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How do I speed up transistor switching in low side power switch? I am working on a project and want to drive a motor with a 12V PWM signal (Motor connected to 12V and the Motor- net on the diagram). The circuit below is an idea I have, however after I built it, it turns out it has quite a slow turn on and turn of time. What are some methods of speeding that up? Thanks in advance, Rob Edit 1: Adding scope plots of input and output as requested. Plot 1 - Between R31 and U4 Plot 2 - Probe at "Motor-" Edit 2: Adding schematic of circuit to the right (Current sensing). <Q> While you do not make clear what the voltage of the PWM is, The voltage on R21 should be half the supply voltage. <S> this will remove the influence of any capacitance at the inputs from slowing down the transition, <S> Next up, would be loose R32, the whole point of U4 is to slam that mosfet on and off as fast as possible, that resistor is preventing it, with a normal op amp it would be recommended, but this device is intended to deal with capacitive loads. <A> At first, before I get to the main point: According to the data sheet, the FAN3100 is not a comparator. <S> The inputs are logic inputs. <S> So, the 3.3V at R21 to the IN+-Pin seems to be all right and should not effect the switching speed. <S> My suspicion is, that the switching time is slow, because the gate charge does not get reloaded fast enough because the current is to low. <S> The driver applies 12V to the gate with a driving capability of up to 3A. The 22Ohm resistor R32 limits this to about 0.5A. <S> Imagine the gate as a capacitor, that must be charged from 0V (off state) to 12V (on state). <S> R32 and the gate capacitance <S> (3247pF according to the data sheet) form a shunted capacitor. <S> You can calculate the pulse response with the usual formulas. <S> As a result, I would change R32 to 4.7Ohms (resulting in 2.55A driven into the gate) and increase the bulk capacitor C10 to a value of 10µF to 47µF. <S> The current pulse to reload the gate, comes almost entirely from this capacitor. <S> It must provide the 2.5A that are driven into the gate. <S> A ceramic capacitor does not work as C10, it must be a bulk polarized electrolytic or tantalum capacitor. <S> Chose the lowest ESR you can get or place two capacitors in parallel instead of only one C10. <S> Having to much capacitance at this point does not do any harm. <S> The capacitors should be placed as close to the driver IC as possible. <S> Be also aware that ceramic capacitors show a DC-bias effect. <S> So C1 on your circuit must be chosen accordingly. <S> Many types of ceramic capacitors have only 10% of the original capacitance at voltages of 12V and above which makes them practically inoperative. <S> Some additional point to think about: Do you really need such a big fat FET? <S> Its capacitance is quit high, which slows down the switching time. <S> Its rise and fall times are also rather slow. <S> I cannot imagine, that you really need 110A in your application. <S> Rather lengthy answer, I apologize, but I hope it will help. <S> Good luck and all the best! <A> The slowly ramping voltage is what I suspect will happen when the MOSFET is turned off - when you deactivate the MOSFET the motor momentarily turns into a generator. <S> Normally, with an inductive load you will get a back-emf that forward biases the protection diode <S> but you also have a snubber that can do the same thing <S> but somewhere between the snubber and the motor acting as a generator when the MOSFET turns off I suspect you get this effect. <S> I suggest you try and model the situation.
| So, choosing a 'smaller' FET with less driving capability and therefore less gate capacitance and faster rise and fall times, should also help to increase the switching speed.
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Why are one channel's transistors in a power amp getting much hotter than the other's? (Context: I don't have a background in e.e., but getting better at basic bench work.) I'm repairing/recapping an early 70s German integrated stereo. The power amp output section has four large heatsinks for its transistors, two each corresponding to the left and right stereo channels. After a very short period of time, even a couple minutes, the left-channel's sinks become extremely hot to the touch while the right channel remains cold. The audio output is correct and expected, a typical radio or aux music signal amplified in stereo. Sound on both channels is totally fine coming out of the speakers. I only noticed the heating asymmetry by accident. I don't know that this is a problem that needs solving, exactly, but the heat in that area seems likely to reduce the life of the new left channel caps somewhat. I'd like to understand if (1) this indicates a problem and (2) what might explain it? Could those power transistors be deteriorating unevenly with age? Could something else that feeds them be behaving strangely? (I've replaced all the electrolytics and a couple questionable looking film caps on this board, for what that's worth.) Any insight or theories/pointers/assistance very welcome. I have a bench oscilloscope and can probe signals in an entry level kind of way if that's useful but not sure what I'd be looking for in this case since output seems fine. Worth noting that the differential in heat remains over long durations-- the right channel does eventually (an hour?) become perceivably a bit warm to the touch, but the left channel becomes scorching very quickly and stays that way. (This is also true if the left speaker is not even plugged in, fwiw) Photo of the board and image of the schematic for this board below. The left channel is the left half of the board, and the two left heat sink sections. Schematic. Input from preamp at left, output to speakers at right. Left channel is top half of the diagram. T708 (and maybe T709? maybe T705?) appear to be the very hot transistors. <Q> See the resistors I've marked with red boxes: - With no signal at the inputs, use a DVM (on DC millivolts) and measure the voltage across each of R721 and R621 - they should be about the same - maybe 5 mV to 20 mV. <S> If the hot channel has a significantly larger voltage then try adjusting its associated bias current trimpot (the other red boxes) to reduce it to the "good" channel. <S> I think R721 and R621 might be <S> these 0,24 Ω resistors marked in green but double check the PCB legends: - Also up for consideration - do the transistors get hot when the speakers are disconnected, connected, or both? <A> Digging down way deep in my knowledge base, but... Looks like what you have is a push-pull (totem pole) output stage, with the two transistors in the top (T706, T708) and bottom (T707, T709) <S> legs of the totem forming a darlington pair. <S> These darlingtons should be biased so that with no signal, both the top and bottom pairs of transistors are conducting, just a bit. <S> This is done to reduce the distortion (crossover distortion) that would occur if the output transistors were operated in a pure B mode. <S> This distortion occurs because with pure B type operation, both sets of transistors would be OFF near the 0 V point of the waveform, creating a couple of volt dead zone in the output. <S> By biasing the outputs transistor pairs to be slight ON at near 0 V, this distortion is rediced/eliminated. <S> The downside is that because both transistor pairs conduct near 0 V, there is higher dissipation in the output transistor due to the current that flows from the top to the bottom pair. <S> The resistors Brian Drummond mentioned (R715, R615), along with the transistor, establish this quiescent operating point, the amount of current that flows through the top and bottom pairs, and so impacts the power dissipation. <S> If this quiescent point is not set correctly, you have excessive current flow (sort of a shoot-thru affect) that increases dissipation. <A> Trimpots R715 and R615 are set differently but they do the same thing for each channel. <S> Try adjusting R715 to look like the setting of R615. <S> Adjusting to match the same voltage across R721 and R621 would be best.
| If they only get hot when the speakers are connected then it might be that C709 (the output DC block capacitor) has gone low resistance.
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do we only use pull-up resistor or pull-down resistors when dealing with switches? I have heard people state that if your micro controller is not reading the signal properly, then we need to use a pull-up resistor. Which I don't get? It makes sense when switch is involved because you don't want a floating signal. What I don't understand is why do my friends mentions the pull up resistor help in better voltage reading?? <Q> It would help if you give more specifics as to where have you "heard" that and in which type of circuit. <S> In short, we don't use pull-up or pull-down resistors only with switches because of their floating state, but also with other components (including, but not limited to, electrolytic capacitors and transistors) because of their leakage currents. <A> Sometimes it does not matter if you use a pull-up or pull-down resistor to keep default state of the data pin, and the pushbutton will overwrite that with another potential. <S> Inside a plastic case you can have buttons wired with pull-downs and the buttons connect to common +5V wire. <S> Some MCUs have internal pull-downs so you can omit external resiators. <S> However, it is far more common to find MCUs that only have internal pull-ups. <S> And, if you want the buttons to be in a metal box, you most likely have the metal case grounded, and so it will be safer and easier to have buttons pulled up and switching to ground, to avoid short circuits of 5V to grounded metal case. <S> Some chips, devices and buses can only pull to ground or let the wire float, as they have open-drain or open-collector outputs, so they need pull-ups for sure. <A> Do we only use pull-up resistor or pull-down resistors when dealing with switches? <S> Yes, but the switch can be an electronic one. <S> We use them wherever the input can only pull one way. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Various configurations requiring pull-up resistors. <S> Each of the examples above can only pull the input low. <S> As a result a pull-up is required.
| I can only say that while with a switch you have a floating input, with other components you have a leakage current which pulls the node up or down, so a pull-up or pull-down resistor serves to counter the effects of that leakage current.
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Is the transistor Q4 in the wide swing cascode current mirror operating in linear region? In the wide swing cascode current mirror shown below, the transistor Q4 is in the linear region of operation according to the voltages at gate and drain. Is it not necessary for transistor to operate in saturation region? Is it ok to operate in linear region as I am not finding any issue? <Q> \$V_{D4} = <S> V_t <S> + <S> V_{OV} \$ \$V_{G4} <S> = <S> V_t <S> + <S> 2V_{OV} \$ <S> \$V_{S4} = <S> V_{OV} \$ . <S> Hence: \$V_{GS4} = <S> V_{G4} - V_{S4} = V_t <S> + <S> 2V_{OV} - V_{OV <S> } = V_t <S> + V_{OV}\$ \$V_{DS4} = <S> V_{D4} - V_{S4} = V_t <S> + <S> V_{OV} - V_{OV <S> } = V_t\$ . <S> Q4 is in saturation if: \$V_{DS4} >= <S> V_{GS4} <S> - V_t\$ what leads to: \$V_t <S> >= <S> V_t <S> + V_{OV} <S> - V_t\$ <S> \$V_t <S> >= <S> V_{OV}\$ . <S> Hence A) <S> if \$V_t <S> >= <S> V_{OV}\$ <S> then Q4 in a saturation B) <S> if \$V_t <S> < V_{OV}\$ <S> then Q4 in a linear region <S> Is it not necessary for transistor to operate in saturation region? <S> No. <S> See explanation below. <S> Is it ok to operate in linear region as I am not finding any issue? <S> Yes, it is ok for a transistor to operate in a linear region. <S> Depends on the situation. <S> Generally, during designing you try to have saturation in all transistors due to e.g. high output impedance. <A> The pre condition for being in the saturation region is the following: $$V_{GS}-V_{TH} = V_{OV} <S> < V_{DS}$$ <S> By applying only \$V_{OV}\$ to the gate of \$Q_2\$ <S> you would be in the threshold between cut-off and linear regions, <S> therefore an additional \$V_t\$ is applied. <S> Based on that you can say that \$Q_2\$ is at least conducting. <S> In order to make sure that \$Q_2\$ is in the saturated region, its \$V_{DS}\$ must be greater than its \$V_{OV}\$ , and this is accomplished with the \$V_{BIAS}\$ . <S> In this case \$V_{BIAS}\$ has to account for the \$V_{OV}\$ of <S> \$Q_4\$ and the minimum required \$V_{DS}>V_{OV}\$ of \$Q_2\$ , hence: $$V_{BIAS} = <S> V_{OV <S> ,Q_4} + <S> V_{DS,Q_2}$$ $$V_{BIAS} = <S> V_{OV <S> ,Q_4} + <S> V_{OV <S> ,Q_2} <S> + V_{t}$$ <S> Assuming that all transistors have the same characteristics: $$V_{BIAS} <S> = <S> 2\cdot <S> V_{OV} <S> + V_{t}$$ <S> Now it is clear that \$Q_2\$ is in the saturated region. <S> What about <S> \$Q_4\$ ? <S> \$V_{BIAS}\$ ensures that \$Q_4\$ is either in the linear region or saturated one by applying an effective \$V_{GS,Q_4}\ = V_{OV}\$ . <S> In order to drive \$Q_4\$ into saturation, you just have to apply a voltage slightly larger than its \$V_{OV}\$ , hence: <S> $$V_{D,Q_4} <S> > <S> V_{OV} <S> + V_{t}$$ <S> Based on that you can say that \$Q_4\$ is in saturation. <A> Condition for saturation is \$V_{DS}\ge <S> V_{GS}-V_t = V_{ov}\$ .For <S> \$Q_4\$ , the condition implies \$V_t \ge V_{ov}\$ . <S> This is usually satisfied in normal circuits. <S> The <S> \$V_t\$ is usually of the orders of 400-500mV (40nm TSMC) and \$V_{ov}\$ <S> is of the order of 100-200mV. Choosing a higher overdrive voltage <S> compromises the swing of your circuit. <S> Your circuit behaves as a good current source only for output voltages within \$2V_{ov}\$ and \$V_{dd}\$ . <S> Thus it would be better to use smaller \$V_{ov}\$ . <S> You may think in this case to use wider devices and choose lower values \$V_{ov}\$ . <S> But there is a catch, your noise performance will get worse (because \$g_m\$ goes up) as you reduce \$V_{ov}\$ for this current biased situation. <S> Thus there is a trade-off between swing and noise. <S> To optimize swing along with noise, you can choose different overdrive for cascode and the tail transistor. <S> Since the noise of the cascode transistor is negligible due to tail degeneration, you can choose wide device for cascode to reduce its overdrive without much noise penalty. <S> And now, you can choose narrower tail transistor to ensure low noise and still have same output swing.
| However, you may want some of your transistors to operate in linear region due to their functions (e.g. resistor behavior) or there is no big difference between behavior of a given transistor between linear or saturation region for a given situation or linear region may be forced by the architecture as is in the case of your current mirror.
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Charge a capacitor with negative pulse What will happen if I apply a negative pulse to the capacitor? Is the capacitor voltage increase positive or negative? The pulse which is shown in the below picture is the output voltage pulse from a photodetector (photo multiplier tube.) The detector is consists of a photoelectric plate (that gives an electron when a light incident on the plate ) and electron multiplier (that multiplies the number of electrons or amplifies the signal.) The multiplied electrons are collected in one end across a resistor as a voltage pulse. My aim is to measure the amount of charge from the detector. So I use this signal to charge a capacitor then I do some other analyses to measure the charge. <Q> It depends on your capacitor and the corresponding circuit. <S> All capacitors have built-in series resistance (ESR), and the corresponding RC circuit (not to mention the rest of your design) will either smooth that spike into a tiny bump, or it'll look almost identical to the input. <S> Different capacitor sizes, types, and even the physical dimensions will impact their transient response. <S> The physical design of the capacitor will also react differently to negative voltages. <S> Standard ceramic capacitors won't care about negative voltages as long as the amplitude is within their design parameters. <S> You might also look into a Coulomb Counter for your design - they're designed to track the charge of batteries, but some are extremely accurate and will factor in their own consumption. <A> Negative voltage means lower potential then reference. <S> So the direction of current flow will be from the reference level towards the negative voltage. <S> So electron flow will be the opposite. <S> It does not matter if the input voltage is positive or negative for a non-polar capacitor (For example ceramic capacitor). <S> Charges will be stored in it. <S> But for polar electrolytic capacitor, you have to be careful. <S> They can not be used in bipolar cases. <S> If the negative terminal is at zero volts, then the positive terminal must be zero or greater than zero volts. <S> If you apply a negative voltage, then it will explode . <S> If the positive terminal is at 0 volts, then the negative terminal must be zero or less than zero volts. <S> If you apply a positive voltage, then it will explode . <S> I have experienced polar electrolytic capacitor explosions a few times. <S> There's also non-polar electrolytic capacitor available in the market. <A> Will the capacitor voltage increase is positive or negative? <S> The capacitor voltage becomes negative. <S> There is nothing surprising or complicated about this. <S> All voltages are measured relative to an arbitrarily chosen reference potential, so negative voltages are entirely common and a normal part of circuit analysis. <A> that detector is consists of a photoelectric plate (that gives an electron when a light incident on the plate ) and electron <S> multiplier(that <S> multiply the number of electrons or amplify the signal) the multiplied electron are collected in one end across a resistor as a voltage pulse. <S> My aim is to measure the amount of charge from the detector. <S> Photomultipliers have multiplication noise due to the electron multiplication process. <S> A single photon might generate on average 1 million electrons, but the variance on that output is also about 1 million electrons (although it varies between devices). <S> For this reason, counting charge is not always the most useful thing to do, since you frequently are just measuring noise in the detector that has no significance. <S> A common approach, since you have a signal of almost 1 volt into (presumably) 50 ohms, is just to count how many rising edges you have, especially if you don't expect photon pile up.
| If your capacitor is polarized - like a tantalum or electrolytic capacitor - a negative voltage for an extended period will make it explode (although a small transient like what you have pictured is probably just fine)
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Logic design for 27 inputs I have 27 inputs on a pcb. I need to build a circuit on this pcb that checks if only one of these inputs is high. So more than one input high is not allowed. What is the best way to make this circuit with as less as possible physical components without making use of a fpga. I was trying it but the circuit gets quite big. <Q> <A> Andy beat me to the wire-OR approach. <S> Here is a variation. <S> Use 27 small signal diodes (1N914, 1N4148, etc.) and one common pull-down resistor. <S> What are the lowest input voltage level to be detected, and the power supply voltage(s) available for the input circuit? <S> UPDATE: <S> This circuit illustrates the concept . <S> It is not complete because many details are not yet supplied by the TS. <S> The 27 diodes and R1 form a wire-OR gate; when any input goes high, the voltage across R1 increases above the transition level of whatever is acting as a detector. <S> The nominal transition level for a TTL gate is approx. <S> 1.8 V, which is far enough below the input voltage minus the diode forward voltage drop to be a reliable detection threshold. <S> Note that if the downstream circuit is a bipolar TTL family (74xx, 74LSxx, etc.) <S> the max value for R1 is relatively low. <S> CMOS families with a "T" in the descriptor (HCT, AC, etc.) also should work directly. <S> Non-T CMOS families (HC, AC, etc.) and most 4000 series gates will not, because the transition level is approx (Vcc / 2), too close to the input signal voltage after a diode. <S> For a non-5 V input circuit, you'll probably need an analog comparator. <S> The schematic shows the basic form. <S> Missing are the operating voltages, a possible output pull up resistor, possible hysteresis, etc., because application details are not given by the TS. <A> The least-invasive way I can think of is to use a microcontroller. <S> This would give you self-test capability with modest software effort on a single, low-cost chip. <S> Something like a low-end STM32 would have enough <S> I/Os to handle this. <S> This raises a question: what is looking at the <S> I/Os, and can that somehow be a scan-capable part? <A> The low hanging fruit has been plucked already, so let's go extra overkill. <S> Build a 32 to 5 priority encoder using the answer from this question . <S> That's a bunch of 74HC148 which take 32 input lines and output a binary code corresponding to the number of the lowest input that is "1". <S> Build another and feed it the same inputs, but in reverse order. <S> Add a comparator: if both encoders output the same value, then there are either zero or one input bit set. <S> If more than one input is 1, then both encoders will output different values since they're priority encoders and the second one has its inputs reversed, and the comparator will notice this. <S> Now I was looking for a really oldskool way to do this, which means obviously that it has to use magnetic cores. <S> I guess by running all the 27 wires through a saturable core that would saturate when more than one wire carries current...
| Use a summing network of 27 plus 1 resistors and a voltage comparator to measure the level at the summing node and produce a logic output if the voltage indicates two or more inputs are at logic 1. Depending on the headroom between the lowest high-level input voltage to be detected and the low-high transition voltage of whatever logic family you are using, it might be enough to simply connect the common node to a gate input.
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What material is used for the PCB traces? Are the traces of a PCB always made of copper, we can use another conducting material?Also, is there is another conducting material that can be used in the cladding layer instead of copper, or do we always use copper clad? <Q> It serves a very niche market: superconducting traces for cryogenic electronics, transparent PCBs for X-ray systems, PCBs with long radiation length materials for nuclear beam applications. <S> Pro: very good for aluminum wire bonding. <S> Con: it is difficult (although possible) to solder to aluminum using fluxes based on fluoroborate. <S> However, it is also possible to have copper plated into the aluminum to make it easily solderable. <A> For extreme low-cost low-power applications as pocket calculators carbon traces printed on plastic sheet are a thing. <S> They also function as keyboard contacts. <S> The high resistance of carbon isn't a problem as the currents needed for keyboard matrices and LCDs are in the 1 µA range. <A> The PCB process generally assumes copper conductors. <S> This is somewhat by definition because all the equipment is designed around a copper etching and plating process. <S> And the technology - the common set of equipment, standards, practices, and knowledge - also generally assumes copper process. <S> This is the "technology brand" known as printed circuit boards aka PCBs, for now at least. <S> The reasons for this are described in the comments, copper conductors are superior for many applications. <S> That being said, other conductors can be printed/plated/etched and otherwise attached to various substrates. <S> For example, here is a silver plated process with ENIG finish on ceramic substrate produced by Remtec <S> They certainly look like PCBs to me, but the manufacturer does not refer to them as such. <A> I have seen flexible printed circuit heaters that used nichrome instead of copper. <S> However, these are pretty specialized and are designed for a single purpose: to act as electrical resistance heating elements. <S> They are common, but specialized in the sense that they are only really good for one purpose: heating.
| There are also "aluminum trace PCBs" (to differentiate from "aluminum base PCB", which use aluminum as a heat-sinking backplane) but they are more expensive than copper PCBs.
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How to make an IR proximity sensor work in daylight? I am trying to make a touchless water tap using an IR proximity detector. I used a simple circuit: It works well inside room but it doesn't work in daylight. It remains active in sunlight and if I reduce its sensitivity, its range also reduces too much. I can use many IR transmitters to overcome this problem but this makes my circuit big and less power efficient. I need a circuit in which the IR receiver can distinguish between the infrared coming from the transmitter or from the sun so, it won't get triggered by daylight? <Q> I suggest you try (you might need a combination): use an IR remote control receiver (for a start, a TSOP ) with an appropriately modulated source (36 kHz + 1 kHz). <S> this will likely be the biggest win. <S> put the receiver at the end of a (black) tube, so <S> your sensor 'sees' only the (reflected) light source putting the source in a similar tube might give a little added benefit because it avoids reflection of the sunlight on your source <A> A common solution to this problem is to use a chopper of some kind. <S> Turn the transmitter off Read receiver voltage <S> Turn it on Read receiver voltage <S> Look at the difference between the 2 readings. <S> This is much less susceptible to ambient light than your circuit. <S> It's easily done with a microcontroller. <S> An analog version of this is to pulse the transmitter at a given frequency, say 10 kHz. <S> Then add a 10 kHz band-pass filter to your receiver amplifier (or maybe just a high pass would be enough). <A> you can actually estimate distances to a (diffuse) reflector, using one LED and two photodiodes IF one of the detectors is offset by a constant distance. <S> This works because one detector has 1 / Range^2 photon flux, and <S> the other detector has 1 / (Range + K)^2 photon flux. <S> If you take log of both, the dynamic signal range is much easier to work with. <S> Diodes in series with the photodiodes may work well for you, if you use a high-frequency drive to power through the sunlight. <A> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Auto-tracking trigger. <S> How it works: <S> The sensor section is the same. <S> R2 and C1 form a low-pass filter. <S> You probably want a time constant of a few seconds here. <S> It has to be short enough that it can track the sun coming out from behind a cloud but slow enough that the circuit can react to human interaction. <S> OA1 buffers the C1 voltage and drives the sensitivity adjustment pot. <S> CMP1 compares the current value from D1 with the long-term trend from C1. <S> I didn't bother to work out the logic. <S> If it's backwards you just swap the inputs on CMP1.
| Another approach is to follow the background illumination level with a low-pass filter and detect any sudden changes in level.
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How to detect 3 way slide switch positions? With one of this 3 position 8 pin slide switches, I want to detect all three positions: Required function: Position 1: Node1 to ground Position 2: None Position 3: Node2 to ground If I pull an digital I/O pin HIGH and connect it after SW2, HIGH means position 1 and LOW means position 2: simulate this circuit – Schematic created using CircuitLab Can't connect another digital I/O pin to any of the nodes (I have spare I/O pins but nodes voltages are higher than 5V. even if I use a voltage divider, I can't get an Ideal HIGH from them because their voltages will vary) and nodes shouldn't be connected to each other. How can I detect all third positions? <Q> I think you're misunderstanding how the switch operates. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Each pole of the switch consists of a common wiper contact, 'C', which can touch one of the other contacts, 1, 2 or 3, at a time. <S> The switch is shown in position 1. <S> I think you're ON-OFF-ON requirement can be satisfied as shown on the right using only one half of the switch. <S> After details of switch added: simulate this circuit Figure 2. <S> Does this arrangement using one pole of the switch satisfy your requirments? <S> Table 1. <S> Truth table assuming pullup resistors on GPIO. <S> Posn 1 2 1 <S> Low <S> High <S> 2 <S> High <S> High <S> 3 High Low <A> Then one A/D converter input on the your MCU can take a reading and directly deduce the switch position from the voltage created by the voltage divider. <S> Here is a simulation example with one less resistor showing some reasonable resistor values. <S> With a 5V supply the A/D readings for the switch detection would be in the ranges of: Position 1: <S> Reading above 2.4V Position 2: Reading between 1.8V and 2.4V Position 3: Reading below 0.8V <S> Switch not connected: <S> Reading between 0.8V and 1.8V. <A> The error is in ignoring the spacing. <S> 'C' is the common terminal. ' <S> 1', '2', & '3' are the positions. <S> The 3 black bands are the 3 positions of the moving contact.
| One way to detect the positions of a switch like this is to arrange a connection like shown here. Careful selection of the resistor values will create voltage levels that are far enough apart that noise or dither in the A/D readings will not be a problem in setting the detection range thresholds. It's a Double Pole 3 Position Switch.
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When do capacitors "decide" to discharge? I am a beginner electronics designer, but I have never been able to grasp "when a capacitor discharges". What causes a capacitor to discharge? Is there a way to calculate a capacitance value to discharge with certain voltage and current values over a specific amount of time? <Q> Capacitors oppose changes of voltage. <S> If you have a positive voltage X across the plates, and apply voltage Y: the capacitor will charge if Y > X and discharge if X > Y. calculate a capacitance value to discharge with certain voltage and current values over a specific amount of time <S> There's the classic RC circuit analysis which should be taught right at the beginning of learning electronics. <A> simulate this circuit – Schematic created using CircuitLab <S> A charged capacitor is always trying to discharge itself and will do so whenever the circuit it is part of permits current to flow between its two plates. <S> An analogy would be a stretched rubber band. <S> It will spring back to its relaxed state whenever it is released from whatever is keeping it stretched. <S> the capacitor is part of has a smaller magnitude than the voltage stored on the capacitor. <S> So in the circuit above if the voltage across the capacitor is greater than the voltage of the voltage source, Vs, the capacitor will discharge through the resistor, R, until the voltage across the capacitor equals the voltage supplied by Vs. <S> At that point current will stop flowing through R, as there is no voltage difference across it. <S> If Vs is greater than the capacitor voltage then current will flow through R charging the capacitor until the capacitor voltage is equal to Vs. <S> Again, current will then stop flowing. <S> The answer by pjc50 links to the equations needed to calculate how long this takes. <A> Is there a way to calculate a capacitance value to discharge with certain voltage and current values over a specific amount of time? <S> $$\text{current taken from the capacitor} = <S> \text{capacitance}\cdot \dfrac{\text{change in voltage}}{\text{change in time}}$$ <S> Or $$ <S> I = <S> C\cdot\dfrac{dv}{dt}$$ Re-arrange as suits your needs. <A> The capacitor discharge when the voltage drops from the main voltage level which it connected to like it connected between (5v and GND ) if voltage drops to 4.1v then <S> the capacitor discharge some of its stored charge ,the drop in voltage may caused by many effects like increase in a load current due to internal resistance of non-ideal source .
| More specifically, a capacitor discharges whenever the voltage in the circuit
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How does the input voltage affect the current consumption of a microcontroller? Is there a relationship between the current used by a microcontroller unit (MCU) and its input voltage? In other words, given a MCU with a wide-input range (e.g. 2.7 to 5.5V for the ATmega328P), will it use the same amount of current for a given operation undepending on the input voltage? I'm not sure if there is a generic answer to the question as it might depend on the particular MCU. I'm asking this question because low-power figures (e.g. in datasheets) seem to always be expressed as current (e.g. 1 μA for deep sleep mode) and never in Watt or similar. <Q> This really depends on the implementation of the microcontroller and cannot be answered in a general way. <S> Some parts will always be affected by the supply voltage, for example the dynamic power consumption of GPIO when they are switching. <S> so they will internally run on the same voltage all the time, rendering the effect of the external supply voltage minimal. <S> Others don't have this and their power consumption will depend on the supply voltage. <S> But still they usually only give figures for current and not power. <S> That is because the tables are for specific use cases which are usually either written on top of the table or below the table. <S> In most cases it will contain something of "Ta = 25 °C, <S> VDD = 3 V unless otherwise noted". <S> For example a note from the STM32G0 datasheet: <S> Some manufacturers will provide detailed plots or tables for different scenarios but as the combinations are incredibly complex, you usually only get one thing and have to measure all the rest yourself if you really need it. <A> Please read the datasheet very carefully. <S> There are different values for power supply current at different VCC voltages of 3 and 5 V. <A> long channel mosfets have a square_law effect of vdd upon Idd. <S> Short channel mosfets behave differently.
| Some microcontrollers have internal voltage regulators for the core or other parts of it
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120v Inverter and GFCI I have a 48vdc to 120vac inverter with a 3 wire output (hot, neutral and ground). I installed a GFCI outlet using those output wires. I tried putting a resistor (I tested with 1k 100 and 22 ohm values) between the hot and ground but the outlet doesn't trip. Using the same resistor on another GCFI outlet connected to the grid, will trip the outlet. When I press the test button on the GFCI outlet connected to the inverter, the outlet will trip. The inverter ground is grounded. I get 120v from neutral to hot and 60v from neutral or hot to ground. It seems like the GCFI won't work if you were to make a circuit to ground. Any ideas of what is happening? <Q> I get 120v from neutral to hot to neutral and 60v from neutral or hot to ground. <S> The important thing to note here is that you \$\color{red}{\text{get 60v from neutral or hot to ground}}\$ <S> This tells me that the output is likely floating BUT loosely coupled to ground with EMI capacitors of a few 100 nF. <S> Something like this: - This means that a regular GFCI won't work because you are just not able to drag the current through the resistor. <S> It's a bit like connecting a battery to ground via a 1 kohm resistor - no current will flow because you haven't made a DC circuit. <S> Your inverter's output is galvanically isolated from ground <S> hence a single point connection from either hot or neutral to ground won't cause a significant current draw that would trip the GFCI. <A> Some (many?) <S> small inverters (and Honda's "inverter" type generators) have a hot neutral. <S> During one half-cycle, the Hot wire will go from zero to +170 V and back to zero. <S> On the other half-cycle, the Neutral wire will go from zero to +170 V and back to zero. <S> This arrangement avoids the need to make a negative high voltage supply, and normal loads can't tell the difference between this and normal commercial power. <S> Grounding the Neutral in this case will cause Bad Things to happen! <S> I don't know how a GFCI would react when faced with this sort of supply. <A> So the normal way we do things, is, at precisely one point: the service point , there is an equipotential bond between neutral from supply/onward, and ground from the Grounding Electrode System/onward. <S> The whole point of that exercise is to a) assure that, say, a transformer leak doesn't cause hots+neutral to float at some wild voltage thousands of volts above ground (which would overwhelm insulation in most appliances). <S> So it pegs neutral to near earth. <S> And b) to assure that fault current that winds up on ground efficiently finds its way back to neutral, with enough current flow to get an overcurrent breaker trip. <S> (of course GFCI makes this second role irrelevant, since it'll trip at 5ma). <S> That bond is called the Neutral-Ground Equipotential Bond due to that first function. <S> Your thing is not doing it that way, so I don't know what to tell you. <S> Normally in a service installation, only hots+neutral are sourced from the power company. <S> And you bond to neutral yourself. <S> So by that definition, your inverter is the supply from utility. <S> You take only hot and neutral from it. <S> You then locally derive safety ground from your ground rods or whatever. <S> Then you bond neutral and ground at your service point. <S> My concern is, this could result in the chassis of the inverter (and the battery system powering it) to now become a bouncing ball, wildly floating as high as 170V+48V above earth. <S> That means all of it would become a shock hazard. <S> Harper's law is that non-UL-listed Chinese cr*p does not ever touch AC mains. <S> Then I'd wire the service as above on the secondary side . <S> 240V/480V-120/240V supply transformers are commonplace, and you could run one at half voltage (and half power) easily enough. <S> Also if you ever upgrade your system, you could go 240V with the bigger system, and re-jumper the secondary for split-phase. <S> At that point, you could feed into a normal service panel like an Eaton CH, do N-G bonding in a normal way, and have circuits come off in a normal way, and the GFCI should work ducky-doo. <A> GFCIs work by sensing the imbalance between line and neutral. <S> If the supply is floating (like it would be with an inverter), making a path to earth wont change the line/neutral currents with respect to each other, it will only shift the floating voltage around. <S> Tie inverter neutral to safety ground, before the GFCI , and test again. <S> It should trip. <A> So I got a response from the manufacurer after sending this picture <S> The response: <S> "Dear, we do not recommend to connect the yellow wire with black one, it will make the inverter shell has the electricity. <S> best regards"
| If I had to make it safe, I would have the inverter output into an isolation transformer, and put both inside a grounded metal box. GFCI trips are only relevant when your supply has normal Neutral-Ground bonding back at the service. You bring your own ground.
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Using a single pushbutton switch for 2 seperate circuits I'm trying to rediscover my technical roots and am in need of some knowledge from experienced hands. I have two toy sirens for my nephews pedal-cart, but I need them to both sound when a single push button is pressed (double the joy for my brother and his wife :) …). Each siren has four types of sounds and the steering wheel only has 4 pushbuttons.Each siren has its own (3Volt) power source so I can't completely take it apart. I am struggling to apply Kirchoffs and Thevenins laws to this problem and want to ensure that I get enough current to the sirens (in case the voltages from the power sources cancel each other), but without frying the circuits (and ensuring that I've not series linked the batteries and delivered twice the expected voltage). Hoping anyone can advise on the dilemma I'm having... I'd like to make sure that my nephew's first memory of "driving" is not one of fire and carnage. Apologies for the simplicity of the problem in comparison to others listed here. Many thanks. <Q> If the sirens are identical, you should be able to simply parallel the switch contacts from each siren. <S> Be certain to match the polarity of the wires. <S> This will work quite well so long as the batteries in both sirens are (about) <S> the same voltage. <A> You just need to source a double-pole switch. <S> I.e. a switch that will close two electrically independent circuits at the same time. <S> These are VERY common. <S> Describe what you want this switch to physically look like <S> and maybe we can point you towards a good one. <S> Something like this should work for you: https://www.ebay.com/itm/19mm-6-Pin-Dia-Double-Pole-Flat-Cap-Momentary-Stainless-Metal-Push-Button-Switch/263101029726?epid=8030268754&hash=item3d420b1d5e:g:Uj4AAOSwmT1Zets5 <A> simulate this circuit – Schematic created using CircuitLab <S> Using a DVM to detect +V on the switch.
| Connect both to allow either switch to activate both or add a 3rd switch externally.
