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Arduino sensor ground isolation on motorcycle I am attempting to use a 5v Arduino pro mini to display the current gear on a 2005 Kawasaki Ninja based on the ratio of the wheel speed and the engine RPM. The display is a single digit 7-segment display. The RPM input is coming from the 12v coil signal wire. The wire provides 12 volts and then goes to 0 volts to trigger the coil to fire. I have no problem reading the 12v-0v-12v pulses with the arduino. The Arduino groud is connected to the motorcycle's ground and the coil signal wire is going to digital pin 2 via a 1k ohm resistor. The wheel speed is coming from a hall effect sensor provided with the TrailTech gauge pack I already have on the bike ( http://www.trailtech.net/digital-gauges/vapor/752-300 ). The voltage difference in between the two wires is 3 volts except when the magnet passes by where is goes to 0 volts. I have been able to read the 3v-0v pulses by simply splicing two wires off of the sensor leads and connecting the positive wire to digital pin 3 and the other wire to the arduino's ground. With the sensor connected like this, the arduino can read the pulses from the sensor and the gauge pack can read wheel speed without any problem. The problem comes in when trying to read both wheel speed and engine rpm. I have the coil signal wire and the wheel speed sensor connected as described above and am able to read the coil pulses but not the wheel speed pulses and the gauge pack is also not able to read wheel speed. I believe this is because of a lack of isolation and the noise from the motorcycle's electrical system but I am not sure. My first thought is to use an optoisolator to send the wheel speed signal. My (possibly naive) thought process and limited understanding of optoisolators lead me to believe that I should be able to connect the two wires from the hall effect sensor to the input for the optoisolator and use the output from the optoisolator to trigger the arduino. Can anyone tell me if my assumptions are correct? Are my issues reading wheel speed likely caused by a lack of isolation? Is an optoisolator the way to go? <Q> Yes I believe your issues can easily be caused by lack of isolation. <S> Yes I believe that is the way to go. <S> I would do it for both sensors. <S> Your title question mentions ground isolation. <S> That may not be a problem, or it might be. <S> Best is if <S> your Arduino's power suppply is regulated if running from the bike's battery. <S> I often use small 12v-to-5v converters <S> (this is a search) to power my arduinos from noisy 12V sources like those types of batteries connected to trickle chargers. <S> Bottom line is that a noisy power source can mess up a perfect program. <S> I did some searching and came up with this optoisolator that does not require the grounds to be the same on both sides. <S> Optoisolator <S> It says long-distance, all that means is that the leads can be long enough for your purposes. <S> We do not give shopping or specific purchasing recommendations on this site, but I am showing this to you just to give you an idea of the kind of unit you should look for. <S> If you can keep the grounds separate then by all means do so. <S> Just running the wires will probably pick up some, so your optoisolator should be located next to the computer. <S> Let it clean up any extraneous signals that come in on the wires. <S> Sounds like a fun project. <S> Enjoy! <A> I read your question just before turning in last evening. <S> When I started to answer it this morning I see that SDsolar hs made nearly all the points I had in mind. <S> I agree with all he has said. <S> You may need some help with selection resistors for use with an optocoupler. <S> The module SD linked to is not well documented so follows an appropriate design for your two signals <S> should you use a common part, 4n25. <S> It draws about 4 ma from your sensors. <S> Make sure your wheel sensor can provide that much current while maintaining nearly 3 volt peak to peak swing. <S> You need a resistor in series with the input of each opto. <S> Use 470 ohms for the wheel opto and 2.2k for the engine speed opto. <S> Use the pullup resistors in Arduino for both opto outputs. <S> They will be 20k or more. <S> Let me know if your wheel sensor can't provide the necessary current. <S> We will find a solution for that. <S> Connections;plus wheel lead to opto input R, 470............. <S> other end 470 <S> to opto pin 1 <S> ................. <S> minus wheel lead to opto pin 2 <S> ................opto pin 5 to arduino digital in, <S> wheel................ <S> opto pin 4 to arduino ground............ <S> plus rpm lead to opto input R, 2.2k.............. <S> other end 2.2k to opto pin 1.................. <S> minus <S> rpm lead to opto pin 2............. <S> opto pin 5 to arduino digital in, <S> rpm.............. <S> opto pin 4 to arduino ground................... <S> With best wishes from the Bitterroot <A> I'd suggest you don't need to optically isolate both circuits or change your grounding to get reliable operation. <S> The sensing of the 12 V signal seems incorrect. <S> If you have a simple single 1K Ohm resistor in series to the Pin(2) on your Arduino, then that is an incorrect way to interface. <S> This means you raise the Pin(2) <S> above the 5 V supply of the Arduino. <S> I'm not sure of your coil drive, but you are likely causing about 7 mA or so current flow into the Arduino Pin(2). <S> Bad Juju. <S> Since your sensing of the Hall Effect sensor is working, and that voltage level is within specs for the Arduino (3 V) <S> then I'd leave that as a solved problem. <S> I'd suggest something like this: <S> simulate this circuit – <S> Schematic created using CircuitLab
| You are correct in your assumption of how an optoisolator works. And direct connection to the bike's electrical will be "noisy" as opposed to a nice clean output from an optoisolator. No good reason to get the bike's magneto impulses into your circuits. For the 12 V sensor I'd suggest that you could optically isolate it if you want, but I doubt it will be required.
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Can I directly connect GPIO to RS232 RXD and TXD? Is it okay to connect a 3.3V GPIO pin to the RXD and TXD pins of the RS232 port and supply it with 12VDC from an external source for serial radio? Or would I need to step up the voltage to 12VDC on the RXD pin to the RS232 and step down the voltage respectively for the TXD pin? I'm using n920 - OEM 900 MHz Spread Spectrum Wireless Modem serial radio. <Q> The specs on the N920 radio say that the serial interface is TTL <S> , it is NOT RS232.There <S> is no purpose in giving either the microprocessor or the radio atransmit/receive RS232 interface, unless they are to be separated bya large distance. <S> The serial <S> I/O terminals on the N920 radio can be directlyconnected to GPIO pins. <S> Unless other hardware needs it, there is also no requirement for 12V. <A> Can I directly connect GPIO to RS232 RXD and TXD? <S> Is it okay to connect a 3.3V GPIO pin to the RXD and TXD pins of the RS232 port and supply it with 12VDC from an external source for serial radio? <S> Simply, no. <S> Or would I need to step up the voltage to 12VDC on the RXD pin to the RS232 and step down the voltage respectively for the TXD pin? <A> I would suggest using a transceiver IC such is <S> ADM3252. <S> Below is a high level schematic of the IC <S> ADM325x support +/- <S> 30V on Rin. <S> Below are the date sheet for you reference. <S> Reference: <S> ADM3251E <S> Isolated Single Channel <S> RS-232 Line Driver/Receiver
| A transceiver IC (e.g. MAX232) is needed.
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Two weaker peltiers vs one stronger peltier for cooling? I'm planning on making a small water cooler for a project. I need to cool approximately 1.5L of water and have two TEC1-12706 (12v 6amp) peltier modules, and hoping to cool to as low as 6 degrees Celsius. Would it be more efficient/effective to instead use a single Pelt such as the 12715 (12v 15 amp) or run the two separate lower rated ones? Or even two 12715s at half power for example?Also any advice on calculating the ideal heat sink size/fan speed would also be very useful <Q> If you wish to use a heat sink for faster cooling, place it on the hot side as this will remove heat from the module and thus allow it to operate more efficiently. <S> From experience, I find that working the module at about 80% of full rated value is best for longevity. <S> If speed of cooling is not a major concern, connect the two modules in series across 12 Volts and each would receive the same current (but lower) at 6 Volts each and thus you would use only 1/2 the power as in the original design. <A> Using two peltiers is going to be better than one <S> : 1) You will get twice the heat flow (equivalent to current) and twice the cooling rate. <S> 2) <S> There will be less self heating assuming that you run them at a lower voltage (ie series). <S> It is impossible to say anything else about your system (like how low the temperature will go, or fan speed) because you need to know what your heat in and out of the system is. <S> The insulation around the tank needs to be known as well as the temperature on the outside of the insulation (or ambient temperature). <S> The insulation (or material if it is just some plastic) will leak heat in from the environment, the pelters need to be able to remove this heat. <S> They can provide 80C of cooling with no heat load at 15 Amp (black line). <S> If you wanted to cool to 5C and ambient is 30C (for a high mark), then you would need 25C across the peltiers, and one would give you 130W of cooling. <A> 6 degrees C might be ambitious for a single-stage Peltier cell. <S> You may want to consider cascading cells: have a large cell between ambient and a smaller cell that actually cools the target. <S> For example, I would seriously consider using the large 15A cell as the outside cooler with both of your smaller cells mounted to that large cell. <S> The smaller cells would be wired in series (as per one of the comments) to better equalize the cooling ratio. <S> Obviously, the cool side of the smaller cells contacts the target.
| If the water is stagnant, it's better to use two modules, one on each side of the container.
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Is it 'safe' to use 200ohm resistor to reduce 9v to 5v I just bought some lasers and resistors from Amazon and I am a bit confused.First of all the seller says the resistors are 200 ohm. I am a bit confused because I have only seen resistors with number (23) with (23k) by it. My question: Can a 200 ohm resistors reduce a 9 v to 5 v so I can power the laser? Is this safe to do without having anything blowing up? Picture analysis of laser <Q> https://www.amazon.com/ask/questions/Tx1M6VEO0CSRRC2/?source=allQuestionsPage <S> The seller says these are self regulating which means it will draw 20mA at 5V. <S> What confuses me is the seller says it draws 30mA at 3V where these lasers are diodes <S> thus they should always have a fairly fixed voltage drop that only changes when current changes. <S> By this logic the diode would be brighter at 3V than at 5V. Anyways we can deduce that: V = IRR = <S> V/IR = 5/0.02R = 250Ω <S> Looking like a simple device, it should be that a 250Ω resistor is already connected in series with the laser diode. <S> If you want to power it from 9V then: V = IRR = <S> V/IR = (9-5)/0.02R = 4/0.02R = 200Ω <S> You would need to add a 200Ω resistor in series. <S> So you're correct. <A> It depends on the current to be drawn by the laser. <S> Using ohms law, where you want to drop 4 volts (the difference from 9v battery to 5v laser), and given the 200 ohm resistors: I = <S> 4/200 = <S> 0.02 <S> Amps = <S> 20 <S> mA <S> so 20 mA should be the current requirement for your laser load for this to work properly. <S> Do you have the voltage and current requirements for your laser? <A> I don't know what your load current is <S> so I can't see how much power you'll be dissipating when dropping your 4 V. <S> If it's less than 250 mA (1 W dropped power), I'd use a 78S05 regulator on a heatsink. <S> This takes in 7..30 V and outputs a steady 5 V, regardless of load current changes. <S> If power efficiency is important to you, there are plenty of switching regulators or pre-made tiny brick DC-DC converters available. <S> The latter are more expensive but much more convenient to work with if you're breadboarding.
| A disadvantage of using a resistor to drop the voltage is that it does so by turning the excess power to heat, draining the battery faster.
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USB device not recognized I have been designing a usb device using a psoc 4L with integrated USB full speed capability. Yet for nearly all of my designs as soon as I plug the usb into the computer I get the following problem. From the research I've been able to dig into, this suggests that there is something wrong on my physical layer (I'm using a generic starter project provided by cypress, and occasionally I have had it correctly connect) I've measured the D+ and D- lines and here are the pictures from my o-scope My question is what appears to be the problem here? My intuition suggests that D+(the blue trace) is taking too long (18 ns which is greater than 12 ns allowed) Therefore it has too much capacitive loading and I need to identify in the layout what could be adding to it. Is this correct intuition, or is something completely different going on? EDIT ******* I redid the scope with measurements on. here it is. Looks like the voltage is consistently around 4.1V EDIT # 2******* I got it to work! I was originally powering the psoc 4 off of the miniprog. As soon as I switched to USB powered it worked. I'm talking with some cypress FAE's so as soon as I get an answer as to why that would be the problem, I'll post here <Q> Your waveforms look OK IMO. <S> You need to check how far you are getting in the enumeration process. <S> You can do that by checking the unknown device in DeviceManager. <S> Look to see if the VIP/PID is being detected in the initial negotiation. <S> If you have a VIP/PID, then you are getting quite far in the enumeration. <S> If the VIP/PID is zero then you are failing early in enumeration, or don't have a recognizable VIP/PID. <S> Your project should have a VIP/PID defined for a HID device which should be automatically resolved by Windows. <S> While there are commercial and free USB analyzers for Windows, you can also simply use the Microsoft Windows Message Analyzer to capture the USB traffic. <S> For this to work the base Phy protocol <S> (things like you USB levels and clocks) must be running OK. <S> However it looks like you are fine here and should be able to capture the traffic successfully. <S> There are online videos within the MMA interface that will show you how to operate the MMA or use the ones here ....and the operating manual is here . <S> It's not a simple tool, but <S> well worth getting to know. <A> The message you see essentially means that your device is plain dead, it can't complete basic enumeration process. <S> It could be that it is either completely dead, or just doesn't return correct descriptor information without protocol error. <S> The fact that the device gets connected and your USB host pulls out frame packets, your device is not completely dead, but the USB engine doesn't function correctly. <S> The signals are looking quite normal, except that the level, as other noted, is a bit too high. <S> Frequency is right. <S> But again, the signals you see are likely coming from your USB host, which has likely nothing to do with your device. <S> (to get a capture of a very sporadic device response you need to work pretty hard, and get a special test fixture with a reference USB device, then use 3-channels to differentiate the packets by their EOPs). <S> The most productive way to debug the USB physical layer is to get a USB protocol analyzer, like this one . <A>
| You can use USBPcap to capture USB packets and see where things go wrong.
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Accidentally downmixed stereo to mono, why does this work? I have a cheap Chinese Bluetooth audio amplifier like this: It has four screw terminals for the speaker connections, namely L+, L-, R+ and R-. I accidentally connected a single speaker to L+ and R-, and was surprised to find that I could hear both the left and right channel over the single speaker. I decided to experiment and found out that if you connect the speaker the proper way (so either to L- and L+, or to R- and R+) it would only play the single channel you'd expect it to play, but if you connected either L+ and R- or R+ and L-, it would "downmix" both stereo channels to a single channel and play them both over the single speaker. I've been trying to wrap my head around how that works, but I can't figure it out. Is the board more sophisticated than I expected and does it output mono when it discovers no load on the different terminals, or is there something else going on? <Q> This can be easier to understand if you look at the waveforms. <S> In a push-pull, or bridge amplifier <S> both lines are driven as shown below. <S> Notice \$L-\$ is literally the inverse of \$L+\$ Similarly the other signal, \$R-\$ is the inverse of \$R+\$ <S> The difference between the +/- <S> voltages is that excites the speakers. <S> Now, if you connect Opposites to one speaker the difference in signal becomes the mixture of both signals. <S> Hey-Presto.. you have a mono system. <S> Note <S> however, the amplitude of each "Side" is now effectively reduced by half. <S> If you can't understand that, consider the case where there is no signal on the \$R\$ side. <S> The difference between the blue lines is now only half what is was between <S> \$L+\$ and \$L-\$. <S> If the original sound was recorded central, that is, an equal waveform on both left and right channels, the \$L+\$ waveform would be identical to the \$R+\$ waveform, same for the negatives. <S> As such, joining \$L+\$ with \$R+\$ would result in no voltage difference for that sound. <S> That is why you need to cross connect them. <A> Yes that will happen. <S> Connecting a speaker to left+ and right- will reproduce a mono version of the stereo signal. <S> After all if you used a mixer to add <S> left and right you get mono but, with a speaker you cannot simply wire together left and right because you'll create a short when there is a left signal and not a right signal. <S> On the other hand, if you fed the speaker left+ and right+ you would get no part of the music that was centralized in the listening field and this would mean bass and (usually) vocal <S> would be severely attenuated. <S> So using left+ and right- to the speaker means you get full mono with a sonic bias towards signals that are straight down the centre. <S> Because the board uses bridge amplifiers to the speakers you get opposite polarized signals for each channel <S> i.e. you get a left+ and a left- <S> and you get a right+ and right-. <S> If this were a conventional single-ended amplifier you wouldn't have access to the opposite polarized signals as each speaker return would be 0 volts. <A> Traditionally amplifiers tie the "-" side of the speaker to ground and drive the "+" side. <S> This works well but it needs either a split rail power supply or a bulky and expensive output capacitor. <S> With such an amplifier it doesn't matter which "-" terminal you use, the signal delivered to the speaker will depend only on the "+" terminal. <S> With a "bridge" configuration both sides to the speaker are driven with inverse waveforms. <S> This allows a higher power output for a given total supply voltage and avoids the need for a split rail power supply (you still need to create a reference at half power supply voltage but that reference doesn't have to carry any significant current). <S> The downside is it means that the board needs four power amplifiers rather than two but power amplifier chips are fairly cheap nowadays. <S> So when you connect your speaker to R+ and L- the "+" side of the speaker is fed with a non-inverted right signal while the "-" side gets an inverted left signal and the overall voltage across the speaker ends up as a "downmix" of the two channels. <S> I expect the falling cost of semiconductors compared to everything else along with the move from main frequency transformers to flyback converters for power supplies has shifted the cost balance in favour of bridged amplifiers.
| The reason it works is because your amplifier uses a "bridge" drive configuration.
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How to determine temperature grade for automotive electronic components? When developing a new automotive electronic component, the component needs to be certified to withstand certain temperatures. There are several standards that list methods of testing, such as ISO 16750-4 and AEC-Q100. However, these standards list a number of 'grades' or 'temperature ranges', and it seems to be up to the reader of the standard to determine which grade applies. In ISO 16750-3, which deals with mechanical loads, each requirement states the purpose, such as "These tests are applicable to equipment to be mounted directly on the engine" or "This test is applicable to devices under test intended for mounting on sprung masses". For temperature ranges, I can find no such guidance. One might say that it is up to the customer to specify the required range, but how is the customer to know? Googling the problem doesn't help , as all web pages do mention 'extended ranges' such as -40°C to +125°C, but omit how they arrived at that number. How do I determine what the temperature range should be for a new component to be developed? <Q> The answer is given in Annex A of the ISO standard on climatic loads in road vehicles (ISO 16750-4): <S> For every mounting location, an operating temperature range is recommended. <S> A dashboard, which is "exposed to direct solar radiation", receives a recommended code G, which translates into a range of -40°C to +90°C. <A> One might say that it is up to the customer to specify the required range, but how is the customer to know? <S> Well, the customer has the whole picture and is the one who knows better from its own thermal analyses. <S> If the component is to be mounted in a place where the thermal interface or the ambient will yield a high temperature (i.e.,located near the engine, etc.) <S> the customer might want to specify a tougher requirement ("grade"). <S> These are the grades according to AEC-Q10, as you probably know: <S> Temperature grades might be already standarized as seen above, but this doesn't mean that the design engineers of your customer are freed from carrying out their analyses and telling you which grade should your component be in order to be used in their assembly. <S> One motor may run hotter than another, even under the same external environmental conditions, due to specification or design differences. <S> But you have to balance that with costs as well. <A> It's really up to YOU, or whomever else is dictating the product requirements, to decide which environment the thing must... <S> A. Survive in, and <S> B. Remain functional in. <S> Those may be different values. <S> For example, a standard LCD is not much use below 0C since the liquid becomes too sluggish to react to the applied voltages. <S> However it will work when it warms up. <S> If however you take it down to -40 the liquid can actually freeze and destroy the unit. <S> However, the above range is also dependent on installation location. <S> Direct sunlight, proximity to other heat sources, poor ventilation, etc. can drastically change those numbers. <S> What you enclose your electronics inside also matters. <S> The electronics inside an unventilated black box left on the dashboard in the sun in Death Valley, USA, may reach temperatures hot enough to melt some devices. <S> In addition, you ALSO need to worry about the enclosure itself. <S> If it is that black box I mentioned above, it has to be made of a material that will not melt, crack or degrade in sunlight. <S> It must also be constructed in such a way that it can expand and contract without stress failure. <S> Sometimes extreme conditions are required. <S> Military specification tend to be much more extreme than domestic, but wider examples exist. <S> For example, devices to be used outside in Antarctica, or a device that will be installed next to a blast furnace. <S> In those cases extraordinary measures need to be taken. <S> Heaters need to be added to the box for the cold example, and liquid cooling piped to the latter from a more distant location. <S> For automotive applications it really depends on what the device does. <S> If it is that little display the kids watch "Sponge Bob" on in the back seat, 0°C to +50 <S> °C may be sufficient. <S> The automotive "standards" indicate... <S> Where cost goes from high for Grade 0 to low at Grade 4. <S> You have to decide which one of those you need to beat for the functionality of your design and to be within your cost budget.
| If your component is not project-specific (i.e., no customer specification applies), then you might want to design it for the wider operating temperature grade actually feasible, so to be able to market it to as many potential customers as possible. If it is critical to the safe operation of the vehicle, e.g. a brake sensor, it needs to work better than the low levels of the standards, perhaps as low as -60°C.
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How to make the connections to the 1 channel optocoupler relay module? I want to light a bulb connected to the socket by using an 1 Channel Optocoupler Relay Module which accepts an input of 5V. From my understanding, when the modul receives DC current (5V), it's going to turn on the other circuit and turn on the bulb. I'm not sure how to connect the module in this context. It looks like this: +----------------------+ | 10A 250VAC 10 125VAC |DC+ – 10A 30VDC 10 28VAC – NCDC- – – COMIN – – NO | SRD-05VDC-SL-C | +----------------------+ It didn't came with any documentation, so I'm not sure what these notations mean. I assumed I have to connect the positive wire from the 5V circuit to DC+ and the negative/ground to the DC- . I'm not sure if this is correct. However, I don't understand how to connect the wires at the other end. I have no ideas what IN , NC , COM and NO mean. Is there any place where I can find the documentation for this kind of relay or are these universal notations? What do they mean? How should I make the connections? The module looks like this: So, I connected the bulb wires to COM and NC and the bulb is turned on when I connect it to the socket. Then I tried connecting a battery to the DC+ and DC- and the power led of the relay component is turned on, but the bulb is not turned off. <Q> The signals on the left are the input side and the signals on the right are the output side. <S> Ok <S> so now it is clear that you have a module rather than just a relay. <S> The module has additional circuitry on it for driving the relay. <S> Based on this the input pins are as I put in my original answer: <S> DC+ - Positive logic supply voltage DC- - Negative logic supply voltage (a.k.a ground) IN - Input control signal - the thing that tells the relay whether to be on or off. <S> On the output side a bog standard switch naming is used: <S> NC - Normally Closed (Connected to COM when relay coil off) <S> NO - Normally Open (Connected to COM when relay coil on) COM <S> - Common pin of switch <S> Your relay operates as a DPST switch as shown in the diagram below. <S> Connecting <S> the IN pin to DC+ will turn the relay on (connect NO and COM ), and connecting <S> the IN pin to <S> DC- <S> will turn the relay off (connect NC to COM ). <S> You may need to add a resistor in series with the IN <S> pin - i.e. don't connect it directly to DC+ or DC- . <S> A 1k resistor will suffice. <S> It is not clear from the description whether or not this is actually necessary as the information on the listing is a very bad translation. <A> Did you bother to scroll down on that page where you bought it... <S> That the last two lines mean is gibberish to me. <S> Try it and see. <S> I'd guess it means you can drive it either with a high or a low signal depending on that jumper. <S> In a nutshell, DC+ and DC- is the power the coil will need, IN drives an opto-coupler to deliver that DC to the coil somehow. <S> What High is, it does not say though, but since it's feeding an opto-coupler, do not exceed the 5mA. <A> Your relay module has a relay plus some interfacing electronics in it. <S> You have to supply the 5V from your batteries to the relay through the DC+ DC- pins. <S> The IN signal is the input control signal. <S> You can connect this pin to an output pin from a microcontroller like an Arduino, or to a mechanical switch connected to the 5V supply. <S> As a plus, you also have a jumper to select whether your control signal is active LOW <S> (i.e., a logic ZERO makes the relay switch) or active HIGH ( <S> i.e., a logic ONE makes the relay switch). <S> The relay internally connects the COM (common) pin to the NC (normally closed) <S> pin when the IN signal is driven LOW (or when it's driven HIGH, if you set your jumper to "active LOW"). <S> Then, when you drive the IN signal HIGH (or when it's driven LOW, if you set your jumper to "active LOW") the relay switches and internally connects the COM (common) pin to the NO (normally open) pin. <A> NC is Normally Closed, and NO is Normally Open. <S> COM is COMmon. <S> What this means is that, when no signal is applied to the coil, there is a short circuit between NC and COM, and an open circuit between NO and COM. <S> When you energize the relay coil, there will be an open circuit between NC and COM, and a short circuit between NO and COM. <S> I'm not sure what IN is. <S> You're right on the DC+/- terminals, I believe.
| You need to apply power to the DC+/DC- pins (such as a battery), and then use the IN pin to control whether the relay is on or off.
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Limiting Current interference in Photomultiplier Amplifier Circuit I've been working on a circuit for a while now. It's a circuit consisting of 4 Silicon Photomultipliers, MicroFJ-60035-TSV, in parallel all going into an op amp circuit. I posted questions about it before regarding separate issues about it, and I learned a bit more about about it by studying how the circuit functions. You can see it in the image below along with one of my simulations. Apologies for the messy part layout. In the circuit, I've connected the cathode to ground, and I'm reading the current from the anode. I wanted to get the correct pulse shape as shown in documentation, so I'm following the recommendation of reading from the anode. I am planning to read from the cathode once the circuit is finalized and not in simulation. I'm using voltage sources attached to the photomultiplier parts to control when they 'fire'. The 1 microOhm resistors are just there so that I can read the current coming from the sensors. I figured a low resistance wouldn't affect the circuit much, but ideally, there would be no resistance there. I'm kind of facing two issues right now. The first issue is with the current coming from the photomultipliers. You can see in the simulation results on the left that when one of the photomultipliers fire, some current also flows back into the other photomultipliers. What can I do about it to remove it or control it so that this doesn't happen? My next issue is with the feedback circuit. Through reading, I found that the op amp circuit converts the current signal into a voltage signal, and the feedback circuit acts as a low pass filter to hopefully filter out high frequency noise. You can see another test below where three photomultipliers fired at once. I've increased the capacitance to try and reduce the noise from the 4 parallel photomultipliers, but I'm having issues with the rise time and clipping. The time where three were firing had a faster rise time compared to the one which fired only once. However, the voltage seems to have clipped, staying the same after a certain point. Is there anything that can be done about it? There always seems to be a trade off between the cutoff frequency and the rise time. UPDATE: I changed my schematic so that I'm now reading from the cathode instead of the anode. The current from the sensors looks a little messier, but the output is now positive as I was planning. The circuit does not require biasing. It's already done in the model of the part. My team is now looking for better signal-to-noise ratio for the circuit, so what can I do in my circuit in order to improve the SNR? <Q> Glad to see somebody else is using silicon photomultipliers! <S> They are nifty devices, I've been playing with them recently on a project. <S> First off, I'm not seeing a bias across the device - remember, you have to bias the silicon photomultiplier past its avalanche breakdown point to get any avalanching in response to photons. <S> While the SPICE model may give you a current pulse when you trigger the fire input, this is not the real behaviour of the device. <S> In your case, the grounded cathode should instead be raised up to about 27.5V to bias the SensL device into operation. <S> Unless cost is a significant issue, I would use four transimpedance amplifiers to read them out, one for each channel. <S> This will solve your backfeed problem. <S> Something like the <S> OPA4354 (4 amplifiers, 250MHz GBWP, <S> $8 in singles) might work for you. <S> Since the SiPMs themselves are quite a bit more expensive than this, it probably won't break the budget. <S> If you need to add the channels together in analog, this can easily be done afterwards with a resistive summer. <S> Keep in mind that if you read out at the anode with a transimpedance amplifier, the signal from your transimpedance amplifier will be inverted. <S> This isn't an issue if you have bipolar supplies or use a virtual ground, but I thought I should mention it. <S> If I recall correctly, anode readouts from the SensL devices are preferred for speed since the cathode is tied to the package ground and has greater capacitance. <S> I would skip the feedback capacitor in the opamp circuit altogether unless you're having ringing problems. <S> As you have likely surmised, adding the feedback capacitor in parallel with the feedback resistor creates a lowpass filter, which is going to directly impact your rise time. <S> So, only do this if you need to combat ringing. <S> Finally, if you need extremely sharp risetimes, I would look at using the capacitively coupled fast output of the SensL device instead. <S> This is described in the C-series manual . <A> Your first issue would probably be best corrected by buffering the output from each multiplier into their own voltage then summing those into the final amp. <S> The second issue would be better handled using a classic RC bypass filter rather than the integrator type you have chosen. <A> You need a voltage buffer (assuming that you really want to isolate the diodes in your model, that may or may not be how the real world works). <S> Use a universal op amp or an ideal op amp found in the Devices\opamps folder. <S> Remeber the characteristics of ideal opamps, they can source infinite current ect. <S> If you just want to 'copy' a voltage to a node, use a B-source or in this case four of them
| Noise should not be a problem with a SiPM setup, since the current pulses are discrete (1 pulse height = 1 photon) and the peaks are well above the noise floor in a properly designed circuit.
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Is my oscilloscope working or not? I bought today a very old second hand oscilloscope (Philips PM 3253), see picture below. I put the probe into a +5v circuit before a LED inside an Arduino breadboard (and checked with a multimeter +5v was on that circuit). However, whatever knobs I turn or switches on the oscilloscope I set, the horizontal line always stays in the middle (except when I change the position knob on the bottom left). It's my first-ever oscilloscope ... what switch/knob should I set to get a reading of +5v ... or is the oscilloscope broken? @Update: It works now (sort of). I noticed there was a multiplication of 10 switch on the probe itself (so the difference between 0 and +5v was not visible), and had to change the main amplitude knob and the inner knob (whatever it means) and also the 0 X B Bal knob ... But the most important is, that it works, now I have to learn what all knobs mean :-) I made a new question, since I got some major problems using it: Oscilloscope makes my mains (fuse box) group go down (maybe problem solved) <Q> See that BNC connector on the left, labeled "Cal."? " <S> Cal." is short for "Calibration". <S> Stick the tip of your probe into that <S> (thanks to Peter Bennet for this clarification). <S> It provides a square wave signal that's intended for calibrating probes. <S> If you see the signal, the scope is working. <S> If you don't see it, the calibration output might not be working, or the scope itself might not be working. <A> Try moving the switch above your probe to DC <A> Been a while since I've used an analogue scope but try the following: <S> Connect your probe to channel B as channel A might be blown. <S> You may have a trigger enabled, use the 'level' knobs on the right to make sure it is within range. <S> It could be expecting an external trigger so test it by applying a voltage to the Trigger BNC input <S> The switch at the top, move it from LF to DC. <A>
| You're looking at a DC signal, you need to change the coupling mode from "AC" to "DC" (switch in the middle).
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How to create a constant current source from a 12V,19AH Battery which is load independent? My intention of connecting a copper coil directly to Battery is to basically make the copper coil red hot to boil water. What I have observed is when I connect the 12V,19AH Battery to the 15mm thickness 50 turn copper coil, the Battery tends to drain quickly and the voltage shown by a multi-meter of the battery confirms the draining. On the other hand, when a 30V,10A Laboratory Power Supply is connected to the same coil, it gives a stable current of 9A at 9V, 10A at 10V and trips OFF above 10V. With 10A of current the heating of the coil achieved is fair enough that serves my purpose. How can I provide the coil a consistent of 10A of current which will also stop the draining of the Battery at such a fast rate which otherwise would had been greater than 100A? <Q> The most straightforward solution is putting a resistor in series of the right value: 12V (nom) / 10 A = 1.2 ohm. <S> Depending on the battery capacity (Ah) and the state of charge, the voltage at its terminals while discharging might be more (fully charged, large battery) or slightly less. <S> Good. <S> But this is your heater itself! <S> So, all the dissipated heat shall go to your water and in this case your system would be efficiently serving its purpose. <S> Just fabricate a wire arrangement that gives you more or less 1 ohm. <S> For a conductor Ohm's Law says <S> R = resistivity * length / squaresection = rho*L/S with copper: <S> hard, because rho when heated to high temperature is still low <S> (something more than 20 mohm/m/mm2; for robustness you may choose a cross section of some mm2, e.g. 4, and you end up with 5 mohm/m, so you need 200 m coil to make 1 ohm. <S> And it's not as hygienic as steel <S> may be steel: steel has a 20-40 times larger resistivity; <S> wikipedia gives at 20°C temperature 1.7 10^-8 for Copper, 6.9 10^-7 for stainless steel, so 40 times. <S> With this material and same cross section of 4 mm2 you need only 5-6 m of length, that fits your need. <S> if you have smaller steel wires you can put them in parallel; if you have a total of 2 mm2 cross section, you reduce length to about 3 m. <S> Of course you might solutions based on DC/DC converters claiming higher efficiency or better regulation. <S> But all the heat you create in the load is going into the water (feeding cables from battery excluded), fine current regulation is not a concern, so the simple scheme calculated above works. <S> Brown wires are copper feeding cables from battery to heating coil, made of the said steel. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> How can I provide the coil a consistent of 10A of current which will also stop the draining of the Battery <S> A regulated power supply is wasted if you just want a heater. <S> A hundred-wattsoldering gun, of course, is the smallest that can match the heatoutput of 10V at 10A. <S> Expect higher current, but lower voltage,and adjust the wire dimensions accordingly. <S> Copper, at red heat, oxidizes and won't last long. <S> Hightemperature wire, like nichrome, might be used instead. <A> If the coil is dunked in water as is, there will also be some conduction in the water (the conductivity of water varies a lot depending on the type of water; tap water, distilled water, salt water etc), which will be fun. <S> For sanity's sake, I shall assume you have insulated the copper coil with something similar to heat shrink. <S> The properties that limit your current then are: internal resistance of the battery and resistance of the copper, two resistances in series. <S> Current through these depend on their value and the voltage across them, as per Ohm's Law (V=IR). <S> This means that you'll need a constant current circuit to drive it. <S> Have you tried any kind of internet search for "constant current circuit"? <S> There are quite a few options out there depending on what you want to do. <S> As for how to get your battery to last longer: you will drain it at a constant current.
| It is not surprising that the current is very high if you have a copper coil, as you're then just making a short circuit. The only way to make it last longer is either to draw less current or to have a bigger battery. Considerusing a transformer (such as a soldering gun) that has a high-current lowvoltage secondary, and run the wire on AC current.
