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heat dissipation in watts of equipment in a box Here is the situation, some electrical equipment will be in low ambient temperatures and I need to make it warmer so that there is no damage to it.I want to calculate how much power in watts is needed to heat up equipment inside a box made of aluminum. The box is 5 inches X 5 inches X 11.5 inches (surface area is then 1.727 ft^2) it is made of aluminum and it is .1 inches thick and insulated. from previous tests I recorded that when the ambient temperature was at -20 Celsius for a long time (at least 1hr) the surface temperature of equipment was steady at -10 Celsius and the air between the box and equipment was -11 Celsius. Two questions:1) How much power in watts is being dissipated by the electronics?2) How much power in watts is needed to bring the temperature of the surface of equipment to 0 degrees Celsius? to 10 C? I did some research and found this: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatcond.html heat conduction formula Q/t = k A (Thot - Tcold)/d where k = thermal conductivity,A = surface area,Thot - Tcold = 10,d = thickness, The issue with this is that when I plug in k for aluminum = 205 or .5 by adjusting the units in the formula I get a large value for Watts either way...am I doing this completely wrong is there another formula that would better model this problem? <Q> Your problem here is that you're looking at the wrong part of the thermal equation. <S> The heat dissipation of a heated metal box is dominated by the thermal resistance of the metal/air interface, not by the thermal conductivity of the box itself. <S> Characterizing the thermal resistance of that interface will be difficult without taking a lot more measurements. <S> From a practical perspective, the simplest solution will be to overdimension the heater and use a thermostat to keep the enclosure at the target temperature -- this will make the exact size of the heater unimportant, and will also mean that the temperature will remain stable even when the ambient temperature changes. <A> You are using the wrong temperatures in your equation, and the wrong equation altogether. <S> For that equation to work you have to measure the temperature of the internal and external aluminum surfaces, not the air temperature. <S> But given the very low differential there, that exercise would be pointless. <S> As a first-order approximation you can simply calculate the dissipation of a block of material with a 10 degree differential and the same surface area of the box. <S> The equation is very similar, but the important factor is not the heat conductivity of the box but the coefficient of heat convection for air and the total surface area. <S> $$ <S> Q = <S> h <S> * A <S> * ( T_s - T_a <S> ) $$ \$ T_s \$ <S> is the surface temperature, \$ <S> T_a\$ <S> the external air temperature, and A the total surface area. <S> \$ <S> But if you use the air temperature this is a gross estimate, as it ignores that the internal temperature also has to be coupled from the air to the box, and that some of the heat is being radiated as well. <S> You would do better if you directly measure the box surface temperature. <S> But in your application I would do two things: <S> Add some non-flammable foam insulation to the inside of the box. <S> Even a few millimeters covering the internal surfaces would go a long way. <S> If that is not enough, add a resistor to dissipate a few extra watts. <S> If you want to regulate the temperature and you need relatively little power you can directly use a positive temperature coefficient thermistor. <S> Or you can couple a thermistor with a transistor to increase dissipation. <A> Thermal conductivity of aluminum walls has very little to do with the problem (thermal impedance of walls is likely negligible). <S> This is a question about thermal properties of a rectangular box under free convection in the field of gravity. <S> As formulated, the problem has no easy solution and needs an involvement of CFD - computational fluid dynamics. <S> There is a vast layer of engineering articles to assess thermal regimes of enclosures, like this one , although it is pretty clueless (fin spacing is too small for good natural convection to develop). <S> Here is a better article on how to access heat transfer across hermetically sealed electronics enclosure, here is the general sketch of the problem components, On external side, you need to account for natural convection around vertical surfaces of the enclosure, from top and bottom horizontal surfaces (the transfer properties are different in all three cases), and radiative exchange. <S> For inner convection the same factors are working and need calculations/estimation too. <S> So, there are a lot of factors involved, and only crude estimations are possible.
| You could save yourself some time if you simply experiment with a couple known power levels (a couple resistors and a power supply) and directly calculate the heat transfer coefficient air to air for your specific box. For a small temperature differential and still air h can be as low as \$ 10 W/(m^2 K) The result will depend on box orientation and whether the ambient air is still or moving.
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Does ATtiny RESET pin need a resistor? I need to make sure that my ATtiny does not reset unexpectedly. To do that, I am planning on connecting the RESET pin directly to VCC. However, I have heard some places that you need a 10k resistor between VCC and the RESET pin, to prevent too much current from flowing into the RESET pin and burning the IC. I have also seen some places that you do not need a resistor between VCC and RESET, and I have also read a few places that you do not even need to connect RESET to anything, because it is pulled high internally. I could not find any definitive answer on this, and I would like to know what the case would be for each of these condidions (directly to VCC, VCC through resistor, don't connect RESET pin at all) is. <Q> First, in the ATtiny25/45/85 datasheet, page 161, it suggested that there's a internal pull-up resistor between 30 kΩ - 60 kΩ on the RESET pin. <S> It means the MCU will almost always work for a hobby project. <S> However, if the requirement is high reliability, I need to make sure that my ATtiny does not reset unexpectedly. <S> Then you have to put external EMI/RFI and ESD into your consideration. <S> I strongly suggest reading AVR040: EMC Design Considerations . <S> Finally, it should be noted, that unexpected reset is inevitable for any MCU facing hostile environment in the long run, sometimes you even have to trigger it deliberately (e.g. through a watchdog timer) to prevent damage from a runaway program. <S> On one hand, you should take precautionary measures in your circuity to defend yourself from spurious reset and ESD, on the other hand, you still need to make your program to be robust enough to handle unexpected reset. <A> As far as I'm aware, there is no function of the attiny that requires the reset pin to be pulled low (for example, the watchdog timer doesn't try to pull the reset line low). <S> Having said that, there isn't really a good reason why you would want to have the line directly connected, and there are quite a few where you wouldn't want it to be directly connected (like programming, and manual reset etc). <S> The attiny does have an internal pull-up to VCC, but its quite weak (~100k iirc), so environmental conditions could cause a spontaneous reset. <S> If you're worried about this, you should use an external pull-up resistor. <S> It means using an extra component, but it's safer, especially if your reset pin is routed to a programming connector. <A> The reset pin is internally pulled-up and glitch filtered <S> so you do not need a resistor. <S> You can leave the pin unconnected (no trace since a trace can pick up noise) and it would take some pretty strong noise to make the chip reset unexpectedly - probably enough that other bad things would happen first. <S> If you really never want the chip to get reset then you can set the RSTDISBL (reset disable) fuse after programming your firmware. <S> Once this fuse is set, the reset pin becomes an IO pin and will not reset the chip even if tied to ground. <S> Note that after disabling the reset pin with RSRDISBL <S> it is harder to reprogram the chip since you might need to use High Voltage Programming (depending on which ATTINY you are using).
| I don't think theres a technical reason why you couldn't connect your reset pin directly to +V, as long as it was already programmed and you never wanted to modify that program.
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How can we understand portable battery for mobile phones having 10000 mAh capacity, and 6200 mAh "nameplate capacity"? In USA, I always see portable batteries for mobile phone stating they have 5000 mAh, 10000 mAh, or 13000 mAh, etc. (One of the popular one is Anker as on Amazon.com). However, in Asia, I saw the batteries having a spec of 10000 mAh "battery capacity", but a "nameplate capacity" of 6200 mAh. Another battery was 10000 mAh but its "nameplate capacity" was 5100 mAh. The word for "nameplate capacity" is "額定容量" and it seemed to mean "effective capacity". It seemed 10000 mAh is relative to 3.6V of the internal batteries in the charger, and the effective capacity is relative to 5V. (and if it is for QC 2.0 or 3.0 and 9V charging, then the mAh drops to close to about half). How can we understand this? Does that mean batteries like Anker stating 10000 mAh is also nominal and it really has an effective capacity that is not stated? <Q> Quite simply - they are lying to you, and including the "nameplate capacity" so that they can try to cover themselves. <S> Batteries are typically rated at 1C discharge - this means at a current level that will discharge the battery to the minimum safe level within 1 hour. <S> So a 10Ah battery will discharge at 10A for 1 hour. <S> If the nominal voltage was 3.7V, we can ballpark by saying that's 37Wh. <S> You can get maybe +5% by discharging a battery at <S> a lower current - so for example 2A will probably get you 38.5Wh because there's less loss due to internal resistance (remember higher current = <S> more lost due to resistance). <S> But you're not going to take a 6Ah battery and turn it into a 10Ah battery by using it differently. <S> I expect if you take any of these battery packs and run your own tests, you'll see. <S> They rely on people not knowing any better and just buying something because it says that it's higher capacity. <A> This is because the amp-hour ratings of these packs are a bit ambiguous. <S> This type of battery pack has one or more internal cells along with a DC to DC converter. <S> The cells generally have an operating voltage of somewhere between 2.8 and 4.2 volts, depending on the state of charge. <S> Usually the 'nominal' voltage is around 3.7 volts. <S> The internal cells are then connected to the USB connector via a DC to DC converter, which (relatively) efficiently converts the cell voltage to 5 volts. <S> However, because of the converter, the current drawn from the pack differs from the current drawn from the cells inside the pack. <S> For example, if you draw 1A from the pack at 5V and your pack has one 3.7V cell, the cell will have to supply at least 1 <S> A * 5 V / 3.7 V = <S> 1.35 A. <S> Also, the converter will not be 100% efficient, generally only 80-90%, so the cell current would probably be <S> more like 1.35 / 0.9 = 1.5 A. <S> This is where the rating becomes ambiguous: are the 'amps' in 'amp-hours' measured at the cell or at the USB port on the pack? <S> If you take your 10000 mAh / 6200 mAh battery as an example and assume a 3.7 V internal battery and a converter with 85% efficiency, 10000 <S> * 3.7/5 <S> * 0.85 <S> = 6290 mAh, which is pretty close to 6200 <S> mAh. <S> A much more useful method would be to specify the capacity in watt-hours or joules, which is independent of the voltage. <S> But this would make far too much sense. <A> USB powerbanks typically consist of a single cell lithium battery (or for higher capcities multiple parallel cells). <S> This has a voltage of 3-4V depending on it's state of charge. <S> To get 5V for powering a phone it has to be run through a boost coverter which increases the voltage at the expense of current. <S> It also has losses. <S> As such as you surmise in the question the capacity in AH measured at the cell is significantly higher than the capacity measured in AH at the output of the powerbank. <S> Also cell manufactures <S> AH claims are often to put it politely "optimistic". <S> My understading is that powerbanks marketed in the west are normally marketed based on cell capacity, not effective capacity.
| Marketing dictates that the larger the number they put on the box, the more packs they will sell, so these numbers usually indicate the amp-hour rating of the internal cells, which IMHO is rather useless and misleading.
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Why does RAM (any type) access time decrease so slowly? This article shows that DDR4 SDRAM has approximately 8x more bandwidth DDR1 SDRAM. But the time from setting the column address to when the data is available has only decreased by 10% (13.5ns).A quick search shows that the access time of the fastest async. SRAM (18 years old) is 7ns.Why has SDRAM access time decreased so slowly? Is the reason economic, technological, or fundamental? <Q> It's because it's easier and cheaper to increase the bandwidth of the DRAM than to decrease the latency. <S> To get the data from an open row of ram, a non trivial amount of work is necessary. <S> The column address needs to be decoded, the muxes selecting which lines to access need to be driven, and the data needs to move across the chip to the output buffers. <S> This takes a little bit of time, especially given that the SDRAM chips are manufactured on a process tailored to high ram densities and not high logic speeds. <S> To increase the bandwidth say by using DDR(1,2,3 or 4), most of the logic can be either widened or pipelined, and can operate at the same speed as in the previous generation. <S> The only thing that needs to be faster is the I/ <S> O driver for the DDR pins. <S> By contrast, to decrease the latency the entire operation needs to be sped up, which is much harder. <S> If you compare CPU caches with RAM and hard disk/SSD, there's an inverse relationship between storage being large, and storage being fast. <S> An L1$ is very fast, but can only hold between 32 and 256kB of data. <S> The reason it is so fast is because it is small: <S> It can be placed very close to the CPU using it, meaning data has to travel a shorter distance to get to it <S> The wires on it can be made shorter, again meaning it takes less time for data to travel across it <S> It doesn't take up much area or many transistors, so making it on a speed optimized process and <S> using a lot of power per bit stored isn't that expensive As you move up the hierarchy each storage option gets larger in capacity, but also larger in area and farther away from the device using it, meaning the device must get slower. <A> C_Elegans provides one part of the answer — it is hard to decrease the overall latency of a memory cycle. <S> The other part of the answer is that in modern hierarchical memory systems (multiple levels of caching), memory bandwidth has a much stronger influence on overall system performance than memory latency , and so that's where all of the latest development efforts have been focused. <S> This is true in both general computing, where many processes/threads are running in parallel, as well as embedded systems. <S> For example, in the HD video work that I do, I don't care about latencies on the order of milliseconds, but I do need multiple gigabytes/second of bandwidth. <A> I don't have that much insights, but I expect it is a bit of all. <S> Economic For the majority of computers/telephones, the speed is more than enough. <S> For faster data storages, SSD has been developed. <S> People can use video/music and other speed intensive tasks in (almost) real time. <S> So there is not so much need for more speed (except for specific applications like weather prediction etc). <S> Another reason is to process a very high RAM speed, CPUs are needed which are fast. <S> And this comes with a lot of power usage. <S> Since the tendency of using them in battery devices (like mobile phones), prevents the use of very fast RAM (and CPUs), thus makes it also not economically useful to make them. <S> Technical <S> By the decreasing size of chips/ICs (nm level now), the speed goes up, but not significantly. <S> It is more often used for increasing the amount of RAM, which is needed harder (also a economic reason). <S> Fundamental <S> As an example (both are circuits): the easiest way to get more speed (used by SSD), is to just spread the load over multiple components, this way the 'processing' speeds adds up too. <S> Compare using 8 USB sticks reading from at the same time and combining the results, instead of reading data from 1 USB stick after each other (takes 8 times as long).
| Most likely, parts of the ram would need to be made on a process similar to that for high speed CPUs, increasing the cost substantially (the high speed process is more expensive, plus each chip needs to go through 2 different processes).
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Is high current in a speaker the same as high voltage? I'd like to build my own Amplifier, and because I know that the current creates the magnetic field, I don't understand, how amplifying the voltage makes it louder. I mean with high voltage doesn't automatically come high current. For example if an Amp can only deliver 100V at 1A (I don't know whether such an amp exists but ok) (the reason for that is the Load Resistance RL, right?) , then the Output Power is 100W. Would that make the same sound on the same speaker in same Conditions, like if I put 100A with 1 V through it? EDIT: Ohh thank you guys for that many answers in that short time :_D How could a class D amplifier amplify with ~90% Efficeny, if 4,5W is the Maximum amount, you can get out of a 12V PSU? (4,5W/0,9 = 5W -> 5W / 12V = 0,417 A) That would mean, the PSU has 0,417A of output current, and so, because the output current can be calculated, the current doesn't play a role... How could that be? Or is that the reason, why you wrote "Rough Calculations"? In case, higher Impedance means the Voltage of the Power can be higher: If i want to build a Subwoofer with Amplifier out of a Car-subwoofer with 4Ω , is it a must, to have relatively low Voltage audio signal and relatively High current? Then, depending on Phil G's Explanation, i have to build an Amplifier with high Current Output? How to calculate the Current and Voltage I need to achieve the wanted Power of 200W through that 4Ω Subwoofer? My plan is, to RC Low-pass filter the Pure Audio Signal, amplify it and put it out on my Subwoofer, any Ideas? Ty for Answers in advance! <Q> For a given voltage, the impedance of the speaker determines the current. <S> In order to get a given current with a given voltage, you need just the right impedance. <S> So you can't drive 100A at 1V into an arbitrary speaker -- you would have to find a \$\frac{1}{100}\$ -ohm speaker (and some really thick speaker cable). <S> So for a single 8-ohm speaker, the power you can drive into it depends on the voltage -- if you want more power, you need more voltage. <S> This is why car speakers are often 4-ohms -- it's an easy way to get twice the power that you could with 8-ohm speakers, from the 12V <S> that's available in the car. <A> Would that make the same sound on the same speaker in same conditions, like if I put 100 A with 1 V through it? <S> In general, yes. <S> Loudspeaker design is a balance between several factors which include mechanical strength, wire size, reasonable voltage (insulation) and reasonable current. <S> The result is that speakers are readily available in 4 Ω, 8 Ω and 16 Ω versions. <S> The system designer now has to decide whether the amplifier will be a higher voltage, lower current or lower voltage, higher current design. <S> This may be determined by the available power supply - 12 V auto systems being a good example. <S> I don't understand, how amplifying the voltage makes it louder. <S> As you suspect, you need to increase the power . <S> This may be achieved by increasing the voltage, the current or, more usually, both. <S> I mean with high voltage doesn't automatically come high current. <S> Correct, though it does if the speaker resistance remains the same. <S> Rough calculations: <S> The max power output of the amplifier into a speaker of known resistance, R, can be calculated from the availableDC voltage in the amplifier. <S> $$ P_{max} = <S> \frac {V_{RMS}^2}{R} = <S> ( \frac {V_{DC}}{2\sqrt 2} ) ^2 \frac {1}{R} $$ <S> So for a 12 V supply and a 4 Ω speaker \$ P_{max} = \frac {V_{RMS}}{R} = <S> ( \frac {12}{2\sqrt 2} ) ^2 <S> \frac {1}{4} = <S> 4.5 <S> W\$ <S> is the maximum available unless some other trickery is used. <A> The power delivered will be the same regardless of the voltage, but what is important is the load that it will be driving. <S> Most loudspeakers (if that's what you are intending to drive) have an impedance of between 2 and 16 ohms, but there are PA amplifiers that do as you say and produce a limited current but much higher voltage output - which allows thinner wiring to be used over long distances, that are intended to drive high impedance speakers, which either have voice coils with many more turns of finer wire, or use step-down transformers at each speaker.
| For a given current, the impedance of the speaker determines the voltage.
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Where exactly appears the channel capacitance and Ron when switch on DG442 switches I have this switch in a circuit, actually there is 30 switches. https://www.vishay.com/docs/70053/dg441.pdf When system is not executing any operation, switches are off. Sx and Dx are not connected. But some system operation will switching them on, connecting Sx and Dx. In the worst case I have to take into consideration the effects of some associated parameters such as capacitances or intern Rs. I am analyzing the circuit in the worst case and I need to know how to take into account the channel capacity when switch ON. The datasheet tells us that there are 16pF for each channel of the switch. My question is: 1.- This capacity refers to a capacity between the switch output dx pin and GND? Does it mean that when say "channel"?It also indicates a Ron = 50Ohms per channel. Is it supposed to be parallel to this channel Capacitor? or is it serial with capacitor? 2.- Where can it be in the simplest circuit where a signal S1 is switched on with CHA = GND and CHB = signal M is switched on with D1, where M has connected RL = 12K serial with V= 12V? VBA is my Vout as seen here: I would like to model this circuit adding Ron and Cp. But I'm not sure how to add them to B and A nodes, specially when switching S1 with GND. That would mean S1 != GND? <Q> The on capacitance is across the switch analog terminals. <S> From the schematic in the datasheet, I have added where it exists: <S> This is also the primary cause for off isolation degrading with frequency coupled with the isolation resistance and load; as you can see it is a classic single pole with 20dB/decade characteristics. <S> When I did a quick calculation, I got a value of about 15pF for the source to drain capacitance, close enough for confirmation. <S> For a really good overview of analog switches, I recommend this application note . <A> The FET likely looks like this simulate this circuit – <S> Schematic created using CircuitLab <A> I got this image from app notes, shared by Peter Smith, I think that CD refers to C between drain and GND. <S> And Ron is R between drain and source. <S> Althought they talk about "channel" <S> I think they mean CD. <S> If not, they would talk about CDS. <S> That's my opinion. <S> If I'm wrong, please correct me. <S> This is what I have just undertsood from app notes.
| The capacitor exists across the analog switch channel (which was not identified in the schematic) so the A and B analog terminals are where I have put the capacitor.
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Why is a laptop power supply output voltage different from its battery? I hope you will not mind answering. This question has long been on my mind. Normally, the adapter output voltage is higher than that of the battery. In my laptop's case, the output voltage of the adapter (or charger or power supply) is 19.2V. (That .2V itself is also a big question for me. Is that so sensitive?) But the voltage of the battery is 10.8V. My question is, why is the adapter output voltage different from the battery's voltage? This is not about asking how it is regulated by the internal circuitry, but asking why. <Q> The voltage on your battery "10.8V" is the "nameplate" voltage, some average voltage that your battery delivers over full discharge cycle. <S> The value of "10.8" indicates that this is a battery of 3 Li-Ion cells in series, giving their standard "nameplate" voltage of 3.6V per cell. <S> Charging the Li-Ion cells requires variable voltage levels, from 2.5-3 V per cell (when in deeply discharged state) to 4.2V (4.35 in some cases) per cell in "constant-voltage" stage of charging process (otherwise the cell won't be charged to full capacity). <S> So the feeding power must have some overhead to provide the charging process (or let internal charger to do so). <S> So, for 3-cell, it comes up to 12.6 - 12.9 V of input. <S> The external power supply must provide this headroom, which includes minimum of "drop-out" (or regulation) voltage for switching electronics inside the external power supply and internal charger, 1-2 V per device, give or take. <S> Eventually it comes up to 12.9+4 = <S> ~ 16-17V. The "19.2" nameplate is a bit of mystery, since it is not that stable in the first place. <S> It is just an industry standard. <S> Any AC-DC adapter in the range of 18 - 22 V will happily charge your laptop, very likely. <A> From your other question, it appears that you have a laptop power supply with power rating of 65 Watts (3.42 A × 19.2 V). <S> As power is voltage times current, it means that if the voltage is higher, the current is lower, so thinner, cheaper, more flexible wires can be used to deliver that 65W to the laptop when charging. <S> That's why it's not a 65 Watt, 5 Amp 13 V power supply. <S> Also it means that when the battery is being charged, a DC-DC converter in the charging circuit converts the 19.2 V down to match the battery voltage so that suitable amount of charging current flows into the battery. <S> So in this case a nominal 10.3V battery could be charged at over 6 Amps with the same 65 Watts if the battery can handle that amount of charging current safely. <S> The charging current will be limited by what is safe to the cells. <A> This is because the laptop is designed for a optional battery you don't have. <S> Laptop batteries usually come in 6 and 8 cell varieties. <S> A 6 cell battery has two strings of 3 series cells. <S> A 8 cell battery has two strings of 4 series cells. <S> The higher cell count provides more power at the expense of weight. <S> (Sometimes a 4 cell option is available consisting of 1 string of 4 series cells) <S> Most commonly the smaller capacity battery will sit flush with the case and the high capacity battery sticks out the back. <S> Ale.. <S> chenski identifies a max charge voltage of around 4.2 V/cell. <S> So a 6 cell battery needs 12.6 V max to charge, an 8 cell battery needs 16.8 V. <S> Then add cable losses and regulator losses to get 19.2 V. <A> Take for example if the battery is designed with 12volts you need at least 13 to 15 volts to charge it, and don't forget your laptop have to stay charging while you perform other operations on it. <S> To my understanding there is a charge controller that monitors the charging. <S> If am wrong please i need someone with a better explanation, you know we are all learning <A> Here is the answer to this question, after I did some research/searching. <S> The laptop power requirement is determined during design. <S> In my case, my laptop's requirement is 19.2V/3.95A. <S> But why does battery voltage is 10.8V or 11.1V? <S> It is because every battery consist of some 18650 dry cell battery which its voltage is 3.6V or 3.7V. Three batteries are arranged in series to produce 10.8V or <S> 11.1V (3.6*3=10.8V, 3.7V*3=11.1V). <S> As the laptop power requirement is 19.2V*3.95A (=75.84 watt), then the charger must be made according to that spec. <S> How about the power from the battery? <S> Need boost converter to convert the 10.8V to 19.2V. <S> And that is inside the laptop or battery. <S> What about the cell battery is not 3.6V? <S> Say if you can increase the voltage to be 10.4V <S> like 9V dry cell 6LR641, and each two of them you arrange in series <S> and then you parallel to another two series, then you will get voltage 20.8V. <S> To meet the 19.2V <S> you need to step the voltage down, also converter but to step down (rather then to step up). <S> So, when we plug that 19.2 charger to a laptop with a common battery we have (10.8V), then the voltage from the charger need to be stepped down inside the battery. <S> Vice versa, when laptop is taking power from the battery, it need to be stepped up to 19.2V <A> It is likely that the voltages are different by design / intentionally. <S> A higher DC voltage enables power to flow with less current (compared to the lower 10.2 Volts). <S> This can be important when pushing DC power through appreciable distances. <S> Battery and voltage is stepped down to 5 or 3.3. <S> V or the CPU. <S> Possibly other voltages for the display and other discrete devices.
| I think the reason why the adaptor or charger would produce such big amount of volt is because, while the laptop is charging, the extra volt will help run the laptop without problem.
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Is it Safe to Plug an Extension Cord Into a Power Strip? Just a bit of context, I am a technology enthusiast. I love everything from software to hardware, Windows to Linux, Computer to Phone, and everything in between. So, I wouldn't be surprised is my electrical usage exceeds 50% of the entire household (8 people). To power all of the equipment I use, I use a pretty expensive surge protector, with 12 outlets on it. Recently, I have been very plain and simply, running out of outlets. I have an extension cord with 3 outlets on it. Maybe, I thought, I could plug some of the things on my surge protector into the extension cord, and then the extension cord into the surge protector. I know, it's pretty dumb of me, but I went and did it, without checking to see if it was safe first. But I did, and everything seems to running fine. It's been running for about a 30 minutes now, and neither the surge protector or the extension cord are overheating. However, after the fact, I am reading on websites about weather or not this is safe, and they're all saying it's not, even though everything is running perfectly fine. Thoughts? Something I'm not understanding? <Q> However, there are some things to keep in mind: <S> Never exceed the maximum power, current and voltage. <S> Voltage will normally not be a problem since this is (more or less) stable in a country. <S> When you go abroad, check if they can be used. <S> current is most important, sum up all devices you connect on that extension cord AND all devices on extension connected to the the first extension cord. <S> Take some margin, especially motors sometimes can use a lot of 'starting' current. <S> Don't forget when you switch on an extension cord (either by plugging it in, or by using a switch if there is on), ALL devices will be powered on at the same time, using possibly some additional extra current. <S> When you go towards the maximum current supported by the extension cord or in a warm environment or radiation (e.g. from the sun), make sure the cable is fully unrolled (meaning not bundled), since warmth cannot spread (updated according comment of FerryBig). <S> Another potential problem is that if the outlet is equipped with an earth-leakage circuit breaker safety device, that may also trip "on it's own" as the current in neutral wire is diminished due to excessive resistivity in the circuit. <S> (addition from Kurja). <A> If you only plug low current devices into the extension cord there should be no problem. <S> In any case, you need to keep the total current requirements of everthing plugged into the surge protector and extension cord below the rating of the surge protector. <S> You also must consider the current requirement of anything else on the same circuit as the outlet the surge protector is pluggged into. <S> Manufacturers warn against plugging extensions or other power bars into a surge protector in case someone plugs a 15 amp heater into the last of three power bars. <A> I'll add some information to what Michael and Peter wrote in their good answers. <S> You are right about lots of warnings being written everywhere on the subject. <S> Manufacturers must design their product and the included instructions around the (possibly dumbest) average consumer. <S> Average people, sadly, have very little understanding of how electricity works: they confuse power, voltage and current all the time, let alone knowing what Ohm's law or Joule's law are! <S> So manufacturers don't take any chances. <S> They tell you the simplest possible rule to use their products with safety: "One extension cord (adapter plug, or whatever) to one socket, no multiple adapter/extension things joined together" . <S> Stop. <S> Simple and hopefully foolproof. <S> Of course, extension cords, plugs, sockets and similar "electrical connection things" follow Ohm's law and Joule's law. <S> This latter is particularly important to calculate heating. <S> I have been running various multiple extension cords setups for ages, without the slightest problem. <S> " <S> Normal" people could watch my setups and cower in terror (if they are sane), or they could do the same and burn their house down, because they can't tell the electrical difference between a 4W power pack and a 1500W toaster! :-D <A> Leaving whether it is safe or not (or widely practiced or not) to the others, I'll mention that in my U.S. jurisdiction the fire martial doesn't like it and made everyone in my company reconfigure practically every office to avoid this or face penalties.
| Apart from heating cables or connectors, a primary safety concern with an extension plugged into an extension of any type is that overall resistivity can become so large that even a full short at the furthest end of the cable may not trip a breaker. To be honest, I do it all the time. If you know what you are doing and know the characteristics of your loads, you can mix and match almost anything by providing generous safety margins . This is some sort of scarecrow tactic from the manufacturers (and rightly so, IMO).
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Resistor in series needed with LED string? I have an adjustable 24VDC power supply. I have 1W LEDs that I was planning on stringing 7 or 8 in series to get approximately 24V across all of them together. They are rated 3.3-3.8V each.Do I need a resistor in series to maintain current? I was thinking so since a lit LED is basically a short? I can't find an answer to what it is really needed for, just that most people say to put a resistor in series to 'limit current'... if so, it would be a small resistor, around 2 or 3 ohm? <Q> This range and your heatsink 'C/W controls <S> the k= dV/dT for NTC coefficient on LEDs Vf and T ['C] so choose a low side R around 0.1 to 0.5 Ohm depending on heat sink k to avoid thermal runaway. <S> Expect Vf to drop no more than 0.5V /24V at constant current from temp rise due to NTC effects. <S> A power change without a series R for a 0.2V change in V+ is ΔP=ΔV²/Rs where ΔV =(3.4{=Vf.nom}-2.8{=Vt})/If= <S> 0.6V/0.3A = 2 <S> Ohms <S> (24² -23.8²/ (7 * 2 ohms) = <S> 0.71W <S> rise in 7 LED stringfrom 1W to 1.1W using 24V. <S> If LED heatsink is 50'C/W this raises junction 5'C and reduces Vf <S> ~ <S> -4mV/deg C or -20mV <S> so this assumption is thermally stable. <S> I suggest you use a 0.1V current sense shunt on low side or <S> 0.1V/0.3A= 1/3 Ω and adjust ΔV=0.1V for 300 mA and adjust while hot . <S> You can now measure your thermal resistance by the change in ΔV from the datasheet thermal coefficient ( <S> -3.4 to -4 <S> mv/'C <S> to ? ) <S> as the string reaches steady state above ambient temp. <S> A good design is case temp below 45'C meaning never too hot to burn finger. <S> But using this method you can measure actual voltage change in Vf. <S> This Vf is the actual change in threshold "knee" voltage Vt not the Rs bulk resistance. <S> 0.3A <S> * 0.1V <S> = 30 mW <S> so you can afford to use 8 LEDs at 3.4V = <S> 27.2V <S> with a 0.5W drop current sense at 0.35A from 29.5V max or R= <S> Pd/I²= 0.5W/0.35²= 4 Ohms 1W with 0.35A*4= <S> 1.4V drop results in 27.2V nom + 1.4= 28.6V <S> So if your LED tolerance is on the high side , you have to bypass 1 LED down to 7 and still regulate V+ for current. <S> I have an empirical formula to determine the heatsink required for CV control to prevent any string from thermal runaway due to NTC Vt effects from temp rise. <S> ( but not included here) <S> But a long string is less prone to thermal runaway since internal bulk resistance stabilizes the Vf voltage as mainly the threshold internal Vt voltage is affected by temp, not the bulk electrode resitance determined by LED power rating and size.. <A> A pre-fabricated LED string may already has resistors integrated into it already. <S> If you need a resistor don't guess. <S> Calculate it out, though you will need to know the number of LEDs, the minimum forward voltage drop, and desired operating current of the LEDs. <S> I'm sure the answer is somewhere else on this site already: led resistor calculation, and resistor wattage? <S> But basically, the LEDs have a pretty fixed voltage drop across them. <S> The balance of the voltage from the supply will appear across the resistor and then you just use V=IR to determine the current flowing through this resistor, and since everything is in series, this will also be the current flowing through the LED. <S> \$ <S> I = <S> (V_{supply} <S> - (V_{LED} \times N_{LED})) <S> / R \$ <S> I recommend you use the minimum LED forward voltage since this scenario will cause the most voltage to be dropped across the resistor, and therefore result in the greatest possible current for the resistor value chosen. <S> Doing it any other way may result in an actual operating current higher than what you calculated due to LED component differences. <A> LEDs have a negative temperature coefficient, meaning that as they heat up, their resistance decreases, causing more current to flow and more heating. <S> Using a series resistor is less efficient than current control as you must use a resistor sufficient that the positive temperature coefficient(PTC) of the resistor counteracts the negative temperature coefficient of the LEDs, preventing thermal runaway. <S> If you're using 7 or 8 LEDs to approximate 24V, you have roughly 3.2V 350mA LEDs(edit your question and add the specs/datasheet), and 350mA is at the top end of what you'd want to run through a current limiting resistor. <S> If you do, you'll want to keep the voltage drop over the resistor small(while still having the temperature coefficients cancel) and use a fairly high wattage one (you can actually use some types of resistor as a fuse as well). <S> You'll likely waste a lot of power regardless. <S> Instead I'd recommend using current controllers for LEDs of this wattage at full brightness. <S> On the other hand, if you underpower the LEDs, using a resistor becomes less of an issue and you'll increase the efficiency and lifespan of the LEDs, although in this case you'd still be better with current control.
| Either a series resistor or current controlled driver is necessary unless the thermal dissipation of the LEDs dwarfs their potential output.
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Rounding up or down with Motor Overcurrent Protection When sizing motor over-current protection, CSA 22.1 28-200 3a states that a circuit breaker can be sized up to 250% of the rated full load amperage (FLA) of the motor. E.g. 100hp, 575VAC motor draws 99A (CSA 22.1 Table 44): 99A * 250% = 247.5A Usually I would've rounded up and selected a 250A circuit breaker. I just noticed today that in the over-current example on page 662 that they always size the the next lowest standard size and specified a 225A circuit breaker. This makes sense since a 250A circuit breaker is past 250% of FLA. However, Rule 28-200 4c goes on to state that if the circuit breaker above isn't large enough for the motor to start that you can go up to 300% of the motor FLA (297A). Therefore, 250A isn't in violation. My question: As a rule of thumb, when sizing motor branch over-current protection, should you round up or down to the next standard breaker size? If rounding up, do you need to prove that the motor couldn't start with the smaller sized breaker? Does the answer to #1 change when considering feeder over-current protection? <Q> It depends on the motor rating and load to full speed time. <S> Start current can be 3 to 8x <S> rated current and start-up time is highly load dependent. <S> Thus they have 3 classes and 3 categories of solutions depending on motor class and "heavy" load. <S> 125/150 175 200 <S> A <A> Don't know about CSA standards, but in EU (machines were exported to US also) <S> we use motor protection switch. <S> It's a device made for motor overload protection. <S> Keep in mind that using a 250A circuit braker will also impose to use wires with a gauge for 250A. <S> Also it may happen that if the motor is overloaded at 250% the circuit braker won't never disconnect it leaving the motor to melt down. <S> Maybe you have read some standart that is refereing at some small motors, but the right way IMO is to use motor protection devices. <S> https://www.amazon.co.uk/Siemens-3RV2041-4MA10-Protection-Screw-Type-Connection/dp/B0796X54RX <A> This rule, and the similar version in the NEC, is based on there being another way of protecting the motor from running thermal overload, either an Over Load Relay (OLR) or the motor having internal thermal protection (TP). <S> The branch breaker is then really only providing the SHORT CIRCUIT protection. <S> Most thermal mag breakers have their magnetic trips set for 10x the thermal rating, so a 250A breaker has a magnetic trip setting of 2500A. More recently many mfrs have taken to making the mag trip settings adjustable, but will still max out at 10x the breaker rating. <S> But if your motor can start with a lower mag setting, that provides better protection for the motor. <S> The later addition of the 300% rule came about because a decade or so ago when "Energy Efficient" motors were mandated, there were issues with some of the techniques used to increase efficiency also causing a significant increase in magnetizing inrush current, and the 250% rule no longer worked. <S> So the allowance is that IF you can demonstrate that 250% will not allow the motor to start, you can bump it to 300%, basically to get a higher trip threshold on the magnetics. <S> In your case because the motor FLC is 99A, the magnetizing inrush is likely to be only around 1500 <S> -1700A max, <S> so no need to go with the higher size. <S> In fact a 225A breaker would work as well.
| The reason for the 250% rule was to ensure that the instantaneous (magnetic ) trips in an Inverse Time (Thermal-Magnetic) circuit breaker would be high enough to allow for the motor inrush current without nuisance tripping.