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Is it safe to touch 200 volts DC? I was watching this YouTube video. The content maker simply declared that 200/300 volts DC is safe to touch. I think the capacitor used in that circuit is limiting the current. That's why nothing is happening to him. I have seen my friend to get injured by touching the supply line of 400 volts DC generator in lab. My question is - Is it really to safe to touch any DC supply (200- 300 volts) without considering line impedance or the maximum current that can be provided by the source? <Q> No, most electrical regs class low voltage “safe” as below 50V DC <S> It is not the max current but where it travels through the body and, also <S> , when in the heart cycle it hits... <S> So, don’t play. <A> The content maker simply declared that 200/300 volts DC is safe to touch. <S> He didn't say this anywhere in the video, and he didn't imply it either. <S> In the first few seconds he says: I want to kick it up a notch and show you at what voltage it hurts over skin He is just showing "when does it start hurting". <S> That's all. <S> He has another video of a live demonstration, hooking up multiple car batteries to produce 120V DC and touches them: https://youtu.be/ZxBF7WC0TQk?t=654 <S> He feels it and it hurts, but not nearly as much as 120V AC. <S> He's not saying DC is always safe. <S> He explains the dangers in a different video where he talks about the relationship between voltage and current in hurting you: https://youtu.be/XDf2nhfxVzg <A> The linked video at 3:48 reveals something very essential of "teacher's" capabilities. <S> He can output nonsense with quite assuring voice. <S> I do not think he believes himself that human body has capacitance which makes easier for AC voltages than for DC voltages to generate harmful current. <S> Do not believe him. <S> I have seen how a car battery made in about 1 second a screwdriver so hot that it burned its way to user's flesh. <S> 100V or more DC can stop ones heart when a person gets it from hand to hand or from hand to feet. <S> There's no exact voltage limit for this. <S> Someone can stand more than another. <S> The moisture and how tight the contact happens to be effects, too. <S> Regulation codes for electric works vary. <S> Check which is the limit of extra low voltage (=ELV) where you live. <S> Staying below the limit does not make the voltage safe, but it's used in a court of law when the judge must decide if someone has caused a danger intentionally. <S> Any electric works with higher than ELV voltages can be considered intentional searching for troubles, no matter did a person know that such limit exists or not. <S> AC causes pain continuously because the current turns forth and back and every pulse disturbs the nerve system. <S> DC hurts that way only when the current starts or stops. <S> But high enough starting pulse can stop one's heart. <S> "Teacher's" circuit was galvanically connected to mains AC. <S> His touching the DC output could have given a different result if he had accidentally a connection to the ground for example through leather shoes and a concrete floor. <S> Having a rectifier would in that case make no difference. <A> Strictly speaking, voltage won't tell you whether something is safe or unsafe to touch. <S> A Van De Graaff generator can create extremely high voltages, and many are designed specifically to be touched. <S> These won't hurt you because the output current is extremely low. <S> It's actually the current that hurts you , not the voltage . <S> People have died from voltages as small as 42V. <S> I've touched a Van De Graaff generator rated at around 1,500,000V and had no negative effects. <S> In your specific case, power supplies are generally designed to output a significant amount of electrical current, so I wouldn't recommend touching a power supply rail at any voltage. <S> Regarding your comment, the capacitor isn't limiting the current. <S> Capacitors limit how fast the voltage can change. <S> When it comes to current, one of the big use cases for capacitors is how they can provide extremely high output currents when they discharge. <S> If anything is limiting the current, it would be a resistor of some sort. <A> No, in general it's dangerous to touch 200V-300V DC. <S> Look at https://en.wikipedia.org/wiki/Extra-low_voltage for details on voltages considered safe to touch by different regulating bodies.
| Even much lower DC voltage can be dangerous.
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How to safely expose one wire of an AC mains circuit? I need to use a rail to connect one moving piece where a switch is, to a fixed piece where a power tool is. The rail is exposed, so I need this connection to be safe, like 12V or below. I'm guessing I will need some sort of MOSFET module, but each and every search result is about Arduino. Is there such a thing, that will let me use small wires and switches in a "safe" exposed circuit to turn on and off a mains powered device? <Q> I'd even go further than what Graham Nye says, and look at mains-rated (UL-Listed, not RU-Recognized) relays, contactors etc. <S> designed to attach directly to AC mains wiring methods and provide low voltage terminals with separation. <S> An example is Aube relays, which are designed to mount in AC mains enclosures (e.g. a junction box and standard 1/2" knockout). <S> They are designed to keep mains on the mains side, and low voltage on the low voltage side. <S> They have a built in transformer, so operating the relay is as simple as shorting two terminals (which have 24VAC on them). <S> That connects the transformer to the relay coil, which picks up the relay. <S> Now it's not necessary to handle mains at all (except in Electrical Code approved enclosures), and all your work is being done on the low voltage side. <S> They make other relays, such as RiB , with coil voltages of your choice if you want to supply coil power. <A> Or you could use a traditional electromagnetic relay with a low voltage operating coil. <S> You can stop words appearing in search results by preceding them with a hyphen ("minus"). <S> So -arduino should clean up your search results. <S> However for safety I suggest you don't try making your own MOSFET switch but get a solid state relay instead. <A> Here's how, using a 12V= Relay.
| Look up solid state relay for a ready-made mains switching block that can be controlled by a low voltage.
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What's a conceptual way to understand ohm-cm? Resistivity is defined in units of \$\Omega \times \textrm{cm}\$ . I don't conceptually understand what is meant by the unit. If it was \$\Omega / \textrm{cm}\$ , that would be easy to understand - a certain number of ohms for every centimeter. How can one understand \$\Omega \times \textrm{cm}\$ ? <Q> Imagine a block of material with a uniform density to it. <S> Something like this: The material also has, let's say, a uniform "resistivity" to it. <S> Now, suppose we cover the entire face pointed at by the arrow, and the face opposite to it that we cannot see, by plating them with silver (which is very conductive.) <S> We then measure the resistance between these two silvered faces on opposite ends using an ohmmeter. <S> There will be some value for that in Ohms. <S> Now, let's consider three modifications: <S> Suppose we doubled the length. <S> Here, since the silvered faces touched by the ohmmeter have the same area as before, but are further apart, we should expect that the resistance we'd measure between the opposite X faces would double. <S> Suppose we doubled the height. <S> Here, since the silvered faces touched by the ohmmeter have doubled in area but are the same distance apart as before, we should expect that the resistance we'd measure between the opposite X faces would be cut in half. <S> Suppose we doubled the width. <S> Here, since the silvered faces touched by the ohmmeter have doubled in area and are the same distance apart as before, we should again expect that the resistance we'd measure between the opposite X faces would be cut in half. <S> So, we postulate the following about the resistance <S> we'd measure: <S> \$R\propto \text{Length}\$ <S> \$R\propto <S> \frac1{\text{Width}}\$ <S> \$R\propto <S> \frac1{\text{Height}}\$ <S> \$\therefore <S> R\propto \frac{\text{Length}}{\text{Width}\:\cdot\:\text{Height}}\$ <S> Now, if we call the length, \$L\$ , the width, \$W\$ , and the height, \$H\$ , and introduce a constant of proportionality, we can say: <S> $$R=\rho <S> \cdot <S> \frac{L}{W\cdot <S> H}$$ <S> Let's now express the above only looking at the SI dimensions: <S> $$\begin{align*}\Omega=\rho \cdot \frac{\text{m}}{\text{m}^2}, &&\therefore \rho=\Omega\cdot\frac{\text{m}^2}{\text{m}}=\Omega\cdot\text{m}\end{align*}$$ <S> Just simple dimensional analysis. <A> To understand this, you must first know that resistivity is basically the total number of resistance per unit length AND cross sectional area. <S> $$\frac{\Omega}{\textrm{cm}} <S> \times \textrm{cm}^2 = <S> \Omega \times \textrm{cm}$$ where \$\Omega / <S> \textrm{cm}\$ : value of resistance per unit length <S> \$\textrm{cm}^2\$ : cross sectional area <A> Another way to think about this repeats essentially the same dimensional analysis as what jonk wrote above, but it starts from Ohm's law which can be written more generally as: $$J = \frac{E}{\rho}$$ where \$J\$ is the current density, \$E\$ is the electric field and \$\rho\$ <S> is the resistivity. <S> This is always true while <S> \$V= <S> IR\$ is actually rarely true. <S> However, if we keep it simple and consider the rectangular prism that jonk describes above, we can consider the material to be isotropic (meaning the resistivity is the same in all directions), and we have: $$J = <S> \frac{I}{A} = <S> \frac{E}{\rho}$$ <S> where \$I\$ is the current above and \$A\$ <S> is the cross sectional area. <S> This can simply be rearranged: <S> $$\rho = <S> \frac{E\times <S> A}{I}$$ <S> Looking at the RHS and doing SI units analysis (fudging dimensional analysis a bit) gives: $$ \require{cancel} \frac{[\frac{V}{\cancel{m}}][m^{\cancel{2}}]}{[\frac{C}{s}]}= \frac{V}{Amp}\cdot m = <S> \Omega\cdot m$$ <S> Here we have used the usual units of volts per meter for electric field, and coulombs per second for amperes. <S> In electromagnetic theory, units are sometimes highly confusing <S> and it's better to focus on what the quantity means through fundamental equations. <S> As food for thought, consider that in Gaussian units, the resistivity is measured in seconds! <S> You could rationalize that as a time required to travel unit length in response to an applied field, etc, but I still think it's better to stick with the fundamentals. <A> So if \$\rho\$ is given as \$1.6 \mu\Omega-\text{cm}\$ <S> (copper) <S> If you consider a strip 1cm long and <S> 1 cm wide, it's the thickness of the strip in cm to make it <S> \$1 \mu\Omega\$ in resistance. <A> If we start with restistance as relating to a particular thing, say a length of wire of length 10cm. <S> It is measured in Ohms. <S> Now, the wire manaufacturer will probably specify the wire as 99% copper, with a specificed resistance in Ω/cm. <S> That is a slightly more abstract concept. <S> If we put two resistors in series, we know the resistance is doubled. <S> For a physicist the interest is more about the properties of 98% pure copper in general. <S> So far we have a quantity that depends on the gauge of the wire. <S> To a good approximation the resistance is inversely proportional to the area: a good way to visualize that is to imagine we have a stranded cable, so a thicker cable is just more strands in parallel. <S> This gets us Ω.cm. <S> It's not easy to visualize units like that. <S> One thing that might help would be to think of the resistance between two faces of a cube of copper. <S> It would decrease as the cube grows. <S> Maybe it's easier to think of the conductivity of a material having units of S/cm (where S = Siemens aka mhos)
| The best way to think about resistivity or conductivity is that it translates an external electric field into a current density inside of a material with free charge carriers.
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My Coilgun Stage Driver Keeps Burning Out Mosfets. Any idea why? Here's my stage driver circuit: The circuit worked as expected at around 100v. But at 330, both mosfets burned out at the first firing, and the entire cap bank discharged through the coil. I thought it might have been an issue with my firing circuit so I replaced the mosfets and tried again. Again, burned out immediately. I can't figure out why. The mosfets are rated for 650V, and a pulse current of 520A, which I don't think I've exceeded. The gate drivers are very strong, so I don't think switching speed is the issue. HV is a 330v 5400uF bank of flash capacitors. The MCU is programmed to turn on both fets for 100uS, and then turn them off. I checked, and it does. I also checked the mosfet gates (without the cap bank connected), and they pulse up to 12v as expected. The rise time is very quick, <100nS. The coil is approximately 686uH, with a DC resistance of 1.8 ohms. The mosfets are normally 20 bucks each on digikey, but I was able to find 10 of them on ebay for $8 each. Maybe they're counterfeit. Update: Here's my layout: Datasheet Links: Gate Driver IXDN614PI Mosfets STY139N65M5 Diodes C3D10065E Optocoupler 6N136 Isolated DC/DC Converter PDSE1-S5-S12-S <Q> Only guesses are possible without having your circuit. <S> Guess1: <S> When the mosfets are turned off the coil current bulldozes its way through D1 and D2 as you have planned. <S> That leads to an accelerated mosfet <S> Vds jump. <S> Too high d(Vds)/dt can trigger parasitic parts inside the mosfets and that current route maybe doesn't stand what's available. <S> Of course, I cannot prove that's the case. <S> The phenomena is described in this paper: https://www.mouser.com/pdfdocs/Impacts_of_dv-dt_Rate.pdf <S> You very likely could use the same ideas how thyristors are protected against too high d(Uak)/dt. <S> Guess2 <S> : <S> The parts this cheap can be fake and do not fulfill the specs as you said in the question. <S> But you hang at the edge of the genuine parts, too. <S> See the safe operating area curve: Your coil can have a substantial capacitance. <S> Charging it too fast needs a current which can be too high because Id and Vds are both high at the same time. <A> Current slope is as little as you reckon to 300V/600uH=0.5A/us in the inductor only. <S> From the power supply point of view you see peak current, say 50A, changing direction from source to sink during MOSFETs switching time. <S> That's the slope you have to cope with. <S> I'd go for a good low ESL/ESR snubber grade capacitor very close to your half bridge supply. <S> Also take care of poorly damped ringing that could arise, some kind of snubber may be required too. <S> A second option would be to slow down switching, that's not an inverter working at 10's kHz rate. <S> In virtually one-off zero duty cycle application switching losses are no concern as long as you don't exit RBSOA. <S> I believe both actions should be taken, together with some layout improvement to reduce high di/dt loops area. <A> So you want to switch 500 amps in about 50 nanoseconds, in a system with 100 nanoHenries of inductance. <S> What will be the spike? <S> V = <S> L <S> * dI/dT <S> V = <S> 100nH <S> * 500 amps / <S> 50 nanoSeconds <S> The "nano" cancel, and we have 1,000 volts. <S> ============================================= <S> OK <S> So what? <S> Given no info on your ground planes, you should assume the worst. <S> But ... lets look at coupling from high current wiring into low-voltage control traces. <S> Assume the high current flows 1cm away from 1cm by 10cm loop. <S> What is the induced voltage? <S> (the high current is parallel, for 10cm) <S> V = <S> [MUo * MUr * Area / ( 2 * pi * Distance)] <S> * dI/dT <S> which for MUo = <S> 4 <S> * <S> pi * 1e-7 <S> , MUr = 1 for air/copper/aluminum/FR-4 becomes Vinduce = <S> (2e-7 <S> * Area / Distance) <S> * dI/dT <S> and we substitute <S> [I don't know how bad this will be; any more than 1 volt will upset control signals] <S> Vinduce = <S> 2e-7 <S> * (1cm <S> * 10cm)/1cm <S> * 10,000,000,000 amps/second Vinduce = <S> 2e-7 <S> Henry/meter <S> * 0.1 meter <S> * 1e+10 amp/second and the prediction <S> (remove the card from the mayonnaise jar, please) <S> Vinduce = <S> 2e-8 <S> * 1e+10 <S> ==== <S> 200 volts <S> ======================================= <S> You do have a Ground Plane, right? <S> or two. <S> And you understand the 10,000,000,000 amp/second transient will cause currents to spread over ALL possible paths, as mother nature desperately attempts to solve the many simultaneous differential equations so as to give you the minimum energy solution.
| That's most probably inductive kickback killing your MOSFETs by overvoltage. So a relatively high gate resistance would help a lot keeping EMI low too.
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Why adding capacitor to a 10x passive oscilloscope probe I'm trying to understanding input impedance and its circuitry of a simple, passive oscilloscope 10x probe. From reading Input Impedance of an Oscilloscope and the video EEVblog #453 - Mysteries of x1 Oscilloscope Probes Revealed , I don't understanding why we add these capacitors in the probe circuitry and make things more complicated. The point is to add enough impedance to a passive probe to minimize loading effect of the circuit being measured. A big enough resistor will do and the with just resistors, voltage divider works the same for both AD and DC source regardless of frequencies. If there are intrinsic reactances in the wiring, then can't we just add a big enough resistor in series with those to make any reactance negligible? <Q> I don't understanding why we add these capacitors in the probe circuitry and make things more complicated. <S> They make things far less complicated. <S> The low-pass filter that would otherwise occur is eliminated. <S> The point is to add enough impedance to a passive probe to minimize loading effect of the circuit being measured. <S> A big enough resistor will do and the with just resistors, ... <S> No. <S> Increasing the probe resistance decreases the high-pass cut-off as it will be proportional to \$ \frac 1 {RC} \$ . <S> ... <S> voltage divider works the same for both AD and DC source regardless of frequencies. <S> No. <S> You have omitted the effect of the scope's input capacitance. <S> If there are intrinsic reactances in the wiring, then can't we just add a big enough resistor in series with those to make any reactance negligible? <S> Figure 1. <S> From Introduction to oscilloscope probes . <S> The trick is to use two potential dividers; one resistive and one capacitive. <S> Since we've got a ratio of 9:1 with the resistive divider we need to do the same with the capacitive divider. <S> Remember that the capacitors impedance is given by \$ Z_C = <S> \frac <S> 1 {2\pi fC} \$ <S> so $$ \frac {R_1}{R_2} = <S> \frac <S> 9 1 = \frac {Z_{C1}}{Z_{C2}} = \frac {\frac 1 {2\pi fC_1}} {\frac 1 { <S> 2\pi fC_2}} = \frac {C_2}{C_1}$$ From this <S> we get <S> \$ C_2 = <S> 9C_1 \$ . <S> With the values shown in Figure 1 we can just about achieve this if C1 is 8 pF <S> and C COMP is wound up to the max to give us 72 pF total (although this model omits the cable capacitance <S> so we have room to spare). <A> The cable has capacitance, and the scope input has also capacitance. <S> If you add resistance, it would just reduce the bandwidth. <S> That's why you add capacitance to the probe, to nullify the other capacitances. <S> Basically it's the same thing with the resistances and the capacitances. <S> If you want a 10x probe, and scopes have 1M resistance, you add 9M so that it divides DC voltages by 10. <S> If you want also the AC waveforms to attenuate by 10x, and you already have some capacitances in the system, you select a correct capacitance to get 10x attenuation. <A> You cannot take off Cs nor Cc. <S> Together with Rp they would make a RC lowpass filter. <S> Elementary impedance calculations reveal that the low-pass filtering effect can be compensated by adding Cp. <S> You can consider it a high frequency boost. <S> The total loading of the circuit under measurements is still lighter than without the probe. <S> General rule for flattest frequency response: <S> RpCp=Rs(Cc+Cs) <S> That's useful as long as the cable is short enough. <S> As the frequency grows the user must finally take into the account that the cable is actually a transmission line which is badly matched at both ends. <S> The compensation becomes gradually worse as the frequency grows. <S> There's no strict limit when this happens.
| Increasing the resistance decreases the signal available to the 'scope to the point that you won't be able to get high enough resolution and ADC noise will become a problem.
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Calculate resistance of coil/inductor made of PCB traces This is to be a NFC coil/inductor of 1.1uH according do ST's eDesignSuite tool, with 7 turns. The drawing above follows these parameters: ST's tool give inductance value as 1.1uH, but also I need to know the resistance of entire track, in ohms.I have measured the antenna total lenght on the layout software and it is 418 mm long. So I have these parameters: 0.25mm width 418mm length 0.5 oz (17.5 micrometers) copper thickness Then I entered these value of this calculator: https://www.allaboutcircuits.com/tools/trace-resistance-calculator/ And the result given by the site is 1.66 Ohms. Can I trust this value? I need the know correct value of resistance to calculate the components of RF portion of NFC filter and matching circuits. The board will have 0.5 oz copper thickness, I can guarantee this. Help please. Regards. <Q> Resistivity of Annealed copper: 1.72x10-8 <S> Ωm Resistivity is defined as: \$ \rho = <S> R\cdot \frac{A}{l}\$ <S> therefore \$ R = \frac{\rho \cdot l}{A} \$ <S> your length is 0.418m <S> your area is \$0.25\,\mathrm{mm}\times <S> 17.5\,\mu \mathrm{m}\$ <S> therefore your resistance = <S> \$ \frac{ 1.72\times10^{-8} \,\Omega \mathrm{m} \,\times\, 0.418\, \mathrm{m}}{ 0.25\,\mathrm{mm} \,\times\, 17.5\,\mu \mathrm{m}} = <S> 1.64\,\Omega\$ at 20C <A> Standard copper foil is 0.000498 ohms per square of foil, for any size square!!! <S> Just to be very clear, this is 500 microOhms per square, whether 1 square micron or 10 square centimeterS. Notice this resistance is substantially lower than typical ESR values in capacitors. <S> BUT THE VALUE IS NOT ZERO <S> This fact is useful for copper resistor design, or for examining hot spots where current crowds, etc. <S> =================== <S> This value is for 1 ounce of copper per square foot, which is the 1.4 mil thickness or 35 microns. <S> And 35 microns used in skin_depth formula will produce 1 neper attenuation to penetrating waves at about 4MHz. <S> Or 10 nepers at 100X the frequency. <S> Which is 88 dB. <S> The 88 dB attenuation is good explanation of why ground planes or power planes are so useful in creating clean systems, despite the presence of MCUs with 400MHz edge rates (1.25 nanosecond trise). <A> You can trust the value of 1.66 ohms given by your calacuator, at DC, at 25 C, if the foil thinkness and etched width are as you've specified. <S> You would be surprised how much in error the etched width could be in practice. <S> The temperature variation is fairly small, +10% for 25 C change. <S> The main error will be if you want the resistance at NFC frequencies rather than DC, as the skin effect will increase it significantly. <A> You could not expect the width and the thickness of the track to be very precise. <S> A track width of 250 µm may vary at least by +- 5 µm or more depending on fotoplotter resolution, resist exposure time, resist development time, under etching, etching time, etching temperature and a lot more. <S> Copper thickness of 18 µm may vary about +- 2 to 5 µm depending on manufacture. <S> So ask your PCB manufacturer about the tolerances for copper thickness and track width.
| There are tolerances depending on manufacture methods.
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Using a LiPo battery or different battery for outside project First I am sorry if this is the wrong section to post this, I have been searching for a clear answer and haven't found one yet. I have a project that I want to keep outside. I have a solar panel, charging circuit and a Lipo battery. It works great, though I have read a lot about how LiPo batteries can fail and catch fire. I am wondering if Lipo is the way to go for something that will stay outside for years. It could get cold and hot from the sun :/ Should I redesign to use a lead acid? Would it be safer??? I know that was a lot, If this is the wrong place for this please let me know and I can remove this post, If this is the right place, thank you so much for the assistance.s <Q> <A> LiPo <S> / LiIon batteries are almost always safe when charged and discharged within specs. <S> To get usefully longer lifetimes Try to charge to less than the usual 4.2V (say 4.1 or even 4.0 V) <S> This gives longer and much longer cycle lives and whole of life capacity at cost of some per cycle capacity. <S> Do not discharge under say 3.2V. Terminate CV phase of charging at high current - say CC/2. <S> Be sure CCCV charging is done correctly and NEVER "float" at 4.2V once charging is complete. <S> With solar you may have variable rate charging. <S> This is acceptable as long as CC rate is within spec. <S> Some cells allow C/1 and some C/2 rates. <S> C/2 is less stressful. <S> LiFePO4 is better again for longevity and safety. <S> Look at manufacturer's temperature limits and observe them. <S> How cold does it get. <S> Sub zero temperatures can cause issues if specs not met. <S> Ventilate to ensure summer temperatures are as low as reasonable. <S> How long is "years"? <A> i recommend you to use dry batteries, a bit more expensive then lead acid batteries but its good for projects that have a low discharge rate like a UPS and not recommended for cars and stuff. <S> You also have to be careful not to over or undercharge it, but i think a proper circuit will solve that. <S> But, on the bright side there are no gases that can damage the solder joint. <S> I think you ought to a research a bit before using it to see if it suits your application
| I presume this is a somewhat sealed project, if you have room for a lead acid then have a look into LiFe batteries, they are similar to LiPo, but without most of the safety concerns of LiPo Lead acid will have some small off gassing, and the hydrogen can cause solder joints to become brittle, this is part of my reason for steering away from that, if you have some mitigation towards that then lead acid should be fine.
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Why CMOS turns on when input voltage is zero? When the input voltage is high, Q1 is off and Q2 is on. In this case, the shorted Q2 pulls the output voltage down to ground. On the other hand, when the input voltage is low, Q1 is on and Q2 is off. Now, the shorted Q1 pulls the output voltage up to VDD. BUT I read that in CMOS, the high is +VDD and the low is 0. How does Q1 turns on when input voltage is zero(won't it only turn on when Vin is negative) ? <Q> Q1 is a p-channel MOSFET (aka "PMOS"). <S> It turns "on" when the potential difference between its gate and source ( \$V_{gs}\$ ) is negative. <S> But where is its source connected? <S> It's not connected to ground (in fact, none of its terminals are connected to ground). <S> It's connected to Vdd. <S> So what turns Q1 on is having the input voltage (its gate voltage) brought substantially below Vdd. <S> Since it has no connection to ground it doesn't "care" what the gate voltage is in relation to ground (i.e. whether it's positive or negative). <A> Q1 is a PMOS, so the "turn-on" condition is Vgs (gate-source voltage) <S> < -Vt. <S> Vgs here is defined as vin - Vdd. <S> So when Vin = 0, the gate-source voltage is sufficiently negative for turn-on. <A> That's all they care about. <S> They don't care about GND or anything else. <S> When you say: (won't it only turn on when Vin is negative) ? <S> You are saying when Vin is negative relative to GND. <S> Just saying "the voltage at the gate" implies you referencing to GND. <S> But as we just said, the MOSFET doesn't care about GND. <S> In fact, it cannot know what GND is since it has no pins connected to GND.
| MOSFETs turn on and off based on the voltage DIFFERENCE between the gate and source terminal.
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Opamp input bias current and current sources I have a current source connected to a load and I want to measure the voltage at the output of the source, and therefore I would like to buffer this voltage. In this configuration, I am under the impression that the opamp will always draw the same amount of current from the current source regardless of what current it is set to output. (Eg an opamp with a bias current of 1uA buffering the output of a current source producing 1000uA will result in 999uA going to the load. If the current source produces 2000uA, 1999uA goes to the load.) If this is the case, it seems like I don't need to necessarily seek out an opamp with a low bias current, but I could simply offset the current source by the bias current to ensure that my load gets the desired current. Is this thought process correct? Or would using a very low bias current opamp still be of significant benefit here? <Q> The data sheet input bias current specification is actually the average of the two input bias currents. <S> The input offset current specification is the difference between the two input bias currents. <S> Input bias currents can flow into or out of the opamp's inputs depending on type of input transistor. <S> NPN transistor bias current flows in whereas for PNP input transistors bias current flows out. <A> The input bias current may be relatively constant, but it will tend to change with temperature, input voltage and supply voltage, and will generally not be well specified to begin with. <S> Take an example, such as the MC33078 . <S> At 25°C input bias current is typically 300nA but may be as high as 750nA (no minimum is given). <S> Over temperature, it could be as high as 800nA. <S> Pretty wide range- <S> but you could measure it easily at room temperature anyway and calibrate it out. <S> Remember that's only with one particular set of power supply voltages, one temperature, and one common mode and output voltage. <S> Now, how does it behave with the other parameters.. <S> (typically, no worst case figures are offered): <S> Changes with supply voltage, but if you have regulated supply voltage <S> it's not too bad. <S> Not great, changes about 50% with temperature. <S> Okay here the current changes with voltage, meaning it appears like a resistancein parallel with your current source output (degrading it). <S> It typically changes 100nA between -5V and 5V, which means it acts like a 100M \$\Omega\$ resistor. <S> Of course it could be better or worse than that. <S> So it really depends on your accuracy requirements, and there is no substitute for a thorough analysis if accuracy is important. <S> JFET or MOSFET op-amps tend to have much lower bias currents, so even if they vary a lot it won't make much difference to a relatively large current like 1uA. <S> That all is at DC. <S> At relatively high frequencies like your 20kHz, other (possibly much larger) error effects can come into play. <A> James already highlighted some of the problems in calculating the input bias current of your measurement device in general. <S> The comments and the answers to your question so far, clearly indicate that this is an unpractical way. <S> To make your thought process mentioned in the question applicable, the current through your measurement device must be predictable. <S> How is this achieved? <S> If you use a voltage divider consisting of e. g. 4M7 - 1M - 4M7 resistors in series and then measure the voltage across the 1M resistor with an instrumentation amplifier, you have what you originally wanted. <S> The current through the resistors can be calculated easily (12V/10,4MOhm = 1,15µA). <S> Therefore you can correct your measurements or increase the current of the source to compensate for the current through the measurement resistors, as your thought process in your question suggested. <S> But the instrumentation amplifier still has an unknown and varying input bias current, that adds to the current through the resistors, you might object. <S> Yes, of course, but this bias current is only one thousandth parts of the current through the resistor divider (taking an AD8421 as an example). <S> Does this really care in your application? <S> If you want to compensate for this, then model the instrumentation amplifier with its input resistors according to the data sheet and then calculate the current through the resulting resistor network including the voltage divider as mentioned above. <S> The result is a single resistance value in parallel with your load that you can calculate with as described in your question. <S> From a practical standpoint, in my point of view, this is the only feasible way to get to a valid result. <S> Good luck, all the best and remember to give us feedback <S> , how you managed to solve the issue.
| The two input bias currents are not equal to each other. In my point of view, you would benefit greatly from using an instrumentation amplifier here, because your thought process is not valid until you use one.
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Activate 3 or more PWM signals in PIC I am doing a PWM with the PIC18f4550 and it managed to do it.But only activate an output by varying the value of CCP1CON.What I want to achieve is to have three PWM outputs and then be able to offset them to control a BLDC motor. I attach the part of the code where I initialize the PWM: public override PWM_Initialize (){ PR2 = ("% # x", (_XTAL_FREQ / (PWM_freq * 4 * TMR2PRESCALE)) - 1); CCPR1L = 0x1F; TRISCbits.RC1 = 0; TRISCbits.RC2 = 0; TRISDbits.RD5 = 0; TRISDbits.RD6 = 0; TRISDbits.RD7 = 0; T2CON = 0x03; //CCP1CON = 0b01001100; CCP1CON = 0b11001100; TMR2 = 0; T2CONbits.TMR2ON = 1;} Attached image of the idea I have of the driver for the BLDC <Q> If you have the 40-pin <S> PIC18 <S> then CCP1 has been replaced by the ECCP and you shouldn't have any problems. <S> Just enable Quad PWM mode and you will have 4 PWM channels: P1A, P1B, P1C, and P1D. <S> That's not entirely ideal though since for a BLDC <S> you want 6 PWM signals since you have 6 transistors and you want to be able to control the dead-time between them. <S> That means that you have to forego complimentary PWM since you can't have independent control of all 6 switches for dead-time control or signal inversion which would be required if you are using NMOSFETs on both high-side and low-side. <S> If you have a high-side bootstrap capacitor, this might need tweaking, especially during startup where RPM is lower so commutation cycles are longer. <S> The reason is that the bootstrap cap needs to be constantly refreshed and therefore requires the low-side switch to turn on regularly enough to recharge the cap and you won't be able to do that regularly during a commutation cycle without complimentary PWM. <S> The easiest thing to do is probably just to oversize your bootstrap capacitor so it can sustain the switching required throughout an entire commutation period without needing to be refreshed. <S> After the commutation cycle is over the low-side switch for that half-bridge will get a chance to turn on during the commutation cycle for another phase so you can recharge the bootstrap cap. <A> If you have the option to choose a different PIC controller, you might consider it PIC16F1786 or 1787. <S> Even though this is a 16F part rather than 18F, it does have a full-fledged motor controller on board. <S> In fact, it is has TWO PSMC modules. <S> PSMC1 allows for 6 discrete outputs. <S> If you really do need an 18F part, look for one's that contains at least one PSMC module. <A> Have a look at the datasheet: Your controller PIC18F4550 have only two CCP modules, so you could'nt use three PWM. <S> Choose a different controller.
| You will probably have to PWM just high-side switches and use regular logic signals to just turn on the low-side switches during the commutation cycle with no PWM.
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why are non isolated buck converters a safety risk with 120V DC? So i've come to understand that buck converters CAN be used for 120V to 5V BUT they are not becuase they are often a safety risk. I also understand that they can be used if they are away from a consumer OR if they are used in an industrial scenario. for example: it can be used with a ceiling fan becauase a consumer does not interact directly or near to the buck converter circuit. why is a non-isolated buck converter unsafe with specifically a 120V connection? It is considered safe to move from 40V DC to DC , seemingly regardless of current flow. looking to understand this better Thanks EDITTED (to specify use of non isolated buck converter) <Q> Because 120V is considered dangerous because of the risk of electrocution. <S> The fact that it's direct current and not alternating current makes it much more dangerous. <S> 40VDC is the limit at which it's considered safe. <A> So if you're working with the low voltage end in the safety of the knowledge it is only 5V and something fails, it could easily end up being a dangerous 120V instead of the 5V you were expecting. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> A 5 V DC output is used for this example. <S> (a) Using a non-isolated AC/DC converter in a somewhat safe manner. <S> (b) <S> Inadvertent reversal of mains connection makes this circuit live. <S> This means that the the voltage between the neutral and earth should be zero <S> but it could be several volts if high currents are flowing in the neutral wire. <S> That means that the voltage on a non-isolated AC/DC converter should be close to 0 V on the COM output and close to 5 V on the DC+. <S> These voltages will vary with the amount of current in the neutral wire caused by other loads in the building. <S> The problem occurs if the non-isolated AC/DC converter is connected to the mains in reverse. <S> The COM wire is now live and the +5 V output is +5 <S> V above whatever voltage the live happens to be at any instant. <S> Your question suggests that you understand that this may be acceptable in a completely insulated system where there is no danger of contact by a user. <S> If a user touches either the DC COM or +5 V in Figure <S> 1b <S> then electrocution is a real risk.
| It's simply because it is an unisolated voltage at high enough levels to be hazardous. Ideally the neutral wire is "neutralised" at the supply transformer or where the supply enters the building.
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How can current be in different phases in the same wire at the same time? We know that either in ideal or practical transformer in no-load condition, the exciting current has two components Ic(core loss component) and Im(magnetizing component). Now since the same current will flow through the winding, how can the two current be out of phase as we learned the current goes out phase when the current is divided into paths, and there is the same voltage between two distinct terminals. You can explain this in an equivalent circuit but in the actual circuit the current division is not possible as there is just single wire winding. Also, the same case is with the compensation current that appears in the full load condition. The two currents are out of phase in the same wire. How is this possible? Is it just a theoretical concept just to make it easy for understanding, or am I missing something here? <Q> Consider this: - ( http://stades.co.uk/index.html#transformers ) <S> The sum of those two currents does of course flow into the transformer primary terminals. <A> How is this possible? <S> Is it just a theoretical concept just to make it easy for understanding, or am I missing something here? <S> You could say that it is a theoretical or mathematical concept. <S> In a circuit, you have the summing of two AC currents into one current where two branches of the circuit join together. <S> Mathematically, the summation is a vector sum rather than a mathematical one. <S> However we can also mathematically break apart a single current that has a phase shift with respect to voltage. <S> We can say that the current has two parts that are mathematically summed together even though it is flowing in a single wire. <S> The same thing is done when a periodic waveform is not sinusoidal. <S> Mathematically we say that the waveform consists of an infinite number of sine waves summed together. <S> That is Fourier analysis. <A> currents <S> Ic and I <S> m <S> are components we derived for our mathematical calculations purpose,resultant of those both currents will flow through the winding.and for clear understanding read about vectors.
| The two currents that flow through Rc and Lm are 90 degrees apart because one is a resistive loss (Rc) and one is the magnetization inductance Lm.
