source stringlengths 620 29.3k | target stringlengths 12 1.24k |
|---|---|
Rotary encoder to 2 buttons Please excuse me if this question is a very obvious one, I have searched on here but my electronics knowledge is very very basic. Here's what I'm trying to do, I have taken an old Android phone apart, which has 2 surface mount buttons (Volume Up/Volume Down). I am trying to integrate this phone into a different case so it can act as a head unit for a car, and what I'd like to do is wire an incremental rotary encoder to these 2 buttons, so when I turn it clockwise, the volume up button gets 'pressed' and viceversa. I'm a bit lost when it comes to rotary encoders, I have read up on gray code and all that but I'll admit I'm still very lost. <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> 2-bit rotary encoder waveforms. <S> A rotary encoder works by outputting a pair of quadrature (90° offset) <S> pulse trains. <S> These are used to run an up-down counter to keep track of the position. <S> A suitable control algorithm would be as follows: <S> Track the current state of 'A'. <S> If the state changes to 'high' then: Look at input 'B'. <S> If 'B' is low then count up. <S> If 'B' is high then count down. <S> You won't need to keep track of the actual count. <S> simulate this circuit Figure 2. <S> Extracting up and down pulses. <S> This schematic is to get you started on a solution. <S> A CD4013 D-type flip-flop will update its outputs on each positive-going clock transition based on the 'D' input. <S> Depending on the direction of the encoder rotation either Q or /Q (not Q) will be high. <S> ANDing (or NANDing) <S> these signals with either A or B will give either an UP or DOWN pulse train. <S> Figure 3. <S> Extract from CD4013 datasheet . <S> Your next problems will be: How to interface these with the phone. <S> Have a look at CD4016 switches. <S> What happens if the encoder stops with the UP or DOWN stuck on? <S> This may not be a problem if the encoder detents line up consistently with an on or off position of one of the A or B contacts. <S> Over to you. <S> I'm done. <A> This simulates nicely in LTspice: <S> and you'll need the following links to get to the files you'll need to run the sim. <S> Download all of the files into the same folder and then start LTspice by left-clicking on the .asc file. <S> https://www.dropbox.com/s/bubgfvl8ewj5v9g/Quadrature%20decoder.asc?dl=0 https://www.dropbox.com/s/lonoh1d3bucnf1j/74hc04.asy?dl=0 <S> https://www.dropbox.com/s/i9tsv0cjzoof21j/74HC.lib?dl=0 <S> https://www.dropbox.com/s/9h28abp72zhlynj/74hc00.asy?dl=0 https://www.dropbox.com/s/42dmlx4enjzu1op/74hc74.asy?dl=0 <A> What you're describing might be simple if the volume up/down buttons directly operated counter circuitry, but it's unlikely that they do so. <S> More likely, the buttons are sensed by software within the phone that attempts to make the user interface work in a manner which would be helpful for a human operating the buttons (e.g. the longer a button is held, the faster it repeats) but would make it harder to control electronically. <S> I'd suggest that the most reliable approach is to probably use a microcontroller to send various lengths of up/down pulses to the phone at various rates until you figure out the fastest rate at which pulses can be sent while getting predictable behavior. <S> Then have the microcontroller keep count of how many times the rotary encoder has moved left or right, as well as how many up and down pulses it has sent, and send out pulses whenever the numbers don't match up. <S> This will be harder than using a simple circuit, but it will mean that turning the knob four clicks to the right quickly will have the same net effect as turning it four clocks slowly. <S> Other approaches are likely to, at best, yield behavior equivalent to holding the up button while you're turning the knob clockwise (but ignoring the speed) or holding the down button while you're turning it clockwise (again ignoring the speed). <S> Not nearly as helpful.
| To get an encoder to do what you want you will need to add some logic to see if it's counting up (clockwise) or down (anti-clockwise) and pass the pulses to the appropriate button input.
|
Altium off sheet connector vs net vs ports I have difficulties to understand what are the differences between a net, a port and a off sheet connector. I particularly don't understand when to use a port and when to use a off sheet connector when there are multiple sheet. I see sometimes people use port and sometimes they use off sheet connector, while I still read that only net can be used in some situations. <Q> A net is a connection name. <S> If you have a signal on one side of your schematic with a net name of "SIG_A" and you have another net name of "SIG_A" on the other side. <S> Those two nets are connected. <S> It's as if there is a wire that ties them together. <S> Nets tend to be local to a schematic (unless you are using a power net - which MAY be global). <S> An offsheet connector allows connections to be made horizontally. <S> What this means is if you have a large design that can't fit into one page cleanly, you can use off sheet connectors to "continue" your signals to another sheet (but on the same level). <S> It's almost like an extension of the same sheet. <S> Ports allows connectors to be made vertically. <S> What this means is that you can create sheet symbols that represent your sheet, and connect them together via ports. <S> I tend to use a multi hierarchical design because allows me to see how circuit sections or typologies are connected together <S> and so I tend to favor ports. <S> But on large designs, I use off-sheet connectors as well. <A> It's a matter of style. <S> Altium supports different styles depending on preference settings. <S> Look for a setting called Net Identifier Scope. <S> The location might depend on which version of Altium you are using. <S> ( source ) <S> In my experience, global identfiers works well for small designs (up to 4 or 5 sheets). <S> Hierarchical scope works well for multi-channel designs. <S> For larger designs I tend to use "Flat" scope. <S> But bondage-and-discipline fanatics might prefer hierarchical style. <S> I believe that off-sheet connectors are just a graphical variant of a port. <S> They're not normally used in Altium designs, but they might be used if you import a design from another program. <A> Net (Net Label) <S> This names a given net. <S> The Net name is only valid in the scope of a sheet. <S> If you have multiple sheets where you give a net the name "SUPPLY", they will not be connected, because they are local. <S> From a programmers point of view you could see them as local variables inside a function (sheet). <S> Port <S> This adds a port to the schematic. <S> If you use multiple schematic sheets, the sheet symbol of the sheet will get a new port. <S> The port can be used to connect signals in a multi-channel system (see this and this ). <S> A port is used to connect signals between sheets in a very controlled way. <S> For a programmer it resembles strongly to function arguments and return values. <S> Off-Sheet Connector <S> Honestly, I have never used them. <S> For me they look like "global" Nets. <S> Something like GND and supply symbols. <S> For better readability I would not use them if I don't have to. <S> But that's something you have to decide on your own. <S> For a programmer they look like global variables (stuff that can be accessed from everywhere in the program). <S> I guess they don't make it easier to reuse schematic modules. <S> Summary <S> I don't know your background, but if you have at least a little bit of experience in programming, you could see child sheets as functions in your program. <S> And you can use the net labels off-sheet connectors and ports exactly as you would use variables. <A> I've used upto 8 sheets in my projects and I never used an off-sheet connector. <S> Keep two things in my mind: <S> Two, use Port Connector for global connections, that is connecting different components/parts on different sheets. <S> This always works well for me, either working on flat or a hierarchical design.
| One, use a Net Label for local connections, that is connections within the sheet.
|
Why would I use 2N3904 instead of 2N4401? 2N3904 and 2N4401 appear to be very comparable parts in all specs. 2N4401 has a higher current rating, but otherwise they look to be about the same in price and everything else. Obviously it's difficult to give a 100% universal answer since both parts are made by multiple manufacturers. But insofar as it is possible to say, is there any reason one would use a 2N3904 instead of a 2N4401? <Q> There may be good reasons for choosing the 2N3904 when working at lower current. <S> The 2N4401's higher current capability comes at a cost - higher capacitance and much longer turn off time at low current. <S> Emitter-Base Capacitance:- <S> 2N3904 = 8pF <S> 2N4401 = <S> 30pF <S> Storage time:- <S> 2N3904 = 200ns <S> @ <S> 10mA <S> 2N4401 <S> = 225ns @ <S> 150mA <A> I use the 4401/4403 as my jellybean go-to transistors when it doesn't matter, and most of the time it doesn't. <S> As you say, the 4401 and 3904 have similar specs, but there are differences. <S> Most of the time the 4401 is good enough, so I "standardized" on it since it's more robust <S> and I run into 100 mA applications more than 100 µA applications. <S> Once you pick something as your jellybean part, you want to use it whenever it's good enough. <S> You don't want purchasing and manufacturing to have to deal with more different parts than necessary. <S> Personally I think the 4401 is a better choice for this since it will be a fit to a few more applications than the 3904, at least for the things I tend to do. <S> I therefore don't use a 3904 at all. <S> In the relatively unusual case where I need a lot of gain at very low currents, I'll use something else even more suitable. <S> The 3904 is just too close to the 4401 that I use in volume to be worth stocking. <S> If you decide on the 3904 as your jellybean NPN, then you might still use a 4401 when you need more current than the 3904 can handle, up to a few 100 mA. <S> That's because the 4401 is a really cheap part in that current range. <S> Beyond that, you're going to use something else, probably a power transistor, anyway. <A> I have used both. <S> But for different applications. <S> For any audio pre-amplifier type of application, 2N3904 tends work better to bring the mic level signal to line level. <S> The 2N4401 I use for higher current amplifier where you want to drive some relay or opto-isolator.
| The 3904 is better at really low currents, but the 4401 can handle higher currents.
|
Can I drive a stepper motor like a brushless motor? If I want to accelerate my stepper motor at maximum acceleration, can I drive it by watching the position of the rotor with an encoder and stepping the motor whenever it completes the previous step? I believe that this is how BLDC's are driven. <Q> Yes, you can run a stepper motor closed loop, and that's the way to get the maximum performance. <S> However, that's not what steppers are designed for. <S> They are optimized for open loop control. <S> You might be better off with a geared BLDC motor than a stepper motor for what you're trying to do. <S> One problem with closed loop control of a stepper motor is that stepper motors have many poles. <S> This means a complete magnetic cycle is only a small angle of rotation, often just a few degrees. <S> A shaft position encoder would need enough accuracy and resolution to reliably indicate the magnetic phase with that few degrees. <S> That won't be cheap. <S> If I really needed to do this for some reason (so far used geared motors for such applications), I'd try sensorless drive. <S> This looks at the voltages generated by undriven coils as the rotor spins. <S> The problem with this is that it requires the motor to move. <S> I think for a stepper the firmware would also need to keep track of where it thought the motor was, and intelligently fall back to traditional micro-step open loop control when going to slowly or loosing track or getting data back that doesn't make sense (which indicates lost track of position). <S> Again, go use a geared motor. <A> To detect what the stepper's rotor is doing, will need an encoder with significantly more resolution than the stepper. <S> So a 200 step stepper will need significantly more than a 200 pulse encoder. <S> BLDCs have many fewer phase-changes, so a suitable encoder is simpler. <S> It'll also need a control system and motor drive capable of driving the stepper at its maximum. <S> These typically do not run at constant voltage during ech phase in order to build up the current more quickly. <S> making it slightly trickier <S> (AFAIK, off-the-shelf parts that do this are not designed for closed loop control). <S> I'll assume it needs to be capable of fractional stepping too. <S> Finally, stepper runs hot under normal circumstances. <S> Trying to drive for maximum acceleration will likely drive it even hotter. <S> So if this is for continuous operation, the control system may need to deal with potential overheating too. <A> You don't have to watch it, steppers work in open loop. <S> It's too expensive and too difficult to control the stepper in closed loop. <S> BLDC is different, but you guessed the difference - a hall sensor detects the rotor position <S> and then the stator is switched, whenever the rotor passes the magnetic pole. <S> There is www.trinamic.com DSP processor for that, but I think is already old, since BLDC has lowered the price in this decade. <S> The stepper is two phase machine, thus you need 8 transitors to control, while BLDC is three phase - you need only 6 transitors. <S> The rest: DSP, encoder, ..etc <S> you need for both closed loop, but BLDC is better in any way: less noise and vibration, better accelration, high peak torque... <S> it really doesn't make sense to convert stepper in a closed loop, as you can get a BLDC for the same price . <A> Many (most?) <S> stepper-motor applications operate perfectly fine as open-loop systems. <S> Because they are very conservatively within the ratings for load, torque, speed, acceleration, etc. <S> etc. <S> But if you are trying to achieve maximum acceleration and/or speed while maintaining control of position, then a closed-loop feedback mechanism is pretty well mandatory. <S> Fortunately, there are closed loop systems (motors/encoders, and adaptive drivers) that are becoming pretty sensibly priced. <S> So closed-loop systems are becoming more common and more cost-effective.
| Yes, in theory a stepper could be driven closed loop, though that will need a special purpose motor control system; a BLCD controller isn't suitable, and stepper controllers are designed for open loop control.
|
Why can't I vary the speed of a dc motor with only a potentiometer? I made a simple circuit with a 9 volt battery, a dc motor, and a potentiometer. The positive of the battery is connected to one side of the potentiometer and the negative to the other. One side of the motor is connected to the negative battery side and the other side is connected to the middle pin of the potentiometer. When I turn it, the motor is either off or on. How come the speed doesn't change since I am varying the voltage? Why do I have to use a transistor in this case with an Arduino? <Q> What you're seeing is probably a combination of two things: the current capacity of the 9V battery and the resistance of the potentiometer. <S> Your typical household 9V batteries are notorious for their very high internal resistance. <S> If you're trying to source more than just a few 10's of mA, the voltage across the battery's terminals will begin to drop considerably. <S> You didn't say anything about what kind of motor you have, but I would guess there's a strong chance the battery is having a hard time turning the motor at all, even with low external resistance. <S> Now for the potentiometer. <S> You have the right idea using the pot as a variable resistor. <S> For the simple experiment you're doing, there's no need to connect the other side of the potentiometer to ground. <S> That's just needlessly wasting current out of the battery. <S> This should be all you need: simulate this circuit – Schematic created using <S> CircuitLab The reason it's not working for you is because the 10k value you picked is simply too big. <S> Turning the pot even just a little bit will quickly introduce several hundred Ohms, which will drop almost all of the voltage across the pot resistance before it gets to the motor. <S> Find the largest resistance that still allows the motor to turn slightly. <S> You'll probably find the value is very low: a few 10's of Ohms or in the low 100's. <S> If you then use a potentiometer in that range, your experiment will work. <A> You CAN vary the speed with a simple potentiometer. <S> BUT you must use a pot of a resistance value appropriate for the circuit. <S> In your example you use a power source with a low impedance, and your motor has a low impedance. <S> But you are using a pot with such a high impedance that it won't let enough current through to turn the motor unless it is essentially at the top of its range. <S> Note, however, that if you DO use an appropriately low impedance pot, you will be burning up most of your power in the pot which will not only kill your battery quickly, but may even be dangerous to the pot, causing overheating and premature failure. <S> Furthermore, controlling the speed of something like a DC motor by varying the voltage (even in an efficient way) is a poor technique. <S> It limits the motor to low torque at lower speeds because of limited power. <S> That is why it is more common to see PWM (Pulse-Width Modulation) used to control motors (and lights/LEDs as well). <S> PWM allows motors to have more torque at low speeds. <S> And it is also much more efficient (less wasteful of power) because the controlling element (typically a transistor) is either full-on, or full-off. <S> So very little of the power is wasted as heat in the control element. <S> This is the common method used with microcontrollers like Arduino, et.al. <A> The other answers haven't mentioned a big part of the problem, which is that the device you are controlling is a motor. <S> The current capacity of the battery may also be a problem <S> but I'll assume that the motor is small and low-power enough that doesn't matter <S> (runs well for a decent time connected straight to battery). <S> Now if you connect a very high resistance load like a voltmeter, you'll find the output is 9V whatever the resistance of the variable resistor : but if you connect the other end to ground (potentiometer connection) <S> the voltage will vary smoothly. <S> So, by varying the resistance, you are controlling the available current. <S> But a motor acts as a very low resistance when it's stopped, and appears like a high resistance when it's running. <S> (Its actual resistance doesn't change but when running freely, it generates "back EMF" which opposes most of the driving voltage.) <S> So to get it to start at all, you need to supply several times its operating current. <S> That means turning the potentiometer almost fully up, so when it does start, it immediately runs up to almost full speed. <S> To control a motor's speed, you want to control the applied voltage, not current. <S> One way is to use the potentiometer to control the base voltage of a power transistor, wired as an "emitter follower" (that is, a high current source at a specific voltage). <S> The transistor dissipates power and gets warm so people generally prefer the "pwm" scheme in Richard's answer.
| In order to figure out how big the potentiometer should be, experiment with static resistors in series with the motor first.
|
Exact meaning of dB? I know that when people say that a gain of amplifier is X dB, they mean that 10*log(Pout/Pin). However, How can I interpret it when they say that "noise floor is -120dB"? How can I derive voltage level? <Q> Typically the noise floor is the ratio of the noise with input shorted to the maximum output. <S> If you have an amplifier capable of 1V RMS output and the noise floor is -120dB then the noise is 1uV RMS at the output with the input shorted. <S> Use \$20\log(\frac{V_N}{V_{MAX}})\$ since it is voltage, not power. <S> Divide by the gain to get the input-referred noise voltage. <A> Know that people often speak imprecisely. <S> In this case they may be conflating (or you are hearing) <S> dB when the actual figure is dBm (where, as Dwayne notes, 0 dBm is one milliwatt.) <S> Alternatively, this may be a "signal to noise" figure, unitless. <S> If the signal is X, the noise is 120 dB less than X. <A> I have always used dB in terms of a known reference level. <S> LED sound level meters usually have 3dB spacing between LED's, representing watts rms. <S> The actual wattage used is known by the sound engineer who set up the mics and amps and speakers. <S> He or she would adjust the sound board and amplifier gain so the top red LED's indicate the limit of the existing hardware. <S> So dB is always a relative term, as it is a comparison of 2 sound or power levels . <S> It is not normally used to represent voltage or current directly (linear), but rather very large changes in voltage or current or wattage. <S> -3dB is a drop to 70.7% in power, another -3dB is 70.7% of that value, or a 50% drop overall in just 2 steps. <S> That's why dB is used so often, but the details of each dB are buried in the hardware it is reading from. <S> This is not the same as dBm, which has already been exlained. <A> Do they mean 120dB below average signal level? <S> Do they mean 120dB below Full Scale? <S> For that reason, you can't really derive a voltage level. <S> Perhaps the context would be helpful to better understand the statement. <S> But as an isolated phrase, IMHO, it approaches "weasel words".
| The statement "noise floor is -120dB" is not precisely defined as we don't know exactly what the REFERENCE level is.
|
Making an ID of a board type I want to make hardware ID of a board without using microcontroller or any kind of pre programmed eeprom. All boards of same type will have the same ID. One ID per board type. I was thinking of using i2c I/O expander and connect different combination of input pins to vcc via resistors to define ID of a board. Is that good way of doing it? Will that be robust enough? <Q> You say a micro in the base unit needs to identify the board type, and that these plugin boards already have a IIC bus going to them. <S> This is a no-brainer. <S> Put some device the micro can detect on the IIC bus at a unused address. <S> In manufacturing you can write whatever information you want into the first few bytes of the EEPROM. <S> You say you don't want to use a EEPROM or a micro, but give no justification for such arbitrary and seemingly silly restrictions. <S> We can only assume this is then for religious reasons, which have no place in engineering. <S> Do your job right and use the best-fitting solution. <S> I just checked, and the Microchip 24AA00 16-byte IIC EEPROM is available in a SOT-23 package and costs 18 cents in singles, 14 cents in volume. <S> All you would need is this chip and a bypass cap. <S> You'd have to come up with a really good reason this isn't a good solution. <A> If you have a spare ADC input then a good way to do this is with a simple resistor divider. <S> R1 on the Main board and R3 on the Daughter board form the divider. <S> Different daughter boards will have a different value of R3 to present a voltage to the ADC which depends on the board fitted. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> You can use a preprogrammed unique serial number chip, some of which have a "family" ID that does not change. <S> Maxim has that type of part. <S> Just ignore the unique serial number part.
| The easiest would probably be the cheapest and smallest IIC EEPROM you can find.