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Cooling Methods for Voltage Regulators I have a NCP1117 5V voltage regulator that is on an Arudino Mega. The Arduino Mega, with all of the necessary electronics parts for the project, draws around 200mA when operational. When using a battery that has a maximum voltage of 12.6 Volts, the voltage regulator gets very hot but it works. (Case temperature of 96 degrees Celsius) The goal of the project is to use a 4S LiPo battery which will have a maximum voltage of 16.8 Volts. This overheats the voltage regulator and it thermally shutdowns (This happens at 175 C according to the datasheet). Is there a good way to cool these devices. It is a SOT-223 package on an Arduino Mega board. I have attached a small 5mmx5mm heat sink on the top, and it did help the temperature but not enough for the required 4S battery voltage. It also does not make the best contact due to the low surface area on the top of the device. I don't like the idea of having a fan so is there a good way to passively cool this device? I was thinking of adding an additional heat sink on the bottom of the board underneath the voltage regulator. Just put it on the bottom solder mask. Is that just a dumb idea or is there a better way to cool these? <Q> Don't stuck into one approach, there are many options available. <S> Don't make yourself difficult. <S> There are tons of it in the online store, some of them are already built-in. <S> And there are people who wants to create it themselves such as GreatScott! <S> The BMS will protect your battery and your device from faults, such as protection from overcurrent, short-circuit, overdischarge and so on. <S> Your main goal is to use a 4S LiPo battery which will have a maximum voltage of 16.8 Volts <S> Just focus on it. <S> You don't need addional heatsink and fan. <S> Your device for the 200mA load will be cool. <A> Using 5V out of 16.8V can only achieve < 30% efficiency. <S> What will it take to convince to use something like a TPS562209DDCR ? <S> with > 90% Efficiency for < <S> 2$ with 81 mm² PCB real estate <S> Why have you not considered a 1" 25mm square 5V 200mA Fan in series with the LDO which only needs 6V input by a cut and jumper? <S> http://www.delta-fan.com/Download/Spec/ASB0305HP-00CP4.pdf <S> They are inexpensive in 6pcs online. <S> option 3 8W LDO replacement $4.31 (1pc) <S> OKI-78SR-5/1.5-W36-C <A> If you don't mind wasting the power and runtime, you have two options (or combination thereof), Put a 20-25 Ohm resistor or 5-6 power diodes to drop the input voltage by 5-6 volts; Attach a heat sink, but do it correctly. <S> The Arduino MEGA has the 1117 IC soldered to a copper pad about 10mm <S> x 10mm in area. <S> You need to make a heat sink to solder it to that area, removing some solder mask around the IC, not just on the top of plastic case. <S> An example of such heat sink construction can be found om Ohmite site: <S> Unfortunately this sink is a bit too big for the available surface, so you have to be creative to adapt it, or to make your own out of 0.5mm-thick copper shim. <S> Alternatively you can replace the NCP1117 with equivalent off-board DC-DC switching converter like Recom R-78W5.0-0.5 , <S> This one should have enough room to fit on AtMEGA with some manual dexterity.
| Running a linear regulator from 16V to 5 V at 200 mA is not generally a good idea, you are wasting 70% of your battery life. Use Battery Management System (BMS) + 5V buck dc-dc converter directly on to the 5v rail.
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Sensor For Continuous Water Level Measurement I am designing water level controller for water tank and want to show the water level in LED bar Graph I know how to transmit data to MCU and show level to LED bar graph but my problem is to find the sensor for continuous level measurement i searched in google and find out that there are floating switch type device which track discrete level.But that will not measure the continuous level of water and i also came across ultrasonic sensors that measure the level continuously Is ultrasound sensors are reliable and efficient method? My other question is what type of other sensor are available that can measure water level continuously and efficiently? <Q> A bit of googling helped: a dozen ways to measure fluid level floats <S> load cells magnetic level gauges capacitance transmitters <S> magnetostatic level transmitters <S> ultrasonic level transmitters <S> laser level transmitters <S> radar level transmitters <S> disadvantages of ultrasonic fluid level sensors With some liquid media, however, these ultrasonic waves are absorbed rather than reflected. <S> If you’re thinking about using an ultrasonic level sensor, be sure that your liquid medium reflects ultrasonic sound rather than absorbing it. <S> ... <S> Agitated liquids, turbulent liquids, foaming, sloshing, and other activity can hamper the performance of ultrasonic sensors. <S> In applications where liquid agitation is common, ultrasonic sensors may therefore be a poor choice. <S> Alternatives mentioned: submersible pressure sensors (hydrostatic pressure level sensor), optical liquid level sensors <A> I find that ultrasonic sensors work great for this type of thing. <S> I'm currently building a water purification plant automation system for the private community I live in. <S> We've got a 55 gallon tank that contains a water/chlorine mixture that delivers this into the main water feed lines (from two submerged pumps in our lake) before they enter the pressure tanks. <S> I use two ultrasonic distance sensors for the measurement (one as primary, the second as a confirmation unit). <S> When the level gets to a low point (measured by exponential average), I engage (ie. <S> open) <S> a solenoid for water (timed with a flow sensor), and at the same time, pump in chlorine to my prescribed amounts. <S> While filling the tank, I monitor the raw data instead of average, then when I hit my high mark, I shut the pump/solenoid off. <S> I also have a float valve, so that if for some reason something goes wrong and the tank goes too full, it'll do an emergency all-stop on the system. <S> For testing, I'm using simple HC-SR04 sensors, but I've also got a couple of waterproof ones that I'll be using in the production system. <S> As another poster mentioned, if the liquid is always in a rather turbulent motion, or doesn't reflect back the sound waves, ultrasonic is not the way to go. <A> This sensor is very useful for capacitive un-touch water level measurement. <S> I wish I worked for TI, but I don't. <S> I use TI FDC1004 for level sensing. <S> This IC has 4 capacitive reading output, and one of them is used for level sensing. <S> The second one is using air capacitance and the third is used for water capacitance. <S> For these reasons, the system can not make an error during level calculation. <S> Also, this IC has a shield for EM filtering from outside. <S> For more information, please look at Liquid Level Sensing with the Immersive Straw Approach , which is why I advise using this IC.
| I think the best way is using FDC1004 (from TI).
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Infrared High Intensity LED with two anodes? I got an infrared led emitter, with model ""L1I0-0850060000000" and the datasheet is plenty of technical data but I can`t find why it has three terminals - 2 anodes and 1 catode. Usually it's used for data transmittion. Datasheet: https://www.lumileds.com/uploads/685/DS191-pdf Here is the datasheet showing the three terminals: I don't have many experience with light data links, but I know that some common leds have three terminals to allow switching the color. As far as I understand, this model is made for one specific infrared radiation - I could be wrong. Has someone any idea on how to figure out this? I tried looking on google but cound't find a solid answer. I plan to use this for digital transmission. <Q> It's indeed as @Transistor suggests. <S> The centre tab is a thermal pad. <S> Check this Assembly and Handling Guidelines , page 3 <A> In thus case the central conductor is shown in the datasheet as "Anode" as is one of the side pads. <S> While this COULD be a datasheet typo, Lumileds usually provide good quality datasheets and it is likely that both pads are electrically connected, as shown. <S> The use of 3 pads is most likely related to this allowing a standardised Luileds pad specification for manufacturing. <S> The pads are NOT labelled Anode1 Anode2 or similar - there is no indication that they are different electrically. <S> The lead frame image suggests that the areas concerned are electrically common. <S> This is not certain but likely. <S> There is no good reason to doubt the datasheet, and a check may be made with a suitably low voltage Ohm-meter without likely damage. <S> No Vreverse is specified, and a footnote states that the LEDs are not designed to be driven in reverse bias. <S> However, the circled device shown below is not mentioned in the datasheet but is probably a diode or zener or (functionally) both. <S> A diode is often used in this location to prevent Vreverse damage. <S> A zener or similar here protects against ESD damage in reverse breakdown mode and against Vreverse damage in fwd conduction mode. <A> They also use the same package for sometimes multi chip LED. <S> You have to pay attention as sometimes for LEDs within the same series same manufacturer and same package the connection may vary with sometimes the central tab connected to the anode, sometimes the cathode or sometimes not connected at all. <S> I ran in trouble once because of this. <S> The central tab is also used a heatsink as the other answer mentioned.
| The reason being, for this type of specialized LED, manufacturers often use the same package for many types of LEDs to simplify the production, same machines, same process, etc... In a number of Lumiled devices the middle conductor is neither Anode nor Cathode and is used for thermal sinking of the die.
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Using the same components for two 555 timers I am building a circuit to operate a Linear Actuator to open and close a lid. The open and close times are the same. The standard monostable circuit works. There is one switch to tell if the lid is open or closed, it shows 12v+ when open and is grounded when closed. I have built one timer unit. It works to specs. I was wondering, since the timers will be the same, would it be possible to use the same components for both timers. Pins 2 and 3 would use different triggers and the outputs would go to different places as well. I would use jumpers between the two 555 Chips. Basically my thought would be to connect all pins but 2 and 3 in parallel with each other. Would this fly, or am I missing something. It would be impossible (I trust) to trigger both timers at the same time. <Q> The discharge output is low before and after the monostable pulse, so the outputs would remain low unless you triggered both 555 timers simultaneously. <A> No, that couldn't possibly work. <S> You need separate (but the same value) components for each timer. <A> If you dig into the way a 555 chip works inside and what the waveforms on the various pins look like you would come to understand why this is not a feasible idea. <S> With that said there is another possible scheme that could be used. <S> Since the open and close operation are mutually exclusive (i.e. would not be happening at the same time) it would be possible to use the time interval generated by a single 555 to cover both the operations. <S> To do this some additional logic would have to be devised with logic gates or transistor circuits to channel the operation of 555 to the proper function. <S> I highly suspect that the extra logic to use one 555 far exceeds what you have envisioned doing with two 555 chips. <S> So just bite the bullet and build your second timer circuit with the added capacitor and two or three additional resistors.
| The two 555 chips will need to have their own timing components. This idea will not work.
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Charging phone battery with a lower voltage, coming from a bike charger? Since travel by bike a lot, and use my phone very often during those trips, I tried designing a USB phone bikecharger that would use my bike-dynamo that also powers my lamp. Yesterday, I did my first tests. Without load, the circuit would give me 9,3V (even when the wheel was not spinning, which was a surprise to me :) ). With a 47 Ohm load, it would still give 9.3V. This is way too much for USB ofcourse, but 47Ohm is still quite big I think. When a 10Ohm load was applied (I assumed 5V USB-voltage, 500mA charge current --> 10 ohm), the voltage dropped to 3,5V (it kept quite stable, it didn't fluctuate all that much. It seems the wheel speed doesn't influence the output voltage, except for when it spins too slow and the voltage is cut off entirely. I assume this is some kind of protection mechanism in the dynamo?). So here's my question: what would happen is I were to attach a phone to this charger? My guess is that 3.5V is not enough to overcome the internal resistance of the phone battery threshold for the charge circuitry of the phone, so it will be blocked a few ms, but because of the open circuit, the voltage will rise to 5V and that will eventually be enough to charge. So in practice, it will just charge with a lower current (same as charging your phone on the USB-port of a netbook. Those often can't provide enough current, and the phone just charges slowly). Is my guess right? What are the risks of charging a phone-battery with too little voltage? Should I worry about the charger outputting 9V when the phone battery is full? Because there won't be any load anymore then. Thanks in advance! ps. I also checked the signal with an oscilloscope to check the purity of the DC, and it was pretty clean. There were no big spikes and the level was quite stable. (I didn't try to pitch pennies on noise-supressing capacitors ;) ) Edit: added my schematic EDIT2: J3 in the schematic is an LM2596 Buck module, using a MP1584 step-down regulator module. EDIT3: I was doing some tests a few minutes ago, and I guess I had misconfigured the LM2596 (doh...). I think the 9.3V issue is not existent anymore (I've only tested it on a bench power supply right now, but I assume it'll be the same when I test it on my bike again). So now the only question that remains: Is there any risk to applying a voltage too low to the phone? And will my above assumption hold true? EDIT4: Currently testing it with some old phones, with my bike on a treadmill. 5V is constantly stable now for every speed above 6km/h, and charge current is 300mA, also pretty stable (100mA on an older phone). The phones themselves also indicate that they're being charged At first glance, everything seems to be working, but... the charge percentage is barely going up. I'm not sure yet if this is because the batteries are very old and haven't been used in ages, or if there's still something wrong with the circuit. More testing will be needed ;) EDIT5: as pointed out by @Ale...chenski the path between the rectifier bridge and the two elco's is wrong and will be removed. <Q> First of all, the USB is not being fed directly to the phone’s battery. <S> It’s actually being fed to the battery charger circuitry inside the phone. <S> You cannot safely directly connect a normal regulated voltage source to a lithium ion battery without the distinct possibility of an explosion and/or fire! <S> Then, your question becomes what happens when you feed incorrect voltage into a phone’s charger circuitry. <S> This would depend on the specific model, but we can generalize. <S> With too low voltage, such as 3.5 volts, the phone will probably ignore it and refuse to charge. <S> Your circuitry, which you have not really described, is totally unsuited to charging a mobile phone! <S> You have a good chance of seriously damaging its circuitry. <S> You should probably find a commercially produced buck/boost module, capable of taking in the maximum voltage from your dynamo and outputting a stable 5 volts at 500 ma, or whatever your phone needs. <A> Obviously you have not any data of how your phone's USB connection is specified except it's USB. <S> Don't risk your phone, the voltage must be 5V if you haven't better data. <S> Too low voltage doesn't harm - except there's no charging, but 9V probably does. <S> If you can get over 5V with the max current that your phone eats from exact 5V, then have a voltage regulator which drops the extras off, even 7805 with a couple of capacitors can be ok altough <S> it wastes nearly half of the energy. <S> A switching converter would be better in that sense. <S> If it happens that you do not get enough voltage ie. <S> when biking at a reasonable speed your dynamo doesn't give 5V or more with the needed load, you can possibly help it by having a boost capable regulator, but I guess it doesn't help, because the current under 5V jumps up If you must produce at 5V a certain current. <S> My phone has a special ability: It sees if the USB connector hasn't the wanted output capablity and warns me: Use a proper USB charger for faster charging. <S> It still seems to charge. <S> You possibly have the same in your phone. <S> You can test it by adding a series resistor. <S> If your phone can reduce automatically the charging current, it can solve the problem caused by an underpowered dynamo. <S> One thing to consider <S> : Carry an external high capacity USB battery pack. <S> It can be quite big if you have a bike. <S> If you can charge it with mains voltage now and then, you do not need a dynamo based charger. <A> There is no room to guess whether AC-DC adapter for phone idles at 9.3V or else. <S> It is not. <S> This is the beauty of standardization: if the phone says "USB" and has USB connector, it is USB, and the default USB voltage is 5V +-0.5V, period. <S> Assuming that you have a standard bike dynamo generator <S> 12V <S> like this one, nearly everything is right, except for the brutal miswire of AC-DC rectifier bridge. <S> You need to remove C1 and C2 completely. <S> As shown. <S> it makes no sense whatsoever. <S> Then, assuming that the J3 (MP1584 step-down module) is wired correctly and set to provide 5-V output, then everything should be just fine, almost. <S> As soon as your spinning wheel will provide more than 8 V DC after rectification on C3, the output +5V should be stable and phone will be happy (the module needs about 3V excess at input over output to operate). <S> The MP1584 also has "undervoltage lock-out" at 3V Vin; below 3V the circuit will shut down. <S> Unfortunately, it is unclear how the MP1584 module will behave if the input voltage is above 3V but below 8 V. <S> So you need to keep an eye on your voltages and experiment a bit at various spinning rates. <S> Now, if you want your phone to charge a bit faster, connect D+ to D- on USB Type-A receptacle side, and leave them floating. <S> In this way almost any phone will recognize the standard charger signature (called DCP in terms of Battery Charging Specifications v1.2, or preceding "Chinese Federal Standard"), and your phone will take up to 1.5A, likely 900-1000 mA and charge itself faster.
| The charger circuitry adjusts the voltage and current fed to the battery depending on the state of charge. Therefore as soon as you provide +5V on Type-A receptacle, everything should be safe. With too high voltage, such as 9.3 volts, the phone is likely to either shut down to protect itself or be damaged by the excessive voltage.
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How can I make a two voltage source, 4-layer PCB in Eagle? I would like to make 2 different voltage sources on one layer (layer 15).In the picture, the left rectangular area is a 5V polygon and the right one is 12 volts. Is it possible? How much distance should be between the two voltages? <Q> Spacing isn't important with such a close voltage difference. <S> 0.3mm should be more than enough although I would go larger just because. <S> It makes a nice visual distinction between board areas when it's larger. <S> I had 5mm between 3.3V and 120V on one my board, and about 0.5mm between 15V and 120V in some places (which was probably too small to be safe, but it was a special gate-drive circuit and was isolated from the user) but still encountered no issues. <S> If you want to use official standards the term you need to Google is "PCB creepage distance". <S> I assume that's an H-bridge? <S> I guess there's a chance something could spark over if the inductive kick was large enough but you should have flyback diodes anyways to deal with that <S> (or the MOSFET body diode but that is nowhere near as good). <A> Yes you can do that. <S> With Eagle, if my memory serves me correctly, you make a polygon and attach it to the net. <S> It should then automatically connect to the pads & vias it's supposed to connect to, and refrain from connecting to the others. <S> @Toor's comments on spacing are quite correct. <A> As said by others, isolation isn't a big deal here. <S> Give each polygon the name of the assigned signal, and give the right polygon a higher rank. <S> The polygon of the higher rank is drawn first, and the other polygon takes the area which is not covered by the first.
| If you want the isolation to be small and the same on each edge without having to adjust it over and over again, just draw the left polygon so that it overlaps the right one.
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What sets the resolution of an analog resistive sensor? When I read about film or silicon strain gauges their data-sheets mentions about their resolution. But in analog world what sets the resolution? For example if you vary a potentiometer the output voltage changes accordingly so one can say there is relation between the rotation and the output voltage which would be a continous function not quantized. Can you give an example which would explain the resolution concept for strain-gauge? <Q> If we define resolution of an analog measurement as the smallest change which can be detected, there are some limiting factors. <S> In this case, we're talking about change in strain, not the resulting change in resistance, though obviously they are related. <S> For the sensor itself, there are factors such as temperature coefficent, hysteresis, 1/f noise (drift) and Johnson-Nyquist (white) noise. <S> The white noise effect can be mitigated by reducing the bandwidth, but then 1/f noise becomes more important, so even if you have no constraint on the time to take a measurement you can't reduce the noise effect without limit. <S> To the extent you can know the temperature (and to the extent that it's consistent over the element) and you know the strain history you may be able to compensate partially for some of those factors. <A> The resolution is determined by the instrument used to measure the effective resistance. <S> Over the range of measurements you could also talk about the linearity of the sensor. <A> If you are talking about strain gauges, then the problem has to be expanded also to the application. <S> Strain gauges are used in weighing scales, pressure transducers,... <S> they are glued on piece of metal. <S> Things may differ <S> if you use some alloy vs other alloy due to different temperature coeficients, moreover the metal body doesn't always return to it's initial position when unloaded, it has some hysteresis. <S> Many constraints putted together makes a weighing scale to have a finite number of counts/resolution. <S> It's not just a property of strain gauge. <S> Most legal to trade scales are C3000 standard, meaning 3000 count over full scale. <A> Here is a system-analysis for 100 microvolt sensor, into 90dB gain low-noise (3-stage) amplifier, into 10 Hertz RC low pass filter, into 32-bit Analog Digital Converter. <S> The amplifier produces 3 volts PP into the 5vPP ADC. <S> The right-hand numbers tell the resolution story: ---- <S> total noise 4.94 milliVolts <S> RMS (all uncalibratable errors, causing Code Spread in the ADC output binary code) <S> ---- <S> thermal noise 140 microVolts RMS (dominated by first opamp), with lower left plot showing how that opamp and the Rg (resistor to ground) dominate ---- <S> ADC quantization noise 336 picoVolts RMS ---- <S> Power Supply noise: 4.94 millivolt RMS (60 and 120 Hz; see the table in lower-right corner) <S> What is the limit to resolution? <S> The power-supply-rejection of the first operational-amplifier (which the tool set to 80dB at low frequencies).
| A given resistive sensor would be characterized by its precision (the repeatibility of measurements under identical conditions) and its accuracy (how well the change in resistance truly reflects the change in strain).
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Spec says 7-24V, can I use 25V? I'm building an IoT device with a Wemos D1 Mini. It should control a relay, and also measure the temperature. The thing is, my only practical power source comes from my garage door opener which I measured to be 25.1V. What would happen if I then use this powershield rated for 7-24V? If it won't work, what are my options? Maybe I could make my own shield with higher input range, but that seems to be a lot of work.. What are my options? <Q> Better than diodes, use a dc-dc buck converter. <S> For 2-3 dollars (from china) you can get one with up to 36V input and 3 to 5 amps output. <A> Put some diodes in series with the power source. <S> Like 4 or 5 you're probably safe in case the power source exceeds 25.1V. Make sure to pick the right current rating. <S> (The current rating is determined by the power shield). <A> This board uses the chip MP2359. <S> This chip is rated for 4.5V to 24V, however, if you check the chip Maximum rating, it is rated 26V. <S> The input Capacitor is rated 25V. <S> Usually, the manufacturer always keep a good 10% safety margin on top of those ratings, so you have good chances that it would just work fine. <S> Given the price it's not a huge risk anyway. <S> EDIT: <S> Following some comments. <S> Justme pointed out that there is a diode at the input which would have a voltage drop of at least 0.3V bringing the input at 24.8VDC. <S> The 25.1V power supply is also likely to be a 24VDC supply that is under no load given 25.1 is not really a standard supply. <A> If you check the schematic you'll find that the regulator is MP2359. <S> but you shouldn't design based on that. <S> Furthermore, there is a schottky SS24 in series with the supply. <S> It has a forward voltage of < 0.5V depending on current drawn. <S> It still leaves you a tiny bit outside the spec. <S> If you want to be dead certain that everything works you could add a 24V zener or TVS on the input line and let it burn away the extra voltage. <S> Putting a TVS there might not be a bad idea for other reasons, depending on how trusty your supply is. <S> It doesn't sound like it is of particularly high quality.
| Reading the datasheet, it doesn't really specify the supply voltage other than in the summary where it says 24V. Absolute maximum ratings is 26V
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Using enameled copper wire as a capacitor for measuring water level in a tank? I'm trying to make a capacitive water level sensor using two enameled copper wires used in transformers. I've used sanitary silicone to insulate the bottoms of the wires. (Coating is insulated by default.) The distance between them is 0.5 cm, they are submerged in a water tank made out of acrylic glass. As the fluid level rises, capacitance should change, the "sensor" is connected to a IC555 timer which is used to generate a square wave with frequency inversely proportional to the capacitance. Then I'm calculating the time which passes between two rising edges of the pulse using an STM32 microcontroller. I've connected the IC555 timer output to an oscilloscope and when the tank is empty it gives a square wave of a certain frequency, but when the water level rises nothing changes. What did I do wrong? What could be the issue? I'd like to hear someone more experienced on the matter since I'm just a student and I lack experience. Any help would be appreciated, thanks in advance. <Q> I'm trying to make a capacitive water level sensor using two enameled copper wires used in transformers. <S> ... <S> The distance between them is 0.5 cm, they are submerged in a water tank made out of acrylic. <S> As the fluid level arises capacitance should change, the "sensor" is connected to a IC555 timer which is used to generate a square wave with frequency inversely proportional to the capacitance. <S> Unfortunately the capacitance is miniscule and totally inappropriate for use with a 555 timer based measurement. <S> You are dealing with pF. <S> There are potentially a range of relatively easy approaches you might take: <S> Use a sonar acoustic sensor to measure the return from the top of the fluid. <S> Back the parallel wires with a magnetic strip (fridge edge magnet) and measure the time for a coupled pulse to return from the air/fluid transition (very similar to TDR but much slower wave front speed) <S> Use stainless steel piano wire and an acoustic signal pulse to measure the time to air/fluid transition (reflection). <S> Measure the pressure at the bottom of the tank <S> (there are some very sensitive pressure sensors). <S> Bounce a Laser off the air/fluid interface and use a cheap camera chip to measure the angle. <S> This might be a good start to you reading on sensors for measuring fluid levels. <A> Unfortunately no. <S> Several other good answers address the capacitance problem. <S> Another problem is that the lacquer on enamelled copper wire is not guaranteed pinhole free i.e. the insulation is likely not perfect. <S> In winding a transformer or inductor this doesn't matter, since the pinholes are unlikely to coincide and cause a shorted turn. <S> Varnish impregnation after winding also helps to stop corrosion by moisture getting in the pinholes. <S> In your application, if there is even one pinhole in each wire, and the water conducts any of the signal, this is likely to swamp any capacitive effect. <A> I have seen very good results by using a plastic straw filled with a conductive fluid (e.g., salt water) as one plate and the water outside (also conductive) as the other plate. <S> A Colpitts oscillator could be used with this capacitor to detect changes in level of a few hundred micrometers (I gave this as a lab assignment, <S> all 30 student teams succeeded doing this). <S> You need considerable surface area so that the capacitance change becomes significant. <S> In the example I gave what is changing is the size of one of the plates. <S> If you use two conductive 'tubes' instead, what you are changing is the dielectric constant by a factor of 80 if you assume the water is non-conductive, or increasing the size of the plate if the water is conductive (unless you are dealing with deionized water there is always some conductivity present). <S> But a 555 circuit (a relaxation oscillator) will not give you the precision required for measuring the minuscule changes as parasitic capacitances and switching delays will dominate. <S> Use a tuned LC oscillator instead, in which the water capacitor is a considerable portion of the resonant frequency. <A> As already mentioned, capacitance is too low to be sensed. <S> Test if timer circuit detects additional capacitance when you connect electrodes (of empty tank). <S> Only in such case the rise of capacitance (caused by water) will also be noticed by the circuit.
| Enameled copper wire is a poor choice, both because of insulation issues and because of the small surface area and thus low capacitance. Use rectangular electrodes of size, matched to the size of the tank (as large as possible). Use a TDR (Time-Domain Reflectometry) measurement of the parallel wires and measure the reflection from the air/fluid transition.
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Why isn't there a non-conducting core wire for high-frequency coil applications Background The commonly known skin effect formulas are derived and only apply to solid conductors. The commonly used "skin depth" only applies in these cases. It is for this reason that in some applications tubes are used, as these are much more weight-efficient than the same diameter wire at a high-enough frequency. At 1MHz the skin depth of copper wire is 65µm which means that only 40% of the volume of a 1mm diameter wire is carrying 95% of the current, with >35% of it in the outside 20%. From the skin-depth formulas it is known that a lower conductivity material (e.g., aluminum) has a skin depth that is considerably larger than a higher conductivity one (e.g., copper). As the formula predicts, skin depth is inversely proportional to the square root of the conductivity. If we carry this to its logical consequences, it should be the case that for a conducting tube (which has an insulation core) skin depth should be larger than for an equivalent solid conductor. As an alternative intuition a thin-walled insulated-core conductor would have nearly twice the surface area of a solid conductor. So it should asymptotically approach nearly half the resistance at a high enough frequency. In effect, as can be seen in this paper from HB Dwight in 1922 (possible paywall) , the increase in resistance w.r.t. frequency for a tube whose wall thickness is 20% of its diameter is more than a factor of two lower than for a solid wire. From the above curves it can be seen that a tube with t=200µm and d=1mm, due to the increased actual skin depth, should have less than 50% of the impedance increase than a solid d=1mm wire (do note that the curves are normalized w.r.t \$ F / R_{dc} \$ , so interpretation is a bit tricky). Similar effects (although not as dramatic) can be observed with individually-insulated stranded wire. Application In medium-frequency applications, as for example switching power supplies, it is common to use Litz Wire a multi-stranded insulated wire which reduces the losses due to skin effect but becomes less and less effective at higher frequencies (~1MHz) because of the proximity effect and the capacitive coupling of the individual strands. Probably more gains (particularly with respect to proximity effects) could be obtained if there were multiple individual strands embedded around the periphery of a non-conducting core. Question Have I missed something in the theory? If not, why isn't insulated core wire (either tubes or strands around a core) being commercially exploited for high-frequency inductor applications? Addendum As John Birckhead answer points out, flat wire has basically the same advantages with none of the disadvantages (e.g., fill factor). But this leads me to ask: Why isn't insulated-core flat wire being used for these applications? It should have the same advantage of flat wire with nearly half the resistance at high enough frequencies. Are the possible gains inconsequential? <Q> <A> Wikipedia entry for Litz wire contains a direct answer your question of "Why aren't hollow tubes used instead?" <S> : <S> One technique to reduce the resistance is to place more of the conductive material near the surface where the current is by replacing the wire with a hollow copper tube. <S> The larger surface area of the tube conducts the current with much less resistance than a solid wire with the same cross-sectional area would. <S> The tank coils of high power radio transmitters are often made of copper tubing, silver plated on the outside, to reduce resistance. <S> However tubing is not flexible and requires special tools to bend and shape. <S> The article goes on to describe why Litz wire provides an alternate solution. <A> Induction Heating (Industrial) commonly uses hollow copper tubing for the inductor. <S> When you are running 1000 kW or higher, you better believe the copper loss needs to be minimized. <S> Additionally, the hollow core is used for water cooling. <S> Sometimes the copper is called "hollow bar". <S> It comes in rectangular or round. <S> It is not uncommon to order a "mill run" to get the hollow bar and thickness desired. <S> Image from luvata dot com
| No, you are correct in the theory, but your approach leads to an unnecessary increase in volume when compared to using flat wire, which is both easier to manufacture and provides a similar advantage for skin effect and the advantage of volumetric efficiency.
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Is that a center tap tranformer just labelled differently? my transformer has a secondary winding of 0 - 12v - 24v Is it just labelled differently and is it really a center tap where 12v would be the center, 0 would be -12v and 24 would be 12v? Thanks! <Q> Not necessarily. <S> If the transformer is rated to produce full output from either the 12V tap or the 24V tap, it could be that the winding from the 12 to 24V terminals is of a finer gage, since the current that could be drawn from the 24V tap would be half that you could draw at 12V. <S> This you can determine by comparing the resistance of the 0-12 and 12-24 windings. <S> Would that matter for using it as a center-tapped <S> +/- <S> 12V transformer? <S> Not much, the terminal voltage at the high side would be fractionally lower under load, but the overall heating compared to drawing the full power at a single tap would actually be less. <A> Assuming that's the case: The outputs will be AC, so -12V doesn't really have any meaning. <S> The 12V here will be 12V AC, and the 24 will be 24V AC. <S> Based on the labeling, I think it's fairly certain that this is a regular transformer, but I would do some continuity tests just to make sure. <A> 0V is arbitrary as it is floating secondary centre-tapped coil. <S> so you can relabel with 3 choices; <S> 0 : 12Vac : 24Vac 12Vac : <S> 0V : <S> 12Vac <S> +180 <S> deg <S> 24Vac : <S> 12Vac : <S> 0V
| If you make the center tap 0, then "LABEL 0" will be 12V, and "LABEL 24" will also be 12V (but they will be out of phase with each other). I would guess that yes, it is just a regular center tap transformer.
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Question regarding ground when using teensy to control a mosfet switch Despite the fact that I'm pretty new to electronics I'm working on a quite ambitious project which is basically a USB-Midi Controller attached to different devices that send different signals. The common power source is a external 24V Battery which powers a pcb with two buck converters (5V 3A) and a USB Power Delivery controller. The latter powers a USB Hub which powers - amongst other devices - a teensy which is attached to and powers a PCB that mainly serves as user interface with different buttons and faders. There are also two mosfets on this second PCB which are controlled by the Teensy and serve as switches between the 5V supplies and other devices. I hope the following graphic will make things a little bit clearer: I was wondering how to build the circuit of the mosfets. In my opinion it's not neccesary to connect Teensy GND and the buck converters ground since basically the minus of the 24V battery is the common reference potential, but people in a Teensy-related forum told me that I have to connect those two because otherwise there would be a chance that the 24V would roast my 3.3V only Teensy. I don't fully understand this and I'm wondering if connecting them could lead to a ground loop since theyre connected at two points in the project. I hope you guys can help me to get a deeper understanding of this matter. Thank you in advance! <Q> You are using enhancement-type NMOS (as source followers) to provide 5V from a gate drive of 3.3V. <S> That just won't work. <S> Try either switching load ground, using PMOS (with some additional circuitry), or an integrated load switch to do the switching. <S> Now, to your actual question. <S> Assuming that all of your DC-DC converters are of the non-isolated type, yes, it will create a ground loop. <S> But it might be a necessary one. <S> Given power consumption you will not really know what is the potential between the Teensy and your switches without directly connecting them together. <S> This does mean that some of the ground current will be circulating through the Teensy. <S> You can avoid this by either: Using isolated DC-DC converters and connecting the grounds together. <S> or: <S> Using isolators (opto or any other kind) to drive the gates of the mosfets. <S> You can also manage transients in the ground loop currents by adding common-mode chokes to the Teensy power supply. <A> There are two issues only one of which you have mentioned. <S> For current to flow between two circuits there must be a continuous loop path. <S> In this case the Teensy ground MUST be connected to the same ground as the 5 volt power modules that it is controlling (or attempting to). <S> You have shown N channel MOSFETs. <S> To turn an N-channel MOSFET on the gate must be driven to some voltage more positive than the MOSFET source. <S> Typically the gate needs to be 2 or 3 volts above source ( but it can be as much as 5 or even 10 volts) depending on the device chosen. <S> Special “logic FETS” are available That accepts low gate-to-source Drive voltages but even these will typically need at least 1 volt of Gate Drive relative to source. <S> As the source here is meant to be at 5 volts when the MOSFET is on, the gate needs to be driven to say 7 or 8 volts positive relative to ground to turn it on. <S> If the Teensy has a 5 volt supply this will not happen. <S> As the gate can be driven to say 5 volts maximum <S> The MOSFET source terminal can only rise to about 2 or 3 lower than the gate, so you will only get about 3 to 2 volts output. <S> However, as the Teensy has a 3V3 supply, Even if logic MOSFETs were used which needed only one Vault of drive you would still only get about 2 volts of output from the MOSFET. <S> In order to obtain 5 volts output <S> the choices are either: <S> To use a P-channel MOSFETs with the source connected to + 5 volts and drain connected to output, and drive the gates low to turn them on and to 5 volts to turn them off, or To use a gate driver to drive the gates of the N-channel mosfet to higher than 5 volts when the MOSFET is to be turned on. <A> Your Circuit a) connects only the gate of the MOSFET to the Teensy. <S> So, there is no way to be sure what the gate-to-source voltage will be for the MOSFET... <S> you won't be able to reliably turn the MOSFET on and off. <S> You must have some reference point to establish the voltage difference between gate and source. <S> All of your grounds may be connected together whether you like it or not. <S> If you really want galvanic isolation between the Teensy and the MOSFETs then you should consider using optoisolators between them. <S> Again, not enough information was given to provide more detailed suggestions.
| So, there is no way to know what the voltage at the MOSFET's source will be with respect to any voltage on the Teensy. There is not enough information to tell whether you might have a ground loop problem if you connect all of the grounds. First of all, your circuit idea does not work. We don't even know if the grounds from the buck converters are isolated from each other or from the "USB PD Controller".
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How does Internet communication work on a coaxial cable? In buildings, it is common for the same cable (coax) to be used for pay-TV and the Internet. It makes sense that I can transmit several TV channels on the same cable, since it can be modulated by the frequency and sends them. However, in Internet communication, data must be sent and received. There is no cable for the transmission and another for reception. So how is Internet communication on a coaxial cable? <Q> Energy moves through the cable in both directions simultaneously. <A> Internet over CATV is called DOCSIS ( https://en.wikipedia.org/wiki/DOCSIS ). <S> It uses several channels separated by frequency for downstream and upstream. <S> Think about FM radio. <S> How can you have several channels on radio? <S> They just use different frequencies. <S> This is called "Frequency-division multiplexing" ( https://en.wikipedia.org/wiki/Frequency-division_multiplexing ) <S> Here is an article that covers it: https://volpefirm.com/docsis101_rf-fundamentals/ <A> Radio frequency communications operate the same when travelling through a coaxial cable as they do through open-air. <S> They are just shielded from outside interference (called ingress) and leakage (called egress). <S> As such, signals of differing frequencies can co-exist, with each travelling in different directions. <S> Amplification, however, is a different story. <S> Since amplifiers work in only one direction, the incoming and outgoing signals need to be separated when amplification is necessary. <S> This is performed by a device called a diplex filter , which is sort of like a splitter/combiner that splits/ <S> combines based upon the frequency of the signal. <S> In legacy CATV systems, downstream signals were generally about 50 MHz (around the bottom of analog channel 2) and up, while upstream signals were from around 5 MHz to 40 MHz. <S> An amplifier assembly would (basically) consist of a diplex filter on one end separating the two frequency ranges, followed by an amplifier for each frequency range oriented in opposite directions, and then a second diplex filter to merge the two frequency ranges to its original full spectrum signal.
| Just as different video signals are modulated on different channel frequencies, incoming and outgoing data streams are modulated on different carrier frequencies, and pass each other without interference.