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How do I amplify a square wave pulse of 4 to 8 periods at 3.5 MHz from +-2V to +-60V or more? I am working with ultrasound sensors optimized for 3.5 MHz signals, and I want to test these using my PicoScope 5000 with integrated signal generator. The PicoScope is able to generate arbitrary square waves with an amplitude of max +-2V, and I would like some circuit to be placed between the PicoScope output and sensor which amplifies the square wave to +-60V or more. The ultrasound sensor has an impedance of approximately 41 ohm, and is connected in parallel with a TX810 evaluation board used as a T/R switch. The square wave pulse needs to be from 4 to 8 periods long, and it will be repeated approximately every 70ms. I've been experimenting with Microchips MD1213DB1 evaluation board, but find that it is not a perfect match as it requires a phase shifted square wave and an Output enable signal in addition to the original square wave. Since the PicoScope only has one output (being the square wave) this became difficult to implement. Also, I just fried my MD1213DB1 board and I'm hoping that I don't have to order a new one. Is there a cheaper and easier way to achieve the pulse amplification? Edit: I want to test how the ultrasound sensor performs using square waves compared to gaussian sinusoidal waves. I have already tested using sinusoidal waves at +-60V, by amplifying them through ADA4870 and PA107 evaluation boards. My hope is that transmitting square waves will result in an equally good result, enabling us to design a simpler pulser circuit for the final product instead of constructing the expensive sinusoidal variant using DAC and amplifiers. I was thinking if there were some digital / pulser solution in which the original square wave might be used to open / close some MOSFET transistors, consequentially creating a higher voltage square wave on their output. <Q> You want to drive a capacitive transducer at 3.5MHz with 120Vp-p, which is a fairly tall order. <S> The peak power is rather high (3A peak at 60V). <S> You can purchase high-voltage power op-amps, some of which have example circuits that show driving heavily capacitive loads. <S> I've done a discrete booster amplifier for driving piezos, but it's tricky to get it to work in a stable manner with that kind of heavy capacitive load. <A> Instead of an analog amplifier, consider a digital approach, in which a 60V source is switched on and off by the lower voltage square wave. <A> At 3.5MHz there has to be some available off-the-shelf component. <S> Though you could have a problem with the power. <S> 60 V on 41 Ω is equivalent to 88 W. <S> Another option is to use transistor amplifiers. <S> There are many devices which withstand 120V.
| You can increase the voltage with a transformer, or a chain of transformers.
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Decode/Analyse the following UART signals I have sent a 2 and a 9 using UART. The baud rate is 9600, there is no parity, and it has 8 data bits. As we know 9 is represented as 1001 in binary. However, this is what I get from my UART signal: A 2 is represented as 0010 but this is what I get from my UART signal: I know I have to pay attention to the start bit, but other than that, I have a hard time seing the logic in this signal. NOTE: The signal works, but I just need help analysing it. <Q> Let's annotate your first scope shot. <S> When we do this take note that the lowest order bit of the binary value is transmitted first. <S> Conventional notation in this field is to order the bits in a byte like this: [Bit 7][Bit 6][Bit 5][Bit <S> 4][Bit 3][Bit <S> 2][Bit 1][Bit 0] <S> So when I annotate and then evaluate the bit positions are flipped from the positions shown on the annotated scope shots. <S> The bit pattern represented there is 0b00111001. <S> That is the same as 0x39. <S> 0x39 is the ASCII code for a '9' character. <S> Let's do the same for your second scope shot. <S> Here the bit pattern is represented as 0b00110010. <S> That is same as 0x32. <S> 0x32 is the ASCII code for the '2' character. <A> You probably sent an ASCII 9, not a binary 9. <S> ASCII 9 is 0x39. <S> Bits are sent LSbit first. <S> 0011 <S> 1001 reversed is 1001 1100 <A> UARTs usually send the least-significant bit first <S> hence you get this for 9 and lower down for 2: - UART character frame: - <A> The bits are also sent LSB first. <S> That's why you see a leading 0 start bit, and then 10011100 for '9' and 01001100 for '2'. <A> The 2 that you sent is actually the character two, not the decimal value of 2. <S> The ASCII code for 2 is 0x32 or 0b00110010, which matches one of your images. <S> Also, the bits are sent from LSB to MSB, so the levels you see on the 'scope would be 0,1,0,0,1,1,0,0 from left to right.
| First of all, those are not numbers 9 and 2, but ASCII character symbols '9' and '2', which equal to 0x39 and 0x32.
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What is the purpose of this disk over a Samtec connector? With beautification in mind I decided to attach 3D-STEP models to elements on my PCB. One of them is a connector from Samtec. They do provide a STEP models for a download, but it was immediately suspicious to me, that the preview had this weird disc over the connector. At that moment I thought it was some kind of glitch and proceeded downloading STEP model regardless. See the image below: Attaching the downloaded STEP model I was surprised by this disc again. Turns out, it was also exported into STEP model along with the connector. Unfortunately, this serves bad for the purpose of "beautification" :) Questions are: What is the purpose of this disk? Is there any way to download a STEP model of Samtec connector without this disk? For a reference, the connector is QSH-030-01-F-D-A-K <Q> That orange disk represents a pick-and-place sticker—it provides something for the pick-and-place machine's vacuum head to grab onto. <S> Without this, you'd have to assemble the connector by hand. <S> Here is an example photo of a similar Samtec SS4 with the vacuum pick and place sticker in place: <A> This hasn't been explicitly mentioned yet, but you got a disc because you asked for one. <S> It's a Polyimide film pad, and specified by the last code in your part: <S> -K. <S> The purpose of this pad is so the vacuum pick-and-place machine has something to hold the component on. <S> It needs a flat surface big enough for the vacuum head, as already shown by Jeroen3 and DKNguyen . <S> For manual manufacturing, you don't need the film pad. <S> But some automated lines can't pick up the parts if there's no flat surface provided. <S> QSH-030-01-F <S> -D-A-K has the pad, QSH-030-01-F <S> -D <S> -A does not. <S> 030-01 says how many positions and the lead style. <S> The other codes: -F: <S> 3u plating -D: double rows <S> -A <S> : alignment pins -K: <S> Polyimide film pad See the QSH configurator . <S> Find your new, padless STP files there too. <S> Polyimide (also known under the brandname Kapton) is stable for a wide temperature range and a good electrical isolator, making it ideal for this purpose. <A> Taken from Adafruit <A> The polyamide pick and place disk is part of the assembly as defined. <S> If you have a 3D CAD program that uses of one of the very many supported formats you can simply open the file and delete the disk, then save it as an AP214 .stp file. <S> Otherwise you are at the mercy of whoever is creating these files and the options they happen to offer. <S> You can probably find a demo or free version of one of the supported pieces of software.
| It is a place for the pick-and-place machine's vacuum head to grab onto.
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Proof of linearity of a transistor When we deal with a transistor we break it up into two parts, 1 The DC bias.2 The small signal perturbations.Then we add the currents/voltages from the both the analysis to get the total response.But what is the proof that we can do this procedure? As an example say I've a DC source V and Ac source W. When I apply V at VCB I get x as a response.Then I use the small signal model and find that when I apply a small voltage W I get a response y. The total response is then x+y.However isn't it possible that if i apply V+W together I would end up with a different response? <Q> You cannot prove the linearity of a transistor, because a transistor is non-linear. <S> However, it's convenient to make linear models to simplify calculations about its operation. <S> By definition, those linear models are linear. <S> This is a 'small signal' model, explicitly stating the fact that it's only valid for 'small' variations around that point. <S> Small signal models cease to be useful when we want to use them over parameter <S> ranges where the non-linearity becomes apparent, that is, for large signals. <A> As an example say I've a DC source V and Ac source W. <S> When I apply V at VCB <S> I get x as a response. <S> Then I use the small signal model and find that when I apply a small voltage W <S> I get a response y. <S> The total response is then x+y. <S> However isn't it possible that if i apply V+W together I would end up with a different response? <S> Of course if you apply V+W together you end up with a different response (NOT x+y). <S> This is because the circuit is nonlinear so superposition doesn't apply. <S> Separating the dc (bias) and ac (small-signal) response is NOT superposition, though it resembles superposition superficially. <S> What you're actually doing in this case is applying a first-order linear approximation. <S> Do you remember something like this in high-school calculus: https://steemit.com/steemiteducation/@masterwu/estimating-2-001-5-with-differentials-and-linear-approximations <S> In the image, x is your dc bias voltage and Δx is your ac signal. <S> The true response is f(x+Δx) <S> (nonlinear), but you can approximate the response as f(x) + f'(x)Δx, which is the dc response, f(x), plus the linearized ac response f'(x)Δx. <S> The term f'(x) represents the small-signal gain and is of course dependent on x, the bias point. <S> Remember that small-signal analysis gives approximate results that are 'close enough' if the excitation signal is kept small enough. <S> Superposition applies only to linear circuits and gives exact results. <S> If you want a proof for why superposition works, that's a whole nother answer. <A> OK - if I understood you correctly, you want to modify the DC bias voltage as well as the signal voltage - both at the same time? <S> Of course, you can do this if you want. <S> However, it will be a problem for you to CALCULATE the timely response of a circuit (output voltage) by hand. <S> It will be too involved. <S> But you can try to use a circuit simulator and perform a TRAN analysis in the time domain. <S> In this case, the computer will do the job for you. <S> However, this procedure has nothing to do with linearity. <A> [ summary: <S> at end of answer, we predict distortion for "small levels of input" ] <S> From what I recall, the IP2 'and IP3 of a standard bipolar transistor, tho slightly different numbers, are about -10 dBv. <S> That means the "Input Intercept" plots, those plots having both the 1:1 and 2:1 logarithmic lines for 2nd order distortion or the 1:1 and 3:1 logarithmic lines for 3rd order distortion will intercept at about -10 dBv. <S> Thus, for either 2nd or 3rd order, the produced_distortion has the same voltage as the input signal, at input of -10dBv. <S> What is we dropped the input level by 50dB? <S> The 2nd order distortion would drop by 50dB, as predicted by 1:1 and 2:1 lines. <S> The 3rd order distortion would drop by (2 * 50dB), as predicted by 1:1 and 3:1 lines.
| The way we make a linear model is to take the slope of the curve at a particular point, and use that as the 'linearised' response, around that point.
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Why doesn't this capacitor short this circuit? I'm doing a super simple Arduino led project. I followed a tutorial online that wired up the LEDs like this: I'm wondering why this capacitor doesn't just short the power supply. Can anyone provide me with some insight? <Q> The capacitor is in fact a short circuit, however only temporarily . <S> When you first turn on the power supply, the capacitor will act like a short circuit during this initial transient phase. <S> There will be a large inrush current as the capacitor charges up (*). <S> Similarly, if you take a fully discharged capacitor, and connect it to a power supply which is already at some DC voltage, you will get a sudden inrush current into the capacitor which if large enough (e.g. very large capacitor) can be enough to overload the power supply. <S> This is especially problematic if switching a large capacitive load into a circuit. <S> If however you have some noise or ripple signal, or a voltage transient, the capacitor will again act like a low impedance or short circuit (frequency dependent) for this noise signal or voltage ripple. <S> This is how the capacitor has the effect of smoothing out the power supply. <S> (*) <S> The power supply may provide some form of inrush current limiting or controlled voltage ramp-up, in which case the current flow will be limited. <A> There is no conductive path from one terminal to another in a capacitor. <S> The symbol shows this. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Non-polarised, polarised and variable capacitors. <S> The symbol represents two parallel plates separated by an air-gap or a dielectric (insulator). <S> Figure 2. <S> An unwound electrolytic capacitor showing the two layers of foil and one of the insulation layers. <S> Source: Translators Cafe . <S> The capacitor can't pass DC current but it can accept charge and if the voltage across it varies current will flow in and out of both terminals. <S> In this manner it is able to pass an AC current through it. <A> It is possible, at least theoretically, to damage a power supply by connecting a discharged capacitor across its terminals. <S> However, in real life the characteristics of the power supply and the capacitor usually prevent this from happening. <S> When you first connect the capacitor to the supply a large current will flow to charge the capacitor. <S> The magnitude of this current is limited by the output resistance of the supply and the effective series resistance (ESR) of the capacitor. <S> Neither of these resistances are shown on your cartoon schematic, but they are there nonetheless. <S> As long as these resistances limit the current to a safe level no damage will occur. <S> Another important consideration is that this surge current lasts for only a very short period of time...once the capacitor is charged to the supply voltage you will see relatively smaller currents. <A> Because as far as DC is concerned, a capacitor is an open circuit. <A> You can imagine the capacitor as a battery. <S> The charge on a capacitor is dictated by voltage <S> and it's capacitance. <S> Since capacitance will be static and charge doesn't teleport, we can see that as charge increases, the voltage increases. <S> The voltage can only go as large as what's feeding it <S> so there's a maximum amount of charges the capacitor can hold, which is again dictated by the voltage. <S> Its purpose here is to provide a current source for the LEDs. <S> There's a few reasons for this, the top one is that large instantaneous current draws cause a droop in the voltage from the power supply, so the capacitor has a lower voltage across it, which makes it lose charges. <S> So it turns into a current source for the short amount of time of the droopped voltage. <S> An additional fun question to ask yourself is why you had to add a resistor to the control signal going to the LEDs! <A> A capacitor is two electrodes separated by an insulating layer. <S> In some cases the insulating layer could be air or vacuum. <S> In other cases it could be a dielectric (such as some type of plastic or ceramic). <S> Unless there is some type of fault, such as dielectric failure, the capacitor is not a short-circuit at DC, by the definition of a capacitor. <A> DC doesn't travel through a capacitor as long as its charged. <S> The capacitor on LED strips is there to dampen sudden Amp changes in the system, when your LED Strip suddenly needs more power etc. <S> The thin cables on these PSUs are most often too high inductance to manage these fast, big shifts well. <S> This is a great answer to that. <A> The capacitor was connected in parallel with the circuit. <S> Current reaches both at the same time and it would be too late for it to protect the circuit. <S> This would have been different if the connection was done in series inline with the circuit. <S> The current would have to reach and pass through the capacitor before it reaches the circuit to protect it.
| Once the power supply reaches its steady-state DC voltage, and the capacitor charged up to the same voltage, the capacitor will no longer act as a short circuit on the supply because it cannot conduct a DC current as explained by the other answers.
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Using a 13.8 VDC 30-50A switching power supply on a 15A house circuit Want to use a bench top power supply to run 2 - 40Watt motorola radios. What is the max amp rating to use safely on a 15A home circuit? <Q> At a rough guess, the radios will each draw under 5 Amps while transmitting (and well under 1 amp while receiving.) <S> With both transmitting, that would be 10 Amp at 13.8 Volts. <S> The power supply will draw a little over 1 Amp at 120 Volts, so there will be no problem powering the system from a 15 Amp home circuit. <S> (and the power supply you suggest is overkill for the task.) <A> Those figures are taken from both a Alinco desktop PS and rack mount of unknown origins. <S> Those supplies are both fairly new. <S> If we are talking about an OLD one, I would cut it down to maybe 13.8VDC @ 20-25A. <A> The recommended power supply for Motorola <S> 25-60W radios is the Motorola HPN4007D. <S> It's a 13.8V DC 15 A Output Switched Mode Power Supply working off 110-220V AC input. <S> Your 13.8V DC 30-50A Switched Mode Power supply would be more than adequate to power your two 40W Motorola radios. <S> It would draw only around 5A from the 120V~ utility supply.
| Considering that a common PS that can put out 13.8VDC @ 30A(con)/50A(int) will have a ~15A fuse, I would say about 25 - 30 amps.
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MOS current mirror region of operation The mosfet M1 is clearly in saturation as gate and drain terminals are shorted. Why we take M2 also in saturation? I know Vgs1 = Vgs2 and they are identical MOS but i cant justify why M2 should be in saturation, as drain voltage of M2 can be any value.. <Q> The drain voltage of M2 "can" be of any value <S> but it's only useful as a controlled current sink when Vgs is high enough that the pair functions as a good current mirror. <S> The compliance voltage for saturation (where the current is mostly constant) is: Vout <S> > <S> Vgs - Vt <S> so if you are operating M1 at a current that creates a Vgs only slightly above Vt (in other words, a very low bias current) you can have Vout almost zero and still get constant current. <S> Here you can see a current mirror compared at 100uA/1mA/10mA and the behavior as Vds increases from 0V to a few hundred mV. <S> At higher currents, the compliance range of voltage is less (you need more Vds to maintain a constant current). <S> Vgs is about 1.41V to 1.73V in this simulation, depending on current. <A> If M2's drain voltage drops too low then of course it won't be in the saturation region but, importantly, neither will it be a good current mirror. <S> The red markers on the curves below show where the diode connected MOSFET M1 sits: - <S> So there is some scope in some places to drop the drain voltage of M2 below that of the diode connected MOSFET and remain reasonably in the saturation region. <S> It all depends on the particular MOSFET characteristic. <S> The 2N7000 for instance has a fair bit of leeway to drop below M1's drain voltage: <S> - No real MOSFETs were harmed in the creation of the above pictures. <A> According to Sedra/Smith text (because textbooks do a better job at explaining things than I do), Microelectronic Circuits , on the subject of basic MOSFET current mirroring: <S> Now consider transistor \$Q_2\$ (or \$M_2\$ from your picture): <S> It has the same \$V_{GS}\$ as \$Q_1\$ ... <S> (from Figure 8.1) For proper operation, the output terminal, that is, the drain of \$Q_2\$ , must be connected to a circuit that ensures that \$Q_2\$ operates in saturation... <S> To ensure that \$Q_2\$ is saturated, the circuit to which the drain of \$Q_2\$ is to be connected must establish a drain voltage <S> \$V_o\$ <S> that satisfies the relationship <S> \$V_o \geq V_{GS}-V_{tn}\$ . <S> – <S> Microelectronic Circuits <S> ed. <S> 7 <S> by Sedra and Smith <S> So in order to have a proper current mirror, we have to assume that the M2 is in saturation. <S> Drain voltage for M2 can be arbitrary (to an extent) with that condition of \$V_o \geq V_{GS}-V_{tn}\$ .
| M2 should have a drain voltage no lower than M1's to ensure it remains in the saturation region but some devices may be able to go lower without entering the triode region too deeply.
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Is it legal to create your own stm 32 board Is it legal to create your own board containing STM 32 controller and advertise it commercially? I have made a custom board and want to advertise it - is it OK? Any legal/copyright issues? If yes how do I address them so that I can advertise safely? <Q> it might not be. <S> And that's without even getting into potential consumer protection laws. <S> Sometimes it's cheaper to give things away than to take money. <A> Sure! <S> After all, ST wants to sell chips. <S> Also most of the time, chip manufacturers don't profit that much or at all on dev boards (some are even sold at a loss) <S> so I don't see why they would reject the free publicity. <S> Assuming ST accepts reuse of parts of the reference design because that sells chips, the only part that could cause trouble would be the name, so make sure the name is not already copyrighted. <S> For example don't call it "nucleo". <A> Of course you can create your own designs using the STM32 components available from the manufacturer. <S> What you would want to avoid is making an exact copy of one of ST Micro's STM 32 evaluation boards or development modules unless you have written approval and agreement from the manufacturer that such direct copy is allowed. <S> Beyond this there is the whole gambit of techniques, design rules, quality, reliability and regulatory requirements that you would have to navigate through in preparing any product for the market.
| Probably de facto legal, but if you copied someone else's layout, infringed their patents or trademarks, infringed copyright on their firmware/software (if any), copied their photos, made it look like someone else's product so that customers might be confused, or are in breach of any of the various laws that might be in your jurisdiction on EMI emissions, mains voltage connection, Pb or other allegedly hazardous materials content etc.
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Why is the soldering surface of this pogo pin slightly rounded? I was curious as to why the bottom of these SMT pogo pins are slightly rounded. Wouldn't it cause them to fall over before they are reflowed ? It seems that a completly flat bottom would make sense like most SMD pads. <Q> The curved bottom will allow for betting wetting when using solder paste, as it gives more surface area on the bottom where the solder paste actually hits. <S> This is less important when hand soldering as you generally use far more solder <S> and it will wet up the sides of the pin, but in large scale manufacturing they use much less solder paste and you want to maximize the surface area exposed to the solder paste. <S> If it was flat the bottom would lay flat against the pad when it is reflowed and give far less surface area. <S> In addition the solder paste is very viscous and and will hold the pin in place, so the pin doesn't actually need to balance on a flat bottom. <A> Interesting that this one is the only one of the 8 similar items in Harwin's line-up that has that well-rounded bottom (the others have a slight chamfer but no radius). <S> For example, this one, which is the same diameter. <S> I might speculate that this one is perhaps not intended to be soldered at all, but to be held in a molding between two PCB pads. <A> T
| I would guess that since this is a pogo and will be pushed against, the rounded surface will allow more solder on the sides of it thereby allowing better solderability.
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Choosing a large, reliable input capacitor for switching regulator A project I am designing has to be reliable for many years. I have a switching power supply using a LM2596S-5.0/TR I am having trouble deciding on the capacitors. I have placed a tantalum 10v 220 uF on the output but on the input I am having trouble. The datasheet says: A low ESR aluminum or tantalum bypass capacitor is needed between the input pin and ground pin. The example application lists a 660uF. Looking on LCSC, I can only find a couple of SMD capacitors at 50V, 660 uF and they don't mention their ESR.It doesn't help that the datasheets are in Chinese. I can't seem to find tantalums of this size and voltage either. Can anyone recommend the types I could use here? Favouring reliability if possible. <Q> For a buck converter like this, it is very important to have a low ESR ceramic capacitor directly at the input of the converter. <S> Otherwise you will have EMI problems. <S> So it is a common practice to put a small (10nF) ceramic cap directly at the input. <S> For the bulk capacitance, you can then use the proposed 680uF (low cost) aluminium capacitor. <S> (I would not recommend a tantalum capacitor for this). <A> The datasheet says — A low ESR aluminum or tantalum bypass capacitor is needed between the input pin and ground pin but the example <S> application lists a 660uF. <S> The data sheets lists a 680 uF capacitor and <S> not 600 uF <S> : - The phrase "660" does not appear in the data sheet at all. <S> The phrase "680" appears in the data sheet 4 times and each is related to the 680 uF input capacitor or the 680 pF feedforward capacitor. <S> 680 uF capacitor are as common as muck <S> but, if you want long life then choose carefully following this answer . <A> 5000 hrs @105° <S> C 90m \$\Omega\$ @100kHz <S> If you really need 50V or better rating and 680uF or more, a Nichicon UPW1J681MHD3 leaded type 18x20mm would make more sense than some huge SMT part. <S> 5000 hrs @105°C 55m \$\Omega\$ <S> @100kHz.
| Given your constraints, I would pick a UUD1C681MNL1GS SMT type from Nichicon, assuming 16V rating is adequate .
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How to combine two resistors with a voltage source simulate this circuit – Schematic created using CircuitLab simulate this circuit Are these two circuits the same? If not, could you please indicate why as well? <Q> When starting out, your brain is easily tricked by the arrangement in which things are drawn rather than what they actually are. <S> Side-by-side doesn't automatically mean parallel. <S> What if I took your "parallel" circuit: simulate this circuit – Schematic created using CircuitLab and just changed it to this. <S> Does it still look like it is in parallel to you? <S> Or series for that matter? <S> Remember, the current can flow into the circuit on those stubs from outside. <S> simulate this circuit Parallel = <S> voltage across all components are the same Series = <S> current through all components is the same <S> Don't be tricked by how things are arranged on a page. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Whatever way you draw it <S> it's a series circuit and the total resistance is R1 + R2 . <S> simulate this circuit Figure 2. <S> R1 and R2 are in parallel in this case. <A> There is a good way to understand either a combination is series or parallel. <S> For example, take a look at the circuit below. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The left side of R 1 is directly connected to the positive terminal of the battery. <S> Let's say the voltage is V a . <S> Same as this, the right side of R 1 is connected to the negative terminal of the battery. <S> Let's say the voltage is V b . <S> So the potential difference across R 1 is (V a - V b ) <S> Similarly voltage across R 2 will also be (V a - V b ) <S> If two resistors are parallel, then they will have the same potential difference between their terminals. <S> simulate this circuit Now for this new circuit, the upper side of the resistor R 1 is directly connected to the positive terminal of the battery. <S> The voltage of that terminal is labeled as V a . <S> Now there will be a voltage drop caused by the resistor R 1 <S> (N.B. Ohm's law V=IR). <S> So a new voltage level V c will appear at the lower side of R 1 . <S> And the lower side of R 2 is directly connected to the negative terminal of the battery, which is labeled as V b . <S> All series combination will cause this type of voltage change. <S> Let's take a complicated example. <S> simulate this circuit Just by inspection <S> , it's quite difficult to understand. <S> But if the node voltages are marked, then you will see that all of them have the same potential difference, which is (V a - V b ). <S> This means that they are all parallel to each other. <A> Sometimes, when you're still new to all this, it just isn't clear if two circuit elements are parallel (or series) connected. <S> If you're not sure, there's an easy 'trick' to test whether or not the two resistors are parallel connected: consider setting one of the resistors to zero ohms. <S> For example, set \$R_1=0\Omega\$ (essentially, replace \$R_1\$ with a wire). <S> It's clear that the voltage source sees the resistor <S> \$R_2\$ as the load. <S> But this wouldn't be the case if \$R_1\$ and \$R_2\$ are parallel connected since, with \$R_1=0\$ , we have $$R_1||R_2 = \frac{0\cdot R_2}{0 + R_2} = <S> 0\Omega$$ <S> That is, zero ohms in parallel with \$R\$ <S> ohms equals zero ohms . <S> Since the source doesn't see zero ohms with \$R_1=0\Omega\$ , it follows that the two resistors are not parallel connected. <A> Maybe your confusion comes from the Thevenin equivalent circuit of the one you show ( <S> the voltage source Vth=V*R2/(R1+R2) in series with the resistor Rth=R1//R2). <S> but you don't have to short circuit the output. <A>
| These are two different circuits because in first circuit the current is same throughout the circuit but in 2nd circuit if we place R1and R2 in parallel than current is divide between the resistors by the rule of KCL.
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How critical is winding a solenoid/electromagnet neatly? We've all seen the expensive toroidal winding machines on YouTube, and manufactured solenoids are always so neat in construction - as in: Each winding is neatly packed to its previous, without overlapping itself: like "-/////////-" There's no explicit numerical answer here - I'm just after an indication of significance - if a winding was to be done in a factory - perfectly rolled, no overlaps, each coil immediately next to the previous one vs. one of the same specifications done at home with a few gaps, a few overlaps, and other imperfections here and there - what is the "ballpark" performance hit of that? Are we talking 0.1% or 10%? If it's actually possible to perform some sort of calculation (with some assumptions put in there) - let’s use the example of: winding a ⌀35 mm annealed soft iron rod with ⌀1.2 mm magnet wire, wound 200 times across a 40 mm length. <Q> Compared to close-packed coils, how much space is inside a random scramble-wound coil? <S> Not 10%, I bet it's more like 50%, and the total volume of the coil is nearly twice that of a close-packed version. <S> For random "scramble-wound" coils, the wire's turns/kg and the turns/km values are low, but also the average skin-depth for the entire inductor is large. <S> That's what we want, <S> so, for low-loss, high-freq RF coils we must avoid the close-packed windings. <S> Buy a fancy scramble-type coil winder, and perhaps also wind your coils "pie-wound," as a stack of pancakes. <S> But for DC or 60Hz, a close-packed coil is much smaller, but with the same gauss/watt value as a big mushy scramble-wound coil. <S> If physical size is an issue ( <S> motors for example, also solenoid actuators,) then those close-packed windings produce strong, miniature devices capable of high-wattage drive. <S> Also: vibration. <S> Cheap, poorly-made motors will fail because the windings weren't tight enough. <S> Some of the turns were vibrating, and this chewed through the wire's insulating varnish. <S> Eventually a short-circuit developed. <S> A motor with internally shorted turns will experience drag and heating, and may even "run away" into internal charring, fires. <S> With motor coils, we want the coil to behave like a solid object, with nothing inside that ever wiggles. <S> Also: cooling! <S> (Usually this wire size is well above 10AWG, intended for large transformers.) <S> Thermally, the resulting rectangle-wire coil acts like a solid metal block, with high thermal conductivity. <S> Fan-cool the outside, and the interior is cooled as well. <S> On the other hand, a scramble-wound coil is full of insulating air: more like a hunk of styrofoam than a hunk of metal. <S> It will have a smaller maximum wattage than a dense, non-scramble coil. <S> Below, as scrambley as possible? <S> A few-mH value, HF tube-amp anode choke . <S> See also: WP: basket wound <S> https://en.wikipedia.org/wiki/Basket_winding crystal radio coils https://www.google.com/search?q= <S> "basket+wound"+"coil" rectangular magnet wire . <A> Any wire looped through a high relative permeability \$\mu_r\$ core works pretty much the same magnetically regardless of position within the core. <S> If there is an air gap in the magnetic circuit then position matters a bit more, but usually not a lot in such parts as gapped flyback transformers. <S> Coils wound in a "basket fashion" as in wbeaty's photo have lower distributed capacitance and thus a higher SRF (self-resonant frequency). <S> So for RF applications, it matters a lot. <S> Coils for high voltage (typically the insulation on ordinary magnet wire is not rated for much voltage and cannot be relied upon for either safety or for functionality to withstand much voltage) <S> may be wound on segmented bobbins and it's necessary to avoid the ends of the coil from being too close to each other or crossing over. <S> Messy winding may mean you can't use as much of the winding window that you would expect, so you're not using the core as efficiently as possible. <A> If we're talking about magnets or solenoids, which don't actually do any significant amount of work when turned "on" in a steady state, then: For a given coil geometry, the efficiency (regardless of wire gauge) is determined by the amount of heat produced per ampere-turn. <S> For a given number of ampere-turns (= magnet strength), that amount of heat is determined by the resistance of the coil material. <S> If your coil is 50% empty space, then the average resistance of the coil material is twice as much as a densely-packed coil, and your magnet is therefore only 50% as efficient. <S> That's a pretty big deal.
| We can buy specialized coil-winding wire with square or rectangular section, which lets the windings pack together with minimal gaps. Depends on your performance parameters.
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Obtaining 4-4.2V rail from 5V output of LM7805 voltage regulator I'm creating a numitron clock with 4xIV-9 tubes (seven-segment filament displays) and have decided to illuminate them using shift registers controlled by an arduino. Each segment takes a voltage of 4-4.2V and around 17-20mA of current. My voltage I have available is a 5V output voltage from an LM7805 voltage regulator IC. I need to power 4 high current shift registers (TPIC6B595), each with an expected maximum output current of 200mA. Owing to this, the 4-4.2V rail needs to be able to handle a maximum current draw of around 800mA. I was considering using a simple potential divider however calculated that the required power losses greatly exceed that of the standard resistors that I have available.I was also considering using a diode to drop the voltage however the forward voltage is a function of the current; therefore as the current can vary from the extremes of 0mA to 800mA I did not think a diode would be appropriate. Any suggestions on how to obtain 4-4.2ish V rail would be much appreciated.Many thanks. <Q> Replace the 7805 with a 1Amp or more adjustable regulator and set it to 4.2 volts. <S> Run both the arduino and everything from that instead of running it off the 5V. <S> The Arduino will be fine with 4.2 volts without any issue, and it makes it easier since you dont have to worry about level translation for the input lines either. <A> Since 4.2V is the end of charge voltage of LiIon cells, there are plenty of off the shelf <S> 4.2V power supplies available. <S> Failing that, you can use a SMPS or wall wart with adjustable voltage, or simply change the feedback resistor in a 5V 1-2 Amps USB "phone charger" to make it output 4.2V instead of 5V. <A> I am assuming that the input voltage that feeds the 7805 regulator is significantly higher than 5V. <S> If so, I would use a small adjustable SMPS Buck Regulator that operates from that input supply. <S> You can either build one yourself or simply purchase one of the very inexpensive boards / modules from your favourite supplier. <S> Set the output voltage to the desired value. <S> The advantage of this approach is that you are not dissipating all that power in the 7805 regulator. <A> Use an adjustable voltage regulator. <A> Use a low-dropout adjustable voltage regulator.
| The Atmega microcontroller in the Arduino will run fine at 4V, so the simplest solution would be to power the whole thing from a 4-4.2V switching power supply instead of using a voltage regulator. Measure the output voltage of the 7805 and make sure you are really getting at least 5V, then look for a regulator that will work with an input-output voltage difference of, say, 0.5V or less.
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Purpose of 1000uF capacitor in input The board connected to power supply with 50-100m cable. There is C1, 1000uF 50V capacitor on input of board, next to power connector. Schematic of power circuit is below. I think it is for bypass or decoupling. It is really big in physically, so i think the purpose of it and any different solution instead of 1000uF cap. Also, 20-30 board can be connected in same power line sometimes. All boards are same. Has any impact on purpose of that cap? Thanks. <Q> One of the purposes of C1 is to help damp-out input voltage overshoot. <S> The 50 to 100m cable has a lot of series inductance. <S> If you were to put, for example, a 10uF ceramic cap in place of C1 you will probably have a very large voltage overshoot when you first connect the regulator. <S> Very possible to blow up the regulator first time you plug it in. <S> I do not know if 1000uF is needed. <S> But if you make any changes, make sure you test for overshoot. <S> Electrolytic caps generally don't overshoot as much as ceramic because their effective series resistance provides damping. <S> It is also possible to add a resistor in series with a ceramic cap to get the same effect (but you may have trouble finding a 1000uF ceramic cap!). <S> The amount of resistance needed for damping may less than an Ohm. <S> Simulation can be done if you have very accurate models. <S> Application Note 88 by Goran Perica of Linear Technology goes into a lot of detail about input voltage overshoot. <A> 30 power supplies like this taking their input from the same about 24VDC voltage source can cause something that you have not expected. <S> Buck converters take current pulses which all are as strong as the load current at the +5V output. <S> The total current is the sum of the pulses of all regulators. <S> They are not synchronized so in theory there can be moments when every regulator wants current at the same time. <S> My suggestion: <S> Do a simulation where the feeding lines are realistically simulated and let there be a bunch of current sources which take realistic pulses. <S> Then you can see if the capacitor can be reduced. <S> Placing an inductor in place of D1 can smooth the input current pulse, but the total effect with cables and 30 regulators must still be simulated. <S> C2 has manufacturer's suggestion in TI's datasheet, do not omit it. <A> a good value for dampening is sqrt( L /C); this produces Q = 2, so has some ringing. <S> With 10 uh and 1,000 uf, you have sqrt (10/1,000) == sqrt (0.01) = 0.1 ohm. <S> And you need several times that, to avoid all overshoot. <S> Plus this assuming 10uH.
| Also, if you add a fuse or PTC to the circuit, that may provide some damping. Otherwise you can test it and monitor the overshoot with an oscilloscope.