|
How to prioritize a current path in a circuit Here's my problem: I have a power supply rated at 5V, 2.1A I need this to feed two devices, both running at 5V Device one is a single-board computer that can consume anything between 0.5A and 1.5A depending on what it's doing. Device two is a battery charger, which is happy to take anything it can get up to (and quite possibly beyond) 2A, and equally happy to take no power at all. I need a way to feed device one with as much current as it is demanding right now, and give whatever is left to device two. In order to stay within the PSU limits, I'm going to be limiting the input current to 2A. I'm a beginner, and from my limited understanding (so please correct me if I'm wrong) the chances are that device one could end up current-starved otherwise. I'm not happy with simply making two fixed current-limited paths (1.5A to A, 0.5A to B) - I really do want any excess current delivered to device B, if there is any available for it. The only thing I can think of is to dynamically monitor the amount of current currently going to devices A and B, and to throttle down device B by opening and closing current-limited gates with transistors (or using a variable current-limiting circuit controlled by a digital pot) with a microcontroller until the sum of the two is less than (2A minus a safety buffer), along with some caps in there to keep things running during switching. However, that sounds a bit complex for something that I have a gut feeling ought to be much simpler, and possibly involves using some kind of specialized IC that I don't know the name of in order to search for it. This is a problem I feel I'm going to run into with other things as well - not just charging a battery, so I'd like to know if there's any way (simpler than the complex microcontroller solution I imagined earlier) to manage the current distribution in a circuit when the supply doesn't meet the demand - some way to simply say "current goes here first, and what is left over goes there" <Q> You can build a circuit that behave as you say. <S> Basically, you need a small shunt resistor that measure the total current consumed by both device, a P-channel mosfet that will limit the power provided to device two, and an opamp inbetween that adjusts the gate voltage of the mosfet depending on the total current (i.e. that will shut down gradually the mosfet as the total current reaches the maximum). <S> The thing is: such a monitoring circuit can only work if you have a resistive circuit as device two (e.g. a set of lights). <S> However, a battery charger is an active circuit. <S> It will probably not work at all if the voltage is below a given limit. <S> So in the end, this will not work for your application, or be completely unstable (when the battery charger uses too much current, the monitoring circuit will shut down the mosfet a little bit, leading the battery charger to go undervoltage, so it will shut down, not using any current at all, then the monitoring circuit will rise again the battery charger voltage, which will draw too much current, etc...). <S> So, either you find a battery charger designed with this capability, or you have to design your own. <A> First, some clarification. <S> There are two loads for one power supply: load A and load B. Load <S> A needs to get power at 5 V with (at most) 10% tolerance. <S> It can take any current in range from 0 (or <S> 0.5 A, it does not matter) to the maximum capacity of the power supply. <S> Load B must get "what remains". <S> There is no common solution for this problem. <S> The design depends on properties of the second load B. Consider these cases: Load B is a resistor <S> Then our circuit can control its power consumption by lowering the voltage on the load <S> B. Then the right design is: make the step-down (buck) DC-DC power supply for load B. <S> It regulates the voltage on the load B in range from 0 V to 5 V in such a way that the common current by loads A and B does not exceed the capacity of the power supply. <S> This is not so simple, but doable. <S> Load B has some external (analog) control input to adjust its current <S> Then the design is similar: the control circuit controls the voltage on this "load B control input" to keep the total current below the maximum one. <S> Load B is a step-down DC-DC controller that powers up a computer <S> There is no good solution for this task. <S> The best we can do is to switch off the second load when the total current by loads A and B exceeds the upper limit. <S> The reason is simple: the step down power supply would increase the input current on any attempt to lower the input voltage (until it shuts down due to undervoltage). <S> This feature makes a positive feedback. <S> Any attempts to regulate the power for load B leads to shut down of this load B. <S> It seems your case falls to the last class: you can not control the current consumed by the battery charger. <S> So you can only switch it off by any kind of switch (N channel MOSFET is the best choice, until you do not need to keep devices on the same ground). <A> Plenty of single-board computers have battery charging chips built-in . <S> If you don't want to design your own charger circuit, the most logical thing for you is to get one such SBC and simply connect a battery to it. <S> This solution will have the added benefit of battery status being known to the SBC, so it can shut down properly when the battery is getting low. <S> You'll also have plenty of settings available, like maximum PSU current, maximum battery charging current, target battery voltage etc. <S> Alternatively, get a battery charger with charge current limited below 600mA.
| The only way is to be able to tell the battery charger how much power it can suck.
|
Can I2C be used with momentary sensors? [edit: clarifying the question] Is it possible to see events on I2C slaves, or a functional equivalent? I'm brainstorming a new project. I'm using an ESP-12F as a main logic board to control and receive input from various connected modules (e.g. pushbutton, beam sensor, solenoid valve, shutter trigger) Each module (slave) is PIC based, and communicates with the ESP-12 (master) using I2C. Sending commands for valves and triggers over I2C is straightforward. I'm not sure how to utilize I2C for slave-triggered momentary events though. Should I just keep sending status requests to those modules and look for changes that way? I'm also considering using XOR logic to flip the level of another line, which I can detect via interrupt to trigger the status request. My device is timing sensitive, I'd like to get a response to a trigger down to maybe 50-100 microseconds. I'm new to all of this so feel free to correct me on anything fundamental I'm missing. <Q> You can poll the slave, if that fits your requirements (latency, for example). <S> You can add separate <S> line(s) for slave devices to alert the master asynchronously. <S> These lines would be out of band with respect to the I 2 C bus (not a part of the I 2 C bus itself). <S> For an example, here's a bog-standard I2C I/ <S> O expander PCA9554 , which has got an alert output. <A> The terms of art here are Polling , i.e. repeatedly checking the sensor for changes, vs Interrupt , being alerted on demand of a change. <S> No, I2C does not provide an Interrupt protocol directly. <S> A subset of I2C, called SMBUS, does. <S> It has the SMBALERT#, which the Host device has a single Interrupt pin which all the slave devices can attach to, and when triggered, the Host will read from the special Alert Response Address (0001 100x), and <S> any slave that has a pending alert will write <S> it's specific address. <S> The Host then reads directly from it. <S> If more than one is pending an Alert, arbitration takes place, and the Host will need to read from the ARA multiple times. <S> Neat trick, you can skip the interrupt pin, and poll from the ARA anyway, and then only if you get a valid address <S> do you try to read from the slave. <S> This can cut down the number of times you try to poll, if you have multiple slave devices. <S> Of course, you can implement this on I2C quite easily, if you can program the slave devices and the host device. <S> Alternatively , you can make all the micro controllers masters and slaves, implementing an I2C Multi-Master setup . <S> Have the modules become masters and write to your main device. <S> Then your main device switches to master mode, to read back. <S> Or just leave your main device as a slave and have the modules as masters write to it on demand as needed. <A> For a interrupt/request line, you wouldn't use XOR, as two simultaneous requests would cancel out. <S> If you want to have a common interrupt/request line, use wired-OR: make the drivers open-drain with a pull-up (just like the I2C lines) fed to an active-low interrupt input on your MCU. <S> This way, when any of your slaves pull this low, your master knows to poll everyone. <A> A chip like the PCF8575 provides I2C input and output expansion with built-in interrupt functionality.
| The I 2 C bus per se doesn't have the ability for a slave to alert the master asynchronously.
|
determining current in individual phase for a 3 phase winding brushless dc motor I have an 3 phase brushless dc motor with no neutral wire. I measured the resistance from phase to phase and found them to be on average 0.255 ohms(between A and B, A and C, etc...all the values are pretty close). These 3 phases are in a star formation. You can see the motor here: http://www.cda-intercorp.com/Products.cfm?cid=1&Data=Design#ScrollTop I am running the motor on 28V. How do I know what current are running through each phase(A,B,C). To add to this, if I stall the motor, only two of the phases would have current right? What would the current be then? <Q> The absolute best way to measure the current in a line is to use a current probe such as the TCP0030A . <S> They are a bit pricey, but totally worth it if you are doing motor control. <S> Of course, if you are doing this in an application, you can use three ACS711 (pdf warning) chips (or similar). <S> These are good enough to use for current control loops or similar, but not necessarily good enough for lab-grade measurements. <S> I have used both successfully. <S> What happens in a stall condition depends entirely on the drive architecture. <S> Most drives have some sort of stall protection built-in so that they don't burn up the motor. <A> First, you said it was a star wound motor with no neutral. <S> That means the 0.26 ohm measurement you made is already across two (0.13 ohm) windings in series. <S> That also means the stall current at 28V is somewhere over 100A. <S> As you suggest in a comment that you have a peak current limit of 7A (what imposes this limit???) <S> that means your motor driver must soft-start the motor, i.e. start it at a much lower voltage (or with PWM at a much lower duty cycle (about 7%). <S> It's not clear which motor from that page you are using, but <S> the typical "performance" data on that page is unusual - it shows a "stall torque" that is only slightly higher than the torque at max power - and a much lower power dissipation at stall. <S> That confirms that the permissible drive voltage must be reduced by the motor controller at stall, and probably also at startup. <S> Either their recommended controller will do this for you, or you'll have to actively measure the current and control the PWM duty cycle to stay within the motor's limits. <A> if I stall the motor, only two of the phases would have current right? <S> What would the current be then? <S> Probably not: <S> - The current waveforms will be phase shifted in each winding but there are only transient ponts when the current in one winding is zero. <S> How do I know what current are running through each phase(A,B,C). <S> You can't know without measuring it unless your drive circuit has some feature to tell you. <A> Since ONLY two phases are energised at any one time you could infer via having a single DC-side (pre inverter, post capacitor) current sensor. <S> you can then infer to a reasonable degree of accuracy what current is in what phase via the present commutation state (NOTE: zone change will result in some inaccuracy as the current commutates from one phase to another). <S> If the motor is stall however... <S> depending on the complexity of the ECS & whether there is a maximum PWM duty you could infer via two series phase resistance and the percentage of the DClink applied via PWM.
| If you want to know the current in the phases of a BLDC machine you really need a current sensor.
|
Finding RPM of DC motor How can we find the RPM of DC motor using the given voltage?.To be clear, I need to find out rpm of a motor using the pwm signal sent from Arduino Uno.In my case i'm using a BO motor, the link is here: http://www.coretechnologies.co.in/index.php?route=product/product&path=64_83&product_id=753 I'm doing this for my tachometer project, for proving theoretical values and practical values are correct. For getting practical values i'm using this: http://playground.arduino.cc/Learning/Tachometer <Q> You can estimate, but the load will directly affect the speed of the motor. <S> There is a way to measure the speed of the motor based on the current waveform. <S> Most DC motors have a 2-pole stator and a 3-pole rotor, so 6 current peaks can be observed for each rotation. <S> Run the current waveform into your ADC and write some slick software or use a comparator circuit to peak detect and you will have your speed. <S> You can see the measurement made and described on for(embed) . <A> Each motor has a BEMF constant. <S> So if it runs with no load, you can say that velocity equals voltage times k. Since each motor has some minimal load always (and usually it's not minimal, motors exist to drive loads), actual voltage has another component, the current times motor resistance. <S> So if you measure the current, you can calculate it too. <S> And the best way to know the BEMF constant is to measure it. <S> The one given in datasheet may vary between motors. <A> Back emf is directly proportional to speed. <S> The back emf constant is numerically equal to the torque constant and this is normally given in the data sheet. <S> Alternatively, calibrate the motor speed vs back emf by measurement. <S> Then measure the back emf during the PWM off-periods via an ADC to get the instantaneous speed. <A> A brushed DC machine can be modeled as simulate this circuit – Schematic created using CircuitLab voltage source stator winding inductance stator winding resistance <S> The voltage source will have a voltage proportional to rotor velocity, <S> the \$K_e\$ voltage constant: Volts per RPM (or rads depending on the datasheet). <S> The faster the rotor rotates, the higher this value. <S> There will equally be voltage due to current flowing through the resistance and <S> equally changing current will cause a voltage across the inductance. <S> Higher load -- <S> > <S> more current draw -- <S> > more additional voltage due to the R. <S> As long as the load & speed is assumed constant (maybe not...) <S> and as long as the PWM frequency is high enough, the L contribution could be ignored. <S> All this can be measured at the motor's terminals. <S> How to use PWM to determine speed? <S> The minimum prerequisite is knowing the \$K_e\$ of the motor in use. <S> If the motor is UNLOADED <S> and you are just energising it with a fixed PWM duty <S> , the rotor velocity can be estimated via: \$ <S> \omega <S> = V_{cc} <S> * D <S> * K_e <S> \$ <S> \$V_{cc}\$ being the voltage being PWM'ed onto the Motor's stator D <S> being the fixed duty <S> \$K_e\$ <S> being the backEMF constant in \$V/\omega\$ <S> If however some form of speed control is required (sensorless) some means to measure the terminal voltage is required. <S> You can then sample this DURING the off period's of the PWM <S> so you are aware of the motor's terminal voltage. <S> How much you then compensate for iR <S> (current sensing or ignore if unloaded) or wL (can it be argued its negligible for a DCmachine?) <S> is down to your system considerations. <S> EXAGGERATED wL
| The RPM of the motor cannot be found using voltage and motor ratings alone.
|
"High Voltage" Logic levels why they are rare I'm continuing to dissect and learn from this old (1974) calculator. One thing that surprised my was that the pins of the main IC have both negative and positive voltages 0 to ±28V. Having worked mainly with ICs like the ATMEGA line, PIC controllers and other popular chips, the voltage surprised me. Am I correct in thinking that modern IC operate at lower voltages so they can pack in more transistors without over heating? Inside view of circuit board of Sharp ELSI 8002 Calculator. The mystery IC: HD3623 has no online data sheet that I could find, but MK14 at eevblog pointed me to this data sheet showing how a similar calculator works. When I once destroyed an ATMEGA128 with 12V, was the nature of the destruction heat that melted or burned the traces in the IC making it inoperable? But this old more robust IC must have thicker, more conductive traces. Is that why it is so big despite doing comparatively less than smaller modern ICs? There is one advantage to this beefy old IC: It can drive the VFD, which seems to have a -28V grid, directly. Maybe, I should look for a programmable IC that runs at higher voltages? Or would that be frustrating? Are there IC that meet this description that can be programmed by a novice? Maybe I should use a logic level converter of some kind instead? I want to be able to repurpose the buttons and the display within the existing case. I like the form factor and quality of the calculator, I want to make it work with a modern micro controller. Sharp ELSI 8002 calculator. I'm fairly certain that the VFD is the Futaba 9CT06 , it's not super well documented, but I can make it do what I want using bench power, so having identified the pins I just need some kind of PWM... I think. Maybe I should table this project until I've studied more! I've found some people with similar projects, but none with the same VFD. I've put notes by my main questions for clarity. <Q> 1) Partly true indeed but it is more that in modern ICs the transistors are smaller meaning <S> the supply voltage must be lower because of the smaller structures. <S> The limit is the field strength (Volts per distance) that can be handled so when the distance goes down the voltage must go down as well. <S> How much power is consumed also relates to this but this can also be tuned separately by the the threshold voltage implant offering a choice between fast but high power consumption or slower but smaller power consumption. <S> 2) <S> If you destroy an IC it is very rarely the wires that melt, depending on what you did wrong there are many ways to damage an IC. <S> Reversing the supply voltage for example damages the ESD protection diodes. <S> These have nothing to do with the functionality of the IC itself but need to be there to protect it against ESD discharges. <S> 3) what you see is the packaging, the IC itself will be only a few square millimeter in area. <S> Fact remains that in the 1970s we could only fit a couple of thousand transistors on a few square millimeter. <S> Nowadays we can easily fit hundreds of thousands of transistors in a few square millimeter. <S> This is due to more modern IC manufacturing processes. <S> 4) <S> the -28 volt is probably needed for the VFD. <S> You will not find modern ICs that can handle such voltages as almost no-one uses VFDs anymore as these are very power hungry. <S> Interfacing with this old technology is a challenge, you really have to know what you're doing. <S> It is not for beginners. <A> There are three reasons why ICs with such high voltage are not produced (anymore): <S> Higher voltage require larger clearance between traces. <S> They don't necessary need bigger traces (trace width depend on the current, not voltage), but if they are too close together, the high voltage will cause electrical breakdown , causing the two traces to be shorted. <S> So, it means the chip die must be larger to account for clearance between the interconnects. <S> And larger chips are more expensive. <S> Higher voltage means higher power requirements (for the same resistance). <S> Today, chip design is clearly oriented towards low power consumption, so chip designers tend to use the lowest possible voltage that still allow convenient external interfacing with the chip. <S> Most high-end MPUs even have an internal working voltage that is lower than the one used for external interfacing (with level translators for each pin) to allow for lowest power consumption. <S> Higher voltages mean that the signal swing is larger when the logical state changes. <S> So, if the slew rate is the same, it takes more time when changing state and thus limits operational frequency. <S> This has to be mitigated, however: this is true when you compare a chip working at 28V to a chip working at 3.3V, but it is not true when you make the same chip working at <S> , for example, 2.0V vs 3.3V. <S> All in all, there is no real advantage in making the chip able to work with high voltages, except in making it easier to interconnect with fancy peripherals. <S> So it is actually a lot cheaper to make logical chips work at low voltage and use specific circuitry to interconnect with fancy peripherals when required. <A> You can actually get CD4000 logic that can go up to 15V. That being said, it replaced <S> transistor logic that operated at 5V. CD4000 was replaced with the familiar 74HC logic that can operate at 3.3V so you can see "standard" voltage was established decades ago. <S> On processors and the like the move is towards smaller and smaller but quite wide variety of ICs work perfectly well with 3.3V. Lower voltages mainly give you benefit on switching losses and leakage current. <S> Those really become an issue only with very high density logic and/or high speed interfaces. <S> The reason 3.3V is popular because very low voltages like 1.2V are a pain in the backside to generate and to work with. <S> Also cutting voltage often means boosting current that may become problematic.
| In this case, usually, the higher voltage can allow for faster operating frequencies (at the cost of a lot higher power consumption) because the higher voltage difference applied to fet gates makes the switching actually faster (the slew rate is a lot higher).
|
Finding the number of poles in BLDC motors I actually have a 3 phase BLDC motor ( The datasheet is not available) with 3 hall sensors. I would like to measure the motor speed (rpm). Therefore I need to know the mechanical revolution or number of poles and the frequency. How can I find out the number of poles? <Q> When setting up for driving a BLDC, it's useful to step thru the drive phases one by one manually. <S> I usually mark a point of the shaft, then add a sticker or something on the stator to record where the shaft is. <S> I like to use a 12 phase drive. <S> In that case, I go thru each of the 12 phases and mark where the shaft is. <S> This not only tells you what order you need to generate the 12 phases in, but the number of poles falls out by seeing how many times the pattern repeats in a full revolution. <S> Another way to just get the number of poles is to connect one lead to the supply, one lead to ground, and leave the last lead floating. <S> That drives the motor to a particular phase. <S> By manually rotating the shaft, you can see how many places the rotor is pulled to in a rotation. <S> That is the number of poles. <A> Turn the motor shaft one revolution and analyze the output of the Hall-effect sensors. <A> If you get the rotor out of it, or just if you can get close you can use a magnet to determine how many North and South poles the rotor has, this is the number of poles.
| The number of poles is the number of magnetic cycles in one revolution of the shaft.
|
Using connectors rated for 250-400V with 5V I have gotten a hold on a really great deal on Connectors for a electrical project of mine. These connectors come in various sizes, but common for them all is that they are rated between 250 volts to 400 volts. My system is using 12V and 5V. The project is to connect a bunch of LED lights together in my garden on various distances. I am aware of current drop in the wires; I am just not sure what kind of issues I can encounter from these connectors. So, can I use these connectors and what could the side effects be? <Q> Connectors are typically rated with a maximum voltage, meaning every voltage up to the rated one is okay. <S> For you that means the 12V and 5V are totally fine to be used on a 250V connector. <S> You still need to make sure you don't overload them with the current you are going to use, e.g. a 5A/250V connector should not exceed 5A, even if you use it only with 5V. <A> Wires and cables as well as connectors work the same way. <S> They have voltage, and current, ratings which are maximum ratings. <S> It is always OK to use less voltage or less current than the rated maximum. <S> The maximum voltage of a wire or insulated connector is determined by the insulation. <S> A higher voltage means the insulation will be thicker or made of better materials. <S> A multi-contact connector might also have larger distances between contacts to prevent arcing. <S> The maximum current of a wire or connector is determined by the conductor thickness and contact area — which determines the resistance, and therefore the amount of heat produced. <S> A wire or contact which has less resistance, or can tolerate higher temperatures without being damaged, (or is in an environment with better cooling) can handle more current. <S> Therefore, the only consequence of using an over-rated connector is that it will be larger than necessary. <S> If they're not too large for your application, then don't worry about it. <S> (Another way being too large could affect things is in the particular case of crimped joints: <S> attempting to crimp a large crimp barrel onto a small wire will make a poor joint likely to fall apart. <S> They should only be used for the wire sizes specified.) <A> Those are typically MAXIMUM ratings, and frequently they are actually called "Maximum Voltage" or whatever. <S> Note that the maximum voltage rating and the maximum current rating are independent. <S> So even though you may be well below the max voltage rating, that doesn't mean you can exceed the max current rating. <S> Connectors rated for higher voltage may be physically larger than you need. <S> This may be a problem in situations where you have a space constraint. <S> CURRENT CAPACITY. <S> Since Power is voltage multiplied by current, connectors rated for higher voltage may not have the current rating you need for a lower voltage of equivalent overall power capacity. <S> In most cases, you could parallel connector pins to increase current handling capacity. <A> A good connector for power at 250V is not necessarily a good connector for signals at 10 uA. <S> It all depends on how the two parts mate together - read up on <S> wetting current - wiki says this: - In electrical engineering, wetting current (sometimes also spelt as whetting current in archaic sources) is the minimum electric current needing to flow through a contact to break through the surface film resistance. <S> "Wetting current" should not be confused with the current needed to generate involuntary insanitary actions in a human. <A> One possible risk is that of misconnections. <S> If someone inadvertantly connects mains voltage to an ELV circuit by plugging together two devices that were never intended to be plugged together then bad things can happen. <S> Unfortunately most high current connectors seem to be rated for mains voltage. <S> So short of having custom connectors designed and built for your devices (which is only practical at large volume) it's difficult to completely eliminate this risk. <S> However it is prudent to take steps to minimise it. <S> You should take steps to understand what environment your device is likely to be used in and what connectors are commonly used for what purposes in that environment. <S> Another option is to use connectors that support blocking some of the pin positions and block a random selection of pin positions to reduce the risk of misconnections.
| Two possible disadvantages of using connectors rated for high voltage in a low-voltage application include: SIZE. For all practical purposes, there is no danger in using a component at below its maximum rating.