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Powering up Arduino + GSM with LiPo rider with a LiPo battery Arduino Mega as a controller (5V) LiPo Rider as battery charger and to provide 5V - 1A to thearduino 3.7 V 10000 mAh Lipo battery EFCom GSM shield (500mA average, 2A pulse ) The problem is, the specs of the LiPo Rider Pro claim it is able to provide up to 1A Max. However, the GSM shield needs 2A (in peaks time), in addition to the arduino power and the sensors (around 100 mA). I tried powering up this system with a 2A powerbank but it didn't work, I powered it up with 12V, 2A adapter and it works fine! I couldn't understand why the powerbank is not able to power it up although the current in both is 2A. I also tried connecting the GSM shield directly to the battery (which is 1C and should provide up to 10 A, but it also didn't work. The main question is, how to power up the arduino with the GSM shield using the battery and the LiPo Rider, do I need to use a capacitor for short term power storage? Is there a way to boost the current? What is the best way to do this? Here is a picture of my circuit: <Q> The key issue is that your GSM shield has its own on-board regulator (the huge 5-pin IC in the top right-hand corner of your photo) and requires VIN > 5V to provide a properly regulated 4.1V to the GSM module. <S> Additionally, the Arduino itself requires Vin > 6V for proper operation. <S> I'm not sure if your version of LiPoRider can be used to charge two cells in series. <S> If not, find a charge control board that does. <S> You could also bust a few moves to improve battery life of your Arduino board. <S> As a side note, a GSM module that was switched off takes around 30s to connect to a base station - don't forget that in your battery life estimates. <S> Safety note: If you circumvent the PWRIN jack by connecting your LiPos directly to VIN and then plug an external power supply into to the PWRIN jack, <S> you could start a LiPo fire. <S> Therefore, be safe and connect the LiPos to the Power jack. <S> (or use VIN but put epoxy in the power jack/unsolder D1 to make the jack unusable) Connecting the LiPos to <S> the PWRIN jack additionally gives you reverse battery protection due to diode D1 (see Arduino snippet below), but on the down side D1 drops a whole 1V while the GSM module transmits. <S> Arduino Power Supply Schematic <A> ability to source current immediately to your Arduino Mega board. <S> The lipo rider proabably doesn't have enough current to source even if you did <S> Here are some options: 1) Take some caps and solder them as close to the input or linear regulator on the Arduino Mega board . <S> You didn't list your board, but I believe its the same as the one listed here . <S> If your soldering on electrolytic capacitors pay attention to the polarity. <S> This will negate the inductance of the USB cable by providing local storage. <S> This may not work because the regulator might only source 500mA anyway (but this will depend on the flavor of board you have that you didn't list) <S> 2) <S> The Arduino Mega board has as MP2307 DC to DC regulator on it. <S> Because of this it has a 7-12V (depending on where you bought it from) range on the J3 input. <S> This will regulate the bus voltage on the Arduino Mega board to 5V with good efficiency. <S> It would be better to use a step up DC to DC and use the 7V input on the Arduino Mega board. <S> Stepping up voltage is a great way to get around resistance problems (power companies do it with HVDC lines) 3) <S> You might be able to get away with the setup you have now if you had a really short USB or a cable that has a bigger gauge wire. <S> The USB spec is 500mA, I believe there are nonstandard cables that will carry more current on the market. <S> 4) <S> Look up the voltage and change the feedback resistors on the board, it can go at least up to 12V. <A> Many GSM module datasheets call for a lot of capacitance to handle the transmit pulse current demands. <S> I would place 220 uF or more as close to the GSM module as possible. <S> Obviously with correct design you can drive GSM with an even smaller LiPo cell but wires are short and the power circuits are designed with the current pulses in mind. <S> Your 1A output is not enough for the GSM module without some help. <S> If the GSM module can handle 3.5 to 4.5V input you could possibly connect it directly to the battery <S> but you would need to check the data sheet and possibly have to make wiring changes. <S> EDIT: <S> A SIMCom 900 datasheet indicates direct battery connection to the module of 3.4 to 4.5V so direct battery connection to a LiPo would be the best for avoiding the converter on the LiPo Rider in the radio supply path. <S> More EDIT: <S> I am suggesting bypassing all regulators between the SIM900 module and the LiPo cell the way it was designed to be used. <S> That way the regulators cannot be blamed for not being able to pass the required current or have the required headroom.
| If your into hacking, you could change the feedback resistors on the lipo rider's U1 DC to DC regulator which is a isl97516 PWM step up regulator and step it up to 7V (or more) (and not use the usb output of the Arduino) and wire that to the J3 input on the Arduino. The main problem is if your trying to run it off of 5V USB, the cable and regulators are going to add a lot of resistance and inductance which is going to slow down the sources (Lipo Rider) An easy way to meet the above two requirements is to use two LiPo cells in series to supply 7.2V to the system's PWRIN ("9V") jack, as shown below:
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Simple way to read analog sensor I have a 5V ultrassonic sensor HC-SR04 and I want to read the value and (if > x) turns ON output 1 and turns OFF output 2 and (if < x) turns ON output 2 and turns off output 1. Outputs are 24V so I would use a transistor or a relay to turn on the outputs with the arduino. The only way I know is to use an Arduino. But dont you think it's overkill? Is there a simple way to do this? <Q> There is no need of an arduino in this case. <S> You can do it with few transistors and one opamp to compare sensor's output to "x". <S> Arduino or µC are use to do complexe operation, read bus, compute some data... <S> Yes it's possible to do it with an arduino but as you said, it's overkill. <S> If you look closer at the cost of your circuit, the design with arduino will be a way more expensive than the opamp (<1$) <A> All you really need is an analog comparator (either made from discrete parts or a dedicated IC, an op-amp or a purpose built comparator), some resistors to set the switching threshold (or a voltage reference, e.g. a zener diode, for more accuracy), a FET or some other buffer with sufficient drive current to run a relay, a DPDT relay, and a flyback diode for the relay. <S> The buffer and relay side of it are the same for both systems. <S> Both approaches have their pros and cons but ultimately it's probably best to go with whichever you feel most comfortable with. <A> The easiest way to use Arduino . <S> Ultrasonic.h library is lifesaver, friendly! <S> You can do this in a few lines on Arduino. <S> Either, You can research arduino-like products.
| The arduino method eliminates the need for the comparator and allows you to set the trigger point in software rather than using resistors making the system a lot more flexible. And also, You can use programmed pic like PIC18F877A .
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TC426 Mosfet drivers keeps blowing out i made this BLDC motor controller, here is schematic But TC426 Mosfet drivers keeps blowing out when i connect a load, after drivers fail, mosfets are still working, only drivers blows out, I checked all voltages, they are fine. I am using 33kHz pwm modulation for each phase and mosfets are IRF44N. I can't find a problem, but i was thinking that i need to put resistor between mosfet gate and driver? Could that be a problem? Thanks for everybody who answered. While you were answering, i tried to do some more testing and i have started to see some pattern, last three times first mosfet driver failed, in schematic it is left driver. I will double check all connections, but I think i will redraw PCB taking all your advice because i am tired going to electronics shop every day and spending money there for drivers! Thanks everybody UPDATEI added capacitors to each driver and change mosfets to more powerfull ones, now everything works, thanks to everybody! <Q> The TC426 FET driver ICs do not need a series gate resistor as they are designed for the capacitive loading of direct connection to a FET gate. <S> However, the TC426 data sheet also makes it very clear that unused driver inputs must be connected to VDD or GND and must not be allowed to float. <S> If this lets the inputs float, it could well be why they are destroying themselves. <S> If you are driving these inputs from a microcontroller (MCU), for example, there will be a delay on start-up while the MCU goes through reset then configures the I/ <S> O pins as outputs. <S> Before that, the I/ <S> O pins will be configured as inputs and be high impedance, leaving them floating. <S> The TC426 input leakage current is +/- <S> 1 uA and lets assume a generous 50 uA leakage from whatever driver you have on this. <S> A 4K7 pull-down would put about 0.235 V on a TC426 input, well inside its 0..0.8 V range for logic low. <A> I'd be looking at the signal rise time at the inputs of the TC426. <S> The TC426 has a mosfet input, meaning it is capacitive. <S> You have not indicated the MCU voltage, or how long that cable is, but if the rise time is slow at the TC416 end, it will leave the TC426 output shorted for a considerable duration. <S> Running the signal ground back through the power ground is also a bad idea. <S> If there is a significant rise on that ground it may drop your digital input into the grey area, again creating a short inside the TC426. <S> You should consider decoupling those control signals, perhaps with opto-couplers to keep the signal ground completely separate from the power ground. <S> Also, more capacitor storage is needed close to the TC426s as indicated in the spec sheet. <A> With your layout, loop area for the currents that will be switching the MOSFETs on and off are large. <S> This means they will have high inductance, which will limit your switching time. <S> Combined with the capacitance of the gate it can also cause ringing, and that ringing can be high enough to damage the MOSFETs or the driver. <S> This may or may not be your failure mode, but it may merit attention anyhow. <S> Minimize <S> that inductance: <S> Move the driver closer to the MOSFET, and making the trace from the driver to the gate as short as absolutely possible. <S> Next, make the trace from the driver ground to the MOSFET source also as short as possible, and run directly beside the gate trace. <S> Lastly, put a low inductance capacitor (chip capacitors are good) as close as possible between Vdd and ground for each driver. <S> This means the currents which switch the MOSFET have a small loop area as they flow through the decoupling capacitor, through the MOSFET between gate and source, and through the driver IC. <S> Since inductance is proportional to loop area this minimizes the inductance, which reduces switching time, reduces ringing, and increases efficiency. <S> Section 3 of the datasheet has a little bit on this, though it's very brief. <S> You might go through this IR paper on MOSFET basics to get some more detail on the issues mentioned there. <A> I've had MOSFET drivers destroy themselves, by oscillating. <S> Why would a circuit oscillate? <S> Huge power gain. <S> If the GND (shared by Vin and Vout, right?) <S> bounces by 1.5 volts, you certainly are in the forbidden region. <S> I'd target 0.5v bounce or less. <S> A mere 10nanoHenries and 1amp/nanosecond produces 10 volts Ground bounce. <S> 10 nanoHenries is about 1cm (0.4inches of wire). <S> I'd provide GND to those drivers like this: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> On circuits like this (for most circuits, in truth), lay out the GND first. <S> And protect that GND plan. <S> Learn how to provide "local batteries" for each IC, particularly when the IC must briefly provide amps of current for a few nanoseconds. <S> A sizable cap by each IC, with a bead or 1_Ohm resistor to the global power, protects each IC from interference with other ICs.
| It looks like you are driving the TC426 inputs from another circuit board. I would add pull-down resistors to all inputs on this board to remove this problem.
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Can hair short an electrical circuit? Taking this as an example - if a strand of hair gets in say in cpu socket of a PC or any other slot of device, what would be the consequences, rather the symptoms if the hair gets in between those pins: <Q> No, not at this voltage. <S> But it could 'open' the circuit and change the impedance of the circuit in the case of a CPU socket. <S> If hair got in between one of the pins and the contacts there could be a problem. <S> Hair resistance is probably in the MΩ range, I couldn't measure anything with my meter and it goes up to 10 MΩ. <S> The pins on a processor need to make contact, hair is not going to help that situation and could open the circuit. <S> Use 'dust off' or something to get it out. <A> First however, define your hair.... <S> Wet/with hair gel/oil/spray/dog/cat/dyed/ <S> etc then use google (or alternative) to find the conductivity of it. <S> (Alternatively, experiment yourself) <A> As others have said, breaking the contact is the biggest issue. <S> Dry hair is a good insulator, but hair is also great at absorbing moisture... <S> It can, under certain conditions absorb up to 45% <S> it's weight in moisture. <S> Add a little sweat salts and you have a conductor. <S> As such a stray hair in a high humidity environment could introduce a current path that is detrimental to the functionality of the device.
| A quick google shows that most of the time hair is an insulator, so a hair getting in will make it mechanically hard to plug it and possibly cause an open circt.
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power 12v fan using laptop charger while running laptop I have a laptop with partially damaged internal fan. Dont want to buy new one. I am thinking of swapping the internal fan to a 12V fan. The 12V fan I have can fit as it is small. Internal fan is 5V. Most solution out there uses external power supply to power 12V fan uses 5V from usb to power 12V fan uses laptop charger to power 12V fan but laptop is not connected What I want power up 12V fan using laptop charge WITH laptop connected. 12V voltage level will be achieved using a step down converter with 19V laptop charger as input. OR connect in parallel to an existing 12V connection in motherboard circuitry. 5V usb is good idea but I want to operate the fan at its full power using 12V. Point of doubt Laptop charger is already used to supply power to laptop. External load (fan) will definitely drain current but will it interfere with the normal laptop operation? Question can i use buck converter to step down voltage to power up 12V fan safely? is laptop charger capable of giving more juice even when it is already used on laptop? Do please share your knowledge. Much appreciated. Thanks! <Q> Yes, you would use a voltage regulator. <S> An extra 100 mA for a fan won't affect it much. <A> Instead of using buck convertor. <S> You can find many small boost convertor modules. <S> Buy it and fix it within your laptop near internal fan. <S> Supply it with original 5V supply of laptop for fan. <S> NOTE: <S> Replace original fan with the fan which have same number of wires. <S> Usually they have third wire for feedback to control the speed of fan and temperature of CPU <A> Depends on the power of the fan, if it is low enough. <S> Powering it from USB <S> and I can even adjust the fan speed.
| Depending on the current required by the laptop + the current required by your fan, you can tell how much the charger could support. What I did: grabbed one of the step up modules from ebay, they even come with a a potentiometer to adjust the output voltage. Regarding your Questions: I dont think there will be a problem using buck convertor with charger.
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Drilling hole through PCB ground plane I want to add mounting holes to a PCB. The PCB is not of my design so I can't change it. There appears to be a large ground plane in the locations that are good candidates for drilling the holes. Is it okay to drill the holes through the ground plane? Once drilled, the holes will be fitted with metal screws and standoffs. The circuit on the PCB does not operate at high frequencies. The PCB has two layers. <Q> If it's a two layer board then you can at least see if there are any tracks there. <S> You still have the possibility that the designer put power on one side and ground on the other though. <S> If it's a multilayer board then all bets are off, on multilayer boards tracks are very often hidden under planes. <A> Use a counterbore drill to make the hole and cut out a circle of copper around it at the same time. <S> I know it's a bit of an angry looking beast <S> but it's much more repeatable and a lot safer than spinning knives! <A> A good method I've used is take an exacto knife and start spinning it in one place. <S> I don't know how large of a hole you can drill, but it sure is awesome to drill holes into the board and cut traces. <S> It cuts traces clean. <S> If you have one layer that is not sheared off right (most drill bits leave copper behind) you risk shorting two layers. <S> Or you might be able to drill a hold and then clean it up with an exacto knife <S> , I've done a lot of work uncovering traces with exacto's knifes and surgical blades <S> work really well for scraping FR4 flat.
| You could probably cut through a multilayer board, but you have to ensure that the cut is clean and that might be hard to do.
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High Current/Voltage will pull people There is few questions about when someone touch the wire accidentally and can't get out from there. Whether is high current will pull people or high voltage? Is it only happen on AC? What is the reason behind? <Q> High voltage leads to high current, so you can't separate them and AC and DC are just as deadly, whatever Edison and Tesla used to claim. <S> Current flow causes muscles to contract - that's how they work. <A> High voltage. <S> No , DC is more dangerous , because there is no zero crossing , hence it is more difficult to de-attach. <S> human body has resistance , when voltage is applied , current pass through , only few hundred milli ampere will cause death <A> Voltage levels of 500 to 1000 volts tend to cause internal burns due to the large energy available from the source. <S> As the voltage increases current increase only when the resistance of the body remains constant (from the basic law V = IR ). <S> Under dry conditions, the resistance offered by the human body may be as high as 100,000 Ohms. <S> So if the resistance of the body is less then even the small voltage also cannot be withstood by human. <S> Electricity which flows through human bodies is the charge, but it's not electrons. <S> Instead, it's charged particles: potassium ions, sodium ions, chloride, etc. <S> Since these particles are always inside our bodies, we can't say that "electricity" is dangerous. <S> Instead, it is the FLOW of charges which causes problems. <S> In general for a normal human 1mA(mill Ampere) or (.001A) of current can be felt. <S> 5mA of current can be painful <S> 10–15mA, the person loses muscle control. <S> If it's above 50mA <S> then it would fatal. <S> 100mA above would be lethal. <S> Also if the voltage is DC then the above figures can be even lesser, because continuous direct current(DC) can harm more than alternating current(AC)
| Sometimes you get lucky and the path the current takes causes the right muscles to contract to throw you away from the source of the electricity, but usually the layout of the human body means that the reverse happens and you get more tightly attached to it, most often by closing your fingers around a wire. Voltage is like electrical pressure, voltage causes charges to flow, and when charges flow through our bodies at more than a certain high rate, bad things happen.
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Lack of documentation on batteries? Trying to understand why LR1130 and LR1131 aren't equivalents Recently I purchased some AG10 button cells for a device I have. The device specifically calls for LR1131 batteries, but someone online told me that AG10 was equivalent (and they are in a way: same voltage and same size . The batteries I got in the mail say LR1130 on them, which I trust is equivalent to AG10 (see previous charts). Basically the device doesn't have enough power. It works for a few minutes and then shuts off, and other odd behavior happens when it's not working like it should. On my quest for answers I've understood now that there's quite a bit more than just voltage, but every single search I try fails to give me good information on why these batteries differ in a way that would keep my device from working (it works with the proper LR1131 batteries, I just have no more left). I've noticed that the batteries appear to differ in their charge, which would suggest that maybe that's why I'm having problems. I have no battery tester to see if I was actually sold good batteries. At this point I know quite well that I just bought the wrong batteries, and will soon buy the correct set. But how can I trust it? How can I be sure that they'll work when other items say that they're LR1130 and LR1131? Edit: Actually called for LR1131 not L1131, even though this likely doesn't change anything. <Q> LR1131 and LR1130 are both 11mm diameter by 3.1mm thickness. <S> That's what the 1130/1131 means. <S> Note that the image below indicates some variance in dimensions, so both numbers are rounded anyway. <S> L indicates alkaline cell, so the same chemistry as in your typical AA battery, just smaller. <S> R simply means round; it's an indicator of the shape. <S> R is by far the most common for this type of battery <S> and I'm not actually aware of any other shape nomenclature, so <S> it's very possible it just gets left off by some manufacturers. <S> So, in short, the two are in fact identical. <S> I think you were sold bad batteries. <S> These use a silver oxide chemistry with a slightly (1.55V rather than 1.5V) <S> higher voltage and slightly longer life. <A> https://www.buyabattery.co.uk/gp-alkaline-battery-l1131-lr1131-1-5v-pack-of-10-batteries.html <S> From Above site I found this:The LR1131 battery is quite a popular alkaline button cell battery. <S> Each manufacturer can have a different part numbering system and so the LR1131 button battery is also known as 189 LR54 AG10 V10GA G10A L1131 L1130 LR1130. <S> Manufactured by GP Batteries, the LR1131 is an alkaline button cell battery and made popular by small electronic equipment including calculators. <S> This pack contains 10 batteries. <S> Voltage <S> 1.5V. Customers often ask why this battery is called LR1131 L1131 and also LR1130 L1130. <S> The L or LR denotes an alkaline battery. <S> The diameter of 11.6mm is where the 11 comes from. <S> The thickness of the battery makes up the end of the part number. <S> The normal thickness is 3.05mm and can be rounded up or down to 3.0 or 3.1 for part numbering reasons. <S> Hence some manufacturers round up and some down. <S> So we have LR1130 and LR1131, both of which are the same battery. <A> Two things. <S> The number part is a sizing in mm. <S> 11 <S> mm x <S> 3.0 mm and 11 mm x 3.1 mm. <S> A single 1/10 of a millimetre is so minor that it makes no difference in most holders. <S> But the designation L and LR are not the same. <S> L is a substitute for CR, which is a 3V nominal lithium cell. <S> LR is a designator for Alkaline, 1.5V nominal. <S> So the behavior you're seeing is because your device is starved for voltage. <S> It's getting half the voltage.
| Incidentally, for a slightly better battery life, you can get SR1130/SR1131s. The answer is in the dimensions of the battery itself.
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Connect push button switch to two circuits I have a push button switch that is single-pole, that I need to independently control two circuits. The button will be connected to a digital pin of two different microcontrollers. So its a low current application. Unfortunately, in this case, a two-pole button cannot be used. Is there a way of doing this? Obviously the first that springs to mind is opto-coupling, however, there is a chance that one circuit is not powered, which would mean there is a chance the opto-coupler is not powered. It cannot be guaranteed that these two circuits share the same power supplies, hence the complications. Technically the two circuits should be +5V, however the grounds aren't necessarily common. <Q> How about a DPST Relay? <S> Ideally, you'll want the COILVOLTAGE to be one of the voltages you are using for 2 circuits (VCC1, VCC2). <S> simulate this circuit – <S> Schematic created using CircuitLab <A> A simple diode isolation will work for you: simulate this circuit – Schematic created using CircuitLab <A> Hmm.. tricky set of conditions... <S> Though if you are willing to detect a modulated signal back <S> you could probably get rid of all that regulator stuff. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Though I am pretty sure there is a way to do this from the primary side, but I cant seem to simulate it. <S> Or, you can use one of <S> these simulate this circuit
| You can do it with transformers and a function generator to make a fake isolated supply to drive the led of an opto-coupler.
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Connecting led fairy lights together powered by ac mains Iv a set of fairy lights, Rated Voltage 3v, 15mA, I would like to connect four sets together in series (12v 60ma).I can get uk ac/dc 12V 1A Power.Question1 : Do i need to connect a resistor if so what one.Question2 :Is 12 volts enough.Any advice would be appreciated <Q> Four 3 Volt LEDs connected in series will work diretly from your 12 Volt supply, with no extra components. <S> One such group will draw 15 mA. <S> You can connect up to 66 such sets in parallel to the same 12 Volt 1 Amp power supply. <S> Correction: I must have been thinking of incandescent lamps when I wrote the first sentence. <S> One group will draw 15 mA.(second sentence is OK) <A> Your fairy lights are likely 2 AA battery powered string of parallel leds and a single resistor. <S> Super common. <S> You could, in theory, wire four of these in series, including the resistors, and it should work just fine. <S> You would be better off wiring them in parallel, and use a USB power supply with 3x 1n4001 diodes in series to drop 2.1V from the 5V supply. <A> If they are LEDs, as they very likely are, then no 12 Volts would not be enough. <S> But it is not likely the set of four LEDs or your supply will be exactly 12v. <S> You would have to first measure the LED string's forward voltage and find out if they are over or under your supply voltage. <S> The supply voltage must be higher (not equal) than the LED string's forward voltage. <S> Once you know the voltages you can calculate the resistor value or use a 50 cent CCR (constant current regulator) like the On-Semi <S> NSV45015WT1G.
| LEDs must always be used with a current-limiting resistor, so you want a group of three LEDs, plus a 200 Ohm resistor to operate from 12 Volts.
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Why are computers not cubes? If you look at the old NeXT Computer, two things jump out about the form factor: it's a cube, and it looks really cool. Ditto the original Connection Machine. If you look at, well, absolutely every other computer produced since then, one thing they all have in common without exception is that they are not cubes. Granted that there is no particular technical advantage to being cubical, I would have expected some manufacturers to do it just for the sake of appearance; manufacturers certainly do plenty of other things to make their products look cool. Is there any technical disadvantage of the cube shape that would account for it being universally shunned? <Q> Yes, I can think of at least two. <S> The natural construction of a computer is to put as much stuff as possible on the motherboard - a flat PCB. <S> You can then have expansion cards sideways (PC-style) plus peripheral disks inside on cables. <S> The "flattish" form factor lends itself to laptops, 1U rackmount systems, and set-top-boxes. <S> The other reason is heat dissipation. <S> For this you tend to want a high surface-area-to-volume ratio and good laminar airflow through the system. <S> A cube is the opposite of that unless you can make one face a fan. <S> Again, if you look in 1/2/4U cases a surprising amount of space is full of ducting to make this work. <S> Steve Jobs agreed with you that cubes are cool, which is why he tried to make the PowerMac G4 Cube. <S> It was generally considered to be a cool-looking failure due to heat and expandability problems. <A> Cubes are an awkward and wasteful shape to have on a desk or under a desk. <S> And with the vast majority of PCs being used with desks, space efficiency will drive it. <S> (I'm talking about desktops and not laptops, with all the irony that carries.) <S> Looking cool is great for about ten minutes until it's subsided into irritatingly inconvenient. <S> Irritation and frustration have a more profound effect on consumers than admiration and pride and manufacturers recognise that. <A> Is there any technical disadvantage of the cube shape that would account for it being universally shunned? <S> Well, motherboards are basically flat (certainly not cube like) and if volume occupancy is a measure of how something might intrude into an office or desktop environment, a rectangular shape will be able to fit a larger motherboard.
| You asked for technical disadvantages but the overwhelming reason will be the practical disadvantages of cubes. Also the amount of air it has to shift in and out (via the fan) will be smaller and this will tend to make it easier to cool.
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Power supply for A6/SIM900 GSM module I am going to buy A6 GSM module for connecting to my Raspberry PI and making a little research here. Will I be able to power up this module purely from PI or external power source is a must ? A6 data specification says it requires from 4.9 to 9 volts.Can it be powered up, for example, from GPIO port, which seem to be provide 5V? Some sources say about 5V battery, but maybe it's just a a recommendation? I am totally newbie in electronics and had no experience with PI before. I have Raspberry PI 2 model B revision 1.1. UPDATE: I found out that module specified in my first link is not actually a SIM900 module so I corrected the title. It is A6 tagged as a replacement of SIM900 and seems to be based on GSM/GPRS A6 chip . Also in the web I found several modifications tagged both as SIM900 and A6 ( 1 , 2 , 3 ), and all of them have different power consumption (1.3mA, 1.5mA, 2mA). As I have no particular preference in one GSM module over another during the purchase, opinions on any of them are appreciated. Real experience is highly appreciated. <Q> It all depends on your power supply. <S> The module seems to use about 1A (from the source you linked). <S> If you are powering the Pi from a 2A microUSB connection as recommended, and you don't have any other peripherals (e.g. a Wi-Fi dongle), then yes, it is okay to power it from the 5V pin on the GPIO header. <S> What another user posted about was the generic GPIO pins. <S> These provide ~20 mA @ 3.3V, and can be switched by software. <S> You would not use these, but the dedicated 5V port on the module. <A> Absolutely not, the maximum current draw for gpio pins is usually in the order of 10-20mA. <A> If you see reference design guide of Sim900 it clearly said Normal operating voltage is from 3.2V to 4.8V. <S> The peak working current can rise up to 2A during maximum power transmitting period, which will cause a voltage drop. <S> So the power supply must be able to provide sufficient peak current, if not, the voltage <S> may drop <S> lower than 3.1V, and the module <S> will automatically power down see <S> 2.1 Power Supply Design Now come to your question <S> SIM900 data specifications says it requires from 4.9 to 9 volts. <S> Can it be powered up, for example, from GPIO port, which seem to be provide 5v?
| You will need to use an external power supply for the module Answer is no, Raspberry Pi GPIO pin allowed 16mA max current per pin & 2.3 VO with the total current from all pins not exceeding 51mA. See this datasheet
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LED driver PT4115: how to avoid arc around IC pins? I built a LED driver with PT4115 IC. The board lighted LEDs up but sparks appeared around SW, VIN & GND pins after 10 seconds and the phenomenon didn't damage the IC. I recorded it by my mobile phone. Spark around pins Below is the schematic and the PCB board: I removed R2, D2 and R3 so the DIM pin is floating. It should help us to concerntrate to this problem. I checked every pin of the IC and didn't find any short or something wrong. I thought that something conductive around these pins and start to conduct after 10 seconds. It is very strange and confused me. Any good suggestion? <Q> Assuming the diode is still good, the layout of the board is really bad. <S> That current sense line loops around the chip, crosses and runs parallel with the drive line. <S> That's bad. <S> The inductor is also a long way from the chip and Vin also crosses v-out..... . <S> The PT4115 needs to be turned round 180 degrees with that part layout. <S> But in truth the whole layout needs redone. <S> More like this... <S> See also "Layout considerations" on page 14 of the datasheet . <S> Vital parts are: Minimize ground noise by connecting high-current ground returns, the input bypass-capacitor ground lead, and the output-filter ground lead to a single point (star ground configuration). <S> The SW pin of the device is a fast switching node, so PCB tracks should be kept as short as possible. <S> ... <S> It is particularly important to mount the coil and the input decoupling capacitor as close to the device pins as possible to minimize parasitic resistance and inductance, which will degrade efficiency. <A> It also seems to me that I see a wisp of smoke drifting over that chip when the sparking stops. <S> That indicates that the problem is the flux. <S> Use some alcohol to remove the flux, then let it get good and dry, then try it again. <S> Given that cleaning the area around the chip fixed the sparking, it seems to have been correct. <S> It would still be a good idea to follow the suggestions in the other answers to improve the layout and to check the type and function of the diode. <A> Your diode D3 is bidirectionally non-conductive or disconnected.
| The sparks are inductor's way to keep the current on if all low voltage drop routes are off. Making an answer out of my comment: To me it looks like there's flux between the pins, and there's current flowing through the flux.
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How to ensure an LED is completely off when pin is set LOW? I have a situation where, when the associated pin is set LOW, the LED still glows faintly. As this is to be used in a darkened area, it is quite noticeable. So far, I have encountered this only with one of my Arduino boards, and only when the LED colour in use is pink, as opposed to yellow or blue, which are the only other colours I have tried. Increasing the resistor from pin to LED to 1K ohm does not resolve the issue. Also, the use of a different pink LED does not affect the outcome. I realize that setting a pin "LOW" does not imply zero current, so how should I handle this problem? Though I am quite familiar with the Arduino, my knowledge of Electronics in general is quite shallow, and ask that any response take this into consideration. <Q> The LED should extinguish completely. <S> You must have a wiring error or are not driving it correctly in the software. <S> One possibility is that you have the pull-up enabled in the processor. <S> The AVR microprocessor has the ability to enable a high-value pull-up resistor (20k) if needed. <S> If you do this when you intend to turn the LED off it will glow dimly. <S> One way this can happen is to set the pin_mode() as INPUT_PULLUP. <S> It will also happen if you turn the pin to an input and write a 1 to the pin output register. <S> The usual way to connect an LED to an Arduino or other processor is to connect one end of the LED to the +5V with the other end connected to the output pin through a resistor in the range of 220-1000 ohm depending upon the LED and brightness required. <S> Some people call this resistor a current-limiting resistor, I prefer to call it a current defining resistor. <S> It is not there to protect the LED <S> it is to set the desired current. <S> You write 0 to the pin to turn the LED on and a 1 to turn it off. <A> Simple answer: <S> Pink LEDs work by using a phosphor to convert the light emitted by an ultraviolet LED to a mixture of visible wavelengths. <S> That phosphor can be excited by ultraviolet light from other sources. <S> I suspect this might be what's happening here. <A> In good old days when microcontroller was not able to provide enough current to drive LED or even BJT and that situation what we usually did was to use invert logic topology. <S> For example when Arduino Pin=1 LED will be OFF and vice versa. <S> Connect Cathod of LED to Arduino Pin and Anode of LED connect to Power supply via current limiting resistor and try logic 1 on arduino it must turn off the LED. <S> Use logic 0 to turn ON. <A> This really isnt possible. <S> The led has a resistor with it as the voltage level is generally wrong for an led, so you either have high and high on either side of the led or low and low on either side to make it "off" lets say low, low <S> , one of those is the system ground the other is a ground connected inside the part (more transistors and resistors). <S> so techincally they are not exactly at the same level, and can never be, close but never exactly. <S> Even if you were to use a relay you wouldnt be exactly zero as there are different length traces (electrically) to a common ground point. <S> now if this is charlieplexed then you could reverse the polarity if high low is on then low high should definitely be off, but I am sure someone will correct me on this. <S> If you are seeing a glow, then you probably dont have one side pulled high or low right so that you have a high high or low low situation to the same potential there is some resistance that is creating a voltage divider making your led a little bit high low just enough to glow. <S> Or maybe you are blinking it and dont realize it... <A> Since I know the 8051 <S> well, I'll talk about it since I ran into a similar problem at the beginning. <S> The port pins (at least the ones I'm using for the answer) are designed to allow input and output under certain conditions: <S> If that pin is set to logic high in code, then the pin itself will be set to high-impedance. <S> If you try to hook up the LED and resistor in series from pin to ground, then you'll get anything from a dim light to no light. <S> If you set the pin to logic low in code, then the LED does not light at all in the same configuration. <S> The easiest approach would be to wire up the LED inverted style. <S> In my circuit, its LED and resistor (LED2 and R3) in series connected between VCC (which is 5V regulated) and a port pin. <S> By doing this, the LED would either light normally (if pin is set to low in code) <S> or it would be off (if pin is set to high in code). <S> The reason the easiest approach works is because the micro has internal pull-up resistors. <S> If pin is set to logic high, then the LED and resistor would be connected from VCC back to VCC through microcontroller's internal pull-up resistor thereby giving LED zero voltage (meaning LED is off). <S> If pin is set to logic low, then LED would light because then you have a normal LED circuit which is VCC to LED with resistor connected to ground. <S> an answer to your way <S> Well it might not be the best answer, but since the micro can provide very little output (via internal pull-up) and zero output, you could run the output through an op-amp and adjust the potentiometer to create a reference voltage and if the voltage of the micro controller pin is bigger than the reference voltage then the LED lights. <S> otherwise it doesn't. <S> You may need to play with the resistor values a bit to get the desired effect. <A> Another potential reason is that you fried your pin and it always has some potential on it <S> (I have such atmega in my drawer)
| Maybe you did not set the correct pin direction (I saw it before if the pin was set as input not output).
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Why is gold preferred over silver for bond wire? Silver is a better conductor than gold and in Integrated Circuit package where exposure to air should not be a concern Gold is used. Any particular advantage of using Gold over Silver? <Q> Apart from oxidisation and tarnishing, silver migration is also a serious concern, to the point that alloys are required to mitigate this. <S> However alloying silver also means that the advantageous increase in conductivity with respect to gold is somewhat lost. <S> This might be the reason why silver has not taken over gold for wire bonds. <S> OTOH: <S> nowadays, palladium-plated copper wire bonds look very promising, with their performance very close to gold wire bonds (even better: the copper-aluminium intermetallic products formation is just 1/100th of that of the gold-aluminium system - the co-called "purple plague"), and their cost below silver wire bonds. <S> Look at <S> this report published by NASA about copper wire bonds. <S> They might even evaluate them for space flight. <A> Gold doesn't corrode. <S> Gold doesn't react significantly with any common component of air or the plastics used in ICs, even at elevated temperatures, and so it's the material of choice for bond wires. <A> There is a large body of knowledge about bonding with gold wire. <S> Much is known about process and reliability. <S> Gold wire can be stored in air and remains bondable for many months after purchase. <S> It is the most convenient type of wire for setting up a process the first time. <S> Some gold wire types are extremely malleable lending themselves to being bent into almost impossible wire shapes.
| Some packages with advanced looping are sometimes only possible with gold wire. Silver will oxidize and tarnish, reducing conductivity and potentially breaking the wire.
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Precision with opamp in transimpedance amplifier configuration with LED as photodiode I found this circuit in searching for a circuit that uses LED to measure absorbance What I want to ask is: in order to "accurate" the readings from \$ V_{out} \$, for example: assume that the current through LED is \$ -0.1 µA \$, \$ V_{R} = 10 kOhm \$, then \$ V_{out} \$ should be \$ 1 V \$ but the actual reading is \$ 1.303 \$ for example. I know this is due to many parameters such as: devices, offset current, bias current, offset voltage, blah blah ... I just wonder how can I reduce the error and obtain the result more accurately? <Q> You start with "how accurate do I need?" <S> then move on to "How do I get it?" <S> Never start the second until you know the first. <S> The example you cite shows a 0.3% error . <S> You can work for a decade without needing that sort of precision. <S> For this particular case, to get better than that, you would need to start with a very precise \$1.000M\Omega\$ resistor. <A> then Vout should be 1V but the actual reading is 1.003 for example . <S> that would be an extremely accurate reading. <S> if that's indeed what you get. <S> I just wonder how a real "accurate" readings could be obtained ? through calibration mostly. <A> There is real problem here: <S> assume that the current through LED is −0.1µA, VR=10kOhm, then Vout should be 1V but the actual reading is 1.303 for example. <S> Apparently you "assume" the current is 0.1 uA, but you measure the output voltage at 1.303. <S> Unless you measure the current as being 0.1 uA, there is no way to tell if it really is. <S> Then you have no way to tell that the output is actually wrong. <S> I'd suggest building the following circuit <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Here the critical parameter to consider is the input bias current current of the op amp. <S> Make sure that it is on the order of 1 nA or less. <S> Also fairly important is the input offset voltage. <S> Since the nominal voltage across the resistor will be about 100 mV for an assumed LED current of 0.1 uA, you'll want an offset of 1 mV or less if you want 1% accuracy. <S> With this circuit, you can establish the actual current being provided. <S> Only then can you start determining what your system error actually is. <S> Note that this circuit essentially does the same thing as the TIA you built for the first stage, and doesn't have the possible stability issues which make the feedback capacitor necessary. <S> So why use a TIA? <S> It's faster for large gains. <S> The feedback capacitor compensates for the LED capacitance which will limit the follower circuit.
| Then, you need an opamp with small bias voltages and currents.