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How do you call wire pins that can be used as jumpers with your breadboard? I can't seem to find these things on electronics distributor websites. There is a bunch of various pins and connectors but I can't find these among them. What are they called? I tried "dupont 2.54mm connector/pin" and didn't get the results I expected. Thank you. <Q> I've always heard these referred to as "Male crimp pins". <S> I won't post any links or recommend any places to buy them as that would be against the rules of this site, but it should give you a good search term. <A> The generic term is "crimp contacts" if i'm not mistaken. <A> 0.1" pitch 0.025" <S> square pins: Once upon a time these things were called "Berg" connectors since then several other manufacturers are making compatible connectors for a while <S> there were just called "100 mil square pin header" connectors <S> Berg was aquired by FCI and the parts were called "Berconn Mini Latch SR" <S> FCI was aquired by Amphenol and the parts are now called "PV™, Basics+ Series". <S> Pin part number 75653-002LF single pin Housing part number 65039 <S> -036LF <S> but it's certaily easier to say than "PV™, <S> Basics+" unfortunnately the real Dupont seems to not be acssociated with this in any way. <S> and major distributors are not using the term Dupont.
| Somwhere along the line someone started calling these things "Dupont" I have no idea why they did that.
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How to detect movement of gadget without electricity I'm trying to make a device which needs to be turned on when it's picked up and might sit in standby for several weeks before getting picked up. Right now I'm considering using a Photo resistor at the bottom of my device or using a piezoelectric sensor for this purpose, but both of them use electricity and might drain the battery before actual usage of the device. Is there any other way for a device to detect its own movement without draining its battery? <Q> Figure 1. <S> A selection of microswitches. <S> The lever types are very sensitive so little force is required to operate them due to the lever's mechanical advantage. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> Wiring is as a simple as using the normally-closed contact (which will be usually open when the weight of the container is applied to the switch). <A> When moved, the spring contacts the pin, closing a circuit. <S> Gravity alone isn't strong enough to close the circuit. <S> They are surprisingly effective in all dimensions, not just in the axis perpendicular to the pin. <S> They are available in a variety of sensitivities and constructions, including bare and enclosed. <S> I have seen versions made of just coiled steel wire soldered to the board. <S> Source: <S> Adafruit Similar to Solar Mike's suggestion <S> , these avoid any energy usage while not triggered. <A> Should the unit be stored in a dedicated holder, a 'NC' type magnetic reed switch would suffice. <S> The reed switch could be mounted in the unit and wired in its supply line. <S> A magnet, located in the holder, would keep the reed switch open. <S> The reed switch would close and the unit would power up, on retrieval from it's holder. <A> If it is a box, you'd need to put some conductive tape or spray conductive paint on all sides on the inside. <S> This adds a feature: the device can know where it is touched. <S> For example if it is a box, it will be able to detect fingers on all sides independently. <S> So it will know if it is touched (one finger detected) or picked up (at least two fingers, on opposite sides). <S> For example, MSP430 microcontrollers implement capacitive sensing with very low power: <S> The few µA required are lower than battery self-discharge. <A> A metal ball that moves and connects contacts - no movement <S> no connection no current drain. <A> I am assuming that the object picked up is an eCommerce package left at the door step or something similar. <S> Base on the above assumption <S> This should address the requirement for no battery, lack of power or low power. <S> Take a look at solutions offered by enocean self powered IOT solutions <S> Below is an example of a push button transmitter module. <S> The following is used as light fixture switch. <S> The system uses on/off motion of mechanical light switch motion to generate sufficient electrical power. <S> The electrical power is generated using the built in electro dynamic generator. <S> Functional Principle <S> The electro-dynamic energy transducer is actuated by a bow, which can be pushed from outside the module on the left or right by an appropriate pushbutton or switch rocker. <S> When the energy bow is pushed down, electrical energy is created and a RF telegram is transmitted including a 32- bit module ID (PTM 210J optional 48 bit). <S> Releasing the energy bow generates different telegram data, so every PTM telegram contains the information that the bow was pressed or released. <S> Alternatively you could use a pressure sensor, but this would require power. <A> A magnet near a coil of wire only generates a current when the magnet is moving. <S> Put a spherical magnet in a short and wide box with a coil on the lid (by "short", I mean similar height to the width of the magnet). <S> If the device is overall flat in shape you might be able to make the coil out of a corner of the main PCB and just clip a plastic cover over that area to keep the magnet in the appropriate area, or even design it into the case. <A> Possibly not quite what you're after, but you could tether the device with an easily removed string. <S> When you pick up the device, the tether comes undone, and the device knows it's been picked up. <S> This would be sort of the reverse of a "tether kill switch" that you might find on a motorbike or jetski: https://www.mpsracing.com/products/MPS/hc01.asp
| When the device is moved, the magnet will roll around, generate a current in the coil, and that could be enough to power a startup circuit that turns on the main one. You could use capacitive sensing to detect when the device is picked. Microswitches are available in a range of formats, are very small, are mechanically simple and very reliable. If you design the circuit for lowest standby power, it doesn't really matter if it is drawing a few µA from a couple AAA batteries or if it is fully off. A vibration sensor switch can be built or bought using a spring, wrapped around a central pin. , I recommend look at solutions that is based on energy harvesting. I suggest taking a look at PTM 215 Push button multi-channel switch module or something similar
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Why frequency of a DC signal is chosen as zero? If the period can be anything, isn't it same for frequency? This is a page from Oppenheim. We can assume a DC signal as a repetition of 10 strips per 10 seconds/10 strips per 5 seconds. Like that, if we choose different periods, we get different frequencies also. Then why is frequency chosen particularly to be zero? <Q> If you want to treat this signal as a periodic one, then you can take its Fourier series. <S> Unlike most other periodic signals, you have free choice of what frequency to consider as its fundamental frequency --- you can calculate the series for any fundamental frequency But regardless of which fundamental frequency you choose, you'll find that all the terms in the series except for the 0-frequency one have zero magnitude. <A> Frequential analysis is normally taken in the context of integral transforms with complex exponentials - such as Fourier and Laplace - which can alternatively be explained as infinite sums of trigonometric functions, with varying phases and amplitudes. <S> The only frequency value \$\omega\$ <S> that will allow you to represent a constant with a trigonometric function, such as \$cos(\omega t)\$ , is the value of \$w = 0\$ . <S> Using any other frequency value will instead represent an oscillation in time-domain, therefore not a constant. <A> As you wrote, "Like that, if choose different periods, we get different frequencies also." <S> So, I think the problem is the author's saying that a constant signal is periodic "with period <S> \$T\$ <S> for any positive value of \$T\$ ". <S> The more normal way to understand the period of a constant signal is, that it is \$\infty\$ . <S> Otherwise, it can lead to a confusion regarding the corresponding frequency. <S> By duality, frequency could similarly be any value we choose, and hence undefined, if we allow the period to be any positive value of \$T\$ . <S> So, we should define the period of a constant signal to be \$\infty\$ , and by duality, it follows that the frequency is unambiguously \$0\$ . <A> First the frequency difinitionfrom wikipedia link <S> https://en.wikipedia.org/wiki/Frequency <S> Frequency is the number of occurrences of a repeating event per unit of time. <S> The relation between the frequency and the period, T of a repeating event or oscillation is given by f=1/T --for the DC (real direct current from battery) there is an infinite repeating event(very big period) <S> so f=1/∞ means that f=0 <A> The fragment you quoted doesn't mean that "the period can be anything". <S> The statement "x is periodic with period T" means that x(t) = <S> x(t+T) <S> for all t. <S> This is not the same statement as " <S> the period of x is T", because being periodic with period T also implies being periodic with period 2T, 3T, etc. <S> (this is easy enough to work out on your own). <S> What we call the period of x is the smallest T for which x is periodic. <S> But if x(t) = x(t+T) <S> regardless of T, because x is a constant function, there is no smallest T to choose. <S> Therefore the period of x is undefined , not arbitrary . <S> As for why that should correspond to a frequency of zero, consider the family of functions cos(2π f t). <S> With decreasing f, period gets longer and the frequency gets lower. <S> In the limit as f goes to 0, we have cos(0) <S> = 1, which is a constant function. <S> This is a reasonable argument for saying that constant functions have 0 frequency.
| The period is the duration of time of one cycle in a repeating event, so the period is the reciprocal of the frequency.
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Help with a PNP Switch simulate this circuit – Schematic created using CircuitLab I'm trying to design a simple switch for a electric valve. I'm using a 2N4033, and I'm providing 12.3V between the emitter and the collector. For the base, I have a voltage divider of a 100k and a 1k, which provides the base with about 12V. This should cause Vce to be 0, as the voltage difference between Vce and Vbe is under 0.7V. However, when everything is powered on, the Vce is 12V, when it should be zero. Can anyone help with what I'm doing wrong? Edit: Added circuit diagram. The SPST is the control mechanism for turning the circuit on and off. The valve is represented by the inductor. The current draw from the valve is 540 mA at 12V. My power supply can supply 2A at 12V. <Q> Thanks for the question and for the schematic :-) <S> With the SPST switch open, the resistive divider settles the base at about 120 mV, which is below the opening voltage of a BJT B-E junction. <S> So the transistor does not conduct from C to E. R2 <S> is too high to have any meaningful effect <S> , you could as well leave it out. <S> Note that you do not have a resistor in series with the SPST switch. <S> If you close the SPST switch = make it conduct, the circuit between your 12V PSU, the switch and the BJT's B-E junction has no defined current limit. <S> If your 12V PSU is strong enough, it will blow the base connection out of the transistor, or fry the chip, or both. <S> Check how much current your transistor can sustain. <S> In a randomly chosen datasheet for the commodity 2N4033, I cannot see Ibe(max) specified, but Ic(max) is 1A, so you can probably consider that. <S> Actually much less should be sufficient. <S> To allow for Ic=1A, considering a conservative Hfe=50, you could ballpark that you need to limit the base current to 20 mA. <S> At about 11 V available differential voltage (the B-E junction will clamp at about 0.6 V) <S> the resistor should be 11 / 0.02 = 510 Ohms (a particular standardized value in the series). <S> EDIT: <S> See Spehro Pefhany's answer. <A> When the switch is closed, the transistor will be destroyed, with the circuit as shown, as the base current is not externally limited. <S> R1 and R2 do nothing of value. <S> The 2N4403 is pretty marginal for switching 540mA <S> and you should have a diode across the valve coil or the transistor will be destroyed when it switches off. <S> Since the load is inductive, I think the 2N4403 will be outside the SOA limits and will fail either immediately or after a short time, even if properly driven with ~50mA base current. <S> This answers your immediate question, but there is not enough information to make helpful suggestions on how to do whatever it is <S> you are trying to do. <S> A MOSFET would be better in this voltage/current range. <A> I suggest looking at a MOSFET circuit And please add a diode in inverse parallel to your inductor to prevent voltage spikes from destroying your components, since inductors can create large spikes when switched off
| I suggest that you put a resistor in series with the SPST switch.
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Why is transistor needed when using a relay? I will start building a relay based on this answer: https://electronics.stackexchange.com/a/464345/56969 Why is T1 needed? Every other component has an important purpose. But does T1 has an important purpose as well? Can I just remove it? What would happen if I remove that transistor? Edit I know Arduino cannot supply more than 40mA on each pin and that is the reason why there is a separate power supply to turn on the relay. I guess my question should have been "Can the optocoupler supply 100mA of current?" If so, that means I can remove the transistor and have fewer components. <Q> Image from linked question: - <S> Basically the opto-coupler cannot provide enough current at low enough voltage drop to turn on the relay coil shown. <S> The transistor acts as a power buffer and it "delivers the goods" with a small input signal power from the opto-isolator. <S> Addition <S> - Pick the DC/AC contact rating you need and the isolation voltage and if it fits your application then you're good to go. <A> A microcontroller pin can't supply enough current to turn a relay on. <S> It can supply enough current to turn a transistor on though, and a transistor can provide enough current to turn a relay on. <A> If you are sure that the relay coil current is less than 100 mA, then yes, you can eliminate T1 IF you can find an opto with a current rating of 150 mA or more. <S> They are available, but cost far more than a 10-cent transistor. <S> Example: LCA710 <A> - Extended comment <S> Your question and a simple answer are guaranteed to cause design errors from vague assumptions. <S> This is due to the lack of details and specs and ought to be revised and deserves to be rejected. <S> How do you know how to read specs to guarantee 100mA output if only the input current max is given and temperature rise is unknown and effects on CTR etc. <S> BJT's unless binned, sorted and purchased with guaranteed CTR (or effectively hFE) with protection for exactly your correct specifications, design failure may occur. <S> The optoisolators eliminate conducted EMI coupling issues but not radiated. <S> Optoisolators vary widely with tradeoffs for speed, CTR, cost significantly yet CMOS and BJT switches are cheap. <S> This is due to the challenges of extra conflicting processes for optical and doped silicon chemical processes. <S> Opto-isolators have inherently wide hFE's and thus CTR typically 50% to 600% on the best over environmental range <S> whereas CMOS switch <S> RdsOn only varies + <S> /-50% typ. <S> yet CMOS <S> Opto's are very expensive, reliable is better than BJT Opto's were designed for high power SSR's, IGBT's and MOSFET bridges. <S> yet mechanical relays can be cheap and can have current gains <S> > 2k <S> An adequate question must have ALL the requirements including cost, qty, ambient As suggested by @Analogkid there are now more modern solutions using CMOS Opto switches intended for high (power & cost) systems. <S> Your question is almost void of specs. <S> Bottom Line or Lesson to learn Learn how to read specs and write design specs in the same depth. <S> This is not a secret but rather essential to good designs and becoming a good Engineer. <S> "the better the question, the better the answer" or the more time you spend in good specs <S> , the easier it is to design fault-free and with more options. <S> It should not contain any specific part numbers.
| You may be able to replace the opto, the transistor and the relay by using Panasonic's PhotoMOS product range:
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Why do I need phase margin if I know the transfer function? What is the point of examining the phase margin (or gain margin) for a closed-loop system if I can just solve for the transfer function. The transfer function will give any poles and zeros, which can be used to know if your system is stable, the step response, etc. In fact, the Q of a two-pole system can be solved in terms of phase-margin, and vice versa. <Q> Well they are margins , as in a margin to (possibly) becoming unstable. <S> Try reading about it in Chapter 10 of Feedback Systems: An Introduction for Scientists and Engineers , they answer your question on their first phrase <S> In practice it is not enough that a system is stable. <S> There must also be some margins of stability that describe how far from instability the system is and its robustness to perturbations <A> Why do I need phase margin if I know the transfer function? <S> Short answer: <S> you don't <S> But if all you have is a real device that may become unstable then a physical measurement may be all you can do. <S> The physical measurement may also hint where the poles might be but for sure, the physical measurement will deliver phase margin or gain margin. <A> You're right in that the transfer function alone determines whether a closed loop system is stable or not. <S> However, it turns out that as we approach instability (when the phase margin or gain margin gets close to 0), a lot of undesirable effects appear. <S> Consider the step response of a two pole system whose DC response is 1. <S> Clearly the response approaches a constant DC voltage with time, but there is a transient portion towards the beginning. <S> There is a time it takes for the system to reach the correct value (rise time). <S> It is possible for the response to overshoot and have to come back down before settling on the final value (ringing). <S> It turns out it is possible to relate these quantities such as rise time or overshoot to the phase margin and gain margin. <S> These results imply (I haven't shown it here <S> but I'm sure you can find something online)that if the margins are smaller <S> then we get larger overshoots, more ringing, and more undesirable transient effects. <S> That's why it's not only important to know whether the system is stable, but also how close to stable it is. <S> In addition, as jDAQ said, having larger margins gives you larger room for flexibility in the case of tolerances.
| Having the calculated transfer function being stable with a very small margin means that any change (or error) in your model or controller will mean that that stable transfer function will probably only be stable in your calculations.
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Would you approve this soldering job? I am new to soldering, and I am not sure if I did a good job while trying to solder some pins into a HX711. My concern is that the solder in SCK and VCC are touching (also E+ and E-), so I thought this might create some electrical noise, leading to mis-measurements. <Q> It could certainly serve as a bad example. <S> Heat applied to the wrong place, no flux, some combination of those. <S> Fortunately there are plenty of videos showing the right way to do it. <A> I'm guessing how this solder-job went wrong... <S> The soldering iron tip was applied to the rectangular pin, heating it sufficiently to melt solder. <S> The soldering iron tip was not applied to the tinned printed-circuit board pad at the same time . <S> Since the pad received no heat, it didn't attract any solder. <S> Both pin and pad must be heated by the soldering iron tip at the same time . <S> After heating for a second-or-two, push solder into the area between pin and pad...rather than pushing solder into the soldering-iron-tip. <S> Solder containing flux should wick nicely into the space between pin and pad. <S> Then pull the tip away from the joint. <A> No, I wouldn't approve the job. <S> The grey frosted look of the two solders near your thumb are a pretty clear indication of a cold solder joint. <S> That could fail unpredictably, and Murphy's Law says at the worst possible moment. <S> The B+ node looks like it might be shorted to the B- node. <S> In general, there's too much solder at each joint, to the point where shorts are likely. <S> It also doesn't look like <S> the plating on the through holes was heated at all. <S> As a rule, heat the joint , not the solder, and then touch the solder to a part of the joint you believe is hot, and not the iron tip. <S> If the solder melts, you're closer to doing it correctly than you are now. <A> No, those joints are not ok in their current form. <S> However it is likely easier to complete the remainder of the joints before going back and cleaning up the bad ones, getting a good solder joint while at the same time trying to stop a part falling out <S> is much harder than doing the two jobs seperately. <A> There is not a single solder point on that image that isn't plain ridiculous, with far too much solder but none actually flowing on any solder pad. <S> You say "I am not sure if I did a good job". <S> That is really hard to believe since comparing with any circuit board should tell you how a sane solder joint looks, not with a ball of solder hanging in the air but with a (usually slightly concave) cone of solder covering the solder pad at its base and climbing up due to cohesion on the wire which forms an alloy at its surface with the solder, like the solder pad does. <S> Insufficient heating of the contact points leads to "cold" solder joints missing that alloying and not having reliable contact. <S> They can be hard to see when they are not as strikingly bad as on your picture. <S> So you may want to have someone help/teach you before improving your technique from producing obviously bad joints to hard to see bad joints.
| As long as you start with a nicely tinned soldering iron tip at some reasonable temperature and use good flux-core solder on fresh boards and parts you should never see anything like this.
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Non inverted OpAmp & Shunt circuit not working I am trying to measure the output from a Hall effect sensor (LEM-LF 310-S) for a welding application. The maximum output of the sensor is 0-0.25A when sensing 0-500A. This range I need to convert to 0-1V for my measurement equipment (Oscilloscope for testing, later RedPitaya). So I am using a Shunt with 100mOhm which should give me an output voltage of U=R*I=0.1Ohm*0.25A =0.025V. This voltage I wanted to amplify with a non-inverted OpAmp (MCP6292) to get 1V as Output-Voltage -> Amplify with a ratio of 40 (a=32.04dB). According to an online calculator 100Ohm & 3.9kOhm are sufficient for the required amplification.The used OpAmp is a dual OpAmp where I just use one of the two. The supply Voltage of my Sensor is +-15V DC and the supply voltage of my OP-AMP is +5V DC. I tried to read 0-1V as output with the following circuit but I am receiving nothing but noise: What I tried already: Measuring the Hall-Sensor output directly with a 4Ohm measuring resistance -> as expected Checked wiring with multimeter Measured output of DC-DC Converter (15V to 5V) = Supply Voltage for OpAmp -> Output voltage is 5.6V (Should be ok according to OpAmps datasheet?) Measured resistance of Resistors and Shunt -> as expected I also thought about adding a capacitor of 100nF to the OpAmps PowerSupply, but I think this is not the problem here? Hope someone can help me figure out the problem. <Q> 0R1 is too low for the secondary side shunt (R M ). <S> If you read the datasheet then you'll notice that all the performance characteristics are given for R M = <S> 10R. <S> You may worry about the dissipation for R M = 10R compared to that with 0R1. <S> So you can use R M = <S> 1R. <S> Besides, with R M = 0R1 <S> the full-scale voltage will be 25mV for I <S> L <S> = 500A. <S> The offset voltage of the opamp can be as high as 3mV or as low as -3mV, so you may have a measurement error of (±3) <S> x 40 = ±120mV which translates into ±12%. <S> But for R M = 1R <S> the full-scale voltage will be 0.25V, for for 0-1V range a non-inv. amplifier with a gain of 4 (R2 = 3k6 1%, R1 = 1k2 1%) is enough and the error will be significantly reduced. <S> I also thought about adding a capacitor of 100nF to the OpAmps PowerSupply, but I think this is not the problem here? <S> It's always a good practice to place decoupling capacitors right next to the supply pins. <A> According to your suggestions I modified the circuit and included Capacitors for the Amps Power supply, changed the measuring resistance to 1Ohm (R1&R2 accordingly) and connected the DC-DC-Converters GND to AC-DC-Converters GND so all OpAmp-Inputs do have the same GND: <S> Please correct me if I'm wrong? <S> Edit: <S> The DC-DC-Converter is of the Type "TMA 1505S" (TRACOPOWER) <A> According to your suggestions I modified the circuit ... <S> Looks like you didn't :) <S> Since capacitors don't allow the DC to pass through, your opamp cannot get supplied and thus can't work. <S> What I just found out is that my Converter is an isolated one. <S> That explains everything. <S> Isolated converters have isolated references (i.e. GND). <S> So, in your first schematic, the sensor outputs wrt AC-DC converter's GND, but the opamp output wrt DC-DC converter's GND. <S> Since they are isolated, the opamp outputs zero. <S> Here's a schematic for you: simulate this circuit – <S> Schematic created using CircuitLab <S> There's no need to use a DC-DC converter, <S> a 78L05 is completely enough. <S> C1, C2 and C3 are decoupling capacitors (look how they are placed). <S> NOTE: <S> 4-pin shunt's measurement pins seem unconnected but it's just a drawing.
| You have placed decoupling capacitors in series to supply pins.
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Is ther a standard circuit/component to control the temperature 500W resistive heating element from a PCB? I have a 500W (230V) cartridge heater element and a K-type thermocouple. I want to use the feedback from the thermocouple to be able to vary the temperature of the cartridge heater. I have no concerns about being able to implement the thermocouple circuitry and software, however I have never used a high power heating element before. I'm looking into the control of the element. I personally don't have an issue with an on-off control (as in, I'm okay with applying some hysteresis and having, say +/- 10°C from the desired value, or using PWM), however maybe there are some other considerations as to why this wouldn't be a good idea (switching life cycles for example). If there is a component to safely vary the power to the heating element, so I can implement PID for example, then I would also be interested in that. As I want to control the heating elements from a microcontroller (preferrably 3v3, but can level shift to 5v if needed), I've been looking into relays. However, I am struggling with the power requirements - from what I find, 500W seems to be a lot of power for a PCB based relay (although since I've never used a relay before, maybe I am misinterpreting the specifications). I'm also concerned about the actual traces on the PCB - I'd much prefer to have the relay controlled by PCB traces but the heating elements just connected to the relay by wires, if such a device exists. I've looked into latching, non-latching, solid state relays, SCRs, triacs, but I feel like the more I look into it, the less sure I am. I'm now at the point where I feel like there must be a standard way to do this, but I'm just getting lost in the sea of options. So, is there a standard circuit/device to safely control the temperature of 500W (230V) heating element using microcontroller-level voltages? The cheaper/simpler method is usually better in my eyes, as long as it's safe. EDIT: Thanks all for the comments and answers so far. Looks like I'll go with some type of relay, and possibly some backup protection if I want to use a latching relay. One of the main issues I'm having with finding a suitable relay is geting the specifications correct. For example, the following is taken from a relay datasheet: The first thing I see is that max switching voltage and max switching current are both high enough - great. Then I saw the rated load box, when confirms this. However, what confuses me is the 'Max. switching power' box. I'm already a little bit hazy on the difference between VA and Watts, but the fact that at has a maximum of 300W, while my element is rated at 500W, confuses me. Can anyone help explain this? <Q> The easiest way, and probably the one that will yield the best control, is to buy an industrial temperature controller with some kind of serial interface, and switch either a power relay or an SSR with the output. <S> That buys you the thermocouple front end, properly isolated from ground so a fast sensor can be used, an auto-tune PID algorithm and a user display. <S> Power relays are inexpensive and reliable, however they wear out after about 100,000 operations at full current. <S> At 20 seconds per cycle, that's a couple years operating 24/7. <S> If you derate the contact capacity you can get considerably longer life. <S> SSRs generate heat (albeit not much at only a few amperes, but will generally require a heatsink) and tend to fail more randomly, and generally fail 'on', which is often an issue. <S> In either case, if the controller being stuck 'on' or 'off' could possibly cause significant injury or damage you'll need a redundant backup such as a thermal fuse or an independent limit controller. <A> 500 watts at 230 volts is only 2.17 amps. <S> If you don't want to have the power continuously on at that power level, you might want to design for 2.5 to 3 amps. <S> An PCB mounted electromechanical relay for that current rating and a resistive load should not me difficult to find. <S> You should also consider using a solid state relay to trial with zero crossing control. <S> The idea would be to pass a certain number of full cycles of AC line current and block a certain number, varying the on/off ratio under PI or perhaps PID control. <S> Regulating to only +/- <S> 10 <S> °C seems like quite a pessimistic expectation. <S> The old-fashioned thermostat suggested by @Hearth would do better than that. <A> There are many ways to skin a cat, as you've found... <S> You can get relays with faston tabs, so that you aren't running the current through your PCB, although 2.2Amps isn't that difficult to do in a PCB. <S> See example image below, though this specific relay may not be what you need, it shows the fast on mounting. <S> Being that it's only 2.2Amps, you can also control it with Triacs. <S> This would allow much finer control, like PWM but cycle by cycle instead. <S> There would be higher losses on the load side compared to a relay, but the relay coil is also a loss, so you'd have to see if one wins out over the other for your situation.
| If your heating load has a large thermal mass, and takes a while to heat, then the bang-bang on/off model is simple and you can use a relay.
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Could Cat5e be used for RS485 I'm working on a project that will be using RS485 to communicate between a bunch of modules and a master controller. Low baud rate, 9600, but a decent distance, like a couple of hundred feet. I'm wondering if I can get away with reliably using Cat5e for this, and whether or not I need it to be shielded and connected to ground? If that is the case, would I be able to use the remaining 3 pairs as the ground instead of getting cable with actual shielding? <Q> Short answer: <S> yes. <S> That should work just fine. <S> More detailed <S> : RS485 requires a common ground reference for all devices. <S> Yes, you can use spare pairs in the cable as ground reference. <S> Proper shielded cables provide better noise immunity. <S> But it is not by any means necessary for RS485 communication to work. <S> Depends on your environment. <S> Lots of high power electrical motors and welders around? <S> Then you may want to use shielded cables. <S> Depending on your exact cable length requirements, you may want to consider thicker gauge CAT5. <S> Up to ~100m/300ft any CAT5 should do the job. <S> Beyond that, the resistance of the copper conductor can start playing a role in attenuating your signal. <S> Thinner conductor=higher attenuation. <S> RS485 cables designed for 1000m/3000ft have really thick copper wires. <A> Definitely yes, the CAT5/5E or CAT6 cable is perfect for industrial use of RS485 network. <S> Pay attention of the shield connection and for long distance networks is recommended to use ISO tranciever like ADM2483 ( Datasheet here ). <S> If the shield is not connected (that is a common mistake) <S> an isolated tranciever will almost solve communication issues (a traditional MAX485 or non isolated tranciever will probably not work <S> and it will potentially be damaged due to compensation currents to earth) <S> In addition to that, the termination resistor on the farthest device can help to balance the entire bus. <S> Without termination resistor and the ADM2483 ISO tranciever i've reached 1350 meters with a CAT6 cable ( shield unconnected ), only A and B communication cables at 57600 baud. <A> The ground goes nowhere, since both devices are connected to the local earth (usually not directly, rather capacitive coupling from environment, the transceiver is floating). <S> The shield goes connected to the earth in a single point or it is divided into sections. <S> This situation may be different if you are using non-isolated RS485 transceivers, a third wire is needed - GND. <S> If devices have GND shared with local earth, then large compensation currents will flow over this tinny wire.
| It is quite common to use CAT5 for RS485. The RS485 forlong distance is better to be isolated one. Yes it could work, but you have mentioned something about the ground and shield.
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KiCad footprint or supplier footprint? I download symbols and footprints from the component supplier into a project's library to go faster but I realized that many of the standard components footprints are slightly different in KiCad, like for example the SOT-23-6 so my question is: is it better to stick to KiCad footprints or just use the suppliers footprints? <Q> There is more to good footprints than just "matching the pads to the pin". <S> Depending on the dimensions and distances the results of soldering can strongly vary. <S> Good footprints can reduce the risk of shorts and grave stones for example. <S> There was a rather prominent repository of a KiCAD library derived from the IPC-7351 recommendations <S> but I cannot find it anymore. <S> However, it seems like there is still a version around <S> https://github.com/alexisvl/kicad-pcblib . <S> I typically use the "Least" but only because I do a lot of tiny crowded PCB with reflow soldering. <S> The "Most" version should be adequate for everyone with decent knowledge about hand soldering. <A> Best approach would be to ask people who will solder these components. <S> For one component, KiCAD footprint may be better, for another - the supplier one. <S> We ended up making our own library of components based on existing one. <A> How different? <S> But usually the supplier knows the part (while KiCad has some generic footprint). <S> You should also check the datasheet to see which one better fits the mechanical drawing (and remember to enlarger them a bit if you plan to hand-solder them).
| From my personal experience, I strongly advise using the IPC-7351 footprint recommendation because the standard library coming with KiCAD did not really live up to the expectations (at least back then I started) and even the footprints described in datasheets sometimes resulted in less optimal (reflow) soldering for me.
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Help Identifying Inductor From a Schematic I’m repairing a laptop but I don’t know what value inductor to use/get. The schematic has values, I just don’t know how to interpret other than it’s a 2mm inductor. It’s labeled 5A_Z120_25M_0805_2P on a 1.35V line. <Q> Looks like something with 120 Ohms Impedance (z120) at 25MHz (25M) that can take 5 Amps (5A) in a 0805 Case (0805) <S> Close but not quite would be a Würth 742792025. <A> I think that is just an obscure part number used by the manufacturer. <S> This was also common practice with certain IC's used in some products. <S> Generally, your "manufacturer-authorized" repair shops will have the decodes of the "part numbers". <S> This is to try to keep others from being able to service these devices. <S> So, your best bet is to look for any color-coding on the part, in order to deduce its inductance value. <A> How do you know you have to replace the inductor? <S> They seldom die. <S> Watch <S> Louis Rossman who does a lot of fixing these things live. <S> The power chips (usually buck switching controllers) have a hard life, and often go. <S> But I do see him replace resistors and capacitors as well. <S> I linked to his youtube videos and did a search on inductor for you -- maybe your inductor did go. <S> He might have done a similar repair. <S> Sometimes he uses alcohol to see where a short is, after pumping some voltage into the board. <S> The hot component gets dry first. <S> That's the one to replace if your board is shorted. <S> Somehow, juice, or coffee, or soda, often finds its way into these boards. <S> Look for signs of this abuse for prime suspects. <S> If your inductor went, and you can find a dead laptop / motherboards of the same type, it should have your exact inductor, and odds are it will still be good. <S> Good luck.
| Otherwise, Your only other reference to that set of numbers, is to be found in the manufacturer's manuals or documentation (if they will let you have that information). The IC's real part numbers would be sanded off, and replaced by a different kind of part number from the manufacturer of the product it was used in. On the internet, buy a few laptops or motherboards of the same exact type you are fixing, and cannibalize them for good parts.
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Custom made PCB problem after lost power supply I have 2 custom PCB which need to communicate via Spi. Communication works fine when I flash code and MCUs have stable power supply. After disconnecting from power supply and then reconnecting slave MCU send some garbage and I need to press reset button on PCB to allow normal operation to continue, basically after pressing reset button on slave everything works fine. I read that wrong connection on BOOT0 and RST pins can make such mistakes. On logic analyzer I see that I have some glitch on BOOT0, glitch intervals are not periodic and take a little time. I figure out one more thing, when MCU don't have power supply and I press reset button and hold that button and then connect power supply to MCU and then release button I got fine result, communication works good. Any idea is welcome. Here is captured start sequence when PCB is connected to power supply. <Q> I suspect the 100k on BOOT0 is not pull-down enough. <S> In the newer F4 series the BOOT0 pin also serves high voltage programming supply (Vpp). <S> On the demo boards they hard tie BOOT0 to ground without pull-down, or use a 10k resistor. <S> Maybe you can measure the leakage from BOOT0 to determine the highest possible resistor. <A> I found what is problem. <S> Problem is cheap logic analyzer and cables for communication. <S> Small raster is used (raster 1) + bad connection on cables + logic analyzer = unpredictable behavior. <S> Thanks all for help <A> Let there be 2 lines connecting Master and Slave other than SPI: <S> One of the lines is Slave's reset. <S> Second line is master's input pulled down on the master side and connects to Slave's VCC <S> When slave is plugged in, the Second line (let's call it SLAVESNS) goes high. <S> Master either polls it occasionally or gets an interrupt. <S> Then master waits a pair of millliseconds and pulls slave's reset low and back high. <S> There may be other ways to make reset happen using small circuits of course, maybe even without interrupt, just analog circuit, but I still find it reasonable to have slave's reset being controlled by the master, maybe you even make reset low by default and only when you detect something connected you let it go high
| Not sure if it's the right solution, but you can try to connect slave's reset to master and detect it being plugged in.