|
Zener Diode: Easing the load on a regulator Summary: I am using a LM317 to drive a liquid pump that draws 1A. Unfortunately my regulator is dissipating way too much power and getting too hot. Currently my input voltage to the regulator is 12V and I'm trying to create 7V so that's 5Watts! So I bought a heat sink BUT I would also like to drop the input voltage of the regulator a little lower, no need for it to be 12V. Goal: To drop the 12V input voltage to around 9V for the regulator by using a zener diode, that way the power dissipation is lowered to about 2W. Question: 1) Could I get a sanity check on my circuit I created? 2) Do I need to worry about the current going through the zener diode because its 1A? 3) Could I just use a 3.3ohm resistor instead of a zener diode at all that way 3.3ohmd*1A is 3.3V drop which should do the same thing. <Q> In principle this will work- the zener would have to dissipate more than 3W, so it will have to be something like a 5W zener- not cheap. <S> The resistor will work if the output current is a constant 1A current, however motors, especially those used in applications such as yours, tend to draw much higher current during starting so it may end up stalling your pump and burning up the resistor. <S> At this level, I strongly suggest you use a switching regulator (buck regulator), such as those based on the LM2596 . <S> It will run cool. <S> Inexpensive modules are available from China which may or may not use genuine LM2596s, but seem to work well enough. <A> Assuming that you are using the TO-220 LM317 without a heatsink, it has a junction to ambient temp rise of 19 degrees C / watt. <S> Given you need to dissipate 5 watts, this would give a temperature rise of about 95 degrees C. <S> With an ambient temperature of 21 degrees C that is 116 degrees (HOT). <S> Since the part has a maximum operating temp of 125C, it MIGHT not burn up immediately but it is too close to the limit. <S> If you got a heatsink that is 5 degrees C / watt and add the junction to back thermal resistance of 3 degrees C / watt that would give 8 degrees C / watt and give a temperature rise of 40 degrees C + <S> the 21 degrees <S> C ambient gives a temperature of 61 degrees C (about 142 degrees F) which is much better for the part but still too hot to hold your hand on. <S> If you use the Zener diode as you suggest, as others have noted, you need a 4 or 5 watt Zener diode, for most of these parts the junction to lead thermal resistance is about 15 degrees C per watt (with .25" leads) which would have the diode at a 49.5 degree C temp rise which puts it's temperature at about 71.5 degrees assuming its mounted to a significant board that can sink heat well through the leads. <S> The Zener should be fine at this temperature. <S> For the LM317 in this instance, it will need to dissipate 1.7 watts which is a 32.3 degree C temp rise and should be about 53.3 degrees C. <S> I suggest that the better option is to get a proper heatsink or use a different method of voltage regulation (i.e. switching). <A> Your reasoning is flawed because you are just moving part of the power dissipation to the zener which is much less tolerant of that kind of current. <S> It will most likely melt in seconds, unless its a large 4W zener. <S> Aluminium is great for heatsinking, and special heat paste for heat transfer. <S> When you use heatsinks its all about surface area and airflow for best cooling.
| The heat has to go somewhere in a linear regulator, adding the zener will move part of it to the diode. You are better off trying to use heatsinks to improve your regulator performance.
|
What's the purpose of crossing over in the middle of a toroid winding? Seeing toroids wound like this is somewhat common in the amateur radio hobby to reduce common-mode current on an antenna feedline: Why the cross over the middle? As far as I can tell, all the turns go around the core in the same direction. Why not just keep winding all the way around? Sometimes they are also seen with two enameled wires instead of coax: Is this any different? As far as I can tell, it's effectively a common-mode choke in both cases. Does the cross over in the middle serve any purpose? <Q> This is an RF split-wound balun. <S> As this says: Some articles and handbooks show a Split winding method. <S> This method is supposed to reduce winding capacitance by moving the ends of windings further apart. <S> The proposed theory is by reducing shunt capacitance that "leaks RF around the balun", balun performance is enhanced. <S> You can find a link to another page from the same author comparing measurements to conventional winding here . <A> I'm answering with observations and a little speculation. <S> I've not seen one of these before but... <S> What you might wish to avoid (at high frequencies) is a separation of the two conductors as this will cause a change in the characteristic impedance of the pair and could give rise to mismatches i.e. a poor VSWR (voltage standing wave ratio). <S> A traditionally wound common-mode choke would not be great in that respect. <S> Neither would you want to continue winding around the core so that input wires and output wires are brought-together and significantly capacitively coupled because, that would negate the usefulness of the common mode choke. <S> At frequencies below several tens of MHz this isn't a big deal but at medium VHF frequencies onwards this could be noticeable. <S> Having said all of that the lower picture in the question separates out the two conductors in order to fasten them to the terminals on the left <S> so maybe this was wound without much thought. <S> And a final observation is that at only a few MHz most ferrite cores are going to be pretty useless magnetically but pretty good as a means for dissipating high frequency energy <S> so maybe that is the intention in the 2nd picture <S> - it's a common mode VHF resistor. <A> One primary advantage is for mechanical reasons, <S> Classically a common mode choke is drawn as shown (picture from coilcraft website): <S> Which would involve two separate winding operations. <S> Winding <S> the wires together is faster (depending on machine) but then leaves the wires all in one location on the ferrite. <S> This way it flows naturally and "looks" like how a choke might look if wound individually. <A> The split winding decreases the leakage (parallel) capacitance between the input and output by increasing the distance between them. <S> Reducing the parallel capacitance increases the impedance at higher frequencies. <A> IMO, the prupose is to have input and output at opposite side. <S> If you continue to wrap in the same direction, you have both input and output close each other. <S> If you wrap separately each wire, then you have larger leakage inductance and you would have also differential mode choke, which I suppose is a non wanted feature, therefore the best way is to wrap two wires together and this is a result.
| Moving the windings apart is intended to reduce capacitance between input and output.
|
Regular (not Zener) Diode as Voltage Regulator I see a lot of information regarding Zener diode voltage regulators but how about a regular diode voltage regulator? What would be the potential drawbacks of using something like this on the picture below: simulate this circuit – Schematic created using CircuitLab Where Rload should get whatever voltage is across the 1N4001 or whatever diode we use? <Q> The choice of voltages is very limited. <S> For a silicon diode it will be around 700 <S> mV. <S> The voltage is not well regulated. <S> The forward voltage of a diode is reasonably invariant with current over some range, but it's not as good as most zeners. <S> The temperature dependency will be higher. <S> This is particularly compared to a zener roughly in the 6 to 6.5 V range. <S> There are two opposite temperature effects that cross over at about that voltage, reducing the overall temperature dependence. <S> In that case the temperature dependency can actually be beneficial. <A> In the circuit you show, the voltage across the load will be about 5mV, with or without the resistor. <S> It won't start to regulate until the load resistor is more than about 135 ohms. <S> In this circuit with a 1K series resistor to +5, the diode acts a bit like a voltage source with about a 12 ohm resistor in series.. for load resistance that is relatively high. <S> That dynamic resistance increases with the load current- with a 270 ohm load <S> it will be more like 24 ohms (calculated from 52mV/Idiode- the diode ideality factor of about 2 multiplied by the thermal voltage at room temperature). <S> That means that with a 270 ohm load the ripple rejection is only about -20dB. <S> Load regulation will be similarly poor, but zeners for less than a few volts are horrible in performance or nonexistent <S> so you can't really compare it with a zener. <S> You won't find a 0.6V or a 1V zener. <S> The diode has a temperature coefficient of about -2mV/°C. <S> So, it's quite usable in some undemanding applications- for example to determine the current through an LED or to set an overcurrent limit on a power supply, but it won't be mistaken for a precision reference. <S> This type of circuit (where the zener might be something else) is called a 'shunt regulator' and is mostly used for relatively low currents and certain other special applications. <S> It suffers from a number of disadvantages such as inefficiency if the input voltage is close to the output voltage and the load is variable (because that series resistor has to be very low and the wasted power with a light load is very high). <S> A popular example of a chip that does this well is the TL431 (or TLV431). <S> Using a couple of resistors you can set an accurate voltage from about 1.25V to some tens of volts depending on the chip. <S> The chip needs 1mA or less to operate. <A> That's not a voltage regulator, it's just a voltage divider until the voltage drop across Rl starts getting close to 0.5 volts. <S> As you've drawn it, the load is tied high side,in parallel with D1, and your circuit won't even start to regulate the voltage across the load until the supply gets to 700 volts or so, like this: More tomorrow... <A> As others pointed out, that simply won't work as your load current is far too high. <S> To get 0.7A to your 1R load you'd need to use a 6R series resistor or less. <S> This kind of regulator would be appropriate if need a few milliamps. <S> I use a zener occasionally when I need 9V or something I need very little current from. <S> I also occasionally use a LED when I need a proscribed voltage drop with small currents as zeners do not work well with very low currents.
| One place where a diode as a shunt regulator can be useful is when you want to generate a voltage that offsets another diode or the B-E drop of a bipolar transistor.
|
Flame sensing using Flame Rod and Short circuit identifying I am sensing a flame using a flame rod. I am using 12V AC at the input and at the output flame sensing circuit which is made of opamp, some resistors and capacitors, I get 1.6-2.4V when there is a flame, and 3.8-4V when there is a short circuit, and 5V when there is no flame. My flame sensing circuit seems to work fine but when there is a flame and short circuit happens at the same time my circuit only detects short and gives output of 3.8-4V. But I want my circuit to give output of 1.6-2.4V and see a flame instead of a short when both occur together. Is there any way I can achieve this?I would really appreciate your opinions on this. Thank you I am using this circuit below. <Q> STOP tampering with safety equipment! <S> If you continue with gas supply in case of a short and the flame is not present <S> the gas will eventually find a flame source and your house will likely be exploded into kindling. <A> It may be possible. <S> But it would appear to make the flame detector inoperable. <S> This could be a very dangerous situation. <S> Not knowing if there was a flame or not. <S> And is probably the reason that grounding the flame detector shuts down the furnace in this video . <A> When I get a difficult problem, I start by writing down the requirement details and then jot down all the possible solutions I can think of, simple, complex wild or crazy. <S> The IC makers data books and online op-amp applications are a big help. <S> I'm an Electrical Engineer with many years experience, but I still go back and study the basic circuits in the text books. <S> this will give you different voltages levels as inputs. <S> You can use one or more comparators (like the LM311) to sense a particular voltage level, or a high low-band. <S> You can use more components, two flame sensors, maybe a thermocouple ckt, shielding one section from moisture, then combine multiple signals for the output. <A> To decrease the likelihood of a short circuit you could increase electrode to ground air gap distance. <S> If there is a short you cannot detect a flame if the flame 'diode' is effectively being bypassed
| Flame detectors are designed to be fail-safe and incorporate redundancy to ensure failures lead to the gas valve closing. The voltage divider circuits can be rearranged, with the flame sensor, anywhere in the series, shifted on the high side or the low side.
|
Interference from motors on PWM signal Am building myself a robot chasis that is RC controlled. There is an RC receiver attached in solution and I have a remote RC transmitter. When the motors are not running, I get a perfectly clean signal on the output of my receiver ... a PWM signal. Here is a quick trace when the motors are NOT running: When I switch on the motors, and then run them up, the signal as measured on the PWM output becomes the following: If you look closely, you can see the good signal (every 20 msecs) but with lots and lots of noise. Here is a picture of my robotics chassis: My scenario was previously working but in this setup, two things have changed: I am using a metal sheet for my chassis I am using 4 new motors that I previously hadn't used before. They are 170RPM motors as found here: https://www.servocity.com/html/170_rpm_econ_gearmotor__638354.html My questions are: What is introducing this "noise" when the motors run? How can I eliminate the noise? Unfortunately I'm a "software" guy and mechanical electronics is not my comfort zone but am willing to learn/study. Looking forward to any assistance. ... later ... I have a new clue, I placed my digital signal analyzer on an otherwise unused GPIO on the Raspberry Pi and when the motors were not running, the signal was constant flat. However, when I ran the motors, interference was found ... see: This makes me think that the "setup" is receiving a ton of interference when the motors run ... but I'm at a loss of how to alleviate the problem. Neil ... later ... after answer proposal from Richard ... I soldered in 3 capacitors per motor. Each capacitor was 0.1uF. One capacitor across the +/- of the motor and two capacitors from the motor terminals to the casing. See (sorry about the blur): I then re-ran the tests and captured a new signal analyzer recording. Sadly, nothing obvious has changed. See: With no motors running, a perfectly flat line. <Q> Brushed DC motors are notorious for throwing out EMI (electro-magnetic interference). <S> AS CLOSE as you can get to the brushes. <S> Sometimes, with larger motors, it takes additional filtering in the form of shunt capacitors and even pi-filter elements with series inductors... <S> These photos are from a good tutorial on reducing EMI: http://www.stefanv.com/rcstuff/qf200005.html Additional bypassing of both power and signal lines is also beneficial. <S> Capacitors can cause sluggish response in sensors if you use too much. <S> It doesn't take large capacitors to be effective in shunting away ("filtering out") EMI. <S> Shielding and grounding are a "gray art" at best. <S> Halfway between engineering and applied magic. <A> Neil, It will be in the inductors/Capacitors to act as a low pass filter that will help keep the Noise down. <S> Regards, Dave M. <A> I just fixed interferences generated by a small rotary vane pump, controlled by a 100Hz PWM with ferrite beads . <S> Wind up a couple of turns with the motor wires around the bead, paying attention to spread the turns. <S> Advantage : easy to install, no PCB change.
| The PRIMARY solution is to use a good bypass capacitor (like a 0.1uF ceramic) right across the power pins on the motor.
|
Purpose of perfboards with pads all connected A while ago I saw a protoboard with the pads all connected; at the time I thought it was a manufacturing glitch, but today I saw another one. I cannot think how these boards may be useful - any ideas? Here they are: <Q> That perfboard doesn't have its pads connected. <S> It has no pads, since there's a clear area around each hole. <S> Instead it has a plane with a void for each hole. <S> You could use this plane as a power or ground plane, while making other connections with wires between the pins of the components in your circuit. <A> Those are usually used as ground and power planes. <S> Through hole devices like wire wrap sockets do not touch the copper traces. <S> The builder purposely connects the power plane to the power pins of the sockets where necessary. <S> This board is a bit messy. <S> But you can see where there is a gap around each pin. <S> So the builder can choose to or not to connect a pin to a power plane. <A> I have seen people take a knife to these kind of boards and manually etch off paths to create a prototype circuit. <S> Generally this only done when they don't have a breadboard handy or ability to get a real board printed. <S> It's a pretty quick and dirty solution but it has come in hand to test components fitting together and circuit designs.
| These printed circuit boards are usually used when prototyping using wire wrap sockets. In addition to distributing power, such planes help reduce noise by acting like shields.
|
50V Ceramic Capacitor for 36V Vehicle Battery In my vehicle Battery powered system which can vary from 12V to 36V , I want to use a Capacitor of minimum 10uF on the input. I have few doubts :- Which is best for 10uF value @ 36V - Ceramic/Tantulum/Electrolytic Can I use a 50V(Do I need to derate @ double the voltage) ceramic for 36V rating?? <Q> When using Tantalum capacitors , manufacturers recommend derating devices with MnO2 electrolytes by 50% and to 80% of rated voltage for polymer electrolytes. <S> Tantalums are popular as their CV product density is very high compared to other types. <S> Aluminium electrolytic capacitors are quite rugged and can be used closer to their rated voltage, but there is a trade-off, mainly in terms of ESR and physical size. <S> I do not know if <S> your vehicle battery powered system is on the unswitched system in a vehicle (i.e. it is connected to a generator for charging); if so, you would need to withstand very large voltages during load dump <A> There are many factors that help decide which kind of capacitor to use. <S> None of those factors are revealed here. <S> In reality, bypassing a large battery is not a particularly rigorous application and the type of capacitor is not of great concern. <S> A 50V (or 63V) capacitor would be quite reasonable for a battery with a nominal voltage up to 36V. <S> That is probably what most designers would choose. <A> Ceramics are good when you want <S> really low ESR/ESL, Electrolytics when you want lots of bulk capacitance and Tantalums occupy some middle ground between the two. <S> Just be aware, ceramics (well, X7R and X5R at least) lose much of their capacitance near their max voltage. <S> A 50V 1uF ceramic might have 1uF at 10V but 0.2uF <S> at 40V, Samsung and TDK have charts of this for a lot of their ceramics (many others don't). <S> Electro's and Tantalums don't have this problem.
| Ceramics with class 2 dielectrics (X5R, X7R, Y5v and so forth) have a significant change of capacitance with DC bias , and for that reason I would normally use a device with a rated voltage of at least double the voltage I am bypassing with this technology.
|
Why is a cellphone charger rated like that? Here is what I read on the nameplate of my phone charger: Output 5V DC 1A. So, I think the Max. Output power is 5 watts. Input: 100-240 AC 50/60Hz 200mA. I assume that the mains voltage is 100 volts. If the input is 5 watts, The input current is 50 mA only. Why is it rated at 200 mA? I think the headroom of rated input current is too much. I would rate it at 100 mA which means 50% of safety. Can transients reach more than that? Thank you very much, <Q> It has more to with how switching supplies are built. <S> They rectify the mains and smooth it with a capacitor. <S> Problem is, current can only flow when the input voltage is higher than the capacitor voltage (peak of the sine wave). <S> So what happens is that the power supply draws energy from the mains in short bursts (while the capacitor feeds a steady trickle of power to the rest of the power supply), so it may very well be pulling 200mA, but only 25% of the time. <S> For small supplies up to a hundred watts or so this is no big deal, but in larger units steps need to be taken to make sure the power supply draws a more continuous stream of power rather than in intense bursts (Power factor Correction). <S> (side note, cheap electronics sometimes skip these steps, there was a no-name led light I tested at my old job that pulled all it's energy in a brief 200W burst despite only being rated to 5W <S> , it hit 135'C on the outside <S> amazingly it kept going). <S> Most power supplies that cost more than a dollar perform better than that particular led. <A> It could mean one or more of several things. <S> Output 5V DC 1A. <S> So, I think the Max. <S> Output power is 5 watts. <S> Correct. <S> This states that the power supply will maintain 5 V up to a 1 A current draw. <S> Note that it doesn't say what happens if you try to draw <S> more than 1 A. Let's imaging the case where the power supply output voltage drops to 4 V if you draw 1.5 A. Power is now <S> 4 x 1.5 = 6 W. <S> This might reach a maximum at, say, 3 V and 3 A (9 W) because the power starts to decrease at some point due to the lower voltage. <S> Input: <S> 100-240 AC 50/60Hz <S> 200mA. <S> If we now take the input current specification as worst case then we should expect this to happen at lowest voltage, 100 V. P = 100 <S> V x 0.2 <S> A = 20 W. <S> This could be: <S> Manufacturer being cautious. <S> Maybe the PSU is very inefficient into a high current load or short. <S> It may be impossible to find out. <A> Many testing agencies and/or international marketing laws for electric appliances require you to specify either the worst case, being inrush or hiccup, current, or fully specify the current drains, powers and power factors. <S> In quite a few cases, in fact, you are required to specify the limiting current (either as an extra line, or as the number if you have only one line), which can be a "by design cannot draw more than" current, or can be the current at which an internal fusing element will blow. <S> Add to that cautious component maths around the power factor of ringing core transformers at varying loads and input voltages and <S> >=300% peak currents surprise me not at all. <S> Apart from that, under-estimating the peak current will mean you fail random tests, over-estimating peak currents of small appliances <S> costs you nothing. <S> As long as you stay under an input power of 50W (and likely even 100W) rated the rules will be the same almost all over the world for a certain device and design. <S> Regardless of whether below 50W means 10W or 40W. <S> So why not estimate a worst case scenario of 25W or 40W, if it means you always pass the requisite tests and have to change nothing about the design anyway? <S> Concluding with the personal experience that a well designed small ringing core can be made to function at upward of 90% efficiency with a no-load/null power of (much) less than 300mW, I wouldn't lose too much sleep over your power bill. <A> The input rating must take into account the huge inrush current that occurs when input filtration capacitors charge up when an SMPS (switched mode power supply) is turned on. <S> This can cause current spikes an order of magnitude larger than the steady-state current. <S> For example, the input ratings must be taken into account for devices powered by a DC to AC inverter in order to determine if the inverter will be overloaded. <S> As such, one can often find discussion of these matters by inverter manufacturers. <S> For example, below is an except from Samlex's Inverters - General Information . <S> See esp. <S> the second paragraph ("The inrush current at turn on is several to tens of times larger that the steady state RMS current"), which is illustrated by the graph in Fig. <S> 2 where it is 15 times larger.
| It could be the maximum surge current on switch on.
|
How to increase the distance of wireless power transfer? I'm trying to make a wireless charger. But I want to make the distance farther.So, How can I increase the distance of wireless power transfer? Does making the frequency higher will increase the distance?Or winding more turns in the coil? Or making the voltage higher? <Q> Beyond a certain distance magnetic flux density reduces with the cube of distance. <S> To avoid this you need bigger diameter coils. <S> Try googling for the formula because unfortunately I'm on an android and haven't learnt to embed pictures using one. <S> Something also to consider is making a passive coil extender if this is practical for your application. <S> Basically it's two sets of coils that take power at the drive end and deliver it to the receiving end. <S> but you will need to consider litz wire to make this effective. <S> Basically you need to maintain Q at lower inductance values. <A> larger coils <S> increasing the power will have a small effect on range, but as wireless power is near-field signal strength will fall off rapidly as you increase distance. <S> So, larger (diameter) coils is the only practical way to increase range. <A> Your range is pretty much limited to one coil diameter. <S> You can stretch this a little by increasing the Q of your coils, and backing/coring them with ferrite. <S> Increase the Q by using Litz wire, and high Q caps.