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How to calculate PCB track width on proteus? The current in my circuit is 60A and voltage is 36V. How wide should the tracks on my PCB be? <Q> You'll need it wide, and thick if you can (2 oz copper vs 1 oz copper). <S> There are calculators like this one <S> https://www.4pcb.com/trace-width-calculator.html that take into account <S> current, copper thickness, allowed heat rise, internal or external trace, trace length into account and calculate a trace width. <S> Other board houses have calculators also. <A> If you do this, you will melt the track simulate this circuit – <S> Schematic created using CircuitLab <A> This is not determined by board houses or PCB design software, but by IPC standards. <S> IPC-2221A <S> Section 6.2 states: <S> The minimum width and thickness of conductors on the finished board shall be determined primarily on the basis of the current-carrying capacity required, and the maximum permissible conductor temperature rise. <S> The formula provided in the standard is: $$ <S> I= <S> k\Delta <S> T^{0.44}A^{0.725}$$ <S> where: \$I\$ is the current handling requirement in amperes <S> \$\Delta T\$ is the allowed temperature rise of the trace in degrees <S> Celsius \$A\$ is the the cross-sectional area of the trace in square mils <S> \$k\$ <S> is a constant which equals 0.048 for outer layer traces, or 0.024 for inner layer traces. <S> If you don't want to do the math by hand, there are certainly plenty of calculators out there on the net. <S> My point is that it is not a custom value depending on your board house, your design software, or anything of that sort. <S> They will all point to this formula, so I simply prefer to skip the middle man and calculate the required trace widths myself from the start.
| The cross-sectional area is determined by your trace width and your copper weight (measured in mils, millimeters, or oz/ft^2).
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Why is my crystal oscillator not working with KSZ8999 switch? I am using KSZ8999 ethernet switch from Micrel and I have problem with 25 MHz external oscillator. There is constant 2,1 VDC on output pin which is VDD for the switch, so crystal is not oscillating.There should be no problem with crystal, because I tried to connect it to other design and it worked fine. I also tried using different values of load capacitors but with no result. There is also no short circuit between VDD and crystal pin.Can you please help me find out, what could cause this problem? Thank you for your answers. R15 is now 510k, R16 is 0R, C41 and C42 is 15p. Crystal MPN is now GC2500081. Sorry for the weird schematic symbol, it was downloaded this way. <Q> In the clip below from the reference design schematic in the demo board user guide , there is a 500K bias resistor across the crystal. <S> To fix the problem, you can probably solder an 0201 resistor across the package pins or a larger resistor across the crystal. <A> It's pretty hard for us to troubleshoot this with very little information. <S> Assuming you are correctly measuring the output pin with low enough loading, there are several possibilities. <S> The load capacitors could be grossly wrong value (eval board uses18pF implying their crystal is designed for maybe 12-16pF load). <S> There could be some issue with leakage (unlikely, but possible) orPCB <S> construction (open, short, bad via etc.) <S> that is preventing the oscillator from working. <S> Shorts under SMT crystals are a possibility. <S> If the leakage was comparable to the bias resistor then it might not start. <S> As crj11 says, you need the bias resistor (some ICs have it internal, this one does not). <S> The IC could be damaged <S> I would start by removing the crystal and load capacitors and replacing the crystal (and load caps) with a different one with compatible load capacitance of a similar level to the eval board, having a good look at the (cleaned) PCB under a microscope prior to re-assembly. <S> Also make sure the solder and flux being used is appropriate. <S> SMT capacitors are unmarked so checking the value with a meter prior to installation wouldn't hurt. <S> Mistakes happen and an 18nF capacitor looks similar to an 18pF capacitor (usually the body is a bit darker in colour). <S> Double check the bias resistor value as crj11 says, if it is too low (or too high) <S> the oscillator will not work. <S> Pierce oscillator (from this document: <A> Even that I didn't want to believe it, truth is that both switches are probably damaged. <S> I measured only 50 and 130 ohms on these chips respectively between X1 and GND which is much lower resistance that it should be. <S> I measured same resistance even when I disconnected pin from PCB, so the problem is inside IC.Anyway <S> thank you all for your advices.
| The crystal could be defective or incompatible with the chip
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Are the AVR GPIO registers used by AVRGCC? Most (all?) of the AVR family controllers provide some special GPIO registers. 4.8.1 General Purpose I/O Registers The lowest 16 I/O Memory addresses is reserved for General Purpose I/O Registers. These registers can be used for storing information, and they are particularly useful for storing global variables and flags, as they are directly bit-accessible using the SBI, CBI, SBIS, and SBIC instructions. Q: Are those registers actually used by compilers, especially AVRGCC? Or would one be free to use those registers in some inline ASM for example without messing things up? After some digging, I haven't found anything conclusive. Background: Controller: XMEGA128A1 In a recent project, I discovered that an uninitialized pointer to a struct (size = 9 bytes) was dereferenced (and "written" to), potentially corrupting data at address 0x0000. This address is located in the I/O memory section instead of the SRAM area. The first 16 bytes are used by the 16 x 8-bit GPIO registers available for this controller. As I discovered this bug (that has actually been present for some time) the question came up, why there were no noticeable ramifications. So it would be interesting to know, if those registers are used at all or if we just got lucky so far. <Q> Peripheral registers are not accessed by GCC unless you specifically write code to access them. <S> There is a difference between CPU registers and peripheral registers. <S> The CPU registers are in the CPU core. <S> Peripheral registers reside in the peripheral blocks. <S> Peripheral registers are only accessed when you specifically write code to access them. <S> //EXAMPLEPORTA <S> = <S> 0xFF <S> ;//set all bits on port A to 1, writes to peripheral register PORTA //EXAMPLEX = <S> ((Y+Z)*76+P);//Calculate something, doesnt use peripheral registers. <S> If there are registers are implemented in your device and they don't connect to any used IO pins then you can probably use them as scratch registers without any problems. <A> No. <S> The register macros are all defined in the io.h headers to point to some location in memory. <S> If you don't expressly use the memory locations, they will remain unused. <S> I have compiled code for numerous AVR chips at every optimization level, and the compiler has never automatically put any of my variable in the GPIO registers. <S> That said, you should manually use those registers in place of any global variables. <S> I've also found it to significantly increase efficiency in terms of program performance and flash memory use, especially when using volatile variables with an ISR. <S> // <S> Do this#define my_global_var <S> GPIO1 <S> // <S> Or equivalent register definition// <S> Instead of thisstatic volatile uint8_t my_global_var = 0 <S> You can actually do this with any of the available registers for chip peripherals. <S> For example, if you aren't going to use any of the hardware pin change interrupts, the pin mask registers are free to use for global variables. <S> // <S> Not using the pin change interrupts, so this mask register is free to <S> use#define my_global_var <S> PCMSK0 Just note the width of these registers, as some of them will only have a portion of the bits available for storage. <S> The datasheet will say something like bits 7:4 will always read 0 <S> If you only need to count to 15, 4 usable bits of a register will work fine. <S> Also note the power-on-reset value of the register; some of them are not 0! <S> Also, this is not the same as using the register C keyword , which is a hint for the compiler to put a variable in a processor register instead of RAM for quicker access. <S> The compiler does this automatically as part of its optimization, using the "working registers" mentioned in the datasheet. <A> In C, an uninitialized pointer is called a NULL pointer. <S> The value of a null pointer must be a constant with the value zero. <S> What you are seeing is not a bug, it is the standard behavior for a pointer in C. Dereferencing <S> a null pointer leads to undefined behavior, so you should be very sure that a pointer is not null before using it. <A> Unlikely that compilers would use these, because these are connected to <S> I/O ports. <S> You could use those registers that are not used for I/ <S> O on that particular CPU or board, <S> but you'd need compiler options to specify which ones they are, they aren't really useful for passing values between functions (because then the function call ABI would become dependent on the list of usable low-address registers), and only a tiny fraction of programs would benefit in a measurable way. <S> So, not worth it for compilers to support. <A> Those registers are like PORTA, PORTB etc. <S> and not like r0 ... <S> r31 CPU registers. <S> So no, C compiler will not use them. <S> However, you can still make use of those registers if you treat them like global variables and map them to a struct like this: <S> #include <avr/io.h>typedef struct{ <S> unsigned bit0: 1; unsigned bit1: 1; <S> unsigned bit2: 1; <S> unsigned bit3: 1; <S> unsigned bit4: 1; unsigned bit5: 1; <S> unsigned bit6: 1; <S> unsigned bit7: <S> 1;}Bits;int main(void){ volatile Bits *bits = <S> ((volatile Bits*)_SFR_MEM_ADDR(GPIOR0)); bits->bit3 = 1; return 0;} This code will result in only one SBI instruction: sbi 0x00, <S> 3 While doing the same (setting bit 3) in normal global variable will result in three instructions: uint8_t global;int <S> main(void) <S> { global |= 8; 228: 80 91 00 20 lds r24, 0x2000 ; <S> 0x802000 <_edata> 22c: 88 <S> 60 ori r24, 0x08 ; 8 22e: 80 93 00 20 sts 0x2000, r24 ; 0x802000 <_edata <S> > <S> return 0;}
| Peripheral registers are not used by GCC to compute any logic or math expressions or to manage the stack.
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Why logic low Inputs to a logic gate need to be grounded For example if we take a 2 input AND gate . And want to apply 0 and 1 to the input terminals ..one terminal will be connected to a voltage source . But the other terminal where logic 0 is to be given , why do we ground it . Logic 0 means low voltage than logic 1 . Why can't we just let the other terminal hang freely in air ....not connected to anything ..Will it not have zero voltage when in contact with air ? <Q> Will it not have zero voltage when in contact with air ? <S> No, it will not. <S> Imagine an isolated node (input of an AND gate). <S> If some charge get trapped on that node it cannot go anywhere. <S> The node will have some voltage and a charge. <S> That charge can be enough to open a MOSFET (and most AND gates use MOSFETs) which will influence the behavior of the AND gate. <S> Leaving logic gate inputs open / floating is an extremely bad habit as that <S> makes the input undefined . <S> The connection can also be made through a resistor as long as the charge is not trapped on the gate's input. <S> But what if I simply discharge the input and then let it float? <S> Won't work either because there are ALWAYS <S> leakage currents and non-ideal isolators (like MOSFET gate oxide) which cannot prevent some charge to accumulate on the input. <S> So after some time there will still be some unknown charge on that input giving it some unknown voltage. <S> Who knows what can happen then? <S> No one knows, so the input needs a connection. <A> The reason is electromagnetic fields. <S> Instead of seeing current flowing in wires, another way to see reality is as a universal sea of electromagnetism which is, um, channelled by wires and components. <S> It's everywhere. <S> Our electrical circuits operate in this context. <S> In fact I have this joke that electronic circuit design is about building radios that may do other things. <S> This is very noticeable with any audio circuit where you can randomly and very easily pick up radio stations just by letting wires "hang" in the air, or having unshielded wires, or just touching bits of it with a finger. <S> High impedance inputs like logic gates only need a tiny current to drive them (in that case, to logic 1 or 0). <S> Dipping into the ocean of electromagnetism as they are, they will easily drive as they pick up EM fluctuations, like little radio receivers (not really "like"- they <S> are little radio receivers). <S> They will oscillate. <S> Current will flow in them. <S> This means the transistor spends most of its time in a state of somewhere between on and off, which means some current is flowing in its output stage, which means two things; <S> Firstly it will produce random results at the output. <S> Secondly, it will dissipate power as heat. <S> So to avoid this, you "lock" it to a steady voltage from a low impedance source to ensure that the stray fields of this ocean of electromagnetism we live in cannot drive it. <S> Broadly speaking, much of circuit design involves avoiding building radios. <S> Unless you actually want a radio. <S> Which is presumably why the point where "electrics" became "electronics" was when humanity started building radios. <A> Unconnected inputs can float, randomly switching between 0 and 1, especially in modern electronics where it takes less than 1uA of current to drive an input. <S> Floating inputs can cause the extra switching after the input. <S> In a CMOS chip, switching states uses current. <S> To keep things at low power, one wants to avoid unneeded switching. <A> Earth is a very big conductor . <S> But now you have a wire that is hanging in the air. <S> The electrons see an infinite potential barrier at the free end so they can't escape into the air. <S> It's like a wall of infinite potential that you can't cross. <S> Now if an electromagnetic wave comes then it induces oscillating currents on the surface and inside the wire , also producing an electric field and hence a voltage drop at base terminal of the transistor making it's input 1 .
| The input does not have to be grounded, it can also be connected to the supply voltage. When you ground the base terminal , all the charge flows to the ground and you have 0 volts at the base and transistor is in cut off region.
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Achieving MPPT of a solar panel with LM2596 im having some trouble with DC converters in a solar system.I have a solar panel of 21V (Voc) and 1.33 (Isc).I have studied that with a DC buck converter with input regulation (Feedback loop taken from the input) would result in a fixed output and equal to the MPP. The thing is that i have a LM2596 adjustable regulator , and i was thinking about soldering the resistor divider taken from the output of the original topology, and take the input as a source.Could I get the MPP voltage and current?If yes, how could i calculate? I attach the datasheet from Texas. http://www.ti.com/lit/ds/symlink/lm2596.pdf Original Topology from Texas\ Edit: The objetive is to charge a battery pack of 3 Li-Ion cells efficiently, firstly mppt step is where my doubts are. <Q> Although MPPT trackers are based around typically a buck converter, there's additional actions from a control loop, in hardware or programming, that actively tracks the point where maximum power is being drawn from the cell - this varies a little with solar intensity, so you might find a fixed voltage that is good enough for most conditions, but it also varies a lot with cell temperature, so the controller does need to be able to compensate. <S> Since the solar intensity and cell temperature can vary, an MPPT tracker constantly adjusts the set point up and down, and compares the power generated, and aims to settle at the peak. <S> To be able to vary the input, the controller changes the duty cycle of the buck converter, but obviously this assumes that the load (the battery pack) is capable of taking the full output, and once the battery reaches full charge the controller needs to limit the output voltage too, and the power available at the solar cell <S> goes unused. <S> Regulating only the input voltage would risk overcharging. <S> Increasing the duty cycle increases the load and decreases the voltage at the solar cell, and a simple regulator wired with the feedback at the input would do the opposite. <S> You're far batter off getting a dedicated MPPT controller. <A> If you want to do MPPT on your solar panel, perhaps the easier way is to use devices that are made for that . <S> One way of doing so is to use a buck topology to which the switching is controlled by a MCU and to write some simple MPPT algo. <A> The MPP is not a fixed point but is highly depending on the amount of light shining on the panel. <S> A MPP-Tracker is always checking the input power, slightly changing the point on the IU-curve and checking the resulting power again. <S> By comparison of the power values the Tracker is able to determin, if the MPP is reached or if the point on the curve has to be shifted a bit. <S> This means, that you are not able do set a fixed value for your converter. <A> If you put no load on a solar panel at all, the voltage is maximum but the current is zero. <S> If you short out a solar panel, the current is maximum but the voltage is zero. <S> The power is the product of the current and the voltage. <S> The point of MPP tracking is to figure out the amount of load on the solar panel that maximizes the power produced by the solar panel. <S> Since different MPPs will result in different voltages produced by the solar panel, this requires some kind of voltage converter that can take whatever voltage the MPP happens to be and make it useful. <S> If, for example, you are using 12V batteries and a solar panel whose MPP tends to be around 26 volts, a buck converter will do. <S> But you need something to adjust the load the converter places on the panel to find the MPP. <S> MPP trackers typically have software that varies the converter parameters and measures the voltage across the panel and the current it is supplying to discover the configuration that maximizes the power delivered. <S> That's what MPPT is. <S> This requires discovery as the light on the panel changes. <S> Sometimes it is brighter, sometimes it is dimmer. <S> This results in changes to the maximum power point that must be discovered by varying the load the converter places on the panel.
| Without any device to actually do the MPP tracking, you will not track the MPP. I doubt that would work, the output will start to oscillate trying to compensate the error.
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How to measure voltage of each bank in a Marx Generator? If I have a 10 stage Marx generator with each stage at 2.5kV, when I erect all the stages the last stage's negative terminal will be at 22.5kV with respect to ground. If I want a voltage measurement of these capacitors what is the best way to do this? I am worried about violating the standoff internal to my differential probes since the probes would all share a common ground on the supply/measurement side. <Q> Make e.g. a voltage divider with 30 resistors that rated for 1kV each. <S> When needed, use a capacitor parallel to each resistor to compensate for parasitic capacitance of the resistors. <S> Benefit of using multiple resistors is that a high resistance can easier (lower cost/availability) be achieved and power dissipation is distributed over more resistors. <A> VR37000003005JR500 <S> 30MΩ <S> 3.5kV working voltage <S> IEC 60950-1-07 PCB Pollution Level 1 creepage 4mm <S> /kV <S> > <S> 1kV <S> Use ~ 1cm air gap around the body to nearest part or above cellulose or dry wood substrate. <A> You can decide on the divider ratio to use used. <S> Typically, a hollow perspex cylinder is used with a dilute solution of \$\text{CuSO}_4\$ using de-ionized water. <S> For example, in this case you might use a cylinder that is \$30\:\text{cm}\$ long and \$1\frac14\:\text{cm}\$ in diameter. <S> (Increasing the cross-sectional area improves the non-inductive behavior.) <S> The cylinder design also suppresses the effect of coupling capacitance with the surroundings. <S> Advantages for using a solution of \$\text{CuSO}_4\$ <S> is that it is easy to adjust to the resistance you want for the divider (given some other value you've also chosen for the non-inductive carbon film resistors used in parallel, perhaps, to make up the other resistor of the divider) <S> , it can withstand many megavolts across it, and you can make it in a variety of convenient shapes. <S> Enclose the non-inductive carbon film resistors (use a number of them in parallel for the value you want -- do not use just one by itself) in an aluminum cylinder that extends partway up the outside of the perspex cylinder to provide shielding from EMI for the non-inductive, parallel carbon film resistor group. <S> Look these up on the web. <S> They are built and used very often for Marx generator voltage measurements in laboratories.
| For measuring these kinds of high voltages, you might consider using a shielded copper sulphate ( \$\text{CuSO}_4\$ ) solution in a cylinder of perspex as one of the resistors in a resistive voltage divider.
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In which way proportional valves are controlled solely by current? Proportional valves are drived by current because we need a steady force and since resistance changes with temperature if they were driven by voltage this force wouldn't be steady. Ok, this is what I read. What I don't understand is how can a device be driven only by current and not by voltage if voltage is the cause of current? <Q> So you should adjust the voltage in order to maintain a certain current if you want the valve "position" to be relatively stable. <S> That adjustment is done automatically by a "constant current" circuit. <S> In the case of an electromagnet as in such a valve (or a relay for that matter) the mechanical force is proportional to the current through the coil <S> **. <S> The voltage, for a given current, will change depending on the temperature of the coil, so a constant current circuit should not only start off with a voltage dependent on the coil temperature, but it should increase the voltage across the coil as the coil self-heats. <S> ** <S> There will also be some change with the temperature of the materials involved in the magnetic circuit, but usually that's relatively small. <A> A voltage regulator allows the load to take the current it wants, while forcing the terminal voltage to remain constant. <S> A current regulator allows the load to determine the terminal voltage it wants, while forcing the current to remain constant. <S> Either can be implemented with a mainly voltage-output stage (like an emitter or source follower) or mainly current-output stage (like a collector or drain). <S> The only difference is which of the output voltage or output current is sensed and fed-back to control the output stage. <S> To address your main point, voltage is not the 'cause of current'. <S> At least, not in the sense that stops 'current is the cause of voltage' also being true at the same time. <S> You set up the conditions, and current flows. <S> While you can calculate numerically what's happening, arguing around the word 'cause' is going to confuse you. <A> A voltage regulator will adjust the current while keeping the voltage constant. <S> A current regulator adjusts the voltage to keep the current constant.
| "Controlled by current" does not mean that no voltage is required (although that might be true in some specialized situations), it means that current is what determines the action of the valve.
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Over-voltage protection system for a 7 - 20 A Load with 4 - 5.5 V output from PSU The power supply used on the product, is a DC-DC buck controller 24 V in (3 - 5.5 V out). I need some suggestions to help protect a load (LED Display) which draws roughly 5 - 15 A the power supply on the product outputs 5.5 V and is capable of delivering 18 A - 25 A, the IC on our PSU can fail from time to time and cause over-voltages that destroy the LED drivers and LED displays. I've looked into Zener clamping circuits, SCR crowbar circuits and Varistors, But I don't know how well and how to calculate whether they would survive with the high current flow on the output of the PSU. Any ideas? <Q> I was thinking along the same lines as Huisman — an electronic circuit breaker for the input that's triggered by the output overvoltage — and came up with the following circuit. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The 555 is a good building block for this, because it contains a flip-flop and a couple of voltage comparators. <S> Here's a brief circuit description: R1, D1 and C1 establish <S> the power supply and reference levels for the 555 at 5.1V. R1 can supply about 5mA. R2 and C2 create a trigger pulse that makes sure that the 555 starts up in the "triggered" state. <S> M1 is the main power switch, and Q1, R3, R4 and R5 drive it. <S> Q1 serves as a switchable 1-mA current sink, which develops 10V across R3 to drive M1's gate, independent of the actual supply voltage. <S> When the 555's "discharge" pin is grounded, this circuit is switched off. <S> R5 and R6 establish the trip point for the output voltage. <S> The 555 will shut off the output if the "threshold" pin goes above \$\frac{2}{3}5.1V = 3.4V\$ , and with these resistor values, that means if the output voltage goes above 6.0V. C4 provides a 22µs time constant to filter out any fast glitches that might cause "nuisance" trips. <S> To restart the circuit, either interrupt the input power, or temporarily ground the "trigger" pin of the 555 ( <S> e.g., with a momentary switch). <A> <A> You have powerful PSU, so protective circuit have to deal with large current and power. <S> Zener diode and varisor are clamping-type solutions. <S> Being activated they will dissipate huge amount of power due to high (higher than nominal) voltage. <S> That is why they are inapplicable. <S> SCR crowbar, on the over side, is very well suited for the purpose for several reasons: low voltage drop in open state (low power dissipation even under large current), load will not suffer from overvoltage SCRs <S> are capable of carrying huge currents for short period of time (easy to select proper part). <S> Of course, there must be a fuse at the input of DC-DC converter, that will be burned by opened SCR in case of converter malfunction.
| Make a "smart" device that senses the output voltage of the PSU and interrupts the 24V input of the PSU in case of an overvoltage at the output.
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why speed is not directly proportional to torque in dc machine? I have a conceptual doubt about torque-speed relation in a DC motors. It's probably a gap in my thinking but I'm posting this question anyway.As per me "in case of dc machine we see that the armature rotates because of the developed torque.so from the physical point of view it leads that if armature gets more torque then it will rotate at high speed; i.e. speed directly proportional to torque."BUT why the reverse thing happens????? why speed is inverse proportional to torque....?/?pls, explain physically...... <Q> All of this applies to permanent magnet DC brushed motors -- different kinds of motors behave differently. <S> Be warned. <S> Speed is not proportional to torque. <S> Rotational acceleration is proportional to torque over moment of inertia, though: \$\alpha = \frac{T}{J}\$ , <S> where \$\alpha\$ <S> is acceleration in \$\frac{\mathrm{radians}}{\mathrm{sec}^2}\$ , \$T\$ <S> is torque in N-m/s, and \$J\$ is moment of inertia in \$\frac{\mathrm{kg}}{\mathrm{m^2}}\$ . <S> Armature current is a function of the motor's back EMF, the applied voltage, and the armature resistance: \$i_a = <S> \frac{V_a - E_{back}}{R_a}k_t\$ , where \$i_a\$ is the armature current, \$R_a\$ is the armature resistance, \$V_a\$ is the applied armature voltage, and \$E_{back}\$ is the back EMF <S> Back EMF is a function of the motor's speed: \$E_{back} = <S> \omega_a k_t\$ , where \$\omega_a\$ is the armature speed and \$k_t\$ is the motor's torque constant. <S> Torque is a function of the motor's armature current: <S> \$T_a = i_a k_t\$ <S> If you put that all together, you'll find that a DC motor with a constant voltage on it will "try" to spin up to a speed proportional to the applied voltage, and that its torque (and current) will be proportional to how much slower it is than that ideal unloaded speed. <S> A DC motor that's driven with a constant current will behave differently than one driven with a constant voltage -- and it's fairly common in servo applications to drive a motor with a current command, rather than a voltage command. <S> But it'll still follow the equations above. <A> in case of dc machine we see that the armature rotates because of the developed torque. <S> So from the physical point of view it leads that if armature gets more torque then it will rotate at high speed. <S> I do not see how the first part leads into the second part at all. <S> You must remember two things: the faster the motor spins, the more back-EMF (a type of voltage) is produced. <S> more current flowing through the motor windings means a stronger magnetic field which means more torque <S> If you apply load torque to the motor to slow it down, the BEMF generated is reduced <S> so there is an excess in supply voltage which pushes more current into the motor. <S> The amount of current pushed into the motor produces an extra voltage drop across the motor's internal resistance. <S> The current stops increasing when this resistive voltage drop plus the new, reduced BEMF equal the power supply voltage so that everything balances out again. <S> More load torque->slower <S> speed->less BEMF->more current->more output torque (which you would need to spin the higher load torque to begin with). <S> Therefore, for a given supply voltage, you get more torque at low speeds and less torque at high speeds. <A> The answers are excellent but if you want a simplified understanding, consider this. <S> For PMDC motors, Torque is proportional to current to charge the electromagnetic field. <S> RPM is proportional to the applied voltage above stiction starting voltage with no load. <S> The drop in RPM as load increases causes current to rise since the RPM drops the internally generated BEMF drops and <S> the difference with the applied voltage draws more current or I= ΔV/DCR, motor coil DCR (Ohms). <S> Accelerate is just Newton's 2nd Law applied to rotational acceleration. <S> \$a= <S> F/m\$ For Field induced motors the field current also controls torque from the BEMF voltage it creates and thus the difference internal voltage, which has various configurations of series and shunt wound motors.
| The speed a motor spins at is the speed where the back EMF perfectly cancels out the power supply voltage.
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How to Turn off a PIC Microcontroller in high capacitance circuit? Below image is the input voltage VDD to PIC16F18326 . I am turning off (Switching off the mains supply) my circuit at point A (see image). Due to large capacitance at load, which is required in my circuit to reduce large current ripples, the VDD to microcontroller is OFF after 8Sec. But I want to turn off my microcontroller immediately (within 1 sec) after I turn off the mains. I have tried to change the capacitance placed across the VDD. No big changes in the time taken to turn off. The VDD is supplied and controlled by a LinkSwitch ( LNK304 ) which is fed from the output. I tried to control the feedback FB of LNK304. But the large capacitors at the output has high influence on this as well. Any simple idea, how to Pull Down this VDD when my I switch off the mains? (Any suggestion on Addition of a component at VDD? for eg. Logic level MOSFET?) Thanks in advance. update: µC schematic (not complete) (VDD is supplied by LNK304) <Q> You don't need to cut off the power - you can simply hold the reset low. <S> MCU will be turned off then. <S> There are monitoring circuits that will do that for you, like ADM6713 for example. <S> It will disable MCU once voltage will slightly drop. <A> Monitor the mains by using high value restive n/w & signal conditioning. <S> Add Series MOSFET switch to cut down the VDD for uc. <A> A simple solution is to just add a resistor in parallel with the capacitors. <S> It will always burn power when the device is on, so it's a no-go for low power applications, but it will discharge the capacitors more quickly. <S> At 1.8 V, a 500 ohm resistor would consume 3.6 mA = 6.5 mW, but it would change the time constant for 20 uF to about 10 ms. <A> Depending on the nature of why you're trying to make sure the uC shuts down quickly, I would try one of three things: 1) Use a diode on VDD and have your sense/ reset circuit before the diode - especially if you can compensate for the voltage drop and not impact stability. <S> This should ensure that once the capacitors are the only thing supplying current, you can still watch the regulator's output drop and shut down accordingly. <S> 2) Use a different regulator. <S> A lot of switching regulators have a "PWR_GD" pin, or something to that effect to signal that the regulator is stable and ready. <S> This might also help avoid issues on start-up. <S> 3) Create an isolated feedback circuit on the mains side. <S> This route would probably be the most tedious since you need to protect the optocoupler, and if it ever craps out it'll keep you from operating.
| If you use an optocoupler/ isolator driven off of the rectifier with some basic capacitance, you should be able to tell when mains power disappears very quickly.
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What can happen if 24vdc is applied to an Ethernet port? We have a device with two RJ45 ports. One of these ports is for receiving 24vdc on pins 7(-) and 8(+) to power the device. The other RJ45 port is an Ethernet connection. The ports are of course labelled to advise against such a thing, but is there guaranteed damage if someone accidentally plugs the RJ45 power cord into the Ethernet connection. Or, would it depend entirely on how robust the implementation of the NIC card is? This strange situation is due to this being an engineered system of several different products from different companies. <Q> Pretty sure that 24V connected to pins expecting 3V or 5V is going to damage something. <S> Starting with the magnetics in the ethernet connector. <A> If it is Gigabit or at least uses a connector including magnetics for Gigabit (which may be the cheapest option), pin 7/8 is a bad idea, because these form a pair. <S> It is completely normal to have 48–56V between different pairs for Power over Ethernet, but pairs are 1-2, 3-6, 4-5 and 7-8, with PoE using either 3-6/4-5 or 1-2/7-8. <S> The resistance between wires in a pair is usually rather low. <S> Your options to make this safe: a soft-start power supply that tests the resistance between the outputs, and turns off if it is too low. <S> a custom board with two RJ45 plugs in the right distance as edge connectors. <S> This way, you get a double connector that cannot be inserted the wrong way. <S> 2a. <S> a custom board with a PoE extractor and 24V regulator. <S> Same, and eliminates the need to ever connect a power supply there. <A> Yes. <S> Using pins 7 & 8 in this way (pair 4 of an ethernet connector) would almost surely guarantee damage to the Ethernet magnetics and associated traces. <S> Unless the device only works at 10/100 <S> BaseT speeds (which does not use, need, or require this pair) <S> there is very little a designer can do to avoid such damage and still remain within Ethernet specifications. <S> In the unlikely case that the designer chose to add TVS diodes on the field side of the magnetics, which is generally not recommended as this counterintuitively would tend to reduce ESD reliability and degrades signal integrity, you would destroy these diodes together with the magnetics. <S> Of course that depends on how much current the power supply can provide, but for the sturdier PoE magnetics, a maximum of 350mA for common-mode current in a pair is the maximum specification but the differential mode is normally specified at less than 10mA. Given the <2Ω winding DC impedance <S> , you could be talking as much as 12A going through the magnetics. <S> Using whole pairs instead (e.g., 7&8 for Vdd- and 4&5 for Vdd+) is closer to the normal way for PoE devices to transfer power ( PoE 10/100 mode B ) <S> and, if the device is new enough, these common modes are isolated by capacitors in the RJ-45 magnetic jack or in components associate with discrete magnetics. <S> But the damage will mostly be relegated just to the RJ-45 jack, or to the jack, magnetics, and associated PCB traces--if using discrete magnetics. <S> Any transients that make it to the other side of the magnetics are unlikely to transfer enough power to destroy the ESD-protection diodes in the PHY ICs. <A> Ethernet magnetics look like this: <S> Source: https://www.analog.com/media/en/technical-documentation/application-notes/EE-269.pdf <S> And like this for regular POE, the difference is the rectifier: <S> Source: POE external Magnetics Alternative to magnetic RJ45 jack <S> ieee803af <S> The difference is in how the powered device is connected to pins 7\8 and 4\5. <S> (Pins 1\2 and 3\6 have transformers on the end <S> so if you send in the same voltage on both pins (like the 1/2 pair), nothing will happen because transformers block DC.) <S> Some devices have a DC to DC converter that aren't true POE devices, the ones that are POE, have a rectifier to avoid problems with polarity. <S> The other RJ45 port is an Ethernet connection. <S> The ports are of course labelled to advise against such a thing, but is there guaranteed damage if someone accidentally plugs the RJ45 power cord into the Ethernet connection. <S> Or, would it depend entirely on how robust the implementation of the NIC card is? <S> If it has a rectifier, then the rectifier should take care of the polarity.
| So it depends on what the implementation of the powered device is, if you send it the wrong voltage it could burn out the components if the polarity was wrong for your device.
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Polarity of Ultrasonic Sensor I see some ultrasonic sensors have legs with different length, such as below. Does it means they have polarity? Are there any differences taking different leg as the signal pin? <Q> As people have already indicated in the comments, from the perspective of the actual ultrasound signal (transmitted and/or received) it doesn't make any difference at all. <S> But my intuition tells me -- and small features in the drawing seem to indicate -- that one of these terminals is probably tied to the outer shield. <S> Without an actual part number <S> and/or datasheet <S> we have no way to be certain. <S> But what we know for sure is that this part has a shield, and the shield needs to be connected to something to be useful. <S> If this is a 2-pin device, then yes it's a safe bet that one of these terminals is connected to shield and the other isn't, so yes polarity matters. <A> Other answer mention that it matter if transducer have shield, but it also matter if it doesnt. <S> When you use one transducer it usually doesnt matter which way you use it, but if you want to use more of them in array it does matter that same voltage will result in same acoustic wave and not inverted one. <A> Regarding why amplitude could be different for different polarity, it can also depend on the geometry of the transducer. <S> Though as an AC driven device it is intended to have symmetric (+/- <S> ) <S> response, it may not be exactly the case. <S> So pushing in one direction (+) may meet greater mechanical resistance/impedance than going in the other direction (-).
| Difference is that when terminal one has higher (lower) voltage than terminal two then piezo element will flex outward (inward) or other way around.
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Floating Source Voltage Difference to Earth I was reading about floating voltage on web but there is something that I couldn't understand. I attached photo of circuit. As you see it shows 60v difference between neutral of flating source and earth but how is that possible since there is no physical connection between them ? What is the reason of this voltage difference. Thanks in advance. <Q> Floating means that there is a high impedance with respect to ground. <S> However 'high', in practice, does not mean infinite. <S> If you attach a light bulb between the neutral and ground it will be unlikely to light and will most likely reduce the measured voltage to close to zero. <S> However with a typical voltmeter that may have a 10M or higher input impedance, the measured voltage may be pretty much anything, depending in part on where that 120VAC comes from. <S> That's because there will be capacitive and possibly resistive coupling to both sides of the 120VAC and possibly to other voltages, depending on the circuit. <S> There could also, in theory, be a static buildup of relatively high voltage. <S> Usually the measured AC voltage on a real meter will be less than half the mains supply voltage. <S> Think of something like this: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> If C1 == C2 and the meters are ideal (and the transformer is ideal) <S> each meter will read half the transformer secondary voltage. <A> As you see it shows 60v difference between neutral of flating source and earth <S> but how is that possible since there is no physical connection between them ? <S> What is the reason of this voltage difference. <S> Strictly speaking the circuit is not-sensical. <S> Conductors are neutralized by connecting them to ground. <S> If the supply is truly floating then neither line can be "neutral". <S> One possible scenario is that the lower line is supposed to be neutral but due to a fault is now at 60 V AC. <S> The other point to notice is that the circuit is incomplete. <S> We have no idea what the supply is or how it is configured. <S> The fact that one line is labelled "Neutral" should lead us to expect a connection to ground at the supply end of that conductor. <A> The reason is that we define ground to be zero volts compared to the V+ or in alternating current compared to the L1, L2 and L3. <S> Volts is the potential difference between two points and makes no meaning without a reference. <S> So any other grounding in another system or even the actual ground can have a different potential to the grounding in your system. <S> In other words there will be a voltage difference between them, for example 60 volts. <S> Your schematic shows that the transformer of your system is not grounded to the earth and the voltage between the earth and your transformers <S> neutral is 60 volt. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> So say you wanna connect two systems together that has their own power source <S> you should usually do what we call <S> equalize ground which is basicly to connect them together <S> so they have the same defined ground at zero compared to their power source. <A> e.g. Line to ground FAULT, thus raising local earth ground voltage relative to neutral which is earth grounded at one or more places or a another ground fault <S> is a dry arid zone with inadequate earth bonding> 100 OHms and ground currents. <S> One cannot say exactly where yet, but it is a fault condition. <S> If that were the residential voltage, you should expect Neutral may reach 5% of line voltage by design. <S> The Neutral-earth bonding at the transformer and distribution wire R = <S> V=IR , at maximum Neutral current. <S> That means the panel rating of say 200A and wire resistance installed to present day codes ought to never exceed 5% of line. <S> (Using only a single line and Neutral being 100A. ) <S> If current for L1=L2 then neutral current would be near 0. <S> If that were simply a small transformer secondary, it would never be called L1 , N , but regardless when floating and measuring with a high impedance meter say 1Meg or 10Meg <S> then the voltage you might see depends on the impedance division from stray capacitance as it floats according to these parameters..
| It's either a measurement error ( stray capacitance) from a floating transformer ) or a line fault condition as Neutral is low Resistance to the earth-ground by design in North American power systems.