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Replacing capacitors in older electronics I was working on a supplementary power supply board in an old arcade cabinet. It has a capacitor with some markings that I was unable to identify and was curious if they were important. Also, I would like any advice that could help me avoid damaging the arcade. The markings on the capacitor that I am confused by is "CE04 W". I don't know if this is just manufacturer specific or what? Accordingly, the rest of the capacitor markings are as follow: 25V 4700uf 85 degrees <Q> It is made by Nippon Chemi-con, so it is not a cheap bad quality capacitor to my understanding. <S> The CE04 W is the capacitor model series, CE04 is very old and <S> at least the datasheet I found on CE04 series does not list capacitors with this high capacitance. <S> Almost any capacitor with matching capacitance (4700 uF) and matching or exceeding the other parameters is fine. <S> So the voltage must be at least 25V, but it can be more if you can't find a 25V capacitor. <S> The temperature rating may be rated for 85 C due to longevity reasons even if the temperature does not go up to 85 C. So for same reasons, you might want to have the same 85 C rating, or if you can't find one then larger temperature rating is fine. <A> The Nippon Chemi Con CE04W is rated for 2,000 hours at +85 <S> degC: - 2,000 hours is 83.33 days but, for every 10 degC the capacitor operates at below 85 degC the life time doubles. <S> So at 35 degC average temperature, the lifetime can be expected to be 64,000 hours (2667 days or 7.3 years). <S> If the voltage actually applied is about 50% of the rated voltage (25 volts) then the operational lifetime doubles again. <S> See this calculator : <S> - So, choose a replacement that is good for many years is my advice - <S> if needs be choose one rated at 105 degC or a higher voltage if it helps finding one that fits the space envelope <S> but, whichever you choose look for the lifetime hours in the data sheet - if the capacitor says it's 85 degC rated but doesn't give a number of hours then it's likely to be a poor capacitor. <S> Don't choose a capacitor that hasn't got a temperature rating of at least 85 degC even if it appears to have a lifetime figure. <S> Also note that the capacitor has a tolerance of +/- <S> 10% or <S> +/-20% and <S> this is usually a sign of reasonable quality because crappy electrolytic caps are usually specified at +80% / -20% in order to overcome their inherent weakening electrolyte over time. <S> So, choose a decent cap - your arcade machine deserves it!! <S> So, <A> It's a 4700uF/25V aluminum electrolytic capacitor. <S> Radial package (two leads - at least- coming out of the bottom). <S> If you go to a distributor such as Digikey and search for capacitors of that value and voltage rating you can find a number. <S> I get 53 that are currently in stock and of those exact values. <S> Look at the bottom and see if there is a third hole for mechanical stability. <S> Try to match the lead spacing so the part will fit down on the board- <S> if the board is really old you may not be able to find one that big without going to a higher voltage rating (higher is fine). <S> Personally I would use a 105°C rated part so it will last longer. <S> On aluminum electrolytic capacitors the polarity is indicated by a stripe on the NEGATIVE side. <S> If you don't get it right, the new capacitor will vent noxious electrolyte and you might blow a fuse or worse. <A> The pins are crooked to make it hold onto the PCB <S> so it doesn't fall off when you turn it around to solder it. <S> These usually last a very long time. <S> On the picture, I see a rectifier bridge right behind the cap, so that means it's a power supply smoothing cap. <S> This means characteristics don't matter much as long as it has a voltage rating and capacitance equal to or higher than the old one. <S> To pick a proper replacement cap, make sure it fits, so check pin spacing. <S> These caps are available in plenty of diameter/height combinations. <S> Since the diode bridge is very close, make sure the new cap isn't too wide. <S> Height may be a factor if some other device in the arcade cabinet is mounted right above the board, so check that too. <S> Also look if the original has more than two pins, if this is the case check which pins are for mechanical stability and which are for electrical connection. <S> Since modern technology squeezes more µF into a smaller size you may have to get a higher value cap to get the correct pin spacing. <S> Get a snap-in cap which will have thick crooked pins as shown on the pic above. <S> A standard radial cap will have much thinner pins, thus plenty of empty space left in the holes, which makes soldering more difficult and less reliable.
| This type of radial aluminum electrolytic capacitor is known as a "snap-in" cap. That's a bulk capacitor for a simple linear power supply, so nothing special is required. When you install the new one you have to make sure you observe polarity.
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How does this circuit in an electric mosquito swatter work? I had an electric mosquito swatter lying around so I decided to open it up and check out how it works. I understood some parts of the circuit however I did not clearly understand how the rest of it works. I have posted the circuit of the mosquito swatter. (I did notice that there was a similar question on the forum but the circuitry differed and it was not yet solved.) Here's the circuit: Things I did not get: Why did they use the capacitor and resistor in parallel as input for the full bridge rectifier? Why is the battery just directly connected to the output of the full bridge rectifier. How is the (6-pin) transformer connected internally? If I can understand this I am guessing I can understand how the transistor works. What is the configuration at the output of the transformer? Note: Below the main circuit there is a diagram that shows the interconnection between the transformer pins when I used a multi-meter to check for continuity, note the transformer was still on the pcb. The 0 and 1 at the output is just to indicate that the 0 goes to outer 2 meshes and 1 goes to inner mesh. The LEDs are red. Some of the components did not have values written on them like the capacitor C6 so I removed them and checked them using a component tester so the values may be slightly off the standard values. I numbered C6 out of order sorry about that. The resistor parallel to C1 is 675kOhm and not 675Ohm. The resistance between Xmer pins 1-2: 2 Ohm; 3-6: 0.8 Ohm; 4-5: 245 Ohm. Additional working note: When the device is not charging the NO switch is closed and the push button is pressed this seems to charge the meshes for the next electric swat. <Q> The resistor across the 240 VAC mains series cap, C1 serves to drain residual charge, to prevent possibility of a small shock. <S> Depending where in the AC cycle the device is unplugged, the charge on the cap could be <S> as much as 1.414... times the RMS (nominal) AC mains voltage, ~340 VDC. <S> Why not? <S> If it's sealed inside a case, double-insulated, there's little chance of shock from touching battery and ground. <S> Oh... <S> do you mean is the 2.4 V battery getting 240 VAC? <S> No, because C1, a 0.49 μF capacitor, has a reactance of ~6,500 Ω at 50 Hz, and ~5,400 Ω at 60 Hz, limiting the voltage and current across the battery. <S> That would produce a charge rate of ~ 36mA at 50 Hz. <S> BTW, I would not leave the swatter plugged in permanently, since that might be more current than good for long-term trickle charge on a small NiCd or NiMH cell. <S> You can check winding connectivity with an ohmmeter, but basically, the emitter and the B+ are connected to pins 1 & 6, the primary winding; pins 2 & 3 are feedback windings, sending signal to base to cause oscillation; and 4 & 5 are high-voltage secondary output. <S> The diode-capacitor combination on the output serve as a voltage multiplier . <S> In this device, it triples the voltage from the secondary to charge C6, which stores enough power to permanently incapacitate an arthropod (give it "a short, sharp shock, from a cheap and chippy chopper", as G&S put it). <S> BTW, the Cockcroft–Walton generator uses voltage multiplication to generate megavolts from lower-voltage AC. <S> No, I don't suggest building a basement particle accelerator or X-ray machine. <A> Why did they use the capacitor and resistor in parallel as input for the full bridge rectifier? <S> It is basically a cheap way to make a low voltage, low current supply directly from the high mains voltage. <S> Why is the battery just directly connected to the output of the full bridge rectifier. <S> It is part of the capacitove dropper circuit and used as a device to limit the voltage for the rest of the circuit to around 2.4 V. <S> In most capacitive dropper circuits a zener diode is used. <S> How is the (6-pin) transformer connected internally? <S> If I can understand this I am guessing I can understand how the transistor works. <S> What is the configuration at the output of the transformer? <S> I suggest that you search for "transformer oscillator circuit" and look at how those circuits are drawn. <S> Then realize that the transformer has 3 windings, possible between pins 1 and 2, pins 3 and 6 and pins 4 and 5. <S> You can measure this with an ohm meter while the circuit is off. <S> Then re-draw your circuit so that it looks more like the examples you found. <S> This circuit is at mains voltage when connected to the mains so don't touch anything when it is connected to the mains voltage! <A> Diodes D1,D2,D3 and D4 have been copied wrongly.they all have to be reversed or flipped over(anode instead of cathode)
| That is a " capacitive dropper circuit "
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LDO output and input capacitor clarification I would like to understand how to size the input and output capacitor of an LDO. What I think to know: About the input capacitor: The input capacitor is used for filtering high frequency input components. Nevertheless in a perfect LDO, the drop voltage across the P-channel mosfet is adjusted so as to maintain the output voltage constant whatever the input signal as long as the input voltage is higher than the voltage drop across the P-channel mosfet plus the output voltage. So if the output voltage is constant, it means that the LDO filters all the AC components of the input signal. So why do we need to add an input capacitor? I think that the bandwidth of the LDO is not sufficiently large to filter all the AC component, especially the high AC components, so an input capacitor is added to filter what the LDO could not filter. Where I can find this information into the datasheet? About the output capacitor: Suppose the load constant at the output of the LDO. The output voltage of an LDO is constant. So the output current is constant. Suppose the LDO is taking no current from the source. So the current taken from the source is equal to the output current and is CONSTANT over the time. So why do we need an output capacitor? The reasons: During load transient, the output voltage will vary during a time depending on the bandwidth of the LDO. So an output capacitor can be used for helping the regulation. The other reason is for stability concerns. Nevertheless, I have already seen 100 uF output capacitor on an LDO and a load around 500 mA under an output voltage equal to 5V. What I do not understand is that the output capacitance seems to be a function of the load, something that I do not understand. Does the stability problem imply to increase the output capacitor at this value and is function of the load? I could fairly understand that the overshoot or undershoot during load transient is dependant on the load, but what transient load need to have a 100 µF capacitor? Here is the LDO. <Q> The regulator you linked to is not LDO. <S> LDO regulator have even more strict requirements for the output capacitor, it needs specific capacitance range and ESR range for stability. <S> But for a normal linear regulator, if you have long wires from regulator input to the power source, you have inductance. <S> When the regulator needs to provide a surge current to load it requires surge current from the input and it can't get the current. <S> So the input capacitor is a local bypass capacitor for stabilizing the voltage and allows current pulses to be drawn. <S> Without input capacitor the regulator can turn into an oscillator. <S> As the datasheet says, input capacitor may not be needed if the input bulk capacitor is close, but it also mentions it might be needed if the output capacitance is large. <S> Sometimes the output capacitor is needed also for stability. <S> The datasheet says it is not required for stability, but it improves output impedance during transients. <S> For example, USB is a hot-pluggable interface with 5V power supply. <S> It requires that a device has a maximum capacitance of 10uF at the input. <S> Surely the regulator output would jump momentarily to zero if you connect an empty 10uF cap. <S> That's why USB hosts need to have enough capacitance at the output, in the order of 200uF, to prevent the voltage dropping too much for other devices. <A> All voltage regulators want a peaceful life; if there is an output capacitor then any fast cyclic changes in load current are largely dealt with by that capacitor and the lazy output from the regulator is only expected to deal with topping up the charge to the capacitor so that the capacitor can handle the next cycle of load current change. <S> There will be a small transient change of output voltage because the lazy regulator can't be expected to keep up with fast load changes. <S> That is due to: <S> - $$I = <S> C\cdot\dfrac{dv}{dt}$$ <S> In other words, there will be a ramp down in voltage depending on how high the current is and how big or small the capacitor value is. <S> And the lazy old regulator will try and deal with sorting out the average output voltage in its own time. <S> In other words, voltage regulators are not as quick to deal with load changes as you might think and therefore, the output capacitor does the main job of keeping dv/dt as slight as possible. <S> Some voltage regulators are better than others of course but they are, after-all, a control loop and won't be as effective in the short term as an output capacitor. <S> Having said that, output capacitors with relatively high ESR are going to cause problems too so, use the capacitor as recommended in the data sheet. <S> So, instead of the voltage regulator having to deal with step changes in current (without a capacitor) it has to deal with dv/dt (a much slower change). <S> As for the input capacitor, if there is a load change that causes the input voltage to fall suddenly, the lazy voltage regulator is not well equipped (speed wise) to adjust its series impedance to compensate for that input voltage fall in order to keep the output regulated. <S> Having a capacitor at the input holds-up that voltage to a higher degree than if it were not present. <S> This allows the lazy voltage regulator time to catch up with the situation that is occurring. <S> The same story if the input voltage changes due to external factors - having an input capacitor can slow down that change and make life easier for the lazy voltage regulator. <S> So, both capacitors are there to make life easier for the voltage regulator and deliver the performance expectations stated in the data sheet. <A> They help you smoothen transient response (when load needs a bit more and a bit less current). <S> Because it takes a little time for the chip to adjust to smaller/higher load. <S> And don't understand this wrong: even when load chip simply works, it still uses a bit more and a bit less current all the time, because its internal transistors are switching and doing stuff. <S> And of course if your load chip is shutdown or disconnected, you get sharper current consumption changes that lead to LDO output voltage spikes (up or down depending on what exactly happened). <S> Small caps on the input of LDO compensate for greater/smaller current from input source (usually 0.1uf - 1uf for common circuits).P.S. More competent people with greater experience are welcome to correct me or add to my response if I'm wrong or incomplete.
| Usually the capacitor values are in the datasheets of the voltage converters, typical output capacitance is usually 1uF-10uF, so you should have some 1uF and 2.2Uf caps around, they will almost always be he right choice for LDO output for common circuits.
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Eliminating an earth / ground loop in an A/V set-up I have an A/V system with two active sub-woofers. One is close to the amplifier connected with a 1m RCA cable. Its mains connection is into the same power strip as the amplifier. There will be an earth loop but it is short and it does not add a significant amount of noise. Update, I now know that the amp is not directly earthed but it may be through one of the other devices. The second sub is further away and is connected with a 10m RCA cable. It is not actually that far but the cable needs to go around the wall and not across the floor. 7m would be sufficient but that is not a commonly available length. I tried a 5m cable with a 2m extension but it was not noticeably better. The mains for this sub is close to the sub and hence not close to amp. This earth loop is much larger and picks up a significant hum which I would like to eliminate. Further evidence that the problem is an earth loop: The hum is stopped if the signal cable is disconnected. I tested by disconnecting the at the sub end. My initial concern was that there was a fault within the sub. I have tried disconnecting the signal cable at the amp end. Initially, I got a worse hum but when I shorted out the loose end, this went away. If I feed a signal from a battery powered iPod, there is no hum. The iPod was connected with a short cable. I'll test with a longer cable along a similar route to the usual signal cable and update later. Temporarily breaking the supply earth while connected to the amp also stops the hum. All plugs are three pin and connected to grounded sockets. It is possible that the earth is not used but test 3 suggests that the sub is certainly grounded. I could repeat test 3 but break the earth at the amp end as an additional test. Most of the plugs are sealed and not easy to inspect. So, a simple and effective but maybe not safe and legal solution is to break the supply earth. I would like to find a better solution. I found this item on Amazon: AV Link Ground Loop Isolator . Cheap and easy to fit. I feared that it might attenuate the signal and cause harmonic distortion but a quick test using test tones from my iPod and my ear to judge suggested that it was not too bad. However, when I connected it with the amp, it actually made the problem far worse rather than better. I guess that it contains a transformer which was picking up more hum than the earth loop. I found other items, e.g. PAC SNI-1 Noise Isolator which did not seem to be transformer based, but the reviews were terrible so I did not even try them. So, I am still looking for a solution and preferably not a very expensive one. The devices above are passive but I would consider a powered device. It is common advice to keep the signal cables away from the mains and that is already the case here but I was considering the exact opposite: power the sub from the same power strip as the amp and the other sub. Hence, it would need an equally long power cable as the signal cable which I would run together. The idea is that both the signal and power cable would pick up the same hum but out of phase and hence they would cancel over. Is this worth trying or have I missed something? I am in the UK so the supply is 230VAC 50Hz. Since this is the UK, I expect that everything in the house is on the same phase. In this case, both sockets are on the same breaker in the consumer unit. The AVR is a Denon X3200W - manual . The subwoofers are BK Gemini II s. I cannot find an online manual but even the supplied paper manuals are very simple. Other devices connected to the amp are: the TV, a Tivo (digital video recorder), two disc players, a digital iPod dock, and the other sub. The sub uses an RCA cable, the iPod dock uses an optical cable, and all others are HDMI. All devices except the Tivo and the distant sub are powered by the same power strip. The Tivo goes to another power strip but both strips go to the same double socket. This is partly just a matter of sockets on the strips but also one strip contains devices that can be off when no one is present and the other is for devices that should remain on: Tivo, internet router, phone base station, etc. I have now checked the amp end of its power cable and it has a two pin connector. So, it is not earthed and my loop must be through one of the other devices, possibly even the other sub. Thanks to the comments, I have more ideas to try. Annoyingly, repeats of tests that I have made before don't seem to entirely consistent. Possible explanations of the inconsistent results are that my hearing varies and other noises vary. E.g. sometimes I hear the hum from a fridge two rooms away more clearly than the hum from the subwoofer. More research is required. <Q> you may reduce the hum, using power cables that have rapid-rotation twists of the hot/return pair. <S> This fast rotation of the magnetic fields will reduce the induced voltage in all nearby wiring loop. <S> The density of twists will increase the self-cancellation of magnetic fields, even for small short audio cables or small regions of your PCB boards. <S> To evaluate this, get apiece of sheet steel, tie it to GROUND with a flexible wire (to ensure it will not become an electric_field radiator) and experiment. <S> Hold the sheet next to aggressor wiring --- hidden power wiring inside the walls, or near power cords running to your AV setup or near power transformers inside your AV boxes. <S> Hold the sheet near your victim wiring (coax cables?) or audio Preamplifier boxes. <S> Ohhh ensure all your audio cables have grounds running end-to-end. <A> You can buy cheap analog (RCA) to optical SPDIF and replace all audio with this (but this is only an alternative path to the grounding problem) <A> Hum in AV systems is often related to antenna or (old coaxial) network, cable TV or other signal cables connected to the system. <S> Those cables bring a 2nd or 3rd ground connection to the amplifiers' ground.
| The best sollution to remove the hum is to use all optical cables for the audio.
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Why are graphic displays still so expensive? As a new electronics enthusiast, I've been able to score a lot of components for very cheap, ranging from beefy MCUs in the 10$ ballpark to heaps of ICs for sub-dollar prices. However, buying a coffee-maker-sized OLED screen cost me roughly 50$. Meanwhile, I can buy a computer monitor for 90$, although they're full HD, 60Hz full color screens. Why are the components relatively so expensive? <Q> It's also to do with supply chain. <S> A component you buy on digikey or <S> RS etc has a very significant price overhead attached to it. <S> You are paying for the convenience of the supplier (digikey or RS) cataloguing items, packing, shipping, <S> A finished LCD monitor as product can benefit from all the advantages of volume production and the efficiency that comes with it. <S> It was somewhat different when it came to 2-row alphanumeric displays, but things like OLED displays are not standardized. <S> When I design a product with a display that will be manufactured in the hundreds of thousands (or even less), I will talk to a display manufacturer directly, and have them built to-order my design. <S> The displays sold by distributors are so expensive because another intermediate company actually designs, orders, and assembles the display as a module, and then sells it to the distributors you buy from. <S> This can also be seen in the price. <S> It is somewhat similar to integrated DC/DC bricks. <S> You can make DC/DC assemblies with the same performance for a fraction of the BOM and assembly cost, but the design cost, and the benefits of having that product now , outweigh the higher cost of a pre-made module. <A> It's partly the difference between OLED and LCD and partly where you're shopping. <A> Note the price curve. <S> One is expensive. <S> But even the 10-off price is considerably cheaper, then it flattens out for a little more saving at 25, 100. <S> Consider it may be worth buying 10 and keeping the spares for future projects. <S> You're investing a lot by writing software to drive it anyway; you may as well make that effort pay down the cost of future products. <S> For larger displays, consider embedding a webserver in your software and building a tablet into your project...
| When it comes to displays versus integrated circuits, there is another difference: Displays are usually made custom to-order. They are targeted at prototyping/low volume, where that cost is insignificant in the scheme of a product-development cycle.
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Increase amps with set voltage? If I had a battery that is 50ah at 30V, what would I have to do to increase the current rating to 150ah, while maintaining the same voltage? <Q> 150 <S> Ah is a capacity rating, not a current rating. <S> You may well find your existing battery can already deliver 150 amps. <S> If you want to increase the capacity from 50 Ah to 150 Ah, then you would need to put two more 50 Ah batteries in parallel with the first. <S> This would also increase by a factor of three whatever the current rating of your first battery was, assuming they were identical. <S> In order to find out what the current rating of your first battery is, you would need to look at the specifications. <S> They normally give a C rating, which can be interpreted roughly as 'hours to fully discharge the battery'. <S> Capacity falls as the output current increases, hence the qualifier 'roughly'. <S> A C rating of 1 means that the current rating is numerically equal to the capacity rating, which it looks like you are assuming. <S> However, few batteries have a maximum C of 1. <S> There are different ratings for current. <S> The two most frequently met are the maximum current that can be drawn safely, and the current for which the capacity rating is valid. <S> With a car battery for instance, the maximum C is often around 10, for cranking the engine. <S> However the maximum capacity C is often 0.1 or even 0.05, requiring a 10 or 20 hour discharge to obtain the stated Ah capacity. <A> With "a" battery, you can't. <S> Because, 50Ah is the maximum electrical charge (from Q = <S> i × t) <S> that the battery can contain and you can't increase it directly "in a single battery". <S> Since they are paralleled you'll get the same voltage (30V) with tripled capacity (3 × 50Ah). <A> Putting it simply, three batteries in parallel. <S> You are basically trying to draw more current (over time) than the battery can possibly handle.
| The only thing you can do is to connect 3 batteries in parallel.
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Why is the PCB part number traditionally on a copper layer? I'm an experienced board designer, having designed 100+ boards over the past decade, and I have yet to hear a good answer for why the PCB part number is traditionally on an external copper layer, and the PCBA part number is on a silkscreen layer. To me copper is always "sacred" and should not have anything unnecessary on it, as this can screw up voltage isolation, RF performance, etc. But for some reason designs tend to have the bare PCB part number in copper instead of silkscreen. Why? <Q> Copper etching is a very early step in the PCB manufacturing process. <S> Etching the board number on the copper layer allows identification of the board at all subsequent stages. <A> I do a lot of multi-layer boards and the board reference number plus layer name gets put into the copper on all layers plus, there is a drawing accompanying the job that states the layer stack up to make sure that the bare PCB is made correctly. <S> No chance of screw ups. <S> If there is enough room on the PCB (say with 4 inner layers), there will be 4 distinct places on the board that if you shine a strong light through you will see the internal numbers. <S> You can therefore inspect the layer stack to see it has been built correctly. <S> I'm not saying that there aren't other reasons but these are mine. <A> My company started designing PC boards in the early '80s. <S> These were single-sided boards with NO solder mask or silkscreen. <S> The only permanent location for the board number and company information was on the solder-side, in copper. <S> Even though we now design double-sided boards with solder mask and silkscreen, we still include board number and company info on the copper-side layer. <S> We also include that information on the component-side layer as well, room permitting. <S> And the silkscreen layer, of course. <A> I asked this same question years ago earlier in my career. <S> In other words it was harder for someone to scrape off the copper number text than it was for someone to remove the silkscreen. <S> Some of the intention of more permanent board markings, I was told, was to make it easier to see if some unauthorized person or company was modifying your hardware somehow and selling it as their own product. <S> A scraped off copper area on the board would be much more obvious than a silkscreen that was removed by some solvent. <A> It's better to have the BOM sticker (with revision) on a sticker, never know if you are going to have two alternatives :) <S> If you are cheap you might skip the silk and have the PN on copper.
| The answer I got was that the copper layer usage for the board number was more permanent than silkscreen. To omit it would risk confusion due to lack of a board identifier.
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Ambiguity about concept of baud rate and bit rate I know that there is lots of questions and answers that is related to this topic, but i can not clearly and distinctively find answers to my questions, yet. In addition i found some answers which seems to be contradictory to each other. I know that baud is number of symbols per second in a data transmitting signal, As here is mentioned: https://electronics.stackexchange.com/a/282382/254779 , Then its unit could not be "bits per second", right? But these two answers bellow say that the baud rate unit is "bits per second": https://electronics.stackexchange.com/a/273817/254779 , https://electronics.stackexchange.com/a/117245/254779 . Further more, I have ambiguity about the way measuring baud rate. In some web sites is said to be measured by number of times line changes per second and some where else as signal units per second. But i guess that they are not completely the same, right? Because this really means that every signal unit should be correspondent to only one signal change. So for example how baud rate is measured for Manchester encoding? I guess that in this encoding number of signal changes per second is not the same as number of signal units per second, right? And the last one is that i found some answers which said meta data bits like start and stop bits were not taken into account for measuring bit rate, however, I saw somewhere else that all bits were used for measuring bit rate. So which of these is correct at the end? For example is baud rate / bit rate = 1 in the UART protocol? I have these questions in my head for long, and i decided to ask them once for all. Thanks for reading. <Q> For binary symbols (like RS-232), it's synonymous with "raw bits per second". <S> Manchester encoding is a poor fit to this model, but if you wanted to squeeze it in, then a symbol period is equal to a bit period, and there's all the weird Manchester encoding rules that need to be followed. <S> For RS-232, baud rate = raw bit rate, and that includes the start & stop bits. <S> For instance, 1200 baud means a bit time of \$833\mathrm{\mu s}\$ , and that applies for all bits. <S> So if you're sending 1 start, 1 stop, and 8 data bits, the fastest you could transmit payload bits would be \$1200\frac{8}{10} = <S> 960\$ bits/second. <S> Typically, in the literature the term "baud" has become clouded because it's often conflated with "bits/second", and with more complicated modulation schemes like OFDM <S> the exact meaning of a "symbol" can be up for debate -- so if you're authoring a paper, it's a good idea to make it clear what you're talking about independent of the term. <A> Baud is symbols per second. <S> This can, but does not have to be bits per second. <S> Example: You have a symbol which can have 4 states (e.g. red/green/blue/white). <S> If you transmit this symbol 2 times each second, you got 2 baud, but 4 bits per second. <S> A real life example: <S> The ancient 56k modems, the bitrate of them is 56 kilobit/s while the baudrate is 8000 baud <S> /s. <S> For UART where you have a binary symbol (0 or 1), bitrate is the same as baudrate. <A> Baud rate can equal bit rate, it just assumes the simple case of one bit per symbol such as for UART comms. <S> Baud rate is the rate of symbols per second and for complex modulation techniques you can send multiple bits per symbol. <S> A modem can send 4 bits per symbol <S> so 600 baud modem can send 2400 bits per second. <S> You can press one key on a piano, and assuming the piano has 64 keys then each symbol is worth six bits. <S> You can type single ASCII letters (symbols) using a keyboard, but each symbol is 7 bits.
| "Baud rate" means "raw symbols per second".
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How to control LCD monitor with micro controller? I would like to control my PC monitor.After disassembly I found the control buttons are implemented like this (just more data line and buttons) I would like to control it (with ESP2866)With the first button which is alone in a DATA line 1 there is no problem I just simply pull down with one of the GPIO like this and it is working flawlessly (I don't use any resistor because expecting in the other side there is) With Data line 2 I’m having trouble with the same approach the button with the smallest resistor is triggered from other button. I try to use resistor, transistor but it just got worse. Please note ESP is working a bit higher voltage than 3.24v it is 3.26vAlso connected the ground wires I wonder how should I try to interface with this? Thank you <Q> Use an NPN transistor or small signal <S> N channel MOSFET. <S> The GPIO drives the base/gate (the NPN transistor includes a series resistor with the base). <S> The collector/drain is connected to the resistor side of the switch. <S> Have one sub-circuit like this for each of the switches. <A> You can make this circuit independent of power voltage of ESP, it could be 25 volts. <S> You hook up exactly there, your GPIO pins are all just INPUTS - <S> that makes them open line. <S> They look like an unpressed button to the system. <S> And when you "press" - you OUTPUT LOW them. <S> Releasing it - making INPUT again. <S> See if you may want to make it LOW for a few milliseconds. <S> I think it should work fine like that. <S> I had an idea to wire up an MCU to a display panel like that for a while now, maybe will finally get my hands on it. <S> P.S. <S> I assumed to did output high when unpressed, you didn't clearly indicate that <A> If a GPIO is wired to Data1, then wire the other GPIO's as follows: <S> Next GPIO to 120Ω to Data2 <S> Next GPIO to 3.3kΩ <S> to Data2 <S> Next GPIO to 39kΩ <S> to Data2 <S> Note all these GPIO's should be "open-drain" type, meaning they should either be open (floating, input) or pull low. <S> Might make sense to implement this as: <S> Set pin to input Set pin value to 0 or low Toggle <S> the pin tri-state (input or output) to toggle between "open" and "low."
| You could also use an IC chip that has open collector/drain output for buffering the GPIOs to the switch connections.
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Is there a limit to the number of inputs that can be summed with an op-amp summing circuit? I'm interested in the limits you encounter if you try to add many inputs into an op amp summing circuit. I've read that the inverting configuration is more natural for adding many inputs since the gain for an input is just determined by the feedback resistor and the resistor on that particular branch. Is there a maximum number of inputs you can get away with? This answer maximun number of inputs on a opamp adder circuit (dealing with a non-inverting amplifier) says that noise is a limiting factor as you add each input. I guess I would assume that the noise would add in quadrature so that the noise would go as sqrt(N) for N inputs. Is this right - and are there any other factors that would limit you from adding inputs indefinitely? Thanks! <Q> A summing amplifier: - A summing amplifier showing the internal noise source of the op-amp: - <S> Now let's show that circuit with all the summed inputs connected to ground: - The noise gain of this circuit is: - $$1 + \dfrac{R2}{R1/3}$$ <S> So, as you add more inputs (more R1 resistors connected to ground) <S> the R1/3 term becomes R1/4 (four inputs) <S> then R1/5 for 5 inputs. <S> This means that the output noise caused by the grounding of inputs gets bigger as you add more inputs. <S> But you might say that the inputs are connected to voltage sources and not grounded. <S> And I would say it matters not one bit because whether inputs are connected to ground or a real voltage source then noise becomes progressively amplified as you add more inputs. <S> Sure, an unconnected input won't add more noise gain. <S> But also, as you incorporate more summing inputs the non-inverting input <S> node acquires more parasitic capacitance to ground and it eventually dominates the input resistors at high frequencies (hence why I've removed the input resistors): - <S> And the noise gain clearly becomes very massive at higher frequencies because Xc becomes very large. <S> are there any other factors that would limit you from adding inputs indefinitely? <S> The inability of the op-amp to provide the current to all the inputs and thus not maintain the virtual earth condition required for proper numerical summing. <A> The current from the output back through the feedback resistor will be limited by the current capability of the op amp. <S> The total current, through all summing resistors, can't exceed this maximum feedback current. <S> Eventually, as the number of inputs gets large, the current through the summing resistors will be comparable to the input bias current of the op amp. <S> The summing function will no longer be determined primarily by the ratios of the resistors... <S> the input bias current will start to cause a significant bias voltage. <S> Furthermore, any noise or variation in the bias current will cause a corresponding noise voltage on all summing resistors. <A> All those solder joints and PCB traces will add parasitic capacitance onto the Virtual Ground node, causing more and more phase shift, degrading the transient settling, and eventually you'll have an oscillator.
| As the number of inputs increases the current passed through each of the summing resistors must also decrease (the summing resistors will be very large compared to the feedback resistor).
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Is it safe to bypass a battery? I have an old camera which Im tinkering with and when I opened it up I found that the battery (liPo) is puffy. As I dont have a spare I can't replace it and Im beginning to doubt myself that its safe to just put a wire there instead. I will be plugging it in to a 5v usb. TL;DR: Can I put a wire in place of a battery? Edit: It has a built in 5v usb port, which is what I intend on using to power the device. I would like to know a safe way of removing the battery, so that it will only be powered when plugged in. <Q> Have you tried just running the device off your USB power without a battery? <S> possibly with an electrolytic cap in place of the battery? <S> If this doesn't work when i wanted to have a wired wii mote, i just soldered a 5v to 3v SOT-223 LDO to the battery prongs (and some ceramic caps) and wired it to a USB cable. <S> This is essentially what @Passerby is talking about with his Battery Eliminator. <S> Also like @Elliot Alderson said, don't short your battery terminals with a wire. <S> If the device doesn't work because it cuts off the USB power if there is no battery, then you can simulate a load with a 1k through hole resistor to make it look like a battery is being charged. <A> No, this is not safe. <S> If the camera tries to charge the "battery" you may get a very high current through the wire, causing it to get hot and possibly cause a fire. <A> It would be better to simulate the battery and provide ~3.6V (for each cell, a once cell will be 3.4 a two cell will be 7.2V) to the camera through the battery terminals (remember to get the polarity right, and make sure if you do simulate a battery, that the usb charger or any other charger is never connected ). <S> I have a DSLR that has an adapter that does a similar thing.
| A wire in the place of a battery cell is dangerous as it would short out the charging circuit inside of the camera.
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Op amp to supply current to 3 thermistors and an ADC voltage reference input I have 3 thermistors in parallel that I need to measure.These thermistors will be fast changing - think up to 100C in under 1 second. I want to do this ratiometric measurement, so I want to supply the ADC reference from the same voltage which the thermistors are excited by. Is it OK to do this using 1 opamp follower from the voltage reference? The LT1499 says it can supply up to 30mA, and there will be lets say max 2.5mA for each thermistor when it heats up (total 7.5mA), and a few 100um for the ADC reference pin. Is this a normal technique? It would mean I can use just a quad package of opamps (+3 for buffering the thermistor output to the ADC input channels) These thermistors are fast moving temperature, so will go from 10k to 500 Ohm in under 1 second, and with the current needed to maintain the 2.5V going up 20x accordingly. So this is effectively a variable load for the op amp? and I may have to put some 10n or 100n caps in parallel with the thermistors <Q> Well, you have 800 ohm in series with each thermistor <S> so, at the minimum thermistor resistance, the op-amp output sees 3 parallel resistors of 1300 ohms = <S> 433.33 ohms <S> and, if as you imply, the reference voltage is 2.5 volts then, the total current to be supplied by the op-amp output is 5.8 mA. <S> I don't think this will be a problem to most op-amps providing that the 2.5 volts isn't close to their lowest output voltage when the negative supply is at GND. <S> The LT1499 can supply that. <A> No, the resistance seen by the opamp for each thermistor only ranges from 10.8k down to 1.3k, an 8:1 change in current. <S> The total maximum current from the opamp for all three is 5.8 mA, well within its capabilities. <S> And changes over 1 s <S> are glacially slow as far as the opamp is concerned. <A> The reference amp output current changes 5.075 mA in under 1 second. <S> Lets call it 6 mA per second. <S> That's pretty slow. <S> Same for the three buffers - 1.35 V in under 1 second, rounded up to 1.5 V or even 2.0 V per second, still is very slow. <S> Focus on the opamp's input errors and their drifts. <S> If the chip is seeing even a fraction of the same temperature swings as the sensors, thou shalt have wandering error terms in the output voltages. <S> Caps in parallel with the thermistors will slow things even more. <S> Update: In case it isn't clear, I am not recommending you use an LM741 in this application.
| Except for the noise, offsets, and headroom issues, a 741 could do it.