| Lowering inductance will cause a higher coil current and this delivers more flux at a greater distance for a certain drive voltage
|
I want to learn how to make a CPU from scratch I am new to electronics, I program from time to time and i know the basics of the logic gates. I have used logisim in the past and I was just wondering if someone could tell me about a good online class i can take or a good site to learn more. Thank you in advance <Q> I don't know about online classes, but I can tell you what I did when I designed my first CPU. <S> So there are three steps: Get a good Digital Design / Computer Architecture book. <S> I strongly recommend the "Digital Design and Computer Architecture" from Harris and Harris, and also "Computer Organisation and Design" from Patterson and Hennessy (2nd edition). <S> Of course there are tons of other books, but these are generally accepted as very good ones. <S> I prefer VHDL, but you can also try Verilog Design your processor on paper. <S> Begin with a relatively small instruction set, and then you can add functionality or upgrade it (pipeline stages, FPU, etc.) <S> After you design your ISA and your architecture, you can start implementing it on the HDL you like. <S> Then you can "download" it on an FPGA evaluation board and see your CPU running. <A> I can recommend this online coursework . <S> You make a 1 bit ALU <S> , then a 4 bit ALU and then an 8-bit ALU. <S> Download Quartus II from Altera and start making a multiplexer, an adder, a decoder and put it together as an ALU <S> and you got a 4 or 8 bit system that can perform your operations. <S> If you get an FPGA you can even execute your own code in your own custom CPU. <S> FPGA is a good way to learn but you can also simulate everything in a program like Qsys in Quartus II. <A> I agree with @transistor's comment. <S> Programming at assembler level might be a much quicker and more effective way to learn initially than trying to start at designing a CPU. <S> I think designing a CPU would be quite a lot of work to get it to do anything meaningful, and without proper test cases or use cases, the design might be rubbish <S> but there'd be no way to tell. <S> To get the thing to do anything meaningful, it'll also need memory and peripherals. <S> To put the scale of the task into context, simple 32bit embedded-style CPUs are over 2000 gates. <S> (Based on recent RISC-V comparisons, I'll try to find the link) <S> I believe that might be without a bus interface, or important peripherals like a communication mechanism or any timers. <S> IIRC the size of the 32bit ALU and registers makes a difference, but a usable 8bit CPU is unlikely to be a lot smaller. <S> Learning to program at assembler first would be necessary anyway, as you will likely want to test your CPU design, so it isn't a waste of energy. <S> I'd recommend program in C and assembler with ATmega, maybe even an Arduino as they are quite cheap and quite well documented. <S> They important value of ATmega for me is the relationship between C and assembler is good enough that you can write in C and understand the assembler code relatively easily. <S> Further, ATmega was designed to be a good target for C in embedded systems, and C/C++ are still the favourite high-level language for embedded development. <S> So you would be learning about a relatively modern architecture designed for a high-level language, rather than assembler programming. <S> Finally, it is not cluttered with lots of mechanisms designed to support full operating systems like Linux, so there is much less to learn. <S> A key question is what is it you want to do with your CPU? <S> However, your question is likely to get opinion-based answers, which is not a good ee.se fit.
| Learn an HDL (Hardware Description Language).
|
What do these dashed/dotted lines mean in this power cord schematic, and how should I ground this device? --Edited at Sparky256's suggestion for better clarity-- Background Information I'm making a power cable for a piece of obscure, surplus military equipment (an AN/UGC-74B teletype, to be precise). I have the manuals that describe the power cable, but the wording doesn't account for part of the schematic, and I can't find a definitive answer that seems to make sense searching online. I'm somewhat familiar with electrical terminology and symbols though not an engineer by any means-- just a hobbyist who still doesn't know much compared to most of the people who will read this. I haven't seen anything like this before. If it's something simple, I apologize. On the right side of the schematic we see the standard three-pronged 120-volt male adapter. BLK and WHT wires connect to pins K and M on the 12-pin Amphenol adapter (not all pins are used) on the other end of the cord, but BRN (or ground) stops at the dashed line that appears to make a circle around the other wires. Then, on the other end of the cord, we see another dashed circle, this time with another wire, also labeled WHT, seemingly coming off it to pin E. Another schematic of the power input on the device itself shows that pin E appears to have no input from the cable but connects to chassis ground via something called E2-- see here: Also note that there is not only this power input connector (labeled J2 in the schematics) to which my cable will connect, but there is also a ground terminal, and I'm not sure if I should bring the ground wire out of my cable and connect ground there or not, either. If the device isn't grounded via pin E through the power cord's ground wire, I'd assume that I'd need to ground via the GND terminal for safe operation instead. This is an image of the power input and ground terminal to show you what I mean (additional photos of various parts of the teletype available here-- not my photos but the same model of TTY: http://imgur.com/gallery/FK5nB ): At this point, I just am not sure what to do about this. I want to make sure I create the cable properly so I don't break anything; I understand everything else about how to make the cord, but not what to do with ground and pin E. The manual can be accessed here. It's pretty hefty, so maybe I've just missed something it instructed somewhere-- but I don't think I did: http://www.nj7p.org/Manuals/PDFs/Military/TM%2011-5815-602-24-1%2015-Sep-87%20USAPA.pdf The full schematics I referenced are on pages 143 (labeled 3-93 in the manual) and 396 (labeled FO-3), as well as additional information describing the power input wiring on page 96 (labeled 3-46). The Specific Questions I Have What do the dashed/dotted lines referenced earlier mean? If the dashed/dotted lines indicate some kind of shielding, does the BRN AKA ground wire connect to the shielding and then the second WHT wire connect from the shielding to pin E, effectively grounding pin E? If the ground wire does not connect to pin E through the shielding, would it be wise to instead connect my cord's ground wire to the GND terminal instead? <Q> The dotted lines that circle the bundle of signals is often used to delineate a cable bundle (e.g. wrapped together in a sleeve), and when signals are connected to it, that generally means that it represents the (conductive) cable shield. <S> The second thing you circled (in J2), is a typical representation of Earth ground. <S> I take it to mean that the chassis is bonded to Earth ground. <A> I see several details in your drawings. <S> 1) <S> The top drawing is of a shielded power cable, and the power can be AC or DC. <S> 2) <S> The J2 connector has a gnd symbol at the bottom to clarify <S> it is earth grounded. <S> 3) J2 also shows built-in capacitors to filter out EMI noise from either direction. <S> I would use pin 'E' as power ground even if you just run a green wire to your power source ground. <S> 4) <S> The dashed lines around J2 refer to the chassis in an abstract way, and show it is grounded to J2 and J3 connector shells directly. <S> 5) <S> There maybe other drawings that tie into these drawings, but for power source and ground these seem to cover the issues. <S> 6) <S> The power cable does not have to be shielded for non-military use. <S> Make sure J2 is wired correctly for the voltage you are using, AC or DC. <A> I would agree, this appears to be a conductive shield around the the 2 leads in the power cable, the clue here is the military spec on this device, this is done generally to try and prevent Emf interference on the operating device <A> The dashed oval outline designates a conductive overall shield. <S> Typically a braid of small, bare copper (or tinned copper) wires. <S> However ordinary power cables just have the safety ground as a third, insulated internal wire just like hot and neutral. <S> Note that back in vintage days (as when those diagrams were drawn) <S> the color standard in North America was white for neutral, black for hot, and green (or some other color) for safety ground (if there even was one). <S> Today the European Community standard along with the international marketplace have de-facto standardized on power cords with light blue for neutral, brown for hot, and green with a yellow stripe as safety ground. <S> For something like an antique teletype, I can't see that you actually NEED a SHIELDED power cord. <S> An ordinary cord capable of the voltage and current rating would be completely sufficient.
| Pin 'E' on the J2 connector is Earth ground for the chassis and power.
|
Why is there always a capacitor on input and output of a voltage regulator? Looking at the datasheet I can see that the voltage regulators are not just a zenner diode inside, they are complex devices. I have noticed that there is always a capacitor at the input and another one at the output. An example is the uA7800 series fixed voltage regulators. I have read that one of them is to "stabilize the circuit operation" while the other is to "reduce ripple on the output". Looking at the datasheet, why do they have this fixed value? And if they do have a fixed value then why not just fabricate them into the voltage regulator itself? e.g for the uA7800 series it is 0.33uF at the input and 0.1uF at the output. It is not explained why they have these values. <Q> Most voltage regulators (especially LDO types) require a capacitor on the output for stability, and it will usually improve transient response even for regulators like the 7800 that may not strictly require it. <S> An input capacitor is usually required to reduce source impedance. <S> It is impractical to make capacitors more than tens of pF <S> (or so) on an inexpensive chip- they take up too much expensive silicon area, and external ceramic or electrolytic capacitors are very cheap in quantity. <S> That is not in the cards. <S> And the capacitors actually provide energy storage <S> so it's not something that clever circuitry can substitute for. <S> The values are compromises that make sense based on the chip stability behavior at different load currents, and also what caps were common when the datasheet was composed (that might be 35 or 40 years ago for the 7800 series). <S> It is almost always acceptable to use a larger capacitance on the input, and usually acceptable on the output, however there may be minimum/maximum values on the capacitor ESR- the equivalent series resistance. <S> Most modern regulators will indicate what values and types of capacitor are acceptable, so reading and understanding the datasheet is all you need to do. <A> The answer to this question lies in practical experience. <S> Omit the input capacitor and, sooner or later, the stabilizer goes into self-oscillation, over-heats and (literally) explodes. <S> These chips are, in fact, series-stabilized shunt-controlled arrangements with enormous gain in the side-chain. <S> That input capacitor controls phase-shifts. <S> As stated the output capacitor is normal with any regulator/supplier circuit. <S> Full details see: http://www.clovellydonkeys.co.uk/kengreen_website/index.htm <A> I thought I would build my latest circuit one component at a time so that I could "debug" and learn what was actually happening at each step. <S> So I hooked up my 12V supply to my Input and Ground and eagerly waited for the 5V regulated output. <S> What I got was 10 seconds of looking for a 5V output followed by smoke!That <S> is what happens if you don't have the capacitors in the circuit. <S> Self-oscillation, huge heat and smoke. <S> Please take my word for it and don't try this at home. <S> Teaches you to read the datasheet properly from cover to cover.
| In some cases a capacitor that is too ideal may cause the regulator to oscillate.
|
How can I do offset nulling for this opamp circuit? Circuit below amplifies 0-5mV signals to 0 5V level. But at zero input there is like 0.5V offset. Is there an easy way to offset null? edit: edit2: <Q> You could add a pot as shown. <S> The LM324 is not a good op-amp to use for mV DC signals- offset can be as much as +/-20mV <S> and this will only adjust for + <S> /-5mV offset (you have observed 0.5mV). <S> simulate this circuit – <S> Schematic created using CircuitLab <S> If you don't like having to come up with +/- <S> supplies (or having it drift around by maybe 5-10mV/ <S> °C at the output), use a better single-supply op-amp . <S> Edit: Since you are using a thermocouple, and assuming it is floating, you can (and should) offset the input by adding a voltage divider to a reference on the (-) side of the thermocouple. <S> Otherwise you'll never be able to measure below the temperature of the instrument terminal block. <S> but that's way beyond the scope of this question. <A> The LM324 datasheet says Large output swing: 0 V to V\$_+\$ - 1.5 V. <S> However it's not very good at close to zero. <S> See my answer to <S> How can LM324 work in dual supply mode when it only has 2 pins for the supply? <S> where this is discussed in some detail. <S> Try connecting the op-amp negative supply pin to a -2.5 volts supply and run the simulation again. <A> You may not need to, and I suspect it would be the wrong approach, upsetting the behaviour during normal operation by trimming the input for an output-related problem. <S> The datasheet (Table 6.5 on page 5) shows that the LM324's output stage can source 20-40mA but only sink as many microamps under similar conditions. <S> Given this information, you can improve the performance of the LM324's output stage in certain applications by forcing the output stage into Class-A, so that it only ever sources current and never sinks it. <S> To do this, simply connect a resistive load from the output <S> o V- <S> (0V in your example). <S> 1 kilohm <S> if you can afford a few mA extra current consumption, up to 10 kilohms if you can't. <S> Start with 1k, and if that fixes the problem, experiment with higher values if you need to. <S> One benefit is the elimination of crossover distortion (to which the LM324 is prone in AC applications such as audio). <S> But here, the benefit you're looking for is improved output swing when the output voltage is close to 0V.
| It's possible to combine the voltage divider with a temperature sensitive element to both offset and cold-junction compensate the input
|
How To Detect a Gunshot Hey everyone I am working on a project that is a lot harder than what I thought. It is a project for school. How can I detect a gunshot indoors without using an array of microphones? I had something going well with using microphones, but I would be breaking patents which I cannot do for this project. <Q> One point to note : patents stop you selling what you build (if it infringes on them). <S> But they don't stop you building one to test. <S> That's deliberate, because they are as much about communicating the patented technique as they are about protecting the inventor - to encourage the spread and growth of good ideas for public benefit. <S> In theory. <S> In practice, the "communication" aspect is encrypted in legal jargon and is usually written to satisfy the letter of the "communication" aspect without actually communicating anything. <S> If an inventor can get the protection of a patent without breaking the trade secret behind it, he has a good patent attorney drafting the claims... <S> Nevertheless. <S> you have the right to build any patented invention to test that the published idea works, as well as to improve on it if possible. <S> You just can't sell (or hire or lease or give away or benefit from) the product without some agreement with the inventor, unless you are prepared to challenge the validity of the patent in court (not recommended!) <S> So legally I think you can use a patented technique for a school project (though I Am Not A Lawyer) - with proper attribution of course - as long as you are gaining no material benefit from it. <S> The school may consider winning a competition or passing a course is a "material benefit", or they may impose their own rules for the sake of the competition - it may be worth asking your tutor or adviser. <A> A microphone is the easiest way to detect shots (well, plus some processing and communication). <S> Gun shots are loud. <S> Really loud. <S> As in, 140dB to 180dB . <S> For comparison, a jet engine is 140dB . <S> They also have very distinct characteristics, such as a very short, sharp rise in audio level. <S> You will need to collect actual data on how different guns sound indoors. <S> When you do, please wear good hearing protection. <S> I usually recommend wearing ear plugs under muffs when shooting indoors. <S> Movies simply do no convey just how loud firearms really are. <S> With that said, you explicitly stated you can't use microphones (although you probably can, but I do understand how school projects can be). <S> You might want to look into creating a "burning gun powder detector". <S> I'm not sure what all is out there as far as "smoke classification sensors" (maybe a spectrometer?), and you'll probably have to borrow a chemist or two to get a good profile on burning gunpowder. <S> Other than a big bang and smoke, firearms really don't produce anything else. <A> I am aware this sounds verry silly, but if you are not allowed to use microphones you could try to build something as simple as a breadboard with a couple of wires attached to some LDRs (Light depending resistors) hook those op to a serial monitor (you could use Arduino for this if you are familliar) and you can monitor diffrences in light... <S> Ofcourse a microphone would be way more accurate. <S> But the flash of a gunshot should trip a ldr. <S> Again, this could be complete nonsense.
| Kind of like a smoke detector, except much more sensitive and only alerts on certain chemical compounds.
|
Is there a technical reason why most touch screens use glass rather than plastic? Most modern touch screens in portable devices are made of glass. This glass often breaks if accidentally dropped. Also, it is very reflective, making it difficult to use in strong light. I know that touch screens without glass exist. For example, the multi-touch screen on my e-ink e-reader has a plastic front. I remember many other examples, such as the personal in-flight entertainment systems on many airplanes. What are the reasons that most modern portable touch devices come with a glass panel on their fronts, rather than plastic or something else? The cracking of glass seems to be a pretty big problem. Edit: I've seen a lot of cracked touch devices, and it's nearly always only the front panel that's cracked. The actual display is usually fine underneath. Even the digitizer usually works perfectly. <Q> Title of question: Is there a technical reason why most touch screens use glass rather than plastic? <S> Note the word "technical" and <S> not "marketing" <S> What are the reasons that most modern portable touch devices come with a glass panel on their fronts, rather than plastic or something else? <S> This makes life easier on the electronics that has to detect finger positions and movement. <S> Taken from this article <A> When decisions about consumer electronics are made, many reasons beyond technical come into play. <S> There is no valid reason for a phone to be disassembled in 7 pieces in order to replace a battery, yet that's how one of the most popular phones is made. <S> Mobile phones are as much a product of marketing as they are of electronics, and many design decisions become clear when you take a look at that perspective. <S> Glass looks good, so it sell good. <S> And when it shatters, people have to pay again - either for a new phone, or for a glass replacement job. <S> Plastic doesn't shatter or otherwise fall apart, unless you try to cut or burn it on purpose. <S> It can also be made matte, which makes the screen much more readable in presence of reflections and glares. <S> Since plastic doesn't have to be hard, it can be made thinner than glass, improving touch sensitivity. <S> Unfortunately, it looks cheap even before it is scratched (and plain terrible after), so you can't make big money selling phones with plastic screens. <S> Worse, people will carry these cheap-looking phones for ages (because the screen won't shatter), projecting that cheap-looking and outdated image of your brand everywhere they go. <S> So you either go out of business, or switch to glass like everyone else. <A> You mention cracking as a downside to using glass, but most touchscreens will encounter far more potential scratch-causing events than crack-causing events. <S> Glass is highly scratch-resistant: <S> at a Mohs hardness of 5.5, it's harder than anything else in your pocket (steel is around 4). <S> Synthetic sapphire is even more scratch-resistant: at a hardness of 9, the only common material that can scratch it is diamond. <S> In contrast, most plastics have a hardness less than 1, and will get scratched up in short order (among other hazards, fingernails have a hardness between 2 and 3). <A> Glass is hard, and therefore brittle, so it shatters. <S> Plastic (acrylic or polycarbonate) is softer, so more prone to scratches. <S> It's certainly a possibility and some cheap phones have plastic touchscreens. <S> But the underlying LCD behind the transparent touchscreen has to be made of glass, due to high temperature parts of the process. <S> So that's still vulnerable to breaking. <S> The ultimate is synthetic sapphire, which Apple were going to use but abandoned for some reason. <S> Much harder and harder to shatter than glass. <A> Resistive touch screens are plastic <S> Capacitive are glass - for capacitive touchscreen to work, there is wires manufactured on the glass itself - this up to just recently was possible on glass only so this is why it is glass. <S> Also LCDs are from glass for the same reason, there are already plastic film LCDs but are pretty new (like flexible amoleds and flexible epaper) <S> Most ereaders uses IR touch sensing(which <S> enables to use plastic covering of the display, but the epaper module itself is glass based again) <A> Here's some history: Back in the day, almost all of the early touch (not-so-smart) phones used plastic displays. <S> It was, in fact, Steve Jobs , who demanded that the first iPhones use unscratchable glass. <S> He said that consumers would carry their smartphones with keys in their pockets and products which were easily damaged weren't acceptable from a corporation like Apple. <S> This was less than 3 months before the iphone's launch date. <S> “I want a glass screen," Steve is quoted as saying. " <S> And I want it perfect in six weeks.” <S> Obviously, other companies followed suit. <S> Source: <S> http://www.businessinsider.com/steve-jobs-new-iphone-screen-2012-1?IR=T
| Glass (as a cheap and common material) has a good dielectric constant (more than most cheap plastics) and this makes the change in capacitance bigger for those devices using that technology.
|
Is it okay to use these kind of batteries with this holder (pictures inside) I have the pictured battery with 1.2 v .. It does not look like regular AA batteries on both ends. My questions: 1) Is it fine to put the batteries in the holder although it does not fit (the batteries are slightly shorter than regular AA). So I have to force the ends to touch. 2) I do not know why the holder specifies 1.5 V on each slot. Would it be problematic if I use 1.2? Battery brand and specifications: Lucky sky AA NI-MH 1200 mAh 1.2 v (made in china) Thanks. <Q> Although the cells fit mechanically, those tagged cells, intended to be soldered, are capable of much higher current output than the spring terminals in the battery holder are intended to handle. <S> If your application is below the fraction of one amp for which the holder is intended, then fine. <S> If your application is capable of drawing the several amps that those batteries can source, you will probably have problems with the quality and consistency of those spring connections. <A> It is OK if it does not have mechanical issues. <S> 1.5V is the voltage for regular batteries, this is rechargeable <S> (NiMH) <S> and their voltage is 1.2V. <S> This is not the issue for holder at all. <S> Please take note that the final voltage would be lower (4x1.2V instead of 4x1.5V). <S> These types of batteries are meant to be soldered on board <S> so this is the reason they have these connectors spot welded on the ends. <A> The battery holder is designed for AA (or Penlite) batteries. <S> The battery holder cannot tell the difference between 1.5 V and 1.2 V so you can still use it. <S> The battery has soldering tabs <S> so it also is not designed to be used in a battery holder. <S> It is designed to make packs of batteries with wires to a connector. <S> Still, you can use it in a battery holder. <S> But since the batteries are slightly shorter than an AA don't be surprised if the connections get loose after a while. <A> Basically, if it fits, you're usually fine. <S> That battery looks like it was intended to be soldered into a big pack, but there's no reason you couldn't remove the tabs and put it in a holder. <S> The 1.5V marking has more to do with common naming conventions rather than a specific voltage rating. <S> A lot pf people refer to AAs as 1.5V batteries (mainly because the non-rechargeable ones are ~1.5V). <S> It's really just 1.5V as opposed to a 9V battery or a 6V battery. <S> It's only a connector so if it fits and whatever you want to run is happy with the slightly lower voltage of rechargeable AAs, then you're fine.
| Yes you can use this holder for this battery even though it is not designed for it. Normal Alkaline batteries are 1.5 V when full so that is why the battery holder states 1.5 V. Usually this is not the issue, but check if that voltage is suitable for the device you want to power. It depends on the current you want to draw.