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Can a 9 volt battery through your bloodstream kill you? I usually try to do some research before asking questions on this site, but everything I've found has been contradictory. If I were to direct the positive and negative ends of a 9 volt battery through open wounds on either of my hands, allowing the electricity to travel through my bloodstream and presumably past my heart, would it kill me? What if I were able to let go as soon as I felt the shock? <Q> This paper makes that statement Electrical Safety in the Operating Room: Dry Versus <S> Wet and my recolle4ction is that the writer has come to specialise in establishing best safety practices in that area. <S> He says: Microshock refers to very small currents (as little as 10–50 μA) and applies only to the electrically susceptible patient, such as an individual who has an internal conduit that is in direct contact with the heart. <S> This conduit can be a pacing wire or a saline-filled central venous or pulmonary artery catheter. <S> In the electrically susceptible patient, even minute amounts of current (10 μA) may cause ventricular fibrillation. <S> Once started ventricular fibrillation will often not stop without external action - usually very high magnitude electric shock - whose purpose is NOT to supply a heart start signal but to FULLY STOP the fibrillation signals so that the normal heart waveforms can re-establish. <A> I wrote a new answer that will be more precise after more resarch. <S> The internal resistance of the body from hand to hand will be something like > <S> 300ohm <S> I = <S> U/R <S> I = <S> 9V / <S> 300ohm <S> = <S> 30mA <S> DC current is about 2-4 times less dangerous than AC current because the AC current will cause faster ventricular fibrillation which is often the cause of death from electric shock. <S> So what rates are dangerous? <S> If I were to direct the positive and negative ends of a 9 volt battery through open wounds on either of my hands, allowing the electricity to travel through my bloodstream and presumably past my heart, would it kill me? <S> What if I were able to let go as soon as I felt the shock? <S> Applying 9V from your hand to hand directly in your bloodstream would then give 30mA DC which is highly unlikely to kill you. <S> What if you applied the battery straight on your heart? <S> We know the resistance hand to hand is about 300ohms. <S> The length is about 150cm+- and we use 10cm for the heart, reduce to ~20ohms. <S> I = <S> 9V/20 <S> = <S> 450mA. <S> This number would get close to possible heart fibrillation. <S> While not pleasant it seems that connecting a 9V battery directly to your heart is in the danger zone but not necessarily going to kill you. <S> Connecting a 9V battery however from hand to hand in the bloodstream is highly unlikely to kill you. <A> "It's the volts that jolts, but the mills that kills" Roughly speaking <S> , humans can feel voltage (via muscular contraction) but death comes from milliamps of current passing through the sinoatrial node near the heart. <S> If you could construct a scenario where you could present a conductive path that includes the sinoatrial node, of resistance less than about 90Ω, to the terminals of a 9V battery, then you might have an electrocution risk. <S> The human blood vessel system with wounds at either end is probably not sufficient. <S> But since the 1930's, experiments of this nature have fallen out of favour so the state of the art is really just a lot of extrapolation. <S> The state of the art is that <120Vdc is low risk in normal circumstances, but you could still come up with a scenario with electrocution risk. <S> Thus <60Vdc is often used as a safer limit in various worldwide standards. <S> Finally, in particular hazardous scenarios like swimming pools (think of underwater lighting), <25Vdc is required to be considered safe. <S> 9V is considerably lower than this already very low limit, so would take some very special circumstances to present an electrocution risk. <S> But I reckon if you worked hard enough at it, you could kill some very unlucky, highly susceptible, high sodium content individual. <S> To answer your second question: yes, it is possible to let go and save yourself - electrocution only occurs if the shock coincides with a vulnerable phase (about 10%) of the sinus cycle. <S> At 9V the muscular convulsion would not be enough to render you paralysed, so if you're lucky, you might be able to sense the shock and disconnect before the vulnerable period arrives. <A> This question stems from the rumor that some navy tech wanted to test the conductivity of their body so they pushed the meter beneath their skin and got electrocuted. <S> (at least the version I heard, I'm sure there are many variants by now.) <S> Yes, it can, it only takes 10-20mA to stop a human heart. <S> A 9V battery can provide much more than that. <S> If the skin is broken the resistance drops significantly. <S> The current must be across the heart. <S> Offhand <S> it would seem that a shock of 10,000 volts would be more deadly than 100 volts. <S> But this is not so! <S> direct current. <S> The real measure of shock's intensity lies in the amount of current (amperes) forced though the body, and not the voltage. <S> Any electrical device used on a house wiring circuit can, under certain conditions, transmit a fatal current. <S> While any amount of current over 10 milliamps (0.01 amp) is capable of producing painful to severe shock, currents between 100 and 200 mA (0.1 to 0.2 amp) are lethal. <S> Currents above 200 milliamps (0.2 amp), while producing severe burns and unconsciousness, do not usually cause death if the victim is given immediate attention. <S> Resuscitation, consisting of artificial respiration, will usually revive the victim. <S> From a practical viewpoint, after a person is knocked out by an electrical shock it is impossible to tell how much current has passed through the vital organs of his body. <S> Artificial respiration must be applied immediately if breathing has stopped. <S> Source: <S> https://www.asc.ohio-state.edu/physics/p616/safety/fatal_current.html
| IF you get cause currents in the 10's of microamps range onto the actual surface of the heart death can occur. Your skin has sufficient resistance that it can stop current. Individuals have been electrocuted by appliances using ordinary house currents of 110 volts and by electrical apparatus in industry using as little as 42 volts
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4.5 V on gate of Mosfet while switch open I have a strange behavior with the circuit below. (the bidirectional logica level should be read as: VL = Voltage Low GNDL = GND Low VH = Voltage High GNDH = GND High L1 = Low channel 1 (I did only draw 1 for simplicity, there are 4 in total) H1 = High channel 1 This is the level shifter: Voltages read: When the switch is open: 1.98 V between GND and L1 (of the shifter) 4.57 V on the gate of the Mosfet LED is on When the switch is closed: 3.30 V between GND and L1 (of the shifter) 5.26 V on the gate of the Mosfet LED is on Of course the LED is always on since the gate voltage is in both cases high enough. How can there be 1.98 V between GND and L1 while the switch is open (I assume the gate does not return any voltage) and 4.57 on the gate? earlier tests When I change the 10 KOhm resistor by a 200 ohm, the LED works as expected (on when closed, off when opened). Without the level shifter and using 5 V for the entire circuit, also the LED works as expected Without the level shifter and using the 3.3 V than the 2N7000 does not get enough voltage on the gate (therefore I want to use the shifters, at least until I receive my better mosfet, IRL44N). <Q> Put the LED on the other side of the MOSFET, be sure to use a current limit resistor. <A> This is unworkable. <S> The "logic shifter" part you show is, for your purposes, nothing more than a 10K pull-up resistor to 5v which can be disabled by the small signal FET which joins its two sides. <S> There are parts sold for what you want to do, under names like "High Side Driver" and they will be explicit that they provide an output referenced to the high side of the load. <S> Mostly their justification for existence is that N-channel devices have historically been enough better than P-channel devices that it is worth the extra circuit complexity to use them even on the high side where they are an unnatural fit. <S> If you were to look at say, quadcopter brushless motor drivers, you might find that some used a half bridged built from a combination combination of N- and P-FETs on the low and high side, while others (especially the more powerful) use all N-FETs with the necessary drive circuitry to apply those on the high side as well as the low. <S> For your purposes, it would be far simpler to just use an appropriate N-FET on the low side of the load. <S> If you need to switch the high side for some reason, you could consider a P-FET, or perhaps consider a USB downstream port power switch chip, which often consists of an N-FET and the needed high side driver packaged together in IC form. <S> While intended for USB they usually work over a range of low voltages. <A> In addition to above answers, a MOSFET can put a voltage out of its gate pin if the internal oxide layer has a short due to either gate overvoltage or severe ground bounce. <S> In this case a real chance of excessive gate to source voltage does exist. <S> With power OFF, disconnect the gate pin and measure ohms from gate to source and gate to drain. <S> Any value under 100 megohm is a short, especially when voltage is applied. <S> If power is applied but gate is open, it should read almost zero volts to source if it is good, else a few volts on the gate is a sign the oxide layer is damaged, and is no longer an insulator. <A> How can there be 1.98 V between GND and L1 while the switch is open <S> (I assume the gate does not return any voltage) and 4.57 on the gate? <S> The first thing you need to understand is the schematic for the level shifter. <S> In all probability it is this schematic: <S> Some boards appear to have <S> 2N7000 while other use the BSS138 . <S> Here I've redrawn the schematic to achieve what I think you wanted to do: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> I've shown L1 as having only two states --> GND or 3V3, but if L1 was left open some leakage current (1-2uA) could flow from 5V to 3V3 via R2. <S> In your schematic you show L1 as connecting to 3V3 or to a 10k Ohm to Gnd. <S> This would explain your strange measurements since the FET Source is then connected to voltage less than the 3V. <S> This will allow a BSS138 to partially turn on and current will flow from the high voltage side to Gnd. <S> This explains why you see 4.57V (about 200ua through R1) and 5.26V <S> (I assume here your 5V supply is actually 5.26) between the two states.
| You are trying to use an N-FET as a high side switch, which means that you need a flying gate driver which can impose a suitable voltage of the Gate above the source in the situation where the Source is not grounded, but rather riding on the varying voltage of the load.
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Is it safe to abruptly remove Arduino power? If I have an Arduino, can I abruptly turn the power on and off without corrupting it? If I am designing a product that has an on/off switch, do I need to incorporate a delay before turning off the power so that the microcontroller can do a clean turn off of some sort? Or is it okay to just connect an SPST switch to the Arduino's power line? (By designing a product, I mean using an ATmega with Arduino code independently on a custom board with other components.) Edit:By Arduino I just mean flashing the Arduino bootloader so I can program in Arduino instead of AVR.The chip I want to use is Atmel atmega328PI do not need any EEPROM usage. <Q> Yes, it is safe to abruptly shut off an Arduino. <S> Well, mostly safe. <S> The processors used in the various Arduinos have three types of memory: Flash - Where your program is stored. <S> Your program can read stuff from here, but cannot write it. <S> RAM - Where your program variables are kept while the program in running. <S> The data here disappear when you turn the Arduino off. <S> Your program reads and writes here constantly. <S> EEPROM - <S> Where your program can store stuff it will need the next time it runs. <S> Usually stuff that changes rarely, but is needed anytime the program runs. <S> Like calibration data for a sensor. <S> Most programs only use Flash and RAM. <S> You can switch the Arduino on and off any time you like with those programs. <S> If your program writes to EEPROM, then shutting off the power while writing to the EEPROM could corrupt the data there. <S> How that affects your program depends on what the data is. <S> If it corrupts the calibration data for a sensor, you would get bad measurements for whatever the sensor is detecting. <S> If you write checksums with your EEPROM data, then you could detect the corruption and your program could shut down instead of using bad data. <S> You, of course, would know if your program writes to the EEPROM - you have to load a seperate library and use special commands to read and write to the EEPROM area. <S> The danger is really only in that short moment when you write to the EEPROM. <S> Since that happens rarely (and usually only under controlled conditions) it will also be rare to corrupt the EEPROM data. <S> Summary: <S> You can turn an Arduino on and off at will with no danger, unless you are using the EEPROM - and even then you will get away with it most of the time. <A> Yes, you can turn the power off quickly without corrupting it. <S> The only reason I can see for putting a delay into the power down of the circuit would be for safety or functional reasons of your external peripherals. <S> E.g. on power off, need to save data to non-volatile memory. <S> Or on power off, need to ensure a mechanism is in a safe position at power off. <S> This would require monitoring of the supply and having sufficient hold-up capacity to run the processor and what ever function that was required to be performed. <A> The term Arduino refers to a very broad spectrum of microcontroller boards that have various different chips, and the effect of power loss thus varies. <S> So, I would recommend you put the exact part number of the ATmega chip being used. <S> But, in a general way for chips like ATmega328, the following holds good. <S> If your code utilizes EEPROM then the flash might get corrupted or the data may not get stored accurately when the device is flashing data and power is removed. <S> The only-way a sudden power loss can do harm apart from EEPROM is to the devices being used with Arduino (e.g. an SD card).
| Yes, you can remove power from an Arduino without corrupting or damaging anything but do keep in mind the following things:
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PNP transistor turns on at 3.3v I have connected a 3906 PNP transistor base to a GPIO pin of ESP32 board. Emitter is connected to 5v and collector is connected to a buzzer. When I program the buzzer (send a HIGH from ESP32), it is permanently on. When I check it against a multimeter, the transistor turns on even when base current is 3.4v (when I though it turns on only if voltage is less than 0.7v). To cross check the transistor and buzzer, I directly connect base to 5v and the transistor is turned off. Connecting to ground turns on. So, it works correctly when connected directly. For those who would like to suggest a NPN transistor, I have tried using 3904 NPN transistor and everything works as expected. However the issue is that when programming, all the pins of ESP32 are high by default and the buzzer is ON until programming done. Can someone tell me how to solve this issue? I need to use PNP transistor only. Thanks in advance. <Q> I suspect you are trying to drive the PNP base with a 3.3V logic signal while controlling 5V to the buzzer. <S> That just isn't going to work. <S> To turn the transistor off (non-conducting) you need to raise the voltage on the base to be close to the emitter voltage, or about 5V. <S> You can't do that with a 3.3V logic signal. <S> You make a PNP transistor conduct by lowering the base voltage about 0.7V below <S> the emitter voltage, not by setting the base voltage to 0.7V above ground. <S> You would also have to select resistor values that limit the current that will flow back into the 3.3V logic output when that output signal is high. <S> This is all just speculation, you haven't provided a datasheet for the device that is driving the base. <A> try this <S> simulate this circuit – <S> Schematic created using CircuitLab <A> This will work: simulate this circuit – <S> Schematic created using CircuitLab <S> The emitter of Q1 is at 3.3V <S> so it is off when the input is floating or at 3.3V. <S> When the input goes low, Q1 turns on, providing almost 3.3V at the top end of R2, providing about 0.5mA to the base of Q2. <S> R3 drains away any leakage Q1 haswhen <S> it is off. <S> If the buzzer takes more than about 5-10mA then decrease R2. <A> simulate this circuit – Schematic created using CircuitLab <S> You may have to play with resistor values. <A> The classic way to do this is with a circuit like this: simulate this circuit – <S> Schematic created using CircuitLab <S> Here, any input signal above <S> ~0.7V <S> switches the NPN transistor Q2 on, which pulls current through Q1's base to ground. <S> When Q2 off, R1 pulls the base high, to keep the PNP transistor off. <S> R2 and R3 are here to limit current through the bases of the transistors.
| You might be able to add a pullup resistor from the base itself to 5V to turn the transistor off when the logic output is 3.3V. Select resistor values so that when the 3.3V signal is low the voltage at the base is less than 4.3V.
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Predictions based on resistance measurements not matching reality? I'm using my multimeter to measure the resistance of my tongue, and leaving it on there for a full minute it seems to average out at around 1.5 Million Ohms. However, when I put a 9 volt battery on my tongue, I can definitely feel an unpleasant shock. According to Ohm's Law, the current going through my tongue should only be (9 / 1.5M) = 0.000006 Amps (0.006 mA). However, this chart indicates that anything under 1 mA should be imperceptible. Why can I feel the battery through my tongue? <Q> more in the range of skin resistance. <S> It will also vary with the spacing (and area) of the electrodes. <S> Edit: I tried it with a multimeter and got more like 50K. <S> You can put the multimeter in series with the battery and actually measure the current from the 9V battery directly. <A> A 'resistor' made from metal film, or carbon, or conductive ink, or a length of wire, is a very simple beast, and you can expect a fairly constant resistance from it. <S> Your tongue has at least two conductive solution interfaces to the electrodes. <S> Electrolytic cells are notorious for behaving in a very voltage sensitive way. <S> As they're water-based, the dissociation voltage of water, around 1.7v, will be key. <S> You can expect voltages below this to drive very little current through the cells, so they will appear high resistance. <S> As the voltage increases above that, much more current will flow. <A> The resistance of your tongue is most certainly not 1.5 mega ohms. <S> My skin resistance is only in the 10s of kilo ohms, and it's dry as the sahara. <S> Your tongue has far more nerve endings than normal skin, making it more sensitive. <S> I will not put a 9V battery on my tongue. <S> 1.5V from a C cell is bad enough - and one end of it is only touching the skin of my hand rather than a second spot on my tongue. <S> I did lick a 9V battery once when I was a little kid <S> - ain't doing it again. <S> I expect you've done something wrong in measuring the resistance of your tongue. <S> Even just doing that must have been unpleasant.
| Your tongue is more sensitive to electrical current than your fingers. You might find the current is higher than 6uA too with the higher voltage (linearity is not guaranteed). Frankly 1.5M seems extremely high for a wet mucous membrane- I can put a 9V battery on anyplace on my skin and not feel it. Add a 10K resistor or something like that to protect the meter. Keeping contact with the probes for more than a second or so seems like it would be a really not fun thing to do.
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Is the positive supply of Arduino USB port connected to 5V pin? You can power the Arduino Uno/Nano using the following: Barrel Jack (for UNO only), it is connected to the input to voltage regulator. VIN pin, it is also connected to input of voltage regulator. 5V pin, it is connected to the output of the voltage regulator and directly of 5V of Atmega MCU. Finally, the USB. Is the positive supply of the USB port of an Arduino is connected to the 5V directly? I tried to use a connectivity test using multimeter but it is not connected. Moreover, it is mentioned in the Arduino documentation around the internet that if you powered the Arduino via USB connection, the 5V pin will be limited to 500 mA only. What if I use an AC to DC adapter (2A output current) that is connected to the USB port of the Arduino, will I will still get only a max of 500 mA? Or if I use USB 3.0 that can supply 900 mA, will I still get 500 mA max for the 5V pin? I read that the polyfuse is the one that is limiting that current. I have used Arduino Uno and Nano for a long time but I just became aware of this basic concepts when I made a project that should compute the overall power consumption of the device. <Q> There are a lot of "Arduino-compatible" boards which are sold as "Arduino" labeled. <S> The schematic is common in general, but can be different in details. <S> You should investigate what do you have on you specific PCB yourself (or post detailed images of the both sides of the board here). <S> I think, 500 mA limit comes from two factors: a) <S> USB 2.0 specification (500 mA max); b) <S> Some of the "Arduino-compatible" boards have fuse on USB power line of the same 500 mA value (but your specific board may not have it). <S> Or, let say, the diode which can handle 0,5 <S> A only. <S> "Normally" when your device (Arduino board + additional modules etc.) does some useful job you should power it from one definite source - USB or power supply. <S> And you have to decide yourself how to power it correctly. <S> The choice is up to you. <S> You can also unsolder fuse or diode from you PCB <S> and then you'll get (almost) <S> no limits to power current from USB input <S> (but you'll get the risk of frying you USB port also, if something goes wrong on the PCB side). <A> Uno has a 500mA resettable fuse similar to this between the USB 5V supply and the 5V pin on the header. <S> A PFET is used to keep 5V from the board from driving back into the PC.It <S> is located near the USB connector https://www.digikey.com/product-detail/en/bel-fuse-inc/0ZCG0050AF2C/507-1762-1-ND/4156148 <S> Nano has a diode between USB 5V and the regulator output so that the PC cannot be driven from the board. <A> Not all of them are, get a Digital Multi Meter out and ohm out the lines, if they are under 1 ohm then there is a good chance that they are connected directly. <S> Some times Eval boards put DC to DC converters between the USB and device.
| Also powering from USB can be made from computer USB port or from power supply with USB output.
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Do modern processors have redundancy in their logic units to compensate production faults? Modern processors consist of billions of transistors and new production technologies often have problems with the yield, at least in the first months, but I guess that even after years there will be faulty chips every now and then. I know that in large blocks (e.g. the cache) there is the possiblity to just disable parts of it and by that reducing the available amount of memory (so you can at least sell the chip at a lower price instead of throwing it away). But is there something similar for the logic units? I am aware that there are multiple ALUs for dispetching, but is this a thing to just disable one of them if there is a production fault? Or are there even additional spare ALUs? Because for me it is hard to believe that fabs just dispose of every chip where there is a faulty transistor in the logic parts, while disabling a complete ALU would proberly reduce the achievable processing power significantly. <Q> Not in the logic. <S> However if there are big memories (SRAM) <S> it is common to use a memory with 'redundancy'. <S> These have special logic which can be programmed to replace a area, often a number of rows or columns. <S> The failing area is detected during testing and then the redundant memory is programmed to replace the faulty location(s). <S> However this 'replacement' must be set-up using OTP (One-Time-Programmable) bits or some other memory which holds its value. <S> Thus these memories are only used in chips which have such a 'permanent memory' feature, or such a programming feature must be added as well, with all the costs this incurs. <A> This is certainly not the case for simple MCUs, or typical single core processors. <S> The cost of having spare blocks would not be worth it, and those processors don't use cutting-edge engraving processes, and don't require huge silicon areas, so the yield is good enough. <S> However, this is done for some multi-core processors, for which the silicon area is rather large, and that uses finer engraving processes which can lead to higher defect rates. <S> The processor is then sold as a lower-end model. <S> Source: https://skeptics.stackexchange.com/questions/15704/are-low-spec-computer-parts-just-faulty-high-spec-computer-parts <A> As others have said, it is difficult to see redundant ALU logic within a core. <S> A core was designed to optimize throughput. <S> Any additional logic for a redundant ALU would impact performance and increased area would slow down the whole core. <S> As technology evolved, the silicon became smaller, making cores faster, but essentially using the same intellectual property. <S> Why have redundant ALU's, when space is available for redundant cores to increase production yields? <S> In 2011, Intel filed a patent for at least 32 cores with 16 active and 16 spare. <S> The patent states failing cores would have higher temperatures allowing a spare core to be switched in. <S> Essentially, dynamic core allocation as required. <S> You could have high-power and low-power cores allocated as required by tasks. <S> Or switch out a bad core detected by higher temperature levels. <S> Operate the cores in a checkerboard manner to reduce heat. <S> Intel Patent: Enhancing Reliability of a Many-Core Processor <A> I certainly cannot answer your question for sure. <S> It makes no much sense to disable units smaller than 1 core, since it becomes a very fine-grained "feature set" that can be enabled or not, and the Cartesian product of all possible features would make myriad of possible CPU models. <S> There are a lot of CPU models already, making them 10-100 times more will definitely not help! <S> Another aspect is that the billions of transistors are used (for the most part) in making caches, and for defective transistors there manufacturers definitely sell CPUs with parts of their on-die cache disabled (e.g. see, AMD Thorton vs AMD Barton). <S> But I can tell you an anecdote which I've heard from a person I trust. <S> A long time ago I was a curious overclocker. <S> In my days, the budget overclockable CPU of choice was the AMD Athlon <S> Thoroughbred: <S> When mounting custom cooling solutions, one had to be very careful while attaching the heatsink, as it presses directly against the die. <S> If you applied uneven pressure, the dies were notorious for cracking easily at the corners, if you applied force at one corner first. <S> This person had done exactly the same thing, a significant portion of one corner was gone, but the CPU was miraculously working fine, albeit at much reduced memory performance. <S> The corner contained L2 cache only, so with that part gone, the caching protocol was somehow working around the now very defective die. <S> It was probably reporting cache misses for all queries in that part, so the CPU was reduced to its L1 cache only (or only part of L2), so it was much slower in most tests, yet had virtually the same performance on tight loops. <S> In the same line of thought, it could be made that if an ALU is defective and is capable of somehow signalling back that it rejected work, the CPU may be capable of falling back on other ALUs. <S> Whether this is being done by CPU manufacturers is unknown (and I doubt it), but the cache example (from 15 years ago) shows that it is definitely doable.
| On these processors, entire cores can be disabled (which are rather big logic blocks, containing much more than an ALU) when they are defective.
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Is induction motor's rotor speed or torque change with delta or star connection? An induction motor's synchronous speed Ns is given as: Ns = frequency * 120 / number of poles This formula is independent of the voltage applied to the windings. On the other hand, I know from basic theory that if the voltage across a solenoid is increased its magnetic flux will increase. And more magnetic flux will create more force on a current carrying wire. Would the speed of the induction motor's rotor change with delta or star connection? And is there a formula explicitly shows such dependency between stator winding voltage(delta or star) and rotor speed? Would a three phase fan motor rotate faster if connected delta instead of star? <Q> As the formula says, the synchronous speed is determined only be the frequency and number of poles. <S> The operating speed is reduced below the synchronous speed by slip. <S> Slip at rated torque is about 2 to 3 percent for a standard squirrel-cage motor. <S> A high-slip motor could have up to about 8% slip at rated torque. <S> Slip with no load is nary zero. <S> From no-load to rated load, slip is fairly linearly proportional to torque. <S> Induction motors are generally designed to operate close to saturation at rated voltage and will overheat if the voltage is too high. <S> If the voltage is reduced, the maximum torque will decrease approximately proportional to voltage squared. <S> The slip will increase. <S> To the extent that the motor can tolerate operation at increased slip, the speed can be reduced by reducing the voltage. <S> That method of reducing speed is of limited use. <S> Normally changing between delta and star connection is used only for accommodating two possible supply voltages and for star-delta starting as a means of limiting the starting current by reducing the voltage for a few seconds while the motor is coming up to speed. <S> If a motor is designed to operate at a certain voltage with the star connection, connecting the motor in delta at the same voltage will cause saturation and overheating. <S> A the rated voltage for the star connection is 1.732 <S> x the rated voltage for the delta connection. <S> The rated current for the star connection is .577 <S> x the rated current for the delta connection. <S> The input kVA is about the same for either connection to the proper voltage for that connection. <S> The peak torque, rated torque, rated slip and other torque vs. speed parameters are the same for both connections. <A> The speed of the motor does not change with delta or star connection. <S> The thing that changes is the voltage accros the stator windings, thus the motor maximum current depends on the delta or star connection. <A> As said, the reduction of torque will cause the motor to increase slip slightly, so there might be a slight change in unloaded speed. <S> With load, the loss of torque can create a much GREATER loss of speed and increase in slip, to the point of possibly stalling the motor, depending on the AMOUNT of load applied. <S> With an effective 57.7% voltage applied to the windings, peak torque is reduced to 33% of rated. <S> If the LOAD on the motor is less than 33% of rated, you may never see the difference in working speed. <S> But if the load ever increases, you run the risk of stalling. <S> In fact this is a technique used by some to "save energy" in motors that have been grossly over sized for the load. <S> i.e. a machine was built for one purpose and later on, repurposed for a task with much less load on the motor, but it would be too expensive to install a smaller motor. <S> So the old oversized motor is connected in Star (Wye) so that with the effective drop in phase voltage, there is less magnetizing energy consumed in the motor (so called "iron losses"). <S> It doesn't make very much of a difference, but I've seen it used in places with very high tariffs or where power is generated on site and every watt counts.
| So, the motors maximal torque is affected depending on the type of the connection.
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UART pins to unpowered MCU? I'm using a USB-UART IC with an MCU, which is powered from a battery. The USB-UART IC is powered from USB connector, not from the battery, so that I don't need to open up a console every time when the switch goes off and on. simulate this circuit – Schematic created using CircuitLab edit: I didn't draw it on the figure but the IC has an internal 3.3V regulator and every VDD is on the 3.3V level when the switch is on. Now I'm worried about when the USB is plugged and the switch is still off. The MCU document says that every input pin's maximum rating is VDD + 0.3, which would be 0.3 V when the MCU is not powered. If the TX/RX pair on the USB-UART side goes high, will it destroy the pins on the MCU side? If so, What do I need between the TX/RX pairs? <Q> It depends on the MCU, but in most cases it'll power up the MCU, and possibly the rest of the board through the MCU. <S> The MCU will try to run, and do odd things. <S> Your board will do odder things. <S> If your board draws enough current, it'll damage that pin on the MCU. <S> You need to arrange for the UART signal to stay at 0V when the MCU is off. <S> If the UART chip (or UART) that you are using doesn't have an enable pin (the USB UART chips that I've worked with can be configured for exactly the case you're describing), then AND the UART outputs with the microprocessor's VCC. <A> This will mean that the USB-UART will only be able to drive up to 3.3 V minus the threshold voltage of the transistor. <S> You'll have to check whether this is still enough to satisfy the V_IH of your MCU. <S> I'm skeptical of solutions that involve a logic gate powered (indirectly) by VBUS. <S> As long as the logic gate drives the MCU input high, the MCU VCC may not drop far enough to switch off the TX driver. <S> With the NMOS solution, the TX pin can only be driven to a lower voltage than VCC, making this sort of feedback impossible. <S> By the way, you should also consider the reverse direction: If the MCU is powered, but the USB-UART is not, you'll want to make sure that you don't accidentally provide some power to the USB-UART, draining your battery faster than you expect. <S> A software-only solution should suffice for this. <A> I have a completely different solution, which, however, depends on which USB chip you use. <S> Some of them have different VBUS and VCCIO. <S> In this case, you might be able to power the IO part of that chip together with the uC. <A> Easy solution is to put some resistors between the I/O pins. <S> This will limit the current flow into the pins so the transceiver cannot power the uC. <S> The resistor value is a balance between limiting the bandwidth between the chips and limiting the current. <S> Logic buffers powered by the USB but with outputs enabled by the uC. <S> This does the same as the transceiver output enable of the other answer. <S> Is there an advantage to having the uC off when it's connected to the computer? <S> If not you can power the uC from the 5 V USB power. <S> Ways to do this are: An SPDT switch that selects USB power when battery power is off. <S> Replaces the SPST switch in your circuit diagram. <S> A diode, diodes, an ideal diode integrated circuit, or MOSFETs controlled by the uC to select USB power when available. <S> Now you need to consider what happens when the battery switch is closed and the USB is connected. <S> Uncontrolled battery changing is rarely a good thing. <A> Have you considered if a couple tri-state buffers would be a good solution? <S> You could tie the enable pin to your switch and then make sure the polarity from tx to rx is correct, which effectively gives you the protection you are asking for.
| The solution might be as simple as an NMOS between the TX pin of the USB-UART (drain) and the RX pin of the MCU (source), with the gate connected to the MCU VDD.
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Why can't I use Ohm's Law on my circuit? In my circuit, I have 5V power supply and I have 200 ohm resistor. When I measure the current, it's 14.8 mA. Why can't I use the equation I = V/R, which is 5/200 = 25 mA? <Q> Ohm's law applies to the resistor. <S> The voltage across the resistor will be about 3V. <S> The remaining 2V or so appears across the LED. <S> To a rough approximation <S> I = <S> Vr/R ~= <S> (5V-Vf)/200. <S> Vf, the voltage across the LED, stays more or less the same at different currents within the normal operating range. <S> So if you increase the voltage to 6V you would expect the current to increase to about 20mA (actually a bit less because the voltage across the LED will increase a bit). <S> From this answer, there's a graph that is similar for most red LEDs. <S> As you can see, at about 15mA Vf is a bit under 2V. At 20mA <S> , Vf is just about 2V exactly. <S> At 30mA it's more like 2.05V. <A> The LED in your circuit is non-linear and so Ohm's Law does not apply to it. <S> Figure 1. <S> The I-V (current versus voltage) curves for various colours of LED. <S> Source: <S> LEDnique - IV curves . <S> From your measurements we can calculate the voltage across the resistor as \$ <S> V = IR = <S> 14.8m \times 200 = 2.96 \ \text V \$ . <S> (Let's say 3 V.) <S> That leaves 2 V across the LED and from the I-V curve in Figure 1 we can see that at 14.8 mA we would get a 2.0 V drop across the LED for a green(ish) LED. <S> The important thing to learn here is that the LED has a voltage drop and that it does not act like a resistor. <S> Figure 2. <S> The I-V curves for an assortment of resistors. <S> Note that these are linear. <S> Source: LEDnique - IV curves . <A> Ohm's Law applies only to resistors. <S> The voltage term must be the voltage across a resistor, and the current term is the current through the same resistor. <S> Your 5V is not the voltage across the resistor, it is the voltage across both the resistor and the LED. <A> This is all about linear regression above the knee voltage, then you can use Ohm's Law. <S> It's only because you don't know how to model a diode. <S> Diodes are nonlinear initially then they somewhat behave as a voltage drop with a small series resistance. <S> Ohm's Law or KVL analysis <S> \$Vcc= <S> I_f*R <S> + V_{f (LED)} <S> = I_f*R <S> + <S> (V_t+I_f*R_s)\$ ...... <S> (1) for knee threshold Vt and diode bulk series Rs and forward current <S> If rated at Vf forward voltage. <S> From looking at the datasheet VI curve you can get a typical ΔV/ΔI = <S> Rs in Ω <S> (V/I slope or tangent) near the current range you expect. <S> As the current drops to 10% of the rated current (not Abs max.) <S> the series resistance rises to about 10x that at rated current <S> so it is mostly linear above 1/4 of the rated current given the wide tolerances on this Rs <S> is +50%/-25% for an open "bin". <S> Here is how Ohm's Law can be used on diodes. <S> e.g. RED LED's Vf = 1.85V <S> + 12*If <S> (Rs = 12 Ω + 50%/-25% ) thus using (1) <S> \$Vcc= <S> I_f*R <S> + V_t+I_f*R_s\$ or \$I_f= <S> \dfrac{V_{cc}-V_t}{R+Rs}=\dfrac{5-1.85}{200+12}=14.3mA\$ <S> which is well within expected tolerances due to Rs. <S> There is also a voltage drop with rising junction temperature which may be computed , but it's not a good idea to get it too hot to touch. <S> This model is far more accurate than the tolerance on parts so changing Rs by your Vdd and Rs tolerance <S> allows anyone to use Ohm's Law for any diodes (including Zeners) with some understanding. <S> You can adjust the model to the Datasheet. <S> Note <S> this LED <S> Vf=1.85V (typ) 2.5V (max) <S> here the typical curve is shown ONLY or Vt=1.75V and slope ΔV/ <S> ΔI=(1.9 <S> -1.75)/30mA = 125mV/30mA = <S> 4.2 Ω = <S> Rs added <S> Still If=(Vcc-Vt)/(R+Rs)=(5-1.75)/(204)=15.9mA. <S> It is not much different than 14.3mA above. <S> Now try 2.5V <S> , what R gives 20mA? <S> 20mA=(2.5V-1.75V)(R+4.3), <S> R= <S> 750mV/20mA-4.3 <S> = <S> 33 Ω <S> (Ohms Law says 20mA*33= 660mV across R) <S> so <S> Vf=2.5V - 0.66 = 1.84V vs 1.86V above, which is what the datasheet says for TYP. <S> Vf@ 20mA <S> Conclusion: <S> One can use Ohm's law while including the resistance of the LED if you choose nom. <S> values from the VI curve for the intercept, Vt and slope <S> Rs, then you can interpolate or extrapolate and get more accurate results than using a fixed Vf.
| You can use Ohm's Law but only on the resistor.
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How can I execute multiple for loops sequentially in Verilog? I'm trying to turn on the LED lights on my FPGA Spartan board one at a time until all lights are on and then turn them off in the reverse order. I could easily do this in other OOP languages by making two for loops and slowly change the values in an array with each for loop reversing the other just like in a simple starter Arduino project. However, Verilog executes in parallel therefore I am confused as to how I could make the loops run sequentially. I know that using a blocking and non blocking statements affects the how the code executes but I do not see the correlation to loops. example: for (counter = 0; counter < n; counter = counter + 1) #7 lights[counter] = 1b'1;for (counter = n; counter > 0; counter = counter - 1) #7 lights[counter] = 1b'0; <Q> A few things to keep in mind about writing HDL: <S> HDLs are not programming languages. <S> Your code ultimately describes a schematic, it is not executed or interpreted. <S> Whatever code you write has to correspond to some physical element on the target device. <S> Things like nested if statements become AND gates operating on signals that represent the conditions, <S> mathematical and logical operations become the corresponding hardware (adders, subtractors, logic gates, etc.), array operations can become RAMs and ROMs, etc. <S> Delays are in general not synthesizable. <S> You have to find some other method to create the functionality you are after. <S> Delays can be useful in simulation for various purposes, but they are ignored by the tools when creating the actual FPGA configuration. <S> Some parts have special primitives for performing delays on individual IO pins, but these are very specialized components with delays on the order of nanoseconds, only usable for signals passing through specific IO pins, and must be manually instantiated and configured as necessary (see Xilinx IODELAY). <S> Loops are shorthand for replicated statements. <S> Loops in HDLs are very limited as the synthesis engine unrolls them. <S> Therefore, you cannot use things like while loops, break and continue, loop limits based on signals, etc. <S> Everything must be constant at synthesis time (module parameters are OK as they are constant during synthesis). <S> For your application, it would make sense to build some sort of a state machine that can turn the LEDs on and off in a specific sequence, then add some rather large cycle count delays so that the changes are visible. <A> I have quite limited knowledge when it comes to Verilog syntax as I'm only writing VHDL <S> but the idea would be the same. <S> To achieve something like this, you could run a very small state machine. <S> So the first state would turn the lights on in the order you want, once the end condition is reached, change state and turn the lights off again. <A> The answers around here are correct, but probably cover the more broader area of how you should approach HDL; I'll try to present a more fast-forward solution. <S> In OOP, the sequencing of what you type in code is implicit, so generally you have your code being executed in the same order <S> it goes line by line. <S> In HDL, you are working with a much more low-level description, practically writing parts of circuits. <S> And then you'd also need the circuit to know the sequence you want to switch your LEDs in. <S> The easiest way for this is to have an overflowing clock counter and pick its values as conditions at which you operate. <S> So all converges to (considering here that we have 8 LEDs): // <S> 3-bit counter from 0 to 7 reg [7:0] lights;reg <S> [2:0] clk_cnt;int i;always <S> @(posedge <S> clk)begin <S> clk_cnt <= <S> clk_cnt + 1; for (i = 0; i < 7; i++) if (i = <S> = <S> clk_cnt) lights[i] <= <S> ~lights[i];end <S> This will make lights switch their state in sequence one after another. <S> Note that the for-loop here is just to compact the code, <S> in reality it's expanded into 8 "if"s <S> that compare the counter value to lights index. <S> To expand this into first switching lights on in one order and then switching them off in reverse order, you'd need two different counter values for each lights index, thus twice the counter. <S> // <S> 4-bit counter from 0 to 15reg [7:0] lights;reg <S> [3:0] clk_cnt;int i;always @(posedge <S> clk)begin <S> clk_cnt <= <S> clk_cnt + 1; for (i = 0; <S> i < 7; i++) if ((clk_cnt < 8) && (i = <S> = <S> clk_cnt)) lights[i] <= 1'b1; else <S> if ((clk_cnt >= 8) && (i == <S> (15 - clk_cnt)) lights[i] <= <S> 1'b0;end <S> Of course, if the clock you supply here is of high frequency, you won't be able to observe the LEDs. <S> You need to either divide the clock or use a bigger counter.
| So to have something executed in order, you need a synchronizing source, like a clock.