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Does higher power wiring go higher in a control cabinet? In looking for a image to answer an Industrial electronics layout and design question, I noticed that: within the cabinets, the higher power wiring seems to be at the top of the cabinet. Simply, the thicker wires are at the top. You can see it in this , this and this example. Have I just randomly picked three examples that show this by chance, or is there a best practice for putting higher power components at the top of control panels? <Q> As you say, the cabling goes high then travels low. <S> If a cable falls out under its weight because its not secured properly, its bare ends will be going to a load and unlikely to be carrying a hazardous voltage. <S> If the routing of power was travelling upwards, the same bare ends would be coming from a supply. <S> There are several by-products of this decision, such as keeping the big on-off switches at the top and so on. <S> But safety was what I understood to be the original idea. <S> I imagine someone took that approach and copying rippled it out over the decades to become commonplace. <A> Generally when you look at circuit breakers and contactors - Line or incoming connections are on top and Load or outgoing connections are on the bottom. <S> In my experience this drives the top down design in control cabinets. <S> The general thought process is if you have a circuit breaker or contactor that is off - the lower part of the device should be dead. <S> Typically the arc flash will travel up splattering copper and carbon on the higher voltage and amperage components. <S> If the design were reversed then your lower voltage and more sensitive components would get splattered and arc over with copper and carbon. <A> Industrial cabinet layout depends entirely on how the facility installation is to be done, and for practical purposes. <S> While it is common to feed power in (and out) of the top of some enclosures, it is also common to feed power from underground. <S> Buried conduits are more expensive, but when excavating to do an installation, it is just as easy to put the electrical services below ground. <S> That also makes for a much neater installation. <S> Control panel layouts are normally done so that components that may require interaction (drives, PLC's, scaling and control modules) are placed at approximately eye level. <S> Things that are not typically accessed are placed either higher or lower. <S> Interconnect terminal strips are typically at the bottom of the enclosures, unless there is a reason to place them elsewhere.
| This is also a benefit if a contactor or circuit breaker fails and explodes. Panels are built as dictated by the installation requirements. The reason I understood for it was safety. In summary, there is no actual code, or design rule about panel layout (aside from isolating high voltage from low voltage controls typically).
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Solution for accepting 3-24v as digital high for microcontroller I'm after any suggestions for converting a digital input in the range of 3-24vdc to a 3.3v signal for a microcontroller. I need to handle a total of 32 inputs (each of which could vary anywhere between those voltages), so density is more important than cost. Someone suggested a non-inverting hex buffer like TI's CD4050 ( http://www.ti.com/lit/ds/symlink/cd4050b.pdf ) might be the solution, but I'm not entirely sure if it will handle multiple different input voltages. Thanks in advance for your suggestions. EDIT: The inputs will be connected to either: Various mechanical switches which may pull the line to GND or to a voltage anywhere in the given range of 3-24v or Sensors of various types with digital outputs. Those outputs could be low (gnd) or high with voltages that could be anywhere in the given 3-24v range, depending on the type, make and model of the sensor. Any of the inputs could be connected to any type of switch or sensor within the limits specified above. EDIT 2: Multiplexers and the like are unfortunately not possible as timings are VERY tight and interrupts of varying priority are to be used extensively which requires direct connections to the microcontroller pins. <Q> You could use dual pre-biased transistors <S> (16 required). <S> If the MCU has suitable internal pullups you would not require any other parts. <S> R1/R2 of 22K/22K or 47K/47 <S> K could be suitable. <S> The input voltage rating is -10V to +40V for those values. <S> They're about 2mm x 2mm, so they don't use up much board space. <A> Since your voltage range is so high <S> I think you could achieve a good result with a single N-Channel FET.There <S> are FET's available that have V(GS) ratings above 24 V .... <S> here's one that has a 40 V rating: 2n7002 . <S> You can even get FETs in an array, though it's a bit tougher to find them with high V(GS) ratings. <S> You can see from the datasheet that the ESD protection is beginning to conduct at the gate rating of 20 V. simulate this circuit – <S> Schematic created using CircuitLab <S> The V(GS) transfer curve is: <S> This should provide an input high of about 1.5 to 24 V.Low would need to be below 1 V input. <S> No matter what you do for 32 inputs you are going to need at least 32 resistors (if you want high impedance inputs) and probably 16-32 SMD packages, so significant board space is going to be required. <S> You could consider an HV5622 and use a serialized reading method as I proposed here . <S> The sense input could be re-arranged to use a single comparator. <S> This reduces PCB complexity, but it's still a layout job with lots of tracks. <S> The nice thing about serial acquisition is of course it/s easy to separate the acquisition board and isolate from the MCU. <A> simulate this circuit – Schematic created using CircuitLab <S> With pull-up resistor input voltage microcontroller is HIGH. <S> No matter what is voltage on 3-24V input. <S> Except on 3-24V input voltage below the threshold microcontroller input. <S> (including the forward voltage across the diode). <S> In this case, the diode passes current and the input of the microcontroller is LOW. <A> the simplest would be to use a large resistor for each of the input. <S> the clamping diode on the input pin + <S> that resistor will do a fine job. <S> alternatively, use a resistor + a 3.3v zener. <S> more complex answers also exist.
| You could also use the newer FETs such as the NTJD5121NT1G , these have Gate ESD protection diodes built in and you could use this to clamp the input with only a series resistor.
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Determining the current from an infrared LED used as photodiode Let me talk in general first: I have two LEDs of the same wavelength (940 nm) placed opposite to each other. Between them is a 140 mg/dL glucose solution (only glucose + water). One LED is used as emitter, the other as detector. Of course, the signal generated from detector will be amplified later. But the problem is: Choosing an opamp depends on how low the current from detector is. Is it in nanoampere, picoampere, or even lower? How can I know the range of the current generated by the detector? Since every opamp has an input bias current that influences the result. <Q> The amount of current that is produced by a photodiode is related to the light power it receives in watts and the "yield factor" in converting watts to amps. <S> All recognized photodiodes have a specification for this. <S> So, you begin by analysing the emitter: if an emitter is producing "so many" milli watts from (ostensibly) <S> a point source, you can calculate the power density (watts per sq metre) at any distance. <S> For an isotropic emitter, light power is emitted in all directions so, the power density at a given distance (R) is related to the surface area of a sphere (\$4\pi R^2\$) and the power originating from the "point source". <S> Given that a recognized photodiode will have a specified active surface area you can converts watts per sq metre back to incident power (watts). <S> This then converts to amps via the "yield factor". <S> So you need to know how much light power is produced by the LED and how this power is concentrated in a particular direction. <S> For instance, it is very unlikely that it will be isotropic and much more likely you can assume that at least 50% of the power produced is in a 3D angle of maybe 60 degrees. <S> The LED spec should tell you that. <S> This "concentration factor" will give you watts per square metre at the distance your receiver is placed from the emitter. <S> The active surface area of your receiver converts watts per sq metre to watts and then on to amps <S> and you are done. <S> Of course, the glucose solution will absorb some power so you also need to factor that in <S> but, I guess that's the whole point of your experiment. <S> However, in your scenario you don't appear to know any of this so, unfortunately, you cannot know how much current to expect and, as others have said in comments, your only option is <S> to try it - find an op-amp with low bias currents and try it out. <S> This also allows you to get rid of the effects of ambient light changing the reading. <A> Since your problem seems to be how to measure a small current - you can typically measure a tiny current with an inexpensive multimeter. <S> You need to know the input resistance of the multimeter. <S> For example, I have an inexpensive Uni-T M890G multimeter. <S> The lowest current range is 2mA (1uA resolution). <S> So we're scuppered, right? <S> Nope. <S> The input resistance of the meter on the 200mV voltage range is 20M, so the resolution when used as a current range is 100uV/20M = 5pA! <S> For higher currents ( <S> >10nA in this case), don't use a different voltage range, shunt the input to get a lower input resistance so that the ammeter voltage drop ('burden') is not excessive. <S> A better longer term solution is to breadboard a transimpedance amplifier but as you point out you need to have an idea of the order of magnitude before you can pick an op-amp (hint: <S> LMC6001 is pretty good <S> but you would need to take some heroic measures such as PTFE standoffs to really get full value from it). <A> How can i know the range of the current generated by detector ? <S> By measuring it . <S> That really should have been obvious. <S> LEDs aren't usually characterized as receivers, so you don't have the necessary information to calculate it. <S> Build a test setup and see what you get. <S> You could use a 940 nm photodiode instead of a LED for the receiver. <S> That should come with enough data to calculate the signal. <S> However, even then, there are things in the setup that are hard to know. <S> The photodiode will probably give you a better signal than the LED, but to really know what the level of that signal will be, you need to measure it. <A> You get one point for using the LED to get a matched wavelength detector however <S> a 'proper' photodiode will get you more gain and most likely more repeatable results. <S> In this instance I would also seriously consider using an off the shelf optical output controller to drive the LED to achieve a constant low output on the detector. <S> This would let you use the LED drive current to determine the path attenuation and make the rest of your circuitry simpler. <S> Sort of like the circuit shown here . <S> - https://www.maximintegrated.com/en/app-notes/index.mvp/id/1769 <S> Here is some interesting reading as well - http://www.mdpi.com/1424-8220/8/4/2453/pdf <S> Here is also another nice application note from Linear Technologies. <S> - http://www.analog.com/media/en/technical-documentation/technical-articles/Optimizing-Precision-Photodiode-Sensor-Circuit-Design-MS-2624.pdf <S> And here is a neat experimental setup. <S> - http://www.ece.rice.edu/~jdw/243_lab/exp6.1.html <A> The current will depend on the led in use. <S> Due to inconsistency and temperature sensitivity, it is best that this being a digital sensor - detecting the presence of light, rather than intensity of light. <S> Mitsubishi lab has a paper on this.
| I would also recommend using a light source that is pulsed because then you can cancel out drift by sampling before light is emitted and then sampling whilst light is being emitted.
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How can I control two motors using only two wires? Yesterday I bought an old RC Car from a junk sale. I assume this circuit has a built in motor driver, which is how it gets the forward/reverse of the motor working. This circuit is running of a 9v household battery, not the car battery for the motors. I want to scale this up so I can use it on an old electric mobility scooter, but here's my problem: The motor is driven by switching the polarity of the driver output wires. In order to control the 12v motors (Powered from a car battery), I need to use a relay device (I'm using an opto-coupler) and know I need the below circuit: However, if I were to complete my circuit using the yellow wires in the above diagram, the other relay device would be activated and both motors would turn. I don't want this to happen, just one motor should turn. Unfortunately, I don't know how I can achieve my expected result, if it's even possible. Perhaps some kind of logic IC is needed? I'm not sure which one. Thanks in advance. <Q> What do you mean when you say <S> However, if I were to complete my circuit using the yellow wires in the above diagram, the other relay device would be activated and both motors would turn. <S> Unless I am misunderstanding something here, it should work, provided you add the required circuitry (limiting resistors for the optocouplers). <S> If you are in the "forward" position, we have the following circuit: <S> The left optocoupler is forward biased and <S> it's LED will conduct current. <S> The second optocoupler is reverse biased, and it's LED will not conduct any current. <S> Hence the left one is turned on and the right one is turned off. <S> If we reverse the supply, we turn on the second optocoupler, and turn off the first one. <S> We can actually get rid of one resistor by doing the following: <S> Now, I would like to point out that if you are using actual optocouplers, it's a bad idea to use them to directly drive the motors. <S> Instead, use them to drive a MOSFET or other switching device, that will in turn actually switch on the motor. <A> As long as both motors only run in one direction, the basic idea is pretty simple. <S> Joren Vaes has priority. <S> However, his answer can be expanded. <S> First, let's say you're willing to drive your motors in bang-bang mode (either on or off). <S> Then simulate this circuit – Schematic created using CircuitLab will do the job, with a few reservations. <S> 1) <S> The motor will not start to turn on until the joystick is pretty far from the zero position, so there will be a large dead zone which will make precision operation difficult. <S> 2) <S> The motor will go from full off to full on over a fairly narrow deflection range. <S> This is actually a good thing, since 3) <S> If you try for partly on and succeed, the MOSFET (M1) will start to get very hot very quickly unless you provide a good heatsink for it. <S> Like, it will die. <S> And if the motor draws multiple amps of current (and a mobility scooter will indeed do that) you will need a much bigger heat sink than you think. <S> Just a warning. <S> On the other hand, if you just use the controller to drive the motors full on or full off, you won't need much heatsink. <S> Also note that I'm assuming you use a single 12 volt battery. <S> This is important. <S> If you use a higher-voltage setup, like 2 12-volt batteries in series, you will produce 24 volts, and you will be able to kill the MOSFET by applying too much gate voltage. <S> If that's a problem, start another question. <S> Finally, keep in mind that, as I stated at the beginning, this will only work if you are willing to drive your motors in one direction. <S> EDIT - <S> And you should also keep in mind that driving one motor at a time is unlikely to do anything you want to do. <S> With this setup, the vehicle will spin in place (approximately) and your only choice is which direction it spins. <A> I could show you how to do what you ask, but the fundaments of this solution are in error. <S> Scooters are driven with two motors for a reason. <S> The speed of the motors is adjusted by the user in order to steer the scooter. <S> Since one motor will ALWAYS drive at a different speed than the other without user intervention, or a "smart" control system, the scooter will continue to veer in one direction. <S> Worse since you are only allowing the controller to apply full power or none, it will do that at a violent and dangerous speed. <S> It also needs to have the ability to drive either or both motors in reverse when required. <S> e.g. One forward one back to turn on the spot, both back to get away from the obstacle you just ran up to. <S> Of course, this is all doable. <S> However, it is far more involved than the simple solution you intimate in your original question and beyond the scope of an answer here. <S> I suggest you look more closely at how your RC model works and how the scooter control system works. <S> Once you fully understand that you may find integrating one with the other is actually simpler than you think. <S> If you have difficulties understanding what you find out during that study, by all means ask more questions on here with sufficient details and I am sure someone will be able to explain it to you. <A> This could be done, but with a lot more electronics. <S> Basically, consider the two wires as providing binary outputs, in which case you have 0-0, 0-1, 1-0, 1-1, which is four possibilities (0 <S> = negative, 1 = positive). <S> So, off/stop, left only, right only, both. <S> Converting those four possibilities into two-motor controls is what the additional electronics are for.
| Your control system MUST therefore provide the ability to drive each motor proportionally, based on the joy-stick position. There is no simple way to allow reverse drive for this sort of setup.
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Can I use the wrong end of a voltage regulator? Let's say I wanted to power two circuits: a 12V circuit and a 3V circuit. Could I use a 15V power supply and a 12V voltage regulator with three pins: Input, Output, and Ground. Input to Ground would be 15V, Output to Ground would be 12V, and presumably Input to Output would be 3V. Can I use the Input to Output voltage difference to drive a circuit? EDIT: It sames the answerers agree that this would not be advisable. I'm trying to think of a good alternative. I could use a 12V power supply to power the 12V circuit and in parallel power a 3V LDO. However, that seems like it would waste a lot of power. Am I forced to using two separate power supplies? <Q> Most positive regulators can only source current, not sink it, so what you suggest would only work if the current being drawn from the 12V to ground was more than the 3.3V circuit. <S> If it was not then the 12V output would rise towards the 15V rail and your 12V circuit would experience overvoltage, and the 3V circuit would not get enough voltage. <S> The same thing would happen in reverse if you tried to use a -3V negative regulator hung off the 15V input. <S> Usually when multiple voltages are required you want the grounds to be common <S> so this seldom comes up. <A> This will not work. <S> A voltage regulator will use some manner of letting energy from the input go to the output. <S> It will measure the output constantly to verify that the output at the right level - if it is too low, it will increase the amount of energy let through, if it is too high, it will decrease it (this process is called feedback). <S> However, it cannot take energy from the output and send it to the input. <S> In fact, trying to do this will break most linear regulators, and will just be inpossible in switching types. <A> This might work if: the 15V input voltage is stable enough <S> (e.g. if you have 15±1V on the input, you'll have 3±1V between the input and the output of your 12V regulator) <S> the load on the output of your 12V regulator consumes more current than your 3V load, in all circumstances. <S> Note that if you don't have a 12V load, supplying a dummy load will result in essentially the same efficiency as you would get with a 3V linear regulator (that is, around 20%), without any actual regulation. <S> So, if you need to provide your electronics with 3V, consider using a 3V regulator.
| It would work provided the 3V load was ALWAYS greater than the 12V load, otherwise the 3V load would see overvoltage and the 12V load would not get enough.
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SMD soldering: tiny components stick to tweezer point When soldering SMD components, the 2-pinned ones (like resistors, capacitors, ...) are the hardest in my opinion. I started long time ago with the 1206 format. Then scaled down to 0804, 0603 and finally I'm working with the 0402 ones. I'm facing a new challenge right now. Once I get some flux on my tweezer points (it happens all the time), they get really sticky. This leads to numerous problems. If I'm really lucky, I grab the part first time right. This means that I grab the part by its middle, and I can move it to the pads for soldering: Misery starts when I grab it wrong first time. "Wrong" can be many things: I grab it by the side, it's not aligned well to the eventual target, ... Anyway, the moment I want to release, I open my tweezers and the part keeps sticking to either side of the tweezer points. Has anyone experienced the same problem? What was your solution? EDIT Here are some ideas that might inspire you.. Tweezer material or coating I've found a pair of tweezers online like this: These tweezers features a "Stainless Steel Body" and "Carbon Fibre Tip". Will this tip material help to prevent the stickiness? Perhaps you're using tweezers with some special coating? Super hydrophobic coating Maybe this is a bit strange, but did anyone experiment with a "super hydrophobic coating" on its tweezers? I believe this should help to avoid any liquid/flux from sticking to the tweezer points. I've discovered some sprays online ( http://www.neverwet.com/ ) but haven't bought any of them yet. Would they work on a pair of tweezers? Tweezer form To avoid stickiness, would you recommend a very fine tip? Curved or not? If curved, curved inwards or sideways? Tweezer tip roughness Some tweezers have polished points, others are more rough. What is best? Demagnetization Some tweezers claim to be "antimagnetic". Does this feature really help? <Q> Pick it <S> right the first time When you un-peel your component tape, the little buggers fall randomly on your work surface. <S> Or more likely, since Murphy's law applies, all resistors fall upside down. <S> Solution is to use double sided adhesive to stick the component tape to something like a piece of cardboard. <S> Then you can pick components from it with your tweezers directly. <S> Clean tweezers <S> Yeah, the flux sticks... <S> I use a simple sponge, it works... <S> Use vacuum <S> I haven't tried this yet, but a vacuum pick and place tool is supposed to be amazing for this. <S> Check on youtube for DIY inspiration. <A> Yes, there is some inconvenience with that in hand soldering. <S> Yes, tweezers got sticky due to flux, and you can't keep them clean all the time. <S> My solution: First put a small blob of solder on ONE SMT pad. <S> Leave the other pad clean (or even you might need to wick the solder up) <S> Then get the part on right side, maybe with a help of another sharp object. <S> Then get the part to that pad, and solder this ONE END while aligning the part. <S> Once soldered, the tweezer will unstick easily. <S> Then solder the second end of SMT part. <S> If the part gets misaligned substantially, I use a lot of flux and a hot-air pencil, so the surface tension of solder will put the part into right place automatically. <A> to clean the flux off of your tweezers. <S> I solder with an iron and wire solder, using a different technique then others have mentioned. <S> I don't know if it's better, but it works quickly for me with 0402's. <S> Melt just a little solder onto <S> one of the two pads. <S> If you use too much, it'll make step #3 impossible. <S> Use tweezers to place the part on the pads. <S> It doesn't have to be perfect. <S> Use a wood (or bamboo) toothpick to nudge the part into exact position <S> , then Use the toothpick to hold the part down. <S> Touch <S> the pre-soldered pad with the iron. <S> The component will settle onto the pad and will be held in place by the solder. <S> Solder the second pad as usual, with the iron and fresh solder. <S> Important Reflow the first pad again with flux. <S> Until you do this, that pad may have a cold joint. <S> Usually, to save time, I don't bother applying flux: I just use a little fresh solder and it does the job. <S> Since I'm not placing the part onto flux, my tweezers seldom get fouled and sticky. <A> There are two problems that I've ran into: 1) Flux on tweezers (or something else) will make the components stick Clean off your tweezers with flux cleaner 2) <A> When I was Ops Mgr for our own quick-turn SMD prototypes, some assemblers used a microscope, others with good eyes direct by hand. <S> All parts were in numbered 35mm film containers from numbered BOM and some assemblers would lay out all same parts and then use either use a syringe for paste with hot air low velocity small nozzle or very fine wire solder and tack one end. <S> It all depends on skill level. <S> They were able to make 10 prototype boards a day with 100 mixed size components. <A> NONE of that stuff will help you if you get sticky flux on it. <S> Pick it up more carefully and your problems should go away- <S> in fact you may lose fewer small parts in the transfer. <S> Depending on your eyesight, you may need a magnifier overtop of where you are picking the parts as well as a microscope over the soldering point. <S> The best approach is to tin one pad and slide the part sideways into the pad keeping it flat to the board. <S> Do NOT apply any downward pressure to the other terminal, apply gentle pressure and heat it from the side only with the tip wetted. <S> Use plenty of liquid flux and solder that is the right size. <S> Clean with IPA (eg. <S> MG Chemicals). <S> Be careful, it's quite flammable, and so may be the flux <S> (the below MSDS sheet has a typo- <S> it's actually Health hazard 1, not 2). <S> I'm afraid that 0402 parts are getting close to the limit of what can be efficiently placed by hand- <S> you may want to consider staying with 0603 parts- <S> they are bigger and sometimes marginally more expensive but for hand placement the cost difference is irrelevant.
| Magnetized tweezers steel in the tweezer gets magnetized Use a magnetizing tool (usually have a slot for demagnetization) or degauss your tweezers. As others have said, you can use a sponge (or a paper towel) with 99% IPA ("rubbing alcohol", "isopropanol", etc.)
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How to power one fan with two different power supplies individually I have two systems that run on separate power supplies and turn on independently of each other.The issue here is that each system needs to be cooled by the same cooling fan. So I need to know if there is a way that I could have either system on and the cooling fan turn on for either one. Also need to be able to have both on without overvolting the fan. I also really dont want to run another power supply just for the fan and have to turn it on independently of the systems. Thank you all for any help. The fan is rated at 12V but can be ran at about 5V. One of the power supplies is 5V and the other is 7V. Also, what would I do with the negative wire in this situation? Is it fine to have it connected to only one of the power supplies or should I have it configured a different way? <Q> Assuming both power sources are withing the fan´s working range, and that they have enough power for the fan and all other loads when running alone: simulate this circuit – Schematic created using CircuitLab <S> The maximum voltage in the fan will be the highest voltage between both power supplies. <S> The voltage will not add to each other. <S> So if you have 12V on V1 and 5V on V2, then when V1 is turned on (with V2 on or off), your fan will run at 12V, and when only V2 is turned on, your fan will run at 5V <S> If your fan is causing interference or noise, use a big enough filter capacitor. <S> I would also put a protection diode, so that any back EMF from the fan has a path to leave the circuit. <A> You'll want to do something like this. <S> Of course, you'll have to get the properly rated relay and possibly step down the power to it's coil. <A> The most general solution would be to connect a SPST relay to each system, and wire the contacts in an "OR" configuration to switch the fan on. <S> This allows all three power supplies (the two systems and the fan) to remain galvanically isolated from each other. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The relay coils need to be rated for the system power supplies, and the contacts need to be rated for the fan. <S> They could be mechanical relays or solid-state relays, as appropriate.
| The easiest here is to read your power signal before the diodes and check for noise when the fan turns on and off, and tweak the capacitor value accordingly.
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Adding a switch to a water boiler I've got a water boiler I want to add a switch to. It has a rating of 120v 60Hz and 1000w. Would I need to find a switch that is capable of withstanding 1000w or is that irrelevant to the switch? Thanks! <Q> Using 1kW at 120v means your boiler is drawing 8.33A of current: 1000W / 120V = <S> 8.33A <S> so this is what the switch much be capable of handling continuously. <S> A 10A rating should be safe & 15A or more would be best. <S> This simplistic formula only works for 'simple' things which turn electricity into heat, not things like pumps & compressors. <A> Switches will be rated either for DC specs, AC specs, or both. <S> You can use a DC switch for AC and you can use an AC switch for DC. <S> But you have to be careful, then. <S> If you can find an AC switch rated for 10A at 120VAC, then you should be good. <S> One reason why it's important to be careful and notice the rating is that AC has zeroes in it, twice per AC cycle. <S> The moment you choose to open or close the switch, with AC, is "random" relative to the cycle. <S> But for example when you open the switch there will be an "arc" drawn out as the metal contacts open up. <S> But very soon after, the AC cycle will hit one of those zeros (your load is resistive so <S> both the current and voltage will hit the zero around the same moment) and the arc is extinguished as the ionized air dissipates. <S> DC doesn't have any zeros. <S> So switch AC ratings are usually higher than the DC rating for the same switch. <S> I don't know how to uprate a DC switch for AC use <S> and it's probably partly a matter of construction details. <S> That way, you can be sure about it. <A> But for temp control a thermostat and a 10A relay. <S> Heaters are essentially linear R with low reactance at 60Hz, with some small positive temp coefficient on Resistance. <S> Other details for future reference. <S> All switches have resistance in mΩ and the temperature rise must be limited by the current \$I^2*R\$ meanwhile the voltage*current is just always a very small % of the load which is 1000W in your case. <S> Mechanical Relays or Switches, in fact, are more efficient than transistors but the tradeoff is speed and MTBF. <S> A Relay can have a current gain at full load of 500~3000 depending on size. <S> Switches must consider if the load is inductive there will be an arc when opened for a period of time > <S> L/R of load until holding current of gap is extinguished. <S> So these relays or switches are de-rated for DC inductive loads. <S> For capacitive loads a surge current when closed must be considered and thus zero-crossing switches are often considered like ZCS Triac. <S> For magnetic loads like large transformers, remanence can be stored from last shutoff and <S> smart grid power switches use that memory of phase cutoff and remanence to reclose at the same phase to minimize magnetic saturation effects of core due to switching. <S> But for heaters are just resistive loads <S> so current is linear and the simple variable, Irms for choosing a switch with Vrms rating. <S> MTBF is often demonstrated is strongly <S> (log) determined by the amount of switching current and derated from >1e6 cycles of mechanical operation to 1e4 cycles at rated current using linear AC <S> rated RMS current. <S> Since the sinewave is almost triangle wave the average V or I is exactly 1/3 the peak and <S> this correlates with a DC resistive load current at 1/3 the Ipk AC current for MTBF. <S> For MTBF reasons, reducing or "de-rating <S> " the max current capacity of a switch or relay or transistor is necessary for high quality designs. <S> ( Ref OMRON site for technical details )
| Typically all you need is 1000/120V=8A or round up to a 10A switch rated for 120Vac. So it's best you search for and select a switch that is actually rated for AC use.
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Using transistors (or relays) on a 12V 3A pump I am trying to use an Arduino Uno to separately control a 12V .5A solenoid, a 12V 1.5A solenoid, and a 12V 3A pump. I am currently using a TIP120 NPN transistor for each one controlled by the 5V output of the Arduino, but the pump transistor is getting hot after only about 30 seconds of runtime. To reduce the heat, should I put two transistors in parallel and have a limiting resistor after to limit the current flow? As of now the pump is not pumping hard enough and I would like to get more current (?) to make it run harder. I know it can run harder if it gets the required energy. My power supply says that I am only drawing 2.6 A total for all three, (2 solenoids and a pump/motor) which are hooked up in parallel to the 12V supply and it seems to just max out at that, although at 12V, it says on the power supply it can supply 6A. The solenoid transistors seem to be fine, the only issue I am having is with the pump transistor getting hot and not delivering enough current to make the pump go hard enough. Questions Does the current or the voltage drive the motor? Or both? Would only having 1.5A flow through each transistor to get 3A to the motor be too much current? Do I need to put a resistor in line with the Arduino signal voltage? (Beta (DC voltage gain) for TIP120 is 1000 on datasheet) Should I get a bigger transistor? A MOSFET? Which one? Should I get a relay instead? Which one? <Q> The reason your transistor is heating too much is because it's a darlington and has a volt or more drop at 3A. <S> The TO-220 package is good for 1 to 1.5 watt without a heatsink; you have at least double that. <S> For example, the IRLZ34 is rated 60V, with 50 milliohm Rds at 5V gate-to-source. <S> A 100 ohm resistor in series with the gate should calm the the switching and 10k from gate to source <S> makes sure it's off when not powered. <S> A 3A-rated diode around the motor (cathode to positive) is good insurance that any inductive spike on turn-off is not absorbed by the MOSFET. <S> The other questions: On a DC motor, voltage determines the motor speed, the motor draws as much current as is needed to run its load at that speed. <S> Voltage drop across the transistor (and wiring) will reduce motor speed. <S> It would be a good idea to have resistors in series with the outputs to limit the base current. <S> You probably only need 10mA or less, so like 390 ohms. <S> I wouldn't use a relay, since you don't need isolation. <S> A MOSFET is probably less expensive than a relay, and the relay might need its own drive circuit if your output is inadequate. <A> Here is the layout for the pump if I do use two transistors. <A> Here is overall setup. <S> The only part that really matters is the right hand side. <S> I just have it shown here as a single transistor, but I could use two transistors or use a relay, or really just whatever you all think would work well. ] 2
| Since the supply voltage is only 12V, I suggest using a MOSFET.
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Electricity through water can injure a person? I have maybe a simple question.I would like to know if current can travel through water and can injure a person. For example as shown below(A little funny picture!!!) a man watering his grass and his garden from rubber tube and his hand is in contact with water(which doesn't appears in the followning picture but we can make this assumption).Also in garden there is nearby a worn cable and has a high current leak.So when this man throw water by mistake on this cable, can be electrocuted?Can current travel through water and injure this man? <Q> I do remember MythBusters asking a similar question: can urinating on a live track on a railway kill a man. <S> First question is how the water is travelling: if it is anything other than a continuous, unbroken stream, there is no conductive path, and so there will be path for the electrons, and so no electric shock. <S> If we assume that the stream is continuous, then, as water can conduct (depending on your water, conductivity varies), we have a path due to water. <S> Then we need to work out how the electricity will complete the loop. <S> The man is one part, the water is another, but the electricity is trying to flow somewhere. <S> If the man (woman or non-gendered entity) is insulated from the "somewhere" you have no reason for them to get shocked. <S> On the other hand, if they're barefoot on damp ground, they could well conduct to ground. <S> So the answer is: yes, if there is an unbroken path. <S> Which is a pretty fundamental rule in electronics. <A> I also think it unlikely in this scenario that much current would flow up the water path from a cable that is already on the ground. <S> Even with a continuous stream, the shortest circuit path would be the millimeter or so from the cable to the surrounding soil. <S> The gardener might still die from a heart attack caused by the shock of seeing the nice sparks and smoke coming from his tulips though. <A> Can current travel through water and injure this man? <S> Yes. <S> That basically the risk utilities warn people to stay away from downed power line for: stray voltage. <S> Large animals like cows suffer from this risk as well. <S> It is also one of the means that lightning kills people: stray voltage caused by ground potential rise (GPR). <S> GPR here acts like a downed power line. <S> With very high potential. <A> For your example the current would have to have a path that led through your hand to your feet. <S> Say you tried to put out an electrical fire from a downed line, and you used a watering hose? <S> with enough water on you and your shoes, you could become the preferred connection to ground and be electrocuted. <S> This is certainly the case with a HV distribution line. <S> 240VAC may or may not have the potential to flow through the water, but 8KV certainly will. <S> Can you get electrocuted with water? <S> Absolutely. <S> It's under-reported, and many of the strong swimmer drownings near docks may have been caused by electric shock from faulty wiring. <S> http://www.boatus.com/seaworthy/magazine/2013/july/electric-shock-drowning-explained.asp <S> It's a freshwater problem, as saltwater has greater conductivity than a human body and will dissipate the current more readily. <S> (not 100%, but it happens less in the ocean) <S> The Mythbusters episode is a poor example. <S> Fence chargers are a 4-12KV impulse every few seconds, and are current limited. <S> I don't believe they tried the experiment in a comprehensive manner. <S> EN-60601-1 Leakage testing would be a better methodology. <A> Yes. <S> Electro Shock Drowning (ESD) is a big problem where docks are not wired properly. <S> Many people have died recently because of ESD. <S> In 2012 there were seven deaths from ESD. <S> This has been documented extensively by BoatUS. <S> http://www.boatus.com/seaworthy/assets/pdf/esd-what-you-need-to-know-presentation.pdf <S> http://www.boatus.com/seaworthy/ESD.asp <A> Yes! <S> Arguably even if the stream isn't quite continuous/unbroken, if there is enough power, you can still potentially get a jolt, because electricity can arc across small air gaps. <S> Edit: I might add "large air gaps too" eg lightning, but I think we're generally referring to man generated electricity, which rarely reaches those sorts of magnitudes.
| Many liquids conduct electricity, and if there is a circuit/continuous stream, you can get a shock.
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Is there an electronic reason to make micro USB cable without data wires? When buying electronic products that use a microUSB plug for powering, we often receive a male USB <--> male micro USB charging cable that doesn't include the 2 DATA lines (Data + / Data -). It's extremely annoying for the end-consumer, because : at the end you have a big amount of USB-µUSB cables at home that you cannot distinguish visually (no distinctive symbol on the cable) : some of them are "power only" (won't work to transfer files from phone to computer for example), some others "power+data" (will work for file transfer). it can lead to ambiguous situations like "My device X doesn't work anymore. I plugged it to my computer and it's not recognized as an USB device" . In fact it's just that the cable is not a real USB cable, but only a charging cable. Question: Is there an electronic justification to produce micro-USB to USB cables that don't include the data lines , except for saving 0.00001$ per cable for the maker because they can avoid 4 soldering points for the soldering robot? <Q> No. <S> They're not USB compliant. <A> Using such a cable when charging your devices on a public USB port or charging via an unknown computer assures that no hacking attempts may be made through the USB port while the device is charging. <S> There are devices <S> that are marketed as "USB condoms" to provide the same function. <A> The answer is NO. <S> There is no electronic reason. <S> Just cost. <S> Never underestimate cost... <S> If it only saves 5¢, when you sell a million of them, that's $50,000 you didn't need to spend. <S> Yes it can be irritating, but a lot less irritating that the former state of the art of having a box full of wall warts, all different voltages and AC and DC with different jack plug configurations on each, and never the one you needed. <S> Though, you are right, it would be nice if they changed the symbol on the power ones so you could tell by looking. <S> Though as someone else mentioned, it should not have the logo if it is not a full USB cable. <S> ADDITION: <S> There may also be a limiting liability reason. <S> The manufacturer may not want the headache of dealing with accepting the signals from whatever the cable is attached to. <S> i.e. Complaint calls of.. <S> "When I plugged your gizmo into the port on my computer, the screen went blank and smoke came out..." or "After I used your cable to recharge the gizmo, my USB port will no longer talk to my hard drive.."
| They're not even allowed to carry the USB logo. One reason to omit the data lines is to make a "charge only" cable.