|
Is it possible to make a plastic touch screen as sensitive and accurate as a glass one? I learned in this answer that touch screens with a glass front are generally more sensitive and accurate than those with plastic fronts. This corresponds with my personal experience. However, is there a way to bring devices with plastic fronts up to the accuracy and sensitivity levels of those of glass? What would this require? <Q> Your questions requires a bit more specification as to the technology behind the touch detection. <S> Touch detection techniques vs. Materials <S> Typically, plastic surfaces are resistive touch panels placed in-front of glass or plastic LCD screens. <S> They have to be made from plastic (as opposed to glass) because they detect touch by deforming . <S> The user's finger bends the plastic film where it touches and this presses an upper film into one beneath it completing a circuit (4-wire design) or enables a complex triangulation of the deformation location (5-wire design). <S> This style of screen is used in many industrial applications because it works when you have gloves on (factories, hospitals, etc). <S> Typically, glass touchscreens use a technique called capacitive touch detection. <S> They emit a constant quasi-static electric field and look for changes to that field caused by the presence of conductive (charge storing) materials nearby (like your finger). <S> Yes. <S> You can make both. <S> It is possible to make capacitive screens on plastic substrates. <S> There are many applications that do, but it's not done on CellPhones (and virtually all consumer applications) because for the higher cost of the capacitive circuitry and the relatively minimal cost to go from plastic to glass, the outcome is much nicer for the finished product. <S> Glass works well because it is optically clear and doesn't electrically polarize easily, this can hold true for certain plastics as well. <S> This is, however, not true of the typical commercial plastic (PE, PET, PTFE, PVC, etc) -- you need additives or alternate chemistry. <S> Source: Geoff Walker, Sr. Engineer, Touch Tech. <S> @ <S> Intel; Lecture slides from FPD China 2013 <S> So in short... <S> Plastic touch panels appear less sensitive because the ones you commonly encounter are a different type of technology (resistive touch panel -- RTP) that requires physically deforming a film, measuring the location of the deformation, and tracking one finger at a time. <S> Glass panels appear more sensitive because capacitive sensing responds to the mere presence of your finger and does not even require that you touch the screen. <A> There are many ways of constructing a PCAP touchscreens and some of them are entirely independent of the glass material. <S> In effect the touchscreen is a film attached to the coverglass (or plastic) <S> so as long as the material does not mess with the electric field projected by the capacitive terminals the touchscreen is perfectly happy. <S> On the other hand glass has better permittivity than most plastics <S> so you get roughly similar capacitance with 2-3x thicker glass than you would with plastic. <S> Whether or not this is significant depends on the application. <S> General trend is to create as thin coverglass as possible and thick glass solutions are reserved more to vandal-resistant public displays. <S> Some touchscreens are actually integrated to the LCD display, which is what Apple does. <S> Without going to gritty details, your touchscreen requires two terminals, between which the capacitance change is sensed. <S> Depending on construction these may be separate films with insulating layer in-between, single layer constructions with some clever routing, exploiting LCD display cells as a terminal etc. <S> Most basic of these are organized as a crisscross pattern with separate layers, more sophisticated use diamond patterns that etch jumper bridges to "hop" over opposing terminals and so on. <S> In fact you don't need the glass at all for the touchscreen to function. <S> Except touching the touch screen would "short" the traces. <S> If you're interested in touch technology in general, Geoff Walker has quite comprehensive presentations on the subject here: http://www.walkermobile.com/PublishedMaterial.htm <A> Plastic screens have resistive touchscreens and glass screens are capacitive touchscreen devices. <S> The difference is fundamental: the first checks the resistance which changes when you press the screen, and the second detects the presence of the capacitance of the finger. <S> Resistive screens must be calibrated and then can be accurate but are usually cheaply made and need to be recalibrated from time to time. <S> You can see more here: <S> https://en.wikipedia.org/wiki/Resistive_touchscreen https://en.wikipedia.org/wiki/Touchscreen#Capacitive
| You can make plastic capacitive touch screens as accurate as glass ones if you use the right substrates.
|
Amplify 0 to 3.3V input to +/-3.3V I'm working on modifying a board, while going through its schematic design I came across a section where a DAC output into the range of [0, +3.3V], which is then supposedly gained to +/-3.3V range, see extract below: U24 is the an AD5662 DAC, with both VDD and Vref at +3.3V. U25D is an OPA4170 opamp, with supply at +/-3.3V. It's in a non-inverting configuration. The transfer function is derived from: $$V_{+}=\frac{1}{2}(V_{out}-3.3)$$, so $$V_{out}=2V_{+}+3.3$$ Two things that don't make sense to me: Input range of 0 to +3.3V would be amplified to [+3.3v, 9.9V] according to this transfer function, since the positive supply is only at +3.3V,that would result in Vout to be constantly at the rail. Since the input voltage is always positive, how can the output of the opamp ever be negative? I would think that this is probably a mistake and that the left side of R61 should be connected to -3.3V, resulting in a transfer function $$V_{out}=2V_{+}-3.3$$, which would work as advertised. However, the production board does indeed follow the schematic, and the opamp converts a strictly positive Vout (sinusoid with offset) from the DAC to a zero-centered signal. This wouldn't surprise me if the DAC output is capacitively coupled to the opamp. But R57 and C51 act as a low-pass filter, so I'm kinda stumped here. Clearly this board's design works as advertised, but it doesn't make sense to me why. Can anyone point out where I'm making mistakes in my analysis? Edit: Seems like I made a pretty silly mistake in my transfer function derivation, should be $$V_{+}=\frac{1}{2}(V_{out}-3.3)+3.3$$ to account for the voltage divider potential as being offset from 3.3 rather than ground. Thanks all the for the help, I wish I could mark everyone's as solution. <Q> It's easier if you think of the circuit as an inverting configuration simulate this circuit – <S> Schematic created using CircuitLab <S> The voltage across R1 is 3.3 - V2, so the output is $$ V_O = 3.3 - 2(3.3-V2) <S> = <S> -3.3 <S> + 2V2$$ Keep in mind that for this to be accurate, the 3.3 has to be very accurately matched to the 3.3 reference of the DAC. <A> The input voltage for R61 is correct. <S> The voltage at the V- pin will be the average of Vout and 3.3V because R60 and R61 are the same. <S> In the case where V+ is at 0V, V- should also be driven to 0V. <S> The output voltage that satisfies this requirement is -3.3V. <S> The equation for V- is \begin{align}V_- = <S> \frac <S> 12*(V_{out}+3.3V)\end{align}The transfer function is then \begin{align}V_{out}=2V_+-3.3V\end{align}as <S> you say it has to be. <A> I don't really know what else to put in this section. <S> Foo bar. <S> $$ \frac {3.3V - V_+}{10k} = \frac {V_+ - V_{out}}{10k} $$ $$ 3.3V + V_{out} = <S> 2V_+ <S> $$ Negative feedback, so V+ = <S> Vin $$ Vout <S> = 2_{Vin} - 3.3V $$ Vin = 0, Vout -3.3VVin = <S> 3.3, Vout = 3.3V <A> The original design may have required only a single polarity output, so U25 is acting as a comparator. <S> The behavior of this circuit is determined by R61. <S> If connected to +3.3v, the op-amp would have put out -3.3v to keep its (-) input at zero volts, or if the (+) input went above zero <S> the op-amp would put out +3.3v, so it would switch output voltages. <S> 1) <S> If R61 is connected to GND, the gain is 2, so that means the DAC will be twice as sensitive to input voltage, but also means that U250 will saturate quicker, but it is an amplifier with a gain of 2, not a comparator. <S> 2) if R61 is omitted, then the gain is unity, acting as a buffer for the DAC, but it is a amplifier with a gain of 1, not a comparator. <S> 3) <S> If the op-amp power was +/- <S> 15 volts, not much would change in terms of its behavior, only the output, because Vo - 2(15 - V2) = <S> -15 + <S> 2V2. <S> V2 is only dependent on the DAC output, and is best defined as V2, or zero volts (for clarity).
| Any mismatch will result in a zero shift.
|
Why decoupling capacitors used for power rails should be very close to the IC pins? I read many experts here recommend the caps (usually 100nF ones) as close to the power and ground pins as possible. Why do they have to be soldered very near to the pins? And I used the term decoupling capacitors. Is decoupling and bypass caps same thing in this context. <Q> The ICs contain fast switching transistors generating RF signals. <S> As you know all electrical signals travel in loops. <S> For these ICs the loop is through the power supply pins. <S> The decoupling caps form a short circuit for these RF signals so the closer you mount the decoupling caps to the power pins of the ICs <S> the smaller the loop will be. <S> This is desirable as it increases the effectiveness of the decoupling because any distance increases parasitic inductance of the wires ( <S> about 1 nH per mm). <S> Also large loops emit more RF signals <S> so you have more chance of violating EMI (Electromagnetic Interference) specifications. <S> Also, a longer distance to the decoupling cap means that the supply voltage inside the IC will be more noisy and polluted with spikes. <S> Worst case the IC stops working because of all the ripple on the supply ! <S> Decoupling caps and bypass caps are indeed the same thing. <A> Decoupling Caps are used to avoid the noise/glitch on the power supply line. <S> Basically when you say noise, it can be of many types. <S> For the decoupling caps, one of the primary advantage is to remove the ground bounce(from Ground plane) & voltage sag (from voltage rail). <S> Inside an IC, using NMOS & PMOS circuit, the switching happens. <S> In the figure, when the signal is pulled to VSS, there will a voltage drop across parasitic inductor as shown which will lead to sagging of voltage & when the signal is pulled down to ground, there will be a voltage drop on the ground side parasitic inductance which leads to ground bounce. <S> As a board designer, you can not do anything about it. <S> So, to remove this issue on the board level, which is caused by parasitic inductance of trace & plane, we add a decoupling capacitor to provide a local path of voltage & ground. <S> The board level figure with decoupling capacitor is given below : <S> - The farther the capcitor is , the more is the trace length & the more is parasictic inductance. <S> So, it is advised to place it as close to the voltage or ground pin as possible. <S> It is a trade off or vendor recommendation to put it near to voltage pin or ground pin. <A> Yes "decoupling" and "bypass" capacitors are the same thing. <S> Ideally the power supply to a chip would have a zero impedance at all frequencies. <S> If the power supply has a finite impedance it will act as an unwanted coupling path. <S> The higher the impedance the stronger this unwanted coupling path. <S> The unwanted coupling path can have various effects. <S> In an amplifier it can cause feedback and hence oscilation in analog circuits. <S> In a multi-channel analog circuit it can cause crosstalk between channels. <S> In a digital circuit current spikes from gates switching can potentially cause glitches in other gates. <S> It is important to realise that the frequencies that cause a circuit to misbehave can be much higher than the operating frequency of the circuit. <S> An amplifier can potentially oscilate at any frequency where it has gain. <S> If you look at a fast edge in the frequency domain you will find very high frequency components. <S> All electrical connections have inductance and the longer the connection the higher the inductance. <S> So to keep the impedance at high frequencies down we place a capacitor as close as possible to the device. <S> This bypasses the power supply providing a low impedance path between power and ground at high frequencies and hence reduces the coupling between circuits fed from that power supply.
| During fast switching , the capacitor acts as a decoupling element to reduce the drop across parasitic inductance.
|
Why are both fuse and earthing are needed? This question arose in my mind today that when you already have the fuse for safety precautions why do you need the earthing? <Q> From the Wikipedia article " Fuse (electrical) <S> ": A fuse interrupts an excessive current so that further damage by overheating or fire is prevented. <S> Wiring regulations often define a maximum fuse current rating for particular circuits. <S> Overcurrent protection devices are essential in electrical systems to limit threats to human life and property damage. <S> However, one could easily be electrocuted without tripping a breaker or blowing a fuse. <S> Thus, earthing a metal appliance, enclosure, plumbing, etc. <S> prevents potentially lethal voltages from developing on these conductors. <A> If the live becomes in contact with any exposed metal part of a device and you touch that metal part it would not be good news. <S> If the exposed metal parts are connected to the Earth then a very low resistance circuit is completed when the live touches the metal part. <S> A very large current will flow, blow the fuse and disconnect the live from the device. <S> Some devices with metal parts do not require an earth because all the metal parts are covered with an insulator. <S> This is called "double insulation". <A> They do different things. <S> Almost opposite, in fact. <S> Earthing prevents potential difference building up between the device and surroundings (which would flow through you if you touched it, or arc through the air, or to the nearest chip or sensitive component, blowing the device). <S> The fuse prevents a large current flowing through the circuit. <S> In a lot of cases, there is a breach from the live wire to the casing, and it's the earth wire that ultimately swallows the current. <S> So, you need the earth to equalize potentials, which prevents the current that would blow the fuse. <S> And you need a fuse to prevent a large current flowing through the earth (and other parts too). <A> Fuses are over-current protection - a 10A fuse might help prevent a fire, since it limits the power that an appliance can receive. <S> What other answers haven't mentioned is earthing or protective-earth wiring (PE). <S> Sometimes a live wire may come in contact with a metal casing of a device - this will not cause a fuse to blow. <S> However the case is "live" and if someone were to touch it the results might be fatal. <S> That's where a residual current device comes into play ( https://en.wikipedia.org/wiki/Residual-current_device ). <S> Current should normally run only in live or neutral wires. <S> If it happens to run in a PE wire, or finds another path to ground (e.g. through a human body) <S> the RCD will break the circuit. <S> This may happen at current of 10-30mA flowing the "wrong way" even though the appliance nay draw 10A and function normally. <S> Personally I consider the RCD to be a simple and beautiful device - go read more about it ;) <A> To add to the previous answers, a sustained current of around 30mA is enough to kill a person. <S> The fuses in house circuits might be in a range around 5A to 45A. <S> So a fuse by itself does nothing to prevent electrocution. <S> The earth (ground) conductor provides a low-resistance return path for current should a live wire come adrift inside an appliance and touch a metal part. <S> This low resistance will allow a very high current to flow, blowing the fuse. <S> This cuts off the power before anyone touching the appliance gets electrocuted.
| Fuses are typically to protect wiring and other components from overheating and perhaps catching fire and/or exploding in the event of a low-impedance fault.
|
How can I protect my MOSFET? I'm using this depletion-mode MOSFET in a high-voltage power-management circuit. My 500V DC supply isn't high grade, and now that I've blown two of these MOSFETs (gate shorts to drain) I'm wondering if I can put any components on either side to protect it? I was getting weird results with the first one before realizing it was blown. I might have blown the second one (not sure at which point in my testing it failed) simply by turning the high-voltage supply on with it wired like this: simulate this circuit – Schematic created using CircuitLab For protection purposes: If it simplifies things I never want to see more than 200mA through the circuit. If no better ideas I guess I could put a fast-blow fuse in front of V+? But I'm not sure if that's the only thing that can blow this MOSFET. E.g., I'm beginning to wonder if having no load on the drain can cause problems. <Q> The common way to protect a MOST gate is to use a Zener diode between gate and source. <S> Your MOST has a max Vgs of 20V <S> so add a ~15V zenned diode in reverse bias and you'll be fine. <S> There are also devices called transils, which are specialized for over-voltage protection but do basically the same. <A> <A> Just wanted to add, for future people viewing this, why the original circuit was failing. <S> If the transistor turns on more than a little then a large portion of the supply voltage is across the resistor, causing Vgs to be a high negative voltage. <S> In ideal conditions this would never happen but was likely occurring due to either pulse effects when turning on, or it began oscillating at some point. <S> This could be avoided with a small capacitor from gate to source and a resistor, if this current limiter does not need to respond very quickly these can be oversized, making the selection easier. <S> And still throw a zener in there just in case. <S> This circuit should both protect the gate and reduce the chance of oscillation: <S> simulate this circuit – <S> Schematic created using CircuitLab
| This datasheet provides detailed recommendations on employing TVS diodes to protect both the more sensitive gate and the entire MOSFET against transients, by connecting the diodes like this:
|
Can I only use a crystal earpiece for my crystal radio? I am building a crystal radio for my science fair project, and I am having some issues. For my earpiece, I am using a telephone receiver which may be the issue. For my ground, I am using a water spigot in my backyard which is completely metal. My diode is a germanium diode, which works. My coil is a bottle wrapped in enamel coated magnet wire that is between 18 and 22 gauge. Instead of sanding a section of the wire and using a wiper blade, I have taps with the enamel sanded off. I have tested all of my parts with a voltmeter, so the circuit isn't the issue. Any advice? What might I be doing wrong? <Q> The problem involves the "impedance" concept, so it's usually not discussed in beginners' crystal radio projects. <S> Crystal radios need a high-resistance earphone. <S> This is the type that produces tiny sounds when driven by several volts, while only drawing a few hundred microamps. <S> For AC signals it appears to be a large resistor, 5K or higher. <S> A standard 8-ohm earphone appears as a much smaller resistor for AC; roughly eight ohms, and it expects a signal of hundreds of millivolts, while drawing a few tens of milliamps. <S> Yet the RF signal from a crystal radio must be high voltage at low current, so it can well exceed the 0.3V detector voltage. <S> It's designed to produce few-volts DC output, not few-tenths. <S> A standard earphone will just short out your receiver, and won't convert very much of the DC output into sound. <S> There of course is a simple cure. <S> Connect a small audio transformer to your earphone. <S> You want to step down the few-volts output by a factor of 20 to 50, converting 8ohms into many Kohms. <S> Small transformers like this are available, called audio matching transformers, or audio output transformers for old-style transistor radios. <S> Connect one of these to your tiny earphone, so the high-volt, high-ohms side connects to your crystal radio output. <S> The above will work, but unfortunately this transformer uses up some energy (as wire heating,) as does the coil in your earphone. <S> A piezo-crystal earphone is far more efficient than a coil/magnet earphone. <S> So, for the same milliwatts of EM energy being received by your radio, a crystal earphone will sound distinctly louder than a coil earphone, even if exactly the right audio transformer is being used. <S> The missing audio ends up as coil-heating. <A> I doubt that a telephone receiver is sufficiently sensitive to work with a crystal radio. <S> It needs a significant amount of power to operate, which you cannot get from a crystal radio. <S> You will have to use a crystal earpiece, such as this. <A> the problem is the lack of power, a crystal radio draws its power wirelessly from the signal that it's receiving. <S> Regular speakers require quite a bit of current to operate (and a reasonable voltage - hence high power). <S> So unless you have an antenna the size of a house, it's probably not going to collect enough energy to drive a regular speaker. <S> If you had a large enough antenna you could power a full hifi system, but the antenna would be impractically large. <S> If you're interested in working out roughly how much power might be available, measure the voltage at the speaker (with the speaker disconnected) using the AC volts range, then measure the AC uA, multiply the two and that should give you a rough idea of how many microwatts you've got to play with <A> The heart of the crystal set (as in any other radio receiver) is a parallel tuned circuit. <S> This parallel tuned circuit is attached to the antenna and the antenna behaves like a very short dipole. <S> This means it looks like a capacitor to earth from the perspective of the tuned circuit. <S> Why does it behave like a short dipole - at 1 MHz <S> the wavelength is 300 metres and half of this is 150 metres and, does your xtal set antenna have an antenna this long? <S> Almost certainly not because nobody has that amount of space. <S> Hence the parallel tuned circuit formed by the coil and parasitic capacitance (or tuning capacitance) has added to it a bit more parallel capacitance due to the antenna. <S> At resonance (the tuning point) the tuned circuit behaves like an open circuit and any current taken from it causes both selectivity and sensitivity to be reduced so, for a reasonable load (~20 kohm), <S> the sensitivity (size of signal) might half and the selectivity <S> (ability to exclude close by unwanted signal) may also half. <S> For 10 kohm load this gets nearly twice as bad and as load resistance drops it becomes worse and worse. <S> In other words, the tuned circuit wants zero energy taken from it by any load in order to be sensitive and selective. <S> Xtal earpieces that I've seen have an impedance quoted at about 20 kohm <S> so there is some loading <S> but, it's not that bad if the tuned circuit is designed/built correctly. <S> Some loading is actually required to avoid selectivity value that excludes the full spectral width of the station being received. <S> Anything below 1 kohm is going to be fairly rubbish and anything below 100 ohms just won't work.
| Crystal earpieces need almost no power to operate (they also need almost no current - which is the harder to generate anyway), that's why they're used, because a regular crystal radio only generates enough power to drive a crystal earpiece.
|
How does grounding the circuit provide safety? I don't understand the concept of grounding a circuit for safety reasons, in this circuit that I simulated, imagine my hand touching the circuit as the closed switch. I am creating a closed loop between my body and the circuit letting current to pass through me. If the circuit was isolated or floating however, it would be impossible for me to get zapped because there would be no closed path for the current to flow in and out. So what is the point of connecting the circuit to the earth ground? And secondly, does connecting any point in a circuit to the earth ground pull the voltage in that point to the ground level? <Q> You cannot rely on an AC mains or high voltage DC feed being isolated. <S> For this reason, any exposed metal (conductive) parts should be earthed because, if an internal wiring fault developed that connected "live" to those exposed metal parts, the fuse (yes a fuse or breaker is required for electrocution prevention) will blow and the live is disconnected upstream. <S> The fuse is much "weaker" than the ground wire <S> so, after a short period of a few milli seconds the fuse blows and the device becomes safe. <S> In that short period of time you can receive a small shock but probability suggests that this is an unlikely event. <S> A floating AC supply isn't that safe even to the touch due to capacitance to ground - a decent "tingle" can usually be felt. <A> Ground used for safety is a completely separate concept from the ground in low-voltage circuits. <S> While the latter is just a reference point with low impedance to power supply terminals, the safety ground is a completely separate circuit which normally carries no current. <S> In modern installations (with diff current protection ) <S> any current in the ground wire above a certain threshold leads to immediate circuit breaker tripping. <S> This protects the person who somehow managed to touch a live wire, like in your example. <S> The only way to get shocked without tripping the protection would be to touch both hot and neutral wires, while NOT touching the ground. <S> This guarantees that metal cases can't go live without provoking a prompt short circuit. <S> Of course, if you open the case and manage to touch a live wire inside, there will be nothing to protect you. <A> If the device is powered by mains it is always somehow related to the ground. <S> No absolute floating potential is possible unless is battery powered, even then there is always a finite isolation strenght capability. <S> At some voltage the isolation breaks, always. <S> The grounding is connected to the metal case, you should make a new sim, and if a faulty component touches the case, the current is sinked into a ground, your body would be just a parallel circuit with much higheer impedance than a wire. <S> And secondly does connecting any point in a circuit to the earth <S> ground pull the voltage in that point to the ground level ? <S> Yes. <A> The safety earth provides protection against faults where a component in the equipment is not acting as it was originally designed. <S> Some fault conditions may be made more dangerous by the presence of the safety earth, but in general a safety earth will be designed to detect latent failures before a dangerous situation occurs. <S> It is true that a human body needs to be exposed to two parts of a circuit before electrocution can occur. <S> These practices are designed by analyzing hundreds and thousands of real-life failure conditions.