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Posssible Delay for a electric circuit needed While I know that it is impossible to slow light down I was wanting to delay an electrical response from a diode using either a resistor or a capacitor but I could not remember, which does which. The idea for the product is to create an electric delay on a row of led lights after a flashlight turn on the photoresitors behind them on so we can see it turn on slowly like a light chaser. I was planning on using a led a diode and resistors in a row with appropriate delays. I'd like it to be compact so no mechanical delays. PS: This is behind a opto resistor and am wanting one diode to turn on the row of led's then go down the row as the delay is making them react as if slowly like a gradient. <Q> You can program any delay you want. <A> A microcontroller is probably your best best For example, microcontroller senses change in resistor → interrupt <S> → LEDs turning on in sequence. <S> PWM could also be used for dimming the LEDs. <S> You also get programmable LED strips. <S> Not sure if a capacitor is the best way to go for a project like this <A> 1Meg Ohm resistors and 0.1uF capacitor, into CMOS 74HC14 <S> (hex schmitt inverters), may work. <S> I' d implement the dimming feature with Pulse_width_modulation into NAND gates and some logic to produce numerous duty cycles. <S> But the MU idea is more compact solution. <S> In either case, you need some high-current LED drivers with 33 ohm or 330 ohm current limiting resistors. <A> For an LED light chaser, have a look here: <S> https://electrosome.com/led-chaser-ic-4017-ic-555/ <S> It uses a one-hot counter <S> https://electrosome.com/cd4017/ which does the chasing, and a 555 timer as a clock, with which you can set the chase speed. <S> In your case you might need to add another flip flop to reset the counter. <S> The flop is SET when the opto sensor is hit by light, and RESET when the last LED is illuminated.
| A 555 timer delay circuit may also work. I would use a microcontroller to detect when the photoresistor is illuminated and then digitally turn the diodes on and off.
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High-Side P-Channel MOSFET Won't Turn Off! I'm using the attached circuit to control the power supply of my battery-powered product. This is the 8th version; the newest addition is Q1 which I'm using for reverse supply protection. All previous versions work just fine. The issue is no matter what I do to the latch node, the second FET will not turn off. I've removed C7 to rule it out, and tried to swap R7 for a 1k resistor. I've also cut the trace so that LATCH is floating and then directly applied both VBAT and GND to the latch node, neither has any effect. Measuring the supply pin of Q4 shows 4.62V, at LATCH it's 4.58. I've also removed the diodes. VIN is going directly into an LDO and nowhere else. Is there something obvious I'm missing here? I've swapped out Q4 for another MOSFET (still an AO3401 but from a different reel). How do I get Q4 to turn off? <Q> On reverse polarity protection: In the schematic in the question, Q1 still conducts in case of negative input voltages (= reverse polarity) and D1 creates a shortcut. <S> I assume, that this was not intended. <S> Q1 must be connected the other way round like this: The Z-Diode is only necessary, if the maximum input voltage (+BATT) can be greater than the maximum Gate-Source-Voltage of the P-FET (usually 20V). <S> [The 47k in my schematic can also be 100k. <S> This should not matter.] <S> On positive input voltages the internal body diode charges the capacitor. <S> The voltage over the capacitor turns the P-FET on (Vgs must be negative; gate must have lower voltage than source). <S> On negative input voltages, the internal body diode is blocking, thus preventing reverse polarity from doing harm. <S> On LATCH: If C7 charges up, Q4 is turned off. <S> So, R7 and C7 must be interchanged. <S> The polarity of Q4 in the schematic in the question is right. <S> These points should fix your problems. <S> Do not worry about defective FETs. <S> All of them are right, when delivered to you from the original manufacturer. <S> If you find one, that is defective, assume you broke it and look for the reason. <S> Good luck with your circuit and have fun! <A> This may be wrong, but if I am not mistaken a P-MOSFET requires negative (+ thresholds..) <S> voltage on GATE relative to source to turn on. <S> Since your voltage measured on source is almost equal to the gate voltage, it may not have "enough" to turn off. <S> Also once you connected GND to GATE, Vgs was negative (0 - 4.5 V = -4.5 V) <S> thus the FET was kept in the "ON" state. <S> After connecting the VBAT (since you measured 4.62 V on the source of FET, I expect the VBAT voltage to be something like that) <S> Vgs voltage was (4.58 - 4.62 = -0.04 ), which is still negative. <A> I tried a third reel of AO3401's and the board now works. <S> I reworked it several times <S> so it's unclear whether the issue was a defective FET <S> (makes sense - they fail to a short) or the issue that Tomas called out where there is some voltage drop causing Vgs to be slightly negative. <S> In the later case, I may have removed whatever unintended path to ground was present, but I find this unlikely since shorting Vg directly to VBAT did not work. <S> TLDR for anyone exploring this same problem: make sure the FET is not defective or in a failure mode and make sure there are no unintended paths to ground.
| There are no defective FETs shipped.
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Does cooling the NAND chips on an SSD negatively affect its reliability? The problem of heat dissipation in high-performance, small form-factor SSDs is well-known, for example, the paper Transient Thermal Analysis for M.2 SSD Thermal Throttling published in 2018 17th IEEE Intersociety Conference on Thermal and Thermomechanical Phenomena in Electronic Systems states: Solid State Drive (SSD) technology continues to advance toward smaller footprints with higher bandwidth and adoption of new I/O interfaces in the PC market segment. Power performance requirements are tightening in the design process to address specific requirement along with the development of SSD technology. To meet this aggressive requirement of performance, one major issue is thermal throttling. As the NAND and ASIC junction temperatures approach their safe operating limits, performance throttling is triggered and thus power consumption would drop accordingly. Naturally, if space allows, adding a huge heatsink is a possible solution to this problem, there are many products available on the PC gaming market. I also see many M.2 to PCI-E passive adapters on the market have built-in heatsinks by adding a huge copper pour with connection to the ground plane under the M.2 connector. But one can find many unsourced posts on random computer hardware forums, which claims that the NAND chips should never be cooled. It is claimed that they are actually designed to heat itself up to an optimum operating temperature, and adding a heatsink to the NAND chips adversely affect its reliability. Here's some examples. One claim reads, Don't cool the NAND dies themselves! They heat themselves up to operating temperature by design, cooling them means they just continually dump out power trying to hit temperature, and will be operating with a lower endurance (simplified: higher operating temperature = lower energy input to set/erase cells = less degradation of each cell per write/erase cycle). Another claim reads, Cooling the NAND is bad. You want the NAND to run warm and stay warm. As its temperature fluctuates, and as it cools down, if you suddenly transfer a large file (read or write, I can't remember) while the NAND hasn't had time to warm back up first, it can significantly reduce the life of the NAND. I tried to fact-check and/or debunk these statement, start by looking for a NAND chip datasheet found in some recent SSDs. I went to Digikey and Mouser, set the filter to the highest storage density and sorted them by prices. Unfortunately, it seems that datasheets are not available (all under NDA? I'm looking at the wrong place?). Are these strange statements have any factual basis? <Q> The paper Influence of temperature of storage, write and read operations on multiple level cells NAND flash memories from 2018 shows the following graph, which suggests that writing to flash cells at a temperature of 25°C or lower results in earlier problems at reading compared to writing at 85°C. <S> In their discussion they deduce the following reasoning: <S> Most NAND Flash memories implement the Fowler-Nordheim tunneling effect 1 in order to inject charges through the floating gate [7] during write operation. <S> During write cycles, the programming circuit controls the charge of cells to ensure a sufficient margin of voltage threshold. <S> It is assumed that the writing management circuit probably drifts with low temperatures. <S> Indeed, transistor parameters (threshold voltage and gain) vary with temperature which in turn induces drain current shifts. <S> And in the conclusion they summarize: <S> Write <S> operations at low temperatures lead to a decrease in data retention time, probably not due to a degradation of the cell but due to parametric drifts of the die embedded electronics dedicated to write operations. <S> This suggests why the comment cited in the question might say that. <S> But in practise I would assume that this effect is not relevant, because a better cooling of the flash will simply give the flash controller more headroom to higher performance while keeping the same temperature (assuming cooling with a traditional heatsink). <S> After seeing the above measurements I would NOT cool my SSD with LN2, though. <S> https://doi.org/10.1016/j.microrel.2018.06.088 <A> First, I don't think (e.g. haven't encountered) <S> flash chips will actively heat themselves. <S> They just get hot when they are used like a CPU will. <S> Power consumption and temperature measurements of SSDs also seem to indicate this (they would use quite a bit of power in idle to stay at optimal temperature). <S> I think the claim is looking only at one thing and not at the whole picture. <S> This article on EEWEB shows two things to consider. <S> Raw Bit Error Rate (RBER) and data retention. <S> There is a definitive increase in the RBER with decreasing temperature, but at the same time data retention decreases significantly with higher temperatures. <S> There are probably different failure modes at work which contribute differently on different temperatures. <S> Seeing that the data retention is significantly reduces at elevated temperatures (above 55°C), I'd try to keep it relatively cool. <S> Data suggests there is no negative effect on staying at room temperature where data retention is still a lot better than at higher temperatures. <S> It's surprisingly hard to find anything on retention and endurance at temperatures below 25 °C. <A> How true is this claim? <S> In my opionion: this is completely untrue . <S> As far as I know there is no "optimum" temperature, the highest temperature at which the chips will work is limited by the temperature control on the chip. <S> Slowing down means less power consumption so less heat generation so the temperature of the chips will stabilize at a maximum value that is controlled by the chip. <S> When a heatsink is attached, more power can be dissipated in the chips while they are at that same temperature. <S> A heatsink will help to remove the heat from the chips. <S> As more power can be dissipated the chips can run at a higher speed. <S> higher operating temperature = lower energy input to set/erase cells = <S> less degradation of each cell per write/erase cycle <S> I doubt that the amount of energy needed to write to a cell <S> depends so much on temperature. <S> Also, this seems to relate only to writing to cells and not reading. <S> It has been proven that cooling an SSD increases its read performance. <S> The writer of the article (you should include a link where you fond it) does not seem to understand very well how flash memory works.
| The chips slow themselves down when the maximum operating temperature is reached.
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Is Gaussian noise equal to white noise? If noise on a signal is randomly fluctuating in time domain around its mean following a Gaussian normal distribution ( Gaussian noise ), would this be equivalent to an constant intensity in the frequency space ( white noise )? Can we somehow relate the frequency distribution to the amplitude distribution? <Q> No, they are completely orthogonal concepts. <S> The probability distribution says nothing about the frequency content, and the power distribution across frequency says nothing about the sample probability distribution. <S> You have to specify both. <A> As Dave (and Brian) said: two totally different concepts. <S> One doesn't imply the other. <S> This is homework, and you should research it well! <S> Getting the difference between (auto)correlation/PSD and amplitude distribution straight is a critical thing. <S> If this isn't clear to, you should probably ask your professor/teacher (if you have one) – it's easier to explain if one has a didactic "framework" to work with. <S> There's one thing that's special about Gaussian noise w.r.t. to correlation, and that if random variables (for example, noise measurements from different times) are jointly uncorrelated (and that's a big restriction!) <S> , then they are independent. <S> For all other distributions, lack of correlation does not imply independence. <S> This is a property about circularly symmetric gaussian ( \$\sim\mathcal{CN}\$ ) noise that allows us to do a lot of mathematical transforms on it (e.g. correcting the phase of a received signal) and still have independent noise components, and that is what's necessary for a lot of estimators to actually work optimally. <S> So, hurray for circularly symmetric gaussian noise! <A> However, contrary to the other answers, there is a sense in which white noise implies Gaussian noise, if the noise is white to arbitrarily high frequencies (arbitrarily small time scales). <S> Or more practically, if our measurements average the noise over time intervals much longer than its correlation time. <S> In this case, the central limit theorem says that the measured noise amplitude, being composed of many independent contributions (with finite variance for physical reasons), converges to a Gaussian distribution. <S> EDIT: <S> How much longer than the correlation time is needed for the central limit theorem to converge depends on the statistics of the noise. <S> John Doty's comment points out that it does not happen quickly for white noise consisting of pulses that follow a Poisson process. <S> In this case, the amplitude has a highly skewed distribution that is mostly concentrated on zero. <S> This is a "worst case" for the central limit theorem. <S> Averaging over a few pulse widths (correlation times) doesn't make it Gaussian; we have to average over longer than the mean interval between pulses. <S> When we do this, we start to get a less skewed Poisson distribution that is approximately Gaussian. <S> So it still holds that if measurements are averaged over long enough times, white noise looks Gaussian. <A> I'd like to add something which the other answers haven't mentioned yet. <S> It is true that a gaussian distribution is not the same thing as a uniform white-noise distribution. <S> They can be related however. <S> True white noise occurs in a resistor as a result of the brownian motion. <S> It kind of implies that the imaginary part of a complex resistance is zero: in a + ib , b has to equal 0. <S> Non-zero imaginary are caused by capacitance / inductivities, and will cause low-pass or high-pass behaviour, and thus, the spectral distribution is no longer uniform, but still "linear" <S> However, non-linear components, such as diodes, can shape a uniform distribution in a way that it becomes a gaussian distributed random value. <S> Or any other distribution, depending on what non-linearity is at play. <S> Because from an engineering perspective, it sometimes is not very important how the spectral composition looks like, we may want to calculate one number to express a noise level instead. <S> This can done by integrating over the whole (or parts of the) spectrum. <S> Afterwards, it is impossible to tell wheter <S> the integrated value came from what was originally gaussian, or uniform, or anything else. <S> Assuming we integrated over a a gaussian distribution, we can always find an uniform distribution that is scaled so that its integral matches the integral of our gaussian distribution. <S> This has immediate merit for our calculations: We can then assume a complex non-linear component is just a resistance after all, which may simplify calculations. <S> I recall that sometimes this is done by the manufacturer already and being given in the datasheet. <A> "White" implies the independence of signals in time, "Gaussian" implies the probability distribution of the momentary value at an individual point of time. <S> Pretty much orthogonal.
| Gaussian noise definitely does not imply white noise, because Gaussian noise can have an arbitrary (not necessarily flat) frequency spectrum.
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What would happen if I used a potentiometer to produce a 5V DC signal for an Arduino from a 9V battery? What makes the potentiometer not a good component to use for reducing the 9V to 5V DC for an Arduino? I thought of using it but I been told it is not a good idea, so I am looking into the reasons why. <Q> The potentiometer is a resistor component. <S> When you try to create a voltage reduction through a resistor the voltage drop is 100% dependent upon the amount of current flow through the resistor. <S> A load like an Arduino is not a fixed current load so it is not possible to use a single resistor value to reduce the 9V to 5V suitable for proper operation of an Arduino. <S> You will need to investigate what a voltage regulator does and why it is applicable to the application. <S> Beware that if your 9V source is a typical 9V battery it is far from a suitable long term power source for an Arduino due to the relatively low capacity of such a battery. <A> When I was young and innocent I tried the same for a bunch of relays, and my potentiometer caught fire! <S> First off, as you allude to, it is in some cases possible to regulate load voltage with a potentiometer. <S> You can use it as a divider (three terminals) or as a series resistor (two terminals). <S> However, this involves you adjusting the potentiometer depending on the drawn current. <S> For constant loads this is feasible, but not so for fluctuating loads from a processors, as others here pointed out. <S> Your circuitry, including it's input/outputs and peripherals, use different currents depending on their state/mode. <S> And their reliable operation depends on a constant load-independent supply voltage. <S> But there are other factors too: heat and efficiency. <S> The current through the potentiometer, and the ensuing I^R heat, generates heat in both configurations. <S> And in the divider configuration you have a current running even if there is no load. <S> Since my pyrotechnic experience, I instead use a "LM7805" integrated regulator for simple cases ( https://en.wikipedia.org/wiki/Voltage_regulator ) <S> These regulators keep the 5V constant practically for all (fluctuating) current loads from your Arduino and any peripherals (sensors, LEDs etc..). <S> They are very easy to use, and quite suitable for a simple table-top home lab, but they are not very efficient. <S> In other applications they can get hot, and so now I only occasionally lightly burn a finger. <S> If power efficiency / battery life matters, look at switch mode power supplies, like the LM2674 https://en.wikipedia.org/wiki/Switched-mode_power_supply <A> You may use as potentiomenter to provide an analog input signal to an Arduino (preferably with an additional resistor in series to prevent pulling the input above 5 volts), but you should not use a pot to provide power to an Arduino, for the reasons given in other answers.
| That said, if the current fluctuation and the ensuing voltage fluctuation do not affect the operation of a (perhaps fixed) load, then the potentiometer or voltage divider can be used.
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Kelvin Source Terminal on MOSFETs On MOSFETs with multiple source pins, is the so-called "Kelvin-connected source pin" connected any different internally? Or is it just like every other source pin? EDIT: Just to be clear, I am not asking what the Kelvin connection is for. I'm asking whether it is physically implemented any differently than the other source pins or whether it is just an extra regular, old source pin. <Q> You could search for "Kelvin source pin" on the manufacturer's web site . <S> Basically it's a Kelvin connection to the source itself, that bypasses any package parasitics that would screw up your control circuitry's measurement of the source voltage. <S> Google "Kelvin connection", and maybe "four wire resistance measurement". <A> The Kelvin connection is basically a dedicated small signal connection straight from the SOURCE/EMITTER of the die dedicated for control. <S> The reason this is done as oppose to using the SOURCE/EMITTER which carries the main current is to mitigate inductive voltage changes on the gate-source/emitter when switching current. <S> As devices switch faster this inductive drop can be enough to turn the device in/off. <S> Consider the image below. <S> A MOSFET and a gate drive as part of a H-Bridge. <S> Essentially this is how you would capture the circuit but <S> this is not the complete picture <S> The package will have stray inductance and <S> equally a classic 3pin FET/IGBT will have trace inductance and poor layout <S> can result in poor control With higher current or higher switching speed, the voltage developed across the stray source inductance will increase. <S> A TO-247 will typically have around 20nH of source pin inductance, but with current of 100A switching in 50ns, the resultant voltage developed across this pin, while the gate is being charge or discharge will interfere with switching and result in shoot-throughs. <S> To mitigate this, manufacturers of devices for power applications provide an auxiliary Source/Emitter connection to permit gate voltage control without transient behaviour due to the main current carrying connection <S> Typically this is done via an additional bondwire (thinner gauge as it does not need to carry current) onto the Source/Emitter pad of the tile routed to the auxiliary Source/Emitter pin routed to a standard pin used in the package. <S> https://www.slideshare.net/Yole_Developpement/yole-cree-cas120m12bm2sample <S> Another example of the kelvin source connection https://www.ge.com/reports/post/129076208340/the-odd-couple-silicon-and-carbon-dont-love-each/ <A> This is speculation, I've never heard of such a connection. <S> My best guess is that it's like half a typical kelvin resistor connection. <S> Aka the source pin (2) has high current carrying capabilities, while kelvin source (3) is same node but isn't designed to source significant current (only for voltage measurement purposes). <S> image source: https://en.wikipedia.org/wiki/Four-terminal_sensing#/media/File:Kelvin_connection_layout.png <A> Not an entire answer to my question, but I did find this which does indicate they are physically different, though it does not say why.
| It can be compared to the Kelvin connections on sensing components (sense resistors, NTC/PTC) where their primary concern is to minimise measurement error due to additional voltage due to harnessing and associated current.
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Considerations for converting a LED Bulb from AC to DC power I was asked by my uncle to convert 22 units of gx53 LED Bulbs that are powered by 230VAC 5W to DC voltage power. The LED Bulb itself consists of 5 LEDs(string) in series, in parallel to another 5 LEDs(string) in series. I have disconnected the LEDs inside the Bulb from its inner driver and through a resistor i connected it to a DC power supply. I learned using a measuring device(voltmeter) that each LED lights up with 8VDC and the current flowing through it is 8ma so the Bulb itself can be powered by 40VDC with current of 16~20ma.(5 * 8V=40V, 2 branches of LEDs * each branch is 8ma=16ma) Now what i need is the right led driver. I learned that LED drivers supply constant current while the output voltage is within a range of voltages. So if i have 22 LED Bulbs that each need 20ma with 40VDC then the driver, to my understanding, should supply current of about 450ma with output voltage of at least 40VDC. I ask for advice whether i'm right or wrong. <Q> Your calculation is incorrect. <S> When you connect LEDs in series, the voltage required increases, but the current through the string is the same as for 1 LED. <S> (The same electrons are flowing through all LEDs). <S> So if 1 led requires 8mA at ~8V, then 5 LEDs will require 8mA at ~40V. Two of these strings in parallel would require 16mA at 40V (if your math for 1 led is correct). <S> An easy check you could have done is to calculate the wattage. <S> 450mA at 40V would be 18W, where it should be around 5. <S> Your initial assumption of 8V and 8mA for 1 led would give a wattage per led of 64mW. <S> So for all 10 leds we're looking at 0.64W, which is not correct. <S> Again it should be around 5W. <S> You should hook up the driver again, and adjust it until 1 led is drawing 0.5W. <S> Then redo the subsequent calculations with the current and voltage you measure. <A> No, I think your 8V 20mA for one LED is wrong. <S> Usually these type of LED diode (a more exact term than "bulb") are 3V 150mA. <S> They will still turn on at 8V at overcapacity. <S> Don't try this for a long time. <S> Apply 5V (or 8V <S> if you don't have a 5V supply). <S> Measure the exact voltage at the supply output or at the LED anode (its plus side). <S> It may be slightly different than indicated. <S> Call it Vsupply. <S> Measure the voltage at the LED cathode <S> (it's minus side). <S> Call it Vdrop <S> Calculate: <S> Vled = <S> Vsupply - Vdrop. <S> If by applying 8V you measure 5V, then the LED is 3V. <S> And you have 5V too much. <S> A better way would be to make this test with two LEDs in series after the 8V supply. <S> measure the voltage drop after each LED. <S> If the LEDs are off when you use two LEDs in series but are on when only one LED is connected, then you already know they are <S> more than 3V. Still measure to know how much exactly. <S> About the amperes, it's a wild guess from my part, looking at the picture. <S> IMO, it must use between 100 and 150 mA. <S> If you have 5x3V <S> = 15V, ideally, you will have to find two constant current power supplies of 120 or 150 mA with a voltage range around 15V, or 3W.If <S> you can't find this or <S> if you want to do it cheaper, you can use simply a classical 15V power supply and add one resistor of +- 30 ohms before each series of LED. <S> These resistors must be rated 1/2W or more. <S> Should be OK too. <S> Make sure the LED are ventilated to avoid overheating. <A> The best idea by far would be to measure the voltage/current from the original driver and then replicate it. <S> However, from your measurements some guesses are possible. <S> If they light up at 8V, then they're probably 9V nominal (3 diodes in each SMD <S> ) 0.5W diodes. <S> Ten of them gives the nominal 5W rating for the whole bulb, so that works out. <S> If there are two parallel strings of 5, then the total voltage is 45V, and the max current per string is probably about 50mA, with a little less than that for safety being a good idea. <S> Times two gives 100mA. <S> I would aim for a little less than that to be safe since you aren't sure what the original driver was giving.
| To measure the voltage of the LED, you have to measure the voltage drop.
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If AA batteries don't supply sufficient power, will adding more of them in parallel help? My device requires ~4.5 V. I'm using 4 battery packs connected in parallel Each battery pack contains 3 metal-encased non-rechargeable alkaline AA batteries. I used voltmeter and verified that each battery pack individually provides ~4.5 V. I have also used voltmeter to verify that all battery packs are correctly connected in parallel. All the batteries are of the same nominal voltage, the same brand and product name (Varta Longlife). (It shouldn't matter here, but for completeness: The batteries are used as a backup to the main power supply, which supplies about 0.5V more than the batteries. Both the power supply and the batteries are connected via a Shottky diode [1N5822 40V/3A DO201] to prevent "charging" each other.) The device runs fine on the battery packs, but sometimes, obviously when a load spike comes, the device suddenly switches off.I measured that the device, when connected to the power supply, draws 1.6 A during spikes. Normally it draws about 300-500 mA. Question 1: Shall I keep adding more battery packs in parallel? Will it help or does it only increase the lifetime, but not max current/power? Question 2: If I use a battery with higher mAh rating, will it help, or does it only increase the lifetime? As a side note: I am also going to try top-notch lithium AA battery instead of alkaline to see if it handles the spikes. Thank you. EDIT: The backup is not supposed to be used much, just very rarely, during a power outage (which we haven't had in years). UPDATE: I've tried the best lithium AA batteries on the market, replaced the alkaline batteries, and found out that they didn't handle the spikes, either. Moreover, they supply ~5.5V instead of the ~4.7V that the alkaline batteries do. So that's not nice, either, because that is higher voltage than that of the power supply.I'm thinking about using the capacitors or the suggested buck converter with 12 batteries in parallel. The thing is, I have no experience with capacitors and I don't even know what the circuit should look like.As for buck converter I don't know anything about them, either, and am a bit worried that they would introduce more complexity to the system, possibly increasing the chance of failure.So adding more battery packs in parallel, and changing to D-type batteries as some suggested, still seems to me to be the simplest (least complex) solution, which I feel capable of designing and building myself. UPDATE 2: 4x3 D-type batteries (instead of the AA-type batteries I original used) did help and the device no longer shuts down during high-drain events. <Q> At 1.6A you're shorting the battery. <S> Maximum load is about 500mA, this is already <1 hour. <S> Internal resistance is about 0.3 Ohm, at 1.6A sets you back half a volt per cell. <S> So, for one pack with 4.5V (real 4.2V ) you go down to 2.7V in real conditions. <S> If equally divided over 4 packs this 1.6A is maybe fine (0.4A per pack), but if one cell fails, your pack cascades into failure. <S> In theory with 4 packs you have 0.1 Ohm internal resistance, which is about 120mV drop during you worst case condition. <S> For the entire pack that is about a drop to 3.72V. (4.2 - (0.12*4) when full . <S> In theory. <S> Can you retry and measure the current of each pack vs the supplied voltage? <S> This should give you insight in the paralleling behavior. <S> I also suggest rethinking this backup solution. <S> Either Put all the 12 cells in series and use a buck converter. <S> (less current, same power, better power point ) <S> Use C or D-cells. <S> Use protected lithium cells eg: 14500 (in series). <S> Maybe even consider lead acid if that fits. <S> Now you're abusing the poor AA's. <A> Yes, adding more battery packs in parallel will share the load and have ability to deliver larger currents and survive the peaks. <S> But it's not necessarily the best option. <S> What about using D cells? <S> They are bigger, and besides larger capacity they can also deliver more current. <S> Always read the available datasheet to get a general idea about current they can deliver. <S> If not available for the particular brand you're using, at least find datasheet for that battery type to provide a general idea. <S> Lithium cells in general can handle higher currents. <A> During a 1.6 Amp spike each of your four battery packs will supply 400 mA, so the voltage from three alkaline cells will be about 3.5 volts. <S> You will lose another 0.35 volts across the diode. <S> You need to identify the minimum voltage requirement of your device. <S> Putting a diode on each pack would also reduce the drop across the diode from 0.35 volts to 0.25 volts. <S> Using Lithium batteries would work much better. <S> The voltage from three cells should be 4.2 volts and with a separate diode on each pack the voltage should only drop to 4.0 volts during a 1.6 Amp spike. <S> If your device can tolerate 6 volts you would be better using a 6 volt supply and four alkaline cells in each pack. <A> Considering your current requirements, number, size and cost of batteries, as well as their capacity and as a consequence lifetime, I would definitely go for Li-ion. <S> First of all, there's plenty of choice in terms of size and capacity. <S> They can hold more charge really well. <S> Besides, you can get a tiny charging-discharging PCB for them. <S> The one from powerbank will do. <S> Google powerbank charger pcb or module. <S> Of course, it will output 5V, If your circuit is ok with 5V, good for you, if not, you can throw in some voltage converter or modify the charger board's boost circuit (most likely replacing one resistor near boost converter). <S> Make sure your li-ion can support current you may need. <S> Look at 18650 batteries first. <S> Make sure powerbank board supports 2A/2.1A output <S> Even if you assemble this whole thing from pre-made modules (charger/boost converter etc) it will still take less space than your army of AA.
| Adding more battery packs will reduce the current demand from each pack, so it would help, but probably not enough. Besides, you'll be able to charge your li-ion without taking them out.
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One LED capable of being turned on by two IC's simulate this circuit – Schematic created using CircuitLab I want to be able to turn on a single LED with either one of these two micro-controllers.Would there be anything wrong with this simple approach? The LED would experience double the currentif both IC pins are grounded, but it wouldn't be enough to damage it; also the LED only blinks for a couple hundred milliseconds in any case, and never stays on constantly. Max sink current is also a nonissue. Please let me know if I am making a mistake or not. edit for clarification:U2 pin is an open drain power good indicator that goes low automatically when usb power is supplied. U1 will turn on LED based on a couple of software events <Q> It's not a good idea!.Do <S> this <S> :Replace the two resistors with two diodes (1n4148 for example), with cathode towards the micros. <S> Add a resistor in series with the LED. <A> Provided both U1 and U2 are powered by the same supply (or supplies that turn on and turn off together), this would work fine. <S> Some current will also flow from one micro's output driver to the other when one is high and the other is low. <S> That's not really a problem either unless you are very power-constrained, but you could avoid it by tri-stating the pins when you aren't trying to turn on the LED. <A> You say that two microcontrollers are trying to drive this LED. <S> If that's the case, there is a high probability this can be solved in software. <S> Most* microcontroller pins that are capable of digital outputs are some type of "GPIO", or General Purpose Input/Output, meaning they can function as either an input or an output. <S> Furthermore, several microcontrollers have the capability of operating in an "open-drain" mode, which means there is a way to make it so that the output pin only sinks current. <S> Here's some pseudocode for how that would work, as well as a schematic: // <S> Microcontroller 1 <S> (U1) <S> U1.P1.Out = <S> 0 <S> // <S> Drive 0V to the output pin of U1. <S> Don't need to change this againU1.P1.Out_EN = <S> 0 <S> // <S> Disable U1 drive// Microcontroller 2 (U2)U2.P1.Out <S> = 0 // <S> Drive 0V to the output pin of U2. <S> Don't need to change this againU2.P1.Out_EN = <S> 0 <S> // <S> Disable U2 drive// Microcontroller 1 Turns on LED// U1.P1.Out_EN = 1 <S> // <S> Enable U1 Drive // <S> Microcontroller 2 Turns on LED (still stays on, no drive <S> contention)// U2.P1.Out_EN = 1 // <S> Enable U2 Drive simulate this circuit – <S> Schematic created using CircuitLab <S> Now, you can add diodes (right example) to make it a bit more foolproof. <S> Both examples require a shared Ground/VSS, and the VDD should be the same. <A> If you operate the pins in open-drain mod e (either the pin has an explicit open-drain mode or you tristate the output - change it to an input) <S> then connecting them the way you have cannot damage the drivers. <S> So what you have is fine if you write the firmware correctly . <S> Worst case is that they "fight" and the LED goes on only dimly or not visibly at all (depending on color and efficiency) since it only sees 1.65V open circuit. <S> The two resistors effectively "isolate" the outputs from each other so no damage occurs if a firmware error or some kind of hardware glitch causes them both to be driven in opposite direction with one actively driving high. <S> Most MCUs can be programmed to do that on their GPIO pins. <S> Diodes are not a good solution, particularly if you intend to drive a blue or white LED since 3.3V is barely enough voltage to start with. <S> You could also use a single gate/buffer chip :( <S> of course in that case you would configure the GPIO pins as push-pull outputs) simulate this circuit – <S> Schematic created using CircuitLab <S> P.S. Note that sometimes bad things can happen if the two microcontrollers have independent power supplies and one can be off ( even momentarily ) <S> when the other is on. <S> If this is a possibility, it might be better to drive two discrete MOSFETs with one resistor.
| As you note, more current will flow and the LED will be brighter when both outputs are low, but that's not a problem (maybe it's even a desirable feature depending on what you're trying to indicate).
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Measure resistance of resistor in cirucit How do I measure accurately with a multimeter the resistance of a resistor that is already soldered as part of a circuit? In a simple transistor as a switch circuit, I noticed that when I tried to measure the collector resistor I got a value of about 170KOhms, but I knew that was incorrect, so I took the resistor out and measured again and I got the correct value, so that got me curious, how can I measure the resistance of a resistor that is already in a circuit and know for sure that it is the resistor I am measuring and not the rest of the circuit. <Q> With a multimeter and no allowances for the rest of the circuit, you can't tell the difference between the resistor of interest and the effect of parallel resistances and conduction paths. <S> So, the answer is: you can't take a reliable measurement without knowing what other components could be spoiling it. <A> If you fully understand and can analyze the rest of the circuit (and know for a fact that it is as it should be with no shorts or damaged parts) <S> That does not mean it's not useful. <S> For example, if something reads higher resistance than expected then something is wrong (assuming no charged capacitors or power sources on the board- if resistance reads negative <S> then it's usually something producing voltage and might damage your meter). <S> ESR (equivalent series resistance) can usually be read accurately enough on electrolytic capacitors in-circuit, which is a major troubleshooting advantage. <S> In the case of (say) a 300K resistor, if you read 650K then something might be wrong since external paths can only lower the resistance. <S> Often (because of the way they are connected) you can check diodes and other semiconductors in-circuit well enough to detect bad parts. <S> Not always. <S> Even if you have not figured out much about the circuit, sometimes comparing a faulty unit to a known-good unit will turn up something of interest to help pinpoint the problem. <A> If you still want to measure resistances without desoldering, you can do at least the following: Take two measurements in the straight and reverse directions of the ohmmeter and take the higher value of the resistance (to eliminate the influence of shunt p-n junctions) <S> Wait a while (to eliminate the influence of shunt capacitances)
| sometimes you can infer or calculate, or simply read the value of a part in-circuit, but in general it's not possible.
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How to wire this 12v switch I need to wire this switch, but I have no idea how. I thought I was buying a normal 3P rocker switch :) The power source is 12v battery and the load is LED flashlight. <Q> It looks like you can control the LED of this switch separately from the actual switching. <S> If you apply +/- <S> on the "3" contacts, then it will light up the LED on the switch. <S> Then "1"/"2" are the contacts for the switch, simply giving you an open/close. <S> Unfortunately, it doesn't look like it is what you wanted if you were looking for a 3 position switch. <S> It can be hard buying switches without fiddling with them first to make sure it's what you want. <A> I thought I was buying a normal 3P rocker switch :) <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> (a) <S> A 3P (3-pole) switch would have three independent contacts. <S> (b) <S> (c) What you've got and how to wire it. <A> It's an SPST rocker switch with a 1.85V 10mA LED as the status indicator. <S> The switch terminals are 1 & 2 and the indicator terminals 3+ & 3-. <S> The value of the LED voltage dropping resistor R would depend on the supply voltage. <S> Here are the cut-out details for mounting the switch.
| A 3P2T (3-pole, double-throw) switch would have three changeover contacts. You've purchased a 1PST (one-pole, single-throw) switch with built-in isolated indicator.
