltage with a probe in either current port . <S> The 20A port should not be thought of as a "one size fits all" port.
| If you select the normal socket for measurement source with the dial, the multimeter may not even be measuring the voltage that is caused by current via 20A socket, or if it is, the measurement will be wrong, since it expects to see voltages in a different range from the normal shunt resistance. This depends on what you are trying to measure and how accurate it needs to be.
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Can I control speed of reduction motor? I am planning to use this motor for a project but I am not sure if I can control the speed of this motor . This is 100rpm motor and I would like to use a potentiometer to control speed. <Q> However the proper way to do it would be with a PWM controller, it switches on the supply voltage X% of the time, proportional to some analog input, say a potentiometer, this keeps the motors speed more consistent over varying loads. <S> Edit:Tinkercad exploration image attached. <A> It would be possible to control the speed of the motor, powered by a LM317-based 0-3V adjustable power supply, using the potentiometer in the power supply itself. <S> A flyback diode would be required across the motor. <S> Here's a reference to build a 0-3V adjustable LM317 power supply without the minimum 1.2 - 1.3V output voltage limitation. <S> https://www.edn.com/use-an-lm317-as-0-to-3v-adjustable-regulator/#id2782607-48-a <A> Assuming you don't want to make a whole project out of this speed control, I would consider some all-in-one solutions. <S> There are many off the shelf motor speed controllers sold on websites catering to hobbyists, for example Amazon, Sparkfun, Pololu and Adafruit. <S> Here's an example: Onyehn-Controller-Control-Switch-Dimmer <S> You could use an adjustable voltage regulator to change the speed of the motor. <S> Because your motor is only 3v this might give you the most options. <S> Some examples: LM2596s Buck Converter <S> Another LM2596 Buck Converter <A> You could use a potentiometer to control the speed. <S> However the potentiometer track must be rated to handle the current, and to get good control at lower speeds it will have to waste a lot of power. <S> The rated current draw of your motor is 0.08 A. <S> At light load it will be less, perhaps ~0.04 A, and at stall 3-5 times higher, perhaps ~0.3 A. <S> The pot must be rated to handle the stall current on any part of the track. <S> \$P = <S> I^2 <S> * R\$ , so a 5 watt 25 ohm pot should handle up to ~0.4 A. <S> The next question is, how much voltage variation can you expect with different loading? <S> This depends on the potentiometer's output resistance, which is highest at the midpoint. <S> The equivalent circuit might look like this:- simulate this circuit – <S> Schematic created using CircuitLab <S> If the motor drew no current then the potentiometer would output 5 V / 2 = 2.5 V , and it would draw 5 V / 25 Ω <S> = 0.2 <S> A and waste 5 V * 0.2 <S> A = 1 watt . <S> The Thevenin equivalent resistance is 12.5 / 2 = 6.25 Ω, so if the motor draws 40 mA the voltage will drop by 0.04 <S> * 6.25 <S> = 0.25 V <S> and it will actually only get 2.25 V . <S> At 80 mA it will get 2.0 V , causing it to run ~ 20% slower than expected. <S> This may be acceptable, but the pot will still dissipate ~1 W, 6 times more than the motor! <S> You could use a higher resistance pot to reduce quiescent power consumption, but then the voltage variation with different loading will be higher. <S> If it is too high the motor might not even start up due to not getting enough current at stall. <S> A higher resistance pot will also have a lower current rating for the same power rating.
| As its a small motor, you could control its speed with a "Rheostat" essentially a higher wattage potentiometer, I would highly recommend using a switching regulator, because it will be much more efficient, and the whole thing won't get crazy hot.