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Was this caused by arc flash or just short circuit without any arc flash? Can the following be caused by short circuit without any arc flash? There was an initial short between lugs and chassis. Did arc flash occur that caused the lugs to melt. Or can it melt just by the huge current during the short circuit? In the above. The live wire was lowered by the electrician to the terminal. Then somehow it shorted (between the 208v wire and neutral/chassis ground) causing the breaker lugs to melt and a hole to be produced at the chassis at the back. This also caused a rat size 2nd degree burn in the electrician arm. Was this caused by short circuit without arc flash or was it arc flash? <Q> Most probable casue was a bad contact, it wasn't tighted well. <S> I can see the healthy connection next to it - it's a wrong connection. <S> The wires has to reside in the bottom of the lug, tightening the screw you compress wires and bottom of the lug. <S> The wire lost contact with the lug, but the distance from lug and wire was so small, that the electric field easily break the air, making an arc. <S> This resulted like everthing is OK, the current was flowing and the appliance worked, but with help of an arc. <S> This is also one of the causes why houses get on fire. <A> Updated: <S> Based on your completely altered question, altered images and new information, it seems highly likely that arc-flash was the cause of the damage . <S> Neither the damage to the panel or the damage to the electrician's arm is likely to occur without an arc being present. <S> A POSSIBLE <S> mechanism is that the short circuit current overheated a poorly made connection at the breaker which released the wire. <S> The wire then swung into contact with the panel with the results seen. <S> Damage to breaker probably due to initial failure and subsequent arc due to swinging lead. <S> Most very major catastrophes require 4+ unexpected or impossible things going wrong 'simultaneously'. <S> In an entry level catastrophe like this it may only have taken two. <S> - Firstly the poor termination of the wire. <S> - Secondly the wire then swinging against the panel. <S> Older: <S> The fact that the adjoining connector shows no damage strongly suggests that it was not involved in the incident. <S> You can get damage that looks much the same as this without arc-flash. <S> Resistive heating increases with the square of the current (Power = i^2 x R). <S> A short circuit current say 10 x normal operating current will result in 100 times as much power dissipation across any resistive components. <S> If the connector was not properly used it would be easy to get enough heating to destroy it. <A> If you see blackening on surfaces above the destruction, it can be caused by long burning, by smoke-soot which lasted many seconds/minutes, without any plasma-explosion. <S> Didn't the electrician report a huge flash of light and a loud buzz sound? <S> If they were burned but not electrically shocked, that was from skin contacting the plasma in the arcflash, which is usually far hotter than any flame. <S> (The arc-plasma mostly made of metal-vapor, incandescent gas heated far above the boiling point of molten copper or aluminum.) <S> In your top picture, the blackening is to the side (so not caused by many minutes of rising smoke,) and it has radial lines from explosive trajectory of gas and metal particles. <S> Almost certainly there was a large plasma (arc flash,) <S> many inches across if not larger. <S> Where it touched cold metal, the vaporized metal inside the flash would condense, forming pure-metal "soot." <S> Metal nano-powder is typically colored black. <S> Copper, aluminum, even etc., creates a light-absorbing black soot when its particles are fine enough. <S> No carbon needed, the black stuff is pure copper or aluminum. <S> You can buy "platinum black" or "copper black" from chemistry suppliers, which is simply ultra-fine pure metal powder. <A> That surely looks like a small arc flash to me, and no it does not take a 400V+ line to do it... <S> The actual voltage dropped across an arc once established can be measured in terms of a few tens of volts, with the rest dropped across the supply impedance (Which is why HUGE currents can flow), given a sufficiently stiff supply a 48V battery bank will absolutely get the job done. <S> and that can then create a path to chassis <S> (Seen that one happen). <S> A modest arc due to a loose terminal can produce a plasma flame that if it touches earthed metal can go from a few tens of amps to the load via the plasma to a few kA to ground almost instantly (With the corresponding increase in the size of the cloud of metal vapour). <S> TBH it looks to me like that conductor slipped out of the terminal and touched the back panel of the enclosure (You can see bit of what looks like ground zero behind that cardboard box (Get rid of it, cardboard in a distribution cabinet, bad plan!). <S> A really stiff bolted short tends to be unimpressive because the circuit protection will open almost instantly, sometimes get impressive magnetic effects making cables jump, but a hard fault is generally uninteresting.
| A High impedance connection can of course develop sufficient heat to eventually damage itself, cause an arc But if you see blackening on surfaces to the side and/or below , that's caused by "metal-black," pure metal-powder which was condensed out of a blazing bright plasma of arcflash. As Marko says, a poor connection can cause this sort of damage. From your description that was a electrician failing to isolate the source of supply and feeding the fuckup fairy by working live in a metallic enclosure (Against most countries safety rules), the guy got very lucky.
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TO-220 MOSFET (or SS relay) for 24VDC with 0V turn on voltage I am in need of a MOSFET (or solid state relay) in a TO-220 package that will switch 24 VDC high side to an inductive load (up to 2 amps) when 0VDC is applied to the gate. I don't even know where to begin to look for such a solution. Any suggestions on specific components or how to proceed will be greatly appreciated. Thank you. Edit: I would like to do it with zero other components, if possible. I have limited space on the PCB to include other components. I am currently using a BTS462T as a switch but that requires 5VDC to turn on. I am looking for a solution that will turn on the 24VDC with a low. Would depletion mode or enhancement mode be best in this situation? <Q> You won't find a PMOSFET that you can use as a high-side switch and turn the switch off (non-conducting) with 5V on the gate and 24V on the source, then turn it on with 0V on the gate and 24V on the source. <S> That's not how MOSFETs work. <S> You could combine a PMOSFET with a few other components, but it won't fit in a TO-220 package. <A> Without more information about the driving circuit and the load, this is a guess. <S> The zener can be anything in the 6.2 V - 12 V range, and the resistor can be something in the 10K range. <A> I have learned that I can pull the output of the PLC high to 24VDC even though the PLC operates at 5VDC. <S> So, I believe I can pull the gate of the MOSFET high for an off condition on the PLC. <S> And, when I want to turn on the load, I can turn on the output of the PLC thus bringing the PLC output low and turning on the high side P channel MOSFET. <S> Confirmation of this scheme is invited. <S> Thank you for all your help.
| Almost any p-channel power MOSFET with adequate ratings can act as the switch, with two small additional components - one resistor go guarantee turn-off when the driving signal goes open-circuit, and one zener diode to protect the gate from over-voltage.
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Potential difference and current flow I've been taught that electrons move from cathode to the anode (negative and positive). But what happens when I apply positive voltages at +10V at one end of the circuit and +6V at the other? How would electrons flow if both terminals are positive? Could some please clarify how we determine the sign that we put before the value of a voltage? Also, why does current not flow in an open circuit? If I connect one end of a wire to a battery and let the other hand hang without any connection, what is the potential at the loose end, the one that is in contact with the air? Is it 0V? The second question may seem like a dupe but I looked it up and it was slightly different from what I've asked, the answer for the other question slightly differs too and it didn't clear my doubt either. <Q> I've been taught that electrons move from cathode to the anode (negative and positive) <S> That's generally true but remember that ions can move in electrolytes or solutions. <S> Instead, for circuit analysis I recommend that you think in terms of conventional current rather than electron flow. <S> But what happens when I apply positive voltages at +10 V at one end of the circuit and +6 V at the other? <S> How would electrons flow if both terminals are positive? simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Current flow between to points at positive potential. <S> The resistors in Figure 1 don't know anything about where your ground reference is. <S> From the resistor's point of view each of the circuits in Figure 1 are the same. <S> There is a 4 V potential difference across it with the left side 4 V higher than the right. <S> Therefore 4 mA will flow from left to right in each case. <S> Another way of thinking about this is think what would happen if you connect a 10 V battery to a 6 V battery as shown in Figure 1a? <S> Also, why does current not flow in an open circuit? <S> How could it? <S> There has to be a circuit for charge to flow. <S> If I connect one end of a wire to a battery and let the other hand hang without any connection, what is the potential at the loose end, the one that is in contact with the air? <S> It is the same as the potential between the battery terminals before you connected the wire. <S> You've only extended the terminals. <S> Is it 0 V? <S> No. <S> This should be clear by now. <A> I've been taught that electrons move from cathode to the anode (negative and positive). <S> But what happens when I apply positive voltages at +10V at one end of the circuit and +6V at the other? <S> How would electrons flow if both terminals are positive? <S> Could some please clarify how we determine the sign that we put before the value of a voltage? <S> What happens when one reservoir of water is at 1000 m above sea level and the other one at 600 m? <S> What changes if we don't use sea level as a reference but the base of the Eiffel Tower or some other place? <S> Also, why does current not flow in an open circuit? <S> If I connect one end of a wire to a battery and let the other hand hang without any connection, what is the potential at the loose end, the one that is in contact with the air? <S> Is it 0V? <S> What happens if water flow is blocked by a dam? <S> At which level is the water at the upper side of the dam, at which at the lower end? <S> (There are some limits to this analogy between water flow and electric currents, but as far as the current question goes, it should help understanding.) <A> I've been taught that electrons move from cathode to the anode (negative and positive). <S> The direction of conventional current was defined to be from the positive to the negative before we know about the electron. <S> So from a physics perspective on the quantum scale the electron flows the other way <S> but it is really not significant information while designing electrical circuits, though it's nice to know about. <S> http://web.engr.oregonstate.edu/~traylor/ece112/beamer_lectures/elect_flow_vs_conv_I.pdf <S> Since the convential current flow is defined from the positive to the negative node you should think about it as flowing this way even though it's the oposite in reality. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The flow will therefore be from the 10V battery to the 6V battery and effectively you are charging the 6V battery with 40mA current. <S> Also, why does current not flow in an open circuit? <S> If I connect one end of a wire to a battery and let the other hand hang without any connection, what is the potential at the loose end, the one that is in contact with the air? <S> Potential to what? <S> You see potential only makes sense compared to something else. <S> The potential to it's ground would still be the same but current would simply not flow. <S> Electricity always want to come back to its ground, an analogy to potential energy in physics can be used. <S> A ball on top of a hill always "wants" to fall to because that's the natural way of things, but it can be blocked from falling. <S> Air is a sufficient insulator to stop current from flowing from your wire the ground - <S> > <S> given that the voltage is low. <S> With high enough voltage current can actually flow through air, just like thunderbolts will hit the ground. <S> This book is great for learning about basic circuit analysis. <S> https://www.amazon.com/Engineering-Circuit-Analysis-David-Irwin/dp/0470873779
| Answer: current will flow from the higher potential to the lower.
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Is base-emitter voltage of BJT transistor (common emitter configuration) constant? For sillicon the value of Vbe is approximately 0.7 V. So why is the input characteristics of such a transistor like this Why not a straight vertical line. Or is this curve plotted for I Vs Vbb. I have seen nowhere plotting this curve against Vbb. So what is wrong here? <Q> What's wrong here is that you have read that 'VBE is approximately 0.7v', and now you're wondering why transistors don't follow that rule exactly. <S> The thing is, transistors do what they do. <S> Then we make up simplified descriptions of their behaviour, called models. <S> A model gets used by a lot of people if it's found to be useful. <S> George Box, all models are wrong, but some are useful <S> For 95% of my transistor use, the approximation 'VBE=0.7v' is enough for me to make a good design. <S> In the remaining 5%, it's a valuable first pass, and a good cross-check on whether a more detailed model gives a plausible answer. <S> To explain the shape of, and the difference between, those transistor curves, you can elaborate the transistor model, with some dynamic resistance, with some current density in the junction, with temperature, with some feedback from the collector voltage. <S> Because all of those complications means VBE varies by as much as 100mV in either direction, a good transistor design will have bias arrangements that allow for that variation in VBE without too large a variation in the collector current. <A> I understand your question - rather often, I have seen that there is some confusion regarding the role of the base-emitter voltage. <S> Let`s start with the general statement <S> : The BJT is a voltage-controlled device - <S> that means: <S> This is formulated in the famous Shockley equation Ic= <S> Io[exp(Vbe/Vt)-1] . <S> Because the base current Ib is (nearly) a fixed part of the collector current (1% or less), the relation Ib=f(Vbe) is also an exponential characteristic as shown in the above graph. <S> For using the BJT as an amplifier, we must bias the base-emitter diode with a DC voltage Vbe which is approximately in the range of 0.7 volts. <S> This defines the so called "operational" point within the "quasi-linear" part of the exp. <S> characteristic. <S> However, due to several uncertainties (tolerances, temperature effects) we do not know the exact value of Vbe for a certain (wanted) collector current. <S> Therefore, we use an emitter resistor RE which provides a certain amount of negative DC feedback - with the following result: <S> The actual DC collector current Ic will be less dependent on actual transistor parameters. <S> Therefore, in most cases it is sufficient - for calculation purposes only (!!) <S> - to use a fixed value Vbe=0.7 volts. <S> This value is used to compute the resistors for the biasing network. <S> On the other hand, of course the base-emitter voltage will vary around this bias point as a result of a signal voltage to be amplified. <S> This variation d(Vbe) is transferred to the output (Ic variation) with the transconductance gm of the BJT. <S> This transconductance gm= <S> d(Ic)/d(Vbe) depends directly on the selected DC current Ic. <S> Ic=f(Vbe) <S> and can be found to be gm= <S> Ic/ <S> Vt (Vt=temperature voltage, app 26mV) <S> In the attached graph it is demonstrated how an emitter resistor RE can remarkably reduce the influence of circuit uncertainties (tolerances, temp. effects) on the resulting DC current Ic. <S> Therefore, it is not too important if we set VBE=0.7 V or VBE=0.65 V during the calculation process. <A> Lets plot the input base-emitter voltage and current, on different axis. <S> In fact, lets use linear-voltage on horizontal axis, and use log-current on vertical axis. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The voltage changes 58 milliVolts, for each 10:1 change in current. <S> Increase or decrease, the sign changes, but the magnitude is 58 milliVolts. <S> From 1 milliAmp to 10 nanoAmps, which is 5 orders of magnitude, we expect 5 * 0.058 <S> , or about 5 <S> * 0.06 = 0.300 <S> volts change in the diode voltage. <S> Or 18 milliVolts for 2:1 change in current (BandGap voltage reference designs will have 18 milliVolts, or multiples thereof, deep inside the design math). <S> Beware of strong temperature effects. <S> And semiconductor processing effects. <S> The 58mV isfor ideal transistors and diodes.
| The collector current is an exponential function of the base-emitter voltage Vbe. This transconductance is simply the slope of the curve
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How can I change step-down my variable input voltage? Ok so here's the deal: I have a variable DC Voltage source from 0-10V. I need to step that down to a variable source of 0-3V. This 0-3V DC will be fed to an Analog to Digital converter in a microcontroller. I know I can potentially use a Voltage Divider (using resistors) but apparently, that's not a good solution. Op-amps don't provide a gain < 1. So I'm just struggling as to how I can accomplish this. The microcontroller: https://www.microchip.com/wwwproducts/en/PIC18F47K42 <Q> R divider works fine as long as R is not too high. <S> If you know the conversion rate and Hold Cap acquisition time EQUATION 36-1: ACQUISITION TIME EXAMPLE in the datasheet provides the formula and example of choosing R values < 10k. <S> Rs is the source equivalent // <S> resistance of the R divider, R1//R2. <A> If your source impedance is high, use an opamp in voltage follower mode (gain = 1) before the divider. <S> If your ADC impedance is low (unlikely) use a voltage follower after the divider. <A> In addition to @Toor suggestion (voltage follower and voltage divider) and in response to OP statement <S> Op-amps don't provide a gain < 1. <S> follows my alternative configuration using a single supply difference amplifier. <S> The output is reversed - if you do not mind correct it in software applying \$V_o=3 - 0.3V_i\$ .
| A resistor divider works fine as long as the source impedance is low and the ADC impedance is high (compared to the resistors used for the divider).
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Can I use USB data pins as a power source? I'm trying to build a circuit as seen below, where a sequencer controls the flashing of some LED sections, and an SPDT switch controls power to either the left or the right section of LEDs. However, I want the PCB with the switch and power source to be located a distance (~1 meter) from the LEDs and sequencer, and I was wondering if I could use USB to connect them. To be clear, I am NOT connecting this to any sort of computer or microcontroller, I want to know if it is possible to have a USB cable provide VCC and GND from one circuit to another, and then use D+ and D- to provide/disconnect power to the LEDs via the switch. All voltages would be 5 V, and I can adjust the current draw of the LEDs as necessary. If this is not possible, is there an existing type of cable that can achieve this (three power leads and a ground), or will I have to solder my own? <Q> This is neither Kosher nor Halal. <S> Some certification standards explicitly frown upon this. <S> Besides the wires probably being extremely under-dimensioned for the delivery of power, using narrowly-defined standard connectors for non-standard uses is a sure way to cause an unforeseen problem down the road. <S> Particularly if this use would ensure the destruction of any standards-compliant device that could be plugged into it, as is this case. <S> I have had apparently standard but mis-wired cables in my hand that have caused the malfunction and even destruction of standard-compliant devices. <S> Standardization bodies actually spend a lot of time designing their connectors so that these cannot be confused/mistaken/interconnected with anything else out there. <S> Don't make our job harder. <S> Some connector types are more generic than others, stick to those. <A> If this is not possible, is there an existing type of cable that can achieve this (3 power leads and a ground), or will I have to solder my own? <S> Cables will do whatever you want them to do, they carry current from one point to another. <S> Most USB cables use bigger conductors for the power lines, so the data lines would best be used for lower currents. <S> At that point it might be easier to just go down to your local hardware store and splice a 4 conductor cable and solder to that. <A> Your best bet is likely to be a more generic connector. <S> There are many common types of cable out there other than USB that will have two major advantages: <S> Larger conductors on all pins, and no chance of damaging USB devices by mistaken connections. <S> Consider using a DIN connector on either end of your cable, for example. <S> These connectors are common enough that they're cheap, and generic enough that the presence of the connector does not automatically make people expect it will work for any one protocol. <S> Molex's mini-fit and micro-fit connectors are also popular for power, but be careful if you use mini-fit jr connectors, as they are commonly used on PC power supplies. <S> JST connectors are commonly used for batteries in RC applications, and may be a good fit for this as four-conductor JST connectors are sufficiently uncommon that people wouldn't have any particular expectation for what they're meant to be used in, yet sufficiently common that they're very cheap, even pre-assembled cables with a four-pin JST connector on either end can go for less than $2 . <S> The links in this post are just what I found in a very quick search of digi-key; if you search yourself you may find more options, and you can tailor to your needs regarding conductor size. <S> Or you can look for just the connectors and make your own cable, which is not that hard if you have a crimping tool. <A> No, USB cable's data lines are significantly thinner than their power lines, and shouldn't/can't be used to supply any significant amount of power.
| The biggest concern would be if someone accidentally plugged the USB cable into a regular USB device which would probably blow out the data lines on that device. But you can use any cable you want as long as you can find ends for it and a way to solder your circuit to the ends. It should actually be explicitly forbidden in the Electrical Engineering code of ethics.
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How does one determine the type (AC v. DC) and operating voltage of small motors? I've become fascinated by Rolling Ball Sculpture (RBS) and plan to start building my first one soon. Typically a RBS uses a low-voltage, low-RPM motor to carry the marbles to the top of the structure where they begin their journeys down the track. I've been scavenging fan and turntable motors from old guitar amps, computers, microwaves, etc., but am at a loss as to how to determine the type of motor (AC v. DC) and operating voltages of these normally-unmarked motors. Thank you in advance for any help you can provide. <Q> Do you have a multimeter? <S> Easiest way would be to probe the motor while it's running in its original circuit. <S> Aside from that, if it truly is unmarked then it'll be tough to tell. <S> I believe DC motors typically have only 2 wires, some AC motors have more wires for additional phases. <S> I'm not an expert on the different types of motors, but searching your question yielded this thread , which seems to have a lot of relevant info. <S> The safest way is the buy a motor with corresponding specs <S> so you know exactly how to drive it. <S> Alternatively, the method mentioned in that other thread of slowly ramping DC voltage to see if it works may also suit your needs. <A> But you could try say 12v dc on some of the ones if they are computer cooling fans - geared down <S> they should be ample for moving marbles. <A> Most modern motors can be differentiated by the color, number and size of their conductors. <S> 2 wires usually black and red for DC motors, black and white for AC, and 3 wires, (red black blue/orange brown yellow)for 3 phase AC motors at low voltages, sometimes with an extra white wire if they are wired in wye for some reason. <S> Extra wires outside these sets may provide individual access to coils of an ac motor, or may indicate a motor with speed or temperature feedback, or provide control functions if they are smaller. <S> Make your best guess and then carefully apply voltage. <S> Very little is likely to go wrong if you connect it to 1.5V <S> , then 3V then 4.5V DC by adding 1.5V cells in series. <S> If it doesn't spin unloaded one way or the other at 4.5V, it's probably not a low voltage DC motor. <S> Just an option if you don't have a power supply to carefully test with.
| The size and color of the additional conductors will often let you identify special motor types like positional motors or steppers. Measure V before removing and write it on them...
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Aluminum electrolytic or ceramic capacitors for linear regulator input and output? I am using this linear voltage regulator . The datasheet indicates the input and output values for the capacitance to use, 1uF and 10uF respectively. Should these capacitors be or a particular type, or does it not matter? <Q> As the datasheet for your regulator doesn't seem to say anything about that at a glance, it should be fine with any capacitors you pick, but see the edit below for a warning. <S> Non-polarized capacitors more than about a microfarad used to be rare and expensive, which is probably why the datasheet shows polarized capacitors being used. <S> Today, you can get 10μF ceramic capacitors for less than $0.30 each. <S> Edit: As @ThePhoton points out, this regulator may be so old that multi-microfarad ceramic capacitors, with their inherent low ESR, may have been a far-off pipe dream to the engineers writing the datasheet. <S> So this may still be unstable with too low an ESR on its output, so unless you want to test its stability under different operating conditions with the ceramic caps, it may be best to stick to aluminum electrolytics. <S> After all, that's probably what the IC's designers had in mind. <A> The datasheet application circuit example schematic shows a 1 microfarad polarized capacitor on the input and a 10 microfarad polarized capacitor on the output. <S> Since the values are in the 1 plus microfarad range and the capacitors are shown as polarized, I would guess that the manufacturer (ST) wants you to use electrolytic caps. <S> I guess you could use a tantalum caps, but unless the datasheet specifies tantalum, that would be a needless expense. <S> Very few ceramic caps are over 1 microfarad and very few are polarized. <A> What matters most is a temperature stable ESR with a nominal value of 0.3 Ohms +-25% for good stability and <S> C>22uF. <S> This must be your search criteria for ganged output caps. <A> In contrary to the previous answers, it does matter. <S> The electrolytic cap is very good at storing a large amount of energy, but bad for high transient. <S> Having large transients on an electrolytic cap will shorten its lifespan. <S> Ceramic capacitor store less energy but are good at coping with large transients. <S> That is why, a good implementation, as shown on the datasheet, is a large electrolytic cap that is there to store the energy, and a ceramic cap, as close as possible to the switching element, to take the switching transients. <S> If you only put an electrolytic cap, the lifetime of your circuit will be impacted, and only a ceramic cap, your circuit might be susceptible to issue when you have rapid power draws or input line unstable feed.
| It doesn't usually matter, but be aware that some linear regulators--the popular LM2940 series, for example--may be unstable if the output capacitor's ESR is too high or too low. The polarized caps shown on the datasheet circuit lead me to believe that electrolytic caps are what are intended.
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Potentiometer warm up issue I have a microcentrifuge which has a DC motor rated 24V, 10 Watt and 7000 rpm. I have attached one 5W, 500 Ohm potentiometer with it to control the speed and it works fine without heating up. This microcentrifuge is supposed to be used with a light weight part. But as I need to use it with some large tubes I have 3D printed some parts and attached on the rotor part. This time with this extra weight my potentiometer becomes very warm within 1 minute. Is it because of the extra mechanical load or it is something wrong with my potentiometer rating? <Q> Controlling a 10W motor with a 5W pot isn't great in the first place. <S> At certain speeds, you could already be getting close to 5W, and that won't be spread evenly across the whole length of the pot. <S> The more mechanical load you place on a motor, the more current it draws. <S> More current means more heat in the potentiometer. <S> If you are turning the pot towards one end to increase the speed to compensate for the extra load, then the heat is now being dissipated in an even smaller part of the pot's track. <A> Theoretically, additional weight should only effect the load during acceleration. <S> However you stated "I have 3D printed some parts and attached on the rotor part. <S> " The additional load may be due to the aerodynamic drag created by that modification. <S> Also, the "large tubes" would cause additional drag if they are not entirely enclosed by centrifuge rotor compartments. <S> Another factor could be the operating speed. <S> If you are operating at a different speed than your initial test speed, the heat dissipated in the pot would be different. <S> Also, 500 ohms is not the optimum value for a 10 watt motor. <S> it should be closer to the effective resistance of the motor, <S> 24V^2 / 10W = 58 ohms. <A> If you have 500 ohm and 24 V, nothing happens. <S> If you reduce the value to 50 ohm, though, Joule losses become 10 times as large and at 10 ohm 50 times: $$\begin{equation}P(R) = <S> \frac{V^2}{R}\end{equation}$$ If you can, you should measure the potentiometer voltage and current. <A> You could just go buy a bigger wirewound pot. <S> You could make a PWM controller using a 555 and a power MOSFET. <S> If you go the latter way you'll also need a voltage regulator such as LM7812 because the MOSFET gate and the 555 won't like 24V. <S> Here is a rough idea, diagram was originally on twovolt.com but that site seems to be down so no link.
| My guess: When you want to use the extra tubes, you are probably changing the potentiometer setting. You could buy a DC motor speed control (a few dollars from China).
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Use a TRIAC to connect/disconnect a capacitor to a PSC motor In the image above, we see a PSC motor, with its 3 terminals, connected to a control board.This board has a RUN capacitor always connected to the motor. When the motor has to turn, the board applies 230V AC between the COMMON line and either the FORWARD or REVERSE line. The relevant winding is powered directly, and the other winding gets powered via the RUN capacitor, with about 90 degrees delay. This makes the motor run in one of the two direction. What I want to do is to connect, in parallel to the first, a second capacitor in order to boost the torque (when the motor is starting). After a little time the motor is started, this added capacitor should be disconnected. I would like to use a TRIAC instead of a relay. Can anyone point me to choose the correct device, and give any warning I may need? I've already tried, but the TRIAC fails in short-circuit without heating or other signs (the real schematic is not simple as the one above, but the concept is exactly that). The motor is about 1kW, @230V; the capacitors should be about 25 uF each. The current is about 1A continuous but it can rise to 4A when the motor starts or when the load gets heavy (I'm using a workbench brake for testing). EDIT: The driving part is like this: Basically, the 2nd capacitor has a series triac; the triac is gated by a MOC, taking current from A2 (MT2); the resistor from gate to A1 (MT1) is optional. The Triac goes short-circuit, but the MOC keeps to work. The capacitors, which are normally used and considered reliable, are rated 470V. My triac is an ST, snubberless, rated 1200V and 16A (continuous). There is something I don't grasp... <Q> I suspect that snubberless means that the traic does not need dv/dt snubber but still may need di/dt protection. <S> Switching capacitors will result in high di/dt. <S> That is probably the mode of failure. <S> You will need to add di/dt protection to the circuit or use a relay. <S> A higher current triac might work without di/dt protection. <A> Because you appear to have a random phase turn-on you have no control of the point of turn on. <S> This can result in very high circulating currents between the two capacitors in the circuit you show. <S> You need to put a zero crossing detector across the permanently in place capacitor and use this signal to tell you when you can turn on the TRIAC. <A> Why are you attempting to re-invent this circuit? <S> What problem are you trying to solve? <S> How are you planning on deciding WHEN to disconnect the capacitor? <S> From the standpoint of the triac/SCR, the charging current of the capacitor looks like a short circuit and the resultant dI/dt can damage the thyristor, then when you turn off the triac/SCR, the inductive kickback from the motor winding can cause a dV/dt issue. <S> Preventing all of that is going to cost more than the potential benefits of changing to a triac/SCR, which is why you do not see this concept used by motor mfrs. <S> FYI, GE developed this concept a long time ago and abandoned it as being actually LESS reliable than the electro-mechanical method. <S> GE's circuit, includes RC snubber and discharge resistor.
| If you turn on the TRIAC when the voltage is low, then there will be no (or at least very low) circulating currents.
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Reasons for having MCU pin-states default to pull-up/down out of reset On many MCUs, pin-states default to tri-stated (a.k.a. analog inputs) when the MCU resets so as to not affect the circuits they are connected to until software configures the pins. The tri-stated pins also allow the HW designer to choose the pull state of each pin on a case-by-case basis in function of the underlying circuitry. However, there are some MCUs (and SoCs) that default their pins to instead activate an internal pull-up/down. For example, the LPC845 defaults all pins to pull-ups coming out of reset. Is there a reason defaulting pins to pull-up/down is preferable to tri-stated (other than the possible incremental power savings when coming out of a reset, or the marginal BOM cost savings)? If anything, I rarely find that pins should be pulled- up coming out of reset (I typically need to pull them down, if at all). <Q> Because it's good practice to never leave logic pins purely floating. <S> Brief TI Overview <S> Detailed TI overview Amusing dramatized war story of a real-world example <A> Other answers have given general reasons why a chip maker might make the choice to enable pull-ups by default. <S> However, in the specific case of LPC845, there is an additional reason: it has specialized FAst <S> Initialization Memory (FAIM) that can be used to set the state immediately after reset: <S> The FAIM contents provide a user-programmable initial configuration for aspects of the microcontroller, which take effect immediately after reset, before code begins to run. <S> For instance, the standard I/ <S> O pads normally come out of reset with the internal pull-ups enabled. <S> In some systems this may cause excess current to flow, until software can reconfigure the pads. <S> However, by programming the FAIM appropriately, every pad's reset configuration can be customized. <S> ( LPC84x user manual section 4.2) <S> Thus they've chosen the safe (from power usage and EMI point of view) default, while allowing more advanced users to customize the setting. <A> Back in the days there were Intel 8051 microcontrollers that only had open drain <S> I/ <S> O pins, so most of the time you needed external pull-ups anyway to do useful things like connecting to pushbuttons or controlling CMOS inputs of other chips. <S> This is most likely to have easy redesign of such boards with a modern microcontroller, or people from that era that are accustomed to designing with pulled-up open-collector <S> I/Os. <S> Back in the day, you mostly needed pull-ups if anything, and rarely pull-downs. <A> Leaving GPIO pins as tri-stated inputs have many undesirable effects: <S> As manufacturing process has certain variance and a lot of other circuitry is connected to GPIO (as output buffer and ESD protection), direction of resulting parasitic leakage is unpredictable, so the state can take either logic high or low; Again, due to process variation and temperature dependence, the pin leakage can be very small, resulting in either very slow change of logic state after, say, several minutes, which might be a challenge to accommodate in code, or it can drift in unpredictable direction. <S> Leaving pins floating might lead to establishing some middle potential, where the pin input buffer may act as linear amplifier with substantial gain, causing either self-oscillations (due to parasitic positive feedback across power rails), or be susceptible to external electromagnetic interference. <S> Oscillations can be somewhere internally, and lead to out-of range power consumption. <S> 4... must forget something else... <S> power-on transients? <A> For example, a motor might be attached that shouldn't be activated without command. <S> Peripherals usually expect their interfaces to be in a certain state, and starting in high-Z may not provide the required state. <S> As an additional note, it is nice to see in the datasheet what the expected behaviour of the pins is, this is sometimes not included..!
| As the internal pull ups/downs in a typical microcontroller are relatively weak, they may be overridden by a stronger external pull up/down where required. From a systems point of view, having the pins start in a defined state is a benefit.
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Help identifying an SMD component I need help identifying an SMD component it is a small(1 mm X 1.5 mm) black part with an "O" written on top if it.I think it is a Schottky diode for reverse polarity protection ?. ( Source , first part on codes beginning with O) I have attached an image if that helps The component in question is marked J2/J3 on the silk screen. Also, if it helps the image is of a remote control pcb. PS: This is my first post so please let me know if i did something wrong :) <Q> Basically it is a solder jumper or jumper wire <S> but in a machine assembly friendly package. <S> On single-layer boards these are often used to assist routing and jump over traces. <S> This appears to be the case for J2 and J3. <S> They are also frequently used for changing configuration pins. <S> You typically have two resistor footprints, one to VCC and one to GND. <S> On this you can place a zero-ohm resistor on either footprint to pull the configuration pin low or high. <S> This is sometimes known as "strapping" and appears to be what OP3 is used for. <A> I think it is a Schottky diode for reverse polarity protection <S> Very unlikely. <S> A diode should have a different package, and a diode would usually have its cathode terminal marked. <S> You have a 0 Ohm resistor here, the label "J" may be because it's used as a j umper (see Peter Karlsen's comment). <A> That is an 0 Ohm Resistance. <S> they also called SMD jumper resistors. <S> They are used as wire links to connect the traces on Surface mount boards, which can be assembled using pick and place machines easily. <S> (same like jumper wires in through holes boards) and as well as changing the routing in pcb's the size 1mm <S> * <S> 1.5mm indicates that it is 0603 SMD package .
| These are 0603 zero-ohm resistors - the "O" is actually a "0".
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How is this circuit a voltage regulator? I understand how the v_o expression is found, but I don't really understand how or why it's a voltage regulator. <Q> The circuit is a voltage regulator because the output voltage <S> \$V_o\$ does not depend on the load. <S> Whether you connect a \$1~\Omega\$ , \$1~\text{k}\Omega\$ , \$10~\text{k}\Omega\$ , etc., resistor from the output to GND, the voltage at <S> \$V_o\$ will remain at the voltage set by the expression you've derived. <S> The current \$I_o\$ , however, will be different in all these cases. <S> \$V_o\$ <S> based on the values of \$R_1\$ , <S> \$R_2\$ and \$V_Z\$ . <S> The "opposite" circuit is a current regulator, which keeps the current \$I_o\$ at some value regardless of the load connected to it (and <S> the voltage \$V_o\$ changes accordingly to make this happen). <A> Because the zener diode on the left provides a voltage reference, and the op-amp <S> + transistor provides a power, 1-quadrant amplifier for that voltage reference. <A> The Zener diode provides the regulation source. <S> Ideally, you would choose a Zener whose temperature coefficient is near zero (around 5V -to- 6V). <S> Two feedback resistors allow you to choose an output voltage (regulated) that is greater than or equal to the Zener voltage. <S> Just like most voltage regulator chips, this circuit sources current, but cannot sink current. <S> It has no "input", so it cannot really be called an amplifier . <S> If you regard <S> \$V_A\$ as an input <S> , then the output voltage is independent of its input (so long as input > output ).
| In other words, it regulates the voltage
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24V DC to +5V and -5V I need to convert 24V DC to +5V and -5V to power two op-amps (opa2134) and one ATtiny85 with minimal heat generated from the circuit. What voltage regulator would you recommend ? //Digis <Q> Note - these usually require additional capacitors on the input and outputs. <S> If you want to grow it from scratch, I would go with a linear regulator for the +5 V (because the total load current is so low) and a non-isolated buck-boost converter to turn the +24 V into -5 V. <S> There are dozens of parts to choose from for this, including ones with the power transistors on-board. <S> Linear Technology (now a part of Analog Devices) is a good company for this. <A> For less than 10mA total current per supply, consider an LM7805 + ICL7660 . <S> At the higher end the 7805 will get warm but it won't require a heat sink in most situations. <S> It will draw approximately 20mA <S> +5mA <S> Iq from the supply at 10mA + 10mA output current, so the power dissipation will be (24-5)V <S> * 20mA + <S> 24V <S> * 5mA = <S> 500mW. <S> For higher asymmetric currents, like 50mA on the +5 and 10-20mA on the -5, you can consider a switching buck regulator and the same 7660 to turn the +5 into -5. <S> Suitable switching regulators are the well established LM2596 or MC34063. <S> If you really need the ultra-low distortion amplifier you mentioned, you may wish to think about how to mitigate noise from a switching regulator if this could be a concern. <A> What voltage regulator would you recommend for 24V to +/-5 <S> regulated ….with minimal heat generated from the circuit. <S> I'd suggest that the cheapest but not quite the lowest power would be to use two of the LM2596 based switching regulators readily available online: <S> These cost about $0.60 each and are a great replacement for older linear regulators. <S> You'd need one for the +5 <S> and one for the -5. <S> You can use one of these to generate a -5V supply simply by using a series diode to the unit, and grounding the OUT+ pin. <S> The regulated -5V is available on the OUT- pin <S> The only downside of these regulators is that they ALWAYS draw a peak pulse current of 3A, so your 24V supply needs a reasonably high output capacitor to supply this. <A> If you don't care about cost, but want low noise, LDO regulated charge pump <S> 100µA Quiescent Current use <S> LTC3260 . <S> But if you don't give any other requirements, how can you search or design or decide? <S> ( collorary to the best question has better specs )
| The most simple solution is to buy a small DC/DC converter with bipolar outputs in an oversized 24-pin DIP package.