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How can I order a custom IC? I'm new to IC design and just wondering if it possible to order IC production if you have all necessary design files (like RTL design, chip layout from Cadence for example, etc.) If it is then what are these companies and what is the approximate cost (for a relatively small IC like for example an ethernet PHY?) \$ \ \ \ \$ <Q> <A> It is possible, but it is neither quick nor cheap. <S> There are several different types of fab you could consider talking to. <S> Near me, I have two options, one is a large university which has their own microelectronics centre. <S> They are a little cheaper and keen to try things out, but you'd have to justify why your design is interesting from an academic point of view. <S> I would probably pick them if I want to make a handful of MEMS devices for <S> R&D. The other is a moderately sized company which operates three small fabs and makes a point of begin able to do small runs and prototypes. <S> They might be a better choice for a small run of chips for products to sell. <S> You wouldn't want to start with one of the really big fabs, not unless you're willing to commit to spending a lot more. <S> Once you have a fab, you have to start the process of adapting your design and their processes to work together. <S> You buy wafers, get masks made etc, test the results, and repeat as necessary. <S> Doing a small batch of wafers, with only a few processes (so probably simpler than your PHY) costs a few thousand pounds. <S> Developing it to the point of reliably producing 8 inch wafers full of mostly-working dies might cost £100k or so. <S> Working up to a fully packaged IC, with all the necessary QA processes etc. <S> might need an investment of a couple of million. <S> These are very rough numbers, treat them as order-of-magnitude guides nothing more. <A> The short answer to ‘how’: contact fab vendors and describe what your requirements are to figure out what services you will need. <S> They will walk you through the process and identify your costs for budgeting. <S> Who are these companies? <S> The best-known wafer fabs are TSMC and Globalfoundaries. <S> There are others that specialize in smaller wafers/older processes, like SMIC in China. <S> In your case you might need to work with a third-party provider who can do all the back-end work for you vs. working directly with a fab. <S> Fabs can identify appropriate partners who do that service. <S> What’s it going to cost? <S> The main up-front cost is the mask set. <S> Three factors influence this: die size, number of layers, and process node. <S> For a small, mixed-signal chip like an Ethernet PHY, it could use an older process like 180 or 120nm. <S> Ballpark for such a mask set would be about $75-150k. <S> The next fixed expense is test fixtures and test development. <S> This can run to multiple tens of thousands. <S> Add to that fixtures for accelerated-life and ESD/latchup testing. <S> You must also do a package. <S> Even if you choose a standard type, it needs to be laid out with an appropriate bonding scheme for your die. <S> There’s other expenses I’m certainly missing here <S> but you get the point: expect up-front costs in the multiple hundreds of thousands even before you get your first packaged part. <S> Different companies can negotiate pricing. <S> In one model, they can bury some of the fixed expense by increasing the cost of the delivered IC, but they will pressure you to take agreed-upon quantities. <S> Finally, if your design is of good quality with unique differentiation, you could possibly sell or license it to an existing company in that market space and defray some of your costs. <S> But for it to have value, it needs to be proven in silicon. <S> A lower-cost way to do that is to put your die on a ‘shuttle’ - a multi-project wafer - to get some sample chips for testing. <A> You might want to check out FPGA for smaller series. <S> Some of them contain quite a few logic blocks, and perhaps even som analog circuits. <S> Wikipedia has some descriptions: https://en.wikipedia.org/wiki/Field-programmable_gate_array
| You'll want to check out multi-project wafer services , the best-known of which is MOSIS . Fab companies provide various levels of service depending on your needs and how much money you have to spend.
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How to parallel 2032 coin battery without removing the dead one? I have device that run with 2032 coin battery but right now battery is dead/empty I want to know how to parallel new one without remove the empty one to avoid wipe out data. or a way to charge the empty non rechargeable coin battery without remove it from it socket.Thanks for your reply NB: it's my first question hope to not be kicked after my first question :) <Q> Like doing a heart transplant. <A> If the battery is at 0V, then isn't that the same effect as removing the battery and putting a new one in? <S> If it is not at 0V and RAM is still running, then the only way to do it is to somehow do a quick swap with a voltage source. <S> Emphasis on the quick. <S> As soon as a higher potential is put on the battery, it will try to charge it. <S> A CR2032 can't be charged for long without something catastrophic happening. <A> In the top picture the battery slides to the right to come out of that clip (which is the positive terminal), but there is plastic blocking this path. <S> In the bottom picture those fingers look like negative terminals for the battery. <S> if you cannot remove the circuit board from the plastic case you will need to connect wires tp the exposed terminals and then connect them to a temporary battery while you cut away the clip and the old battery and find a fay to fit the new battery. <S> usually connecting a new battery in parallel with an old one is a really bad idea, but CD2032 batteries are so weak that this should not cause any problems because it will only be temporary. <S> red: positive wires, connect all three, blue: negative wires connect both green: cut here black temporary CR2032 battery. <S> once you have the old battery out you'll have to decide how to connect the replacement battery. <S> the negative terminals should be accessible from the back of the device now so soldering a wire here is now an option.
| Just connect a power source (like a new battery) in parallel to keep it alive, then remove the old battery, then stick a SECOND new battery one into the slot of the old one.
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What kind of circuit do I need with a 6N138 optocoupler to get proportional output? I have 0-5V input that I need to isolate and still output ~ 0-5V to an isolated circuit. I purchased some 6N138 optocouplers to do this but I'm not getting a voltage output with a 5V input. There are some transistors built into the optocoupler that are complicating things. How do I create a circuit with the transistors to get an output of approximately the same magnitude to the input? <Q> It's not easy to do with a simple optocoupler, because the CTR is not well-controlled. <S> The usual solution is use a "precision optocoupler" like TIL300 or HCNR200. <S> These have one transmitting LED and two matched receiving photodiodes. <S> This may look like a lot of circuit complexity, but it's likely to be simpler in the long run than calibrating a single-receiver optocoupler, compensating for temperature variation, etc. <A> If you don't want to shell out the $$ to buy a commerical isolation amplifier, <S> you can digitize the signal and send it over the isolation barrier digitally. <S> For example, you could use a small MCU with built-in ADC + UART and use serial communications. <S> Rather low-speed serial communications because that opto is a bit of a slug, or pick a faster one. <S> Then convert it back to analog using PWM or a DAC on a second MCU <S> (could be another of the same type). <S> Analog opto-isolation is possible, and we used to do it years ago, but the variation with temperature and LED aging make it difficult to guarantee good performance such as stability and linearity. <S> Analog isolation amplifiers can also use PWM directly (which places some constraints on the opto performance), frequency, or using analog signals chopped and sent over an isolation transformer (typical of some Analog Design process control oriented hybrid products). <A> Or, convert your 0-5 volts in to a voltage controlled oscillator. <S> On the other side of the opto, convert the train of pulses back in to a voltage with an RC circuit. <A> Google results: <S> Use two parts, one in a feedback loop to linearize the other. <S> https://www.electroschematics.com/linear-dc-signal-opto-isolator/ EDN, using FET optoisolators; different part, same process: https://www.edn.com/use-a-photoelectric-fet-optocoupler-as-a-linear-voltage-controlled-potentiometer/ <S> HP/Agilent app note 951-2, "Linear Applications of Optocouplers" , same process: <S> https://linearparts.com/documents/AN-HCPL2531-5954-8430.pdf <S> Vishay's app note 50 "Designing Linear Amplifiers Using the IL300 Optocoupler" for @ThePhoton's answer above: https://www.vishay.com/docs/83708/appnote50.pdf <A> This is the circuit to obtain the output signal with unity gain. <S> The solution is based on AMC1211 from Texas instruments, which is an isolated unity gain amplifier. <S> The input signal is stepped down to a range of 0 - 2V using R1 and R2. <S> The output signal from this IC will be in the range of 0 - 2V. To convert this signal to a range of 0 - 5V, a non-inverting amplifier consisting of an op-amp, R3 and R4 with a gain of 2.5 is used. <S> Thus, for an input signal in the range of 0 - 5V, an isolated output signal of the same magnitude is obtained. <S> R1 and R2 are <S> 1.5:1 R3 and R4 are 1.5:1
| You use one photodiode to deliver the signal to the isolated side, and the other one to provide feedback to an op-amp circuit on the sourcing side to enable the signal amplitude to be controlled accurately. Maybe in the 100-1000 Hz range (don't use 0 Hz as 0V, thats a bad practice), then as your voltage goes up or down there will be a train of pulses going through the optocoupler.
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Relay pins too close on PCB by design? I'm designing an intermittent wiper motor controller to be fitted on a classic car. As I'm selecting components, I came around this relay that seems to be fitting my needs (PCB mount, automotive, high current). When looking at the PCB layout, I find pin 3 and 5 way too close to each other to allow for drilling plus some copper around the holes for soldering. This makes me think I can't use this relay as I need wide copper path for the current demand (20A). I'm not familiar with relays, am I missing something ? I find it strange that it is designed that way. <Q> If you use the recommended 2.1mm holes (in my experience the recommended sizes tend to be pretty sloppy and designed for ease of automated assembly rather than being optimal in other ways) then you can have pads with annular ring <S> 0.36mm (14 mils, quite acceptable) and space <S> 0.36mm. <S> Then you can apply traces as so: <A> Even if you need a wide trace to carry 20 A over long distances, you can still use a narrower trace for a short distance to connect to the relay pins. <S> The trace width is limited by the resistive self heating of the traces, and the values are usually calculated assuming the trace is infinitely long, so each segment of track must dissipate all the heat it produces itself, without any heat flow axially along the track. <S> But in the case of a short narrow segment, connected to a wider track, or to a big bulky relay, axial heat flow becomes significant and the track can be narrowed without over-heating. <A> The pins are quite close together, but given the voltage rating of the relay is pretty low, it is relatively simple to route. <S> Here is an example using 2.2mm holes, 0.3mm annulus on the pin, giving a clearance of 0.3mm. <S> You could also use 0.2mm annulus and get 0.5mm clearance. <S> There is plenty of room to make the traces entering the pins actually very large, by placing the pins in the corner of the entering traces rather than in the centre. <S> This allows for trace widths of nearly 10mm at the pins. <A> You can drill a non plated hole with no pad for pin 3.
| You can make this wider still by using multiple layers if needed, and connecting to planes. Or no hole and cut pin 3 off of the relay.
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What's the benefits to using analog-to-digital conversion like PCM? I was reading a textbook, Data Communications And Networking , by Behrouz A. Forouzan, which says: A digital signal is superior to an analog signal. The tendency today is to change an analog signal to digital data. I don't quite get it, how is a digital signal superior to an analog signal? What's the benefits of changing an analog signal to digital data? <Q> As with most things in engineering, it depends on the trade offs. <S> If you want to add complexity in order to gain noise immunity, then yes the book is correct. <S> However, maybe you have a circuit where the analog source is only 0.1" away from the A2D, would I put a PCM circuit for that? <S> No! <S> Maybe you have an analog sensor that requires an op-amp for some reason, now you have a nice low impedance driver, which can withstand noise better, so again, PCM isn't needed. <S> There are more applications for PCM than what I've described <S> check Wikipedia: Wikipedia on Pulse-code_modulation <S> To specifically address the questions: <S> how is a digital signal superior to an analog signal? <S> This typically has to do with noise immunity. <S> An analog signal with 100mV of noise on it is most likely not very usable. <S> However a digital signal with 100mV of noise on it will work just fine. <S> Look at the Vih and Vil of almost any digital chip as a starting point. <S> What's the benefits of changing an analog signal to digital data? <S> This usually has to do with a storage requirement like in CDs(check the Wikipedia link), or transmitting over long distances or on a medium that only supports digital signals. <S> In the later two cases, it really boils back down to noise immunity. <A> You can also apply error detection and correction to digital data. <S> Of course, that's assuming you're willing to live with the limitations in terms of bandwidth (dictated by the sample rate) and resolution (dictated by the quantization). <S> Does the textbook not cover these points? <S> If you want a more specific answer, we need more context. <S> What is the specific book you're looking at? <A> This is a big topic and you'll find plenty of existing text on the subject on the internet. <S> On aspect is noise rejection. <S> Digital signals benefit from relative ease of noise rejection over an analogue waveform. <S> In an analogue waveform, it is not readily apparent what is the wanted part (signal) <S> and what is the unwanted part (noise). <S> With a digital signal, a noisy signal can be more easily restored to a true logic voltage level. <S> If the application warrants it, the digital signal bandwidth allows it and the latency is acceptable, then error check and correction (ECC) algorithms can be applied to blocks of the digital data to allow noise disrupting numbers of bits to be corrected. <S> This improves noise rejection yet further. <S> Digital signal processing is a subject in itself and there are many advantages, and many disadvantages such as quantisation errors/noise, higher bandwidth requirements etc. <S> Again, this is something to investigate further for yourself. <A> Nation security drove the move from analog telemetry, to digital telemetry. <S> Analog telemetry could not achieve 50dB SNR (about the point where the width of the INK PLOTTER ink line set the resolution) at the distances needed to monitor the (mis)behaviors of the missile systems. <S> By using Analog_Digital_Converters, the data_link limitations were greatly altered. <S> If your rocket mis-behaved in mid-trajectory, you had a better chance of diagnosing the cause, because you had better reporting of the minute variations of the fluids and the vibrations and the temperatures. <S> I remember my first Chief Engineer staying late, several days in a row out in the lab. <S> I went over, curious, and he explained "This is a 10-bit ADC from the Saturn Instrumentation Segment. <S> Has a problem. <S> Since I designed it, 10 years ago, I'm the best person to diagnose it and then ensure its properly calibrated after the NASA_spec repair_people replace the failure." <S> The ADC was a 10_bit current_steering successive_approximation circuit, with 10 current sources, and at least 10 discrete FlipFlops to implement the binary-search algorithm. <S> Plus 10-bit output data holding register. <S> I recall a Cordwood Module FlipFlop of size 1cm by 1cm by 2cm, using capacitive input coupling to create the toggle-pulse behavior with diode steering logic and 2 transistors of 2N706. <S> When Fairchild brought out their UL923 logic, the size was about (1/2 cm) cube, but used enough (on_chip) transistors to function down to DC without re_design. <S> All in 8" by 8" by 24" chassis. <S> By the way, in trying to develop (debug, stop explosions from broken piping) <S> their Soviet N1 moon-launch rocket in the 1960s, with their 13 engines needing lots of large piping for oxidizer and propellant with shared turbine pumps, the Soviets kept adding more and more telemetry channels as each version rose from the pad ----- and then exploded at some distance aloft. <S> The final version (#5) had 13,000 telemetry channels and over a GigaBit of downlink data streaming. <S> After that 5th explosion, still not diagnosable (too much vibration in long piping?) <S> , the Soviets gave up their moon efforts.
| For long-distance communications, it's easier to amplify a digital signal, because you don't have to worry about issues like linearity and noise (to a point).
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How do I figure the I2C timings for a peripheral? The I2C timing specs are confusing me a bit. Here's a few timing characteristics from my peripheral : It's basically the same as I2C timing specs provided by NXP. I am looking at specs at 100khz mode. The max rise and fall times for example are given, but not the minimum. Should I program my I2C controller on MCU for 1000 ns rise time only or could I go lower? Similarly the specs for,say, START condition Hold time is given by a minimum 4 us. The max value is left as blank. Do I go just by min and max values? Or if I can program differently, how high/low can I go? <Q> This is not something you'd typically need to adjust away from defaults. <S> In ordinary open-drain operation rise time would be governed by the pullup resistor chosen and bus capacitance. <S> You basically would not be able to make it "too fast" without requiring excessive drive down current, so really would just need to check that the rise time isn't approaching the maximum. <S> If it is you may need smaller resistors, within reason. <S> Should you find you need to a pullup firmer than say 2K2 to get a reasonable rise time, you probably have something wrong with the bus - often you can use at least twice that resistance if not more. <S> The fall time would depend on the strength of the drive, and is limited by physics. <S> Trying to achieve very short fall times is both pointless given the slow clock speed to be used, and perhaps counterproductive as hard edges could lead to the device being a source of radio interference extending to higher frequencies. <A> I would use the smallest rise time that my system could produce, which is likely limited by the resistor pullup and wiring capacitance. <S> Likewise, you will usually only see a minimum limit on hold time or clock high time because any value larger than that is guaranteed to work. <A> This sounds like the MCU (STM32?) has programmable timings in order to allow setting the I2C mode (speed grade) simply by setting the correct timing parameters. <S> The 100kHz standard has no minimum rise time specified, so it could in theory be infinitely fast. <S> However, all chips adhering to 100 kHz standard must allow for the maximum rise time, and thus the timings must be based on that, so that the minimum specified high time is still satisfied. <S> It is safe to put 1000ns there. <S> However, as all timings are programmable, and you know if you have a bus with fast rise time, feel free to set it lower. <S> If you know you have a non-standard bus with 2000ns rise time, then you should use that value so clock high time is still within specs. <S> So by default, put the standard values for a given speed grade there. <S> If you know something about the bus and all the components you can fine tune the values. <S> The MCU can't affect the rise and fall times, but it can just be set to be aware of them, so the fastest speed can be obtained. <S> Some parameters are left without min or max because some delays like the start hold time can be infinitely long if you want, but it must be 4us or more, or long enough so all chips recognize it, even the slowest ones. <S> The rise and fall times being 0 or infinitely fast is also acceptable, but not achievable in practice, it just means it can be as fast as reasonably possible as long as it is not too slow. <S> The faster standards do limit the slew rate so it must be slow enough that it will not cause much EMI.
| If minimum values are not given then you should assume that the rise times can be zero.
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Using capacitors to reduce reactive power I have been trying to understand the concept of reactive power in power systems. From what I have read so far, I understand that any inductive load like a motor needs some reactive power to function (set up the magnetic fields to work etc.). Now I also understand that, while transmitting the current required for this motor to work, the transmission line also has to carry the current required for the reactive power. Now this increases the loss, as it is given by \$I^{2}R \$ . Now, capacitors are used to help generate this reactive power, (as they dissipate power when the inductor consumes it) and are hence placed near the load to reduce the reactive power that needs to be transmitted. I have the following questions: Is my thought process correct? Am I right in my understanding of reactive power? If we already know the value of the inductance of the motor, why don't motors come attached with capacitors that make them less of an inductive load? If we can approximate the inductance of the coils can't we find a capacitance that cancels the effect and makes the net impedance value close to zero? Why do we need to have separate stations built for this when we can try to eliminate this at a device level. <Q> Is my thought process correct? <S> Am I right in my understanding of reactive power? <S> You are essentially correct except for: (as they dissipate power when the inductor consumes it) <S> Should be: "as they store energy when the inductor releases it and release energy when the inductor stores it." <S> There are several reasons that motors don't come with capacitors attached. <S> The motor does not have a convenient place to put them. <S> They would add cost to the motor that the purchaser may not need to incur. <S> Having the capacitors connected all of the time could cause the motor not to stop as quickly when shut off. <S> The stored energy can help keep the motor running. <S> It may be more cost-effective to have just a few larger capacitors for many motors. <S> That is the way it has been done for the last century, from the time no one realized that adding capacitors could save energy and free up system power capacity. <S> Some motors are electronically controlled and do not need capacitors. <S> The need for capacitors must be handled a bit differently if there is harmonic distortion in the power system. <S> Harmonic voltages in the distribution system can cause resonance oscillation currents between the capacitors and the motor and transformer inductance. <S> There are probably some additional reasons. <A> Reactive Power is simply power into a non-resistive load. <S> That could be either inductive or capacitive. <S> If you are driving a motor, that is an inductive load. <S> The power grid will have to deliver current (energy) to create the magnetic field required by the running motor. <S> This magnetic energy is "stored" in the motor, until it is turned off, then the field collapses and returns the current to the grid. <A> Is my thought process correct? <S> Am I right in my understanding of reactive power? <S> Yes. <S> It may help to consider the reactive "power" as sloshing back and forth between the inductive and capacitive loads and, when fully balanced, never leaving the building. <S> That leaves only the real power travelling on the supply. <S> If we already know the value of the inductance of the motor, why don't motors come attached with capacitors that make them less of an inductive load? <S> Cost and control. <S> If we can approximate the inductance of the coils can't we find a capacitance that cancels the effect and makes the net impedance value close to zero? <S> You can for a given load. <S> Changing the load will change the current and the required correction will therefore change. <S> Why do we need to have separate stations built for this when we can try to eliminate this at a device level. <S> It can make sense to do this with motors some distance from the supply - on a large site, for example. <S> If the load is fixed - a pump, perhaps - then the amount of correction required is constant and a fixed capacitor can be attached. <S> In most cases there is variable loading and "load discrimination" is applied to calculate the worst-case PF correction required with certain assumptions for the maximum likely load to be on at any one time. <S> (e.g., Only one of a pair of pumps - run and standby - would run at any time.) <S> Usually a capacitor panel is attached to the distribution board with a controller to switch in and out capacitors to maintain a certain power factor. <S> This results in less capacitors being required - both in quantity and total capacitance value - and hence is more economical to install.
| One remedy is to add resistors and inductors to groups of power-factor correction capacitors so they can also serve as harmonic filters.
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Charging a protected 18650 cell I am new to electronics and just getting started with a project to make a BT speaker. I bought some quality protected 18650 Li-ion cells from Orbtronic . I would like to build a charger myself for these batteries using a 5V,3A wall adapter that I have. Most people use the TP4056 to charge unprotected cells, which is fine, but Do I need a TP4056 to charge my cells (one TP4056 per cell)? If I don't need them, am I okay to hook up the 5V,3A wall adapter directly to the 18650 cells and rely on their protection circuit to charge them? Is this okay? It seems like the more dangerous of the 2 options. I would appreciate any help or, better yet, any links you can provide. <Q> The protection circuitry on a Li technology battery is not a charging profile controller. <S> The protection circuit only tries to protect the battery from over voltage, too deep of discharge and over temperature detection in some cases. <S> The web page that you linked to specifically specifies the protection features and applicable parameteric limit. <S> Protection specs: <S> Over-current <S> (Activated @ 10A) <S> Over-charge <S> (Activated @ <S> 4.29V) <S> Over-discharge <S> (Activated @ 2.45V) Over-temp. <S> and short circuit protection Protection circuit designed by Seiko Devices like the TP4056 are charging profile controllers what you adapt to your system design to properly design the charging process for your selected battery technology. <A> TP4056 is a charger chip for one (1) cell. <S> If you plan to charge multiple cells simultaneously, each charging slot for a cell needs it's own TP4056. <S> Please don't even think about connecting a 5V 3A power supply directly to a lithium cell. <S> Even though the cell has built in protection, it is not a charger. <S> It is the last line of protection to prevent catastrophic failures by disconnecting the cell under abnormal conditions - unless the protection does not work. <A> I created a LIPO battery from many smallers cells for a bluetooth speaker last year. <S> I used a 2S4P configuration to ensure I had a high enough voltage as well as a large enough capacity. <S> I used a different chip to yours <S> but I can't imagine it would be much different.
| You will need a charge controller for your design.
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How do I separate ground planes on a PCB with two different voltages? I am using a four-layer PCB with the following stack up: top signal, ground, 3.3V VCC, and bottom signal. All of my components require a 3.3V input except for this sensitive optical sensor (MAX30102) that needs both 1.8V and 3.3V inputs. In the MAX30102 datasheet, the 1.8V needs to run to a separate ground from the rest of the system. How should I go about separating the ground planes? I am using two linear voltage regulators in parallel and have attached an image of my design. Pad 12 is the ground for the 1.8V input and pad 4 is the ground for the 3.3V input on the MAX30102. How should I connect pad 12 to ground to minimize noise? Furthermore, how do I tie both 1.8V and 3.3V grounds together? I'm guessing that I make a separate ground plane underneath anything that connects to 1.8V and then connect that to the ground of the 1.8V linear voltage regulator. Or maybe, I have to connect the 1.8V ground to the main ground source at only a single point with a track? Currently, the second layer ground is tied to all of the other components powered by 3.3V. Please let me know, I'm a student. <Q> Simple answer is you don't split planes unless you know what you are doing and why you are doing it. <S> Separate the ground planes and route signals over the the gap in a split ground planes at your own peril. <S> The lowest inductance path for return currents is on the ground plane directly under the signal trace. <S> This forms the smallest possible 3D loop. <S> But the return current from such traces cannot cross the gap in the split plane so <S> will flow around the gap instead making a big fat loop. <S> Instead, what you do is just use a single unsplit ground plane but partition components to different areas on the board so ground currents of the noisy parts do not flow through the ground plane under the sensitive parts. <S> The only cases I know where you might split the plane <S> is you need really, really low noise but can't partition sensitive section of the PCB far away enough. <S> Current flowing on a plane kind of spills out and smears to the sides of the linear path it is taking. <S> Split plane can stop the spillage from leaking onto areas of the plane that are under other components because you cannot space them far enough in your partitioning with a single unsplit plane. <S> But it's still contentious whether this is actually needed. <S> Images taken from: http://www.hottconsultants.com/pdf_files/june2001pcd_mixedsignal.pdf <A> A local battery is a resistor PLUS large capacitor, acting as a LOW PASS FILTER for trash from the outside, and providing almost all surge currents needed by your sensor from that LARGE CAPACITOR. <S> You place the LARGE CAPACITOR across the sensor's GND and VDD pins. <S> Bring in the 1.8 volts through a 10 ohm or 33 ohm or 100 ohm or 330 ohm resistor. <S> This placement of the LARGE CAPACITOR and the series resistor is a fine isolator of trash coupled onto your 1.8v regulated voltage; these LDOs are poor rejectors of high frequency trash, such as MCU spikes or switch_reg spikes; adding the LARG CAPACITOR and series resistor provides a guaranteed low-pass-filter of high-speed trash and spikes and switch-reg ripple. <S> ================================= <S> The datasheet shows up to 20mA from the 1.8v supply. <S> So I'd make the resistor be 3.3 or 4.7 or 5.6 ohms, and use 470uF or 1,000uF for the LARGE CAPACITOR. <S> With 4.7 ohms and 1,000uF, the time constant is 4.7 milliSeconds, or about 90Hertz F3dB; this provides only 1 dB attenuation to 60Hz ripple, though the LDO should attenuate that frequency. <S> However, this R+C will greatly reduce any high frequency trash that the LDO is unable to attenuate. <S> All LDOs have internal servo (regulation) feedback loops, and the LDO allocates a small amount of current for the transistors in that loop. <S> Small amounts of current tell us the speed, of controlling the large internal (on_chip) linear dissipation transistor, will be slow. <S> Additionally, that large transistor has large junctions and large parasitic capacitances that provide a direct path from Vin_raw to Vout_clean; thus our LOCAL BATTERY is essential to correct the high_frequency weaknesses of LDOs. <A> Actually reading that datasheet seems to indicate that you are overestimating the sensitivity of the MAX30102 to noise by a lot . <S> There are Arduino compatible eval boards plugged into breadboards that work fine; it's actually remarkably insensitive to noise. <S> LED currents max out at 200mA and flow between pins 10 and 4--a standard 1.0oz plane won't even notice this drop. <S> 1.8V current is limited to 20mA. Move that SCL line <S> so it actually goes out the right and move the SDA line away from pin 4 and things should "just work". <S> Common grounds should be more than sufficient for this design. <S> However, do pay attention to the fact that pin 10 and pin 4 can be carrying 200mA, you probably need to tie them to the GND/3.3V planes with a couple of vias, not just a single via. <A> I agree - Don't cut planes <S> What I'd do is connect the 1.8 regulator ground to your IC pin 12 with a nice fat trace. <S> Then connect pin 4 and pin 12 to the ground plane with vias right at the pins. <S> So the two grounds meet at only one place ( <S> pin 12's via). <S> The idea is prevent currents from either voltage domain from mixing with the other. <S> Physics will drive the return currents back to their sources. <S> https://resources.altium.com/p/how-to-use-a-star-point-for-analog-ground-digital-ground-connection#:~:text=A%20star%20ground%20is%20a,as%20to%20eliminate%20ground%20loops.&text=In%20the%20same%20way%2C%20your,only%20at%20the%20star%20point .
| You do not want to split a plane without justification because you then can only route traces over the bridge between split planes or else you end up with the aforementioned big fat loop. This is a take on a 'star ground' scheme, very very common in audio to prevent unwanted noise. Following up on the excellent answer of DKNguyen, you can best localize the Ground currents of that 1.8volt super-delicate sensor ---- by using a LOCAL BATTERY. Place that series resistor right against the LARGE CAPACITOR. If you have only 1 place they connect, they won't be able to mix.
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Do I need voltage regulator? I power my board with 5V and I am working with ATMEGA8, so I don't need to reduce the voltage. Do I still need voltage regulator for stabilization? And if I do, would this work? I used here 5V regulator and I saw this scheme in datasheet. As I understood, first capacitor is for noise reduce and the other is regular bypass capacitor. I am using USB cable. So do i need any stabilizator there? Is there a chance if i connect this to power source it will burn the source? On ADC pins i have connected IR sensors. <Q> I am using USB cable. <S> So do i need any stabilizator there? <S> Is there a chance if i connect this to power source it will burn the source? <S> If you are concerned that you might inadvertently short something, double check that your supply has protection or add a fuse. <A> As you are using USB, so you won't need any voltage regulator. <S> Mobile phone adapter or laptops provide a constant 5 volts. <S> Use a good adapter and cable. <S> And 7805 has nothing to do in a circuit where the power is supplied from a 5 volts adapter. <S> 7805 works for supply voltage greater than 7 volts. <S> Use a Schottky diode for reverse voltage protection. <A> That depends on what you are connecting to CON1. <S> If it is fixed 5V, then you can't (and don't need to) use a 7805. <S> Usually 7805 needs at least 7 V at its input to deliver 5 V. <S> If CON1 has more than 6 V, you should use a 7805. <S> The output of 7805 will be 4 V with 6 V input, but ATMEGA8 can usually comfortably run down to 3.6 V. <S> It will be 5 V for any voltage at CON1 greater than 7 V up to 45 V. <A> If you use a USB cable I assume you'll be using a 5V USB power source <S> so there is no need to add a voltage regulator. <S> The only gotcha would be if you use the microcontroller's ADC and use the power supply as voltage reference, as the accuracy of the ADC would depend on the accuracy of the 5V power supply. <S> This only applies if you use the ADC to measure a voltage that does not depend on the power supply itself. <S> If you use it to measure the output of, say, a potentiometer or resistor divider, or anything else that has an output which is proportional to the supply voltage, then the ADC will give a correct reading even if your supply voltage is inaccurate.
| USB is reasonably well regulated, and your load isn't very sensitive to minor changes in voltage, so you don't need a regulator. That depends on the specific device, but most USB source or host devices have internal overcurrent protection.
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How can I prevent microwave radiation absorption by a camera lens/electronics inside the microwave (behind the shield)? Let's say I put in an endoscope like camera with the lens at the inside edge of the box and the wire behind metal sheet. Here's a mockup: Possible idea: The lens is behind a holes but the hole is directly in front of lens. The lens would be facing inwards to the center from the edge of the cube. This does not seem sufficient as the radiation is being reflected off the walls so it's fairly random. Is there anyway to make it microwave safe for recording while microwave operates? I watched some microwave tear down videos and it seems like (on a surface level view) most of the electronics are safely just hidden behind a wall. That to me suggests the heat build up isn't too high or the electronics are designed to be heat resistant. In this case the lens/camera is exposed to the radiation.. If you have any ideas/thoughts on how the lens/electronics of the camera can be protected from microwave radiation while having full view of the plate, I would appreciate hearing them. The thought driving the question is: Would an endoscope like camera be safe from microwave radiation/energy behind the inside microwave box? <Q> This is a very dangerous idea. <S> Your questions and your diagram show the potential for you or someone to get badly hurt. <S> The only safe way is to place the camera outside the unmodified microwave oven . <S> Make a hood to screen out ambient light. <S> Buy a microwave with a black coloured screen instead of a white one. <S> But don't go modifying the microwave. <S> Don't do this! <S> A quarter wave, or 3 cm of wire, will form an effective antenna and transfer most of the power of the microwave, to the cable! <S> This will not just damage the camera and the phone/computer, it could injure you if you're holding it, or standing very close. <S> You will notice that there's a gap between the punched metal window and the clear plastic window. <S> This is because there are evanescent waves extending about one hole diameter, out of each hole. <S> The plastic stops you getting too close. <S> Don't put the camera touching the surface of the metal. <S> Microwave ovens are full of dangerous voltages. <S> What happens when your glue comes loose or a child trips on the USB cable and it falls and touches the high voltage terminals inside? <S> The whole phone or computer could be charged to a dangerous voltage, you could kill yourself (Darwin award) or someone else (manslaughter). <S> Don't go adding wires to the inside of a microwave oven. <S> Metal doesn't magically shield the camera from microwaves. <S> It needs to be continuously welded or cleaned and bolted down with an RF gasket. <S> You can't make a shielded box for the camera, inside the cavity, without some way of welding it to the wall of the oven. <S> If you are looking for a way to design an oven with a camera port, for manufacture (after full safety testing) <S> then there are ways of doing it safely. <S> The 12 cm waves cannot pass through a waveguide smaller than cutoff so you could install a port or tube, perhaps <S> 40 mm diameter and 150 mm long down which the camera could safely peer. <S> There are ways of making the hole larger, even bigger than a wavelength, using specially designed chokes. <S> These are used for conveyor belts running through microwave ovens for continuous cooking. <A> Don't place the camera in the chamber. <S> Place it outside with other oven electronics and let it peek through one of those holes in the wall. <S> But I would say keep an open heart towards burnt cameras. <S> You may roast up one or two cameras before finding the way that works. <A> Microwaves can't pass through the holes. <S> It should be fine. <S> Be careful when you take the microwave oven apart. <S> There are capacitors in there that are charged up to several thousand volts. <S> They can hold a charge even when the oven is turned off and disconnected from the outlet. <S> Avoid putting your camera near the high voltage parts.
| The holes in the metal walls of the microwave oven are intentionally made small enough that the microwaves cannot pass through them. Arrange your camera so that it looks into the oven through one of the holes. If your camera is too close, it might get zapped by high voltage. Some more detail if you are not convinced: Your diagram shows the camera poking through a hole.