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What does it take to short a battery I understand that low resistance through a circuit will create a short circuit but what combination of voltage and amps which causes this? For example, if I had a circuit with just a wire and a battery and it had a high voltage and low amperage, would this cause a short circuit in comparison to a circuit with low voltage and high amperage? <Q> 'Short Circuit' gets used in two different ways. <S> In the context of a battery (or any power source), we usually mean it to be a load that is far too large for the source. <S> A short circuit usually produces damaging conditions for the battery, and the load, if maintained for enough time. <S> At best, the battery will be run down quickly. <S> At worst, the battery may catch fire, burst itself or its container, or the load start a fire. <S> The wiring to a high current battery, like a car battery for instance, will invariably be protected by a fuse, which opens in the event of a short circuit. <S> The wiring to a low current battery may not need protection, if the short-circuit current is low enough for any practical wire. <S> Given this, there may be some sense, hinted at in your question, that for high current batteries, a short circuit is an issue, where it is not for low current batteries. <S> For instance a PP3 or CR2032 battery, while it will be run down by a short circuit, is most unlikely to start a fire as a result. <S> In circuit analysis, a short circuit is an ideal zero resistance, that will support any current with zero voltage across it. <A> Outside of formal circuit analysis, there is no universal definition of short circuit in absolute terms of voltage, current or resistance. <S> In most contexts, a short circuit is not a mathematical function or limit, it's just a general term for a type of failure or behavior. <S> Current tends to prefer the path of least resistance, which can also be thought of as the "shortest" path. <S> If a significant portion of the current is able to take an unintended or intended "shortcut" between the poles of the supply, and therefore never reaches the intended load, then that situation may be called a short circuit. <S> If the total resistance of the circuit is "too low", or "abnormally low", or "much lower than intended", then you may also be justified in saying, at least informally, that the circuit is "shorting out" or "shorted". <A> For example, if I had a circuit with just a wire and a battery <S> and it had a high voltage and low amperage <S> , would this cause a short circuit in comparison to a circuit with low voltage and high amperage? <S> To answer this we need to look at what limits the battery's ability to supply infinite current. <S> The model of internal resistance is useful for this. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The short-circuit current of a battery will depend on its voltage, chemistry, size and internal structure. <S> We can usually simplify this to a simple model of an ideal voltage source and an equivalent series resistance. <S> It should be clear from the model that the voltage at the battery terminals will droop with increasing current. <S> With the values I've made up for Figure 1 you can calculate that the 9 V battery will limit the current into a dead-short to \$ <S> I = \frac <S> V R = <S> \frac 9 {20} = 0.45\ \text A \$ . <S> Running the same calculation on the car battery will give us \$ <S> I = \frac <S> V R = \frac {12} {.025} = <S> 480\ \text <S> A \$ . <S> To recap: the short circuit current is a function of several variables but is mostly determined by the nominal voltage and internal series resistance. <A> If the positive and negative terminals are connected by a wire then the battery is by definition shorted. <S> What the voltage of the battery is does not really matter. <S> If the current is very high then that means that the battery has a very low internal resistance. <S> If the current is low then that just means that the battery has high internal resistance. <A> If you can draw and solve a circuit "properly," the line between a short circuit and a really high amperage line is very blurry. <S> As others have said, there's no official line, but I find the most useful line is the point where the high-current effects in the circuit exceed the modeling of your circuit. <S> If you have an ideal voltage source hooked up with a wire across both sides, that is a "short circuit" because that model cannot properly describe what really happens. <S> On the other hand, if you have a battery and a wire with a specified resistance, then that is merely a "high current" circuit. <S> That holds up until the thermal effects of that high current start to change the real wire's resistance. <S> So on and so forth. <S> Put a wire across your car battery, and we think of it as a short because we really don't have a good sense of the resistance of the wire and the internal resistance of the battery. <S> But quantify that resistance, and add a shield of argon, and you have a MIG welder that relies on those properties!
| Any battery, whether a high voltage or low voltage battery, will be 'short-circuited' by putting a low or zero resistance load on it.
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Why would this microphone only work while touching the volume control? My son bought this USB microphone for his computer. It only works if you keep your hand on the volume knob. I assume this is a grounding problem. I took it apart. It looks like the copper wire attached to the housing is there for grounding - it attaches near the rear of the volume knob. But touching that wire or touching any part of the interior of the knob does nothing to get the sound to work; only the front of the knob makes the sound come through. What can I connect to what to get it to work without having to put a finger on this thing? Thanks! <Q> If it is a grounding issue, touching it could change the amount of hum, buzz or interference. <S> However you say "It only works if you keep your hand on the volume knob" which I interpret as "there is sound only when the hand is on the volume knob <S> , otherwise there is no sound." <S> If this is the case, then I'd suspect a bad contact perhaps at the pot wiper, or a cold/broken solder joint that makes contact only when the volume pot is messed with. <S> You can check this by touching the volume pot with a non-conductive object like a plastic rod. <S> If wiggling the knob around with a non-conductive object results in the sound turning abruptly on or off, then you can rule out any influence from your conductive finger, it'll probably be a bad contact. <S> If the failing contact is at the pot wiper, it will depend on the pot's position. <S> So if the problem does not depend on the pot's setting, it is more likely this is a bad solder joint on the board. <A> Repair questions are off-topic on this site <S> so let's make it an education question. <S> Figure 1. <S> Potentiometer terminals 1, 2 and 3. simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> A schematic showing the potentiometer arrangement. <S> As the wiper (2) is adjusted from (1) to (3) <S> the volume goes from zero to maximum. <S> If pressing on the potentiometer causes the system to work then it's possible that the wiper has lost contact with the resistance track. <S> You can test this safely by turning the knob to mid-position and connecting (2) to (3) with the tip of a screwdriver. <S> If that works the potentiometer is faulty. <S> If you're feeling brave you can desolder the potentiometer, bend the tabs folded underneath and disassemble. <S> You might then be able to bend the wiper a touch so that it springs onto the resistance track more firmly. <S> A spray with some contact cleaner would be good if you have access to it. <A> The First Photo "looks" like it's broken... <S> But assuming it's not, this behaviour looks like the potentiometer (variable resistor) is broken. <S> The pressure of the finger pushes the wiper contact on to the track completing the circuit.
| It's also possible the solder joints are "dry" and need to be reflowed Otherwise replace the potentiometer or the whole device would be the best advice. You can use the shell of a ball point pen, or a toothpick for example.