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12 v dc led lights in parallel I have a 12 v dc power supply 36 watt 3 ampattempting to run 5 12v dc 4 watt led lights in parallel. at first i tried running 5 hallogen lights but figured they were drawing to much power so switched them out to the leds. my question is i am being told i need to put in a resistor to keep from blowing my power supply. i do not know the led parameters such as voltage drop or how much power will flow through the led. and need to know what size resistor needed? my led are GY6.35 G6.35 Bi-pin Base LED Bulb 4Watt AC DC 12V Silica Gel Crystal Daylight White 6000k Landscape Lighting,JC Type, Equivalent 25W- 30W Q35/CL/T4 Halogen please note i am creating a circuit from scratch every time i try power up my lights my power supply kicks off <Q> Actually, I'm not sure that Tony's idea will help, because the surge that is tripping the over-current fold-back of the power supply is occurring when the string is first powered. <S> At that time, the PTC resistor will be cold, and have low resistance; by the time it heats up, the supply may have already shut down. <S> Actually, a simple small-valued resistor may be all that's necessary. <S> A resistor of 4 ohms (3.9 is the closest standard value) would limit the supply to its maximum rated current, but would also reduce the operating voltage of the LEDs to around 6 volts, which would no doubt result in unacceptable dimming and wasted power (about 10 watts). <S> However, you might try just using a 1 or 1.5 ohm resistor; this would limit the maximum current to 12 amps while the LED capacitors charged, but only for a few milliseconds; the power supply might just ride that out fine. <S> I had a similar problem with a switching adapter, but one where the LED string didn't even have the capacitor across the line. <S> In my case, the problem was that the (cheap) switcher couldn't start up under near-full load. <S> The solution in my case was to put the on-off switch on the 12-volt side rather than in the 120-volt line. <S> This of course results in the supply being powered up all the time, but once fully started, it had no problem handling <S> the load of the LEDs being turned on. <A> Your lights appear to be direct replacements for 12 volt Halogen lamps, so won't need any additional components when operated from a 12 volt power supply. <S> "Bare" LEDs (individual electronic components) do need a current-limiting resistor or circuit in series. <A> This 12V LED string is designed to draw 333mA (12V*333mA=4W ) and thus 5 such LEDs will draw 1.67A steady state. <S> But I see a 10uF cap on each one which usually have an ESR of 0.1 to 1 Ohm thus <S> the surge current may be tripping your DC over current protection (OCP) and latching off with a surge much much greater than 3A. <S> One solution is to insert an ICL rated for <= <S> 3A steady state in the wiring. <S> This goes in series with 12V and reduces resistance as it heats up so initial inrush can be handled by 4A PS.
| If the bulbs are really LED assemblies designed to operate from 12 volts, then you do not need a current-limiting resistor in series with them.
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What is the best way to get 7V from a benchtop 10V supply? I have a 50 ohm load I am trying to drive at 7V. I currently have a 10V supply with a 10kohm potentiometer but I am expecting to run into trouble when I connect it as a voltage divider. I looked online for a 7V supply but I couldn't find much. What is the easiest way to get 7V that can drive this load? I can purchase some equipment but would like to keep it to a minimum. <Q> Best is hard to determine.. <S> but here is one way for your 140mA (7V/50 \$\Omega\$ ) <S> load current: <S> You can use an old-school LM317 regulator. <S> In your case you can use 249 1% for R1 and 453 1% for R2 which should get <S> you close to7V out. <S> The chip's power dissipation will be 0.15 <S> * 3V + <S> 0.005 * 10V <S> = 0.5W, so a TO-220 case regulator should be okay in most circumstances without a heatsink. <S> The somewhat newer LM1117 could be used but you have to pay a bit more attention the the output capacitance ESR to ensure stability (and the resistor values will be different because the reference voltage is 1.25V nominal vs. 2.5V nominal for the LM317). <S> It does give you lower dropout voltage (it's a semi-LDO regulator), but since you have 3V to play with that's not required. <A> with a 50 ohm load? <S> if you add add 21.5 ohms in series.the 21.5 ohms will get 3V and the 50 ohms will get 7V <S> 22 ohms is a standard size, and may be close enough for your purposes. <S> else add the 10K pot in parallel (with 100 ohms in series with te pot) and use that to trim the resistance. <A> I have a 50 ohm load I am trying to drive at 7V. <S> I currently have a 10V supply <S> , what is the easiest way to get 7V that can drive this load? <S> It depends on whether you need a regulated 7V or not: <S> If your 10V is regulated and your 50 Ohm load does not vary then all you need is a series resistor of 21.4 Ohms. <S> If your 10V power supply varies then you need some form of regulator to provide a regulated 7V to the load. <S> If you do need a regulator (load varies or Vin varies) then there are multiple solutions: <S> A shunt regulator such as the TI <S> Almost any linear regulator such as the LM317 would provide for a wide variation in load resistance and power supply variations from about 5.5V to 20V or more. <S> There are endless varieties of linear regulator replacements through to modules on small PCBs. <S> Be careful that some modules have a small adjustment range (5V +/-5%) while others have 3-15V adjustment. <A> If a 70% efficiency adjustable LDO is acceptable, that is the easiest. <S> There are many types with different current ratings and adjustable. <S> Voltage and current display at low cost.
| TL431 along with a series resistor can allow for a small range of load (50 Ohm) and supply voltage (10V) variation. You can even use a Darlington NPN pair with a pot and a cap on the base as long as you allow for at least 1W heatsink. Any form of switching regulator that works with a relatively low Vin - Vout differential. You can also buy variable DC-DC boards with many options incl. One modules I've had good success with are the ones based on the MP1484 chip such as this .
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Op Amp: signal measurement that is below its common mode input range I was wondering, how is possible to use an op amp, which its common mode input voltage is (V-) + 0.2V to (V+) - 0.2V to measure signals as low as 10mV for example? I'm looking for an instrumentation amplifier and most of them I see have very narrow Vcm range with respect to their supply. <Q> I'm looking for an instrumentation amplifier and most of them I see have very narrow Vcm range with respect to their supply. <S> Consider using the AMP04 instrumentation amplifier : <S> - Although not tested and guaranteed, the AMP04 inputs are biased in a way that they can amplify signals linearly with commonmode voltage as low as –0.25 volts below ground. <S> This holds true over the industrial temperature range from –40°C to +85°C. <S> It operates from a single 5 volt rail too <S> and, the output can swing down to a couple of mV. <S> However, a limitation is that from a 5 volt supply, the upper input commonmode range is limited to +3 volts. <S> There may be better alternatives - I suggested this one because I use it in a design. <A> an inverting DC/DC converter is what you're looking for. <S> Assuming your Opamp doesn't draw a lot of current, that's typically solved with an inductorless inverter for cost and space reasons - a switched capacitor charge pump. <S> These things are rather cheap . <A> Consider the INA28x , which has an input common mode range of -14V to +80V, independent of supply voltage. <S> It is, however, fixed-gain. <S> There are variants with gains of 50, 100, 200, 500, and 1000, but if you need a different gain you will need to add another stage after it. <S> I've used this part in a design before; it's easy to use and doesn't require many external parts. <S> It even has an internal resistor divider for optionally generating a reference of half the supply rail--though you can also feed it any arbitrary reference voltage. <A> If you use an inverting amplifier configuration with the positive input terminal fixed at a reference voltage within the allowable common mode input range, then op amp never sees anything outside the common mode range! <S> Of course, you're then amplifying the difference between the signal and the reference. <S> There are also some rail to rail instrumentation amps that accept inputs outside the rails on the low end, to about -100mv, like the AD8226
| You'd just have a negative supply rail (which is "negativer" than the voltage you want to sense) –
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Parallel Power Supplies with Ideal Diodes I'm working on a project that operates at 5v - and can draw UP to 10 amps. Under normal conditions, it won't be doing that - probably closer to 3 amps, and (if configured for low-power) only 450mA or so. It's designed to be powered from USB when in low-power mode, but I also want to have 2 barrel jack to add external power supplies (5V, 8A from aliexpress). Obviously I don't want the power supply backfeeding the USB, or the USB backfeeding the power supply. I want the option to plug in two of these power supplies because most of the time, I won't need to activate the 10A mode - so just one power supply would be sufficient. Considering these power supplies are not designed to be paralleled (no master/slave setup, etc), how effective would an ideal diode be in enabling these various sources to all operate in parallel (to increase max current)? Note that I don't need the load sharing to be perfect, but I'd like to avoid serious issues. I originally thought I could just use a P-Channel MOSFET, but after some research have learned that when they are "on" they allow current to flow both ways, which makes them non-ideal non-diodes (less than helpful). <Q> You need two back-to-back P-channel MOSFETs for each power input. <S> Here is an example from an Atmel eval board, though you'll want to scale up the MOSFETs considerably, and maybe reduce the resistor values to something like 10K: <S> The VCC_USB input is dominant if both are present (in other words, if only one supply is present, that one will be used, if both are present, then VCC_USB is used). <S> The back-to-back MOSFETs are "ON" when the common gate connection is at ground (assuming the left source is high) and are off with the gate connection equal in voltage to the source. <S> This is the common configuration you'll see in hot-swap controllers and similar devices. <S> The connection to R613 turns Q109A/B on if VCC_USB_EDBG is low. <S> If both supplies are present Q110A/B are off (because their gates are high) and Q109A/B are on. <A> I'd like to take a different approach here... <S> Why not design an active-power supply system? <S> By active I mean, you can use a microcontroller + PMOS setup. <S> Note that I don't need the load sharing to be perfect, but <S> I'd like to avoid serious issues. <S> I originally thought I could just use a P-Channel MOSFET, but after some research have learned that when they are "on" they allow current to flow both ways, which makes them non-ideal non-diodes (less than helpful). <S> True, NMOS will also allow both ways current transmission if on. <S> So we turn them off if we don't need them. <S> Here is what we can do: Supply 1 > <S> > <S> PMOS <S> > <S> > loadSupply 2 <S> > <S> > <S> PMOS <S> > <S> > loadSupply 3 <S> > <S> > <S> PMOS <S> >> load... <S> Supply N <S> > <S> > <S> PMOS <S> > <S> > load <S> We can use as many power supplies as we want. <S> We tie them to the load using a PMOS. <S> Now, we also get a microcontroller, and read load voltage . <S> Whenever the load voltage drops a certain level, we switch to the higher powered supply, using the aforementioned microcontroller. <S> (You better get some capacitors on load side if you don't want that voltage dropping too hard) <S> There is another method as well! <S> We can tie hall-effect current sensors in each of these paths. <S> Then we read the currents with our microcontroller. <S> Whenever the current limit is reached, microcontroller will read this and switch to higher powered power-supply. <S> There is also a third option; your load gives information to our microcontroller! <S> Whenever you press a switch/slider in your load, microcontroller acts beforehand and switches the power supply that you know and have tested that it works. <S> So in this setup, you have no voltage or current sensing. <S> I also have a fourth option (Bonus): Let's say you want load sharing. <S> Since you have a microcontroller, you can basically PWM those PMOS' (with certain deadband of course) and create load sharing. <S> If you will try this out, don't forget to use interrupts, because you want to be as fast as possible. <A> You may want to re-consider your requirements with assumptions in specifications. <S> Current sharing multiple power supplies can be unstable with a light load on the source that is barely supplying current when there is a load disturbance on 2 shared sources regulating on the same load. <S> The effective loop gain drops when a regulator is driven by another source slightly higher voltage from mismatched voltages. <S> This loss of feedback gain can result in overshoot to a sensed disturbance. <S> Often a preload of 10% is needed to limit the variation in loop gain and the resulting loss of phase margin. <A> LTC4370CDE would be my first choice to use to tie two power supplies together, but I wouldn't even bother with the computer's usb port power because its going to be poor quality. <S> Even on a $2000 motherboard, they are still junk.
| You can combine this with a temperature sensor on power supplies to even balance/switch off some of them if need be. Because of the intrinsic body diodes you need two MOSFETs to be able to block in either direction.
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Can a Current Limiter Power Thermistor be used in line to reduce the brightness of a 120v incandescent bulb? I have an application were I do not have the room for a dimmer switch but need to reduce the lumens of a 120v 25w incandescent bulb just a bit. There used to be a product called the Bulb Saver Button which was placed in the light bulb socket to reduce power consumption and increase the life of the bulb. It also reduced the brightness of the bulb. This would work for me but this product is no longer available. I'm wondering if I simply put, in line, a Current Limiter Power Thermistor, would it do the trick? I found on ebay this 10D-15 Inrush Current Limiter Power Thermistors 10ohms 5 AMP. Would this work like the Bulb Saver Buttons? <Q> You could use a common diode in series with the bulb, which will reduce its brightness significantly. <S> The advantage of a diode is that the amount of power wasted in the diode is much less than if you used a resistor to do the same thing. <A> No, this Thermistor would change almost nothing. <S> For your bulb when directly on the mains, the current is: 25 W / 120 V = <S> 0.21 AAnd <S> that means the bulb will have a resistance of 120 V / 0.21 <S> A = 580 ohms. <S> Adding a 10 Ohms thermistor to that is insignificant as you'd get 590 ohms. <S> Then the thermistor would heat up and get an even lower resistance. <S> So in total, nothing significant changes. <S> That thermistor is used in power supplies which need to charge large capacitors at startup. <S> It will do nothing for your lightbulb. <S> Better ideas: <S> use a bulb with a lower power. <S> use two bulbs in series <S> A bit more "nasty" <S> but it works: use a diode in series with the bulb. <S> Since mains is AC, the diode will block one direction of the current. <S> Your bulb will be somewhat flickery but also significantly less bright. <S> Use a 1N4007 diode, these are common and cheap. <A> The thermistor only acts to limit inrush current (which does save the life of incandescent) and is not intended for use as a dimmer. <S> You'd be better off using a series resistor. <S> A better way would be to use a solid state dimmer switch which varies the power by use timing, these are available everywhere. <S> Or get a 20W or 15W bulb.
| If you do use the thermistor it might dim a bit, but you will waste the power as heat and end up with a very hot thermistor.
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If a lithium cell is said to have a capacity of 20 Ah, does that mean all 20 Ah can be used? Toshiba's SCiB 20 Ah Lithium titanate cell is said to have a nominal capacity of 20 Ah. The cutoff voltages are 1.5 V and 2.7 V. Does this mean that all 20 Ah of charge exist between those two cutoff voltages? Otherwise to get all 20 Ah of charge one would have to completely discharge and thus damage the cell. <Q> It will tend to decrease with each cycle, and previous fast charging, exposure to temperature extremes or fast discharging may have negative effects over time. <S> If a reputable manufacturer says it is rated at 20Ah that means that under the specified conditions you can get approximately 20Ah from a new battery before it discharges to the point where further discharge may cause damage. <S> The details of how it behaves should be described in the datasheet and application literature. <S> You may wish, for example, to charge slower than the maximum or discharge only to a higher voltage than the minimum in order to extend the battery life. <S> The latter will, of course, reduce the available capacity. <S> If you'll forgive an analogy, your question is a bit like asking about a car manufacturer that claims 7 liters per 100km (33.6 miles per US gallon) fuel efficiency and whether that means you'll actually get that efficiency. <S> It depends on a lot of factors, such as your driving habits and the condition of the automobile. <A> The available current capacity is listed in the data sheet. <S> This is typically listed as current over time at a specific discharge rate, from fully charged to the dictated cut off voltage. <S> If the operating voltage range is 2.7V to 1.5V, then the stated available capacity when new should be within that range. <A> Batteries still have nonzero voltage when they are completely discharged and can supply no more current. <S> They do not discharge down to zero volts. <S> An empty battery reads higher than 0V. <S> It's not like the fluid level in a gas tank. <S> Therefore, all 20Ah are between 1.5V and 2.7V. <S> But even then you can still only use about 80% of that without damaging the battery. <S> If it drains so far that it gets to 1.5V then the battery already is completely discharged (and probably damaged too from being discharged so deeply).
| The available capacity of a fully charged battery varies with temperature and discharge rate (and duty cycle if it's not constant).
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Soaking phone motherboard in isopropyl to fix potential corrosion Is it advisable/safe to soak my phone motherboard in an isopropyl alcohol bath? I dismantled it after water damage occurred and it now powers on fine but my cell signal isn't consistent. I was advised to soak the MOBO in a bath of isopropyl to remove any potential corrosion on the antenna/board. Is this sound advice? Here is the component in question. <Q> The IPA then evaporates quickly. <S> This works much better if you use 99% IPA instead of the "standard" 70% IPA you find in grocery stores. <S> (The 99% is often in grocery stores, too, but is in a smaller container that costs more!) <S> You could try using IPA and a toothbrush to clean up the corrosion (since it will evaporate before causing more), but I've never had that work very well. <S> Another thing: Once I used IPA on a cheaply manufactured PCB and it softened the solder mask! <S> Another reason to avoid long soaking. <S> Here is a related answer describing what to do if the board is only recently wet. <A> Other thing that people haven't mentioned, if it's an intermittent problem, it might not be an issue with the main board. <S> If it's your antenna, it could be a problem with your flexes or even the chassis of your phone, which you might have caused when you disassembled it. <S> You could try adjusting re-seating the main board and flex connectors and see if it helps. <A> This is a small step up from the advice you to put your phone in a bag of rice. <S> Actually, it's probably worse because you get about the same chance of improvement with much more effort. <S> Corrosion is not usually the problem with a wet piece of electronics. <S> The danger is more that there was power applied to the board while it was wet, causing current to flow in unexpected ways, potentially damaging components. <S> IPA will not fix damaged components. <S> What it is good for is getting rid of residual water. <S> Isopropyl alcohol evaporates very easily. <S> Therefore any residual water will be evaporated as well as long as the IPA mixes with it. <S> If you go down the IPA route, avoid using the 70% IPA you find in most stores - the remaining 30% is probably water. <S> Look for 90% or higher. <S> In the end, an IPA soak is probably a waste of time. <S> Just take it straight to a repair shop and find out how much it will cost to fix. <S> The time and effort of trying to remove the motherboard coupled with the small chance of it actually fixing anything make this seem not worth it.
| Isopropyl alcohol ("IPA") can be used to rinse water from a PCB before the board corrodes. Once the board is already corroded, IPA won't do much.
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Earth Fault in a TT Network: Current loop, or charges flowing from one grounding to the other? There are serveral things I (yet) don't understand about the T-T-earthing system. The most important one is that I don't understand which way the the current takes in case one of the conductors (for example L1) touches the chassis of the device. There are two options I can think of: a) There is a flow of charge out of the earth at the first grounding, through the Resistance RB, through the coils of the transformator, through L1, through the chassis (at the location of the fault) , through RA, into the ground at the location of the grounding of the device. There is not a closed current loop. Instead, ground just works like a really big capacitor (with a very big capacity), so draining charges out out of the earth, or putting it in, will not change the potential of the ground at all. b) There is in fact a current loop in case of a faulty connection between L1 and the chassis of the device, starting at RB, through the coil of the transformator, through L1, through the Chassis, through RA and back to RB. This would mean there actually is current flowing through the earth (which I find hard to believe since there are parts of the earth that have a very high resistivity. Which of the two options I suggested is true? Or does it indeed not matter how to model this, because the values of RA and RB can account for the different ways of modelling this? Edit: I come from another field (physics), and while I have a good understanding about electric potential an electrodynamics, I'm neither familiar with the concepts used in electro-engineering, nor with the terminology. That's where this question stems from. <Q> It's essentially option (b). <S> The fault current travels through the Earth between the two grounding rods. <S> Parts of the Earth may have a high resistivity, but at the same time, it is a really fat conductor - about 12000 km at its widest! <S> But, the resistances where the earth rods meet the ground, modelled as RA and RB, may be quite substantial. <S> This means that the fault current can be quite low, often not even enough to blow a fuse or trip a breaker. <S> That is why a TT installation should have an RCD (or GFCI). <A> Answer: b) <S> The demands for low R require a depth with ground moisture such that the moisture dielectric gives a relatively low impedance limited by the ESR of the conductive salts in the earth and the dielectric moisture. <S> This is often measured by the effective series resistance in Ohm-m. <S> Here are some stats for HV grid dynamic fault current vs required detection time and Resistivity for different soil moisture content. <S> ref Notice <S> my RC estimates below, <S> by coincidence, correspond to an RC time constant just shorter than the maximum response time to reduce Fault damage to the secondary and earth bond network , for a remote fault. <S> As the fault gets closer to the source, these times reduce. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> In addition, thermal and mechanical stresses to the customer's ground grid and ground grid connections can increase the grid's resistance to ground and, at the same time, fault potentials. <S> In order to prevent these problems from occurring, a ground grid assessment, utilizing field and utility updated data, should be carried out on a regular basis. <S> This paper will illustrate a European Committee for Electrotechnical Standardization (CENELEC) approach to ground grid design, aimed to maximize the electrical safety under ground fault. <S> In addition, case studies will be included, showing how high fault currents have damaged ground grids and what repairs are possible. <S> REF <A> There will be fault currents through the earth, for this reason RCDs are almost always used to protect users on TT circuits.
| There is in fact a current loop thru ground.
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Unexpected results when measuring MPU6050 sample rate with RaspberryPi MPU6050 datasheet says that maximum accelerometer data output is 1Khz, so in theory when measuring time it takes to receive 1000 data samples through I2C, so I should receive all data in no less than 1 second, but somehow i always get sub 0.3 seconds. Does CPU clock halts when waiting for data incoming from I2C slave? How to have constant MPU6050 1khz output data rate, and measure it? #include <wiringPi.h>#include <wiringPiI2C.h>#include <stdio.h>#include <stdlib.h>#include <time.h>#define Device_Address 0x68 /* i2c-MPU6050*/#define PWR_MGMT_1 0x6B#define SMPLRT_DIV 0x19#define CONFIG 0x1A#define GYRO_CONFIG 0x1B#define INT_ENABLE 0x38#define ACCEL_XOUT_H 0x3B#define ACCEL_YOUT_H 0x3D#define ACCEL_ZOUT_H 0x3F#define GYRO_XOUT_H 0x43#define GYRO_YOUT_H 0x45#define GYRO_ZOUT_H 0x47int fd;void MPU6050_Init(){ wiringPiI2CWriteReg8 (fd, PWR_MGMT_1, 0x00); /* Disable sleep mode*/ wiringPiI2CWriteReg8 (fd, SMPLRT_DIV, 0x07); /* 8khz/(1+7)=1khz */ wiringPiI2CWriteReg8 (fd, CONFIG, 0x00); /* FSYNC and DLPG_CFG disabled for maximum bandwith*/ wiringPiI2CWriteReg8 (fd, GYRO_CONFIG, 0x00); /* +- 200°/s, and acc by default +-2g's*/ } short read_raw_data(int addr){ short high_byte,low_byte,value; high_byte = wiringPiI2CReadReg8(fd, addr); low_byte = wiringPiI2CReadReg8(fd, addr+1); value = (high_byte << 8) | low_byte; return value;}void passtime(int uSeconds){ // Storing start time clock_t start_time = clock(); //time in microseconds. // looping till required time is not achieved while (clock() < start_time +uSeconds);}int main(){ double time_spent = 0.0; float Acc_x,Acc_y,Acc_z; float Gyro_x,Gyro_y,Gyro_z; float Ax=0, Ay=0, Az=0; float Gx=0, Gy=0, Gz=0; fd = wiringPiI2CSetup(Device_Address); /*Initializes I2C with device Address*/ MPU6050_Init(); /* Initializes MPU6050 */ int count =0; passtime(10000); clock_t begin = clock(); while(count<1000) { //Maximum output rate of MPU6050 is Accelerometer+Gyroscope = 1khz => 1 data read per milisecond /*Read raw value of Accelerometer and gyroscope from MPU6050*/ Acc_x = read_raw_data(ACCEL_XOUT_H); Acc_y = read_raw_data(ACCEL_YOUT_H); Acc_z = read_raw_data(ACCEL_ZOUT_H); Gyro_x = read_raw_data(GYRO_XOUT_H); Gyro_y = read_raw_data(GYRO_YOUT_H); Gyro_z = read_raw_data(GYRO_ZOUT_H); /* Divide raw value by sensitivity scale factor for +-2g's, and +-250/s */ Ax = Acc_x/16384.0; Ay = Acc_y/16384.0; Az = Acc_z/16384.0; Gx = Gyro_x/131; Gy = Gyro_y/131; Gz = Gyro_z/131; //printf("\n Gx=%.3f \tGy=%.3f \tGz=%.3f \tAx=%.3f g\tAy=%.3f g\tAz=%.3f g\n",Gx,Gy,Gz,Ax,Ay,Az); count++; } clock_t end = clock(); time_spent += (double)(end-begin)/CLOCKS_PER_SEC; printf("Time spent %lf after %d data reads\n",time_spent,count); getchar(); return 0;}/* OUTPUT: * Time spent 0.2312s after 1000data reads*/ <Q> First of all, my time measuring was faulty. <S> I've noticed that during data sampling intervals systemcalls do not increase my measured overall time, wallclock time was much higher. <S> After raising i2c port speed from 100khz to 400hz <S> i was able to get 4 times as much samples. <S> So i assume,that I've encountered i2c bottleneck <A> This is different to the rate at which the data can be read out of the FIFO, which is what you are measuring. <S> In the datasheet, it says: An on-chip 1024 Byte FIFO buffer helps lower system power consumption by allowing the system processor to read the sensor data in bursts and then enter a low-power mode as the MPU collects more data. <S> The MPU6050 is sampling data at 1 kHz (set by you) and stores that in it's FIFO. <S> Then, you read that out when convenient. <S> It is a power saving feature, as the system processor doesn't haven't to be awake constantly, it just has to wake up and quickly grab the data that was accummulated while it was in low power sleep. <S> The MPU6050 has I2C and SPI interfaces, so you can "download" the data even faster than you are already. <S> EDIT - not using FIFO... <S> In the while (count < 1000) <S> loop, you are not waiting for new data to be sampled by the MPU6050. <S> You should use the data ready interrupt on the INT pin (pin 12), which will be asserted when new data are ready. <S> Assuming you've set it active high, you could put a stall in with while (INT == 0) , or put it in an ISR depending on your application. <A> In addition to awjlogan answer - you don't have to use INT pin of MPU6050 to watch for a new data - the INT_STATUS register shows the interrupt and cleared once data registers are read. <S> Of course mcu time will be spent on reading the status register over the I2C link. <S> Also, seems your reading from mpu6050 is a bit inefficient - it uses 16 bits reading (guessed as no source for read_raw_data() is present in OP. <S> Better to read data in burst for all 14 registers (temp included) at once in one single sequence. <S> Each read from mpu6050 results in sending of start condition + slave address + register address + repeated start + slave address then actual data. <S> When reading data in burst there will only be one such sequence followed by all 14 registers. <S> The drawback is - 2 temperature sensor registers read in the middle of the burst sequence. <S> For fast reading from mpu6050 arduino must handle mpu's INT connection as awjlogan said. <S> If there is a need to go even faster and to gain time to process the data - ISR for I2C including state machine transition logic (the standard ISR handling for I2C is bulky due to many conditions to check from TWI status register) for 14 register read sequence. <S> Note that the purpose of INT pin or interrupt status pin in mpu6050 is to differentiate between different samples. <S> If you don't check for that then you may get 2 reads for the same sample. <S> of course you may compare accelerometer values from sequential reads (thanks to noisy mpu6050) and drops those whose measurement are same. <S> But if the job you are doing requires strict timing related to sample rate - the mpu6050's INT pin interrupt handling is the ONLY way to go.
| The data output rate refers to the rate at which data are accumulated in the MPU6050's FIFO.
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Is there a standard definition of LPF stop band starting point in terms of dB attenuation Below illustrates a LP filter bands: I can say that the transition band starts at -3dB.But the transition band ends at some point and there the stop band starts.In this example it is -70dB they call this Stop Frequency above. It seems that the transition band always starts at -3dB, but is there a consensus about the location of the beginning of the stop band(end of the transition band)? How is that quantified?(In the above case -70dB. Is that always -70dB? Who decides that? Or is this something about mathematical reasoning?) <Q> Sometimes people will define something different for the start of the transition band. <S> For years in the last century, Tektronix specified it's spectrum analyzer bandwidths at the 6dB point. <S> But it's rare (and a lot of Tek old timers insist it hurt sales) <S> There is no consensus. <S> It pretty much depends on what your system needs are. <S> It is decided by a systems engineer, who looks at both the consequences of signals leaking through the stop band, and the expected expense of having a lower stop band (70dB gets difficult with analog electronics, and lower stop bands require wider data paths in digital filters), and makes a decision. <S> There isn't a mathematical basis in the sense that you can start with a few basic axioms and reason your way to The Perfect <S> Stop Band. <S> It's more a case of a knowledgeable systems person with a spreadsheet, juggling numbers and arguing with the implementation people on one side and the marketing people on the other. <A> Just like there is specification that how much there can be passband ripple and <S> what the passband frequency is. <S> It also defines the -3dB cutoff point frequency, but it is not always defined. <S> Usually any filter that fits the passband and stopband template is acceptable, and least computationally expensive is selected. <A> There is no standard for where a filter stopband starts. <S> This is because there are so many different applications for filters that almost any stopband will find a use somewhere. <S> A deep stopband is expensive, to design, to make, and to test, so designers and filter buyers will always tend to use the shallowest stopband that meets the overall system specification. <S> For instance, if I had a signal in wideband white noise, and I wanted to reduce the total noise power, then I might be happy with only 10dB stopband. <S> If I wanted to visually remove harmonics from a display, 20dB might be plenty. <S> For telephony, a signal to noise ratio of 30dB is essentially indistinguishable from perfect. <S> Building a spectrum analyser, I might specify anything from 60dB for a hobby box to 90dB for the big name manufacturer. <S> Note that there's a difference between a theoretical filter frequency response, and the performance of a filter that you actually build. <S> Although a Butterworth filter (for instance) goes down indefinitely in the stopband, once you build one, you find that stray coupling between input and output connectors, and stray inductance in the shunt components, limits how much stopband you actually get. <S> Part of the difference between a cheap filter and an expensive one is (a) improving the layout and the components to minimise these problems and (b) testing with a good quality signal to prove that you have actually done it. <A> A filter's floor might be due to signals bypassing the filter, or due to its design. <S> A simple analog example of a filter with a "floor" of about -40 dB... <S> here, the term "transition band" has meaning (transition is from about 2Mhz - 10Mhz): <S> simulate this circuit – <S> Schematic created using CircuitLab <S> If L2 was insignificantly small, the floor would be very much improved. <S> If L2 was eliminated entirely (very difficult in the real world), the attenuation slope would continue downward from the pass-band edge, and "transition band" would have no significance: For FIR digital filters, filter floors are somewhat similar to this form, very often with multiple points of high attenuation in the stop-band. <S> Transition band has significant importance for these.
| No there is no standard, but like in your example the filter is well defined - there is a specification how much it must attenuate at stopband and at what frequency the stopband starts.
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How do I implement a feedback to keep the DC gain at zero for this conceptual passive filter? Regarding an example passive LC ladder filter, when I sweep the source resistance, R s , the filter characteristics changes as follows as expected: On the other hand, if I sweep the load resistance, R load , it seems the characteristics does not change, but the DC gain changes as follows: For ease, I showed the above plots in linear Bode plot instead of dB. So 1 V corresponds to 0 dB. I'm not trying to build a filter so this is just out of curiosity. The R load resistance forms a resistive divider and causes attenuation. If R load was known to be 1 ohm then I could add a gain stage with a gain of two and compensate for the attenuation. But if R load is not known and there is no buffer, can there be a feedback between the input and the output which would prevent any DC gain attenuation? In other words such a feedback which would set the DC gain to zero regardless/varying of R load so that the frequency response will start from 0 dB at DC. How could that be realized with any behavioral elements (like VCVS) or op-amps in LTspice or any other simulator? I have written at the beginning of the question that sweeping R load does not change the filter characteristics (besides DC gain), but am I actually wrong? Because I noticed that the phase and group delay plots vary with R load , and below is the group delay for different values of R load : I thought the load resistance has no effect on any filter characteristics besides DC gain. Could you also expound on this? Buffering solved both the DC gain, phase and group delay's dependence to R load : <Q> Your filter has an output impedance. <S> The load impedance interacts with that output impedance to create a voltage divider. <S> If you want to eliminate that dependence, then you need a simple voltage follower (buffer). <S> The output will be independent of whatever load you put at the VCVS output. <S> In the real world, use an opamp voltage follower (unity gain). <A> Simply put, passive filters interact with loads and sources (as you have found). <S> Adding feedback makes it (overall) an active filter. <S> The suggestion of putting a buffer amplifier on the output is probably a good one. <S> It'll be easy at frequencies (like the 0.1Hz of your filter) where you can use op-amps, far harder at microwave frequencies. <A> If you wanted a lossless filter at DC then the source impedance must be 0 at DC for a fixed load OR no load, but then not matched impedance. <S> If you wanted a maximally flat input impedance from DC to almost f -3 dB <S> then it must be a -6 dB lossy Cauer (aka Bessel aka elliptical) filter with the impedance matched at source, filter and load. <S> If you compute a ladder filter , you can see it is not maximally flat and has ripple without AND with load. <S> Top = <S> Bessel Output Response <S> Middle - Bessel Input response <S> Bottom = <S> Ladder Filter <S> No Load and with load and ripple <S> Left= <S> No Load <S> Right = <S> with load Active filters have effectively 0 source impedance so they can be made into lossless at DC or with gain. <S> Bonus Question <S> The grid has almost zero source impedance, so what does this say about impedance effects on the network? <S> Is it an accurate model? <S> Final Question <S> Do you need a flat input impedance and lossless at DC ? <S> If so, then you choose an active Bessel Filter.
| Connect the "nominal" load impedance to the filter, then use a VCVS as an "ideal buffer", controlled by the voltage across that load.
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Varistor? Purpose and principle In the AL9910 LED driver datasheet it shows a part, which I assume is a varistor, is it so? Varistors are shown to be connected across L and N lines, but here it is connected in series. Please explain the operation principle. <Q> This is an NTC current limiting resistor. <S> When plugging in the circuit, the NTC is cold and a bad conductor - this is limiting the charge current to the input capacitor. <S> When the NTC is hot from the power dissipation it will become a good conductor with low losses. <A> It’s quite a common thing to implement. <S> Once the device warms up its resistance lowers and thus the peak current at initial switch on is significantly reduced. <A> It is a negative temperature coefficient thermistor (NTC). <S> It maintains constant resistance at room temperature. <S> It's resistance will Decreases as temperature rises. <S> These are commonly used in inrush current limiting to the input capacitors, protects the fuses and breakers from tripping. <S> Always the most confusing part while referring to these kind of circuits is the symbol, many people represent both (varistor and thermistor) in same way. <S> so we have to decide the device whether it is MOV or NTC based on its operation and application in the circuit <S> (i mean whether it is series, parallel).
| It’s likely a negative temperature coefficient device used to limit inrush current into the bulk storage capacitor and prevent fuses blowing or breakers tripping.