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Why do cell phones sometimes interfere with amplifiers I have an old Sony amplifier (TA-F117R) next to my desk where I keep my cell phone. Sometimes I put it ontop the amplifier, because it is pervecly alighed with the surface. When I hang some call up and land the phone close to the amplifier the loud beeping noises started to burst from the speakers, something like the "modem calls" in the old ages of dial-up internet connection. I suspect the phone - transmitter negotiation is responsible for that. The radi waves transmitted from the cell phone are strong enough to induce parasitic currents in the amplifiers circuits leading to the noise. What makes me wonder is that this noise is generated only when I hang up the call and put the phone close the amp. I can send/recieve SMS, e-mails and browse the internet using the phone. I have bluetooth, NFC, GPS and Wi-Fi on all the time. I have Sony Xperia L now, but I could observe the interference with all phones I used to have (Philips Fisio, SonyEricsson K770i, Samsung Galaxy-Y,...) and amplifiers (Old stereo, PC speakers). What exactly causes such interferences and why SMS and others are transmitted without such interference. The answer here blames the GSM protocol and the fact 800 - 900 MHz signal interfers with the amplifier's structures. However it dosn't address why SMS and data transfer do not cause such interference. <Q> GSM uses Time Domain Multiplexing to share the RF frequency with other users - it only turns the RF signal on for a short time at a 217Hz rate with other users and the base station using the remaining time. <S> This short relatively high-power RF can be rectified by any semiconductor junctions to cause a slight change in the bias level of the amplifier for the duration of the RF pulse (~0.5ms). <S> This is a buzz at 217Hz that can then get amplified by the amplifier to appear at the speakers. <S> It is not necessary for the amplifier to have any response at the RF frequency which can be up to about 2GHz. <S> The cure is to provide filtering to avoid the RF signal getting to the first stage of the amplifier. <S> The filtering may be as simple as a small high-frequency capacitor across the inputs, an RF choke in series or a more complex filter. <S> Bipolar junction transistors are more sensitive to this effect than FET front ends so it will be affected by which devices are used. <S> It's not obvious why SMS and data transfers would not cause noise as well - they use the same mechanism for transferring data. <A> Even though audio amplifiers won't typically process any signals above 40kHz or so in usable fashion, the early stages of such amplifiers may not completely filter them out. <S> If high-frequency signals are strong enough, they may overdrive an amplifier stage, causing distortion. <S> Applying harmonic distortion to a signal that contains multiple frequencies may yield new spectral content at any frequency which can be formed by adding or subtracting any integer multiples of any frequencies present in the original. <S> Some wireless protocols use a variety of frequencies which are distributed in somewhat random fashion. <S> If such transmissions are picked up by an audio amplifier, the sum and difference tones will be distributed over a wide range of frequencies, and the result will simply be a slight increase in the level of background noise. <S> The problem with GSM is that its spectral content is distributed in a way that causes many of the sum and difference frequencies to coincide, thus concentrating their energy at a few discrete audio frequencies. <S> If the designers of GSM had foreseen the extent to which it would wreak havoc on audio, they could have easily adjusted the design slightly to avoid such frequency concentrations. <S> Unfortunately, by the time the problem was discovered it would have been impractical to adjust the design. <A> The chirps you hear from inactive speaker coils or amplifiers is the AM demodulation of Broadcast sync tones from cell phones exciting the coils. <S> This carrier burst sync to nearest towers is modulated quite different from the constant amplitude frequency hopping schemes used. <S> Even some car speakers nearby causes the coils to rub on the magnets from lateral forces of microwave energy on parasitic resonances tilting the coil while moving, which adds some sharper tones to pattern of "dit-da-da-dit" repeated several times.
| The usual problem with interference from GSM cell phones is caused by rectification of the strong RF signal at the semiconductor junctions at the input of the amplifier.
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How much applicable is raspberry pi at industrial point of view? As a student I have to design a project for automation of greenhouse. I can use single board controller as Arduino or I can use Raspberry Pi as well. I understand difference that one is microcontroller based and another is CPU based board. If I talk in terms of industrial application what experience will give me a better skill from perspective of industry's demand. <Q> No. <S> They do not feature the ruggedness required for an 24/7-365 industrial application. <S> You can add these features with extra boards. <S> But you will still have a weak core you've build upon. <S> This is not entirely valid for Arduino, since you can use only the software Go for an PLC instead. <S> Look at CodeSys for suitable brands for example. <S> (eg: wago) <S> I can link some greenhouse controllers I've found on Shodan.io, but that feels wrong. <S> Please secure your automation! <S> But , I understand you're a student, and a PLC might not be available or within course context. <S> In that case, I'd suggest you go for Raspberry PI. <S> Why? <S> - It has Ethernet by default. <S> - It runs linux. <S> Which means it has no probleem running an HMI . <S> - It has USB, which means you can add any external interface for IO or communcations. <S> For examples, to use Wago CANopen Fieldbus Device. <S> Or their Ethernet variant. <S> Instead of using the weak unprotected GPIO's on the processor board. <A> On the microcontroller side, the arduino is only really useful to you as an atmel dev kit. <S> Program the microcontroller using C. <S> The only times the arduino software gets used by actual professionals is when they need to hammer out some quick thing for a prototype or testing system as a quick bandaid. <S> On the raspberry pi side, it's good for learning how to use linux. <S> Learning how to use linux at a lower level, configuring the device tree, and interacting with other devices is a useful skill. <S> There's lots of other small form factor linux computers such as the wandboard which get used in industrial applications, or you may find yourself developing your own system running a linux <S> os. <S> I've seen instances of people actually using raspberry pi's for monitoring or running test systems. <S> They're quite versatile devices. <A> I think your should consider using both a Raspberry Pi and an Arduino (I really recommend the Arduino Nano). <S> The reason for both is that the R'Pi gives you a Linux base OS with lots of storage, ready Internet access, advanced peripherals such as High resolution display, Ethernet and Wi-Fi with readily available drivers. <S> You can even get a power fail/BBU daughterboard if needed. <S> The Arduino allows you to develop your low level interfaces and present them as simple Serial or I2C slave connections to the R'Pi. <S> Here you have cut down software, but easily developed in their GUI. <S> You can program the Arduino directly from the R'PI over the USB interface.... <S> read this . <S> What's not to love..... <S> and both are good experience to have.
| In engineering point of view both Arduino and Raspberry PI's are bare minimum example boards.
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What is the largest inductance value ever attained(in Henries)? For example, are there inductors with values in terms of kilo-henries or even mega-henries? 1 Henry inductors are not all that easy to find. <Q> If memory serves it was 1000 Henries running continuously at 1400 DC Amperes. <S> Energy storage must have been right around 1 GigaJoule. <S> They had a "quench resistor" designed to burn off the energy in case the magnet would develop an open circuit. <S> The resistor (designed by my boss!) could be seen from space on Google Maps until not long ago. <A> What is the largest inductance value ever attained(in Henry's)? <S> The magnetic permeability of free space is 1.2566370614…×10−6 henries per metre and the universe is pretty big. <S> "What" you might possibly say and I would say that radio waves (and light) are carried vast distances between far-away galaxies so this "inductance" is being used to convey energy we can see and detect therefore it exists. <S> Between earth and the sun (93 Million miles or 150,000,000 km) <A> I can't compete with the Universe (that's cheating anyway), but it's fairly easy to buy 25H inductors for power supply smoothing. <S> The audiophile guys do this all the time for their valve amps. <S> Inductance is a way to transfer power from one bit of metal to another. <S> So look at transformers. <S> Clearly power rating is closely related to inductance. <S> I can't find concrete evidence of anything bigger than 1200 MVA, but see this Wiki page. <S> It lists the largest inductance at <S> ~1300H <S> for a 3000MW transformer. <S> ABB do some large ones. <S> So we need another question: <S> What's the biggest single transformer, and what's the power /inductance equation? <A> Not sure about those <S> but I just measured the inductance of a Tesla Model S rear drive AC inductance motor <S> and it's somewhere between 20H and 25H across one phase. <S> I got 23H <S> but it was a really rough measurement. <S> This means it would be around 70H for the 3 phases. <S> It's not the largest admittedly, but it's interesting to me <S> so I thought I'd add it as a real world thing.
| The largest inductor ever known to me was Fermi National Accelerator Lab's Tevatron magnet. I estimate the inductance to be 188,000 henries.
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Driving piezoelectric crystals, low to high or matched impedance? When designing an interface between a fixed frequency RF generator and a piezoelectric crystal, should the aim be to match the load impedance to the source impedance (for maximum power transfer). Or should there be a low to high impedance (maximum voltage transfer). It seems intuitive to me that a matched impedance should be required, as ultimately I am aiming for maximum acoustic power. However when researching the subject, I only seem to read about electrical energy mentioned in terms of voltage, not power. Which leads me to think that I should be aiming for a low to High impedance. What should I be aiming for, maximum power transfer to the transducer or maximum voltage transfer to the transducer? <Q> It depends on how the amp is connected to the load... <S> If the cable is short enough relative to wavelength not to be a transmission line, and its inductance is small enough, then you will get maximum power into your load by using a low source impedance. <S> However if your cable is long and becomes a transmission line at your frequency, then you have to match impedances, or else power will be reflected at the load. <S> So, both answers are right, and the deciding factor is the cable length. <A> It depends how hard ou want to drive it, and over what bandwidth. <S> Generally, piezoelectric crystals are capacitive. <S> Generally, the current needed to charge that capacitance at the desired frequency of operation is higher than we'd like. <S> If we only need a single frequency of operation, and need a lot of power, say to drive an ultrasonic cleaner, then we tend to resonate the transducer with an inductor, and tune it to the operating frequency. <S> This allows the whole circuit to present a resistive impedance. <S> If we want to drive a range of frequencies, then we need to design the tuning components as a bandpass filter. <A> Matching impedances for maximum power transfer only applies for a fixed source impedance and a variable load impedance. <S> If the load impedance is fixed, then the maximum power will be transferred with a zero source impedance so that the full source voltage will be applied to the load. <S> However, if other considerations exist, such as the need to match to a long transmission line, then this does not apply.
| In your case, with a fixed load impedance, your crystal, the maximum power will be transferred with the minimum source impedance.
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Can a TRIAC fail closed? I have a circuit that uses a TRIAC that controls when 120VAC passes to the output. Never really had any problems until recently when I noticed that I had continuity through the TRIAC without any power going to the gate. I have 100 other boards that don't have this problem. So my main question is, is it normal for a TRIAC to fail closed and if so, how does this happen? Manufacturing error perhaps? My knowledge on TRIACs is pretty limited. The only reason I am so concerned is because I use this TRIAC to control power to a heating element. Without any control on the gate the temperature just skyrockets. Oh, the TRIAC I'm using is: MAC4DLM. Schematic: https://ibb.co/f4QPAk Thank you! Derek <Q> Fail-closed is the usual failure mode for power semiconductors, triacs, diodes etc. <S> An overcurrent event will tend to overheat the semiconductor, which melts everything into a highly doped and therefore conductive mess. <S> An overvoltage event will tend to punch through insulation layers, with much the same result. <S> This is why you always use a fuse in the input to a bridge rectifier supplied instrument. <A> If the condition of a failure of one component has a follow on undesirable result, then you have to prevent that undesired result. <S> In this case, you are controlling a heater, so the undesired result is excessive temperature. <S> The best way to prevent that from occurring is by using a thermal cutoff. <S> This should always be installed in heating appliances anyway. <S> I'm speaking with 35 years of experience in product safety. <S> The fuse is not the solution. <S> The thermal protector can be a one-shot, self-resetting, or trip and hold type, the selection would be based on whether or not the product is attended during use or whether it would continue to operate in the fault condition without any user intervention. <S> Other sensors should be provided to alert someone of the device's inability to maintain temperature control. <A> That really depends on the rest of the circuit and the supply. <S> Initially power conductors usually fail shorted as others have mentioned. <S> What happens after that will be one of three things... <S> The output will be driven continually with whatever internal or external consequence that might be. <S> Some other part will be over driven and will fail. <S> This could make it worse or better... <S> The original part will literally burn or melt open. <S> When designing control circuits that drive loads that can cause unfortunate consequences if let in the wrong state, it is critical to ensure that any design is "SINGLE-FAULT-SAFE". <S> Totally fault safe is not possible, but your system should at LEAST be able to stop driving for any SINGLE fault. <S> Whether that be by complex hardware/software, series gates, fuses etc. <S> depends on the design and the consequences of failure. <S> ADDITION <S> In this particular case you could add a 2nd Triac in series with the current one, perhaps on the low side. <S> That will save you if one fails, but you really need to detect that one or the other has failed and stop driving or it will only delay the problem till the second one fails. <S> Feedback is your friend.
| If one diode fails, it's likely to go short, which will then take out its opposite number, also short, shorting the input.
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How to avoid infinite on/off loop of a voltage controlled (with a photoresistor) switch? I have designed this circuit to switch on a light (R2) when it gets dark. The circuit works fine, however I have a small problem. simulate this circuit – Schematic created using CircuitLab As you can probably guess, when it gets dark the voltage at the gate drops and the mosfet is turned off, the relay is turned on and the light is turned on as well. Now as you might have guessed, the light shines on the photoresistor, makes its resistance drop, turns on the mosfet, turns off the relay and turns the light off. This is pretty much an infinite loop... Of course I could put the photoresistor in a place that cannot be reached by the lamp's light , but I was wondering if one could devise a circuit to avoid such an infinite loop, perhaps a circuit "with memory". What possible solutions could I try? I thought of a timer that prevents temporarily the mosfet to change state after it has just changed its state, say for the next 12 hours, but I'm not sure how to do such a thing or if simpler options exist. Then again I might be overthinking things and I could just put the photoresistor in a place not reached by the light of the lamp... <Q> Your state machine has two states right now: "on" and "off" and only one way to transition out of a state (the light sensor) <S> You could add another sensor (time) and introduce another state or two to delay the transitions but the machine will always cycle. <S> This might be ok if your relay was fast and quiet (so no one will notice) <S> then having it check by transitioning from "on" to "off after an hour or whatever. <S> When it's in "off", if the ambient light is bright enough then it'll stay off. <S> Otherwise it'll pop back into "on" then wait for an hour again. <S> A very simple way to make a delay is to discharge a large capacitor with a resistor in parallel and use a comparator to trigger the next state. <A> Issue with twilight is ambient + <S> lamp = oscillation. <S> It needs to get a lot brighter to turn back on. <S> I think I would have designed it like this with some added hysteresis. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Of course if the light is bright and close to the sensor even that will not work. <S> Some shielding from the light may still be warranted. <S> Note, the light should not go on and off when a cloud passes by either so some playing with values is warranted. <S> Maybe add some trim-pots till you get it the way you need it. <A> What you need is a little hysteresis in your circuit. <S> That means the on and off thresholds aren't the same. <S> This kind of circuit has been discussed many times here before. <S> See: https://electronics.stackexchange.com/a/191625/4512 <S> https://electronics.stackexchange.com/a/53681/4512 <A> This will ensure that when the light turns on, the signal at the light sensitive potential divider has to rise significantly higher than the falling (lamp on) trigger point in order to deactivate the lamp circuit. <S> You might be able to do it with an NPN BJT. <S> When the lamp activates the BJT turns on and partially reduces R1 by shunting it with a 10 kohm resistor in parallel. <A> Try shielding the photosensor with a tube that prevents light from falling onto it, and add some deadband to the sensor as so: You can experiment with the value, try something in the range of 20K to 200K. <S> This should also extend the life of the relay by making it switch less frequently and with more authority.
| You set the band between the two thresholds large enough so that the extra light into the sensor from the light that was just turned on doesn't bump it back to the off state. Between the gate of the mosfet and the light sensitive potential divider fit a comparator and ensure there is a little bit of positive feedback aka hysteresis.
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Zener-Resistor Combination at the Input of 7805 I have to drop 14.5 ~ 12V to 5V. Using 7805 will dissipate much heat. So,using below circuit will ease 7805 to drop less voltage and will generate less heat. Using 7.2V 1W zener and 100 ohm 1W resistance : What are the negative impact of adding zener-resistor combination at input stage of the circuit? Will it add any efficiency in circuit. I have read that the 7805 are already replacement for zener-resistor combination, but I have seen people using these at input stage of 7805. Thoughts? <Q> Why bother with a zener.... <S> The zener itself consumes current <S> so you are in fact increasing the power usage not reducing it. <S> That's even MORE power to dissipate and lose. <S> If you MUST take some of the load off the regulator, and your load is 150mA, just use a suitable rated resistor to drop the 12V down to just above the minimum voltage required by the regulator at that current. <S> Perhaps a 1W 30R. Or use a high current zener in place of the resistor. <S> It won't make it more efficient, and you will still have to get rid of as much heat, but it won't be worse and heat-sinking may not be required. <A> ( <S> 0.005 * 12 + 0.125 <S> * (12-5)). <S> A bit of copper on PCB area will dissipate that much heat, depending on your maximum ambient temperature. <S> You can use the TO252 version. <S> Or you could add a zener in series with the input. <S> A 2.7V 1W zener will dissipate 350mW worst case, leaving less than 600mW so a TO-220 would typically not need a heatsink at all (depending on the maximum ambient temperature). <S> In general it's better to use a switching supply where possible, however your 125mA output current is not all that much, so a linear supply might work out a bit cheaper (ignoring lifetime energy costs). <S> You could use a 34063 for example, though the inductor will be larger and more expensive than for a more modern chip. <A> The overall power creating heat in your circuit is MORE with the Zener diode added. <S> With just the series resistor and the 7805 (no zener) <S> the overall power dissipation will be the same as it would be without the resistor. <S> If you really want to be more efficient and reduce heat generation deploy a switching regulator or module design. <S> With careful component selection you should be able to achieve efficiency of around 90% or better.
| Further, the ZENER would need to be a high current device so that the series resistor does not have to be so high that it affects the regulator. Without a zener your 7805 will dissipate about 0.94W worst-case. Usage of either or both of the components will NOT make the overall thing more efficient.
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Why does a commercial PCB need this much rework? I found a PCB that has had some rework done. When I saw it, I thought that someone had actually repaired it after purchase:- It's got additional wiring (white & brown) and under the hot glue are two ceramic capacitors. The capacitors have been sleeved with insulation (difficult to see). They smell like decoupling capacitors. I can understand altering my own boards, but this is a commercial speaker from a successful company (Labtec):- The work seems to be based around the main amplifier chip (TDA2005). I've seen other post design changes to boards, but that's only been a wire here or there snaking across a PCB. Why does a commercial board get sold with this state of rework? PS. Also note the hand thickened tracks in the top left. PPS. The two circles also show hand soldered surface mount capacitors added to this side of the board which is clearly a through hole design. <Q> This looks like it's due to a combination of bad engineering, bad management, and cheap labor. <S> Circuits don't always perform as expected when first designed. <S> Even experienced designers make mistakes occasionally. <S> Usually these are caught in the first prototypes, then the board respun. <S> Not all engineers have the experience and skills to get a circuit mostly right the first time, and the discipline to test it properly before committing to production. <S> Couple <S> that with management that doesn't understand the engineering process, and a junior engineer hired into a position over his head who won't or can't stand up to management. <S> It's exactly those types of managers that hire a junior engineer for such a role in the first place. <S> "After all, it's just engineering, and all engineers are plug-replaceable, so I might as well hire the cheap one right out of school that won't give me all this crap about can't this, and test that. <S> " <S> You've got 10,000 populated boards and someone finally discovers that this thing oscillates when the volume knob is turned to 60% and you use one of the wall warts from the last shipment of 5,000 <S> you just received. <S> Your junior engineer doesn't understand what's happening, but determines that the wall warts are within spec. <S> Now you've got a big problem, so you hire a consultant to look over the design and fix it. <S> The consultant shakes his head, tells you the whole design is a mess. <S> You don't want a new design. <S> After all you already paid for one. <S> You tell the consultant you absolutely need a fix for this design. <S> He comes up with the kludge you see above. <S> You have the same factory in the far east that made the boards rework them. <S> The labor cost is cheap, so it's better than scrapping 10,000 finished boards. <S> Good engineering is expensive. <S> Bad engineering is even more expensive. <A> You also have to consider things are normally manufactured in batches. <S> Sometimes, especially for things like this, VERY LARGE batches. <S> That is a huge investment cost. <S> When an issue is discovered during production the design is usually fixed in the next run, but that can leave you with half a million dollars in inventory that is defective. <S> At that point the bean counters calculate whether the loss associated with throwing them away is worse than the loss incurred in reworking them. <S> With cheap labor, the latter can often be the better alternative. <S> As for how a rework like that can end up being needed, See Olin's answer and my comment therein. <A> Most commonly this is a result of a new revision of the design and they simply didn't want to spend the extra time or money to completely re-do the board layout and have it respun. <S> It's much cheaper to bodge extra components or wires onto existing boards. <S> Generally new revisions may come up after the previous revision has been out in the field, and reports of failures, sub-par performance, or undesired operation are given by the customers. <A> Sometimes product is returned because its defective. <S> Sometimes those defects are repaired or reworked. <S> Sometimes those repaired boards pass test and end up in shipped product. <S> BestBuy offers lots of notebook PC's that are refurbs. <A> Wire lengths for decoupling capacitors can be very critical, so soldering them straight to IC pins like that might be the easiest solution to place them very neat the active parts without having to use a double sided PCB. <S> Alternatively, various brands of the same IC can have differences in how they deal with decoupling capacitors missing or unsuitably mounted - the original design might have been tested with a more tolerant type and without the capacitors mounted that way <S> , then a change to a different IC supplier was done that needed the kludge to work reliably.
| Another possibility is that a nearly but not completely pin compatible IC type was fit to an old board design due to the originally specced type becoming unavailable or too expensive.
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Choosing LEDs/resistors to drive directly from 3v3 HC CMOS logic I'm building a home-brew CPU and I would like to add some blinkenlights, to show the logic state of various signals. (It will run at 1Hz - 100kHz.) Can I directly drive an LED from a 3v3 74HCxxx output (with a fanout of 1-3)? If so, what can I get away with regarding LED colours and current-limiting resistors? <Q> simulate this circuit – <S> Schematic created using CircuitLab <S> There are a number of benefits to this method including. <S> Ability to use larger voltage LEDs. <S> Reduced demand on the 3.3V supply. <S> Possibly allowing a much smaller regulator. <S> Increased resistor size provides more uniformity in LED current over LED forward voltage variations. <S> Gives you the option to turn off ALL the bells and whistles when in sleep mode. <S> In battery operated circuits, LEDS get dimmer as the battery begins to die giving you a free bonus feature. <A> Since you are driving the LEDs with the same output that is connected to CMOS inputs you need to stay within the limits of current. <S> You will do better to have the LED on with the logic level low. <S> From the 74HC00 datasheet : (the different columns are different max temperatures) <S> If you sink 4mA the output voltage with a 4.5V supply (think 5V minus 10%) is 0.33V maximum. <S> If you source 4mA the voltage drop from the 4.5V supply is 0.66V maximum (double), so the noise immunity will be reduced.. <S> 1.02V vs. 0.69V minimum. <S> It would be better if you could use something like 2mA rather than 4mA (still plenty for a modern indicator LED- don't buy the crappiest LEDs you can find). <S> Say you design with 2mA- if the supply is 5V and the LED drops about 2V (red) or 3V (most other colors) you would use a resistor of 1.5K or 1K. <S> Those are approximate, but should be fine. <S> CMOS inputs don't draw significant current, so the fanout is pretty much irrelevant at DC but you will get slowing of the edges as you add loading plus capacitance. <S> Edit: I see from the title that you're planning on using 3.3V- the drive capacity of 74HC logic is much less at 3V than at 5V-- and the resistors get more fiddly since there is little voltage to work with for the resistor. <S> Suggest you rethink the supply voltage. <A> As you should know, the current driving an LED is limited by Ohm's Law the total series resistance in the loop I= <S> ΔV/R where ΔV, the voltage difference is between Vdd and Vth or what you may call the dim threshold of an LED ( e.g. 10% of rated I) where ΔV/ΔI becomes somewhat linear. <S> You can apply Ohm's law to CMOS logic or any semiconductor when conducting towards rated current and determine its equivalent DC driver impedance from the datasheet specs for Vol and (Vdd-Voh) vs Iout to get the incremental ESR. <S> \$ ESR = <S> Rs = <S> \Delta <S> V/\Delta <S> I= <S> RdsOn \$ \$ <S> _{( for \ any\ MOSFET \ <S> or\ <S> CMOS\ or \ diode\ or \ <S> LED \ or <S> \ BJT(saturated) = r_{_{CE <S> }} \ \ ) <S> } <S> \$ <S> (By design standards) I use this rule of thumb: <S> 74ALVCxx is 25 Ω @3.3V <S> +/-25% or so for worst case <S> 74HCxx is 50 Ω @5V + <S> /-25% worst case and 100Ω @3.3V <S> 130 <S> Ω at 3.0V <S> I'll leave you to verify this on your own. <S> Using @Spehro 's datasheet @3.0V <S> using typ values to check my memory :) <S> for \$ESR=ΔV/ <S> ΔI= \frac{V_{OL}}{I_{OL}}= \frac {0.33V}{2.4mA} = <S> 135 <S> \ <S> Ω \$ @3.0V <S> \$ <S> \frac{0.33}{4.5mA}= 73 <S> Ω \$ @4.5V <S> \$ <S> \frac{0.33}{6.0 <S> mA}= 55 <S> Ω \$ @6.5V <S> Next I know that ESR of all diodes and LEDs ~ <S> k/Pd with k=1 <S> (+/-25% typ) for 3V LED's for example 5mm white/blue/green LEDs have an ESR ~ <S> 15 Ω because Pd rated =65mW and 1/65mW ~ 15 Ω (and verified from decades of experience this is a rule of thumb) <S> So what is your LED current using <S> 74ALVCxx @3.3V? <S> (depends on colour and LED power rating Pd) <S> if you use a 74ALVCxx @3.,3V with 25 Ω and a 5mm White Led with 15 Ω ESR and a threshold voltage of ~2.8V ( where Vf = Vth + ESR*I) <S> example simulate this circuit – Schematic created using CircuitLab Assumptions <S> Only using High Bright HB good quality LED's not old ones with Vf from 3 to 3.6V RdsOn for CMOS logic is fairly well matched for Pch and Nch ESR <S> you know how to use Ohm's Law
| With 3.3V systems it is often a good idea to supply bells and whistles like LEDs with their own power rail and drive them with open-collector outputs or discrete transistor drivers.
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How can keep output voltage constant in pv system with MPPT (FOCV method) and buck converter? I simulate pv array (P=213 W , Voc=36V , Isc=7.84) with MPPT (fractional open circuit voltage method + PI controller) by using buck converter , and it's work very good with power efficiency 95% with varies load. BUT voltage not constant at output ! now I need keep output voltage constant at 12 V to connect 12 V battrey . it can be connect battery direct like the load ? and how can keep output voltage constant Ex:12V for varies load with maximum power? Thank you :) <Q> Generally, the output voltage in a solar charging system is held constant by the battery itself. <S> For a first order model, the output of your system is connected to a voltage souce! <S> The MPPT regulates the input voltage, without even considering the output voltage, much as a traditional voltage regulator regulates the output voltage and doesn't care what the input voltage is. <S> Of course, a complete (and safe) charger design should include secondary loops or controls to disable charging when the battery is charged all the way up, and also possibly to limit the current into the battery if the input power may exceed the maximum charge rate of the battery. <S> Edit: I should clarify that, as Tony details in his response, there are many implementations of MPPT, some of which might monitor output voltage and current as part of their sophisticated slow-loop that optimizes the bias point of the cell. <S> However, the feedback for the primary DC-DC converter loop is generally at the input. <S> Unsophisticated MPPT solar chargers simply use a pre-programmed input regulation voltage that is reasonably optimal for the expected conditions. <A> A 12V battery is NOT optimal for a simple Open loop charger from 36Voc , but 80% +- would be or 24 to 28.8V. <S> Thus both input AND output (current and Voltage ) regulation is needed (4) for MPT control. <S> You need to sense output & output current and voltage in order to measure and control efficiency. <S> There are many methods of MPT PD solar sense and compare estimate MPT from k*Isc (momentary Isc test) P+O Perturb and Observe ΔESR or InC, incremental conductance FOCV fractional open-circuit voltage FSCC fractional short-circuit current for k1 Inc hunt realtime Vin*In <S> microsteps <S> combinations of above. <S> The best methods come with multiple stages of MPT and known characteristics of given PV .... <S> a) Measure momentary Isc without C <S> added to PV, estimate MPT b) <S> P+O loop with microsteps c) track ESR changes with ΔVin/ΔIin and goto a) if limit exceeded. <S> Regulate output current available to regulate battery voltage during PWM for each cycle using energy in choke. <A> You should use a buck boost controller for when there is insufficient sunlight. <S> If you are using two stage charging, output voltage should be between 14 and 15.4 V during the absorption stage and 13.4 and 13.7 during float stage. <S> If the battery is not fully charged it will pull down the output to its discharge voltage level. <S> From my Morningstar, TriStar-30 MPPT manual: <S> I use a Diode OR circuit to switch to a power supply when the solar voltage is insufficient. <S> I will likely add a 48 volt regulator on the input of the Diode OR board.
| Your output voltage should be set according to the chemistry of the battery and the charging stage. If you need a regulated 12V output you need to add a regulator to the MPPT output.
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Do caps with a higher max voltage have lower ESR? Does a capacitor of a higher max voltage, all other factors the same, have a lower ESR? If so, is the any relationship between the voltage and ESR differences? <Q> There's no reason to think so. <S> ESR is primarily a function of how the conductors (wires, plates, etc.) are constructed, while the voltage rating is primarily a function of the dielectric thickness. <S> There's no correlation between the two; they are independent parameters. <S> Leakage (parallel) resistance would be related to the dielectric, however. <A> High voltage capacitors do not generally have low ESR. <S> The importance of low ESR is that it is critical in low voltageand high current ripple filtering, and a high voltage capacitor will oftenhave no ESR specification at all (there's other 'dissipation' or 'Q' measures <S> used),both <S> because it is unusual for HV supplies to have a tight voltage tolerance,and because at few-watts power levels, a HV capacitor handleslow currents. <S> An 80 watt CPU that is powered at 1.8V can exceed ten percent terminal voltagetolerance if the ESR of its filter capacitors is over 0.004 ohms. <S> The same 80W from a 200V capacitor will exceed ten percent terminalvoltage if the ESR is over 50 ohms. <S> Capacitor construction technologies differ greatly in their character,and only electrolytic capacitors are likely to bear an ESR specification. <A> The answer is "it depends"... <S> Most cheapo general purpose capacitors (ie, not specified for very low ESR) will have a lower ESR with higher voltage rating. <S> Let's have a look at this datasheet . <S> It gives \$ <S> DF = \tan \delta \$ instead of ESR, but <S> you can get it with: \$ <S> ESR = DF*Xc <S> = <S> DF/(2 π f C) \$ (with f = 120 Hz). <S> Now, the datasheet dives DF=0.2 for a 10V cap, and 0.1 for a 50V cap, so the 50V one will have lower ESR. <S> Notice as DF (and ESR) climbs again at much higher voltages like 200V. <S> You can use this formula to calculate the ESR of your caps if it is not specified. <S> Remember it is not an accurate spec and depends widely on temperature. <S> However if you measure by physical volume of cap, a lower voltage cap will have more capacitance in the same physical size, so if you are space-constrained and need lower ESR, adding more µF is a solution too. <S> The rule kind of applies to low-ESR non-polymer caps, too. <S> When in doubt, check the datasheets... <S> It does not apply at all to ultra-low ESR polymer caps because these are designed specifically for low voltage applications (ie, PC mobo, CPU decoupling, etc). <S> Higher voltage ratings (like 25V) for this kind of product are still relatively new, therefore less optimized. <S> You can get a 5 mOhm cap for 6.3V but for 25V it would be much harder... <S> Now, remember liquid electrolytes don't work well at cold temperatures. <S> Electrolytic caps have much higher ESR at -20°C. <S> ESR of other types of caps (like solid polymer, ceramic, etc) is much less (or not at all) affected. <S> So watch your temperature range.
| Many HV capacitors which are not electrolytic would have low ESR.
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Do I need to terminate my oscope in a specific way for viewing video signals? I'm a novice and have used my Rigol scope to look at small circuits from kits and attempt to diagnose issues on my vintage computers. I'm at a point where I want to check out composite video (for learning purposes) and I keep reading that a video signal expects a 75 ohm load. This makes sense, but given my scope only has 50 ohm and 1M ohm input resistance, and also taking into account my probe, do I need to impedance match to 75 ohms, and if so how do I do it? In my case I would be looking at the composite output coming out of the device. <Q> If you are looking for accuracy and (more probably) avoiding reflections then the \$50\:\Omega\$ oscilloscope inputs will need a \$75\:\Omega\$ feed-thru terminator of some kind. <S> You can buy them or make them. <S> The T connector <S> I'm sure you are familiar with. <S> But you can see the T, the termination, and both hooked together here: <S> Also, you will really wish you had triggering that recognized parts of the signal you will look at. <S> Assuming AC coupling, hopefully you can find (under the triggering menu) something that talks about "video" and allows you to trigger on odd or even fields and even specified horizontal line numbers. <S> Note: <S> Image borrowed from the page located here <A> You can get in-line 75 Ohm BNC terminators, or use a BNC T and 75 Ohm termination at the scope input. <A> It depends if you want to terminate the signal with 50 coax into scope, then add 25 Ohm series R. <S> It is far better to use 10:1scope probe calibrated to square wave test pattern then remove clip and ground wire and only test between a terminated signal ( either by LCD VGA cable, TV SCART or on board) <S> This can be home made with spring wire or purchased and is how I prefer to measure video unless I have many signals to compare. <S> But you can go between two adjacent pins video, gnd. <S> close together using resistor wire as test pins.
| Yes, you should terminate the video signal in 75 Ohms at the scope input.