| The intention of a safety earth is to prevent even one past of the external (touchable) equipment bridging to the live circuit - stopping the problem before it becomes fatal. In older installations, ground wire is simply connected to conductive parts which could be touched, like metal cases (most often TN-S earthing , but other earthing systems exist).
|
Are there any standard FPGA internal buses? Are there any standard FPGA internal buses?I've always used some sort of bidirectional bus between my internal blocks, but is there a standard way of doing this? <Q> Here is a little overview on chip internal buses, which are suitable for FPGAs: <S> Advanced Microcontroller Bus Architecture (AMBA) from <S> ARM Ltd. Current version: <S> 5 Specifications Further reading: Wikipedia Commonly known buses in that family: Adavance Peripherial Bus (APB) <S> Advanced High-performance Bus Advanced Extensible Interface (AXI) Variants: <S> AXI-Lite, AXI-Stream <S> Avalon from Altera now <S> part of Intel Current version from 12.2015: Avalon Interface Specifications <S> CoreConnect from IBM Current version: 4.6 Specification (no official link found) <S> Further reading: Wikipedia Commonly known buses from that family: Processor Local Bus (PLB) On-chip Peripheral Bus (OPB) <S> Open Core_Protocol <S> (OCP) from OCP International Partnership Association Current version: 3.0 Specification <S> Further reading: Wikipedia WishBone from <S> OpenCores.org <S> Current version B4: Specification <S> Further reading: Wikipedia <A> Yes, there is the wishbone bus. <S> Most modular IP cores you can find (e.g. on OpenCores ) use it. <S> Altera uses a variant, which is very close: the Avalon bus. <A> Newer FPGAs that include hard-core ARM processors pretty much use AXI exclusively.
| Most new IP uses the AXI bus , which is derived from the Advanced Microcontroller Bus Architecture introduced by ARM to support SoC designs.
|
Choosing accelerometer or displacement sensor for system I am working on an electro-servo motor system which can be seen as a similar in here ! It is a shaketable system which converts acceleration input to displacement output. My question involves a bit theory; what kind of sensor should be use on these systems for feedback? Accelerometer or displacement sensor as LVDT? Question is totally based on sensor specifications but if we assume all specifications are same, is there any advantages to use an accelerometer instead of LVDT. In my opinion it is pretty sensible to use an accelerometer for feedback but i can not sure. I want to know your opinion. <Q> You will need some displacement sensing for sure, otherwise the bias error will accumulate and your table will eventually slam into the end stops. <S> If you are using a commercial control system you may be feeding it a displacement signal directly. <S> Of course the servo will have limitations in force (acceleration) as well as slew rate and you will want to filter and pre-process the file before sending it through to the actuator(s). <S> Fortunately the kinematics for this setup are simple (as opposed to, say, a hexapod) <S> so it should not be too hard to compute. <A> Shake tables need accelerometers, since these can be used as feedback to control the force exerted on the test article (or, equivalently, acceleration), and to determine response and resonances. <S> If all you control is displacement, it's entirely possible to break the sturdiest component. <A> Of course, the specs are not the same. <S> Further, you can get acceleration from displacement and displacement from acceleration. <S> That said, accelerometers are <S> dirt cheap, and don't require the level of signal processing that LVDT's do. <S> I'd go with an accelerometer, and if you need displacement as well (or instead), I recommend running through the specs to see if a ratiometric Hall effect sensor might be good enough. <A> You would need a linear displacement sensor, like optic scale,etc..or an encoder mounted on a motor shaft (motor with encoder). <S> Without it you can't position the table. <S> The extra sensor is an accelerometer, to have a trace of acceleration/shakes.
| I would suggest looking at displacement feedback via an encoder on the servomotor and monitoring with an accelerometer if desired/required.
|
Reducing the brightness of Nixie tubes I have a Nixie tube clock based on IN-12. I assembled the kit, it works great and it was great fun. But the tubes are just too bright in dark conditions. I would like to modify the clock to be able to change the brightness. I have included the full schematic bellow. I feel that a well placed potentiometer on the high-voltage power-supply (lower right of the schematic) could do the job. UPDATE I change the voltage, as expected, I got some non-uniform lighting. reprogramming the microcontroller is more work than I was willing to put in this project. <Q> Don't mess with the voltage, that will cause uneven lighting. <S> Generally for glow discharge tubes like nixies you can get much better control by PWM than by messing with voltage, so I would suggest programming the tiny to PWM the A1-A4 pins. <S> You will want to make the frequency high enough to avoid quench effects (still a thing even in glow discharge regime) and audio frequencies (Actually it looks like 100-500Hz also works because the time constant is really long). <S> An explanation is given at Wendt's page . <S> He used a 555 to generate the PWM, but the tiny probably can just generate it internally. <S> As a bonus, you can change the brightnesses on a per tube basis. <A> I think the simplest way would be to alter the on time of the segments by changing the microcontroller code. <S> Clean and simple, but you need to be able to tell the chip what brightness you want (spare ADC pin?) <S> and of course it may not be easy to change someone else's code. <S> Trying to PWM with a 555 as Paul suggests should work if you can get it to work with the frequency high enough, but if not it would tend to beat with the scan frequency to cause distracting ripples of brightness in the display. <S> Your suggestion of a series pot in the HV has a chance of working- <S> the individual neon tubes may not track well enough, and you should take care to avoid contact with the high voltage. <S> I think I'd give that a try first. <S> Try 50K or 100K. Just try some resistors if you don't have a suitable pot on hand. <S> Basically cut the connection at "HV", leaving all the power supply parts on one side and the neon drivers on the other side and add the variable resistance. <A> This simplest way to reduce the brightness given the existing circuit is to reduce the high voltage. <S> The brightness will be non-linear with that voltage, but you should be able to find a point that has the brightness you want without it being too dependent on small changes in the high voltage. <S> The high voltage is generated by a boost converter run by a 666 555 timer: <S> It looks like when the voltage gets high enough, T12 is turned on, which kills the oscillations. <S> That's pretty crude, will have a strong temperature dependency, and will be touchy to adjust. <S> However, P2 is already there for the purpose of adjusting the high voltage. <S> The more you turn the wiper to the R4 end, the lower the high voltage. <S> If the existing adjustment range doesn't allow the brightness to get low enough, make R3 a little bigger and/or R4 a little smaller. <S> If the adjustment is too finicky, put a zener diode in series with the base of T12 and add maybe a 1 MΩ resistor to ground right on the base of T12. <S> That will change the setting range of the high voltage significantly, so R3 and R4 need to be adjusted accordingly. <S> Make sure that the maximum voltage the tubes, D5, and T1 can handle can't ever be attained no matter how P2 is set.
| By maintaining the same period for the scan of each digit but turning the digit 'off' for part of the time you could change the brightness without any hardware modifications.
|
What's needed to put a PC fan into reverse? I've got a PC fan here (I've peeled off a sticker). And although this one doesn't come with the usual 3 or 4 wire connectors, it has enough electronics in it to only operate when power + is on red, and - is black but/and it won't operate when power is connected the other way round. I don't know anything about how these (brushless?) PC fans are controlled, but I assume one of the basic/first steps in the circuitry is to make sure polarity is right. Right? So, now, is it possible to reverse the motor with a simple hack, or is is not possible without heavy modification of the controller or stuff... Ideas how to trick it into spinning the other way round? (Notes: Sorry for not providing the exact model of the fan. Let's assume it's pretty generic. Also, I know the fan blades are designed to spin this way round, and reversing it would mean having a less optimal fan. I can't just flip the housing. It's a long story why, but that's not an option. I couldn't get the casing open or the fan blades off, it's all pretty sturdy) Update: Taking all the feedback I had so far into account, I first tried to get that darn thing open (no success) and then had a closer look from the outside: What we see here from the side is (4 legs) a hall sensor, right? That means/would mean: it's the "sensored type" of fan, meaning that even breaking it open and swapping motor cables would have no effect (no cables btw., the motor seems soldered to the controller board). As I have a hard time deciding on the accepted answer, I think I have to check Olin's as he was the first to point that out, although pericynthion was first with an answer. <Q> Reversing the direction of rotation will be difficult. <S> Since you only supply power, there is a controller in the fan that senses rotor position and commutates the motor accordingly. <S> That won't be easy to do since these are all nicely integrated onto a small board. <S> Supplying negative power won't work, just fry the electronics. <S> However, you can still "reverse" the fan by simply installing it backwards. <A> Most PC fans are two phase with 4 pole stators, PM rotor and a single Hall sensor. <S> I think I've seen 3-phase but not for more than 15 years. <S> Eg. <S> (from here ) <S> You may be able to just reverse the two coil wires to make this work. <S> There is more than one way of ensuring that the two-phase motor (which otherwise would be happy to spin in either direction) starts in the correct direction. <S> It's also possible a small magnetic field would bias the sensor so it would spin in the opposite direction, but modern hall driver chips use edge detection and won't be fooled by that. <A> The circuit is a three phase inverter. <S> Ultimately there will be three wires going from the PCB into the motor windings. <S> Swap two of those three. <A> No. <S> You cannot reverse the direction of the fan externally by simply reversing the polarity of the power. <S> The built-in electronic circuit manages "commuting" the magnetic field that causes the rotor to turn. <S> The sequence of the poles are hard-wired to make the fan turn in the designed direction.
| If you can get in there, you can probably reverse the direction by flipping two of the three sensors and two of the three windings.
|
Can you load Gerber files back into a PCB layout designer such as Eagle? There's a PCB I may need to change a solder pad, but the schematic and layout are gone and were not added to source control. Is it possible to load the Gerber files back into a designer to change the pads so that I don't have to try to draw out the entire schematic and layout again? <Q> As of Eagle 7.5, this is indeed possible. <S> It's a simple case of <S> File->Import->Gerber from the layout editor. <S> There is a video demo on YouTube. <S> As others have mentioned, there is no real DRC checking possible of the imported file as the Gerbers contain no information on nets and connectivity. <S> Eagle simply imports all of the various shapes and lines onto the copper layer. <S> It does however allow you to make modifications manually (with great care) and then re-export as a Gerber again once you are done. <A> No. <S> A standard Gerber format file contains only primitive shapes and positions. <S> It contains no concept of WHAT any of the shapes represent, or even that it is an electronic printed-circuit board. <S> However, the Gerber format is a quite simple and strightforward text file. <S> It can be edited in any low-level text editor. <S> So, if it is worth the time and effort, you COULD "tweak" the contents of the Gerber files to make a change as simple as modifying a pad. <A> Essentially, the Gerbers are vector drawings (that PCB fab equipment understands). <S> In a Gerber, you can have a rectangle, but the Gerber doesn't "know" that this rectangle is actually a pad, and it's a part of footprint that has other pads. <S> Gerbers don't carry the schematic information: all traces are just lines (or polylines) for x1,y1 to x2,y2. <S> Afaik, Gerbers can't be loaded back into Eagle PCB layout designer. <S> When Gerbers are generated from an Eagle layout (from the BRD file), a lot of information is left out from the Gerbers. <S> They work independent of the software from which a Gerber was generated. <S> They provide a lot less DRC (if any), because a Gerber Editor doesn't "know" the schematic. <S> (@Matt had already mentioned that manual editing of Gerbers is risky.) <S> You would have to compensate for this lack of DRC with additional human diligence. <S> Related Similar question about Altium. <A> Eagle paints many of the gerber features with a raster of 0.3mm lines, making it difficult to seamlessly identify features such as pads or tracks, versus polygonal pours, ground planes and so forth. <S> If your version of Eagle supports import, you are mostly done and dusted. <S> If not, other options would include converting to another file format which you may be able to import into eagle, i.e. the following utility can create a gEDA PCB compatible footprint with a gerber file, but has to make guesses about which polygons are pads versus ground pours, and like any gerber->layout conversion tool, will not be 100% reliable in reproducing all of the features. <S> https://github.com/erichVK5/translate2geda <S> (disclaimer: my WIP conversion tool)
| There are Gerber Viewer and Gerber Editor programs, which allow to edit Gerber files.
|
What's the difference in a digital optocoupler and a non digital I'm looking at 2 different optocouplers.The Avaago ACPL-K44T and the ACPL-227. The ACPL-K43T states that it's a digital opto-coupler while Avago ACPL-227 doesn't. Schematically, they're a bit different. The K43T needs a Vcc and has parameters like Vol,Voh,Ioh,Iol, etc. I'm wondering what exactly are the applications you'd need the K44T for compared to the 227. http://www.avagotech.com/products/optocouplers/industrial-plastic/other/phototransistor/acpl-227-500e#documentation http://www.avagotech.com/products/optocouplers/automotive/ipm-interfaces/acpl-k43t-000e <Q> Phototransistor-output optocouplers are cheaper and suit for slow digital applications and sometimes for analog applications such as switching power supply feedback. <S> When you calculate the actual switching speed with a reasonably high load resistance (See figure 16) they can be quite slow, but still okay for many applications- such as isolated switches and relay contacts in a PLC. <S> There are also photodarlington optoisolators which are even slower, and a few other types. <S> Digital output optocouplers are much faster and are specified in terms typical for a logic part, mostly. <S> They use a photodiode on a chip rather than a phototransistor. <S> They also draw power even when off. <S> They are also relatively easy to use (in terms of guaranteeing that they will work under all conditions). <S> You would use them to isolate digital signals, such as for an isolated ADC/front end in a data acquisition system. <A> Even though most opto-couplers have a digital 'nature' to them (the exception being an H11F1, which has an fet output and is analog over a limited range), there are those that work over wide voltage and current ranges, sometimes referred to as high-voltage digital. <S> They are designed as a 'logic' interface between 5 volt to 24 volt digital logic, with outputs that can work up to 200vdc (H11D1). <S> Data rate is limited to a few hundred KHZ. <S> They are use for machine control and slow servo control loops. <S> Your ACPL-227 is a typical general purpose opto-coupler. <S> A digital opto-coupler works at standard 3 volt to 5 volt logic voltages, and expects a clean logic type drive current in order to output a clean logic signal. <S> Some can run as fast as 10Mbps, and will likely get faster over time. <S> Your ACPL-K43T is like this, for fast logic (data) transfer or a fast isolated power switch in terms of machine control. <A> ACPL-K43T has a digital output which means you can directly connect that output to any logic circuit operating in TTL levels . <S> ACPL-227 has a bare transistor output: you will need to provide an external resistor and voltage source to match your output requirements. <A> Digital optoisolators are designed for speed and on/off outputs, regular optoisolators are designed to be more analog so you can send a range rather than just an on/off signal (speed is a secondary concern). <S> That's the main difference.
| ACPL-227 offers multiple optocouplers in a single package that can be customized while ACPL-K43T offers convenience for a specific application.
|
The use of fuses in small electronics projects I am new to electronics so I hope this question is acceptable. I have been playing around with an Arduino. I have use the USB output from a computer as a power source. I also have a 9V battery. I am a bit concerned about creating a short somewhere and damaging the computer or the battery. I am guessing that this is what a fuse is for, is that right? For small projects (not necessarily involving an Arduino) what type of fuses should be used? Is it, for example, possible to get a 1A 9V fuse? I have searched online and I only see things like 1A 250B fuses. Will these work well for small electronics projects? <Q> A 1 Amp 250V fast fuse should blow even at lower voltages. <S> The important number here would be the Current rating . <S> For example, my multimeter has a 500 mA 250 Volt fast fuse, with a recommended/listed max current measuring of 200 mA. <S> It has blown when I shorted out a 5V 1 Amp (USB) supply, and a 12V 750 mA supply by mistake. <S> You can test this with a spare fuse, and a resistor. <S> Take your 9V supply and a 100 ohm resistor. <S> At roughly 90 mA, the fuse won't blow. <S> Short the fuse out, and it should. <A> Look up 'Pico Fuses' on the web. <S> They are small PC mount fuses, in a package that looks like a 1/4 watt resistor. <S> They are rated for 50 to 125 volts max, with currents as low as 1/20th of an amp to 5 amps. <S> These are fast-blow uses, so be sure to use them in places where the in-rush current is low. <S> Digi-Key and Mouser supply them in small lots or in bulk. <A> Fuses have voltage drop. <S> A garden-variety 1A <S> 250V 5x20 <S> mm fuse will drop about 200mV, so not too bad, but it might put the USB voltage out of spec with a long cable. <S> 'Most' modern USB ports are protected against shorts (and the computer knows if they have been overloaded), but I have seen old motherboards with soldered in one-time fuses. <A> These glass tube <S> 250V fuses are fine. <S> Make sure they are fast-blow fuses. <S> Some have lower voltage <S> but it's less common. <S> Prior to choosing a fuse, evaluate approximately the extreme maximum current your project may take. <S> If you think it will never use more than 500 mA, then use a 500 mA fuse. <S> Some projects may work with 250 mA or even 125 mA fuses. <S> The lower the safer. <S> A fuse will help you debug your circuit on top of protecting it.
| To answer your question, yes, those 1 Amp 250 Volt fuse will work fine in small projects, as the Voltage is a maximum.
|
Why exactly do you need a LCD controller board to interface LCD to a microcontroller? I've dealt with interfacing a very simple LCD display with a microcontroller, so I get the basic concepts. I'm trying to interface something a bit more involved. I'm a bit confused on why you would need a LCD controller board to interface a LCD that displays nothing more than simple characters (no graphics). Take this LCD from Electronic Assembly for example. It has a table of command for the LCD controller SSD1803 . Does this mean I cannot interface this LCD screen without a lcd controller board to my microcontroller? <Q> The LCD module that you're referencing includes the SSD1803 controller. <S> You do not need an additional controller. <S> Generally speaking, direct LCD interfaces have a number of features that make them difficult to work with: <S> They are analog interfaces, and often require unusual operating voltages. <S> They involve a lot of pins (typically, one for every row and every column for monochrome displays). <S> They must be switched at somewhat high frequencies for correct operation. <S> They do not inherently provide any graphical rendering features (such as alphanumeric characters). <S> Combined, these features make it impractical to interface directly with an LCD. <S> Instead, most small LCDs are packaged with a controller which drives the LCD and presents a more "friendly" digital interface. <S> In this case, that controller uses a parallel interface compatible with the HD44780; in other small displays, SPI and I2C interfaces are not uncommon. <S> Larger LCDs tend to use different interfaces, such as MIPI, LVDS, DVI/HDMI, or DisplayPort. <A> You don't need an external controller for this kind of LCD. <S> This is the controller. <S> In general, character <S> LCD modules will include the controller, usually with some variation of the Hitachi HD44780 controller, perhaps with a different data interface than the ancient 4/8 bit interface. <S> Graphic LCD modules, especially color ones, tend to need an external controller with a frame memory. <S> In some cases that may be part of your MCU or FPGA <S> so it can share memory and respond quickly, in other cases there may be a controller as part of the module so it can work with a much more modest micro (at a cost in complexity, dollars and speed). <S> In the former case, the interface is sort of a digital version of a video signal- color digital values refreshed regularly at some frame rate. <S> The pixel rate will be beyond the ability of a small micro to handle even if it did nothing else. <A> I'm a bit confused on why you would need a LCD controller board to interface a LCD that displays nothing more than simple characters (no graphics). <S> The basic ones allow you to control the LCD from a serial/uart or a i2c/SPI connection. <S> The goal is to save pins or memory . <S> A direct HD44780 connection will need between 7 to 12 pins. <S> The backpack may only need 2 or 3. <S> It may also allow for animations or automatic controls. <S> You can definitely connect directly to that LCD, but a controller middleman may help save time, money and effort.
| The controller boards you see are normally backpacks that provide additional features.
|
A question on using potentiometer as a variable resistor I have a 10k poti. I use it as a variable resistor by soldering two of the three terminals. It is connected to a 1/4 watt 220 ohm resistor in series(this is the load). The voltage across this circuit is 12V DC. Is the poti safe even it is set to zero ohm?Would 220 ohm prevent it to burn in this case?If so, what should be the quantative method to calculate the minimum value of series resistor which would prevent this poti? <Q> A good rule of thumb is that the maximum wiper current (if not stated on the datasheet) is the lesser of the current when passed through the entire element would reach the maximum power dissipation for the pot and 100mA. <S> So if you have a 10K 1W pot, then the maximum current would be 10mA, and the minimum resistor value that would be safe (at 12V) would be 1.2K. <S> That's a conservative figure suited for carbon and cermet pots- pots that designed for use as rheostats (eg. <S> some wirewound types) can handle more current because they have a thermally conductive core attached to the element to spread the heat. <A> So to ask your question another way, "is it safe to connect a \$220\Omega\$ resistor to a \$12\mathrm{V}\$ supply?". <S> I'll let you answer that. <A> If you know the power rating for your potentiometer, you can start with that as a guide as to how much current you can pass through it. <S> Little trimpots adjusted by screwdriver or mini ones that you turn with your fingers are often 0.5 watt, but may not be so you'd best check. <S> The power rating for a potentiometer tells you how much power it is rated to dissipate if the current passes along the entire resistance. <S> For example, say the power rating is 0.5 Watts for your 10k pot. <S> Then, the current that would flow to generate that power: $$P = I^2R$$Rearranging...$$I = \sqrt{P/R} = \sqrt{0.5 / 10000} = 0.0071 <S> A$$ <S> This is the most current that should be allowed to flow through the 10k 0.5 watt potentiometer at any position. <S> Let's go to ohm's law and see what series resistor can prevent that: $$R = <S> V/ <S> I = <S> 12 / 0.0071 = 1690 <S> \Omega$$ Substitute the actual power of your potentiometer, and maybe use one a bit bigger to be safe, and note that it could heat up a fair amount if used at the rated power. <A> Let's use the power formula. <S> So in the circuit: The total current through the circuit is P = <S> (12^2)/220P = 0.654545 <S> Watts <S> Now all you need to do is check whether the 220Ω resistor and the 10KΩ Potentiometer can both handle the wattage ( 0.654545 Watts ). <S> As you said, the 220Ω resistor is 0.25 <S> Watts <S> rated so that will definitely blow up. <S> If the potentiometer is not rated for that Wattage, they both blow up. <S> Now let's calculate what resistance you need to replace the 220Ω resistor. <S> If the potentiometer has a higher wattage rating than the 220Ω resistor, it makes no difference. <S> I am going to assume the Potentiometer has a wattage rating of 0.1 W Since the Potentiometer is 10k, max current is 3mA due to <S> I = <S> root(P <S> /R)I = <S> root(0.1/10000)I = 0.003162A <S> We are going to replace the 220Ω with something better. <S> Assuming your other resistors are rated 1/4 Watts, let's find the minimum required resistance at 12V R = <S> (V^2)/PR = (12^ <S> 2)/(1/4)R = 144 <S> *4R = <S> 576Ω <S> That's the minimum resistance for the resistor. <S> Let's see about the potentiometer. <S> R = <S> V/IR = 12/0.003162R = 3794Ω Jeez a high value... <S> Now although the resistor only requires about 600Ω for itself, the potentiometer requires a whopping 4K resistor, so you have to choose the higher one. <S> Now we don't want our resistor to be on the plain boundary of the maximum Wattage rating, so get like a 4KΩ resistor or more (1/4 watt rated). <S> Also pay attention to the tolerance. <S> For eg if the tolerance was 10%, then the resistor could be 3600Ω which is very dangerous.