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What linear sensor for a keyboard? I'm trying to make a velocity-sensitive keyboard for playing music. I have to measure the position of each key in order to know how loud the sound should be. The volume of the sound is a function of the velocity of the keys at the end of the descent. I know the position can be recovered from the velocity by integrating it. What would be the cheapest and easiest, yet still reasonably precise way to do it? I don't know exactly what “reasonably precise” would mean in the context. I'll have to test it. The length of the maximum displacement is about 2 cm. EDIT: not as some answers seem to suggest, the loudness is NOT function of the mean velocity of the key. EDIT2: The loudness is function of the hammer speed at the moment it hits the string, but the key isn't pushing the hammer until the end of its path. It is really like throwing a ball on a wall: the ball leaves the hand at one moment, before it hits the wall. <Q> Most keyboards simply use two contacts per key, configured so that they close (or open) at different positions in the key's travel. <S> They estimate velocity from the time that elapses between the two events. <S> Even the fancy weighted "piano action" keyboards use this basic sensing method. <S> Trying to measure position and/or velocity directly sounds like massive overkill. <A> I have to measure the position of each key in order to know how loud the sound should be. <S> Normally key velocity is calculated by measuring the time between the normally closed up-switch breaking and the normally open down-switch making. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> A break-before-make keyboard contact is typically used for velocity measurement. <S> The volume of the sound is not only function of the velocity of the keys at the end of the descent. ... <S> The length of the maximum displacement is about 2 cm. <S> In this case you need to mechanically arrange the normally closed contact so that it breaks in the last few mm of travel. <S> simulate this circuit Figure 2. <S> The mechanical switching arrangement. <S> It seems to me that your best bet would be to modify an existing MIDI bass pedalboard to suit your purposes. <S> OP's comment to Dave Tweed: <S> I think this method is insufficient. <S> Here's why: if I slowly push the key down half of the path, and then quickly push it down to the end, the sound would be loud on a real piano, and this method would think the key was pushed really slowly and therefore output a really quiet sound. <S> In addition, if I want to repeat a note, I would have to let the key completely return to its original position, which is not at all what I would do on a real piano. <S> That is why I proposed activating the changeover switch in the last few mm of travel. <A> You should take a look at analog keyboards. <S> Take a look at this video . <S> This is just an example of concept. <S> Edit:Also, if optical switches are not working for you, take a look at this: <S> There is a PCB printed coil under each key cap. <S> As you press the button coil measures the inductance change caused by key spring compressing and decompressing, therefore, you can get pretty accurate reading of the switch position. <S> Source: Reddit Imgur <A> If the important variable is the impact of a "hammer" that you're trying to emulate, consider piezo sensors which allow you to measure this directly. <S> They produce a pulse whose amplitude depends on the impact impulse. <A> One option could be Hall sensors, which sense the strength of a magnetic field. <S> There are different types of Hall sensors for digital and analog sensing, you need one that has analog. <S> For example SI7211 costs about 0.80 USD. <S> You also need a small magnet underneath each key. <S> When the magnet gets closer to the sensor, the magnetic field increases which increases the sensor's output voltage. <A> You can treat this as a hybrid digital/analog instrument by placing microphones[1] inside the keyboard, tuned to the sound of striking the keys. <S> Signals from the microphones augment the traditional switch-based input, so the switch tells you which key is pressed, and the microphones tell you how loud the most recent keypress was. <S> 2 or more microphones along the keyboard's length would permit decoding a chord with loud/low notes and soft/high notes and vice versa. <S> [1]: I keep saying "microphone", but I mean "generic vibration sensing device". <A> How about capacitive sensing? <S> Tape some aluminum foil to the bottom of the key, ground the bottom of the keyboard, and measure the rise time through a 100k resistor. <S> This method can be made almost arbitrarily precise, as long as your processor is fast enough to discriminate the change in capacitance. <S> I didn't explain the circuit very well. <S> You connect a digital output to the key via a high value resistor, set it low to discharge any stray charge, and then set it high. <S> You also connect a digital input directly to the key. <S> The capacitance will slowly charge through the resistor, and you time how long it takes before the digital input turns on. <S> This time is equal to the RC time constant of the circuit. <A> You may want to research spring return variable resistors, spring return potentiometers, or linear position sensors. <S> Here is one example: http://ecatalog.beisensors.com/item/linear-position-sensors/linear-position-sensor-9600-series-compact-spri/9610r3-4kl2-0 <S> The encoder would output one or more sets of pulses as the shaft rotates. <S> A higher pulse rate would indicate a higher velocity press. <S> The encoder position could be tracked directly if the pulse groups are sent to additional digital logic. <S> Here's an example part: https://www.mouser.com/ProductDetail/Bourns/PEC16-4220F-S0024?qs=6FD5PBp7ZtQte%252Bg7b%2FiMUw%3D%3D&gclid=EAIaIQobChMIrKbIjOSa4QIVCEsNCh3JAAKuEAQYAyABEgKM_fD_BwE <A> If you want super-accuracy at a reasonable price, how about using a linear encoder , attaching a Gray-coded strip (which you print yourself using a laser printer onto a transparent sheet, then cut up and attach one to each key) - more details of how they work under rotary encoders here . <S> This way you need two (perhaps 3 <S> so you get an accurate indication of end position) <S> digital lines per key. <S> This will allow you to measure velocity with very good accuracy and even position if that's relevant. <S> The advantage of this is you could retrofit it to an old keyboard (even an acoustic keyboard). <S> Caution <S> : when I was a teenager (a long while ago) I wondered for ages whether I could make a realistic keyboard more cheaply than buying them. <S> It seemed unlikely then and it seems even less likely now. <S> The cost effective method is therefore probably "buy a velocity sensitive keyboard and take it apart" which is no fun.
| Another possibility may be to use a small rotary encoder (and mechanically convert the linear key press motion to rotary motion).
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What are differences between neodym motor wheels and copper motors? I want to learn about electrical wheels, typically used in electric bicycles. But most of information I could find is related to marketing, and nothing on theory. This is probably due to "relative novelty of this field", and each manufacturer want to keep its secret. But to become one - one should learn motor construction theory at first. Afaik, design of the vast majority of these wheels implies multiple coils on the inner stator, which (when fed by AC of rectangular form) forms magnetic fields, that pushes neodym magnets on the rotor, thus creating motion: But also I heard about (magnetless) squirrel cage rotor, with construction principle, illustrated by following image. Despite it is not a actual motor rotor, it has some similarities in its construction (bars, short circuiting rings) : Image from: pnpntransistor.com So I have a question - why even bother with rare and expensive neodym magnets (that could easily break on the road), while always can use latter variant? I dont understand well, how latter works, so it would be nice, if you could compare and explain operating modes of these two types of motors, "side by side" (to show difficulty or impossibility of creating latter variant). Thanks. <Q> Because of the limitations of this forum, it is only possible to provide a brief introduction to the types of motors used for vehicle propulsion. <S> Squirrel-cage induction motors have been around for a long time. <S> Their main advantage is their simple construction. <S> That makes them relatively inexpensive to manufacture and very reliable. <S> The rotating part (rotor) receives power from the stationary part (stator) by induction. <S> The magnetic field in the rotor is produce by the induced current rather than by permanent magnets. <S> All of the electrical power that is converted to mechanical power is transferred to the rotor by induction. <S> In order for current to be induced in the rotor, there is a speed difference called slip between the speed of the rotating magnetic field and the mechanical speed of the rotor. <S> That slip causes quite a bit of power to be lost as heat. <S> Since the rotors of permanent-magnet motors are magnetized by permanent magnets, no current is required in the rotor and their is no slip. <S> That means less power lost as heat and higher efficiency. <S> Although induction motors and simple in their concept, there are certain complexities in their design that make it difficult to construct induction motors with many magnetic poles. <S> Speed is proportional to frequency divided by the number of poles. <S> That means that a motor that operates at the wheel speed of a vehicle needs to have many poles or a very low frequency. <S> That makes it quite difficult to make an efficient induction motor for a wheel motor. <S> Permanent-magnet motors can be designed in several different ways. <S> They can be DC commutator motors, but those are not very good for wheel motors. <S> They can have the magnets on the surface of the rotor to make surface permanent magnet (SPM) motors. <S> They maybe called brushless DC (BLDC) motors or synchronous permanent magnet (SPM) motors. <S> The difference between BLDC and SPM motors is somewhat semantic, but also a design subtlety. <S> As you can see, there is quite a bit to study to really answer your question. <A> There are two concepts you are discussing here in-hub electric machine Asynchronous vs synchronous machines <S> The novelty here is the in-hub electric machines. <S> By producing an inside out motor where the rotor is on the outside and rotates around the inner stator, the drive chain is simplified, maintenance is simplified etc.. <S> There are technical challenges associated with this and hence where there are per-manufacture specific trade secrets. <S> As to the type of machine: Induction vs PMSM, horses for courses. <S> The reason that there are about 6 fundamental types of machines ( How do DC motors work with respect to current, and what consequence is the current through them? ) is because each one solves the need to convert electrical energy into rotational energy in their own way. <S> Fundamentally all need to produce rotating magnetic fields <S> Squirrel cage induction machines are extremely robust and relatively cheap to produce. <S> HOWEVER, they are not that efficient and equally their displacement power factor <S> they draw puts a higher VA rating on the inverter. <S> PMSM solve part of this problem by already providing a rotor flux. <S> If you are after power density you will be hard pressed to beat a PMSM. <S> If you are after efficiency you will again be hard pressed to beat a PMSM complete assembly. <S> If you want a cheap motor, robust motor and can handle the additional losses and oversizing of output stage then induction machines are viable Below is a blog from Tesla as to why they chose induction machines over PMSM. <S> They however were comparing Induction vs BLDC control. <S> A BLAC would provide smoother torque ripple which https://www.tesla.com/en_GB/blog/induction-versus-dc-brushless-motors?redirect=no <A> It's an induction motor, invented by N.Tesla. <S> Also the Tesla company uses this kind of motor in their roadsters. <S> You can learn about this principle in many sites/books. <S> It's not so efficient as permanent magnet synchronous motor (PMSM) or brushless DC motor (BLDC). <S> It's also heavier than PMSM/BLDC but is much cheaper than permanent magnet motors. <S> P.S. : <S> By the way, your second picture is a fan blower, not a squirrel cage rotor.
| They can have the magnets in the interior of the rotor to make interior permanent magnet (IMP) motors.
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Should I get shocked? If not, why? Let say they are two different closed circuit. One circuit uses 400 V and another uses 12 V. Let say I touch the positive terminal of 400 V circuit and the negative terminal of 12 V circuit. Should I get shocked because I myself can be considered as some wiring although I have some resistance? <Q> Let's draw the schematics for a few possible scenarios. <S> The ammeter in the schematics represents your body and its resistance. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Let say I touch the positive terminal of 400 V circuit and the negative terminal of 12 V circuit. <S> Should I get shocked ...? <S> It depends. <S> In Figure 1a both circuits are floating with respect to ground and to each other. <S> In theory no current can flow from one to the other. <S> In Figure 1b one circuit is grounded but the other isn't. <S> Touching the 400 V grounded circuit would be a really bad idea as any path to earth through your body could cause electrocution. <S> In Figure 1c both circuits are grounded so touching +400 V and the negative of the 12 V is, effectively, touching both terminals of the 400 V supply. <S> Electrocution is likely. <S> All of the examples are using DC. <S> When AC is involved you have to take capacitive coupling between each circuit and ground and between the human body and ground. <S> This can be high enough to allow dangerous current to flow. <A> It depends whether the two circuits share a common point. <S> If they are totally isolated from each other, then you would not feel a shock touching any point of one with one hand, and any point of the other with the other. <S> If the reference points on each circuit share a common ground, then there would be 400-12 = 388 volts between the two point you've touched. <A> Ok - you have a weird question <S> but I'll attempt to answer it. <S> The first thing you need to consider is referencing - if you have two closed circuits, where the voltage on the positive terminal isn't referenced to anything other than the negative terminal, then no - you wouldn't get shocked because the 400 volts on the 400 volt circuit's positive terminal is a potential difference of 400 volts with respect to it's own negative terminal and not anything else. <S> This would be the case in the case that you had - say - two batteries with one producing 400 volts and the other producing 12 volts, you can safely hold the negative of one and the positive of the other without getting shocked. <S> This is because there is no potential difference between the two. <S> INFACT - you can connect the two terminals with a perfectly conductive path and it would just effectively put the two batteries in series. <S> This changes when you're talking about voltages that are earth referenced, such as those coming out of your wall. <S> Even after rectifying them to be DC voltages - they are still referenced to the earth ground and would have to Galvanically isolated (using a transformer) in order to convert it into an effectively isolated supply. <S> Such supplies are also sometimes called "Floating" supplies because their reference isn't tied down to any level and may be raised to whatever potential you desire.
| If you were perfectly insulated then no current would flow because there is no return path to the 400 V negative. Note that it's quite easy for two circuits that you think are isolated to accidentally share a common ground, perhaps standing with bare feet on a damp floor, or an unnoticed cable plugged in.
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Diode in opposite direction? Its really bothering me that the diode is shown in the opposite direction here and i don't understand why its been put like that? the input voltage will come from the left side of the circuit then why is the diode's cathode connected to the output of the op amp? does the polarity even matter here? <Q> At the point just before the first amp, the radio waves have been filtered to a particular frequency by L1, C1 & C2. <S> That first amplifier is not an op-amp, it is an RF gain amplifier. <S> It amplifies the incoming signal by a number of dB. <S> The signal before and after will be an AC signal, equally biased around ground. <S> The diode or detector gets rid of one half of the signal (either the positive half or the negative half, depending on which way the diode is). <S> Description of AM Detector. <S> The next stage filters off the carrier signal with a low-pass filter. <S> As the next amp, an op-amp has a bipolar power supply so it can handle either the positive or negative signal. <S> It amplifies the sound waveform for the speaker. <A> If you reverse the diode, you simply track the positive part of the envelope rather than the negative part. <S> Either will give you the demodulated signal. <A> The signal into the diode is a (presumably) AM (amplitude modulated) <S> signal. <S> The variations in amplitude are what is of interest to the listener and are what the amplifier chain is seeking to recover. <S> As shown the diode rectifies negative going halves of the signal. <S> If reversed it would rectify positive going half cycles. <S> Either way , Cd provides a filter that smooths out (and so removes) <S> the RF variations and results in a voltage that varies with the amplitude of the incoming signal. <S> As shown you get negative variations which are smoothed. <S> Reverse the diode and you would get positive going variations. <S> The two are the same except inverted. <S> In either case the resultant "envelope" is AC coupled by Cb, and is DC ground referenced by Rb. <S> So EITHER way "Amplifier" "sees" an AC signal centred around ground. <S> This is amplified and, again, AC coupled vi C3 to the headphones. <S> So, either way the result is much the same to the end user.
| In this particular circuit application, the demodulation of amplitude modulated RF, no, the polarity of the diode does not matter.
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Point of contact with a capacitor plate & rolled design? With a simple parallel plate design for a capacitor, does point of contact(to the external circuit) matter? I imagine any point(a - e) would yield the same feature in storing/discharging charge. However, I'm curious if there is an optimal point of contact relative to to it's performance. In addition, if I were to wrap these two plates parallel to one another to conserve some space, if I had the point of contact at (a) and wrapped the ends such that point(e) meets the point of contact(a) with a gap in-between would it function as normally? <Q> Performance specifications on Capacitors; C [pF to F] <S> Tolerance of C [%] Vmax [V] leakage current @ <S> Vmax <S> [uA] <S> ripple current <S> [Arms] some temp rise. <S> dissipation factor ( % of 120Hz power lost for rectifiers) SRF (self resonant frequency due to inductance) Joining "a" to "e" are in the same conductor plate and thus have the same E-Field <S> so it is not a variable for capacitance design. <S> When the other plate is joined to itself we now have a coaxial capacitor. <S> This is dependent on the gap and dielectric constant to yield some C value in picoFarads per meter. <S> e.g. coax cable ~ <S> 100 pF/m E fields between 2 points decay rapidly with gap will depend on this geometry of equal potential, namely by ; 1/r³ from a point, 1/r² from an edge 1/r from a large area. <S> Getting to a "tiny centre point " now degrades the contact resistance and with a skinny wire adds inductance. <S> So this is counter-productive to "performance". <S> Although the ideal conductor has the same voltage at every point, there is conductor resistance but it still does not affect a constant E field between the plates . <S> The Effective Series Resistance, ESR is limited by the area to dielectric total area contact. <S> This conductor surface may be etched to increase the roughness and effective surface area to thus drop the ESR. <A> To minimize series inductance and resistance (particularly important when the plates are thin metalization) you normally want to connect to the entire edge of the plate. <S> To do that, the plates are wound so that they each overlap the dielectric a bit one side (either foil plates or the metalization on the film) and then each entire end of the roll is glomped together with something conductive (referred to as "schoopage" in the Wiki article linked below) to connect the plate to a lead wire or terminal. <S> The Wikipedia article illustrates stacked film caps and shows a complete uncoated film cap. <A> "Performance" is usually defined in terms of ESR / ESL . <S> In this regard, having contact points in the middle of the plates helps reducing ESR/ESL. <S> So does having contacts points in mirror-symmetrical locations (rather than on opposite sides) and reducing the plates' perimeter. <A> Integrated Circuit designers have this decision to make, for every capacitor they use. <S> If the capacitor plates are POLYSILICON, the ohms-per-square are (perhaps) 50 ohms/square. <S> This inherent resistance (10,000X higher than an 35micron thick piece of copper foil) makes the contact point-or-side and aspect_ratio become key for circuit <S> SPEED and for the dampening . <S> When the designer places on-chip capacitors, to supply the 100 picosecond demands for charge as the logic system changes state, the inductance of metallization and of the bondwire+leadframe_package will interact with all the onchip parasitics and intentional (e.g. the polysilicon or even Metal_insulator_metal) capacitance and will resonate. <S> The designer of the silicon system has the responsibility to design, to dampen, the L+C energy storage network. <S> On a PCB, the designer has the same responsibility to design. <S> Or they can just punt and hope.
| The contact point should be at least the entire edge.
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Detect 230 volt main supply in a cheap and easy way I need to detect by if one o more pumps are really connected to the 230 Volt main supply. Each pumps is activated by means of a triac isolated by a MOC3021. Arduino commands the optoisolators and I need to detect if really the triac has switched on (e.g. the triac didn't switched due to a fault). The solution has to be possibly cheap and reliable. I found this simply circuit from a site, and I would like to know your opinion. The optocoupler is a 4n35 or pc817 or similar. What also I need to know is if the 390K resistor should be dimensioned more than 1/4 W and if has to be a special resistor for high voltage. Thanks a lot. <Q> With the circuit above, the main thing is sizing the rectifier and the resistor for the amount of power and current they would need to handle. <S> The 390k resistor looks ok, even with the LED shorted it only consumes roughly 1/8W max which is well within the heating tolerances of a 1/4W resistor, One thing to consider is creepage and clearance of the 220V circuit, the traces need to be spaced adequately to meet IPC requirements if there going to be used in a product (and if there not, they still need to be spaced correctly for saftey). <S> A hall effect current sensor might be considered in this application instead of the circuit above. <S> I'm not sure if it's cheaper, but it's easier <S> , I'll let you do the pricing. <S> The circuit would have lower loss (no rectifier), and fewer components, and still have isolation. <S> If you had an ADC you could actually report the current to a microprocessor or use a comparator with a set voltage level to detect the current. <S> Source: <S> https://makerselectronics.com/product/current-sensor-module-acs712 <A> The 390K resistor value is too high. <S> That works out to less than 1 mA <S> peak of opto LED <S> current - and only 700 uA at 200 Vac "low line". <S> That is not enough for a predictable current transfer through the opto. <S> For 5 mA peak current and a normal 220 V line, R works out to 65K at 1.63 W. <S> A good solution is two 33K / 2 W resistors in series. <S> You can cut the power dissipation in half by using a single diode (half-wave), but this increases the detection lag. <S> Add a small rectifier diode (1N4006, etc.) <S> in anti-parallel across the LED to prevent excessive reverse voltage. <A> The circuit that you have shown can be made to be much more reliable by simply rearranging the elements. <S> The input current-limit resistors (R1, R2) should be metal-oxide. <S> These have significantly-better transient power capability. <S> simulate this circuit – <S> Schematic created using CircuitLab Pick the resistor value (R1, R2) to ensure that the output photo-transistor is saturated. <S> This current depends on the particular opto-isolator chosen. <S> This can be enhanced by adding zener diodes in series with each of the input leads. <S> That can stop the circuit from detecting leakage currents. <S> Adjust the resistor in parallel with the LED accordingly. <S> The advantage of this circuit arrangement is that the bridge rectifier is not exposed to high-voltage transients. <S> The maximum steady-state voltage applied to the bridge <S> is less than 5V. <A> There are many options, you can probably find one with a darlington output that will work at microamp levels. <S> I can't think of any part numbers off hand but have used them for just this purpose. <S> You will still have to use two resistors on the input probably. <S> Put one of these across each triac and you can tell if it is on or off.
| You can use small-signal diodes (1N4148) or use a packaged bridge rectifier. The simplest way would use a optocupler that has two LEDs back to back.
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Is it common practice to connect power supply to a DB9? I got a project where many sensors which uses CAN were connected to a DB9. And sensor power (24V and GND) too. I want to connect the DB9 terminal to power however how can I do it in a safe way? If I connect using tiny wires like the ones I use with Arduino might burn. Above image was Lidar sensor which I had to buy the correct cable to work and worked. Now using Radar (Delphi ESR 2.5 24V) there is only one terminal and power input goes to a DB9 (like the one in the image) Does anybody have ideas? Estimated current is around 900 mA. Please check two last pictures for terminal information <Q> It is common, but only in systems which will never be connected to a serial port. <S> If you do that, you get what you deserve. <S> Probably there is an exception in CANBus systems which are connected to a PC. <S> Fortunately, most PCs these days don't have a serial port. <A> D-Subminiature connectors are just that, connectors. <S> They can be used for a variety of purposes besides computer serial ports (although that's probably their most familiar application). <S> These connectors have a set of larger power contacts among the smaller signal contacts. <S> According to the first datasheet I pulled up, NorComp rates these connectors at 40 A for the power pins and 5 A for the signal pins. <S> Of course, you have to use the right size wire inside your cable to pull that much current. <S> If you're limited to using DB-9 <S> (it's actually a DE-9) <S> and you're only drawing 900 mA, the pins seem to be rated anywhere from 3 to 5 A according to a quick parametric search on Digi-Key. <S> The high-density D-Sub connectors are rated somewhat lower due to the smaller pin size. <A> There is an industry standard DB9 pinout for CAN which is set by CANopen DS303.1 6.1: <S> This is a widely accepted standard. <S> Anyone deviating from this pin-out for CAN applications is making non-standard crap. <S> The intention here is clearly that this should be a pure data traffic connector. <S> CAN_V+ is dedicated for supplying opto isolators, that use the signal ground CAN_GND. <S> It is required that pin 3 and optional GND on pin 6 are connected internally. <S> None of these were ever intended for noisy supply currents. <S> That being said, the least bad thing you can do if you do need to provide a supply through the same connector, is to do so by sacrificing pin 9 and 3. <S> This given that you don't actually use galvanically isolated CAN or it will obviously not work. <S> And you give up the signal ground, which could mean worse signal peformance. <S> By using 9 and 3 you at least ensure that you won't burn up other devices, but signal and/or EMC performance might suffer instead, depending on the quality and currents of your supply. <S> And well, obviously check the current ratings of connectors and cables. <S> The average DB9 should be able to handle 0.9A just fine, but check the datasheets to be sure.
| Some companies make "power" D-Sub connectors, such as NorComp's Power-D series.
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Does (stray) capacitance depend on voltage? While thinking about how to reduce stray capacitance on my pcb, I got triggered by the question: Should the stray capacitance depend on the voltage difference? Acccording to the equation of capacitance(parallel plates:) $$C = \dfrac{\varepsilon \cdot A}{d}$$ Where \$\varepsilon\$ is the dielectric constant , A the effective area and d the distance. So based on this equation the (stray) capacitance is entirely independent of the voltage. To me this seems counterintuitive. If I have a two traces on top of the ground plane (the two traces are far at a inifinite distance of each other, so lets discard trace to trace capacitance), where trace 1 is at 1000V and trace 2 is at 10V. My intuitions tells me that the parasitic/stray capacitance will be higher between trace 1 and ground compared to trace 2 and ground, basically because the electric field strength is higher. Is my intuition right? <Q> Should the stray capacitance be dependent on the voltage difference? <S> No, it shouldn't; it's a physical property related to physical dimensions and the electric permittivity of the materials. <S> The electric field being higher is immaterial. <S> capacitance is entirely independent of the voltage <S> Correct with one caveat - certain parasitic capacitances associated with reverse biased PN junctions will vary their capacitance with applied voltage <S> but, this is because the depletion layer grows and, in effect, the gap between the "plates" increases thus lowering capacitance. <A> Normally the capacitance is independent of voltage. <S> It's a function of geometry and material properties - as can be seen in the equation for a parallel plate capacitor that you cited yourself: $$C = <S> \varepsilon_{r}\varepsilon_{0} <S> \frac{A}{d}$$ <S> Such materials do not only exist, but are quite common. <S> In most cases the effect is still negligibly small. <S> One example of a material for which the effect is highly relevant is liquid crystals. <S> These have two important properties: 1.) <S> They are liquid. <S> 2.) <S> They are dielectrically anisotropous. <S> This means that you can control the effective capacitance by changing the orientation of the molecules by applying an electric field. <S> Remark: <S> Sometimes it's the geometry that changes with voltage <S> (example: loudspeakers). <A> No, the capacitance will remain constant for a fixed geometry. <S> What will change is the charge. <S> Since \$ <S> Q = CV \$ <S> the total charge stored will vary in proportion to the voltage.
| The capacitance can only depend on voltage, if the relative permittivity depends on electric field strength.
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Voltage of sounds of MP3 files What is the maximum voltage for common music files? When playing back on a smartphone, if the volume of the smartphone is set to maximum, how much voltage is sent to the earphones of the electric signal? thanks so much for commentI’m japanese.I’m not good at English.I use google translator.I want to use a low frequency therapy device on my smartphone. I thought it was dangerous depending on the voltage. And I asked this question. any precautions or dangers? <Q> Amplification <S> It does not depend on the music file, but at the amplifier sending the music to a speaker, or to a headphone/earplug connector. <S> Music file <S> A music file only contains 'values', and mostly in a compressed form. <S> After uncompressing, you get values which has a certain amount on bits per value (typically 16 or more). <S> Conversion <S> A microcontroller can send these values to a speaker via <S> an pre- and/or amplifier, which converts it into actual voltages. <S> So it depends on the amount of amplification. <S> Protection Also note that sending continuous 'max values' do not result in a loud sound, actually it would breaks the speaker (to prevent this, in the amplifier a so-called DC speaker protection is present). <S> A wave is needed to let the speaker move outwards and inwards very fast, and this is done by sending changing values to the speaker. <S> Measure <S> If you want to measure, you can use an oscilloscope to measure the audio output while sending a sine wave with a maximum amplitude. <A> These days a lot of DAC chips for consumer equipment use voltage output of 2 Vrms. <S> The supply voltage is largely irrelevant, as a lot of DAC chips also use internal charge pumps to boost up supply voltage and to generate negative supply voltage for the audio output stage, so they can drive DC coupled loads directly. <A> Normally it's limited by the cellphone battery voltage, which is about 3.7V. <S> However there are other limits in play like the EU volume limit of 100dB - which is specified as SPL rather than a voltage. <A> "Line level" is the term for the standard used between a whole bunch of devices, namely your cell phone. <S> Be sure to read the other answers as they will make you aware of the specifics of your question. <A> The music files like mp3 have only digital logic levels (0's and 1's), First the analog signals get converted into digital logic using ADC's. <S> The ADC works like a charm here, they slice the analog signal into some simple form, the slicing frequency (sampling frequency) depends on ADC.Now, these sliced pieces get converted into bit's (each bit represents some voltage) so the processors/controllers can understand slices and store them, usually these slices has huge data (0's and 1's) <S> so we compress this data using mp3 compression or some other audio formats. <S> When you play a music file the 0's and 1's go to DAC which convert them back to an analog form, the voltage here depends upon your DAC's reference (this limits the swing of analog <S> signal)Here the signal is weak <S> so we amplify it by an amplifier and feed it to speaker. <S> The voltage range from earphone speakers varies with the device its connected to normally 0v-3v.
| A line input level electrical signal typically has a voltage ranging from 0,3 to 2 Volts.
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Do line level circuits need shielded enclosures? I made a quick volume pedal, just a button, 50k pot, input and output jacks I used a plastic enclosure for it since it was convenient, but I realized almost all guitar pedals and other stomp boxes are metal. Not only does this help with durability, but it acts as a Faraday cage. Connected to ground, it reduces noise, just like a shielded cable. Is this necessary? I can line the inside with aluminum foil, but it would be annoying to do so, and not sure if it's necessary. Moreover, I am curious about what is the best practice here. Does a line level circuit like a volume pedal need a shielded enclosure? (I don't plan on ever using this for something pickup level also.) <Q> Does a line level circuit like a volume pedal need a shielded enclosure? <S> No, <S> But if you put your cell phone right next to a plastic box with a high impedance signal loop, poorly designed on the high impedance input, you might be able to pickup the tower re-sync transmissions. <S> dah-dit-da-dit... <S> This RF often comes from input diode carrier demodulation to baseband and can be suppressed with an RF shunt cap. <S> Most poor SNR situations come in from radiated noise induced into conducted noise on the signal cables and from poor HF ripple on floating DC supplies. <S> This current transfers into a voltage from unbalanced impedances for signal and return. <S> A range of supply capacitors for DC ought to be included from 0.01uF to 10uF to suppress AC hum to RF induced noise currents. <A> I'm just a hobbyist at best, but you mentioned lining the inside with aluminum foil, and I have done this before. <S> It wasn't a volume pedal, but it was cheap little stereo. <S> It was about as cheap as they come: it probably came from a drug store and sold for about US $20 (about 15 years ago). <S> And this thing wasn't just inexpensive (which doesn't necessarily mean low quality), but this thing was CHEAP! <S> It had an FM radio that sported a digital readout of the station it was tuned to, which was purely for looks, After opening it, I saw that it was just and old style mechanical tuner! <S> I can't say if that technology was bad in general, but this thing was. <S> Anyway, it's not surprising that it had an annoying hum coming out of the speakers. <S> So I decided to shield the enclosure from the inside like you mentioned. <S> I didn't use aluminum foil, but I used adhesive backed foil tape which is fairly common (used for duct work maybe?) <S> You're right <S> , that can be a tedious task <S> , I think I spent a good week or two on the project. <S> The metal can't be soldered, which meant mechanical connections throughout! <S> The CD player was broken, so I had line inputs free, so I gave it 1/4" jacks for guitar/musical gear. <S> It turned out that a soup can was the exact height for the enclosure so that was what the jacks were mounted on! <S> Anyway when I finished the project, I was totally AMAZED at how drastic the reduction in hum was! <S> I really thought I had killed the thing when I first powered it up because at Vol=10 I couldn't hear anything. <S> There of course, was some, but by comparison it was silent. <S> I wouldn't spend so much time doing that again, but it was an extremely helpful learning experience. <A> Yes, it is a high impedance single ended input, which is extremely susceptible to noise. <S> This noise will typically appear as hum. <S> The wire and or case is actually an antenna which in its own right is very good at low frequencies. <S> Watch out for ground loops they are hard to find and can be very nasty. <S> This is one of the major reasons the professionals went to balanced inputs typically 600 ohm instead of approximately 10K on a line input. <S> The balanced mode puts the noise in common mode and basically eliminates it. <S> Consider when doing your design keeping it at the signal source using as little signal line as possible and use a remote to change it. <S> This can be wired or wireless both will work. <S> In a home environment you may not notice it but it will be there. <S> For more information look on the Rane web site <S> they post a lot of great information. <S> Start Here: https://www.rane.com/library.html#gpm1_2 <S> I have been playing with this stuff both as a hobby and as a professional for over 60 years.
| I would say that if you're not noticing any hum, don't bother with shielding it.
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Best practice: Should I charge a battery while also pulling load? I have an off-grid solar panel setup. The solar panels send current to an MPPT charge controller. The charge controller controls current to a lead acid battery. On these same battery terminals I have wires leading to my load, an inverter and then AC compressor/motor. I’m most concerned about the absorption charge phase, when it dumps most of the current into the battery at an elevated voltage. My controller operates in the absorption phase for 3 hours. It then drops into the float stage sending current at a lower voltage. Should I avoid pulling load from the battery during the absorption stage so that the battery receives all of the current during this crucial phase? I worry if the load is also pulling current during this time that the battery will enter the float stage not having fully recharged. My goal is to maximize long term battery capacity, through shallow discharges and daily (full) recharges. # Update Some people raised some very good questions/comments. I'll add some more details here to give a fuller picture. And yes, because this is an off-grid solar setup, the variability of the weather/elemeents make it hard to say with certainty that there will be enough sunlight to complete the absorption phase uninterrupted. Here's a diagram of my setup. I have 2 100W solar panels (200W total) wired in series. I have 2 55Ah AGM batteries wired in parallel. I use this charge controller I use this inverter I use this compressor for pond aeration in 10ft of water. On sunny days I’ve measure the voltage from the combined solar panels totaling 45V. The absorption phase charges the battery at 14.8V. The float phase is at 13.7V. From my calculations I can only run the compressor for 4-5 hours before depleting the batteries to 70% of their capacity, which is the lower limit I’d like to sustain. I was seeing if I could maybe cheat and run it a little longer once the sun came up. From some of the answers it seems like that may put the battery at risk of not fully being recharged daily. I’m not sure it’s worth the risk <Q> It is very common to have a charging source, battery, and load connected in parallel so that it may look like you are charging and discharging a battery simultaneously. <S> What actually happens is that if the charging source can supply more current than the load demands, the excess current will go to charge the battery. <S> If the load demands more current than the charging source can supply, the extra current will be supplied by the battery (discharging the battery). <S> The battery will change between charging and discharging automagically as the load demand and charging current vary. <A> The answer depends on your solar setup, charger, and inverter. <S> If your solar setup can't do that then you must inhibit the load from coming on while the battery is charging. <S> Second, the inverter must be capable of accepting the higher voltage. <S> Most likely it is fine but it does not hurt to check. <A> This is a very specific case of the more usual "can I charge and operate a load simultaneously" <S> In this case the answer is clear: If you value correct absorption phase charging then <S> The solar system must be able to supply at least enough current at the desired voltage to allow this. <S> You need to be able to specify what this requirement is. <S> An additional load can only be operated if the PV output is able to provide it without preventing 1. <S> The above is "almost obvious" but introduces some extra questions. <S> If the available PV energy falls below that needed to perform absorption mode charging in the middle of an absorption cycle, what should you do? <S> You can continue at a lower "float" rate, or stop charging until enough energy is available to continue absorption charging, or ... . <S> If you stop and then recommence absorption <S> charging are the periods simply additive, or is there a more complex algorithm involved? <S> If so, does it matter where in the absorption cycle the pause(s) are or how many there are or how long between them? <S> What fun! <S> If you can 'spare' surplus energy for another load, is it able to handle the variation from eg clouds, or complete cut off if the charging demands all the energy. <S> eg <S> a water heater would usually happily consume excess energy as available. <S> A television (or its viewer) would not be so happy. <S> An AC compressor is intermediate. <S> Unlikely - food for thought: You MAY be able to split the battery into two (or more) portions to allow largely uninterrupted boost charging in two stages - with the risk of having an unbalanced battery if the solar output fades sooner than expected on a given day. <A> If the charger is able to reach 14.8V with load connected then this means that there is an excess of solar energy. <S> In such case the battery will be filled up to 100% while supplying the load. <S> If the higher voltage can't be reached it means that there is no excess of solar energy and the battery is being charged and drained simultaneously. <S> As I understand your question: would it be better to switch of the load while charging? <S> In case 1 certainly not, in case 2 you won't gain any benefit. <S> If you switch the fridge off, then you just postpone the load demand, you will fill the battery and then drain immediately after you switch on the load. <S> The energy will be wasted, because the efficiency charge/drain is not 100%. <S> In my opinion it better to supply the load while charging. <S> Looking at your equipment I have an opinion that the fridge is an important element of the system. <S> Unfortunately I would say that your choice of using inverter + AC compressor is not the optimal one. <S> If the budged is less important than overall efficiency, then you should go with 12/24V compressor with built in inverter. <S> Such compressor has a integrated inverter that turns on only on a demand. <S> See Danfoss 35F, 50F.
| To charge the battery during a load condition the charger must supply enough current to satisfy both the load and the current demanded by the battery.