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Help needed... NPN current flow with 0v at Base I'm having a head scratcher: simulate this circuit – Schematic created using CircuitLab In my mind this should work flawlessly. The 5v/GND through R1 to the base of Q1 is controlled by a microcontroller (but for testing I am literally connecting to a 5v source or GND by hand) Applying 5v... The Relay closes, great... but upon applying GND to the base the relay remains closed, meaning there is still a current flow. Only upon disconnecting GND from the Emitter of Q1 does the relay open again. What am I doing wrong? I thought my theory was sound. <Q> Try connecting the base directly to the emitter. <S> If that doesn't open the relay, then either you got the pinout wrong or the transistor is faulty. <S> If it does open the relay, then there's a problem with your ground. <S> Update: <S> As others have pointed, you should put a flyback diode accross the relay. <S> This will prevent the relay coil from creating a spike of overvoltage on the collector when the transistor is turned off fast. <S> If the transistor is faulty, that might be what killed it. <A> So I've taken the relay out of the case (was glued in). <S> Ive managed to fry either the diode or PNP by mistakenly putting 12v and GND the wrong way round <S> so I cant experiment any more... <S> Any way here is an updated schematic with the relay module intact. <S> So can anyone explain why the relay remained closed even after putting the BASE of the NPN to GND? <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Applying 5v... <S> The Relay closes, great... <S> but upon applying GND to the base the relay remains closed <S> This is to be expected when you try to get away without a flyback catch diode across the relay. <S> The flyback voltage pulse does bad things to the transistor, probably causing avalanche breakdown in this case. <S> Go back and do it right. <S> If the relay current is low enough, then a small signal diode like a 1N4148 would be fine. <S> Otherwise, use a Schottky, especially if you eventually intend to apply pulses to the base of the transistor, or quickly attempt to turn the relay back on when the flyback current could be circulating. <A> ( and negative spike if PNP "high side" was used.) <S> If coil L was 200mH and Ic was 10mA and R <S> leakage=1M then current decay time constant is T= <S> L/R 0.2mH/1 <S> M = <S> 0.2us = <S> then Vpk=LdI/dt 0.2H <S> * 10mA <S> /0.2us = 4kV. <S> But since above Vce(max) is a zener-like breakdown it would absorb the spike but <S> the high dV/dt causes EMI, so a reverse diode clamp is connected to the collector to supply to drain the current with <S> it's much lower ESR in a longer duration defined by T=L/R(coil) where the diode Rs or ESR is similar to the transistor if both have the same power rating can be neglected as the coil resistance might be around 1k (depending on 12V relay) while transistor and diode ESR <<10 Ohms @10mA. <S> The area enclosed of the coil current to ground and the diode current loop to Vcc and gnd both become loop antenna for EMI radiation <S> so close coupling of these tracks is important to remember in future for noise issues with decoupling cap nearby on Vcc. <S> For the advanced reader, I included the above details and below simulation with the diode switched open during the trace and included an internal 100V zener for the Vce breakdown voltage. <S> Upon close examination below, one can see the std. <S> glass clamp diode exists, next to relay. <S> Although my 5V clock was 1kHz to show the rise time and decay time are the same limited by the L/R ratio of the relay coil, one would expect with <S> **no load* <S> the relay would last 1e6 to 1e7 operations max( Ref OMRON) and would fail after 1e3 to 1e4 seconds and with a reactive load fail in minutes at this rate with severe arcing on contacts. <S> buzzzzz
| Add a diode in reverse across the relay. In reality there is some leakage R in the transistor ("Early" effect) thus at turn off, the inductor current develops a large +ve spike from a "low side" NPN switch turned off.
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SPI Chip Select for different slaves 1)I need to design hardware for SPI communication with 2 different slaves. Slaves chip select pin is being pulled from high to low by the GPIO pin(port output)of the Master. But on the hardware design, I have only one port pin left. Can the same port pin be used to enable/disable the chip select pin of both the slaves? Response from both slave are obtained at different timing. Both slaves are not identical. Clock for both the slaves are derived from the system clock.The connection is as shown below (Fig1). Can this design possible for SPI communication? Also can anyone tell which design (shown below) is recommended. Fig 2 or Fig3 Is it necessary that different slave should have its own SPI unit? 2) Which factor decide if daisy chain type SPI communication possible or not? Because in datasheet of both slaves it specifies, daisy chain type SPI communication is not supported. Does slave decides about the possibility of daisy chain type SPI communication?Can someone please explain how daisy chain communication happens? Thanks <Q> The usual arrangement is to share MOSI and MISO between the slaves, and have a separate CS for each slave. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This works if, like most SPI devices, your slaves are designed to not respond to inputs or drive their MISO outputs when CS is not asserted. <S> If Fig 1 is your current set-up, changing to this arrangement would actually free up GPIO lines. <A> 1) Put a not-gate built with a transistor (output taken from collector) driven from remaining port pin. <S> Use output of this gate for one CS and use port output for the other one. <S> Note that you won't be able to de-select both chips. <S> Which means, if you un-select one then the other will be selected. <A> You could use a single I/O pin to drive two chip selects, routing the I/O pin level directly to /CS1 and through an inverter to /CS2. <S> It means that one device is always selected and it depends upon the SPI devices you have as to whether they'll like that. <S> Some SPI devices communicate commands/data when /CS is asserted then action those commands/data when /CS is negated. <S> You'll have to assess the devices you have yourself. <S> Generally, avoid it if at all possible, it complicates your software. <S> The benefit of each SPI device on its own SPI master versus a shared SPI bus depends on the application. <S> You'd favour a shared SPI bus when you only have one master or when the PCB is very dense <S> and you want to route fewer tracks. <S> Or when your SPI devices need fast transfers or transfers at precisely-timed intervals which stop you just alternating between devices.
| You'd favour an SPI master per SPI device when the bus is going to a PCB connector, causing a long bus length that also could could be shorted off-PCB and kill the whole bus.
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24v truck battery - voltage when engine is running? What's the actual voltage when engine is running for 24v truck? I know that for regular cars 12v battery voltage is above 13v, but I was wondering what's the voltage for 24v truck battery when engine is running...? <Q> I am familiar with buses (especially Mercedes Benz) <S> -they are 24V-vehicles as well. <S> Anyway, most of buses I've seen have around 28VDC (Actual value is between 27.6 ~ 27.8V) while the engine is running. <S> Shouldn't be different from trucks. <S> PS: <A> A float charging voltage for 12V lead acid battery is 13.8V <S> (2.25V to 2.3V per cell). <S> In a 24 system you have to multiply by two, which gives 27.6V. <S> However the battery can be charged also with higher voltages for fast charge, which is not implemented in automotive alternator, so it should output the float charging voltage. <A> Minimum: 24V -30%. <S> Normal: <S> 24V +15%. <S> Maximum 24V +25%. <S> Surge: 24V <S> +/- <S> 40% (max 1 second) <S> Then there are all sorts of guidelines regarding switch-over, load dump and crank duration. <S> Buy the standard if you want to know all the details.
| ISO-16949 and ISO-16750-2 suggest that electrical tests should be performed with 28V supply for 24V-products (and with 14V supply for 12V-products) since the battery voltage is 28VDC while the engine is running.
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Step down voltage regulator uncertainties The step down voltage regulator I'm currently looking at is the 5V, 5A Step-Down Voltage Regulator D24V50F5 from Pololo ( https://www.pololu.com/product/2851 ) In the description it mentions an EN (standing for enable) pin that can be used to power down or power up the regulator. This sounds like a neat functing I can certainly find use for, however, I'm not sure on the explanation on using it: The regulator is enabled by default: a 100 kΩ pull-up resistor on the board connects the ENABLE pin to reverse-protected VIN. The ENABLE pin can be driven low (under 0.6 V) to put the board into a low-power state. The quiescent current draw in this sleep mode is dominated by the current in the pull-up resistor from ENABLE to VIN and by the reverse-voltage protection circuit, which will draw between 10 µA and 20 µA per volt on VIN when ENABLE is held low. If I understand this correctly, if i connect the EN pin with GND it will disable to regulator and thus cut the VOUT? Thanks for your suggestions! <Q> I think you understand it well. <S> They are basically telling you that without an added connection (made by you, somehow), their power supply will be ON by default. <S> They then add that if you add a connection to over-ride their default behavior, which they anticipate you may want to do, then you need to be aware of the fact that to over-ride it you have to "sink" (or pull-down) enough current to achieve it. <S> It's a very modest amount of current. <S> But since they don't entirely know exactly what \$V_{IN}\$ you will be providing, they instead tell you that they have a \$100\:\textrm{k}\Omega\$ resistor there and also that there may be some added current you have to sink (beyond that resistor's requirements) for their added protection circuitry. <S> The upshot of this is that you should be prepared to sink at least \$20\:\mu\textrm{A}\$ per volt and that you should consider the input to look like this: simulate this circuit – <S> Schematic created using CircuitLab <S> (Here, you can easily compute \$R_1=\frac{1\:\textrm{V}}{20\:\mu\textrm{A}}=50\:\textrm{k}\Omega\$.) <S> If you want to disable the module, you need to make sure that the ENABLE line is pulled below \$600\:\textrm{mV}\$. Just grounding <S> it works fine (of course.) <S> That pretty much guarantees it. <S> But they know that no one is really just going to ground that input. <S> That would be pointless. <S> So they also want to make sure you understand the details enough that you can make sure that whatever you use will be able to "pull down" enough. <S> Suppose you didn't just ground it but instead hooked up a \$10\:\textrm{k}\Omega\$ resistor from ENABLE to ground. <S> And suppose \$V_{IN}=35\:\textrm{V}\$. <S> Then it's pretty clear that the voltage would be perhaps as high as \$35\:\textrm{V}\cdot\frac{10\:\textrm{k}\Omega}{10\:\textrm{k}\Omega+50\:\textrm{k}\Omega}\approx 5.8\:\textrm{V}\$, which obviously won't work. <S> In this case, you'd need to use a resistor of less than \$872\:\Omega\$, which will sink near \$690\:\mu\textrm{A}\$, to meet the specifications. <S> This is what they are trying to get across to you. <S> You might need to sink almost a milliamp in some cases and they want you to be aware of it. <S> They also want you to know enough so that if you do want to control it, you can successfully design a reasoned circuit arrangement. <A> You're right, but "cut" isn't the term I'd use - <S> the output is still connected, it's just that no voltage will be output on it. <S> It won't be isolated from your load as if a switch were opened. <A> TTL medium scale logic always used active low input to control something, like Register load, Master reset, because it didn't matter if one used an open collector switch or totem pole or now CMOS. <S> You don't even need a power source to disable the output. <S> Just a switch to ground, NPN or any logic Vol that can go below 0.6V with 100K pull-up, which includes all CMOS and TTL. <S> for sleep or wake-up functions. <S> PC's however use a low logic level from the front panel to power up the ATX supply main outputs.
| This is the most common method to control outputs in supplies for either powerup sequencing or going to sleep mode with a small Aux supply left on.
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Charging Batteries from DC Bike Generator I have salvaged a motor from a treadmill to make a stationary bike generator. Motor Specs:180 VDC -4500 Rpm 1.3 Hp 6 amp I mocked up a bike trainer and using the rollers from treadmill and was able to output 100-160 VDC unloaded, and using a 50 cm piece of Nichrome wire I was measuring 20-30 VDC @ 2-3 amps. But the current fluctuates. I am trying to work out what I would need to safely charge a deep cycle battery which I understand I will need a controller or circuit to reduce voltage and manage battery charging, but not sure what that would look like. The generator will be placed in a communal space for people to help power the campsite, so it would need to be idiot proof. Thanks Edit: 1: Battery hasn't been finalized most probably a 120 Ah 12v AGM Deepcycle Marine battery 2: Disregard what load will be coming off the battery as it should not interfere with what can be put into the battery <Q> Please confirm how you are going to rotate the motor. <S> And at what speed/rpm?The motor is of quite high power & you can get the desired voltage range by rotating it slowly also. <S> My suggestion is to use a 24v battery or 2x12v as the nominal voltage is touching 30volts. <S> I suggest to use a solar charge controller of 10A which supports both 12 & 24 volts (can be used for multiple applications if not giving desired output). <S> If you can share what is going to run in the battery, i can suggest better solution. <S> How much watt is the load or how much current the load required?Edit: <S> You can use dummy load (search online) so that over voltage can be avoided. <S> Or hook up a LED with zinner diode & some resistors so that light will glow if the voltage is exceeding charge controller's input range & you can stop pedaling. <S> Best option is to choose a charge controller which has auto shutdown feature for your ease & project safety. <A> What you need is a converter that can accept, <S> as input, the voltage range that the generator can produce and provide a constant output voltage that is appropriate to charge the battery. <S> If you are charging a 12-volt battery that would be a buck converter / charge controller that will reduce the voltage to about 15 volts. <S> If you have a battery that could be damaged by the maximum current that anyone can generate, the charge controller would need to be designed to limit the current. <S> You may want to experiment with a variety of loads to see if you have the optimum speed ratio between the pedals and the generator shaft. <S> It is unlikely that you can power fridge, pumps, lights, etc. <S> with just human power. <S> It is also impossible to determine much about the charging system without the battery voltage and Ah capacity. <A> Your current at 30V was 2A, so the current into a 12V battery will be... <S> what? <S> If, as seems likely, you can maintain that same power into thegenerator, it'll be 5A and will work just fine. <S> A 120 A-h batterycan take 12A and only achieve full charge in ten hours, <S> so you have lots ofsafe margin. <S> Connect it and go! <S> The only issue, really, is overcharging (and thatcan be corrected by monitoring the battery charge, and the output, withenergy meters). <S> If the stationary bike has gears, it's best to keep the tempo ofpedaling at about one revolution per second (a good athlete might managetwice that), for comfort and efficiency. <A> An average cyclist should be able to keep up about 150 watts continuously. <S> (125 W for an eight-hour shift is 1 kWh - the amount of work you can get out of a servant in a day.) <S> Factor in the efficiency of your drive chain, generator and charger losses and <S> you'll be lucky to get half that into the battery. <S> I mocked up a bike trainer and using the rollers from treadmill and was able to output 100-160 VDC unloaded, and using a 50 cm piece of Nichrome wire I was measuring 20-30 VDC @ 2-3 amps. <S> P = <S> VI = 30 <S> x 3 = 90 W max. <S> You were getting 60 to 90 W which agrees with my comments above. <S> I am trying to work out what I would need to safely charge a deep cycle battery which I understand I will need a controller or circuit to reduce voltage and manage battery charging, but not sure what that would look like. <S> Possibly very little is required. <S> Your generator power is limited by human effort and will be intermittent.
| Be sure to check input voltage range before buying charge controller or try to add a LED voltmeter/rpm meter for input so that you can avoid over voltage. That would be a pedal speed at which most people can perform well. However, without actual load power requirements, and usage times, that is impossible to estimate.
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Using an MCU with built-in CAN controller vs. external CAN controller I'm planning to use the Atmel ATSAMC21J18A microcontroller in an automation project which has a built-in CAN controller. The CAN bus network may contain 32 devices, each generating data packets in 1 second intervals. There will be 16 such CAN bus networks connected to a single gateway node which has an NXP LPC1768 MCU (Cortex M3). What are the factors and parameters that should be considered when deciding whether to use the built-in CAN controller or an external CAN controller? Are there any pros/cons/constraints while using the built-in CAN controller? Since the CAN gateway node will be handling traffic from all 16 networks, would it be a good idea to use an external CAN controller to avoid buffer overflows? Note that I'm anyway planning to use an external isolated CAN transceiver. <Q> What are the factors and parameters that should be considered when deciding whether to use the built-in CAN controller or an external CAN controller? <S> Mostly these are a thing of the past, since there are plenty of MCUs with built-in controllers nowadays. <S> For example if you must use some sort of specialized MCU for other reasons. <S> Are there any pros/cons/constraints while using built-in CAN controller? <S> It is superior to the external controller since you get faster register access, no overhead interface (no need for SPI, I2C etc) and no external components. <S> Cheaper BOM, less things that can break, fewer EMI concerns etc etc. <S> Since CAN gateway node will be handling traffic from all 16 networks, would it be a good idea to use an external CAN controller to avoid buffer overflows? <S> I really don't see how you can connect 16 different CAN bus networks to one node... <S> that sounds like a complete nonsense system design. <S> Are you sure that you don't mean 16 <S> CAN nodes on a single network? <S> Because there exists no sane way you can ever build that system with just one single mainstream MCU. <S> Assuming you mean 16 nodes, you should consider how capable the CAN controller is. <S> There's generally two kinds: the simple ones that only have a rx buffer FIFO of a certain amount of CAN frames, and the "mailbox" ones that have both rx FIFO and dedicated tx/rx mailbox slots for higher prioritized messages. <S> If you expect to receive a lot of different kinds of CAN frames, you should pick a CAN controller with "mailbox" functionality. <S> Your particular CAN controller seems to have the mailbox functionality so you should be good. <S> "Up to 64 dedicated Receive Buffers and up to 32 dedicated Transmit Buffers" <A> An external CAN controller is similar to the internal CAN controller, but connected over a slower bus (higher latency, slower bit rate).There is no reason to use an external when you have a capable CAN controller on-chip (SAM C21): <S> Support for CAN-FD Up to 64 dedicated receive buffers and up to 32 dedicated transmit buffers Direct Message RAM access for the CPU <S> With some chips, using the CAN peripherals reserves a chunk of SRAM. <S> But there isn't a real downside to using an on-chip variant. <S> Only maybe when you need more than two controllers it would make sense, for software design point of view, to get many external controllers. <S> But you'll have extra latency since the frames will not be memory mapped. <S> If your gateway has to connect to 16 independent CAN buses, yes. <S> Otherwise, if it's only one bus, no. <S> The LPC1768 should have plenty of memory. <S> Looking at your story you'll be having an average of 512 messages per second . <S> Which is achievable at 100 kbit/s. <S> The only catch in this is that you can't have two identical IDs within a short period, else there will be an overrun in the messagebox. <S> But the higher level protocol you're using should already have rules regarding the minimum time between two frames of the same ID. <S> The only thing to look for in on-chip controller is the hardware FIFO . <S> If you only have a three-deep hardware FIFO, as on most STM32 controllers, you can overrun the buffer with a short burst of frames with too much interrupt latency. <S> You'll have to add an additional software FIFO on top. <S> If the gateway is to connect to a PC, please don't reinvent the wheel. <S> Get something from Kvaser or IXXAT. <A> Since the MCU you're using doesn't have 16 CAN controllers on it, you'll need to do one of the following: <S> Use one MCU per bus, and then somehow network them together. <S> Connect 16 external CAN controllers to a single MCU. <S> Use an FPGA to implement 16 CAN controllers in one chip.
| The only reason why you would use an external CAN controller is if your MCU does not support it on-chip. Use a multiplexer to switch your CAN controller between the 16 CAN transceivers that are connected to the 16 CAN buses. Direct memory access is a big advantage of on-chip CAN with 32 bit controllers.
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Need a soft latching power on/off that starts OFF I have spent a few days looking for a soft latching power on/off switch. I have breadboard-tested several designs, including at least these three: Design #1 Design #2 Design #3 Testing: I tested all 3 of them with and without load. As load I used a lux meter project (Arduino Pro Mini, BH1750 lux sensor and Oled display). The 555 alternative (design #1) is on the breadboard at the right. Power cables are at far right: All of them work (with/without load), but ... all of them start frequently in ON position when connected to the power supply (with/without load, tested physically inserting/extracting the positive power cable from the breadboard). Just think what may happen when main power returns after a blackout. Question: Do you have a switch design that guarantees starting at OFF when power is apply, just like a mechanical on/off switch? Bonus: One that does not oscillate when you keep the button pressed? (I really like two separate buttons, if possible). <Q> Have you looked at this circuit. <S> If you wire it in the auto-off position, you are guaranteed to startup with no power. <S> It uses one push button to turn on(momentary), turn off(long press) and reset(momentary). <S> If you look throught their website you will find different circuits for different needs <S> but i think this should do it. <S> I have simulated it in spice and it works just fine. <S> The 10 uF cap adjusts the time you need to press the button to shutdown the system(10uF @5v supply=3seconds). <S> Pretty neat. <A> Separate ON/OFF simulate this circuit – <S> Schematic created using CircuitLab <A> Here is a simple circuit. <S> The S-1009 gives you bulletproof reset regardless of startup glitches or slow rise time- <S> it is an open drain output that parallels the OFF switch. <S> The gates can be 74HC or 4000 series depending on your operating voltage ( <S> 74HC is good to 6V absolute max). <S> With 4000 series the circuit will operate to 10V. Quiescent consumption is of the order of 300nA, which should offer some improvement over a mechanical relay. <S> R3 slows the turn on to help reduce glitching if there is some capacitance in the load. <S> Preferably use switches with grounding straps or add some resistance (a few hundred ohms to 1K) in series with each to help with ESD. <S> simulate this circuit – <S> Schematic created using CircuitLab
| You could also use a digital pin of a microcontroller to latch it to off.
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Where do I put digital ground in a dual-rail mixed signal circuit? I'm working on a project for school (a discrete dual-slope ADC) that will combine digital and analog circuitry and operate off of a ±15V dual-rail supply. The "analog ground" will be at 0V, as expected. However, I'm not sure where I should ground my digital circuits. simulate this circuit – Schematic created using CircuitLab Should I put the digital ground on the same node as the analog ground? Or should it be connected to my -15V supply with VCC regulated to -10V? Since I'm dealing with "precision" analog circuitry, I want to minimize switching noise. My Vref regulators are referenced relative to ground, and control +Vref and -Vref via feedback loops (not shown). The digital output from the analog block will be either open collector or open drain, so I can adjust the output level as needed. The first option is more intuitive, but will the second offer significantly better performance? In the digital section, I plan on using a mixture of 74HC and 74AHC chips. Will using the AHC chips at HC speeds introduce more switching noise than using all HC chips? What degree of decoupling is necessary? Will 100nF caps on each digital chip plus something like 10uF between the supplies and ground be sufficient? <Q> Of course all grounds (analog and digital) most be electrically connected, as not connecting them will be the road to endless trouble. <S> As Neil_UK suggests, use a star-grounding scheme, see this article . <S> This is to prevent the currents running through the ground connections introducing voltages in the grounds of other (unrelated) circuits. <S> Minimize switching noise generated by your circuits by keeping the loops short , that means adding supply decoupling capacitors across the supply pins of all individual ICs and especially the logic ICs. <S> Using 2 different size capacitors like 100 nF and 100 pF <S> in parallel is generally a good idea. <S> The higher the speed of a chip, the faster its transitions will be and the more harmonics it will produce. <S> If you don't need the 74AHC for performance reasons then I'd stick with a slower version like 74HC. <S> I'd put 10 uF here and there in the supply lines, 10 uF caps do not do so much at high frequencies (above 1 MHz). <S> 100 nF are better so also use those and 1 nF or 100 pF are best for high frequencies so there's usually no harm in adding (a place for) those as well. <S> Note that on a PCB you can just add a place for a capacitor and later decide not to use it. <S> Adding a capacitor in a place where there's no footprint for it is much harder. <A> In fact, you should use the precision analogue ground as the 'star point' for all other grounds. <S> This may seem weird, but what it does is ensure that all currents that flow on any of the grounds, particularly the power supply grounds, and the digital returns, do not create voltage drops that get picked up anywhere. <A> Fired up an IC some years back, with lots of onchip crosstalk between the SPI configuration registers and the various analog functions being configured. <S> We soon ended the crosstalk, by added 100Kohm resistors (polysilicon) and 10pF onchip caps between each config bit and the analog circuit being configed. <S> We used CMOS interfaces, so the 100Kohm was acceptable. <S> Summary: I suggest you create a 3rd region between the precision analog and the noisy digital. <S> That region contains buffers with private power filtering, and the signals running back and forth between analog and digital get buffered in the new 3rd region. <S> The buffered signals, running to the analog, get 100Kohm resistors and 10pF capacitors tied to the analog ground, to slow down the edges of the control signals and not inject trash. <S> For examples of this mindset, check out the Linear Technology 24-bit ADC datasheets, particularly their application/PCB section showing the buffering and isolation resistors, etc.
| The digital ground should certainly be at the same potential as your analogue ground.
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What is state in a sequential circuit? I understand that the output of a flip flop is also called as the state and the output of the flip flop before the clock pulse is applied is called as present state and after the application of clock pulse is called as next state. But what is state in a general sequential circuit, is it the output of the logic gate from which the output is taken as feedback? <Q> The state of a general sequential circuit is no different to the state of the flip flop in your specific example (a flip flop is simply a state machine with two possible states). <S> It shows the state the circuit is currently in. <S> The outputs of the various gates and circuits that feed the flip-flops determine the next state the machine will occupy, but until the transition actually takes place they are just provisional and may change at any time according to changes in inputs or other factors driving state changes. <A> A sequential circuit consists of combinationial logic elements (which are said to be stateless) and memory elements (which are stateful). <S> The state of the circuit is the combined state of all the memory elements. <S> If the memory elements are flip-flops, then the state of each flip-flop is the same as its output. <S> Note that this definition assumes the combinational logic part does not contain any loops - all loops need to contain a memory element. <A> 'State' is one stable configuration a circuit can take. <S> In a general binary sequential circuit with M memory elements, there are \$2^M\$ states. <S> It's possible that some of these could be transient given appropriate asynchronous gating, so that there could be fewer stable states. <S> A common way to handle the state of a sequential circuit is the 'state vector', a vector that simply lists the state of each memory element. <S> That could for convenience be encoded into some other alphabet, but that's only convenient. <S> For instance, a 4 bit shift register might have a state vector of 0110, but it could be easier to think of that as '6' when tracking what its state is, or where it's going to go next. <A> The state of a circuit is purely the collection of all of the current signals within that circuit. <S> It is the status of the circuit. <S> For a purely sequential circuit the full state can be calculated purely from the inputs to the circuit. <S> For a complex circuit (e.g. a one with internal memory) knowing the state requires either knowledge of some of the internal signals or knowledge of the history of the inputs.
| For a simple circuit with memory such as a flipflop the state can be fully described by a combination of the inputs and the output.
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What is a load resistor? I am not able to understand what a load resistor is and how does it relate with a load. Can anyone explain how the load resistor works and how is it different from the general resistor. <Q> A load resistor is actually a bit of an abstract term... <S> If you consider an electrical circuit is intended to act upon some other device in order to perform "work" then that external device is the "LOAD" of the circuit. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> However it is not that simple, since load must have a reference. <S> Consider the circuit below. <S> simulate this circuit Notice <S> this time there is are two resistors \$R1\$, and \$R2\$. \$R2\$ is connected across the terminals of the left circuit that includes \$R1\$. <S> As before you can say \$R2\$ is the load for this circuit. <S> However, you can also say the load across the voltage generator is \$R1 <S> + R2\$. <S> So you can see, they are strictly speaking BOTH loads depending on where you look. <S> However, generally speaking we say the thing that does the intended work of the circuit is the load. <S> Loads can be simple linear resistances or can be complex impedances as shown below. <S> simulate this circuit <S> As such load resistor can also have several meanings. <S> The load on that circuit is the effective impedance of all those components on the right. <S> \$R1\$ in this case can legitimately be called the "Load Resistor" since there is only one out there, but as you can see, that can cause confusion. <S> Just to make things more confusing, sometimes we use another meaning for load resistor. <S> simulate this circuit <S> In the circuit above, the voltage regulator circuit is intended to drive the load resistor \$R1\$. <S> However, because of the way this regulator works, it must have something attached to it to draw a minimum current in order for it to regulate properly. <S> In order to comply with that requirement, an internal "load resistor" \$R2\$ is included. <S> In Summary Load, and Load resistor in particular, is a vague concept intended to focus function on the objects in question and is always referenced back to something that is driving said load. <S> Load resistor in particular is heavily used during education to allow you to mathematically model circuits. <S> Just like I have done above. <S> In reality the load is seldom a resistor. <A> A 'load resistor' is simply a resistor that is being used as a load. <S> It may be very small, or may need to be physically big, depending on how much power it has to dissipate. <S> When you see descriptions of circuits, various resistors might be qualified by what they do, so you may see resistors called 'feedback', 'damping', 'source', 'bias', 'potential divider', 'isolating', they are all general resistors. <A> A load resistor is well... nothing but a resistor : a 2-terminal component that complies with Ohm's law and whose impedance is real (purely resistive, no reactance of admittance whatsoever). <S> What makes it a load resistor is the fact that it is placed at the output of something. <S> The key here is understanding that, actually, a load resistor (or a resistive load) makes more sense as a modelling/analysis thing than as an actual thing . <S> It's used, for example, to model the current draw you expect when you connect something to (i.e., when you "load") your circuit output. <S> Actual resistive loads are rarely called "load resistors". <S> The widest used real-world mostly resistive <S> loads are light bulbs, and nobody calls them "load resistors". <S> The generalization of this concept is the load impedance . <S> A load impedance can be complex (not purely resistive, thus with reactance of admittance), so as to model the transient and/or frequency-dependent behaviour of something you connect to your circuit. <S> Inductive loads are widely used to model motors, for example. <A> It's just a normal resistor. <S> It's called a load resistor because it is there to add a load to the circuit. <S> There is an implication that it will be dissipating a reasonable amount of power (otherwise it wouldn't be much of a load) <S> but this isn't a requirement.e.g. <S> early linear regulators required a minimum load to ensure voltage regulation, you would often add a small load resistor to ensure that this condition was always met. <A> It's not a special type of resistor. <S> A load is anything that consumes power, whether it be a resistor, a capacitor, an inductor, or any combination of these three. <S> A load resistor is supposed to be a pure resistive load that dissipates power as stated by Ohm's Law: and where P is the power dissipated, I is current, V is voltage, and R is the resistance in ohms. <S> In the case of inductive and capacitive loads, replace R with impedance Z <S> which is a combination of resistance and reactance. <A> A very short answer: <S> Each electrical/electronic circuit or device has a specific purpose: It shall deliver a signal or some kind of energy to a "consumer". <S> Such a consumer will always draw some energy (current) out of the driving circuit/device. <S> This current depends on the input resistance of the consumer. <S> Hence, as seen from the driving circuitry: This input resistance of the consumer acts as a load resistance. <A> In a circuit, an element which consumes electric power is considered as a load. <S> Resistor also consumes power. <S> So, resistor can be represented instead of load, or, every load is consuming power as the same way as resistor consumes. <S> Example of load in electric circuits are appliances and lights. <S> As the load can be any appliances, universally, it is represented as a resistive element.
| A load resistor is simply a resistor being used as a load.
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Why doesn't my multi-meter diode tester work on this LED? I own a TENMA 72-7720 Multi-meter with LCD/continuity tester: I've used this on other LEDs to test, no problem, using the diode tester setting (approximately 2:00 on the dial by the hold button) I purchased a large quantity of Purple (UV) LEDs that arrived yesterday. When I tested them using the above tester, I got nothing. I tested about half a dozen of the 600 I ordered and none worked so I decided to test them using another means. I hooked one up to my current project (8x8x8 LED cube) and the LED worked fine. Since I'm going to be making over a thousand solder connections with these things I want to make sure they're good before I use them and to test periodically during assembly. So why isn't my tester working? <Q> Those LEDs have a typical forward voltage of 3V. <S> Your multimeter's Diode/Continuity test feature reads anything above 2.8V as an open circuit. <S> from <S> this link <S> The Diode Test function on multimeters isn't really intended to test LEDs. <S> Normal diodes have a forward voltage under 1V. 'Old' Red/Yellow/Green LEDs would typically have forward voltages in the 1.2V to 2.5V range, but these new-fangled white/blue/purple/pink thingies are often 3V or more. <A> LEDs typically have a much lower maximum reverse voltage than ordinary diodes. <S> Apply a higher reverse voltage than it's spec'd for <S> and you are likely to damage the LED. <S> Your LED would appear to have a Vrrm of 5 V, a min/typ/max forward drop of 3.0/3.4/4.0 V and typical continuous operating current of 20 mA. <S> (I couldn't find an official data sheet on your part <S> but I found this info' on a dubious-ish source. <S> But they are typical figures for such an LED <S> so I would go with them.) <S> So you should build a test fixture with a 4.5 V supply in series with 100 R to put 10 mA through an LED with the typical 3.4 V drop. <S> Consider putting the output of a 7805 regulator through a 1N4001 diode then through 100 R then to your LED. <S> Don't build a tester running off a 9 V or 12 V supply or anything like that. <S> That will damage your LEDs. <A> As Trevor & Finbarr's comments indicate it's probably a voltage issue. <S> The diode mode on a meter puts a current limited voltage across the probes and looks for a current flow. <S> It then indicates the voltage drop measured. <S> It'll probably be using around 3V for this bias voltage. <S> A normal diode needs 0.6-0.7 and a red or green LED normally needs 1.5-2V. Which means the diode mode will test them just fine. <S> But your LEDs will need a forward voltage in excess of 3V <S> and so the meter can't test them. <S> To test them put a 5 V voltage source and 1k resistor in series and then measure the voltage across the LED. <S> If you see your supply voltage or 0V <S> then the LED is dead, if you see just over 3V the LED is good. <S> Supply voltage or around 0.6V could also indicate the diode is backwards depending upon the exact nature of the part. <S> Potentially a higher voltage could be used if more convenient however as indicated by TonyM <S> this may risk damage to the part if inserted the wrong way around.
| It depends upon the reverse voltage characteristics of the part which are unspecified.
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Zero crossing detection and Light dimmer I am confused by zero crossing detection of opto-triac. I know what is the zero crossing detection or switch. (ZCS) Zero crossing detects what is the point that wave is going to positive to negative or negative to positive. But, I cant explain function of below circuit with help of zero crossing. Can it control the load using PWM signal at terminal named signal ?And explain How can dim the light bulb using PWM signal. <Q> Devices such as the MOC3041 will only turn on as the supply goes through zero - this is desirable for controlling loads such as heaters or motors as it can reduce the interference they cause and avoid getting partial cycles. <S> The load will be either fully on or fully off. <S> Any PWM that is desired can only be performed at a very slow rate - this is acceptable for heaters where the response time may be many minutes. <S> The load can only be controlled for complete cycles (or half-cycles) of the AC supply - this is not what is required for controlling the intensity of a lamp by using phase control where a fraction of a cycle is desired. <S> The lamp would flicker unacceptably <A> ZCS and PWM cannot be used together. <S> You can soft start with ZCS or soft start with increasing PWM duty cycle then adjust final duty cycle to get average voltage. <S> The reason for ZCS control is that light bulbs with tungsten have a cold resistance about 10% of hot and thus draw 10x <S> the current if started at mid phase or peak voltage. <S> Dimmer triacs have a limited gain and the cold load resistance often has a higher dimmer threshold to get started, then you can reduce the level. <S> Other types with higher current pulses are better for low dimmer starting. <S> But you cannot use a ZCS with a PWM. <S> I suggest a small pulse transformer and a Triac instead if you are driving a tungsten bulb <S> and you want a smooth start. <S> A simple bridge from AC to R divider to CMOS inverter can make an effective ZCS then use a variable one shot to adjust phase (555) then drive a small pulse transformer. <S> This prevents heat in Triac from excessive gate current using a pulse for PWM at 100Hz or variable phase. <S> or start to use LEDs that are designed to be dimmable... <A> To make a light dimmer you have to use a non-zero crossing triac driver. <S> For example, the MOC3051/52 . <S> These are sometimes called "Random Phase Triac Drivers". <S> You typically also need a zero crossing detector. <S> To control the brightness you vary the time delay between each zero crossing and triggering the triac, from close to zero (for close to full power) to almost 1/2 cycle (for almost no power). <S> You cannot realistically smoothly control the brightness of an incandescent bulb with a zero crossing triac driver, though you might be able to get a few discrete steps, especially if you don't care about putting harmonics on the power line- for example you could turn the lamp on or off during 3 consecutive 1/2 cycles (then repeat) giving you 0/3, 1/3, 2/3 and 3/3. <S> And get 1/2 by turning on and off for 1/2 cycles like a diode. <S> You might still see undesirable flicker, especially at 2/3. <S> PWM as you suggest can be used to control something like a slow heater or other device that has a long time constant relative to the PWM timebase, which in turn should be many AC cycles. <S> For heaters we often use a timebase in the 2-20 second range. <S> Obviously that would cause undesirable flicker on an ordinary light bulb.
| This is called phase control , and it causes continuous EMI from the switching (which should be filtered) and sometimes the filaments 'sing' audibly with the current pulses.