| If you set it to \$0\Omega\$, then it is no different from a piece of wire.
|
Why is resistance across my arm-span so much less than across my face? This sounds like a strange question (it is) but, from a purely biological standpoint, why is the measured resistance across my hands (I measured 120k\$\Omega\$) so much less than across my face? I thought to ask this because the distance across my face is much less than my arm span. More specifically, what makes the tissue in my face so much more resistant to charge (ion) flow than my entire arm-span? P.S. I used a standard multimeter with resistance set at 2M\$\Omega\$ to get my readings - across my face doesn't even give a reading (too much resistance) <Q> It will be affected by things like sweat, salt, and skin oils. <S> edit: Skin oils can add a more resistive layer. <S> edit: Try different distances in the same parts and like distances in different parts. <A> As Skaperen says it is all about the skin resistance. <S> With an electrode in the meat under the skin, the resistance across the body will be much lower. <S> My theory is that you are able to obtain more contact pressure holding the meter probes in your hands, than when pressing them to your face. <S> Rather have someone else poke you with the probe, in the same way, in different places. <S> You could also put one probe in your mouth, which would be a low resistance connection to your body, and you'd have your hands free to probe all around. <S> Doing this, I think you'll find it's only the skin type and pressure <S> that matter, not the distance across the body. <A> Is it possible that since there is likely more skin oils on your face, and more sweat on your hands that the results were affected? <A> \$ <S> R = <S> \rho * \frac{L}{A_c}\$ <S> Added to that <S> (like what others have said before) <S> a 'sweaty'surface <S> like your hands likely gives excellent probe contact,while the less-moist (and hairier) surface of your face givesless complete contact. <S> All these factors result in much higher resistance across yourface compared to your arm-span
| Your skin surface resistance is what varies greatly. Hopefully this helps:Because resistance is proportional to material length and inversely proportional to cross sectional area (see below), I thinkit's safe to say that because your face has significantlylower cross-sectional area compared to your arm-span (probably by a couple orders of magnitude) that the lengthdifference is negligible.
|
What should be the watts of solar panel and battery size I want to light one 70watt bulb for 12 hours at night time. I want to know what should be the Watts of the solar panel and how much Ah battery will I require? I searched many of the questions here and found this one relevant. But the question is how will I know the volts which you assumed here to be 12 V? Powering 10watt light bulb for 24 hours using solar power May I know if I have done the right calculation as below Panel Watts req = 70 * 32 / 10 Sunshine hours=10 Assumed 12V, Battery Ah= 70W / 12V * 12h = 70Ah <Q> 70 W <S> * 12 hours = <S> 840 <S> Wh. <S> That is how much energy you need to take out of the battery every night and put back in every day. <S> If you want the battery to last, you should pick a battery with twice that capacity. <S> So that is around 1700 <S> Wh. <S> 1700 <S> Wh / <S> 12 <S> V = around 142 <S> Ah at 12V. Pretty big battery. <S> You cannot discharge a battery 100% every day. <S> It will not last long. <S> The solar panel needs to produce 840 <S> Wh in about 5 hours. <S> So that is 840 / 5 = 168 Watts. <S> This is just a rule of thumb. <S> The daily output of a solar panel is around 5 hours * rated power. <S> Obviously there is some variation depending on location and season. <S> You might want to look into data specific to your location. <S> So there you go. <S> A 12V 142 Ah battery. <S> A 168W solar panel. <S> And a good solar charge controller with maximum power point tracking (MPPT). <A> 70W <S> * 12h = 840Wh <S> 840Wh / <S> 12V = 70Ah <S> This is the minimum capacity of your battery needed. <S> Batteries don't like to be discharged to the bitter end. <S> All this has nothing to do with the solar panel. <S> Charging a battery needs always more power than you get back. <S> Normally around factor 1.2 to 1.4 840Wh * <S> 1.4 <S> = 1176Wh <S> 1176Wh / 10h = <S> 117W <S> Your panel as to deliver at least 117W during your sunshine hours. <S> How much energy is delivered by your panel is depending on the mounting position and is (almost) never the amount rated. <S> But with a panel rated for 150 to 200W it could work. <A> You need 70 W * 12 h = 840 <S> Wh from the battery. <S> You are correct that for a 12 V battery, this is 70 <S> Ah. <S> To get 840 <S> Wh into the battery over the course of 10 hours, you need a 840 <S> Wh / 10 <S> h <S> = 84 Woutput from the solar panels. <S> All assuming no loss, perfect sunshine, etc. <S> Have you considered using LEDs instead of a 70 W bulb? <S> That way you would get away with a much smaller battery and much smaller solar panels, and it will probably be much cheaper. <A> I didn't see if you are using a 120 V bulb for an AC lamp. <S> I assume it's an incandescent bulb. <S> If this is the case you will need to step up the voltage . <S> The filament they use glows white hot to produce light (and heat) which is wasteful. <S> I agree with @Dampmaskin that LEDs are more efficient.
| I would suggest to use at least the 140Ah to get longer live time of your battery.
|
Additional requirement for checksum over CAN bus? Im working on a safety system project and sending messages over CAN bus among systems. Since the CAN protocol already includes a hardware CRC for messages sent over the bus,I was wondering if additional checksum for the data bytes is required or is the hardware CRC sufficient. For ex: For 8 data bytes sent in CAN, should 1 byte out of 8 bytes be allocated for checksum of the 7 bytes being sent ,or, can I send 8 data bytes and rely on the hardware CRC to ensure the integrity of message? <Q> It's a probability game and a tradeoff of costs. <S> What's the cost of getting bad data? <S> What's the cost of decreasing the chance of getting bad data? <S> You can't ever absolutely guarantee that data is received correctly. <S> All you can do is decrease the chance of that happening by throwing ever more resources as it. <S> At some point it becomes not worth it to spend more resources to decrease the chance of bad data a little more. <S> Only you can answer what that point is. <S> CRCs can be arranged to always detect a single bit error, but it is possible to change multiple bits so that the result looks correct. <S> The 15 bit CRC in a CAN message has 32768 possible values. <S> If you throw random message content at it, there is one chance in 32768 of it resulting in any particular CRC. <S> At first glance, you can therefore say the probability of a bad message looking correct is the probability of getting two or more bit errors within a message, divided by 32768. <S> It's actually less than that because some random errors will make the whole CAN message structurally incorrect, but the above is a good start. <S> You compute the probability of getting bad bits in the first place from the bit error rate. <S> The typical CAN bus is implemented as a symmetric differential signal, and is quite robust even with lots of external noise. <S> Consider that CAN was originally designed for automobiles, which are electrically very noisy environments. <S> The point of this is that the bit error rate will be quite small. <S> For example, 1 in 10 6 would be a huge and unlikely bit error rate. <S> Unless you are doing something very unusual and critical, just using the CAN checksum without any additional integrity checking at the application level will be good enough. <S> If you're in such a critical application where it's not good enough, then you should be looking at multiple independent and redundant busses and other higher level schemes anyway. <A> The CAN bus already got a 15 bit CRC for just 8 bytes of data + overhead bytes. <S> That's quite a high safety level. <S> The CRC will detect all single-bit errors and most larger errors too. <S> In addition there is also numerous other safety measures present in CAN: <S> bit stuffing, the high sample rate of each bit by the CAN controller, the bus arbitration method, the fairly high signal current, the differential signals and so on. <S> Discarding part of your data for additional CRC seems a bit paranoid, especially since CRC is one of the weaker safety measures here. <S> There are better things you could do to increase safety (if needed), such as implementing a software sanity check of the data. <A> In CAN, there are two systems to check the data integrity: <S> As you mentioned, there are the checksum . <S> The CRC implemented in the CAN ISO can detect 5 errors. <S> But there is also something during the protocol, if a ECU sees a bit that shouldn't be here on the CAN bus. <S> The frame is corrupted and the error frame is sent instead. <S> An ECU can corrupt its own frame if it reads a different bit that what he has written. <S> You can add other mechanism in upper OSI layer <S> but I don't think that it is necessary for a normal use.
| That said, the 15 bit CRC checksum CAN uses on every message is good enough for the vast majority of applications.
|
Ripple flowing through probes connected to the same point! While trying to filter the ripple of my switching PSU, I've come to a phenomenon that is probably well known to professionals, but seems absurd to me.Admittedly, my scope is not of the highest quality, but it can sample at 100MHz and gives usually coherent results. Furthermore, it is only connected to a floating portable computer working on batteries; so, no parasitic current can flow through the ground orthrough another connection. The phenomenon is the following:I took an ATX power supply like the one in your desktop computer, and connected the TWO probes of my scopeto the ground wire (the black one) of the PS, AT ITS EXTREMITY A (that is, point A is connected to nothing, except the two probes).In my scope (set in AC mode), I could observe a small ripple. As this seems absurd, I first thought this is an artifact, so I connected the first pin of a toroidal inductor (5mH) to A, the ground probe of the scope to A, and the other probe to the other pin of the inductor (connected to nothing except the probe); surprise ! my scope now show a ripple of up to 40mV, ranging from 10KHz to 20MHz (centered at 4MHz).I was trying to explain this to myself by invoking wave propagation, when the next experience mystified me completely: I connected the ground probe in the other side of the inductor, and the other probe to point A. Then an incredible ripple of up to 0.5Vwas shown by my scope ! These experiences contradict everything I know on currents, and in particular, why is there an asymmetry regarding the connection of the probes with respect of the two sides of the inductor (the ripple is AC no?) I would greatly appreciate any insight. EDITS: If I shut down the PSU, while maintaining everything connected as is, the scope shows nothing. If I don't shut down the PSU, but I disconnect the two probes from the ground wire, while maintaining them connected one to the other, the scope shows nothing as well. Here is what I believe to be a beginning of explanation: the electrical wave is exploring the ground wire, and most of it is reflected at the first pin of the inductor. So, between the PSU and the inductor appears a standing wave. At the other pin, few of the AC components of the wave has passed the inductor, hence the ddp between the two pins of the inductor. If there is no inductor, a similar effect appears (but much less strong) because the two probes have a somewhat different inductance. As convincing as this may seem, this does not explain the asymmetry of the strength of the ripple with respect to the pins of the inductor. <Q> They are picking up magnetic radiation from the transformer in the SMPS (Switch Mode Power Supply). <S> You could conduct a couple more experiments with the probes inter-connected but NOT connected to the SMPS. <S> And another experiment where you simply power down the SMPS and see what your scope shows without changing the connection. <A> The main concern of the third item in your edit, namely the dependence on the polarity, is explained by the asymmetry of the input impedance of you battery powered scope. <S> While at DC, both the ground connection and the signal connection have practically infinite resistance to ground and can be treated equally, this is no longer the case at RF. <S> Even if your scope is not grounded, the surroundings of your scope are at a certain potential, even if it is just air. <S> The scope has a capacitance to the environment which is notably different from zero. <S> You might have learnedvin physics education not only how to calculate the capacity of a capacitor made out of two concentric spheres, but also, likely even earlier, how to calculate the capacitance of a single sphere. <S> The signal input on the other hand is designed to be extremely high impedance even at higher frequencies, and thus much better suited to pick up stray RF. <S> This makes your floating single-ended instrument different from a real differential instrument, in which the impedance of the positive and negative input are matched. <A> This man seems to have the solution: the shorted scope probe problem <S> it's just a wire <S> isn't it? <S> switching power supplies... <S> common mode effect and many other more or less related articles
| Now, years later, you observe that this is not just strange theory, but actually a real thing that matters in practice: the scope ground is likely connected to a big shield in the case and has considerable capacitance to ground at high frequencies, shorting them away. Your two probe wires connected together at the "far end" (farthest away from the oscilloscope) form a loop antenna.
|
Where does an op amp's input bias current come from if capacitor is used? In the case of a non-inverting amplifier, if I DC couple the input signal with a capacitor only, \$C_1\$ (i.e. no resistor to ground), sources (as well as an experiment in the lab) show me that the op amp input will be saturated as the input bias current deposits charges on the terminal without having a return path to ground. This is solved by using connecting a resistor (\$R_1\$, as shown in the image below) to ground to create the return path. [The image is reproduced from an answer by "Neil_UK" on one of my previous questions, thanks Neil!] My questions: 1) How is the input bias current able to continuously deposit charges and saturate the input if I don't use the resistor? The resistor should not allow the DC bias current through, so where is this current flowing from/into? 2) Why is this not a problem for the feedback path to the "-" terminal? There is no resistor to ground in that case, and capacitor \$C_2\$ seems to be causing no issues. The PDFs I've been reading suggest that the feedback behaves similar to when I have a resistor to ground by acting as the return path, how is this? <Q> 1) <S> In the case that you do not have R1 in place you would expect the + input node to be floating. <S> But that is not the whole story. <S> All the pins of opamps and in fact every chip have ESD protection diodes between the pins and the supplies. <S> So if you apply a signal at the left of C1 there is a way to charge and discharge C1 if the input signal goes below or above the supply voltage. <S> Also leakage currents can raise or lower the voltage of the input. <S> The resistor should not allow the DC bias current through* <S> Actually it should <S> and it must . <S> You found that R1 is needed, even though the DC input bias current can be extremely small, the resistor is still needed to provide a path for it. <S> 2) <S> The - input's input bias current flows from the output of the opamp through R3. <S> If C2 was not there that current could also flow through R2, or it could flow partly through R2 with the rest through R3, it depends on the DC voltages at each point how the current flows. <S> I suggest you browse through the excellent and free Opamps for everyone to learn more about how to use opamps. <A> How is the input bias current able to continuously deposit charges and saturate the input if I don't use the resistor? <S> The resistor should not allow the DC bias current through, so where is this current flowing from/into? <S> A capacitor appears to be an open circuit at DC, and the input bias current is a DC current. <S> It's the capacitor that doesn't allow DC bias current through, not the resistor . <S> Without \$R_1\$, there is nowhere for the input bias current \$I_{\text{IB+}}\$ <S> to flow except to charge up the capacitor (which creates an undesired voltage on the capacitor). <S> Adding \$R_1\$ provides a DC path to ground for the input bias current. <S> There is a small voltage offset introduced: \$I_{\text{IB+}} \times R_1\$. <S> But this is better than the case where the input bias current charges \$C_1\$ until the supply rail is reached. <S> Why is this not a problem for the feedback path to the "-" terminal? <S> No DC current flows through \$R_2\$ and \$C_2\$, but the op amp's output provides a path for the input bias current to flow. <S> Since there is already a path for the input bias current to flow, you don't need to add another resistor to ground. <A> The resistor should not allow the DC bias current through, so where is this current flowing from/into? <S> Incorrect - if the bias current is 1nA and the resistor is 1Mohm <S> the voltage across the resistor will be 1 mV i.e. pretty much at ground potential. <S> If the resistor were instead connected to (say) 2V the voltage on the top end of the resistor would be 2.001 volts. <S> Regards the feedback path, the bias current flows into the op-amp output. <A> A bit of electronics 101 here. <S> 1st, TL081 has very low input bias current of 30pA. <S> On average giving the opamp <S> sideways look induces more current. <S> Theoretically this bias current will eventually discharge your capacitor but you're in for a good long wait. <S> After one minute 30pA current would reduce capacitor voltage by 1.8mV. <S> Conversely that 100k resistor has time constant of 100ms <S> so the capacitor would be discharged after about half a second. <S> Your second guestion is answered by simply considering the fact that there's the opamp output that's perfectly capable of sourcing and sinking current. <S> that's what's keeping your capacitor in check.
| The input bias current is a current sourced or sunk by the op amp input in order for the op amp to operate (ideally, an op amp has infinite input impedance and no input bias current, but no op amp is ideal).
|
How to make perfboard circuits more compact So, I've been working on my prosthetic limb for quite some time, and I've sorta bumped into a few problems, while the circuits I've made work, they don't fit inside the arm. Mostly because I have tons of opamps (around 28 for a state variable filter, it has settable Q, so I couldn't resist, but I've used quad opamps), and the solder job has been a nightmare. I need to fit it in a small space, and I'm okay with components protruding up and down the board. Is there some way I can me my circuits more compact? I'm thinking of using smd insted of DIPs, but I want a few more options.. <Q> I understand your reluctance to use a PCB because it is more work and more expensive initially - however <S> , in the long run it sounds like the only way you will get what you want. <S> You can use perfboard and wire everything together manually, but debugging those circuits is very difficult. <A> You may not thank me for this, but you can look into a technique from back in the dark ages, cordwood construction . <S> This will allow you very high component densities, and if you're just doing signal processing you should be able to get away with the thermal issues which arise with this approach. <S> The reason I say you may not thank be for it <S> is that, once assembled, it's a stone b**ch to troubleshoot. <S> You must be meticulously organized and focused in order to put one of these things together, and your spatial perception must be first rate in order to figure out how to make the connections. <S> But it does make very compact assemblies. <A> You want fast, cheap, and small. <S> If you are dead set on using perf board, pick two of those. <S> If you want all three, you will very likely be forced to get a PCB, whether you like it or not. <S> That's just a fact of engineering. <S> I understand that PCB money doesn't flow from one person's pocket as easily as it might from another, but unfortunately, that's the way it is. <S> If your circuit works, then just strap the big clunky thing to the outside of the prosthetic. <S> Its a prototype, right?
| The ultimate answer is to use SMD components on a PCB - the parts are smaller, and using traces instead of wires will reduce stray capacitance; this can have a large effect in sensitive circuits.
|
DC Bus Conditioners in VFD Drives In VFD Drives, DC Bus conditoners are used to protect the DC link supply from transients. Can you please let me know the different scenarios DC Bus of a VF Drive gets these transients and their relation with different drive parameters ? <Q> There are several issues involved in the design of the DC link supply of a variable-frequency drive (VFD). <S> The reactive current requirement generally determines the capacitor selection. <S> When power is initially applied to the VFD, the inrush current can damage the rectifier and the capacitors if it is not limited. <S> There are a number of pre-charge schemes and other measures that are used to limit the charging current. <S> During steady-state operation, there is a pulse of current into the capacitors that occurs at the peak of each half-cycle of each phase of the supply voltage waveform. <S> That is not a problem for the VFD components, but it represents harmonic distortion of the power current waveform. <S> That can cause distortion of the power line voltage and have negative effects on power distribution components and other connected utilization equipment. <S> It also results in a reduced total power factor. <S> VFDs often have inductance added to the incoming AC lines or the DC link to reduce input current distortion. <S> There are two basic scenarios involving line voltage transients. <S> High-voltage, short-duration voltage transients can damage the rectifier and possibly other components. <S> Metal oxide varistors are often used to protect against that scenario. <S> Line inductors are also useful in that regard. <S> Lower-voltage repetitive or ringing voltage transients may have sufficient energy to charge the DC bus capacitors to a sufficiently high voltage to damage the inverter components. <S> Input line inductors can also mitigate that problem. <S> Sometimes, connecting a dynamic braking resistor across the bus can help. <S> Ultimately, it may be necessary to eliminate the source of the problem. <S> The source is often the switching of power factor correction capacitors somewhere in the power distribution system. <S> Of course motor power regeneration can also raise the bus voltage. <S> Either dynamic braking resistors or deceleration rate limitation can be used in that situation. <S> In some cases, braking energy may need to be returned to the supply using a regenerative input section. <A> Probably the most likely is hard braking. <S> All that energy has to go somewhere, and it usually ends up in the DC bus all of a sudden. <S> I've never seen a DC bus conditioner that claims to do that, but I have seen and used braking resistors, which are connected automatically by the drive across its own DC bus as required to bleed off the extra charge. <S> Lightning is also a possibility, or maybe a non-VFD motor being back-driven faster than its rated speed so that it raises the line voltage (never actually seen that), but I'd say that the most common is active braking by the VFD itself. <A> Your DC Bus conditoner's are inductor/capacitor/varistor combinations that form a 'backfeed' filter. <S> The transients can be measured in amps or volts, but amps is a better representation. <S> It is equal to the total DC current being consumed and is always present if the VFD is active. <S> As an example, a 10 amp electric drill produces 10 amps of noise transients on the AC line (noise and transient are the same thing, but under different context. <S> Noise is a continuous event, while transients are a single pulse of noise selected or created for detailed examination. <S> Fast repeated transients could be called 'noise'). <S> So the VFD noise transients are equal in current to the current consumption of the VFD while driving the motor. <S> At 100 amps of current consumption the VFD would have 100 amp noise transients, before any filtering. <S> The frequency of the noise is equal to the drive frequency of the VFD x 3, for a 3-phase VFD. <S> Most all VFD's use PWM to improve efficiency, but this noise is filtered internally. <S> The fine details of harmonics vs. frequency, etc, would have to be measured under real-world conditions. <S> My answer is general observations from building and using them.