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Why cap across shunt resistor, input bypass caps and small resistors on current sensing inputs? I want to build a sensitive current sensing circuit using shunt, and since I also repair MacBooks occasionally, I have some contact with the schematics of real working expensive devices (although whoever follows Louis Rossmann, knows there are issues there too sometimes, but it's not the point here). While browsing through the schematics of a macbook, I've noticed an interesting thing: most current sensing circuits are simply shunt resistors with each of their sides connected to current sense amplifier such as INA210. All cool and easy, no magic outside Hogwarts detected. But when it comes to PPBUS (main power line) and charging power line, there is a 0.02R SHUNT , and sides of the shunt go through 10R RESISTOR each, then there is a CAPACITOR between the inputs of 0.047uF and a 0.1uF CAP on each input line to the ground. And only then the lines go into an IC . Why so much mess, while on other lines it's simply shunt straight to amp. Should I have something similar? Here's a screenshot of that part of the circuit: There is some info here about the small resistors, but it doesn't exactly compare what happens with resistors and without, like with: this happens. Without: this happens. I still didn't understand how some 10 Ohm resistor is supposed to make things better if input impedance is like a megaohm or greater. Besides, there is also a capacitor question (why cap across? why bypass caps there? what value? what for? should I do it too?) <Q> The basic purpose of a pair of resistors and a capacitor across the differential amplifier is to filter out the noise . <S> Measuring current is often noisy, and measuring critical current path requires clean input. <S> If you follow the design recommendation for Current Monitor IC INA219 in the datasheet, you'll find the exact same configuration: <S> I'll copy the verbatim as-is from the datasheet below: <S> The internal ADC is based on a delta-sigma (ΔΣ) front-end with a 500-kHz (±30%)typical sampling rate. <S> This architecture has good inherent noiserejection; however, transients that occur at or very close to thesampling rate harmonics can cause problems. <S> Because these signals areat 1 MHz and higher, they can be dealt with by incorporating filteringat the input of the INA219. <S> The high frequency enables the use oflow-value series resistors on the filter for negligible effects onmeasurement accuracy. <S> In general, filtering the INA219 input is onlynecessary if there are transients at exact harmonics of the 500-kHz(±30%) sampling rate (>1 MHz). <S> Filter using the lowest possible seriesresistance and ceramic capacitor. <S> Recommended values are 0.1 to 1 μF.Figure 14 shows the INA219 with an additional filter added at theinput. <S> Edit <S> : Is it a good idea to always include a noise filter? <S> It depends on several factors: <S> Your amplifier input bias current. <S> If your amp has a relatively high input bias, then some current will leak into the amp, reducing measurement accuracy. <S> The situation worsens if you're trying to measure low current (μA or nA). <S> INA219 has 100 nA input bias current, so measuring tens of mA to several amperes will be good enough. <S> Your noise frequency. <S> The low pass filter cannot, well, filter out the low-frequency noise. <S> But yeah, adding low-pass filter is generally a good practice. <A> The 10 Ohm with 100nF create a low pass filter with a cutoff frequency of \$ <S> f_c = 160 kHz\$ . <S> This is too high to be meant to filter the PWM of the current control (this will be something like 20kHz), but it filters out noise above this frequency. <S> Without the filtering the current control loop might become instable when there is noise coupling in. <A> This kind of RC is not always a filter but is sometimes used to compensate the zero of the shunt (R-L). <S> The shunt has a very small parasitic inductance that becomes significant when the shunt value is low (<5mOhm). <S> The parasitic inductance is usually between 500pH and 5nH depending on the size and manufacturing of the shunt. <S> The voltage at the shunt terminals displays some spikes when high di/dt occur (e.g. current chopping application), that can trigger a short-circuit detector. <S> The RC filters here have much smaller frequencies than the zero of the shunt (160kHz common-mode and 80kHz differential mode due to C7120) <S> so it is used as a filter. <S> Edit: The small filter resistors are necessary for fast convergence at the analog inputs (parasitic input capacitors). <S> 10 <S> Ohms is much higher than 20mOhm <S> so the resistors won't alter the signal.
| In some applications, the filter feeds an ADC that can have a significant input capacitance (e.g. SAR) or the small shunt voltage signal needs a high amplification so that low input resistors reduce thermal noise.
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Keep circuit on after momentary trigger I want to keep my circuit on after a push button is pressed and released. Trying to do this I stumbled upon soft latch circuits, but they are too complex and I don't need the turning off feature. What is the simplest possible way to keep my circuit on after a momentary trigger, like that of a push button? <Q> You can use MOSFETs to build a soft latch circuit . <S> Q1 is a P-Channel MOSFET. <S> Q2 is an N-Channel MOSFET, resistor values really depend on your requirement (you want a strong or weak pull-up/pull-down resistor?). <S> When you press the switch SW1, the Q1 will be driven low allowing current to flow, thus driving Q2 up and keeping the Q1 Gate low. <S> Please note that Q1 and Q2 both have Rds(on), giving a burden voltage. <S> So for a voltage-sensitive load, this may not be suitable. <A> I stumbled upon soft latch circuits, but they are too complex and Idon't need the turning off feature. <S> This is the simplest relay contact latch circuit I know of. <S> Basically the closed relay contact self-latches the relay by shorting out the start button. <S> Picture from here . <S> A simplified schematic of the same idea is this: - <A> Have you looked at using a one-shot? <S> Also known as a monostable multivibrator, this is a circuit that outputs a single pulse of a set length when it detects a rising or falling edge. <S> You can make it yourself with discrete components (lots of examples on Google), but you can also buy ICs that make it much easier. <S> For example the Analog Devices LTC6993 can output a pulse of a duration from 1 us up to 33 s, after detecting a rising edge. <S> The duration of the pulse can be determined by resistors. <S> See the datasheet here . <S> In the example below the IC is set to output a 16 us pulse on detecting a rising edge. <A> Search for 'compact locking push-on/push-off switch for flashlight'. <A> I don't know if you have a clock in your circuit, but perhaps a T flip-flop might be what you're looking for. <S> On a clock edge, it switches state when there's an input, but if there's no input, then it holds the previous state.
| If your application involves a microcontroller, you can also connect the switch directly to a digital input on the microcontroller and debounce in software (although this probably doesn't meet your "simple" requirement). Use a readily available, latching 'push-to-on/push-to-off' switch.
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Can I run a 24V motor with a 12V PSU Im using one BTS7960 driver to control a motor. The motor i have is 24V 500watt. If i use a 12V psu will it draw more current? <Q> If the control in implemented properly, you should be able to run the motor at half the speed that would be possible with 24 volts. <A> This question is a bit tricky, because it depends on the application. <S> The BTS7960 is a brushed DC motor driver with PWM speed control, so I presume you have a permanent magnet brushed DC motor. <S> Compared to using 24 V, running the motor on 12 V will:- <S> Draw <S> half the stall current and half the initial startup current. <S> Have half the torque output at stall. <S> Spin half as fast at the same PWM ratio. <S> Draw the same current for the same torque output. <S> Draw twice as much supply <S> current at the same speed with the same load. <S> For example if it needs 50% PWM to get the required speed on 24 V then it will need 100% PWM on 12 V, drawing twice the power supply current for the same power output (torque*rpm). <S> This is necessary because power = voltage * current, so at 12 V the current must be doubled to get the same power input. <S> If the motor is geared and you change the gear ratio to get the same speed then it may draw twice the current or more , depending on where on its efficiency curve it is operating. <S> If the motor is driving a load with very high startup torque (eg. <S> a piston pump or compressor) then it may not have enough torque to start up - and so will stay stalled drawing much more current than it would at 24 V. <A> Most likely the motor current will be same as 24V (not doubled). <S> The motor speed will be half of rated speed. <S> Power will be half of rated power. <S> But there are lots of details. <S> If you have the motor and power supply just try it. <S> If you don't have the motor and power supply, select components with the same voltage for best results.
| For the same load torque, the current would be the same as with 24 volts.
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What's the use of neutral wire in single phase home supply? I am aware that the neurtal wire provides a return path for the current to the power source. But that neutral wire too, is connected to the ground at somewhere between the load and the source.My question is: Why use three wires Live, Neutral and Ground? Instead we can use two wires Live and Ground which does the same work, as Neutral and Ground are joined. Another question: Does two grounds of different localities have same potential?If they have different potentials, is this the reason why we use neutral as the common reference for both the load and the supply? <Q> Why use three wires Live, Neutral and Ground? <S> With only two wires, you would have to use the neutral for "earthing" the metallic part of the equipment. <S> I'm thinking of cookers, refrigerators, Hi-Fi, iron etc.. <S> You do this because should the internal live wire break and unexpectedly touch the appliance metal case, you want a significant "earth" current to flow and blow the fuse thus revealing a dangerous situation. <S> However, with only two wires, if live and neutral got swapped the problem <S> would not be self-revealing and would be highly dangerous to anyone coming into contact with the case of the equipment. <S> The equipment would appear to work but be highly dangerous because the exposed metal parts of the case are at live potential. <S> In all likelihood the equipment would still operate correctly but is still safe. <S> If neutral and earth got inadvertently swapped, again there should be no real safety issue because the neutral wire that is incorrectly connected to the exposed metal parts would still be able to "earth" those exposed metal parts and, safety is preserved. <S> If live and earth got messed up <S> then this is a self revealing fault because the equipment could not operate with neutral and earth connected to where neutral and live would normally go. <S> It's not great <S> but it is self-revealing and relatively safe. <S> Of course with modern day gizmos like residual current devices (RCD), those exposed metal parts that are improperly connected to live would likely trip the RCD. <S> First and foremost it's about safety. <A> In typical north American house wiring, ground and neutral are tied together back at the panel, and also to an earthing rod. <S> Nevertheless, the neutral and ground wires have different missions: <S> Neutral carries return current. <S> It is always needed for a single-leg feed (like a 2-wire plug.) <S> It is optional for a 2-leg feed to carry any imbalance current. <S> More about that below. <S> Ground carries current only in case of a fault. <S> It provides a path for a line-to-chassis short back to the panel, preventing or reducing the possibility of electric shock. <S> Examples of how neutral and ground are used in <S> normal 120V/240V wiring: <S> 2-prong 120V receptacle: L1 and neutral; all return is on neutral <S> 3-prong 120V receptacle: L1, neutral, ground; all return is on neutral <S> 3-prong 240V dryer outlet: L1, L2, neutral; neutral carries 120V unbalance return (e.g., dryer motor) <S> 4-prong <S> 240V outlet: L1, L2, neutral, ground; neutral carries 120V unbalance return (e.g., dryer motor) <S> 3-wire <S> 240V oven feed: L1, L2, <S> ground: <S> load is balanced so no neutral is needed. <A> It is for safety. <S> The dedicated ground wire is not used for anything else except fault currents. <S> It is always available and never switched. <S> Sort of like an emergency lane on the freeway. <S> All exposed metal in appliances is bonded to this ground wire. <S> It is like they said hey, we should run a separate neutral wire all the way from the fuse box (breaker box) to every place that electricity is used. <S> We will connect it to all the exposed metal on every device... <S> refrigerators, fans, heaters, etc to make sure nobody ever gets a shock. <S> That neutral will never be switched and won't carry any current (except for short circuit or fault current). <S> Then they said we should give that wire a special name, and also use a different color insulation on it <S> so it is easy to recognize. <S> We can call it protective earth, or ground or something like that.
| With a third wire (specifically an earth wire), if the wiring to the plug got messed up and live and neutral got swapped, the third earth wire is still connected to the exposed metal parts of the case and there wouldn't be a safety issue.
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Wiring an emergency shutoff I want to wire an emergency shutoff button to switch off the circuit of one room. While I understand what to do in theory, I am not quite sure how exactly the button needs to be wired. It is one of these that needs to be pulled out manually and stays there until pressed or power runs out, so the machines also don't restart if the power was cut externally and comes back. It has 4 terminals, and I suppose I need to connect it to neutral, incoming phase and the outgoing phase (which is going to the switch,) but am unsure of the exact wiring. I suppose to do it safely I need a 5-wire cable between the button (which is on a central location in the room) and the switch (which is at the position where the power is coming out of the wall,) so I can send both phase wires over it without "relabeling" a ground wire as phase. I made some pictures of the switch, maybe it's more clear from them.Also, I am from Germany (I know there are some local laws/rules regarding that.) That combination of switch/button has worked before, but unfortunately I did not remember the exact wiring. <Q> The emergency latching push button switch, also known as a mushroom push button switch (Pilz Schaltgerät), is used to cut off, through it's NC contact, electrical power to a machine / electrical equipment under an emergency situation, to avoid injury to personnel or damage to the equipment. <S> The mushroom button S0 is released, by rotating it, when the fault is cleared. <S> To ensure that the equipment does not restart immediately after the mushroom button is released, a 'zero-volt safety' feature is provided. <S> The 'zero-volt safety' feature ensures equipment restart only after a start button S1 is actuated. <S> This feature also prevents the equipment from automatically restarting after power is restored after a failure. <S> The function is performed by an electromagnetic relay K0. <S> The NO contacts of K0 switch on the power to the equipment. <S> In equipment where there is a higher risk of injury to personnel, the single relay K0 is replaced by a 'Safety Relay' (Sicherheitsrelais), with dual redundancy, to ensure fail-safe operation. <S> For your purpose, it would suffice to wire the NO contacts of K0 in the lines powering the circuits of the room. <A> Your switch has one set of normally open contacts ("schliesser") and one set of normally closed contacts ("öffner.") <S> The switches are marked "NC" for "normally closed" and "NO" for "normally open." <S> You should use the NC switch to switch the hot wire to your relay. <S> The relay coil gets neutral and the switched hot from your emergency button. <S> You use the NC contacts so that power gets switched off if something breaks the emergency switch circuit - broken wire, fire, etc. <A> The sketch below shows one possibility. <S> The emergency stop shutoff button is marked "Lockout E-Stop. <S> " <S> When pressed, the top contact opens and the bottom contact closes. <S> The top contact stops the machine and prevents it from being restarted until the button is pulled out. <S> This function is optional, but is one explanation for four wires on the e-stop rather than two. <S> Only one additional wire is required for this function as shown. <S> It might be wired somewhat differently, or the forth wire might have another function. <S> The start and stop buttons are momentary-contact switches. <S> When the start button is pushed it energizes the contactor coil. <S> The small normally open contact of the contactor then bridges the start button connection the keep the machine running after the button is released. <S> If the power goes off, the contactor opens. <S> It does not close when the power returns because the start button is no longer bridged. <S> For normal stopping, the stop button needs only to be opened long enough for the contactor to open the bridging contact.
| The bottom switch closes when the button it pushed and remains closed to turn on a light to indicate that the machine has been locked out. The button does not need a neutral wire, but it might need a ground wire.
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Using 2N3904 as an amplifer for 3.3 V PWM to 10 V PWM I was thinking of a circuit to amplify my 0-3.3 V PWM (500 Hz) signal to a 0-10 V PWM and the following circuit came to my mind: simulate this circuit – Schematic created using CircuitLab I tried this circuit on a breadboard and I noticed when I decrease the DC level of the PWM on the input the DC and amplitude of the signal on the probe node both decrease which I can't understand why! Because I expected that changing the DC amount of the input signal doesn't affect the amplitude of the output. What is the reason? <Q> As I mentioned in the comments, you'll want two-quadrant drive. <S> Assuming that your "load" is \$R_3\$ and \$C_1\$ , which are just a simple RC low-pass filter for the output, then something this simple might be okay for hobby purposes: simulate this circuit – Schematic created using CircuitLab <S> There are a lot of reasons why the above isn't good enough for commercial use. <S> But it should be okay for your light load. <S> But one quick and reliable modification would be to drop about \$100\:\text{mV}\$ at the maximum draw required by \$R_3\$ (which is, at worst, \$1\:\text{mA}\$ .) <S> So adding two emitter degeneration resistors for the two output BJTs might be a good idea. <S> And cheap enough. <S> Also, just in case, a bypass capacitor and resistor to make up a modest filter at the base of \$Q_3\$ : simulate this circuit <S> The above circuits should be okay at \$500\:\text{Hz}\$ . <S> Also, the \$22\:\text{nF}\$ capacitor between the base of \$Q_3\$ and ground should use leads as short as possible. <S> Just as a matter of good practice if nothing else. <S> There's a small issue of shoot-through, though the emitter degeneration resistors will nip that <S> reasonably well while also offering some output short-circuit protection. <S> But it doesn't cost a lot more to add some reduction of shoot-through, as follows: <S> simulate this circuit <S> I might tweak like this, just to learn/play more. <S> But the first schematic at the top is probably fine for your use. <S> Finally, Bruce is absolutely right about the fact that a lower PWM ratio (more LO than HI on your PWM output) should increase the analog output voltage, not lower it. <S> So either you have things set up where 0% PWM to you actually means HI on the output pin or else you've got something wrong in the circuit or the way you are measuring things. <A> The problem with this circuit is that the amplifier's output impedance is asymmetrical. <S> When PWM input is low the transistor is turned off and the capacitor charges through R1 and R3. <S> But when the PWM input is high and the transistor is turned on the capacitor discharges through R3 only. <S> This has two effects. <S> Firstly the voltage at the probe point does not go up to 10 V, but to a lower voltage that depends on the capacitor voltage (which varies with PWM ratio) as R1 and R3 act as a voltage divider. <S> Secondly the capacitor voltage will not track the PWM ratio linearly. <S> For example with 50% PWM input the capacitor only charges to 2.5 V (not the expected 5 V) and the 'high' voltage at the probe point <S> is only 5 V (not the expected 10 V) <S> with 5 V dropped across R1 and 2.5 V across R3. <S> To fix this problem you need to make R1 much lower than R3, or use a 'push-pull' circuit that actively pulls the voltage up to 10V. when I decrease the DC of the PWM on the input the DC and amplitude ofthe signal on the probe node both decrease That should not happen. <S> With lower PWM ratio on the input, the output voltage amplitude should increase because the capacitor voltage increases (due to the inverted PWM output). <S> At 0% PWM the probe point should go to 10 V, and at 100% PWM input it should go to 0 V. <S> If this doesn't happen then your circuit is wired wrong, or the input PWM is actually inverted. <A> The problem is the 30K resistance for charging vs 10K for discharging the filter capacitor. <S> Jonk’s circuit is a good and instructive example of a roll-your-own circuit. <S> If you just want to get something working, you could increase the resistor ratio greatly, for example to 4.7K/1M and proportionally decrease the filter capacitance. <S> A CMOS rail-to-rail buffer amplifier could be used. <S> which is just one wee <S> SOT-23 that replaces your transistor and both associated resistors (available in inverting or not). <S> It would provide output resistances in the 16 ohm range (and fairly balanced) so you could even lower the filter resistance if you wanted. <A> By your shematic, you are using inverting amplifier meaning when you increse the PWM on the input you actually lower the PWM on the output. <S> You can use AND/OR gate to achieve your goal and it is simple and efficient.
| Another alternative would be to use a suitable buffer such as the MCP1400/1401
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How to make an audio 5.1 switch? I'd like to make a 5.1 audio switch to enable my 5.1 audio system (+mic+headphones) to be available for any of the 4 computers it can be connected to. So I guess I would have : INPUT : 4 (computers) x 5 cables (x2 wires (stereo) ?) - 5 cables are 3 for the 5.1 audio, 1 for microphone, 1 for headphone OUTPUT : 1 (audio system) x 5 cables (x2 wires (stereo) ?) And I don't see the beginning of a way about how to do that properly. By properly I mean, first do it, second avoiding current issues like current going back to the system or interferences that could damage the electronics. Note : I'm used to solder electronic components, but I know nearly nothing about electronic theory. I found this as a starting point / reference, but it is only for 2 inputs, and it uses RCA connectors instead of jacks. Could you help me ? <Q> Hm, while you can technically do this with analog switch ICs, you'll have a lot of fun with ground loops, reference levels, and control. <S> That's costly. <S> You can get network sound infrastucture that works on basically any operating system (including exotic ones like microsoft windows) as client and on linux on the pi as server. <S> I'd look into JACK audio . <A> "ABCD" boxes use wafer switches to enable the user to operate a single control to choose between 4 options for large numbers of conductors. <A> As Marcus mentioned, you will have lots of troubles with hum and buzzing and hissing (high frequencies, down-converted) because of the several power supplies involved, each with their non-audio-friendly switching power supply. <S> Thus avoiding the ground loops is key. <S> Instead of trying to use differential amplifiers, that must handle 50 volts common_mode of high_frequency spikes <S> , I suggest you consider audio transformers. <S> You will still have spikes/trash coupled from primary-to-secondary, through the overlapping winding capacitances. <A> Your link shows the right approach - relays. <S> I think this will be more reliable than something with a gigantic multi-pole rotary switch. <S> An ABCD box might work, but I think crosstalk between the 5.1 signals and the mic signals will be a problem. <S> Also, you will need one meant for printers, with 25 pin connections. <S> That's 20 signals per computer. <S> Here is one that is a 4 x 2. <S> If you don't want to build anything, a stack of these would handle all of the signals. <S> Because of its open-frame construction, one could come up with a control circuit to drive multiple board's relays in parallel with a single switch or switch array. <S> Search ebay for "stereo audio switcher". <S> https://www.ebay.com/itm/Audio-Signal-Switcher-4-Input-1-Out-hifi-stereo-RCA-Switch-Splitter-Selector-Box/114144550330?hash=item1a938b9dba:g:bNAAAOSwbLpeZnZJ
| There are many audio switch products on ebay. Sounds like a job for a lot of relays, that you control with a power source connected to a rotary switch. Or: really just a get a RaspberryPi or similar, a decent 5.1 USB sound card to attach to that, and network cabling instead. For true isolation and to prevent the ground loop problem mentioned in another answer, you need to switch each signal and its ground. You have 10 individual signals from millivolts to volts going in two directions, so an all-electronic solution probably is beyond your skill set.
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emulator vs simulator?in context of EE? What is difference between a simulator and emulator? I have tried to google but i see there responses on SO that are much close to CS. How can i understand difference between a simulator and emulator in context of EE? Does it means that emulator is hardware just like dsp kit or micro controller or fpga kit and simulator is software like matlab or pspice? <Q> The terminology is used a bit inconsistently: <S> When you talk to software people , an emulator is software that offers you a simulated system. <S> For example, there are many "Gameboy emulators", with which you can play Gameboy games on your PC. <S> A simulator can pretty much be anything from a physical simulation of thermal properties to a software stub. <A> I disagree that the terminology is used inconsistently. <S> The key is what you are trying to achieve. <S> A simulator tries to simulate (in electrical engineering this is usually based on our mathematical and physical understanding of underlying physics) how a certain device, code, machine, etc will behave. <S> It uses the internal structure of a model to give us an estimate of what that model will do based on the inputs provided to the simulator \$^1\$ . <S> An emulator doesn't model the inside (necessarily, you can do hardware emulation by simulating the hardware, with something like Cadence's Palladium series of emulators).What an emulator seeks to do is to model to the outside world what we think <S> /know the inside of a black box will do. <S> A processor emulator will not necessarily simulate every transistor inside to see how it behaves. <S> It takes a known behavior and emulates it for the outside world to interact with. <S> A software emulator might emulate all the responses of a different operating system or architecture, so you can a program compiled for OS A or architecture B on a different OS or architecture. <S> A CPU such as seen in hardware design will model how the outputs and inputs of a CPU/micro controller might behave, without having to simulate the insides. <S> \$^1\$ note that these inputs must not be inputs in the classical sense of 'current flows in here' or 'a force is applied this way'. <S> You can also do things like eigenmode analysis in a lot of types of simulations to study the resonances of a device, without requiring an external input. <A> In a context of the two previous answers' explanation of what are simulators and emulators and how these are used for modeling, I'd like to add an example of the intertwined use of both. <S> A modeling system can include both simulating and emulating capabilities. <S> The NS family of discreet-event network simulators was primarily conceived to be able to simulate behavior of packets in networks, both wired and wireless. <S> Using simulators of this kind, students and researchers can study packet propagation, queues, routing etc. <S> with existing protocols and develop new protocols. <S> A NS-3 simulator is special in that it most closely combines the two modeling flavors, simulation and emulation. <S> In a simulator code, a file descriptor FdNetDevice represents a net device; it can also be used to provide emulation capabilities when it becomes programmatically associated with an underlying packet socket and receives a capability to send data on a "real" network. <S> Another net device, a TapBridge NetDevice, allows a "real" host to participate in an ns-3 simulation as if it were one of the simulated nodes. <S> An ns-3 simulation may be constructed with any combination of simulated or emulated devices .
| When you talk to processor designer people , an emulator is hardware that you can plug into a system instead of a real processor, and that simulates the processor hardwarewise (and is typically controlled by an external computer)
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Material that doesn't let microwaves through, but is transparent to IR Does a material exist that is transparent to infra-red light, but is opaque to microwaves (S-Band, about 2.5 GHz)? I am basically looking for a filter I can put in front of a infra-red sensor. <Q> For bulk material, you might try a slab of germanium. <S> It's not a metal, but a metalloid, resistivity <S> about 1 Ωm in its pure state, but will improve dramatically with doping, which may not affect its IR transparency in the 8 to 13 µm range. <S> However, germanium is quite expensive and hard to come by. <S> For 2.4 GHz with a wavelength of 125 mm, holes less than 5 mm should provide a reasonable attenuation. <A> I think the screens in the doors of microwave ovens block the microwaves, but obviously wouldn't block the IR completely. <A> Microwaves can't pass through conductors such as metals. <S> Infrared transparent conductors exist (see : https://patents.google.com/patent/US6761986B2/en or https://www.spiedigitallibrary.org/conference-proceedings-of-spie/4375/0000/Infrared-transparent-conductive-oxides/10.1117/12.439187.short?SSO=1 ) <S> But they're space and defense engineering <S> so they will not be readily available. <S> If you're not taking images with that sensor you could consider a metal sheet with holes similar to those on a microwave oven door. <S> Be careful not to make those holes too small or you will block the infrared too. <S> From 3 to 8mm should be enough to be safe from microwaves while still letting infrared pass through <A> if you want substantial attenuation, then plan to locate the sensor well behind the wire mesh <S> ---- <S> if the holes are 5mm square, then have the sensor at least that far behind/inside the shielding cage. <S> If a 1:1 ratio, you will get 6.28 nepers of attenuation. <S> According to the lecture of Richard Feynman. <S> This stackX answer <S> Why are many IR receivers in metal cages? <S> provides lots more details and theory on mesh-shields. <A> There are a lot of possible solutions here. <S> Without knowing more, my preference would be to run an optical fiber from wherever your mixture of microwaves and IR light is to a sensor either far from the microwave source or sealed in a metal box. <S> Diameter of an SMA or APC connector is <<< your microwave wavelength, so you will not have ingress if you put a hole in your enclosure for a fiber.
| If it's not an imaging sensor, then you could simply use a metal mesh, with hole sizes much less than the wavelength of the RF.
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Should a mechanical power switch float or ground main power when the system is off? Whenever I've put a mechanical switch on a design to allow a battery to be connected / disconnected, I've always just floated system power whenever the switch is open, like this: However, a colleague asked during a design review whether it makes more sense to ground system power when the device is off: From a basic functionality perspective, I believe that either should work. However, I was curious if there is a particular reason to choose to float or ground the main supply in the "off" state? Edit: Several (very helpful!) initial answers and comments discuss potential harms or considerations if we do choose to ground the input. Are there any extra risks or considerations for leaving the input floating (the first design)? <Q> Shorting the input to ground can damage some regulators because the output capacitors and multifarious bypass capacitors which may be connected to the output discharge through the regulator. <S> It's not necessarily easy to tell from the datasheet whether that is true. <S> You could add a resistor between the switch and ground to reduce the chances of that happening. <S> Some regulators have a pull-down transistor on the output to quickly discharge the output capacitors so you can reset (say) <S> a micro by quickly cycling the power. <S> This particular one does not seem to. <S> Discharging the input capacitor with a switch takes you part of the way there, but there is no guarantee <S> the output capacitors will be discharged quickly (though they probably will). <A> Generally, I connect input signal pins to ground, even internally unconnected pins. <S> Open pins on active devices (which is not your case <S> - yours is a power pin) can cause undesired noise or even signalling at device internals. <A> I would ask what you are trying to achieve by switching the DC in to GND. <S> No obvious benefit in most/typical applications, but with all the risks mentioned by others (back-driving the regulator, contact arcing due to C60 discharge, etc).
| If you are going to ground it, perhaps then through a high resistor, to avoid any instantaneous current spikes upon switching, or any severe back draining.
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Current rating for 12v plug on 300W in-car inverter I have an in-car inverter which is rated 300W. This is the only labeling: I need to source a 12V cigarette socket plug to connect to the input. I have found the following one which is labled: Carpoint Universal plug 6-24v max 8amp Assuming I have a 12V car power supply, would the 300W rating on the inverter, suggest that this plug would have an insufficient current rating: P = IV = 8 * 12 = 96 Watts Or should I be calculating based on the 230V output device connected: I = P/V = 300/230 = 1.3 Amp Of course, with the latter calculation I'm unsure if this applies exactly to the inverter's input. <Q> In your case your inverter will take 300/12 Amps or about 30 Amps allowing for losses. <S> The cigar <S> lighter is totally inadequate for this <S> the inverter should be hardwired straight to the battery with a suitable fuse in the wiring near to the battery. <A> That inverter will draw 300W <S> / 12V = 25A at full load (assumed maximum). <S> So that plug is not sufficient. <S> You will need wires sized for 30A, go higher if necessary and a suitable fuse. <S> I would connect direct to the battery so as not to overload the car's existing electrical wiring. <A> The connector knows nothing about your inverter. <S> It only knows about the current through the contacts so it's 8 A at 12 V <S> is the limit. <A> The 8A is the current rating for the plug (based on gauge of wires/connections, maximum heat, any fuses inside), so 12V <S> * 8A. <S> This is for a vehicle? <S> Do you know the max current that can be supplied? <S> If you plan to use less than 300W you could use your actual rating instead. <S> These inverters are efficient enough to scale down almost proportionately. <S> So for 300W you'll need to supply 25A, but for 100W you'll only need 8.3A. <S> You'll need a wire gauge of 12AWG for 20A up to perhaps 15ft <S> (if you need precise values I can look that up) <S> Not only is voltage drop a concern, but also heat generated in the wire.
| Power at the input is the same as power at the output minus inefficiencies or losses. You are not only limited by the plug but also by the wiring harness, for which the current is limited by a fuse, usually in the fuse box. Not all your "12 V auxiliary power outlets" in the car can supply enough power, even for a cigarette lighter.
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What is the max current for a monophase RCCB? Images tagged with 1 are before the incident, and 2 after the RCCB burned off. I am guessing that this has happened because of overload. What is the max current for this monophase RCCB? I could also use some tips to prevent a future incident like this. Whole breakerbox was replaced and the burnt cable cut off. The burn in 2nd image went halfway through the RCCB unit. Thats why I thought it could be because of overload. Am I wrong in assuming the max current should be lower than 40A? Replacement was done by a professional, dont worry. I appreciate the good-will though. Thanks. <Q> The problem was probably not due to an overload. <S> It looks more like the terminal screw was not tightened properly. <S> Get someone who knows what they are doing to look at the installation. <S> Asking people on the other side of the world to look at pictures on the internet is no substitute for competent workmanship. <A> The minimum torque on that family is 2.5 Nm. <S> (maximum 3 Nm) <S> Burns and green copper oxidation like this are often caused by people not using torque wrenches on these terminals. <S> Causing the wire to be loose, or too tight. <S> In time due to thermal cycles heat damage will occur. <S> Which eventually causes the wire to burn due to connection resistance. <A> I agree with other posters that this damage was due to long-term overheating of the phase connection. <S> You mentioned the breaker box has been replaced; this should have been done by a professional electrician, qualified in your jurisdiction. <S> However, to answer your specific question, yes, this RCCB does have a maximum load of 40 A, shown as <S> \$I_n \quad <S> 40A\$ on the case. <S> Note that this RCCB offers residual current protection only (residual = the difference between the phase and neutral currents). <S> Overload/short circuit protection needs to be provided separately. <S> Typically this means your RCCB will feed MCBs (minature circuit breakers) with a total load not exceeding 40 A. <S> There should also be a supply side fuse/ <S> MCB not exceeding 63 A. <S> The model number is partially missing. <S> It is 5SM1 314-0. <S> The data sheet is available here <S> (via RS).
| There may also have been some damage to the wire that caused the terminal to overheat.
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Meaning of certain symbols and labels in Circuit board PCB I have a schematic of an electronic circuit board or PCB. I want to know the meaning of certain symbols and labels in the diagram. I have highlighted the diagram below with green, yellow and grey coloured boxes. Question 1:What is the meaning of symbols, >> , < <> > and << ?. See green box. Question 2:What is the meaning of n in HDMICLK_DISn ? See yellow box. Question 3:What is the meaning of 6, 10, 11 in grey box? Question 4: Why the resistors have values as zero at certain places? What does it signify? For example see the 2 resistors on top of the schematic. They have value 0. I would appreciate answer(s) to any/some/all of above questions. <Q> >> , <S> < <> > <S> and << : These are called "off-page connectors". <S> > <S> > means that the signal originates from this page (it's an output from this page) and goes to another page. <S> The numbers "6,10,11" are the pages that this signal goes to. <S> On those pages you'll find those signals next to the << symbol. <S> <<>> means this signal is a bidirectional signal. <S> The signal can originate from this page or the other page it is present on. <S> The n in <S> HDMICLK_DISn typically denotes an active low signal. <S> I dont know about HDMI's specs but from the signal name I assume it has a clock (CLK = clock). <S> Some name + _dis (HDMICLK_DIS) typically means that this signal's purpose to to disable some function, in this case the clock signal for HDMI, make this signal high and the clock will be disabled. <S> Now add the "n" to then end ( <S> HDMICLK_DISn) <S> and again it's an active low signal. <S> A high signal on this line will keep the HDMICLK <S> enabled but now a low signal (typically GND) <S> will disable <S> the HDMICLK <A> I cannot answer Q2 and Q3 without seeing the full schematic. <S> Answer-1: <S> > <S> > , <S> << and << >> <S> indicate the signal direction. <S> For example, in your schematic, USR0...USR3 are outputs, and USB1_OCn is input, and GPIOs are bidirectional (i.e. can be either input or output - most likely they are left for the user's choice). <A> Zero ohm resistors are also used for hardware configuration. <S> When connected between a GPIOpin and GND with the GPIO pin pulled high by a suitable resistor. <S> 2 <S> x GPIO pins gives 4 different configuration options. <S> (00 01 10 11).
| This symbol means that that signal originates from a different page and enters this page (its an input to this page).
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How can you tell, remotely, if someone put their SIM card in a new phone? On a recent television show, on ID Discovery, 'The Case that Haunts Me' (season 3 epi 4 'Cruel and Unusual') the lead detective finds out that the prime suspect has taken the SIM card out of his phone and put it in his murdered wife's phone... I thought only the SIM card could be traced/tracked... How could someone tell, without ever having seen (or seized) the phone(s), if SIM cards have been switched? <Q> The SIM card has an international mobile subscriber identity ( IMSI ) that is used to tell who is calling. <S> SIM cards identify the owner. <S> The phone has an international mobile equipment identity( IMEI ) that is used to identify the phone. <S> Both are used at various times when a phone is connected to a network. <S> That information will be stored for some length of time for billing and other purposes. <S> If you have the authority (court order and such) then the network operators can tell you which IMSI was used with which IMEI, and when (provided they still have the records.) <S> You could then match the IMEIs of the two people to the IMSI of the one person and see when each combination was used. <S> If all you have is the IMSI, then you can still get the records. <S> If the SIM is used for a long time with one particular IMEI, then changes to a different IMEI at about the time <S> the murder (or whatever) occurred, then you have a hint that the killer may have swapped the SIM. <A> <A> Law enforcement can get access to the mobile network operator's infrastructure. <S> (laws governing how that's regulated differ wildly.) <S> Without that access, none can be done remotely. <S> With that access, you get access to everything the cellular network. <S> That of course includes the subscriber data (IMSI) as well as phone data (IMEI etc).
| The SIM contains the IMSI and the phone contains IMEI , if the phone registered to a network, this combination is logged.
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