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Why is modifying (by adding extra gates) clock inputs undesirable? I am taking a course in digital electronics at university. I am second year mechanical engineering student, but I felt like it was important to understand some electronics. The lecturer in one of the slides, when talking about making a binary synchronous counter out of T-flip flops, said It is undesirable to modify clock inputs I don't understand why this is the case. I have thought about it for a bit, and I can only think of the propagation delay affecting the frequency!? But surely for nMOS and pMOS it's < a few nano seconds. If you only have a few MHz clock the difference is not even worth thinking about. Can anyone explain why this is undesirable / correct me if I'm wrong. <Q> There are probably more reasons, but I will list three that come to mind: As you stated, the propagation time of the gates will affect the placement of the clock edges, and can reduce timing margin of the circuit. <S> This may not matter with slower clocks like the few MHz you mentioned, but will matter if running in a faster system. <S> Second, depending on what the other inputs to the to the gates are, the gate outputs may end up being partial clock cycles, generating edges that are far removed from the expected clock edges (or runt pulses or glitches) that can clock incorrect data into the flip-flops. <S> And third, if the circuit design is used in a production integrated circuit, the design may be modified so that it can be tested for defects using scan test. <S> In this mode, the clocks of all of the flip-flops on the chip must be tied together. <S> Clock gating can be problematic in this case. <A> By "modify", I'm pretty sure he meant some sort of gating controlled by feedback from the counter outputs. <S> If you are, for instance, turning off the clock when two outputs change, the two may not change at exactly the same time. <S> This produces what's called a skew error (the two signals are "skewed" with respect to each other), and this may produce a false edge in the clock. <S> In turn, this false edge will produce an unwanted transition in the flip-flops, and may cause the circuit to move to an unwanted state. <S> These sort of error conditions can be very hard to spot, and may depend on things like the temperature of the circuit. <S> So it's a very bad idea even to try it, unless you know what you're doing. <A> Circuits that are designed around a synchronous clock often assume that all edge triggered storage elements, like flip-flops, are clocked simultaneously. <S> The reason for this is that immediately upon being clocked the state of all of the flip-flops change simultaneously, and their new states propagate through the wiring in a couple of nanoseconds in preparation for the next clock/state change. <S> If the clock arrives at a flip-flop sufficiently late it's input data can change before it has a chance to process it. <S> So a design intended to run on a single synchronized clock will fail because some devices are receiving what is essentially a different clock signal, derived from the original clock, but a delayed version. <S> If you modify the path of the clock to some of the flip-flops using some type of additional logic you could delay it sufficiently to cause this problem. <A> By modifying clocks, he means gating them in any way to produce a new clock coming from a logic gate. <S> That's often tempting to do in an FPGA/CPLD: <S> ANDing off clocks for control or to get low power, running a counter and taking the output of a flip-flop as a new CLKIN-divided-by-n clock. <S> An example might be in making a communications controller (I2C, UART, SPI) by producing a clock near bit frequency. <S> The problem is it produces a new clock (call it CLKG) that changes just after the original CLKIN does. <S> This is shown below, with DFF1's output changing just as DFF2 is clocking in its level. <S> This can cause DFF2 to go metastable or to take the wrong level because DFF1's output voltage hadn't transitioned far enough. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This might be the situation in your comms controller circuit, passing data bytes from the fast CLKIN logic to the CLKG shift register for transmission. <S> It is possible to design gated clock circuits perfectly well, circuits that design out these problems or get round them. <S> You'll see them routinely in ASICs. <S> But the circuit is more complicated to design and maintain by others. <S> The simplest and most reliable clock scheme is a single logic clock going from a PLL or input pin to all flip-flop clocks, the synthesis software will require the least effort from you to produce a reliable circuit. <S> Use more than one clock domain sparingly and because you absolutely have to. <A> First, you seem to underestimate the importance of propagation delays. <S> A few nanosecods of clock skew is a lot , even on low frequencies. <S> Having your clock pulse <S> advanced by 5-10 <S> ns is not so bad, because you will be violating your setup times , and derating your circuit e.g. from 10 MHz to 9 MHz will counter that. <S> Having your clock delayed is much worse, as you will be violating hold times , and simply running your design at a slower clock rate won't help. <S> Of course, if the clock pulse arrives in advance for one trigger, the clock on the next trigger will appear to be delayed, so the worst case is almost guaranteed to happen. <S> Second (and more importantly), combinatorial circuits are prone to hazards or glitches , as described <S> e.g. here : Hazards will produce additional rising/falling edges on your signal before it reaches a stable value. <S> That doesn't matter too much for data signals which are sampled at the end of the clock cycle when they are stable. <S> In a clock circuit however, such glitches will result in undefined behavior because they will produce additional clock edges and violate timing requirements of all your triggers.
| The problem with doing this is that you must be very, very careful that your gate signal is clean.
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My teacher said a BJT phototransistor doesn't have a base But I find it to be absolutely incorrect. How can a BJT transistor even work without a base? Shouldn't a transistor without base be just a semiconductor (PP, NN)? Does there exist some special NASA-army-grade experimental BJT transistor without a base? So who is right, me or the teacher? <Q> I think you and your teacher are both right <S> - you are just not using the same definitions and thus your wording disagrees. <S> From your description: Shouldn't be transistor without base <S> be just a semiconductor (PP,NN)? <S> In this regard you are right: A phototransistor still has this PNP or NPN structure. <S> Your teacher might define the base differently - the terminal you put current through to achieve a collector-emitter current. <S> When you look at it this way, he is also <S> right - most phototransistors (I know of none myself that break this rule, apart from some optocouplers) have no "base" lead. <S> The only way to trigger a current flow is through photons hitting the base region, and causing the release of electrons by doing so, turning the device on. <S> Both make sense, although I would argue that you need to watch out when thinking that "no base" means "pp or nn" semiconductor. <S> Many complex semiconductor structures exist that have many regions, yet have few leads. <S> Look at Triacs or IGBTs! <A> The base connection is often not used and would thus serve mostly to pick up noise. <S> You can find optocouplers (4-pin) which have no base connection and others <S> (eg. 4N35) <S> which bring the base connection out (where it is sometimes used). <S> Similarly, individual phototransistors are often packaged in an LED-like packages with lens and 2-pin leadframe. <A> Your teacher right. <S> BJT photo-transistors rarely have base connection. <S> It's the light hitting the base region that controls the current flow from collector to emitter. <S> The more light, the larger the current flow. <S> That's what makes it a photo-transistor and not just a transistor. <A> Clearly the source of confusion here is what base means: you assume it's "base region", and your teacher seems to assume "base terminal". <S> What defines a BJT transistor is the NPN or PNP structure. <S> How many of these regions are exposed through terminals <S> is irrelevant: most transistors will have three, but phototransistors may have only two, and parasitic transistor structures may not be exposed at all, but they still behave similarly and are considered to belong to the same device class. <S> But since the structure still behaves quite similarly to a regular BJT, this fact is usually ignored and photodiode current generated by light is considered to be base current. <A> Base is already present. <S> Connection to base is present too, but it works not from electricity - from the light. <S> No matter how the energy comes into transistor, but it MUST come someway to make it work, so you can think of base connection as the wireless to the Sun ) <A> I use the base pin on a phototransistor to handle DC currents from the sun, or 60Hz currents from incandescent lighting. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> By nulling out the DC error, you can use high value collector resistors thus achieve high gain. <S> An alternative is that current source I drew in parallel with the collector resistor, to be activated by Vout lower than e.g. VDD/2
| You define the "base" to be the semiconductor in the middle of a normal BJT - the P-doped region in a NPN or the N-doped region of an PNP. Phototransistors are a bit special in that they have no base current in the strict sense, as carriers are generated in the base itself.
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What does this Op Amp Circuit in the Voltage Regulation Section Do? This is the mystery circuit (the Op-Amp part): It appears in the voltage regulation section of a stepper controller board (stepper driver and micro not pictured). At a glance, it seems like it loads the 12v rail through Q1 if the rail starts to exceed PB12V, but why would the designer even want to do this? <Q> When input power reduces to 0 volts (or drops significantly) <S> the comparator output switches to 0 volts. <S> This turns off Q2 and biases on the clamping MOSFET. <S> This discharges the 12 volt output capacitors. <S> If the input supply voltage has only partially dropped, the circuit won't fully discharge the capacitors. <S> I have no idea what the function is, but that's what it appears to do. <A> I think Andy and Enric got it with the supply cap discharge explanation. <S> Based on that I can think of 2 reasons this circuit might be necessary. <S> One, like others have mentioned is that makes the system power down quickly when the power is removed. <S> The other, is this: If the user manually turns the stepper motor, it will send voltage into the motor driver, and back-power the 12v rail through the protection diodes. <S> I've witnessed this on other designs that don't have this circuit, (displays flicker, etc). <S> The diode and mosfet arrangement means that the 12v rail effectively has a 6 ohm load whenever the power is unplugged, that would probably bleed of any back-power from the motor. <A> LM311 is a comparator, not an op-amp. <S> The designer has configured it to have some hysteresis, probably to avoid any false triggering of the circuit. <S> In that configuration, the output transistor of the LM311 will switch ON when the input voltage from the connector goes below a certain threshold (667 mV). <S> Another thing that happens when the input voltage goes down, is that diode D2 will isolate the capacitor bank of the 12V rail from the input voltage line, but they will still be charged at 12V. More on this later. <S> Going back to the LM311, we find out that switching off its output will switch off Q2, which in turn will turn on Q1. <S> When Q1 turns on it will switch a 6 Ohm load to ground, which will generate across it quite a high current from the 12V rail. <S> That current will quickly discharge all the capacitor connected to the 12V rail, provided the fuse doesn't blow in first place. <S> If an adequate fuse is selected, the initial 2 A peak current will not last enough to blow it. <S> This way Q1 and the resistor will be protected from a faulty operating condition (continuous 2A conduction) but still be able to do its work under normal conditions. <S> So, what's this circuit intended for? <S> Probably for discharging quickly the 12V rail capacitor, as Andy aka says. <S> Why such a thing is needed is a question that can only be answered by the designer of by careful analysis of the broader design. <S> Note that the designer could have opted for connecting a bleeding resistor between the 12V rail and ground, but that would have greatly increased the current consumption if the discharge time required is short. <S> What the designer does here is connecting a strong bleeding resistor when it's required only (at the expense of a greatly increased complexity). <A> There seem to be two 12V-ish rails, one the 12V in the circle, one the PB12V which makes me think "car battery" but according to D2 this is supposed to be above 12V, so the comparator senses if this is above the (capacitor buffered) 12V rail, and if not the comparator (powered by the 317 and the capacitors) stops driving Q2 which stops pulling the gate of Q1 to GND, making it conduct. <S> This will then discharge the capacitance of the 12V rail, putting the capacitors into a safe state. <S> The fuse will not blow as even if it exceeds the rated 1A, fuses usually don't blow unless their rated current is either way exceeded, or mildly exceeded for an extended amount of time, which certainly it is not for discharging the capacitors.
| It looks like it discharges the 12 volt rail capacitors when the input power is disconnected. The designer must have concluded that if the input voltage goes below that threshold then it is an actual power off, and not just a temporary droop.
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Need for diodes in opamp comparator I was reading through basic op amp comparator wherein 2 clamping diodes are used. This limits the differential voltage to 0.7V. How does it help? <Q> The circuit shown is actually a bad example of a sometimes good idea. <S> If you look up the data sheet for the LM741, you'll see that the maximum allowed differential input (the difference between the two inputs) is +/-30 <S> volts, while the maximum input voltage on any input is +/- <S> 15 volts. <S> So as long as the inputs are within the maximum range their difference is also within the maximum limit. <S> And I can't help but add that the circuit is not good practice in that it uses a general-purpose op amp as a comparator. <S> For any sort of high performance, you should always use a dedicated comparator - but then again, for any sort of high performance anything <S> you don't use a 741. <S> But I digress. <S> Using protection diodes is actually a good idea for op amps which are expected to saturate, that is, drive their outputs close to the power supply voltages. <S> Op amps depend on their input transistors being well-matched and well-behaved in order to produce good results. <S> Driving the inputs to a large voltage difference is bad in both departments, and some op amps can require milliseconds to fully recover when the voltages get back to normal. <S> So high-speed, high-precision circuits which experience large transients can benefit. <S> A good example of such a circuit is the output amplifier for a high-speed DAC. <S> Real comparators, in contrast, are generally designed with large input differences in mind, so input diodes are not usually needed. <S> The exception to this rule is a circuit which may produce input voltage higher than the power supply voltages, such as any circuit which monitors AC power lines. <A> Some op-amps, the bipolar type with super-Beta transistors in the front end, already have those diodes internally. <S> As such they will conduct significant current if you apply a differential input voltage greater than a diode drop or two. <S> This is another reason why op-amps in general don't make good comparators (though in some cases they are justified- such as when you just need a low performance comparator and an appropriate op-amp is there for free as part of a quad or dual). <S> The consequence of breaking down the B-E junction is that the beta of the transistors can be permanently reduced, and thus the bias current will increase. <S> An example of such an op-amp is the OP97. <S> The datasheet says: The input pins of the OP97 are protected against large differential voltage by back-to-back diodes. <S> Current-limiting resistors are not used to maintain low noise performance. <S> If differential voltages above ±1 V are expected at the inputs, series resistors must be used to limit the current flow to a maximum of 10 mA. Common-mode voltages at the inputs are not restricted and may vary over the full range of the supply voltages used. <S> Some other op-amps have internal resistors and they may not provide much of a warning that significant current will flow for large differential input voltages. <S> In any case, in the particular example shown the diodes may shorten the response time because they prevent some internal nodes of the op-amp from saturating completely. <S> The response time of a unity-gain compensated op-amp is going to be horrible anyway. <A> By clamping, you minimize the stored charge at that node, what with Q = C*V. <S> That minimal charge is important for fast decisions. <S> In past decades, with custom-built successive-approximation ADCs, using an input resistor and feedback current-mode DAC (custom built), the junction of resistor and DAC showed the frantic binary-search waveform. <S> Clamping that waveform as close to zero as possible was great trick for decision speedup. <S> I even investigated active-summing-node clamps, using matched arrays of transistors, to hold the summing-node within 10millivolts (not 600 milliVolts) of zero; unfortunately I was not yet aware of the Cmiller of my nulling circuit and put the project aside. <S> In an ADC, you know the timing needed, and a special active-clamp is usable; you can get (almost) <S> the same benefit, with common-base clamps, in non-synchronous circuits, thusly: simulate this circuit – Schematic created using CircuitLab <S> If you want to save power and board area, but give up several hundred milliVolts of tight clamping, try one of these diodes <S> We used a lot of HP5082-2835. <S> If you can keep the diode HOT, it clamps at 0.2 volts. <S> Unfortunately, the diode leads have HIGH PERMEABILITY ~~ 2,000, thus have high inductance. <S> Dumet or Kovar is the brand name. <A> Some Vbe reverse voltages are limited to -5V which can cause breakdown and possible thermal damage from low Z drivers to input. <S> Low capacitance diodes can result in differential V protection when output is saturated and input delta <S> V may not be 0. <S> Comparators are a different design than OA's and have internal Vbe reverse diode protection built in and the Absolute max V differential input will reflect this as in the LM358. <S> Learn to understand all the specs in any datasheet!
| In this case, the diodes protect the circuit if a pathological input condition occurs.
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Time delay on circuit I'm trying to build a circuit with the following properties: When a switch (SW1) is pressed continuously, it should supply a particular load with 5V (modeled as R4 and LED D2) after about 20sec When this switch is released, to preserve battery it should not draw any power After looking at schematics on the internet I've tried the following circuit, unfortunately without any luck. I've tried simulating with LTSpice, it seems no matter how I change the values the NPN tranistor doesn't want to turn on. My circuit is probably seriously flawed, could anyone push me in the right direction? simulate this circuit – Schematic created using CircuitLab <Q> The circuit as given will stabilise with Q1's base around 600mV, which scarcely turns it on. <S> That explains why you get no significant power to the load. <S> Reducing R1, R2 and R3 to a tenth of the given values solves that problem sufficiently to light the LED, though it would also shorten the time constant. <S> This can be compensated for by increasing C1 by the same factor, to 1000µF. <S> If a further increase in load power is needed, connecting a second transistor in Darlington configuration might work. <S> You'll need to adjust the timing components for the higher base threshold. <A> I wouldn't say it was seriously flawed, it's just that your resistor values are a bit high to get Q1 sufficiently turned on to light the LED. <S> If you remove C1 from the circuit, you can do a bit of simple analysis to show that you'll get just under 1.6V across R2 and about 50uA of base current into Q1. <S> With a gain of 100 that gives you just 5mA through the LED. <S> Try lowering some resistor values to get the right base current to turn Q1 on enough for the load you're trying to drive, then determine C1 to get the right delay. <S> Better still, use a logic level MOSFET in its place, then you can stop worrying about the current it needs to switch it on. <A> simulate this circuit – Schematic created using CircuitLab <S> Another approach is to use CMOS Logic for Analog designs.either SOT23 SMD or DIP14 with hysteresis inputs (noise free).
| In general, timing long delays using purely discrete, analogue components is fraught with difficulty, especially if you then want a sharp turn-on of a relatively heavy load. Consider using an IC specifically designed for the purpose, such as the 555.
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Why is Ib proporional to Ic in a bipolar transistor? I am really puzzled why the collector current is considered proportional to the base current in a bipolar NPN transitor, ie $$ I_c = \beta I_b $$ Where I_c is the collector current, I_b the base current and below I_e the emitter current. I can't find any explanation in the books, it seems to be just stated as an empirical result. Is that how I should treat it, just an empirical result? I know that it can be derived by assuming that the proportion of (conventional) current that enters the emitter from the base is a constant fraction of the collector current, ie $$ I_c = \alpha I_e $$ and that $$I_b + I_e = I_c $$ but that begs the question why is the fraction of base current entering the emitter stream always assumed to be a fixed proportion? <Q> The collector current is only approximately proportional to the base current. <S> The rate varies with collector current and can vary wildly between individual units of the same type. <S> See BJT Large signal model <A> You would need much more training to understand semiconductor physics. <S> It is only approximately constant and the tolerance is often a wide range over 3:1 range around nominal value unless binned or from the same batch. <S> For now try to learn the basic functions and variables to understand how things work from a logic and analog behaviour rather than the physics model. <S> If you feed charges in a high electric field, they accelerate. <S> The rate of change of charges is called current. <S> The acceleration ratio of these charges in a high electric field , we call current gain. <S> It changes somewhat with many variables, so you only show the simple formula and the more accurate model is here . <A> Think of a vacuum tube: <S> Electrons are emitted by the cathode, attracted by the grid, then pass through the grid and are accelerated to the anode. <S> However, some small fraction of the cathode current will be "stolen" by the grid. <S> The same thing happens in a bipolar transistor. <S> The base-emitter voltage sets the emitter current. <S> In normal operation, most of that current continues to flow to the collector but the base steals some hopefully small fraction of the emitter current. <S> The exact amount varies between devices and operating conditions. <S> At a given operating condition, the base current is approximately proportional because each electron (or hole in a PNP transistor) acts nearly independently -- it has a certain probability to continue onto the collector, and a certain probability to recombine, in which case it won't be able to traverse the collector-base junction, and will instead leave via the base electrode. <A> A device-physics guy explained BETA to me thusly: the base current charges enter the base region, and the emitter provides opposing charges that try to HIT and recombine with the base current charges. <S> Fortunately, most of the emitter charges miss and continue on into the collector. <S> We thus benefit from those charges that happen to miss, which provides power gain and current multiplication. <A> Unfortunately the rationale quickly goes deep into semiconductor physics. <S> The DC current gain \$\beta\$ <S> is related by the forward transit time of a carrier to cross the base region and the average lifetime of a carrier in the base region. <S> If you inject some base current \$I_B\$ into a transistor, a charge in the base region builds and eventually the collector current reaches approximately <S> \$I_C = <S> \beta I_B\$.
| It is just that both the collector current and the base current are exponential functions of the base voltage.
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Eliminating ground loop in Bluetooth speaker circuit I have built a Bluetooth speaker (based on this Instructable ), which works except there is a faint clicking noise in the background. This is a basic diagram of the circuit for the speaker: I am almost certain that the problem is caused by a ground loop, because I have tried using a separate power supply for the Bluetooth adapter and this fixes the problem. However, I need to be able to power the speakers and the Bluetooth receiver from the same battery. I read online that someone else fixed a similar problem by adding an audio transformer to the left and right channels. I tried using an NTE1 1:1 audio transformer , and this did not help. With the transformer connected the speaker was actually worse, because it only made static noise and did not play audio. If the problem is indeed a ground loop, I would not really expect an audio transformer to fix it because the problem is with the power wires rather than the audio wires. I think a ground loop isolator would solve my problem, but when I researched them online, the only products I saw were adapters made to be used with audio systems like this . This is probably a really noob question, but I am not sure how to fix the ground loop problem with just small circuit components. How would I go about this? <Q> Probably RFI rectification by the amplifier. <S> 2) Use inductive ferrites to increase the inductance of the cables, high frequency currents take the path of lowest impedance. <S> Using a ferrite on the cable can block high frequency signals 3) Shield your wires or your project (this is more expensive) <S> Shielding provides a faraday cage which could block most of the high frequency signal from getting close to your electronics. <S> 4) Shunt the high frequency currents back to the source, use capacitors (RC filters) to shunt the high frequency currents back to the source before they reach sensitive parts of your circuit. <A> Putting an isolator (such as CME0303S3C) inline with the power input on the bluetooth module should fix this problem: https://www.mouser.com/ProductDetail/?qs=XKx0tEJeiQ3UXTrSkJqzTw%3D%3D <A> Connect ferrite beads series with Vcc power GND and analog GND, preferably close to the BT-module! <S> (50-100ohm @100MHz will do)You also can add big (10uF) <S> X7R ceramic capacitor between bluetooth module's power supply and GND pin, right at the module. <S> It helps a lot with my BT-module. <S> This buzzing noise is not due to ground loop. <S> Ferrite beads and bypassing stop these spikes. <A> Bluetooth earpieces can suffer from the 577Hz RF pulse rate. <S> Manufacturers have learned to prototype these circuits and pick OpAmps that are adequately immune to local RF fields.
| There are a few things you can do: 1) Twist your wires into twisted pairs, this will lessen the effect of magnetic field loops. The problem is that 577Hz BT pulse rate is audible, but it is come from RF current spikes on the ground.
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Connect signal ground to safety ground? We are designing a product which has to meet EMC directives. It consists of metal backplate folded into a chasis (the front is open). on the backcover there is a SMPS mounted. and on top of this we have our main PCB which has a signal ground plane on top of the PCB. the whole thing will be surrounded by a plastic cover with some space for user interface on the PCB. See image for clarification. All SMD components will be on the back of the PCB, the front has the GND plane. The product will interface in a master-slave configuration via RS-485. because the units can be far away from eachother and be on different mains circuits, there can be big voltage differences between individual grounds so we also connect signal ground with the RS-485 connection. My question is: (how) do we need to connect signal ground to safety ground (metal backcover)? I understand if we connect the signal ground to the safety ground, and also connect the individual signal grounds between units, we create a ground loop. <Q> OK you have multiple issues here. <S> Communication <S> EDIT: <S> Despite being differential, RS-485 needs a "ground" return path to handle unbalanced signal currents. <S> Since the distance is long and grounds can be severely different isolation is your best bet. <S> Since you have not defined the requirement values, then it is impossible to tell you exactly how. <S> Isolated can mean the device Andy mentions, or opto-coupling the inputs. <S> Cabling Again, you have not intimated what style of cabling you plan on using to carry the RS-485, specifically, whether said cable is shielded or not, and more importantly whether the connectors protruding from your board need to be earth grounded for safety reasons. <S> If the connectors are metal visible, or rather, metal touchable, they need to be safety grounded to the chassis. <S> If the cables are shielded, the shield should either be hard connected to safety ground, or loosely grounded through 10nF capacitors in parallel with 1Meg resistor. <S> The latter may not be possible if the connector in question has the shield attached to the connector shell which in turn is connected to the unit's connector which you already grounded earlier. <S> That is unavoidable. <S> Internally <S> The grounding system designed such that no currents will flow through said grounds from the system itself. <S> That is.. <S> NO SIGNAL RETURN CURRENTS. <S> This grounding connection provides two functions, it prevents static build up in your electronics, and it provides an escape path for EMC and EMI currents. <S> All safety grounds should be star-connected individually to a single hook-up point. <A> The simple and sensible thing to do is use <S> isolated RS485 transceivers. <S> Connect grounds (for EMC and communication integrity reasons) via 10 nF capacitors that are sufficiently rated to accommodate power AC voltage differences. <A> This is a very large topic which you have given <S> no where near enough information to answer. <S> If you want a real answer, you'd need to hire an engineer and give them at least a week to understand it all. <S> There are two options considered by most people: ground everything with large confident paths, or isolate everything (as mentioned in another answer) and have differential signalling. <S> My response is: design the system properly. <S> You describe a "safety ground" and a "signal ground". <S> As I don't know the system, I don't know what these both are. <S> You need to make sure you have a low inductance return path. <S> That means keeping the distances as short as possible for the full path for every signal. <S> I assume the safety ground is your ESD return path? <S> Make sure there is a path from the ESD diodes to this safety ground is as short as possible, keeping the loop small. <S> The signal ground will need to be connected to the signal source (make sure the route is as small a loop as possible, you may befit from using the safety ground, or you may not, depending on set up and the rest of the system). <S> But really, there is way to much to go into, and too little information given. <S> The only quick answer I can do is agree with Andy aka and suggest going for differential signalling to remove the need a signal return path.
| The system ground should indeed be connected to the safety ground, either through a hard connection or via a capacitor resistance combination.
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Slayer Exciter : Aiming for Larger Arcs I was recently building a larger slayer exciter. I've been trying out different coil ratios, coil materials, different transistors and many circuit variation, most of them just excited a fluoroscent bulb, just at different a brightness. But the last one gave me small arcs. The most recent one I made had about 850 turns in the secondary to 4 turns in the primary. I ran it at 40V with an old power adapter[I made that by soldering two 20V DC power bricks in series](I don't own/can't afford a Lab Bench power supply). I was using a TIP41C, and burnt 1 out, because careless me left it powered on for a long time without a heat sink. This one gave me half cm arcs when I leave the free end of the coil in air. Then I switched to a MJE3055T. It worked pretty well, and added 2mm to my 5mm arc, but Looking at all those pictures of 2 inch arcs on youtube, I was not that happy with mine. I recently bought 1500 meters of 37AWG magnet wire(for about 9$) and I really need some advice from experts/professionals/engineers, to build another one, which can give me large arcs with my 40V supply, a 40KV ceramic capacitor, 18AWG primary coil and a TO-220 type transistor I guess(TIP41C, MJE3055T, and I've got a few other high voltage transistors that I salvaged). Thanks a lot for any help guys! Would really appreciate some advice! <Q> Slayer exciter circuits are by design very inefficient. <S> It is nothing more then a simple resonant circuit operating in a lossy manner. <S> I would suggest you spend some time reading up on Tesla coils. <S> Basically you are switching FETs or IGBTs on and off, not partially driving them in a class <S> A style like the slayer exciter does. <S> That being said, if you are not aware of the dangers of working with high voltages, building a Tesla coil is a good way to toast yourself. <A> Try 4000 turns on a secondary coil 1.5 inches in diameter, use an ultra fast switching diode like a byv 600 and an audio amplifier transistor such as an nte36, happy sparking p.s. up the resistance alot to save the transistor from overheating, use a large heat sink and try a primary of 12 turns at a 3 inch diameter and insulate between the primary and secondary to avoid losses. <A> The last guy said more windings make for more resistance... <S> this is true but it's not a Tesla coil and as a lossy circuit in the first place makes little difference. <S> Try 4000 turns of your 36 AWG wire around a 1.5-2 inch pipe, loosely coupled with a primary of 10-12 turns of 12 AWG at a 2.5-3 inch diameter respectively. <S> The transistor's gain is very important. <S> The higher gain and switching frequency will bring larger arcs because the more breaks per second mean an easier path for the discharge to flow, since previous arcs are still present. <S> 3 more tips: <S> Tuning can be done on the primary coil (try bare wire spread apart enough that you can move an alligator clip along the coil to try different settings, when it is tuned well wind that coil from insulated wire). <S> With two equal transistors, try a dual primary coil sharing power supply and feedback (but with their own primary coil and diode). <S> On that note, throw out your diode unless it's of the ultra-fast switching type (try a BYV600). <S> Unless it's a variation of the slayer exciter, you don't need a capacitor (in circuit drawings, a cap between the HV end of the secondary and ground is a natural occurrence, as air is the dielectric). <S> Good luck.
| Adding more turns will add resistance and reduce the resonant Q factor of the circuit, eventually giving you even less voltage gains for exponentially increasing amounts of power. To get bigger sparks you need something more like a Tesla coil operating as a class D amplifier. A small half bridge Tesla coil can get you 1-2 ft arks and cost under $50 to build if you keep it simple.
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Does USB connector do any thing special like impedance matching or it's just a way to connect two devices to each other? I'm designing a board that both USB Host and device are in the same board. The host is a microprocessor and the device is a HID device which has a stm32f107 as usb device controller. What I need to know is: Is it possible to connect my host and device data lines as differential pair on the board without any extra components like USB connectors, resistor and ...? As in antenna connectors in telecommunication board that matchs impedance of the board and the antenna, does USB connectors do the same thing? I mean does omitting the USB connectors from the circuit affect the operation of USB communication? <Q> Yes, it is possible to connect a host and device on the same PCB simply by running a differential pair of the appropriate impedance between them. <S> In practice you can get away with some mismatch, so you can do this on FR4 with no problem even without requiring special controlled-impedance fabrication. <S> Many embedded or laptop computers do this. <S> I've never had any difficulty with USB 2.0 high speed (480 Mbps) <S> just by following basic high speed design practices; USB 3.0 would require more care and probably simulation. <A> USB 1 and 2 <S> USB 3 <S> A/B connectors are also just connectors but they are reaching speeds where controling parasitics in the connector, maintaining characteristic impedance etc starts to become really important. <S> As long as you maintain the correct impedances going directly from chip to chip should be just fine. <S> Even if you don't a "Full speed" interface will almost certainly work. <S> USB C connectors can have some inteligence inside but afaict <S> it's mostly related to the "power delivery" stuff. <A> Connecting USB devices with host microprocessors on the same board without any intermediate connectors is called "embedded USB". <S> Many laptops and desktops embed USB devices on-board. <S> In some sense the signal integrity can be even better, assuming HS differential traces are implemented with 90-Ohm impedance, since any connector is always a deviation from PCB trace and thus an impedance mismatch is introduced. <S> However, there is one caveat that is frequently is overlooked. <S> Normal USB connect protocol involves VBUS detect on device side of connection, prompting the device to start "connect" protocol by pulling up D+ (or D- in case of LS device) line AFTER the VBUS is received (after cable plug in). <S> The problem is that in embedded design the VBUS (and device power) is assumed as always on, prompting USB devices to assert the connect event (pull-up) BEFORE the host is powered up, booted up, USB controller configured, and USB software stack is loaded making the host ready to communicate. <S> In many legacy host hardware this premature assertion of connect state can disrupt normal host set-up functionality, and likely some additional software work-arounds might be necessary. <S> It might work, but in some number of power-up cycles the embedded USB connection can fail. <S> If a designer doesn't want power-on surprises, the host should implement an additional GPIO line (analogous of VBUS) that would hold embedded USB devices in unconnected state until the host boots up and is ready to communicate via USB. <S> Or use that extra GPIO to power-up embedded USB chips only after the host USB stack is ready. <A> I think that having a good cable and proper connectors, in most cases, is a matter of minimizing loss and signal degradation when your application REQUIRES you to transfer a signal over a distance. <S> Shorter is usually better. <S> I strongly doubt that the cable provides any essential function beyond transferring the signals from point A to point B with as little degradation as possible. <S> That being said, I note that numerous comments express concerns over USB 3.0, so I expect there are other considerations if that is being implemented.
| A/B connectors are just connectors, there is no special magic inside them.
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Set output DC operating point of Op Amp I want to set the output DC operating point to 2.5 V. Here is my schematic showing DC operating point. If I replace the opamp with ideal opamp, the output DC operating point is set as 2.5V. With the real world device, I am unable to set the DC operating point. Could anyone help me with this? <Q> The output is clipping the negative rail trying to sink the 10 uA of input bias current through R1. <S> Try using smaller resistors for R1,R2,R3. <A> What do you man by unable to set the operating point? <S> That opamp has an input bias current of about 10uA ( LT6200 datasheet ) <S> Why use that opamp? <S> - it is a fairly exotic high bandwidth one. <S> It has very low noise <S> but if you wanted that you should be be using much lower resistance values. <A> Your schematic shows no VDD bypass cap. <S> Add one right across the VDD/GND pins, with at most 1mm leads. <S> Of value 0.01uF, SMT.Add 1 Ohm resistor between that Cbypass and some remote cap, of value <S> 10uF. <S> The 1 Ohm is to dampen and prevent ringing of that CLC network. <S> You need a ground-plane, to minimize the feedback from output back to input. <S> You have 4 ""GND" nodes, With the bypass cap, there are 5. <S> Use a 3" by 3" GND PLANE. <S> What is the bandwidth of the SCOPE you've used to characterize the Opamp output voltage? <S> The OpAmp may be oscillating.
| The differential voltage across the input terminals of your opamp is way too high to be in regulation.
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Can I expect SMD components from a same manufacturer to be identical? I'm working on a project which requires at-least 2,000 ambient light sensors, and instead of using a microcontroler to manage and monitor the whole system I just want to amplify the output of sensors with a transistor array and use it for actuation of solenoids. As a beginner I'm a bit confused, and I don't know whether I can treat each individual cell(from sensor to solenoid) identical to other. <Q> So are all parts "identical"? <S> No. <S> Note too that many parts come in various "grades", where essentially the same part is available with better tolerances, a wider temperature range, etc. <S> All of this information can be found in the parts' datasheets. <A> Even the same component from the same batch from the same manufacturer cannot be expected to be identical . <S> The solution is to make your product/system able to cope with these small variations. <S> When a certain accuracy is needed despite component variations, often a form of calibration is introduced. <S> This means that each product is exposed to a certain amount of light (for example) and then you program it telling it that this is the switching level. <S> You could achieve the same with a variable resistor (potmeter) which you trim to the required value for that product. <S> Then every product will "know" that level (you taught it what that level is, the product is calibrated) and then all products should perform the same. <A> SMD components are no different from any other. <S> If the component is specified at +/- <S> 5%, roughly 2/3 of the components might be +/- <S> 2.5%. <S> Depending on the component, some batches might behave similarly (the compounds mixed into them clearly consist of batches, these may vary from day to day), and the machine calibrations might also drift over hours/days/weeks. <S> If you have only 10 components in a circuit, there is already a huge potential difference in performance between nominal and each component half-way to the end of it's specified range. <S> You typically find that most examples of your circuit work, but a few percent are 'unlucky' and the tolerances can cause bad performance. <S> A good design might compensate for the expected variations - or you might use an MCU to perform the compensation (generally far cheaper than specifying more precise components).
| Within the parameters of the data sheet, you can expect every individual component to fall in approximately some sort of bell curve (or normal distribution). It depends on the tolerances of each part, as guaranteed by the manufacturer. But as long as your circuit design allows for the min/max variation of each part, there's no reason why you can't do what you're suggesting.