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Strange LED flicker when powered via buck converter I have a 12V supply with a buck converter stepping down 12V -> 3.3V to power an esp8266 microprocessor. The esp is controlling a 12V LED strip switched via 2n2222 transistors. When all LEDs are switched off I can see this constant flicker going on. How can the transistors switch when the base is grounded, or what else could explain this flicker? I don't have an oscilloscope unfortunately. edit: When I remove buck converter and power esp separately the flicker is resolved. <Q> Using a pullup resistor of 10k to 100k from Collector to V+ should be enough to disable this effect by shunting the voltage across the LEDstrip when off. <S> The base current needs to be at least 5% of LEDstrip current when ON. <S> You can also apply low ESR caps to the 3.3V ESP load to reduce ripple and use twisted pairs to reduce EMI. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Suppose your buck regulator has ringing at 100MHz, with amplitude of 0.1 Amp. <S> And that current flows thru 2" of wire, or 50 mm or about 50 nanoHenries inductance. <S> If, using V <S> = L * <S> dI/dT, we show 0.5 volts peak ringing, then your bipolars may be turning on.....just a little bit. <S> Math: <S> V = 50nH <S> * 0.1 amp /2.5 nanoseconds <S> V = <S> 50/2.5 <S> * 0.1 <S> * nano/nano = <S> 20 * 0.1 = TWO VOLTS <A> Sometimes off is not really off. <S> It only takes about 650mV to turn on a small signal transistor like a 2N2222 or 2N3904. <S> Transistor leakage is possible, but that is more of an issue at high temperatures. <S> So the problem is that the base of the transistors is not going down close to zero volts, so they will be in a clean 'OFF' state, or <S> some type of noise is getting into the transistor base. <S> Check make sure the MPU pins are at zero volts in the OFF state, and not just floating in a tristate mode. <S> Check for ground noise, as ground bounce is the same as putting a pulse into the base of the transistors. <S> It will show up on an oscilloscope or a good DVM set to read AC volts. <S> Just to be sure that off is really an OFF state, install 10K resistors from base to emitter. <S> When ON they will only consume 65uA from the MPU pins. <S> By the way, if the MPU pins connect directly to the base of the transistors without a 1K current limiting resistor, the MPU output can damage the transistors, or damage the MPU pin driver. <S> Please double check this.
| There is a high frequency common mode ground noise causing spurious conduction.
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Mitigating touch current in two-pronged AC without dedicated ground I'm trying to work on a simple circuit where (simplified) I plug in a 240V 50Hz AC through a resistive coil to convert to heat. The coil is insulated and placed under a metal plate. I'm noticing that if I touch the plate, I would feel a slight buzzing/tingling sensation, AKA touch current. I have some questions about that: Is this effect due to induction, or is it a capacitance phenomenon, (or something else??) How should I design to ground the plate to eliminate/reduce the touch current? It's easy with 3-pronged plugs with a dedicated earth pin, but old EU plugs are round two-prong and indiscriminately reversible making grounding tricky. The solution would have to work when I plug it in either direction. <Q> To be easily perceptible the current has to be fairly large (at least hundreds of uA). <S> To get, say, 500uA by capacitive coupling with a 240VAC difference would require a capacitance of around 7nF, which is pretty huge. <S> If the plate is like 25cm <S> x 25cm and the insulation is rather thin (like 0.25mm) <S> it's possible. <S> UL considers 0.5mA to be the limit of perception <S> and it's also their limit for allowable leakage in many situations. <S> The other possibility that that you are using an insulation type that leaks at high temperatures and <S> you are allowing the insulation to get very hot. <S> This is not uncommon with sheathed heating elements. <S> Your best bet might be to make the insulation a bit thicker, and find a dielectric material that transmits heat well enough that provides some separation. <S> Even 1mm would be a big improvement. <S> Lower dielectric constant is also better. <S> I'm not addressing issues of safety or standards here <S> , that's more-or-less a separate issue. <S> But you definitely do not want to connect the plate to your 2-pin plug. <A> The various UL documents will talk about this, but the gist is two indepenent ways of assuring current can't get on the metal plate. <A> No matter how it is coupled, the current you feel is a result is a potential voltage difference between you and the plate. <S> You are coupled thru stray e-fields in the room thru the moist air <S> and it goes thru your body's much lower imepdance without any sensation. <S> But the impedance of the plate close to the coil increases the capacitance and puts a time-varying (high impedance) voltage on the plate. <S> This difference permits some xx to xxx uA of touch leakage. <S> Earth-bonding like metal stoves and ovens could be the prudent fix with a 3 wire cordset. <S> A photo may allow more detailed suggestions how to connect the ground to the plate.
| Given the polarityless nature of those reversible EU plugs, what is required here is double insulation . An afterthought approach ... is a thermal BBQ grill , grounded via the cordset to shunt stray E-fields from reaching the plate, yet allow heat to go thru in terrahertz band ;)
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Reading T/C voltage when Heater is present in series I am trying to read the thermocouple voltage of a Hakko T12 tip which has the heating element present in series with the thermocouple. The problem is that 24V DC will be present on T+/T- of the thermocouple when the heater is being powered. I have seen people get around this problem using OP amps that have overvoltage protection like the ADA4177. The problem with this is I want to have good temperature accuracy and that would mean that I would have to add an external ADC and also add cold junction compensation. Ideally I would like to use something like the MAX31855 which already does everything I want but I can't come up with a solution that would prevent 24V going to it and also maintain its accuracy. Would something like diode clamping work or should I be looking in another direction? <Q> You won't likely be able to get a good reading with the heater on, so that means turning the heater off long enough to get a reading, then on again, which means you have to have a circuit that settles very fast and recovers from an overload very fast. <S> If it's a K type thermocouple it will have maybe 10-15mV output at soldering temperature (and a tempco of +42uV/K roughly). <S> If you ground the end of the heater that's connected to the thermocouple, you'll have a maximum signal of maybe +/-20mV wrt ground, and with the heater on it might go to a few hundred <S> mV. <S> So, an op-amp circuit shouldn't be too difficult to deal with that, though it will have recover from overload fairly quickly. <S> You could "blank" op-amp input by shunting it with an analog switch, which would make it settling time rather than overload recovery time. <S> Use a low bias current op-amp and series resistance to protect the inputs. <S> Cold junction compensation requires you to know the temperature at the point where the thermocouple materials transition to copper. <S> Edit: If you have only two connections available you could do something like this: simulate this circuit – <S> Schematic created using CircuitLab <S> In practice the half-bridge is going to need a bit more complex drive circuit to deal with 24V, but this is the concept. <S> Getting accurate cold junction compensation will be almost impossible in this configuration unless the wires to the handset are actually dissimilar metals matching the thermocouple uV <S> /K. <A> There is a circuit called "common mode stripper". <S> With all 4 resistors being Equal, you can take +24 and +24.01 volts into an opamp that uses +-15 volt rails, and produce a 0.01 volt output. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> So, you'll want to sample the thermocouplevoltage at special times, and either hold the analog value or latch the digital value. <S> Such all-in-one items as MAX31855 are not suitable, unless you use trickslike strobing the power supply, because they lack time-related cycleawareness. <S> It is easier to use a microcontroller to control the sample/hold/convert and display functions, than to work around the timing problems. <S> The 'linearization'and cold junction compensation features of the Maxim device, while useful for mass production,are easily implemented with some calibration work, using a forward-conductingdiode (or B-C shorted transistor as a diode) to sense the temperature, andsome easily calculated tables to translate to a number for the temperature display. <S> To keep the 24V away from your sensitive circuitry, simple diode clamps and resistorsto current limit will be suitable; when power is OFF, a diode with less than 100mV applied is effectively infinite resistance, and a noninverting op amp amplifiercan scale a thermocouple value (for something much hotter than room temperature)without much trouble.
| If your thermocouple really IS in series with a driven heater, the only way to getan accurate reading is to suppress the readout unless the heater drive is OFF(disconnected, hopefully). The disadvantage of this is that the circuit will be sensitive to noise, because you won't be able to add an aggressive low-pass filter, but it should be do-able in a controlled situation like this one.
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What is this part of the wire called? I have no idea what this part is called. I want to figure out if I can remove the wire out of it and slide in the other broken end before soldering it the board. Is it reusable, or do I need to buy another one? (e.g. because it's heatshrinked or something) <Q> Found it, it's a "strain relief grip" and this one can't be replaced since it seems bonded to the wire. <S> Source: Securing electrical cables to holes in enclosures? <A> Actually, there are standard strain relief parts offered, see example at Digi-Key, with the same function and without epoxy or other resin molding: <S> They also go under "cable gland" moniker. <A> Strip <S> the insulation off the cable above the strain relief and then pull out the copper wire strands. <S> Then use a small drill bit and drill out each of the two halves of the cable. <S> Work your way up in drill size. <S> Use pliers to turn the drill bit. <S> Don't use a power drill unless you are very skilled with it. <S> The strain relief and the cable are usually made from different plastic, so it is easy to scrape out the cable insulation. <S> Once the strain relief is cleaned out, then pull new cable through. <S> (or the undamaged old cable) <S> I have done this procedure few times myself.
| You can reuse that strain relief.
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Does the order of components matter when connected on the same track in a PCB? In this LT3652 solar charger application: the BAT pin first connects to a resistor of 30k, then to the Schottky diode's anode, then to a filter capacitor and finally to a current detector resistor and a battery. How crucial is it that we maintain this order while designing the PCB? <Q> It really depends on the circuit. <S> In good schematics the components are arranged in such a way that it exposes their intent. <S> You can gather this information from the datasheet. <S> For example, C3 is said to be a decoupling capacitor. <S> The closer you put this capacitor to the battery, the better decoupling you will have, since the loop for RF to ground will be shorter, and therefore parasitic inductance of the PCB trace will also be lower, which in effect will increase your decoupling efficiency. <S> So it definitely does matter. <S> The hard task (for me personally) is to recognize the reason why the component is on a board, and design the layout in a way that maximizes the usefulness of this component and its purpose. <A> You want to minimize this area simulate this circuit – <S> Schematic created using CircuitLab <A> Read the layout considerations section from the datasheet carefully, and follow it.
| Order and placement of components does matter, a lot.
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powering an old 9V multimeter with a 6x AA battery pack I have an old Nimex NI2100 multimeter that is working fine with 9V batteries. Btw i'd like to power it with 6x AA batteries instead so i've bought a battery holder and wired like this: I've checked with another multimeter the voltage beetween the +BAT and -BAT pins on the PCB is 8.4v.However it does not turn on! Is it possible it has some kind of protection for using with different kind of power sources? <Q> As others have said, you've undoubtedly got the polarity wrong (hopefully not fatally, but it's possible), but also note that you have a bit of a dangerous setup there. <S> The batteries are tied to the unknown voltage being measured so the battery is normally contained completely within an insulated box. <S> For example, a cheap B&K meter I have in front of me carries this warning: <S> WARNING: To avoid electrical shock remove test leads before opening case. <S> Your case, as shown, is effectively open all the time. <A> The AA cells will power the meter if your polarity is correct. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Standard 9 V battery and clip arrangement. <S> You are using a clip as a battery so the terminals are reversed and you must reverse the wires so that the terminals are the correct polarity. <S> Figure 2. <S> A close-up of your clip by the guys in the forensics lab indicates that you forgot to reverse the polarity. <S> Quick check. <S> Disconnect from the meter and measure the voltage on the clip. <S> The bump terminal should be positive. <S> Why this is a bad idea <S> The meter is designed so that the user is fully isolated from all conductive parts. <S> Sockets are recessed, battery is totally enclosed and the case is, most definitely non-conductive. <S> By using an external battery you risk electric shock should the meter leads come in contact with mains voltage. <S> Buy the correct battery. <S> They last for months or years of normal use. <A> 8.4V should be enough since a typical 9V alkaline battery runs closer to 8V in use. <S> From: LINK <S> Make sure <S> you have the polarity correct.
| As well as shock, you could damage something electrically if the battery connections shorted to something.
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Interfacing a button to a microcontroller (and PC) with a 50 m long cable I am designing a board that will be plugged into a computer and will read the status of a button ~50 m away in an office environment (it's actually a lot closer, but the cable is long). I think it's a good idea to galvanically isolate the button wiring from the computer, since the PC will be grounded. I don't want any faults on the wiring to be able to damage the computer. I'm assuming less than 100 ohm resistance for the cable, and while a simple series resistor would work, I think having a constant current sink for the opto LED is safer (i.e. if the cable has to be a lot longer, or shorter, etc.). Is this a sensible approach to it? Cost/space is not much an issue, so I could add some protection/filtering circuitry, but I'm not entirely sure where/how to do it, so I'd be happy to hear some suggestions. simulate this circuit – Schematic created using CircuitLab <Q> Looks like too much circuitry, which leads to more cost, complexity, failures. <S> There is nothing in the question that indicates anything more than series resistors are required. <S> Adding components, like isolated switching power supplies, adds components with much higher failure rates than a few resistors and diodes. <S> There would need to be a specific, compelling reason to add all that circuitry in the question. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> The two transistor current sink might be slightly better and maybe 100K is a bit on the high side for the resistor. <S> Eg, simulate this circuit – Schematic created using CircuitLab <S> You could also flip the current limiter and put it on the other rail. <S> Right now the opto sees a lot of common mode voltage change when the switch is pressed. <S> Grounding the photodiode would reduce that because of the coupling capacitance of the DC-DC. <A> A simpler way would be to use a shielded cable (to shunt noise and ESD away), and then protect the microcontroller inputs with diodes to V CC and ground. <A> I would make a current loop. <S> Simple, cheap and reliable. <S> You can connect the transistor in a common collector configuration if you want a non-inverted output. <S> The optocoupler LED must be rated for at least 20 mA.
| That looks fine to me, but you may wish to put a diode across the optoisolator LED in case you get some ringing in the choke or wiring. The circuit below is well protected, simple, reliable, and goes high/low when switch is closed/open. The resistance of the cable is most likely to be between 1 or 10 ohms (as long as the AWG is more than 30 gauge).
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Eliminate AC noise from DC supply? I have built an automatic water tank using the circuit given below: But this circuit was working well for 2 days. But after that, it triggers falsely. When a fan in another room is switched on, this circuit trigger the relay and motor will turn on. I think this is due to AC noise. I was using 12v AC to DC converter supply, i.e, 12DC adapter. Can anyone help me to avoid this situation? <Q> Decouple the IC as said in the remarks above: 100 nF close to the IC across Vcc and GND. <S> A 10uF or 100uF cap next to it will help too if the 12V power supply lines are long. <S> Moreover make sure the 100 ohm resistor (connected to up) is close to the transistor and both are close to the 555. <S> And most important, add a 100 ohm between Low and pin 2 of the 555. <S> The resistor should be close to the 555 again. <S> Adding a resistor close to the 555 willl attenuate that noise. <S> If that resistor is far away, noise can still couple into the line and triggger the 555 incorrectly because it won't 'see' the 100 ohm resistor. <A> Your water tank has lots of capacitance to ground .Your <S> high and low sense wires are antennas .Deal <S> with these wires with 100n bypass caps to shunt any AC noise .Your <S> SMPS makes lots of common mode noise .Try something thats properly grounded or a linear supply . <A> Pin 2 is very sensitive, as stated add a 100R is series, I would also try a large pull up to Vcc, 220K or higher. <S> this means gnd (0v) <S> to you tank not Vcc and a pnp for the reset transistor.
| It is most likely the wire to pin 2 of the 555 picks up noise and makes the 555 trigger incorrectly.
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Is there a way to bypass a component in series in a circuit if that component fails? Imagine a couple of speakers connected in series. Could there be a way to build such circuit with the ability to close the circuit if one component isn't working anymore? Like a different path that ignores the malfunctioning component, maybe using a voltmeter to trigger a switch in the circuit when it reads zero. (just a question I'd given my electronics teacher in class. He failed to give me an answer) <Q> It's possible in certain circumstances but generally not. <S> In your speaker example imagine what would happen when there is no sound being transmitted: all series speakers would short out and the amplifier is now driving a short-circuit. <S> Truebeard's Stumper explains this mechanism: <S> [Inside the bulb there is a shunt resistor.] <S> It consists merely of a piece of OXIDIZED aluminum wire, wrapped around the lead-in wires, just above the bead in the lamp. <S> At normal operating voltage (2.5 volts for 50-100 light sets ...), the oxide coating acts as an insulator, and the current goes through the filament. <S> But when a lamp burns out, There is an OPEN CIRCUIT, and, in all series wiring, that puts the FULL LINE VOLTAGE across the defective lamp, and the 120 volts will "BURN" through the extremely thin oxide coating on the shunt, causing the shunt to actually short the lamp out. <S> (This is exactly the same effect as twisting the lamp to short the wires together!) <S> This completes the circuit, and the set lights. <S> Note that this increases the voltage applied to the rest of the set and an accelerating cascade of bulb failures <S> will (eventually) follow. <S> Another example you can research is runway lighting. <S> Rather than parallel all the lamps, which would result in gradual voltage drop along the runway, the lamps are fed from transformers and the transformer primaries are series connected and a controlled current sent down the line. <S> You can research this yourself to see how faults are handled. <S> In general your scheme isn't going to work. <S> Shorting out series connected loads means that the remaining loads get higher voltage than they should and damage will ensue. <S> In addition there is the problem of how to energise the switch to reset the device on power-on. <S> If the device has shorted itself out there is no way for it to power itself back on. <A> Is there a way to bypass a component in series in a circuit if that component fails? <S> mini bulb strings, a spring-loaded bypass-switch closes when the filament burns out. <S> battery management system <S> (BMS) chips bypass excess charge current for each cell due to mismatch to extend life and prevent failure. <S> some LED drivers have string open detection to detect and/or bypass individual without excess current. <S> "diode OR" logic is also "current rectifier" to allow standby power from the battery during a power failure AC grid fault detection systems have re-routing switch methods for some fault conditions <S> The internet was designed with redundant paths where a "switch" is just one unit in a system. <S> Yes/ <S> No/Maybe That depends on a lot of unstated assumptions , which one can specify by questions. <S> What kind of component or failure? <S> Open? <S> , short? <S> or in between? <S> Will it reduce reliability? <S> e.g. bypass a fuse <S> Is it worth it? <S> Cost/benefit for an added detector <S> , multiplexer to bypass Is it better to choose a more reliable part, design or process in the 1st place? <S> What if bypassing that component damages the previous or next? <S> What if the circuit detecting a failure is less reliable, fails and bypasses in error? <S> What is the expected mean time between failures MTBF? <S> and to repair MTTR? <A> Sure, if you are able to detect when a device fails and you have the possibility to switch it off (e.g. with a mosfet) you can achieve this by actively controlling the device. <S> But how you would do this in concrete depends on the kind of device we're talking about. <S> In case of (e.g.) an led this might be trivial as a failure will usually result in an open circuit which is easy to detect. <S> Concerning a speaker it will be more difficult to detect, speakers naturally have a low resistance/impedance and might become a short circuit if the fail completely. <S> But even before an entire failure the sound quality might decrease without a significant change in the basic specs of this speaker. <S> As stated, in generally you only need to detect a failure and be able to switch off the device.
| There exist some examples of this already where it is feasible. Filament bulb series-connected Christmas tree lights have a mechanism whereby failed bulbs are shorted out. This depends on the cost of failure, the failure mode and the ease of detection and safety consequences and the cost of solutions.
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Is it OK to reduce the charging current for a Li-ion 18650 battery? I am using NCR18650 batteries with the tp4056 charging module. When I connect the charger at 2.8 V (fully discharged), the tp4056 IC is getting hot at a 1 A charge current. Is it OK to reduce this charge current by changing the charge current selecting resistor? So the tp4056 IC heat would be reduced, wouldn't it? For my application I don't need a quick charge. With a low current, charging time will be increased, but it does not matter for my application. Will reducing the charge current affect the battery's lifetime? Or will it damage the battery? I am trying to reduce the heat of the tp4056 IC by reducing the charging current, maybe from 1 A to 500 mA or less. The 18650 battery never heats for a 1 A charge current. <Q> If you halve the charging current, charging will of course take longer. <S> But for an 18650 cell, 1 A charging current is reasonably normal. <S> 2.8 V is a very low voltage for a Li-Ion cell, if you discharge often to such a low voltage the battery's lifetime might be shorter. <S> I would prefer not to discharge below 3.4 V. <A> The regular NCR18650 Panasonic battery has the recommended charge current at 0.7C, or about 2A. <S> And the discharge threshold is 2.5V, according to manufacturer's discharge curves. <S> So you are charging the battery already at half permissible rate. <S> Formally yes, charging a Li-Ion battery at slower rate doesn't do any damage and might be even better for battery's SOH (State of Health). <S> It will just take longer. <S> The TP4056 is a LINEAR charge controller , so it will dissipate a lot of heat and will be hot, if the board is designed with improper heat sink. <S> If you are concerned with IC overheating, you should either make a better heat sink (at least over the top if IC), or reduce input voltage to, say, 4.5V. <S> It will still dissipate 0.3- 0.5W and should be warm. <S> Or get a different TP4056 charger board with a better constructed PCB and with bottom heat slug soldered down to it. <S> Or get a better charger system. <S> Reducing charging current of TP4056 will be the least optimal solution. <A> I use TP4056 charger boards a lot, both on their own and embedded in some devices. <S> In addition to existing answers, I'd like to mention this - it's OK for TP4056 to heat up, it's a linear charger and it heating up <S> doesn't impact its performance/reliability all that much. <S> As long as the board has proper heatsinking, you should be fine - i.e. the popular "blue PCB with USB port" chargers from China have good enough heatsinking, as they tend to connect the ground pad of the TP4056 to copper-filled areas on both layers (sometimes only one, bottom one, with vias). <S> If it's a custom board that you yourself designed and it doesn't connect the ground pad to anywhere useful, only then I'd be worried - in that case, swapping the current set resistor for ~2K (that's what you need for ~500mA IIRC) should do the job. <S> Otherwise, you should be fine as you are now, with 1A charging.
| Yes it is OK to reduce the charging current and that is what the TP4056's charging current setting resistor is for. Charging slower (but not too slow) should actually increase battery lifetime.
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Connecting transformers in parallel.Will this be a problem? I am creating a power supply as a small project. I am planning on using two transformers (in parallel) to create a positive rail and a negative rail. Will this cause a problem ? I have researched online and found that the transformers can be connected in parallel given that they have the same number of turns and are in phase. Both the transformers will be connected to the same AC outlet thus I assume they would be in phase ? Also both the transformers are 220VAC to 12VAC , but one of them is rated for 1 A(planning to use that for +ve rail) while the other is rated for 500mA. Will this cause a problem ?. I assume not since they would both have the same no of turns (the 1 A transformer may have thicker windings ?) If there will be a problem , is there any way to connect two transformers to the same AC source ?. I have attached the schematic below if it helps. Thanks for your help :) simulate this circuit – Schematic created using CircuitLab PS : Is my schematic correct? I think this is how to create a negative voltage rail. Edit : Fixed typos, reworded a part of the question <Q> I am planning on using two transformers (in parallel) to create a positive rail and a negative rail. <S> Will this cause a problem? <S> No. <S> The primaries are in parallel as they should be. <S> The secondaries are not. <S> I have researched online and found that the transformers can be connected in parallel given that they have the same number of turns and are in phase. <S> That would be a requirement if you were connecting the secondaries together at both ends. <S> You're not in this case and it would not be unusual to have different voltage requirements for the positive and negative rail and so different transformer secondary voltages would be used. <S> Both the transformers will be connected to the same AC outlet <S> thus I assume they would be in phase? <S> That depends on internal and external wiring of and to the coils but in this case it doesn't matter. <S> At the output of the bridge rectifiers is a 100 Hz (assuming 50 Hz supply) full-wave rectified signal. <S> This will be the same regardless of the input or output phasing. <S> Also both the transformers are 220 V AC to 12 V AC , but one of them is rated for 1 A (planning to use that for +ve rail) while the other is rated for 500 mA. <S> Will this cause a problem? <S> I assume not since they would both have the same no of turns (the 1 A transformer may have thicker windings ?) <S> Your assumption is correct but your reasoning is not. <S> As discussed above, your two rectified supplies are independent and only connected at one point. <S> An analogy is a 12 V car battery for your positive supply and a small 9 V battery for your negative supply. <S> If the 9 V battery can provide the required current on its load then that's all you need. <S> Is my schematic correct? <S> I think this is how to create a negative voltage rail. <S> Yes. <S> You'll probably be adding in some capacitors to smooth it out. <A> That schematic looks fine. <S> I assume you are going to add reservoir capacitors between the rails and ground to hold up the output rail between peaks of the input waveform! <S> Note that with a diode/capacitor supply, you should derate the transformer's rated current for output DC current, to about 85% IIRC <S> , I'm sure someone can look it up and correct me. <S> While you have got the primaries connected in parallel, that's no more than connecting them both to the same mains supply. <S> You have not connected the transformers 'in parallel', the secondaries are totally isolated from each other. <S> They could safely be different voltages, you could invert the phase on one, with no problems. <S> If you had connected the secondaries directly in parallel, then you would need to make sure that not only were they the same nominal voltage, but also had exactly the same number of turns, primaries and secondaries, and were phased correctly. <A> There's nothing wrong with your proposed circuit, but if the transformers are identical, and if you make sure that they're connected in the correct phase, you can improve on it a bit by using just one bridge rectifier. <S> See the schematic. <S> You could also use a center-tapped transformer, if you have one of the appropriate voltage. <S> simulate this circuit – <S> Schematic created using CircuitLab
| The fact that the transformers are different current ratings is not a problem, as long as you don't exceed the respective rating on either rail. Everything connected to mains is connected in parallel.
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Purpose of level-shifter with same in and out voltages I am looking at a schematic for the Pixhawk 2 board (page 3). The UART ports on the microcontroller are sent through a level-shifter chip ( TXS0108 ) where the input voltage is the same as the output voltage. What benefit does this provide? <Q> On the full schematic there is an interesting note that starts:- due to the serial lines being able to back power the cpu... <S> The STM32F407 MCU is only rated for an absolute maximum of 5mA injected current per pin (-5mA, +0mA on 5V tolerant pins), so if it was powered through a UART pin it could easily be damaged. <S> connecting an RS232 serial device (which might put out +-12V at 15mA or more). <S> Another reason for buffering the serial port lines could be to make the MCU more tolerant of EMI. <S> Autopilots are often used in harsh environments with nearby rf transmitters and high power brushless motors. <S> An EMI induced glitch that caused the MCU to freeze or go crazy could crash the drone before it had time to recover. <A> The only obvious thing (other than buffering) I see is that the I/O can be disabled via a control line Level_Shift Enable . <S> When that line is low the outputs are high-impedance, so it acts as a bidirectional tri-state buffer. <A> The level shifter is used not only to "shift" the logic levels of a signal, but also to lower the impedance of its source and increase its current drive capability in order to drive heavier loads without going out of specifications: <S> in this case, the higher impedance output UART is sent to out the microcontroller board via the TXS0108 probably to rise its output drive capability. <A> The other answers do not make sense. <S> The output impedance of a microcontroller pin is about 30 ohms: <S> perfectly capable of driving the 120 ohm series resistors. <S> Also, I have not seen a microcontroller where you don't have tristate control in the chip. <S> I suspect historical reasons: in the past it was level shift to 5V,which has evolved to levelshift to 3.3V, but to reduce impact on the layout the levelshifter was left in place.
| The TXS0108 level shifter is rated for 50mA per input, so it is less likely to be damaged by eg.
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Mounting finned flat-base heat-sink with no tabs or screw-holes The heat sink has a flat base with no tabs or holes. How is it supposed to be mounted on a PCB? Wire brackets soldered to the PCB come to mind but then how does one ensure there's enough pressure between the heat sink and the PCB for good heat conduction? <Q> Thermally conductive adhesive is often used between small heatsinks like this and the chip they're cooling. <S> There are various silver, aluminium or ceramic loaded epoxies commercially available for the purpose. <S> Clamping is an alternative. <A> These types of heatsinks are quite common on computer motherboards. <S> I assume by "mounting on a PCB" <S> you mean mounting it to an IC <S> soldered onto a pcb. <S> Once glued on they cannot be removed without possibly damaging the chip underneath. <S> Here is the link for some thermal glue : <S> https://www.amazon.com/Conductive-Silicone-Adhesive-Compound-Heatsink/dp/B019MSKTX8 To use the glue simply trace the outline of the chip with the glue and add a small drop of thermal paste in the center(if you want) then push the heatsink on and hold it until the glue sets. <A> [In addition to what others have suggested already.] <S> Wire brackets soldered to the PCB come to mind <S> but then how does one ensure there's enough pressure between the heat sink and the PCB for good heat conduction? <S> Heatsinks like that are often pressed to the IC with a purpose-made spring clip. <S> The ends of the spring clip attach to anchors soldered to the PCB. <S> The middle of the spring clip presses the heatsink to the IC. <S> , more drawings here <S> I've seen an arrangement like that on some of the desktop motherboards. <S> related reading: How to choose a heat sink attachment method to secure your heat sink and for optimal heat transfer. <S> Part 1 . <S> Part 2 .
| They are glued directly to the IC using special thermal glue.
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Ultra-low power sound/audio sensor for LOUD noises I'm searching for a sound/audio sensor with ultra-low power consumption to detect the presence of a really loud noise (100dB+). The controller runs on battery and there is no option for a different power supply. For me as a non-expert there seem to be two major types of sensors: Electret microphone (with potentiometer): Here I had a battery life of not more than a week. But the device works as intended due to the right sensitivity. MEMS microphone: These do look like the real power savers but the byproduct seems to be some way too high sensitivity. I read about that Vesper VM1010 ( https://vespermems.com/products/vm1010/ ) which looks like a perfect fit (if the noise threshold can be set high enough) with about 10us current but as a hobbyist I'm totally unable to do anything with PCB and I can not find any (cheap!) breakout boards. The common use case for sound sensor seems to be the detection of (very) low noises and I have been searching for a while now to find my product. And yes I even considered to design my own PCB circuit but I have a lack of electronic knowledge and I just don't want to believe that there isn't any smart solution for simple people. I really appreciate any help or suggestion you can give. Thanks a lot! <Q> Try using an ordinary loud speaker as a microphone. <S> And it doesn't require any applied power. <S> The moving coil and magnet generate a voltage. <S> Loud speakers come in a variety of low impedances. <A> The VesperMems product seems to be ideal for your requirements; amplitude detection, variable threshold and ultra-low standby power. <S> All you need is about 80 dB of attenuation, across the whole audio frequency which is harder than it sounds ( pun intended ) as lots of materials behave more like a low pass filter, so you can still hear the bass. <S> This may require some further research on your part to identify the spectral sensitivity of 100dB which to humans becomes an almost flat response. <S> Courtesy <S> WIki <S> When you use a speaker or microphone, you can imagine how loud it is when you tap the diaphragm. <S> So protection from these near-field vibrations must be isolated or prevented. <S> Acoustic Pressure is logarithmic (dB SPL) <S> meaning it spans many decades of power levels and choosing a calibrated 100dB can be done, but this is another task for you to define , as well as defining the minimum event duration and s tretching the event to stay ON or to some interface device. <S> So there is more to your design specs than simply choosing the sensor. <A> However, these are high-impedance devices, and to preserve the large output voltage, it must see a load resistance that is large. <S> As a test, a piezo element was measured with a 10Mohm oscilloscope. <S> A "hand-clap" was the source of loud noise, about six inches away. <S> When changed to a 1Mohm load, results were similar. <S> It produced the following voltage, whose peak-to-peak maximum amplitude reached about six volts: This particular piezo transducer was surrounded by a plastic shell, which provides a large resonant peak at about 3.4 kHz. <S> That results in a sinusoidal wave that swings above and below zero volts. <S> There are about seven cycles within each (dotted) division of the oscilloscope scale (0.002 seconds per division). <S> You could quite possibly hose this transducer up to a microcontroller's analog-to-digital converter directly - no amplifier needed. <S> A few large-value bias resistors might be necessary to establish a known DC voltage at the input pin. <S> This might even work hosed up to a CMOS or HCMOS logic gate input.
| A piezo transducer might suffice to provide a voltage needing no amplifier.
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Trigger modi of an oscilloscope I'm sure this belongs to the "every once in a while asked again" questions, but I've read several posts on this site and also on other sites and haven't found an answer to two of my questions, yet. LINE-Trigger The LINE-Trigger is supposed to trigger whenever the signal has 50 Hz as frequency.It makes sense to me that 100 Hz, 200 Hz, and so on are triggered, since instead of one period it does 2.However, when I put in an 75 Hz signal into the oscilloscope it still triggered and I dont' understand why. The points where the 50 Hz signal is 0 are also points where the 75 Hz signal is 0, but while the 50 Hz one is increasing (if it's a sine-signal) the 75 Hz isn't.So does it really only matter that the signal is 0 at time 0 and then after 1/50 seconds it's zero again? EXTERNAL-Trigger The second question I have is about external triggering.According to the sources I read "the external signal is used to trigger". For me, this sounds as if the oscilloscope triggers, when the input signal is exactly the same as the external trigger signal.However, I used a function generator and used the external trigger output and also let the oscilloscope display the signal, it was a square voltage.I then gave a sine input (which I wanted to use the external trigger to trigger for). (For me) surprisingly, it still triggerd even tho the sine signal had a different form than the external trigger (square voltage).The sine signal was increasing when the square voltage was positive and the sine signal was decresing when the square voltage was negative.So not only the shape was different but also when the sine signal reached it's peak, the external trigger jumped from negative to positive, so the 0 points didn't match.The only way I could explain this to myself is that the oscilloscope only triggers when the frequency of external trigger and input signal are the same. Sorry for that much text, I just wanted to show you that I have already thought about this and I'm not just to lazy to read other posts. <Q> If you use the line trigger or the external trigger then the frequency and form of the displayed input channel does not matter . <S> The scope doesn't trigger when the line or external signal are <S> the same as the displayed signal, it triggers when the chosen trigger signal passes through your selected trigger level with your selected slope. <S> You should only use these modes when the signal you want to observe is derived from or strongly related to the line or some external signal. <A> The scope has a special box inside: the Triggering. <S> The Trigger box has Trigger-Source selection, usually just a switch (rotary switch, to accommodate the numerous Trigger Modes). <S> Once a strong enough input signal arrives at the selected Trigger Sources ----- internal trigger (from one or both of your channels you might be displaying) <S> ----- line trigger (from the Scope's power supply) ---- <S> external Trigger (usually a BNC coaxial input connector) <S> the Scope then SWEEPS across the screen ONCE . <A> Normally, the external trigger of an oscilloscope, if it has one, is connected to another piece of test equipment that has a digital pulse to trigger the oscilloscope. <S> The external trigger signal is normally not related at all to the input signal.
| If you are looking at a signal that has a different frequency from the line or external signal then the scope will seem to be triggering at random.
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300VDC to 24VDC voltage converter - Cheapest implementation In an application, 100+, 100W 24V halogen lamps need to be individually controlled and run simultaneously. They stay on for a few seconds at a time and need to have very constant irradiation. Furthermore, It needs to be as cheap as possible, but have fine control of the lamp with at least in the 0.5% granularity of the maximum power output. This system is enclosed and is for lab use. The first attempt was to rectify and filter main (230VAC) and use a DC/DC buck converter to go from 300VDC rectified to 24VDC being controlled by a dsPIC33. However, I miscalculated the inductor required and the main issue is, with the correct inductor, the Duty cycle of about 8% which at a frequency of 20kHz only leaves very large granularity for the speed of the dsPIC to generate the PWM. Another option is to have a large (or several) AC/DC supply to go from the main to 25V or so and then a buck on each lamp, but that would increase the cost significantly, about 2-3k$ for the supplies and large currents will have to be handled. Is there a cheap, more appropriate topology for this need, or is there a way to increase the nominal buck duty cycle without wasting power and having reasonably sized coil ? EDIT: Perhaps a possibility would be to have directly 230VAC and regulate the lamp directly with a Triac although the ripple would be too big at 50 (100Hz) so there would need some sort of circuit behind as well. EDIT 2: Someone came up with the idea of using lead-acid batteries which would lead to the use of a much smaller power supply and drive the lamp directly in PWM without buck, not sure about EMI and cold start. <Q> How many lamps are you running simultaneously? <S> I ask because dropping the mains to somewhere in the 24-48V region makes your control problem easier, and if for example only say 20 of those lamps are on at any one time <S> then you are into the place where cheap surplus 'telecom rectifiers' are a thing. <S> A 2kW 48V telecom rectifier is not expensive on the surplus market, and 48V would increase your duty cycle in a PWM arrangement significantly, possibly even allowing simple PWM (Watch the difference between RMS and average here) without much in the way of inductors. <S> TH lamps have significant thermal time constants so flicker is negligible by the time you hit even a 1kHz switching rate. <A> use a Basic phase controlled light dimmer .Now replace the dimming pot with a LDR .Tie a LED to the LDR to make an optocoupler <S> .This <S> trick has been used for making a crude lead Acid battery charger. <A> "However, I miscalculated the inductor required and the main issue is, with the correct inductor, the Duty cycle of about 8% which at a frequency of 20kHz only leaves very large granularity for the speed of the dsPIC to generate the PWM." What about modulating the duty cycle?Take a running average of your duty cycle and compare it to your target duty cycle. <S> If the average is higher than the target decrease the duty cycle and vice versa. <S> The trade off is that you increase resolution but will experience more overshoot and undershoot around your target. <S> How much that is a problem I suppose depends on how much halogen lamps behave like low-pass filters and <S> how much light variation you can tolerate in your application. <S> There is another problem however <S> and I don't know if you've considered it;The voltage you are supplying the halogen lamp with <S> does not have a linear relation with its power consumption. <S> A given voltage at any time does not always correspond to the same power consumption (even if warmed up). <S> If you want precise control over the amount of light you need some feedback at the very least.
| A given voltage does not necessarily equate to similar power consumption with another (warmed up) lamp. Cheap and dirty is to use a standard 50 HZ 24V Halogen lamp transformer .Then
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