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Current Sensing on positive and negative voltage rails I'm working on a design that uses two separate buck regulators to generate a positive and negative voltage, and I'm trying to implement current limiting with a current sensor. However, I am running into issues with the way the sense resistor is integrated into the feedback loop as shown in my block diagram below: simulate this circuit – Schematic created using CircuitLab With the way the regulators are set up, the current sense resistors have to be included in the appropriate feedback loops in order to maintain the correct output voltage. Otherwise, the regulator won't take into account the voltage drop, and the output voltages will be less than what is expected. The issue is with the way the two regulators are configured. When you add the sense resistor to the negative voltage circuit, it seems to create a different 'return node' for the negative voltage. If it wasn't there, both regulators can be tied to the ground node at their returns. What other methods of current sensing can be used for the negative voltage circuit such that the appropriate returns can be tied together? EDIT: As Hacktastical said, there is a way to have a buck regulator output a negative voltage. I based the reference circuit off of an app note from TI: Positive to Negative Buck-Boost Converter <Q> Without going into too much detail, your approach violates the ESR requirements with an external sense R and also that is redundant with the internal Rs. <S> There is a separate port to provide gain Ri for the OCP threshold with internal current sensing. <S> If you want to sense current on your own, use a 2~ 10 mOhm max ground sense resistor and high gain balanced Diff Amp with low ESL and a good layout to reject CM noise. <S> Total ESR on the load cap plus Rsense must be between 10 and 130 mOhm (?) from memory. <S> So your 1 Ohm will result in instability and poor load regulation. <A> I think what you've shown is fine. <S> What will happen is, when there is more IR drop on NEG_SENSE, this will cause Vf to shift down, making the regulator try to compensate by increasing its output voltage. <S> NEG_V increases as a result. <A> You are adding a series resistor into the output and expecting the feedback loop to offset the effect of it. <S> The chip does not have enough response to be able to do this. <S> The LM2673S already has short circuit protection built in, please read the datasheet . <S> The current protection is NOT in the output circuit, it limits the current through the switch and therefore the peak current that flow in the inductor. <S> This can be set to provide just enough energy into (c1,C4) and (C2,C3) to allow the PWM to maintain your required output voltage. <S> If your output current exceed a level that can be maintained by the maximum PWM on time, then the voltage will begin to fall. <S> You must also remember that in the case of a fault current (ie a sudden short circuit on the output) the current is defined by the output capacitors. <S> They will deliver a large pulse of current under these conditions. <S> Note <S> : You should also include a capacitor from the Vin to -Vout as shown in the application note on using the LM2673 for negative supplies. <S> Without this capacitor, startup will be distorted.
| The schematic you show will most certainly NOT work in the way you want.
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Using diodes to wire a battery pack in series and parallel for different voltages This is my first post on this site.I have an EV design were I want 900V for a traction motor, but I need 450V for auxiliary equipment. The first thing that came to mind is DC-DC converter, but it is expensive, and possibly has a notable power loss. I thought about alternatives and about splitting up battery packs: For example, lets say I need 36V for a BLDC motor, and 12V for a Arduino.I use 3 12V batteries wired in series for 36V, and use diodes to wire them in parallel for the 12V.The diodes stopping the batteries from shorting. I know diodes have a considerable voltage drop, and for the EV application I would use ideal diodes. By using the diodes, all batteries should drain equally, avoiding the battery pack unbalancing. In the EV, the 12V batteries would be separate modules with their own monitoring. Is this a crazy idea?Would the battery explode for some reason?And should I just use a DC-DC converter instead? <Q> The diodes stop the batteries from shorting to each other, but they also deliver 36 V to your '12 V' output. <S> If your low voltage drain is very, very small, say less than 1% of the drain on the whole pack, then you could maybe supply it from one battery, and rely on the charger to rebalance the cells when you recharge. <S> Best to use a DC to DC converter running from the whole pack to generate your low voltage supplies. <A> Figure 1. <S> The diodes marked with an 'X' are reverse biased and do nothing. <S> The output from this circuit will be 36 V. <S> (Note that it's a good idea to give component designations so that we can refer to D1, etc. <S> rather than "the third diode from the top".) <S> ... <S> and for the EV application I would use ideal diodes. <S> Ehm, you do realise that there is no such thing? <S> No, this won't work as explained above. <S> I don't think you're ready for 900 V work. <A> Compared to the coast of the vehicle, a 36 VDC to 12 VDC converter is negligible, less than US$20 for this one found on a quick search. <S> Not only will it drop the voltage, but t also ensures that there is not too much drain on one group of cells, so that charging can be more predictable. <S> You must also know the answer to some questions to know if this is suitable: Must the output be isolated, or can it share a common ground or B+? <S> What is the current requirement?
| By using the diodes, all batteries should drain equally, avoiding the battery pack unbalancing.
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