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Keeping pest deterrant device running during power outage I have a holiday home in a remote area subject to frequent power outages. I use pest repelling devices powered from 9 Volt mains adaptors but when an outage occurs the little furry rascals rush inside and create havoc. Can I rig up a 9 Volt charger and battery and connect the load in parallel so the battery takes the load during outages. I have heard of chargers which provide a power path but can only find them for industrial use and very expensive. <Q> The $1 solution: simulate this circuit – Schematic created using <S> CircuitLab <S> How it works: AC/DC output voltages tend to be a little higher than rated and batteries tend to be a little bit lower. <S> The 2-diode wired-or configuration means that the higher voltage supply will "win" (sourcing most of the current to the device). <S> That means the AC power supply will provide the device power when available, but after the voltage falls (no AC power available), the battery will take over and provide power. <S> The advantages of this configuration are that it's cheap, easy to make, and make-before-break <S> (there is never a point where power is interrupted while switching over). <S> The disadvantages are that it's not the most efficient possible solution due to the voltage losses in the diodes. <A> How long are the power outages? <S> Are you sure the problem isn't that the pest repellent doesn't work? <S> Unless you are talking about outages lasting weeks at a time, I think it is much more likely that your pest repellent is not effective than that squirrels are waiting outside your house for the noise to stop so they can rush inside. <S> With such a light load, it should last quite a while. <S> If you provide more information on your requirements there may be a better option. <A> Thanks for all the helpful ideas. <S> I particularly like the $1 diode solution. <S> I don't think that my devices have yet become ineffective. <S> My problem is mice and they only seem to invade when power is off. <S> I visited the house in April for the first time since October 16. <S> The first thing I found was a notice in my letterbox advising me that for maintenance purposes power would be shut off for eight hours on a date in February. <S> I then found that there had been a short rampage through the kitchen but no further and no ongoing presence. <S> This suggests that they came in as soon as the repellers stopped and vacated when they resumed. <S> On a previous occasion two years ago I discovered on arrival that due to a fallen cable power had been off for an extended period. <S> That time they had really done a number on the place. <S> Chewed cushions and bedding. <S> Even chewed my antenna cables. <S> I use three small domestic devices which seem to be enough when powered but they need to be live even when the house is occupied so rigging up 12 Volt batteries is not practical. <S> What about my first thought. <S> Using 9 Volt Li-Ion batteries while being simultaneously trickle charged. <S> The devices use less than 200mA.e
| Anyway, since you didn't provide power or runtime requirements, I recommend using a standard computer UPS and plug the 9V adapter into that.
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Creating a circuit with 3 lights and 4 switches (1 master switch) I'm a high-school junior in AP Physics 1. My class was assigned an optional, grade-boosting project where we could make a circuit under certain guidelines. I figured I'd give it a shot, but I'm really struggling at the design. Here are the guidelines: You can only use wires, 4 switches, 3 lights, and one power source, and apparently now diodes Switch 1 is a master switch, and must turn on all three lights while the other 3 switches are open Switch 2, 3, & 4 must turn on an individual light while switch 1 remains open. I have been using the Circuit Construction Kit from PhET found here , and I have really struggled. No other components are allowed. I eventually have to physically build the circuit and get the calculations of the current and voltages for each scenario, but I wanted to just get the design down. Any solutions or tips would be greatly appreciated! EDIT: Here is the original assignment page given. I am going to try to clarify with my professor about what components are allowed. EDIT 2:I have just talked with my teacher, and it seems that diodes are allowed, but NOT multi-way switches. He was very explicit about that. <Q> So something like this then: simulate this circuit – Schematic created using CircuitLab <S> When SW1 is closed all the lights turn on. <S> When SW2 is closed lamp1 turns on but D1 prevents the other lamps from getting power. <S> The downside is that the lamps will get a slightly different voltage depending which switch has turned them on. <A> OK, my original answer was made in haste, and I overlooked a very obvious problem. <S> This can't be done without multi-throw or multi-pole switches or diodes. <S> Are you sure the master switch was meant to turn all three ON and not OFF? <A> I'm assuming that 3-way switches are allowed. <S> Currents through individual lamps will be additive. <S> Voltage across any lamp should be equal to the voltage across any other lamp. <S> Currents and voltages can be calculated using V=IR. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> There is an answer to this question, and it doesn't require resistors, diodes, SPDT switch etc. <S> Yes, 4 batteries. <S> You will notice that it doesn't say you can only use a single battery. <S> I'm not going to give you the solution because this is a homework question afterall, but I will give you a big hint: <S> First think about each lamp individually. <S> Using a battery, a switch, and a lamp, fulfil the requirements for switches 2, 3 and 4. <S> Now using another battery and a switch, connect up your sub circuits to fulfil the requirements for switch 1. <S> In fact, the approach I have in mind <S> could be argued to be one power source. <S> All of the batteries are placed in series in a ring, but with switches between them. <S> Technically it is still one power source. <S> Essentially you have 4 batteries in a circle with a switch between each one. <S> Then attach the lamps. <S> Just never close all 4 switches at the same time (boom).
| You can do it using 3 Lamps, 4 Switches, Some wire, and 4 Batter ies .
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Run A 3 watt LED chip off a 12v Source I have tryed look at other topics on here, but i cant understand what all they are talking about need something simple and to the point please.So i got thease 3 watt LED chips, Forward V 2.2 And Forward Amps (Current) 700MAI want to hook this up to a car battery. I tryed a resistor of 15ohm 1/2watt and i dint think about the wattage power coming in effect so it got hot in a hurry.So a calculator reccomends 7.6 watt capable resistors. There expensive and hard to find and big. Is there any other way to get the voltage and current were i need it? Please use simple terms. Country Boy Here. <Q> Typically, you want to use a switching or buck based led driver for this. <S> Or you can use multiple 5W resistors in series/parallel. <S> These will be cheaper. <S> Especially if you want to use a 12V supply. <S> A linear or resistive solution would be dissipating 9.8V at 0.7 Amps = 6.86 Watts. <S> In this case, a cheap car usb charger would work best. <S> It does the 12V to 5V with a switching regulator, and the rest can be done with a smaller resistor. <S> 5 - 2.2 = <S> 2.8V <S> * 0.7 = <S> 1.96W. <S> A 2 or 2.5 Watt 4 Ohm resistor will work. <S> As mentioned 2.2V <S> * 0.7A = 1.54W, half of the supposed 3 watts btw. <A> If you want to go one step above a ballast resistor (dropper resistor), you might consider an LM317 based current source. <S> Such as the following: Taken from: LM317 <S> The following article has a down to basics explanation of an lm317 led driver <A> The 15Ω 7+ Watt resistor is correct for a single 2.vV LED and 13.5v (2.25v x 6) battery. <S> What is incorrect is using a single LED with a 13.5 V battery. <S> You should have five in series with a 3.5Ω 2-3W resistor. <S> But just because an LED has an I <S> max of 700mA does not mean you can drive it with 700mA without significant thermal management. <S> Especially an orange AlGaInP LED. <S> At 700mA the LEDs are going to get very hot. <S> When a Cree or Luxeon 3W red-orange LED exceeds 100°C, it loses about 50% (or more) of its brightness. <S> You may be able to drive the five LEDs with 300mA (8.2Ω 1-2W) but they will still get hot. <S> Somewhere there will be a happy medium with a balance between the number of five LED strings and a thermally manageable current can be found. <S> If efficiency is a concern like of the LED will be on when the engine is off then a CCR (constant current regulator) would be appropriate for a current of less than 350mA. <S> A CCR acts like a dynamic resistor and adjusts to the dynamic voltage of the LED string. <S> An On-Semi 150-350mA CCR <S> NSI50150ADT4G costs under $1.
| A better solution is a 5V or USB supply.
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Easiest way to find matching li-ion cells (3.7V, 18650) I recently acquired 50 used li-ion cells (18650). I'd like to efficiently determine which cells are good matches (i.e. which cells have similar: capacity, charge times, & discharge times) so that I can put them into battery packs that will perform optimally (e.g. they don't punk-out early because one or more cells discharge too fast or over-charge or over-heat as slower charging cells lag near the end of charging). What would you suggest would be the best way to test multiple cells to get such information? <Q> This is obviously a lot to produce <S> but it's how we once solved the problem you've described... <S> I helped another engineer design and develop a cell matcher, which he went on to sell. <S> This was 25 years ago with NiCd cells used for remote-control vehicles. <S> Our system profiled cells using a constant-current discharger and an ADC in a data logger (actually a home computer back then). <S> Cells were selected and matched by examining the discharge graph and key parameters for each cell. <S> The latter were figures such as time to Vout dropping to something like 80%, time to fall from 80% to 20% and a few others. <S> The discharge circuit was simple enough: a current regulator to ground with an op-amp controlling it. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Our discharger actually had about 5 x TIP31A connected in parallel to dissipate the heat. <S> The whole system of computer interface and discharger worked very well, he sold enough of them. <S> The customers were RC enthusiasts who already had their own chargers. <A> Analyzing the charge and discharge curve characteristics is the most reliable but also the most time consuming. <S> Measuring the internal resistance of an Li-ion offers little value as Li-ion keep a low and consistent resistance throughout their life cycle. <S> It may identify the cells that have little or no life left. <S> A 3.7V, 18650 cell is an Li-manganese which is used in power tools because it can withstand short heavy discharge rates. <S> Some cells can withstand up to a 5 second 30C discharge. <S> This lends itself to testing its capacity by measuring the electrochemical dynamic response with the Pulse Discharge Test method. <S> This measures the ion flow between the positive and negative plates. <S> The Pulse-Discharge method is the quick and somewhat simple method. <S> It is important the cell not be fully charged and ideally at about a 40% charge level. <S> A resistive load with between .1C <S> and 2C discharge is applied for between 1 and 6 seconds, not to exceed 6 seconds. <S> A strong cell recovers quickly, weaker cells are slower getting back to the pre-pulse voltage . <S> Measure voltage before discharge pulse. <S> Measure discharge voltage within one second after falling edge. <S> Measure open circuit recovery voltage within one second after therising edge. <S> This is a patented process Patent Number 7,622,929 <S> The patent has much more detailed information. <A> Just like car batteries, use Cranking amp with a constant voltage sink pulse (7.5V for 12V car CCA and ( 3.0V from 3.7 for LiPo) and open cell voltage <S> Voc <S> after a xx seconds of fully charged cells. <S> This can be done with a pulse with sample and hold on current sense using a diode driven PNP darlington TO220 on a heat sink such that Ve is 3.0V at 10 A more or less for Isense using 0.01 Ohm non-inductive rated power resistor on collector to ground for 100mV sense at 10V. Pulse width = <S> TBD <S> Then record I load and Voc on label and sort into bins of 1% or less . <S> Batteries and load transistor should not get hot in <<1/10 second with 30 Watts more or less being transferred. <S> But the iMax B6 is cheap and does similar thing. <S> more details <S> http://www.mpoweruk.com/chargers.htm
| Measuring the impedance curve between 1hz and 10hz is a reliable quick method but a bit complex.
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In 802.11 (CSMA/CA), does a higher data rate imply more number of collisions? In a wireless environment working with the 802.11 protocol suite, where the MAC protocol is based on CSMA-CA (collision avoidance), a higher data rate would mean that data is sent more quickly, which in turn implies that more number of packets are sent successively. Wouldn't a greater number of packets being sent mean a greater chance of collision? Note that even if powerful error correcting codes or higher modulation schemes are used, collision can't be prevented, as that entirely depends on how many packets are being transmitted 'over-the-air' at any point of time. Overall, I'm confused how high data rates would work well in a wireless environment, if it resulted in more number of collisions (which would be counter-productive). <Q> The premise that a higher raw data rate implies a higher packet rate is false. <S> The packet rate would only go up if the network were already saturated with traffic, and the application actually transmitted more packets as a result of the increased available bandwidth. <S> All other parameters being equal (packet size, payload bandwidth, etc.) <S> a higher raw data rate means that the individual packets are shorter in time, resulting in fewer collisions, not more. <A> In a wireless environment working with the 802.11 protocol suite, where the MAC protocol is based on CSMA-CA (collision avoidance), a higher data rate would mean that data is sent more quickly, which in turn implies that more number of packets are sent successively. <S> No. 802.11 allows different modulations, with more bits per symbol <S> the better the S(I)NR gets. <S> So, same number of packets is able to transport more data in a good signal scenario than under a bad signal condition. <S> Wouldn't a greater number of packets being sent mean a greater chance of collision? <S> Yes/ <S> No. <S> Notice the Carrier Sense in CSMA. <S> Admittedly, assuming stations are trying to access the channel more often, the likelihood of one missing the fact that the channel is currently used within a fixed duration becomes higher, but: Wifi cards (and drivers) are clever. <S> They tend to adapt their MAC strategy (and: aggressiveness) to the occupancy of the channel. <S> as that entirely depends on how many packets are being transmitted 'over-the-air' at any point of time. <S> No. <S> Not entirely. <S> It also depends on how you sense, and what arbitration scheme is actually in effect. <S> Overall, I'm confused how high data rates would work well in a wireless environment, if it resulted in more number of collisions (which would be counter-productive). <S> Even with more collisions, you'd get more packets across – that's like saying "in a city with 100,000 cars, there's gotta be more traffic jam than in a city with 10,000 cars. <S> I don't see how having 100,000 cars helps getting 100,000 people to work". <S> Well, it does. <A> a higher data rate would mean that data is sent more quickly, which in turn implies that more number of packets are sent successively. <S> A higher data rate is one factor. <S> The number of packets sent by a node and the time allocated to a node for exclusive use are other factors. <S> Wouldn't a greater number of packets being sent mean a greater chance of collision? <S> collisions are not associated with the number of packets. <S> On the other hand if a single node is the only one transmitting lots of packets, there is no change of collisions collision can't be prevented, as that entirely depends on how many packets are being transmitted 'over-the-air' at any point of time. <S> At any point in time only one packet is being transmitted otherwise there is a collision. <S> how high data rates would work well... if it resulted in more collisions High data rates will not result more collisions. <S> The faster the packets can get from point A to B, the less time for a second node to want to use the networks at the same time. <S> Less time for the second node to wait for a clear channel means there is less time for a third node to come along and want access. <S> Similar to lower speed limits creating more traffic congestion.
| The more nodes wanting to transmit simultaneously will increase collisions.
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LED burning out due to high current? So using an Arduino I fed 5v to a white LED, nothing happened to it. I fed 5v to a yellow LED, and within a few seconds it started to smoke and burn itself out. My first question is how toxic are these fumes? Second, why do different colored LEDS have different voltage/current maximums? <Q> I have a feeling that you've connected your LEDs directly between the Arduino's output pin and ground. <S> You're not supposed to do that - most microcontroller's output pins aren't designed to drive the current required to light a LED. <S> You might end up burning your microcontroller too. <S> You might have been lucky to have a microcontroller that somewhat limits the output current to a level that was safe for the white LED, but not for the yellow one. <S> Regarding your second question, your white LED is as close to your yellow LED as as a Chevy is to a Ford. <S> They're just built differently, and will behave differently. <S> That's why each electronic component has a datasheet - and why each car has a manual. <S> Regarding your first question, exactly what's burning depend on each manufacturer's process. <S> In case of doubt, assume it's toxic. <S> But you're alive, so it wasn't probably that bad :-) <A> Toxicity is in the dose. <S> I have never seen an LED manufacturer warn about breathing the fums of a smoking LED. <S> I have smoked enough LEDs to know what it smells like. <S> Not likely acutely toxic. <S> White, blue, and green LEDs are usually Gallium indium nitride with about a 3v forward voltage. <S> red, orange yellow (amber) are usually Aluminium gallium indium phosphide with about a 2v forward voltage. <S> Current for AlGaInP range from a few mA to 700mA. GaInN from a few mA to 1500 <S> mA. <S> When you say "I fed 5v" I would need to know HOW you fed the 5v in order to guess as to why one burned and the other did not. <S> Both should have burned if they were connected to a 5v source capable of more than an amp. <S> UPDATE <S> Well I connected the 5v and ground to the LED, and for some apparent reason this one started burning... <S> Trying different colors except yellow, none of them had the same issue. <S> Manufacturing differences? <S> The power supply for your Arduino must be low wattage. <S> Connecting an LED to a 5V power rail should make them all go up in smoke. <S> My guess is that the LEDs you used were not high brightness with a low maximum current. <S> I took a 20mA red indicator LED and connected it to a 600 mA variable power supply and slowly increased the voltage. <S> It turned on at 1.8V and dimmed at 2.4v. <S> As it approached 5V it dimmed even more. <S> On the other hand these two Luxeon Rebel Blue LEDs burned running at their max rating of 1 Amp due to no heatsink. <A> Toxicity of LED Smoke <S> According to the journal named Environmental Science and Technology LEDs contain lead, arsenic and a dozen other potentially dangerous substances. <S> Especially red LEDs contains 8 times the led found in a yellow LED. <S> When converted to smoke is a neurotoxin and exhibit significant cancer and noncancer causing properties due to the high content of arsenic and lead. <S> According to The Scientific American Ogunseitan adds that while breaking open a single LED and breathing in its fumes wouldn’t likely cause cancer, our bodies hardly need more toxic substances floating around, as the combined effects could be a disease trigger. <S> If any LEDs break at home, Ogunseitan recommends sweeping them up while wearing gloves and a mask, and disposing of the debris — and even the broom — as hazardous waste. <S> Furthermore, crews dispatched to clean up car crashes or broken traffic lights (LEDs are used extensively for automotive and traffic lighting) should wear protective clothing and handle material as hazardous waste. <S> Why do different colored LEDS have different voltage/current maximums <S> For example: For a red LED the material used is Aluminium gallium arsenide (AlGaAs) orGallium(III) phosphide (GaP) which has a voltage drop of 1.63 < ΔV < 2.03. <S> Similarly for a Violet LED <S> the material it is built up of is Indium gallium nitride (InGaN) and the voltage drop is around 2.76 < ΔV < 4.0.
| Just about everything is toxic at a high dosage. LEDs are currently not considered toxic by law and can be disposed of in regular landfills. An aticle at https://en.wikipedia.org/wiki/Light-emitting_diode#Colors_and_materials explains that the materials that a LED is made up of effects the color and thus the voltage drop / requirements.
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Why can't solder stick onto the tip of my iron? I can't tin the iron properly, because feeding solder onto the tip has it flow downwards and collect into a blob before splitting off and creating a mess. My iron may not be calibrated. And I don't know how to measure its temperature without a device to do so (I only have a multimeter). It is supposed to go from 250 Celsius to 450 Celsius (840 °F), but I can only solder onto the pad quickly at 450 °C. The solder melts at around 300 °C, but it doesn't flow quickly. The iron I use is the T12 type. <Q> It's possible you've damaged the plating on the tip. <S> Try wiping the tip on a damp (not sopping wet) sponge and tinning again with fresh electronic grade flux-core solder. <S> It should look like this before you add more solder: <S> The site <S> I scarfed the above photo from mentions using sal ammoniac from your nearest Indian grocery store as a last resort. <S> Even if it fails, a nice chaat, bhel or paan can't hurt. <S> If the tip remains stubbornly black and not shiny tinned <S> you may need to replace the tip. <S> You should keep a few different sizes of spare tips on hand anyways. <S> Don't use a good soldering iron tip for things like melting plastic- <S> it will ruin the tip and may result in toxic fumes. <S> Most modern tips are copper, plated with iron, plated with tin (the metal). <S> You can file off the iron plating in an emergency and use the tip for a while but the solder will erode the copper away in no time (relatively speaking). <S> You can measure the soldering iron tip temperature using a type K bead type thermocouple- <S> many multimeters will accept such a thermocouple. <S> Just immerse the bead in the solder, it will be close enough for this purpose. <A> Spehro Pefany has good advice <S> but I'll add that there are " tip activators " that will clean off the oxidation surprisingly well. <S> This Weller is the only one I've tried <S> but it works wonders. <A> My soldering station includes, besides a Weller iron with temp control & a fat roll of solder: flux, and a gold R2-D2-looking thing that has steel wool. https://www.amazon.com/Welding-Soldering-Solder-Cleaner-Cleaning/dp/B00PQ32CUW/ref=sr_1_1?ie=UTF8&qid=1494968675&sr=8-1&keywords=gold+soldering+iron+cleaner
| If the tin plating is damaged, the tip is 'done', since iron does not readily wet with solder.
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OP-AMP circuit for voltage to current conversion It is required to design a circuit for voltage to current converter to have a current output with 4-20mA. So, I wonder if the above circuit will work. Suppose that the input ranges from 1 to 5V. <Q> No, that circuit does not produce a current output that maintains current amplitude across a decent range of load impedances. <S> Your "output current" is dependent on the load you have connected. <S> Try the "Howland current source" instead: <S> - Or maybe this type of constant current sink: - <A> The mistake you are making is that a 4-20 mA sender (the sensor) does not supply 4-20 mA into the loop .... <S> it limits an externally supplied loop to the signal current level desired. <S> In the other answer, the first option (Howland) is clearly not suitable since it tries to supply current to the loop ..... <S> however the second option would work since it limits loop current irrespective of load voltage and impedance <S> (it does have a problem in the required ground continuity, but that is a minimal annoyance in most installations). <S> You should read up on 4-20 <S> mA current providers such as the TI <S> XTR115 or XTR117 which shows the correct methodology. <S> The majority of industrial setting use 24 V or above for the loop source so whatever you stand up on the loop has to be able to withstand this voltage. <S> Most op amps are unsuitable for this task unless specifically designed for the job with the correct output buffers/drivers. <A> Here is a simple circuit that should work for a grounded load, and does not depend on resistor matching to give a high output impedance as the Howland and improved Howland circuits <S> do: (op-amps must be rail-to-rail I/O types) <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The shown circuit has a compliance of about Vdd - 2.1V <S> so it will work with a load resistance of up to (Vdd-2.1V)/0.02A. You can divide the input voltage down to get your required 1-5V input voltage using a 3K/2K resistor pair.
| It won't have the desired functionality because the load resistance for such circuits can vary over a wide range.
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I2C communication with different power sources Will I have some trouble if I try to establish a I2C communication between 2 µC powered from two different source? Both µC run on 3,3v but this voltage is provide from a different regulator (LM317). Exact 3,3v isn't guarantee due to resistor tolerance. Since both regulator have same ground, there is no big voltage "gap" between the µCs. I'm pretty sure it's gonna work but I'd like to have the EE community acknowledgment. simulate this circuit – Schematic created using CircuitLab <Q> Yes it will work even there is a few voltage gap between the two regulators. <S> However, this gap need to be smaller than 0.7V in order to prevent inner protection diode to conduct and cause (little) damage to the chip over time. <S> Supply has to accurate to prevent this to append. <S> Resistor with low tolerance value should be use to provide to configure the LM317 regulator. <S> When startup, both µC won't be powered at the exact same time. <S> The bus line will eventually be powered before both µC (through R1 and R2). <S> But this shouldn't cause any damage due to the low current flow (thanks to high pull up resistor value). <S> Thanks to JimmyB, glen_geek and Janka who helped me in the comment. <A> Of course that supply needs to be some margin above the minimum guaranteed high threshold for IIC. <S> I don't remember what that is, but 3 V is fine. <S> Keep the maximum sink current to hold a line low in mind. <S> That's 3 mA if I remember right (your job to check). <S> That means with 3 V supply, the pullups can't be less than 1 kΩ. <A> If you look up how I2C actually work <S> you will quickly see that it is not that difficult to use with multiple voltage levels. <S> Until the delta becomes too large (eg: 1.8V and 5V) <S> All I2C can do is pull the line low. <S> This is why you need the pull-up resistors. <S> Knowing this, you can use the lowest voltage level as pull-up, then you make sure the higher voltage MCU recognizes this as logic high with enough margin. <S> For this margin you have to look at the bus capacitance and the required timings. <S> The opposite is also possible, but as @M.Ferru explained you have to stay below the threshold of the internal protection diodes. <S> This methods it not recommended. <S> Thanks to sparkfun for the image.
| Yes, what you propose will work as long as the pullups go to the lower of the two supply voltages.
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Automatic overflow killswitch for watering pump Project I'm building a controlled watering system for my plants because I keep killing them by forgetting to water... I'm using an ATMega88PV for the logic and a resistive probe (voltage divider) to get a measurement of the moisture content which is working fine. I'll be using an aquarium pump controlled by a relay from the MCU and a large water reservoir for the watering. I would now like to implement an emergency overflow protection. My pot is sitting on a tray that will collect any excess water that drains out of the pot so that it can evaporate. I would like to have an automatic kill switch that kills the pump if any water makes it into the collection tray in order to prevent a mess on my floor. I'm not very good with analogue electronics and I've never used a relay before so I'm asking for your comments on the design. Here is the design that I have come up with: simulate this circuit – Schematic created using CircuitLab I couldn't find a symbol for a terminal so I used the normally open push button (labeled OVFD for OVerFlowDetector) symbol for my overflow detection switch. It will basically be two electrodes at the bottom of the tray which are normally open and will close when water collects in the tray. I have a weak pull-up to 5V when OVFD is opened. I then AND this with the control signal from the MCU GPIO pins (5V, 50mA max) and use this to drive the relay which is a SPST (I think?) with Ucoil = 5v and Pcoil=200mW (40mA). The relay is normally open (pump disconnected) and will connect the pump to the 230VAC 50Hz from the wall socket when closed. Is this the proper way to use a relay? Now when I look at it, it seems odd... Do I need a SPDT relay? I'm also a bit worried about driving the relay. The relay needs 40mA to switch and according to the datasheets on the 74LS07 it can only source 40mA. Do I need some kind of current amplification or will it work anyway? <Q> You've made a great attempt, but I'd suggest the following: <S> You need a diode across the relay to protect from the inductive kickback when turning off the relay damaging any driver. <S> You get better drive if you connect the relay from +5 to the gate output since it's a saturated transistor. <S> You just need your water level detector as an input to your MCU, you don't need to gate it with a signal from the MCU through an external gate. <S> You have software to do all the logic. <S> You can buy very cheap Arduino peripherals such as this relay board from various sources such as Amazon or Ebay. <S> They have reasonable output connections for the AC side and you are likely to find it much easier and safer to hook up. <S> they almost universally have an on board transistor and hook directly to an MCU port. <S> I'm assuming you think you need a kill-switch to override the system in case your MCU fails to turn off the relay. <S> However this would only seem to avoid a software crash .... <S> which seems unlikely in such a simple system. <S> If you really are concerned, then you should research ways to implement both brownout and watchdog timer events to ensure your MCU is always running correctly. <A> Your problem is quite simple, and you don't need the AND gate, it's more suitable to use your microcontroller to perform the action to stop the pump. <S> That would be my circuit to do so, sorry for the mess. <S> The GPIO_1 is responsible for the input from the OVFD, this works exactly as a button, when there's water, the electrodes short and the current flows to the resistor (5V), if there's no water, then, there is no current in the resistor (0V);Concerning the relay, the most usual is a SPDT, and to activate it, I recommend a NPN transistor to protect the microcontroller. <S> I guess that's it, pretty basic, but you will protect your circuit and avoid using another IC. <S> simulate this circuit – <S> Schematic created using CircuitLab EDIT: <S> Changed R2 value according to @Tony Stewart. <S> EE since '75, but, be advised, that this value depends on your components. <A> Do you really need a µController for this app? <S> You should use a 12v pump. <S> Like this one: 12V, <S> 4.2 Watt, 240L/hr Pump <S> There are many of these being sold with similar specs, they are all the same. <S> All you need is two moisture detectors, one for the soil, and one for the drain tray. <S> When NEITHER detects any moisture, the pump turns on. <S> If and when either detects moisture, the pump cannot be not be on. <S> You have a comparator for each detector. <S> One turns on a 12v driver to the pump and the other drives a circuit to provide ground to the pump. <S> You have your fail safe <S> but you have single points of failure, so you may want to use the µController to monitor the moisture as well and add an alarm. <S> The water supply should be minimal <S> so there is not enough water to go rancid. <S> When the water is gone the µController detects both the soil and tray are dry and activate the alarm. <S> The alarm is configured so it defaults to on and the µController is required to periodically disable the alarm. <S> If the µController does not disable the alarm the alarm goes off.
| Driving any relay that has 120-240 V on it is always hazardous, I'd suggest you buy a module rather than build your own circuit. This board already has a transistor driver and protection diode built in so all you need is a port signal, 5 V and Ground from your MCU to turn it on an off. There are also plenty of water level sensors for Arduino such as this moisture sensor ...
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Why do drones use Li-Po instead of Li-Ion? According to every source I've found the only advantage of Li-Po is their shape, but they are more expensive, have lower energy density and are more dangerous compared to Li-Ion. Why do drones use Li-Po batteries instead of Li-Ion batteries for example 3s2p 18650 Li-Ion with 40A protection circuit. <Q> For this answer I'm going to define LiPo as flat packs, and Li-Ions as 18650s. <S> The reality is more messy than that, but I believe this way of defining the terms will help answer your question as you seem to have intended it. <S> Traditionally, 18650s didn't support high current applications, a continuous current rating of 5A was considered to be high. <S> 18650s <S> that support 20A continuous current were more or less unheard of until high power cells like the Sony VTC3 came on the market. <S> (As a general rule, 18650s that claim to support more than 30A continuous, even today, are simply overrated. <S> There is also a trade-off between capacity and continuous current capability; For example the 30A Sony VTC4 is "only" 2100 mAh.) <S> A lot of the reason why drones use lipos are probably tradition, due to this. <S> When you say that LiPos have a lower energy density, you mean a lower energy to volume ratio, right? <S> I'm pretty sure they have a higher or at least as high energy to weight ratio, as Li-Ions. <S> 18650 <S> Li-Ions have a metal casing. <S> That's unneccesary weight. <A> the reason why people don't use 18650 as power source for drone is the C rate problem or discharging current. <S> for most 18650 battery packs, a 5C rate is sometimes hard to reach for most of the small manufacturers in china not to say 10C or 20C(technology is a problem, pricing is another problem, if more chinese companies can make batteries that reach 20C, pricing should be lower.) <S> but things are quite different for lipo batteries, 30C is common for most of the manufacturers so again, lipo are more cheap and mature for high rate discharging applications. <A> Weight and size also play a role.
| LiPo packs don't use a metal housing (=less weight) and the way they are build makes that use a lot less space compared to 18650 cells. The flat form factor of many LiPos probably also helps with cooling.
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Three phase 230V input to single phase 5V/5W power supply How can I design a 5V/5W DC power supply derived from 3 phase 230V AC input supply? My application need to work, even if, only one phase is available. <Q> SMPS operated by AC normally first do rectification of the input voltage, i.e. they work also with DC or pulsating supply voltage. <S> So you can simply connect all three phases by diodes and use that as input to such a SMPS. <S> simulate this circuit – <S> Schematic created using CircuitLab added note: But make sure you use such a SMPS. <S> It will definitely NOT work with a simple 50/60Hz transformer supply. <A> My application need to work, even if, only one phase is available <S> Use a 3 phase rectifier like this: - You'll still need a smoothing capacitor because if one wire is lost, the ripple voltage will worsen. <S> Of course there will be a significantly higher peak voltage using this method. <S> If you have 3 phase and neutral wires, single half wave rectifiers will suffice for a 5 watt converter: - <S> And, providing you have a good neutral wire, you can "lose" two phase wires and it should still operate. <S> This produces the same peak DC voltage as a conventional single phase SMPS. <S> However, if you "lose" the neutral wire then this will stop working. <A> The circuit being something like: OR STMicroelectronics Provides a three phase SMPS AN2264 <S> You can find the data sheet http://www.st.com/content/ccc/resource/technical/document/application_note/16/e6/c6/e4/08/c8/40/ff/CD00074286.pdf/files/CD00074286.pdf/jcr:content/translations/en.CD00074286.pdf <A> I think, I can use 3 Hi-link modules ( http://www.hlktech.net/product_detail.php?ProId=60 ), to derive 5W un uninterruptible supply. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The problem here is, I need to use 3 Hi-link modules( Cost is around $9 ) and it require more footprint.
| You should use a three phase rectifier with capacitors rated for approximately 500V.
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What is the point of having a level-based interrupts? Wherever I have searched about the practical implementation of the level-based interrupt, I have found only one suggestion that people have given i.e disable the interrupt as soon as it enters the ISR so it doesn't keep triggering back. Another thing I have read is that it is used to create a loop i.e as long as interrupt is there, serve the ISR, but that could be achieved with a while or do while loop. And the advantages a level edge interrupt would provide might be the luxury to run one instruction of the main programme in between serving the ISRs and the latency. I guess. So is there something I am missing when it comes to understanding the level edge interrupt? A great answer would be to show me some sort of practical usage of level-based interrupts. <Q> Level-based interrupts can safely be shared and cascaded easily and reliably; by contrast, reliably sharing edge-triggered interrupts is often difficult and sometimes impossible. <S> When using level-based interrupts, an interrupt handler can simply ask each possible interrupt source in turn "Do you need attention", and service it if so. <S> Once it's done, the handler can return. <S> If an interrupt source which was polled early in the sequence decides it needs attention while a later source is being serviced, the processor will notice that the IRQ pin is still active and re-trigger the interrupt handler, thus allowing the late-arriving interrupt to be serviced. <S> When using non-cascaded edge-triggered interrupts, things get more complicated. <S> After an interrupt handler has gone through and serviced everyone, it must then go through and re-poll everyone to find out if a previously-polled device has decided that it needs service. <S> Only after every device has consecutively reported that it does not need service would it be safe for the interrupt to return. <S> Note that if devices that want service keep their "need service" indication active, returning from an interrupt handler at a time when a device needs service may render that interrupt permanently useless. <S> It may be useful for an I/O pin to have some edge-capture logic which, when an edge arrives, sets a latch and outputs a "need-service" indication until software clears it. <S> Such a thing might appear on the very front end as an edge-sensitive interrupt. <S> At any downstream point, however, it's better to have interrupt logic demand that an interrupt be serviced at all times when any upstream point is not satisfied. <A> Let's consider a typical example ... <S> Signal: " <S> Case_Over_Temperature" Goes low when the ambient in the box is too high for normal operation.. <S> Obviously, this signal could go low at any time, either because we are making too much heat, or because the box is installed in a hot location. <S> Obviously, on power up that line could be in either condition. <S> Let's assume for the moment that the power-on code doesn't just go look but relies on the interrupt instead. <S> If the interrupt is edge triggered and the signal is already low, when the interrupt gets enabled the appropriate code will not be executed. <S> Level sensitive interrupts would be prudent here. <S> Similarly, if the processor is put to sleep and is not set to wake on that interrupt, that line can go low at any time. <S> When whatever else wakes it up happens, you want the interrupt to fire at that time. <S> Indeed, arguably, with the prevalence of sleep mode processors, level based interrupts have become more useful. <S> However, as with all things code related, there is always more than one way to "skin a cat". <S> If you don't use level based, the wakeup code needs to go poll the interrupt pins if they are not automatically queued by the processor. <S> Obviously, level triggered also comes with it's own set of issues in that the code has to handle knowing that it has already handled the condition etc. <A> It's somewhat the other way around: why have awkward edge-triggered interrupts when you can have the easier level-triggered? <S> Edge triggered interrupts are more susceptible to noise spikes and harder to filter. <S> They're then riskier to run off-board or up cables. <S> The interrupt can't be withdrawn by the source. <S> Level-triggered interrupts stay on until the CPU acknowledges the source. <S> So there's the solid base of full handshaking. <S> The CPU can filter noise out of the interrupt signal in almost any way it wants, it's just increasing the interrupt response time. <S> If the application requires and allows for well-filtered signals, level triggering is adaptable. <S> I first saw edge-triggered interrupts used for NMIs on CPUs like the Z80 and 6502, while maskable interrupts used level triggered. <S> The NMIs used edge-triggered simply to stop a stuck pin or stuck driving circuit from keeping the CPU re-entering the NMI ISR forever. <S> The NMI must show activity to get another one. <S> The answer, of course, is that they both have their applications. <S> But level-triggered is the starting point and edge-triggered gone to because there's a special case. <A> Level based ISR's support Ack/Nak which is useful when you have many sources. <S> If you have many ISR's and many priority ranks in both SW & external HW with ranked priorities. <S> If all sources were edge OR'd you would still have to poll each to find which source set and then clear the interrupt. <S> So both edge and level have advantages in different architectures. <S> To avoid missing an Edge IRQ, it has to be enabled then tested immediately after at the end of the ISR. <S> Some level IRQ's can last long than the ISR and detected if expected. <A> It is when a "party line" is being used, together with open-drain or open-collector, when the level mode of operation makes more sense. <S> In this case, each "party" on the line activates their output when they need "service. <S> " <S> As the interrupt handling software goes out and polls (and then handles) <S> each device on the party line in turn, these devices release their hold and go inactive. <S> Eventually, there are no more (level) interrupts remaining, the interrupt line itself goes inactive, and the software handling interrupts goes "quiescent" again.
| One obvious situation where level based interrupts are useful is for the situation where the signal is already in that state when the code begins to monitor the signal.
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Is it a good idea to use a single heatsink for two IC's? I am using a 7805 IC along with a diode bridge Br805D (bridge rectifier) IC on a general purpose PCB, with a 220V / 50Hz AC input. Both the components tend to heat up really quick and I plan to set up a heat sink. My question is: Is it correct to use a single heat sink / fan combination for both the ICs or should I separate heat sinks for both? What are the potential concerns I should know about? <Q> Using a single heatsink is very common - sometimes it can even increase performance (thermal matching of discrete transistors for helping against thermal runaway or better matching). <S> However, there are a few things to watch out for: How big <S> does your heatsink need to be? <S> Are all your parts going to be putting out the maximum heat they can during normal operating conditions at the same time? <S> If not, you don't need as big a heatsink. <S> However, the math on figuring out what size you need can be a bit trickier. <S> What about isolation? <S> When using multiple devices on a single heatsink, you need to take care that you are not shorting out bits of your circuit. <S> Many packages have metal tabs to connect to heatsinks (such as the TO220 package). <S> However, these metal tabs are often also connected to a certain pin internally. <S> Check your part's datasheet to see how it is connected. <S> Take a look at the following schematic: simulate this circuit – Schematic created using CircuitLab <S> If we were to connect these two parts to the same heatsink, without isolation pads, we would be shorting the input transformer! <S> This is clearly not a good thing. <S> Therefor, we often use either separate heatsinks, or isolation pads. <S> These are pads commonly made out of thin pieces of ceramic (Mica is common) or some other electrically insulating material. <S> Make sure that when you use these pads, you also use the propper screws or protection sleeves for the screws! <S> If you isolate the metal tab, but don't isolate the metal screw, you will just short it out through the screw. <A> The other answers here are great, however I would also like to add... <S> CONSIDER EXPANSION EFFECTS <S> When mounting multiple devices to a heat-sink some care is needed to account for the fact that the heat-sink itself is going to expand at a different rate from the PCB substrate which your devices are presumably soldered to. <S> If that difference is significant it will exert a LOT of stress on the legs and solder joints of the devices. <S> This is particularly detrimental if the devices are surface mount devices where the mechanical connection is already poor. <S> Best case the joint will fail. <S> Worst, the pad and track will rip out of the board. <S> These effects can be alleviated somewhat by keeping the parts really close together. <S> Putting them at different ends of the same heat-sink is generally a bad idea. <S> However, I would still not rely on proximity for SMT parts. <S> A better solution to mount the devices to the heat-sink using a softer mechanical connection that allows an appropriate amount of slippage between the device and the heat-sink. <S> For example, heat-sink-clamps. <S> Use a sufficient amount of heat-sink compound to act as a lubricant to ease the slippage while maintaining the appropriate amount of thermal contact. <A> You can use a single heat sink with two components if The mounting tabs of both components can be electrically connected, or you explicitly insulate at least one of them. <S> The total power budget and max temperatures are still met. <S> Add up the worst case power dissipated by both devices, and see what temperature rise the heat sink will cause. <S> Make sure the result is acceptable to both components. <S> If you end up electrically insulating one of them, be sure to take the extra thermal insulation caused by that electrical insulation into account.
| Using multiple parts on a single heatsink could make thermal calculations a bit more tricky.
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Question about sensitivity list in VHDL I have a simple question about the sensitivity list. I read it on some books. It said that the sensitivity list is only important for the simulation. I don't quite understand. Does that mean, if I remove the sensitivity list, for the emulation on a actual FPGA it won't matter. If I remove the sensitivity list, how does the FPGA know when a process is triggered? <Q> Usually, the simulator cares about the sensitivity list. <S> But the synthesizer does not. <S> The purpose of the sensitivity list is to let the simulator know that a process only will change its outputs if one of the signals (A, B, C) changes. <S> That lets the simulator run more efficiently. <S> However, the synthesizer usually doesn't look at the sensitivity list; it just looks at the process's body. <S> That means your final FPGA or ASIC will behave differently than the simulator. <S> If you put extra signals in the sensitivity list, then the simulator will give you the right answers, but run more slowly. <S> So, the only thing the sensitivity list is for is to let the simulator run faster. <S> It's a hint to the simulator, so it can run more efficiently. <S> But the simulator and the synthesizer can ignore the sensitivity list, since it's just a hint. <A> Keep in mind that the "process" does not actually exist in the FPGA. <S> The process is merely an abstract description of the behavior of a chunk of hardware. <S> The synthesis tools create that hardware based on the behavior specified in the source code, including any process blocks. <S> The actual hardware will do what it does continuously . <S> In VHDL, anything that must be controlled by an edge on a signal (e.g., a clock) must be explicitly specified using 'event <S> qualifiers on signals as conditions inside a process (possibly hidden inside a function such as rising_edge() ). <S> This is completely independent of the sensitivity list of the process itself, which is indeed used only to trigger the evaluation of the process in a simulator. <S> Note that this is very different from the sensitivity list associated with a Verilog always block, which is a conflation of the two concepts that VHDL keeps separate. <A> Sometimes sensitivity list is only important for the simulation,if you lost a signal in sensitivity list you may get the wrong simulate answer. <S> But the real FPGA function will right if your logic is right. <S> You can called it ‘pre-sim and post-sim inconsistent’. <S> The real scared thing is your pre-sim is right but your post-sim is wrong. <S> That will cause your FPGA function wrong. <S> fpga <S> verilog <S> vhdl
| If you simulate a circuit, but leave important signals out of the sensitivity list, then sometimes the simulator will give you the wrong answers.
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How to differ output & input resistance of small-signal model BJT? I understand how input and output resistance of small-signal model for bipolar transistor is defined and how draw one.What I don't understand quite yet is how actually input and output resistance of BJT are separated. How do you know which resistances should be included for input and which for output resistance? Because, if you ask me, models like this one could be identified as one whole resistor, whose value would be considered as input and output resistance simultaneously. *For the record: ru - base-collector resistance, rpi - base-emitter resistance, ro - collector-emitter resistance, rb - base contact resistance, rc - collector contact resistance, rex - emitter contact resistance. <Q> Because, if you ask me, models like this one could be identified as one whole resistor, whose value would be considered as input and output resistance simultaneously. <S> Your model has three terminals, B, C, and E. A resistor has only two terminals. <S> You can't reduce a three terminal device to two terminals and consider it "equivalent". <S> And the output resistance is, by definiton $$ <S> R_{out}=\frac{{\rm d}V_{ce}}{{\rm <S> d}{I_c}}.$$ <S> You can work out what these values are from the model you showed. <S> If these two derivatives don't have the same value (and they don't), then you can't say the input resistance is the same as the output resistance. <A> The concept of input and output impedance is not just replacing a circuit with a single resistor. <S> To find input impedance, you connect a power supply to the input, say Vin and find how much current(say $I_in$) passes through your input $V_in$. <S> You get the ratio Vin/Iin and call it as input impedance a.k.a the impedance that the input voltage source Vin feels. <S> You mathematically (using KVL and KCL) represent this ratio in terms of circuit parameters ( Thevenin's equivalen t). <S> Whatever circuit parameters (resistors and capacitors) are in this equation, they become part of input impedance. <S> Similarly, you find output impedance. <S> Few circuit parameters may be a part of both inputs as well as output impedance at the same time. <S> This varies from circuit or circuit. <A> Input impedance is defined as impedance as looked into input terminals when output terminals are open circuited while output impedance is defined, as impedance looked into output terminals, when AC input is short circuited. <S> Both will not be same in case of BJT. <S> Consider image below.. <S> Thus, even if you reduce equivalent circuit to somewhat similar i shown in pic, ip and op impedance will still largely differ owing to difference in conditions in which they're defined.
| If you are using the BJT in common emitter configuration, then the input resistance is, by definition, $$R_{in}=\frac{{\rm d}V_{be}}{{\rm d}{I_b}}$$
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Altium Designer, using signal layers as power planes In cases when a 4-layer board is space limited, and there's a lot signal traces in the whole board, "power plane layers" is not very usefull to distribute power rails. Because it is used only to draw solid regions in a negative format. Therefore it is very hard to place some signal traces along the power plane. I've been thinking, what's the difference to use a "signal-layer" instead of "power plane layer" which offer the capability of non negative format? You can also draw as solid region as you want but you are still able to put a few traces that they are difficult to stand on the external layers due to complexity. <Q> While there's a lot of good answers already, I felt I should add some distinctions to "when should you use which, generally" and "when should you move from the ideal, because Altium." <S> As came up before, it allows programs to assume solid copper in complex analysis. <S> It's also more space efficient when defining large copper areas. <S> In some cases manufacturers can use more reliable negative-defined processes, though these are somewhat outdated and inverting a plot is negligible work anyway. <S> When you send a plane definition layer the entire supply chain assumes high levels of copper coverage, which is key in getting impedance control right at the least amount of cost. <S> However... <S> There are programs, amongst which Altium, in which you cannot always set vital parameters on your plane layers, that you can set on signal layers and/or for copper pours. <S> For one example, Altium removes via pads by default on power plane layers, without any option to force it to place them anyway. <S> This is all well and good if you're designing a power-supply or an audio amplifier (a manufacturer that needs them for processing stability will just add a 0.05mm ring anyway). <S> But if you're designing signalling for high MHz and GHz ranges you may actually want full control over all bits of your signal path, including the way vias couple on plane layers. <S> Similarly you cannot easily force the program to pull back a power plane from cut-outs or edges by a configurable distance. <S> This leads me to the advice, rather than just condoning, to always consider signal layers instead of plane layers in Altium specifically (not as a general argument!) <S> , unless you need trace impedance analysis or other plane-tied features. <S> But if you're working professionally enough to want full analysis of done designs, you are likely to be in a domain of frequencies that someone should arrange a license for a tool like ANSYS or Microwave Studio, which works fine for signal layers as reference plane. <A> The plane feature is good if you have arrangements that are best drawn in negative. <S> Splits are handled nicely, for example, and highlight when the net is selected. <S> If you must run signal traces you might find it easier to do polygon pours on the layer in question for the power regions. <A> The way I route my power planes is using signal-layer and draw power planes with polygon pours. <S> If you want to add a section for signals or put two or more power planes in the same layer, its as simple as drawing another pour. <S> The only layer that I have as a power plane layer is a ground layer because this one is going to be solid and will almost never have traces running through it. <S> I've been thinking, what's the difference to use a "signal-layer" instead of "power plane layer" which offer the capability of non negative format? <S> The difference is if the plane is going to be mostly solid on that layer, then use a power plane layer otherwise use a signal layer. <A> While I don't know the complete list, here's one example. <S> Altium can calculate trace impedance only when the trace is next to a plane. <S> At the same time, if you don't have to use Altium features that rely on designated plane layers, then your approach is just fine. <S> I've done this <S> used it with OrCAD: <S> designated inner layers as routing layers, then used copper pours for ground and power planes. <S> related <S> Can I lay traces on internal layers of 4 layer PCB?
| Normally Power Plane layers offer a nice and well defined inherent separation of rules and restrictions.
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How can I reduce the current draw for a dimmer with a triac for an immersion heater? I've been working on variable control for a immersion heater. The problem is that the heater consumes a lot of current, about 10 to 15 Amps. I already have to circuits to control it, the problem is that both circuits overheat to the point that they already burned some components. The heating comes from resistance in the triac circuit while the components support more than 25 Amps (Triac and the Mosfet). Here are the two circuits that I've tried. I need a way to avoid the overheating of the one or both circuits. How do I avoid overheating in these circuits? <Q> Current draw is determined by the heater power you need. <S> If you can reduce the heater watts, do that. <S> I'll assume that you have already sized the heater. <S> Both triacs and IGBTs have relatively high losses, and you can expect roughly 1W/A of losses for a triac. <S> The IGBT may have a bit more, and the required bridge will have still more. <S> Your best bet is to use an adequately rated part (for example a 40A triac for 25A use) and put a large enough heat sink on the part to keep the temperature rise above the highest anticipated ambient temperature reasonable. <S> For example, if losses are 25W and if you need to allow for 50°C ambient and you think 100°C case temperature is acceptable <S> you will need a heatsink that has a thermal impedance of less than 2°C/W. <S> With natural convection that is a fairly large (and fairly expensive) <S> heatsink . <S> The below photo is one that is 5 <S> " x 3" <S> x 1.5". <S> Get the maximum power dissipation for a given current from the datasheet as with the typical 40A triac <S> linked: <S> For this one, as I said above, the dissipation at 25A RMS is maximum 25W. <S> If your application can deal with a bit of variation, it may be better to use a relay or contactor and switch the heater with a on/off cycle of appropriate times. <S> A mechanical contactor will have little in the way of losses but you have to be concerned about lifetime, as the contact will wear out after perhaps 100,000 operations. <S> It's possible to use a triac to do the switching and short it with a relay (hybrid switching) with appropriate timing. <A> The BUZ41A is rated at 4.5 A so that circuit is totally inappropriate for your application. <S> For the Triac based circuit you need a heatsink as the forward voltage drop will be in the 1.3 - 3 V range resulting in a potential power dissipation up to 45 Watts at 15 A. <A> In any semiconductor switch the junction temperature rise ΔTj is; a product of two resistances; electrical <S> On resistance sometimes called Rs, Ron, ESR, Rce or RdsOn <S> [Ω] thermal resistance from junction to ambient Rja = Rjc+Rca <S> [°C/W] <S> Such that the rise above ambient; > ΔTj = <S> Pd <S> * Rja = I² <S> * ESR <S> * Rja <S> < Since Triacs are composed of 2 BJT's the drop is <S> 1 P-N jcn + Vce/Ice where the diode drop has an ESR and the Rce is another ESR. <S> All transistors can be modeled as bulk resistors in saturation <S> Vce/Ice=Rce , ~ plus some offset voltage <0.1V) <S> MOSFETS are rated for RdsOn in a similar way. <S> So you have 3 choices ; lower either R or both. <S> Always start with a spec: <S> like ΔTj=25 <S> °C then choose your design accordingly. <S> let ΔTj=25°C= <S> I² <S> * ESR * Rja <S> so for Imax=25Arms then ESR*Rja=25/25²= 0.04 consider cost of Rja=0.1 <S> ESR= 0.4 <S> vs Rja=1 ESR=0.04 <S> then decide. <S> since Rja=0.1° <S> C/W is like a "big $" CPU heatsink, ignore & try ESR=0.01 <A> I have reduced power to a heating element controlled by a PID/SSR by inserting a suitable power rated diode in series with the heater which reduced the average current to the load by half. <S> If your application allows that large a power reduction then it is an easy change.
| A smaller heatsink will be possible if you use forced convection (a fan).
|
4 Wire Strain Gauge to 2 Wire Conversion/Connection I have the WK-06-12BT-350 ohm WK125BT Strain gauge from Vishay the Gauge comes equipped with 4 wires ( High endurance lead wires). The following document describes a similar gauge: Strain Gauge is it possible to convert the following gauge to a 2 wire connection? are the additional wires negligible? or they might cause measurement errors if left open? The acquisition machine can accept 2 or 3 wiring scheme but not 4 furthermore the strain gauge will be part of a quarter bridge connection for a fatigue test running at am excitation voltage of 5V Thank you Much appreciated <Q> The WK125BT Strain Gauge you have is only a single gauge, not a bridge (half or full). <S> So if it has 4 wires, then these would allow you to use the 4-wire, single-gauge method described in your second link if you want to . <S> Using the 4-wire method lets you achieve greater accuracy by allowing you to (mostly) ignore the resistance of the wiring. <S> In this case you would use a constant-current source to energize the gauge through the outer wire pair, and take your measurement across the inner pair. <A> The most basic connection using only 2 wires gives you a current path through the strain gage with no compensation for added wire resistance. <S> Four wires gives you 2 wires for each connection, one set for the excitation current and one set to measure the voltage drop across the strain gage. <S> If all 4 conductors are the same length and wire gauge then you can use a 4-conductor cable in a 3-wire connection by leaving one conductor unconnected/open. <S> If your instrumentation has the ability to 'zero' the reading to offset the wire resistance then 2 wires should work ok. <S> In that case you could use 1 wire to each side of the strain gauge, and leave unused wires disconnected. <S> The electronic instrument would basically be providing a differential reading (the amount of change from your zeroed condition). <A> A strain gauge bridge is usually a Wheatstone bridge with 4 elements, or it could be a simple half bridge with only two elements. <S> In both cases there is no such setup with two wires. <S> You could also make a Wheatstone bridge with three fixed resistors and one single gauge, but it is recommended to put everything close to the gauge, so again you have 4 wires. <S> Note that power supply has two wires and the voltmeter in the pictures <S> is DAQ input 2 wires
| Three wires gives you one extra conductor of the same length and gage as the other two, which is used by your instrumentation to "measure" the lead resistance of one lead, then predict and compensate for the lead resistance in the other two leads during the data acquisition.
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Maximum rotation rate for digital compass How well do digital compass ICs (e.g. HMC5843 ) cope with when rotating rapidly? If I were to rotate one at 1000 RPM, what kind of signal could I expect to get from it? Would the measured bearing lag considerably, or would the chip simply be unable to calculate a heading at all? Added: please note, I'm not asking about the sample rate. That is mentioned in the datasheet for every digital compass. What no data sheet seems to mention is if there is a fundamental bandwidth limit for this magneto-resistive technology. <Q> The sampling rates available on magnetometers such as the HMC5843 do not come anywhere close to the bandwidth of the underlying anisotropic magnetoresistive element. <S> For proof, we can look at the datasheet for the HMC1001 , an analog AMR magnetometer, also from Honeywell. <S> The datasheet states that the bandwidth is from DC to 5 MHz typical. <S> Incidentally, the HMC5843 is obsolete. <S> You'd be better off with something like the HMC5983 (which is technically obsolete, but widely available) or the MMC5883MA. <A> The datasheet you linked specifies the maximum output rate as 116Hz. <S> 1000RPM = 16.666 revolutions per second. <S> 116Hz sample rate / <S> 16.666 rev/second means you would have 6 samples per revolution (6.9 roughly, but the next sample would be into the next rotation). <S> I would recommend using a gyroscope in conjunction with the magnetometer. <S> You can integrate the rotation rate to get a total angular displacement and use that to calculate the bearing, referencing the magnetometer when for corrections. <A> This isn't an answer per se, but rather an answer as to how you can find out. <S> Try it by experiment. <S> Connect together a small battery, micro controller and the compass. <S> Assemble on a stout reinforced pcb, laid out symmetrically <S> Add <S> a spindle of something like M10 studding or bolt and nuts. <S> Mark and record bearing of PCB wrt a proper compass. <S> Rotate in drill for brief time. <S> Suggest some form of eye protection. <S> Record finish bearing. <S> Repeat several times from different starting bearings. <S> The micro controller is programmed to read the compass and store the readings in persistent storage. <S> After each test you off load the readings for analysis. <S> You should be able to correlate the start and stop positions recorded electronically with the manual ones. <S> You should also be able to see if the intermediate results make sense. <S> You could further refine the experiment by incorporating an optical encoder disc onto the studding. <S> That would allow more accurate intermediate calibration. <S> This might all seem like hassle, but might be the only way to satisfy yourself of the device's capability. <S> Call it product R & D. Are you developing a better Sidewinder?
| With the 116 Hz sampling rate of your magnetometer, you're not going to be able to detect any bandwidth limitation of the sensor itself.
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Designing 0- 300 mA current source with 12v Voltage supply (load = constant 40ohm) I'm trying to design a 0 to 300 mA controlled current source, the problem is that the supply voltage is a normal adapter with a 12v output. I'v already tried several op-Amp based circuits but the voltage drop over the load exceeds the power supply in all of them. suggestions for a circuit?Load is a constant 40ohm device.Thank you all <Q> You say you want a 0-300 mA current source, but that the load is a constant 40 Ω. <S> There is therefore no distinction between a 0-300 mA current source and a 0-12 V voltage source. <S> The biggest problem you have is that your circuit can't drop any voltage at full output since you only have 12 V to start with. <S> That's not possible, so what you asked can't actually be done. <S> However, you can get reasonably close. <S> To minimize the voltage drop across the pass element, use a MOSFET. <S> It's not hard to find devices that can go down to a few mΩ with full gate drive. <S> 20 mΩ, for example, is 0.05% of your 40 Ω load. <S> That would be within measurement error for most purposes. <S> You didn't give any constraints on how this current source is supposed to be controlled, so we can pick something simple. <S> A pot setting the gate voltage of the MOSFET provides control over the current. <S> Here is a circuit that meets all your specs, except that it doesn't quite put out 12 V or 300 mA with exactly 12 V in: This particular MOSFET can handle up to 20 V and goes down to less than 10 mΩ with 10 V on the gate. <S> Only you can say whether that is close enough. <S> Again, your specs are actually impossible to meet. <S> The control is achieved by rotating the pot (R1). <A> This would be my starting point: - Choose R1 to be low in value hence it won't drop too much voltage in sourcing current to the load. <S> If you chose 1 ohm then to get 300 mA through the load, Vin would be 0.3 volts and the drop across R1 would also be 0.3 volts. <S> The drop across the transistor would be about 0.2 volts at best so, to deliver 300 mA into a 40 ohm load requires a supply of 12.5 volts. <S> So you might as well bite the bullet and generate 15 volts from the 12 volts using a dc-to-dc converter and there are plenty around. <S> If you pick an isolating one it gives more options because you could use a converter that produced 2.2 volts at the output and float that output up to produce a rail of 12 volts + 3.3 volts = 15.3 volts. <S> Plenty of options. <S> If your load can drop significantly below 40 ohms then you might need a heatsink on the PNP transistor. <S> If you want to look at other options regards generating a constant current these google images should help. <A> Any circuit you can build will have some voltage drop. <S> 300mA <S> * 40 ohms is 12V which is your nominal power supply voltage. <S> So, you have two main choices here- <S> Maybe that ends up being 0.5V after you take into account tolerances. <S> That is tractable- <S> the maximum drop across your pass device (transistor, probably) plus the voltage drop across the measuring resistor must not exceed that 0.5V. <S> If the transistor needs 0.25V <S> you can use a 1 ohm resistor and meet the spec. <S> The lower the resistor value the less voltage for your feedback, so more errors will creep in. <S> The trade-off will depend on what your required stability and accuracy spec is. <S> The second choice is to boost uo your 12V using some kind of switching supply to something high (eg 15V), in which case you can choose a nice big feedback voltage such as 3V (use a 10 ohm resistor) or 2.25V or 2.5V to match a reference voltage. <S> Such a circuit will produce more EMI, consume more power and will be more complex than a simple current regulator, but it will meet your stated requirements. <S> Edit: <S> Since others have supplied (perhaps dubious) schematics, I'll throw one or two in here for good measure: <S> Boost converter: <S> If you follow the 34063 data sheet calculations you can easily determine the values of the components not shown. <S> Current source ( note source, not sink): <S> The op-amp <S> I show is an unusual high-voltage rail-to-rail input type. <S> There are other ways of doing it <S> but this is simple. <S> An ordinary LM324 will not work, and certain FET op-amps will only work if you are lucky. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> If your load is actually a constant resistance and you don't need to adapt to any changes then you could build a 0-12V supply, but it would not be significantly simpler than what is shown.
| Of course a better answer would be to start with a power supply a little above 12 V so that the pass element is allowed to drop some voltage at full load. If your power supply is a bit low and/or your load resistance a bit high you will never get to 300mA even with a direct short. the easiest is to compromise the spec and limit the current to maximum (say) 275mA which gives you a nominal 1V to work with.
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What is the function of the capacitor between ground and 12V? This is a switching circuit used for a switched reluctance motor. In this circuit there is a 47μF capacitor on the right side. What is the capacitors function in this circuit? <Q> The 47-uF capacitor is not used to supply the inrush current, which would be needed to start the motor: it's value is too low to do the job (the IGBT handles 7A, so the motor will be quite powerful). <S> It is likely used as a noise suppressor. <A> Put this cap as close to your motor as possible, or at least the connector for the motor on the PCB. <A> This capacitor is used to keep the 12V supply stable as possible. <S> According to the equation P=VI, when the motor draws a high current, the voltage across the motor drops, when the voltage drops, your motor speed reduces. <S> Therefore if you have a capacitor in parallel to the motor, in times of voltage drops, your capacitor provides that necessary current to the motor so that your motor voltage remains stable, thus keeping a constant speed.
| This capacitor is used to keep the 12V supply 'stiff' -- i.e. reduces voltage dips when the motor draws current -- to prevent fluctuations in torque/speed under load.
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Why do laser diodes need temperature controllers? Why not just cool them? High quality laser diodes often require temperature controllers plus a "TE" (thermoelectric unit) that cool the diode and maintain it at a particular temperature using a Peltier cooler. Why not just cool the diode without a controller? Is there some problem with the diode being too cold? <Q> It depends. <S> Some lasers will indeed get less efficient if they're too cold. <S> In some systems, you may want to keep the laser at a fixed operating wavelength, and this requires operating the laser at a fixed temperature. <S> Cooling the laser below 0 <S> C could cause ice to form on the laser (if it's not in a hermetic package), impacting performance and reliability. <S> If none of those apply, excessive cooling will simply consume power unnecessarily, increasing the power consumption and the demand on the cooling system for the whole system or plant. <S> A TEC controller can be as simple as a 6-pin microcontroller, so it doesn't add much cost to the system, after you've already budgeted for the TEC module itself and the components to drive it efficiently, so even if the benefits are small it might still make sense to add the controller. <S> The controllers I am talking about are ones like the Newport 325 which can cost as much as $1,200 new. <S> The $1200 mostly pays for the display and user interface and a remote control interface (GPIB, USB, or ethernet), as well as the amortized cost of R&D to develop the design and its software, NRTL listing, etc. <S> You'd buy one of these instead of building your own if you're only building a small number of systems (maybe only one). <S> $1200 is less than the cost of designing, building, testing, and debugging a new controller design, even if the cost of materials for the new design approaches $0. <S> Many laser users (say, biology researchers, or even electrical engineering researchers in other specialty areas) may not even be skilled in PCB design, control systems, embedded programming, mechanical engineering, and other skills needed to develop a controller on their own. <A> The main purpose is to get a constant power and wavelength output. <S> For example, the commercial DFB laser diodes have a typical wavelength to temperature coefficient of ~ 0.1nm/degree, for some applications that are wavelength sensitive (DWDM, coherent applications using narrow linewidth lasers), the temperature must be held constant with very low noise (<1e-3 Kelvin). <S> For these applications, an ultra-low noise current driver is also needed (CW most of the cases). <A> for CW solid state lasers the reason is to maintain a constant current flow in a device that is not linear in current draw
| The temperature controller is not only for "cooling" the laser diode to get higher efficiency, but it is also used to set the laser diode (actually the NTC near the laser chip) at a constant temperature.
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Why does placing resistors in parallel decrease the resistance? This question has been asked before but I don't understand the answers to it. They say that by adding a resistor, it provides an additional path for current to flow and decreases the resistance. I understand how an additional path decreases resistance but how is adding another resistor adding an additional path? For example, say a parallel circuit branches off into two paths. You then add resistors to those two paths. How does that decrease the resistance? It isn't adding a new path? <Q> There's a waterfall pushing 600 gallons per second. <S> You divert a small section through a small water hose giving you 5 gallons per second. <S> Your current is 5 gallons per second. <S> Add another of the same size hose. <S> You more have 2 currents that add up to 10 gallons per second. <S> It's like a single hose that does 10 gallons per second. <S> This is an approximation of how parallel resistors work. <A> The second resistor is a second path for the current. <S> Consider the following circuits: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The first circuit, with V1, is your initial circuit. <S> It's one volt across a one ohm resistor, with one amp of current. <S> Next, you add a second one ohm resistor, as in the circuit with V2. <S> Both resistors have one volt across them, so both resistors have one amp through them. <S> That's just ohm's law. <S> Since there are two resistors, that's a total of two amps coming from the source. <S> The circuit connected to V3 shows an identical circuit from the source's point of view: one volt over half an ohm is two amps, just like with the two parallel resistors. <A> A parallel circuit decreases the resistance compared to a circuit where there are no parallel branches. <S> Say you had 2 100 ohm resistors in parallel. <S> Then you add a third 100 ohm resistor in parallel with the first two. <S> Now the current has three path to go through compared to two, so it increases. <S> In the example you provided of adding resistors to a parallel circuit with no resistors initially, that would decrease current.
| Adding parallel resistors only increases current if there was not initially a parallel branch at all.
|
Is MPPT used with very small panels Say for example an inexpensive solar calculator or other such similar device with what appears to be lets say a single 2.4V 6uA (microwatt) panel? Is an MPPT used to get the most panel rather than using the panel directly? Is there another solution for this class of panel? <Q> When you're talking about "inexpensive" consumer devices like solar powered clocks and calendars the answer (in general) is no. <S> However, there is a new generation of energy harvesting devices that are very high efficiency designed to extract much more of the energy from devices like a small solar panel. <S> You can see an example of such a device here: <S> TI Energy Harvesting Device <A> You have to balance: <S> Price of the MPPT circuitry (which is not that cheap) versus... <S> Price of a larger solar cell <S> For other devices which use energy harvesting, like a dynamo-crank flashlight, this is also likely... because you're going to buy the whole thing for $2 on aliexpress or in a supermarket, which will purchase it at a cost of about $1 from china. <S> And their manufacturing cost does not change if you got to crank it a little harder because it is inefficient. <S> Now if you have $200 worth of solar panels, then it is extremely likely that the addition of a MPPT circuit worth a few $ in quantity will be highly profitable. <A> To add to @peufeu's answer in addition to cost issues there is the fact that MPPT can only improve the efficiency by a limited amount - say 10-20% <S> and it takes power to run. <S> If it takes more power to run than the additional power it provides there will be a net loss. <S> With high power arrays the overhead of the MPPT circuitry is acceptable, I doubt if you could run any MPPT circuitry on the 10uW <S> or so you mention in the question.
| For a solar calculator, which draws very little power, the extra solar cell area required is likely much cheaper than a smart MPPT circuit.
|
Do wires need to share similar characteristics in order for crosstalk to occur? My understanding of crosstalk is how a radio transceiver works. You basically tune your transmitter and receiver to a similar frequency in order to get a cross-coupling of energy (the transmitter's energy gets coupling over to the receiver, hence the receiver is able to listen to what is being broadcasted). Is this analogous true for crosstalk in parallel unshielded, untwisted wires where the 2 wires have similar inductance and capacitance? <Q> Your question is a bit confusing since your title doesn't quite match what you asked. <S> You also makes few statements that are incompletes. <S> You basically tune your transmitter and receiver to a similar frequency in order to get a cross-coupling of energy Receiver and transmitter are often complex system including many subssystems. <S> They includes modulator/demodulator, antennas, filters etc. <S> So "tuning" them is a complex task that requires many parts to interract together. <S> Is this analogous true for crosstalk in parallel unshielded, untwisted wires where the 2 wires have similar inductance and capacitance? <S> A capacitance involve two elements. <S> A single wire doesn't have a capacitance. <S> It has a capacitance with something else. <S> This being said, to answer your title question, no, two wires doesn't need to share similar charateristic (I assume here that you refer the resistance and inductance) to have crosstalk. <S> See Wikipedia's definition : <S> In electronics, crosstalk is any phenomenon by which a signal transmitted on one circuit or channel of a transmission system creates an undesired effect in another circuit or channel. <S> Crosstalk is usually caused by undesired capacitive, inductive, or conductive coupling from one circuit, part of a circuit, or channel, to another. <S> You need some sort of coupling between the two elements. <S> In the case of long unshielded, untwisted wires, there is severals sources of coupling : <S> Capacitive coupling. <S> The 2 wires act as a capacitors. <S> Capacitor have low impedance at high frequency. <S> A fast time-varying signal like an aggressive digital edge or a high frequency radio signal in one conductor would affect to other conductor. <S> Capacitive coupling has nothing to do with electromagnetic waves. <S> Inductive coupling. <S> Long wires next to each other also have an inductive coupling, so they act as transformer. <S> Big currents in one conductor will create a magnetic fields that will induce an electromagnetic force in the other conductor. <S> ElectroMagnetic coupling (or interference). <S> EMI will moslty be present if you have big changes of current in time. <S> The length, material, shape of the wire will all affect its effectiveness to transmit/receive EM waves. <S> Altough you seemed interested in "tuning" your wire to avoid interference, EMI is not restricted in bandwidth. <S> A current surge in a wire will generates noise in a very wide frequency range. <S> Even if you have an antenna tuned for a specific frequency, it will receives noise anyway. <S> Hope <S> I could make things a little clearer. <A> Lets put some numbers into an example of magnetic coupling, between an rectangular loop of some size: Area, and a long straight wire in the plane of the loop. <S> Vinduce = { MU0 * MUr * Area / (2 <S> * pi * Distance) } <S> * dI/dT. <S> The loop Area can be 1cm long with 1/3 of 1/16th inch height above GND plane; this would be any 2 adjacent layers of a 4 layer PCB of total thickness 1/16 ". <S> The "long straight wire" could be ANY trace on the PCB, containing the aggressor (transmitter of interference) energy. <S> Suppose your loop is 100cm from the 2,000 amp bus of an electric speed controller, switching in 1uS. <S> Thus dI/dT <S> = 2e9 amps per second. <S> Consider the entire PCB..... <S> the GND structure (chopped up badly, or a clean plane) to be 4" by 4". <S> Vinduce = <S> 2e-7 <S> * Area/Distance <S> * dI/dTVinduce = <S> 2e-7 <S> * <S> 0.1meter <S> * <S> 0.1meter <S> / 0.1meter <S> * 1e+9Vinduce = <S> 2e-7 <S> * 0.1 <S> * 1e+9 = <S> 2e-7 <S> e-1 e+9 = <S> 2e1 = 20 volts induced in the GND plane........... <S> massive eddy current. <A> There doesn't have to be any similarity between radios, either — radios include tuned circuits or other such mechanisms not because it is necessary but because it is useful, so that multiple transmitters can avoid interfering with each other and receivers can select just one transmission. <S> (Strictly speaking, any possible radio is somewhat frequency-selective, if nothing else due to the size of its antenna. <S> But this is a much, much weaker effect than deliberate filtering.) <S> You specified untwisted parallel wires, but I'll mention anyway that there is an interesting case in twisted pair wiring. <S> As you may know, the purpose of twisted pair is to ensure that the two wires in the pair are (approximately) equally influenced by reasonably uniform outside fields, so that the differential signal carried on the pair is unaffected. <S> But if multiple twisted pairs are bundled into one cable, there can be crosstalk because both pairs are alternating. <S> Therefore, each pair in such a cable is twisted at a different rate, so that the same effect applies (the lineup of the conductors goes from A+ A- B+ B- to A+ A- B- B+ and back) though over longer distances than the basic twist rate. <S> So, that's a case where a deliberate dissimilarity prevents crosstalk. <S> But it only applies when you have wires.
| No, there does not need to be any similarity between wires.
|
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