| In the most common VFD configurations, there is a DC link capacitor bank that limits the DC ripple voltage and supplies the reactive current required by an induction-motor load.
|
Why do we need sheet entries and sheet symbol in Altium I don't understand why we need to define some sheet entries and sheet symbols in Altium with multi sheet design ?!Indeed, nets are already connected to each other by their ID from sheet to sheets as I understand or via Port. Why do we need to add on the top of this to describe Altium about sheet entries and sheet symbol architure ? <Q> Indeed, nets are already connected to each other by their ID from sheet to sheets as I understand or via Port. <S> This depends on the Net Identifier Scope you choose for your project. <S> Global (Net labels and ports global) <S> This is the scope you refer to. <S> Ports and net labels connect across all sheets throughout the design. <S> This is similar to what you may know from "Eagle" layout editor and other "simple" design tools. <S> It is not suited for a clean modular design. <S> Flat (Only ports global) – ports connect globally across all sheets throughout the design. <S> This gives better control of the sheet inter-connections since you have to specify ports for signals that should go "off-sheet". <S> Hierarchical <S> (Sheet entry <-> port connections, power ports global) <S> This requires manual inter-sheet connection and usually uses a "top sheet" with only sheet symbols on it as a base. <S> Note that power ports (12V, GND etc) are global for convenience. <S> Strict Hierarchical (Sheet entry <-> port connections, power ports local) <S> This one requires you to also connect power ports between sheets. <S> There may be two different 5V voltages in one design for example, which could be connected by accident when using the normal hierarchical design. <S> " <S> Strict" has the advantage that you would manually check every single connection. <A> To Support Reuse ...to support multi-channel design -- the reuse of sections -- either by having multiple copies of the block in one design or reusing it across multiple designs. <S> Global nets are generally a problem as they can lead to forgotten/unexpected connections when moving across designs or when multiple engineers work on the same design -- especially when the design is very large. <S> Ports objects by themselves, do not create a reusable block. <S> They are just half of the solution. <S> The ports manifest as ports in sheet symbols to allow you to attach nets to them. <A> Don't assume that a net label on one sheet is connected to the same net label on another sheet. <S> Its an option in the project settings which I think is off by default. <S> The pcb layout has no problem creating 2 distinct nets with the same name. <S> (I speak from bitter experience) <S> It also allows channel based designs, with repeated sheet symbols which have separate inputs and outputs but the same circuitry
| Sheet symbols with sheet entries allow you to place functional blocks in a sheet and then show how the blocks are connected together on the top sheet which serves as a nice block diagram.
|
(Capacity 20hr: 65AH) Does this mean 65amps per hour for up to 20 hours? Cranking Amps: 675Im trying to find a good battery for a project.That part where it says "20hr" keeps making me rethink what i sorta hope it means. <Q> No, it means that you'll get a total of 65 amp-hours out of the battery when you discharge it over a 20-hour period. <S> This would correspond to a current of $$\frac{65 \text{ amp-hours}}{20 \text{ hours}} = <S> 3.25 \text{ amps}$$ during the 20-hour discharge. <S> Battery capacity usually varies with the discharge rate (current), so for comparison purposes, a standard testing period is specified — in this case, 20 hours. <S> With higher currents (shorter periods), you'll measure less total capacity, since more energy is wasted in the battery's internal resistance. <S> With lower currents, you'll measure somewhat higher capacity, up to some limit determined by the the construction of the battery. <A> Ampere for 20 hours. <S> So you actually need to discharge it with 65 A / 20 hrs = <S> 3.25 <S> Ampere <S> and then it will last 20 hours. <S> At a higher current batteries become less efficient and the amount of energy you can extract goes down. <S> So 65 <S> Ah usually does not mean that you get 65 Amps for 1 hour. <S> You have to discharge at a modest rate so 10 or 20 hours is more reasonable <S> and then you do get the specified capacity. <A> 65A per hour is written as 65A/h, not 65Ah. <S> That's Ampere-hour, or the product of time and current. <S> The 20 hour just means that you can take more energy out slowly. <S> If you drain 65A it'll only last 30minutes, probably, because the higher the current, the lower the amount of energy you get out. <S> The 20 hour is just an indicator for how to get it out or in.
| What it means is that if you discharge this battery for a period of 20 houres, it will have a 65 AmpHour battery capacity. No that does not mean 65
|
How would I go about converting a change in capacitance into a voltage? Just starting out with electronics here as a physics student- I appreciate the help! In a simple LCR circuit with a variable capacitor, how would I go about detecting the change in capacitance and converting that change into a measurable voltage signal? What kind of circuit/ components will I need? <Q> Unless your capacitor is very small in value (less than 100 pF) you usually do not need to use an inductor to use that capacitor in an oscillator. <S> An oscillator can also be made with an RC circuit, this is used in the 555 timer chip. <S> The frequency of oscillation would change as the capacitor changes value. <S> Then you only need to measure the frequency of oscillation which is is often as easy as counting pulses for a certain time (like 1 second) and then displaying how many cycles were counted. <S> Please state between what values your capacitor will vary <S> so we can give you more practical advise in the direction of a usable solution. <A> With a capacitance of 10pF to 100pF, the easiest thing may be to make a simple filter and measure the output: simulate this circuit – Schematic created using CircuitLab <S> For 10pF, the resistor and capacitor will have equal voltages (at \$\frac {V_s} {\sqrt (2)}\$) at 159kHz and for 100pF it will be at 15.9kHz. <S> Change the value of the capacitor and then adjust the input frequency such that \$V_{out}\$ = <S> \$0.0707V_s\$. <S> You could measure both voltages and simply change the frequency until each are equal. <S> The capacitance at this new frequency is easily definable. <S> As <S> \$ X_c = \frac {1} {2\pi f c}\$ we can rearrange so that the capacitance must be <S> \$C = <S> \frac {1} {2\pi f 100k}\$ ( <S> because for equal voltages, the capacitive reactance and the resistance must be equal and I used 100k for the resistor). <S> An interesting way to watch the phase (and therefore relative voltages) is by using a lissajous figure <A> I'm assuming you want a DC like voltage refrence that changes based on the capacitor's current value, on something of the realtime-continuous order, and not an offline capacitance tester <S> I would try using a PWM controler. <S> Some use an external RC circut to set the switching frequency. <S> Which of course is useless in a DC world. <S> But the C of the RC circut also changes the Dead Time of the output. <S> So asuming you have a steady duty cycle feed to the controller(like 100%). <S> A changing dead time duration would effectively change your output Duty Cycle. <S> The PWM signal would read as DC voltage level, who's value is based on the output PWM's duty cycle. <S> So, Change in capacitance - <S> > change in Dead Time - <S> > change in efective output duty cycle - <S> > chang in DC voltage value. <S> This has elemets of RC circuts, and frequncy, however unlike the other current answers this soultion doesn't need to worry about them, or any time based mesurements. <S> Might not be the best part, but as an example look at LT1241
| The easiest way that I can think of is using that capacitor in an oscillator circuit.
|
Are capacitors floating power supplies? I'm struggling with the concept of floating voltages. For a grounded supply I can connect one side to an effectively infinite ground and get full power. It may vary depending on its construction, but if the grounded supply provides X volts then it is necessarily true that: [A] (V+ to G) - (V- to G) = X , correct? Now suppose I want to "ground" a floating supply. My understanding is that the supply's V+ and V- separated behavior with respect to some "ground" of practically infinite capacitance must be characterized by the source of the voltage: A floating driven power supply cannot send power through a path that does not directly connect its V+ and V- terminals, as roughly explained here . A chemical power supply, like most conventional batteries, cannot send power that doesn't connect its V+ and V- terminals, because the driving chemical reaction will not occur without electron transport exactly matching the needs on both terminals simultaneously. But a capacitor should be able to send power from either its V+ or V- terminals, right? It depends on the charge of the G, but not only does the relationship [A] above hold, but also the capacitor can dump all of its power into an arbitrary ground if V+ and V- are independently and separately connected to that ground. I.e., there is no such thing as a "floating" capacitor? (Evidently I'm assuming that a "floating" supply is defined to be one in which power only flows between V+ and V-. Is that accurate? And are there any other "floating" power supply categories than #1 and #2.) <Q> To add to Barry's answer I'd also make these points: <S> A floating driven power supply cannot send power through a path that does not directly connect its V+ and V- terminals, ... <S> A chemical power supply, like most conventional batteries, cannot send power that doesn't connect its V+ and V- terminals, Within lumped circuit analysis, no device can deliver power to any other device that isn't part of a circuit connected to two of its terminals. <S> But a capacitor should be able to send power from either its V+ or V- terminals, right? <S> Current always flows in complete circuits. <S> This means that whenever there is current flowing in one terminal of a capacitor, an equal current is flowing out of the other terminal. <S> Because of this you can't say "the V+ terminal delivered power" or "the V- terminal delivered power". <S> The capacitor delivers power to the rest of the circuit when current flows out of its more positively charged plate and in the more negatively charged plate. <S> The capacitor absorbs power from the rest of the circuit when current flows in to the more positively charged plate and out of the more negatively charged plate. <S> It doesn't matter which terminal is labeled "+" or "-", it matters which one is at a more positive potential and which way the current is flowing at the instant when you are measuring the power flow. <A> The term "floating", be it a power supply, battery or other circuit component, simply means that neither of the component's two leads are grounded. <S> This is true whether the device is floating or not. <S> The amount of current (power) that can be drawn from a power supply does not depend on whether the supply if floating or not. <S> It only depends on the load connected to the supply's 2 output leads. <S> Grounding floating supplies is done for safety reasons and/or circuit requirements but not to allow power to be drawn from the supply. <A> Voltage is relative (to some potential). <S> Usually for simplicity we pick certain potential and call it Ground. <S> It's possible, that in system several grounds will be present. <S> And they are not necessarily connected. <S> If they are not connected, they are called floating. <S> The term itself refers to anything that is not "constrained" to any known potential in the system. <S> For example floating input is one that it's voltage <S> may be anything and you can't guess. <S> This is why sometimes you will see pull down or up resistors, so instead floating node you will get known voltage. <S> To your question: capacitor indeed may be a floating supply. <S> Actually, there is at least one very common case when it is used this way: boost capacitor in switching systems is exactly that.
| To draw power from a floating device does require that the load be connected to both leads of the device.
|
why does sources with same voltage drive different current through same load? I have three mobile phone chargers with rating 5V 500mAh, 5V 1Ah , 5V 2Ah. When I charge my phone Mi 4i with these chargers , the charging time is different. It means all these chargers drive different current through my phone's charging circuit. So my question is, how can same voltage sources(5V) drive different current ( here 500mA , 1A and 2A) through same load(my phone)?? <Q> There are 2 types of charging IC's used in cell phones. <S> The 'dumb' IC uses a standard 1 amp charging current if the wall adapter can supply it. <S> If you use your PC to charge the phone, it may only allow 1/2 amp of charging current, so that is all the charger IC has to work with. <S> It will not use so much current that the source voltage drops below 4.0 volts. <S> This includes a drop of 1/2 volt in the cable going to the wall-pack. <S> The 'smart' IC in some smart phones will draw as much as 2 amps to charge the phone, if the charger can supply 2 amps and still maintain close to 4.5 to 5 volts under that load. <S> It is generically called a 'fast' charge, and is usually controlled by a check box in the phones 'Battery' control directory. <S> A phone that charges at 2 amps will come supplied with a 2 amp charger (wall-pack). <S> My Samsung Note 2 has such a charger <S> and I have it set for 'fast' charge. <S> It is a very busy IC when charging. <S> NOTES : 1) <S> The LM7805CT has a limit of 1 amp with heatsink, but there are simple booster circuits you can find on the web. <S> A 12 volt 7.2 amp-hour battery will work, within the constraints of the LM7805CT IC. <S> 2) <S> My answers in terms of current are in amp-hours or mA-hours. <S> 3) I am not implying a charge current > 2 amps or a charge voltage (at the phone) <S> > <S> 5.1vdc. <S> I have not yet seen any phones capable of going out of these ranges. <A> The current control is implemented by the power management logic implemented in the phone. <S> In particular, according to the Mi 4i FAQ, your phone uses Qualcomm's original QuickCharge 1.0, which consists primarily of two components: <S> AICL = <S> Adaptive / Automatic Current Control, and APSD = <S> Automatic Power Source Detection. <S> APSD incorporates BC1.2 = Battery Charging Revision 1.2, an extension of the USB standard that allows chargers to communicate that they can deliver higher current, e.g. by shorting the D+ and D- lines. <S> AICL includes heuristics for automatically detecting the current limit of the power supply by slowly stepping up the input current (e.g. in 25mA steps). <S> Usually this is limited to about 2A. <S> You ask in the comments if you can safely charge your phone using a DIY charger using a 5V regulator. <S> Generally this should work fine as long as you include the appropriate BC1.2 signalling to the phone. <S> To do that you can either construct your own fast-charge cable with the D+ and D- wired as needed (see below), or you can buy various fast-charge USB adapters, e.g. search on "CW3002" on eBay to locate adapters using a common CellWise CW3002 controller . <S> Below is a bit more info on the technical details, first on on AICL, from the NXP (freescale) BC3770 datasheet , and second on BC1.2 from Maxim's Overview of USB Battery Charging Revision 1.2 and the Important Role of Adapter Emulators . <S> Now that you know the appropriate buzzwords you should be able to dig deeper if need be. <A> I think there is a fundamental misunderstanding here. <S> Are your 3 "chargers" connected to a wall <S> are they just yanother battery? <S> If your charger is rated at 500mAh or 1Ah <S> these are ratings of CHARGE, not of CURRENT. <S> Backup batteries are rated in terms of charge. <S> Many of these are on the market. <S> You charge the backup battery first and then use it later when your phone battery is low. <S> An AC adapter that changes 120 VAC into 5V DC would be rated in Amperes and not amp hours or milliamp hours. <S> 1 <S> Ampere <S> = 1 <S> Coulomb / <S> Second which is a flow rate of charge <S> 1 Ah = 1 <S> C <S> /s <S> * 3600 s = <S> 3600 C of charge <S> Obviously if the phone can draw more current, it will charge the battery faster. <S> This article explains it much better than I can. <S> So in order to answer your first question, what were the charges on each of the backup batteries? <S> What is the maximum charge of your phone battery? <S> If the backup battery cannot provide enough charge, your phone battery will not be fully charged.
| The charger IC in the phone makes all the decisions about how much current there is to charge with, is there a 'fast' charge setting for this phone, is the supply voltage at least 4.5 volts, plus monitoring the charging of the battery and reporting charge status to the CPU.
|
Use transistors for controlling two LEDs I'm sure this is a pretty basic question, but I wanted to run it by you guys before creating my PCB. I want to have two LEDs on the board to indicate the global state of the board. When the board has power, but the MCU is not in an active mode I want the Red LED to light up, and when the MCU puts a pin high I want the red led to be replaced by a green led instead. I want to solve the problem using simple electronics, and not with RGB led, etc. My idea is to use two transistors. One which is normally closed (the green one), and one which is normally open (the red one), and then toggle them on/off using a signal from the MCU. Does the following circuit make sense? Edit: Lots of good suggestions here. The MCU is a Teensy 3.2, which has 3.3V output, and 25mA Max current on the digital GPIO pins. I have both 3.3V and 5V powersources available on the board. <Q> Something like this should work: simulate this circuit – <S> Schematic created using CircuitLab <S> An NPN transistor (such as 2N3904) will conduct when the base is more positive than the emitter. <S> A PNP transistor (2N3906) will conduct when the base is more negative than the emitter. <S> Resistors are required in the base lead to limit the base current. <A> When the MCU goes HIGH (5V) <S> the NPN transistor is turned ON, the PNP is turned OFF and the red LED comes ON. <S> When the MCU goes LOW (0V) <S> the PNP transistor is turned ON, the NPN is turned OFF and the green LED comes ON. <S> With NO input from the MCU (mid point floats to 2.5V) <S> both transistors (and LEDs) turn ON. <S> EDIT ADDITIONAL: <S> For a 3V3 MCU reduce the top 4k7 resistor (E-B of the PNP) to 2k2. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Transistorless option. <S> The 0.9 V reading is obtained with GPIO turned off. <S> 1.1 V reading is with GPIO turned on. <S> Figure 2. <S> Negative rail switching version. <S> I haven't tested this but the theory is ... With GPIO off D1, RED, lights. <S> The voltage drop across D3 and D1 will be about 0.6 + 1.8 = 2.4 V. <S> When GPIO pulls high D2, GREEN, lights. <S> The voltage drop across the green led will be about 2.0 V. <S> Since this is less than required by D1 / D3 the current will fall in the RED led and it will go quite dim. <S> I've assumed some voltage droop when the GPIO is supplying current. <S> This circuit can be inverted if GPIO pull low is preferred. <S> No transistors. <A> This is an inverter circuit. <S> I have used this circuit to answer other questions and it works perfectly. <S> The Resistor in the middle should be about 1KΩ . <A> You can use a transistor as a switch here. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> So when the GPIO is low or zero then the transistor would be in cutoff region and the green LED glows and when the GPIO is 3.3 V the red LED glows since the transistor goes in saturation state the Green LED would be switched off. <S> The resistors values are so chosen that the transistor work as a switch. <S> EDIT <S> : Writing <S> KVL's will let you know the values of resistors. <S> Let 0.7v be across the base emitter junction when 3.3V is given at GPIO <S> pin,1.8V <S> across red LED, Let the voltage at collector emitter junction be 0V when saturated so at this voltage the green LED will not glow. <S> Also let Ib be the current and is generally 20mA for a normal 5mm LED KVL at B-E of transistor : <S> $$3.3V-0.7-1.8V-I_bR_1=0 <S> $$We require about 20mA in the red LED <S> so R1=40 ohm <S> When the transistor is in cut-off state <S> i.e., no voltage at GPIO pin & the voltage at collector is around 5V so KVL at C-E junction of transistor : $$5V-2V-I_3R_3-I_CR_2=0$$ 2V= <S> forward voltage of green LED and let 20mA of current is required for the green LED so R2+R3=150ohm, I3= <S> Ic=20mA <S> assuming no current flows through the transistor since Vc=0V. <S> So choose R2=100ohms and R3=50ohms <S> Note <S> : we choose R2 to be 100 because we do not want large current to be flowing through the circuits at collector when transistor is cutoff. <A> I tried Peter Bennett's circuit, and like Antonio, I too had the red LED on at all times. <S> At first, I tried Antonio's modification to the circuit, but then I realized that it was wrong (even though it worked). <S> Below is my modification of the circuit. <S> I substituted a BC547 NPN and a BC557 PNP because it's what I had. <S> I changed R2 to 8.5k, and added the 2 pull-up resistors(R5 and R6). <S> When the MCU pulls the bases of both transistors to ground, the green LED turns off and the red LED turns on. <S> I know this was an old post, but I figured I'd add to it since it somewhat helped me. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Wow the anwsers to this are complicated. <S> But your circuit is already pretty close. <S> What you're describing is actually a simple differential / ltp circuit. <S> Just disconnect one of the bases and connect it to a voltage divider making 2.5V. <S> When the transistor connected to the MCU is low, it will not provide an current and all current will be diverted to the other transistor. <S> When the MCU is high, that transistor will consume all current and the other transistor will turn off (because the common emitter resistor of the differential / ltp will present a voltage higher than Vbe of the second transistor). <S> However, you may also need to move the LEDs to the collectors because the voltage drop may cause issues with the differential action. <A> The following circuit should work nicely provided that the MCU pin has sufficient current drive. <S> Both Microchip PIC and Atmel AVR chips have pins with sufficient current drive. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> I am assuming that you are using Superbright LEDs. <S> A typical Green Superbright LED drops about 3 Vdc and is blinding when run from 5Vdc with a 10k resistor. <S> A typical Red Superbright LED drops about 1.7 Vdc and is reasonably bright when run for 5 Vdc with a 2k2 resistor. <S> Operation is extremely simple: if the MCU pin is either high-impedance (set as Input) or is Low, the Red LED is lit and the Green LED is NOT lit. <S> When the MCU pin goes Hi, the Green LED is lit and the Hi level raises the cathode of the Red LED high enough that it is NOT lit. <S> This circuit works well with the PIC 16f microcontrollers that I use.
| If the signal from the MCU is Low, the red LED will be on, if High, the green LED will be on.
|
Why do charge pumps use capacitors? https://en.wikipedia.org/wiki/Charge_pump A charge pump is a kind of DC to DC converter that uses capacitors as energy-storage elements to create either a higher- or lower-voltage power source. Is it possible to create a charge pump without using capacitors? What characteristics of capacitors make them useful in charge pump applications which are normally seen in level shifters? <Q> The name gives you the answer : you need to store charge and move that stored charge around. <S> Charge storage devices are either batteries (not very practical) or capacitors (practical, cheap and common). <S> The only alternative I can see is to store charge on an insulator, move the insulator, and reclaim the charge from it elsewhere. <S> A rubber belt makes a suitable insulator for the job. <S> Thus a Van de Graaff Generator <S> (the machine not the band ) is a charge pump of sorts. <S> But capacitors and rectifiers (or even spark gaps used as switches) are a more practical approach, as in the Cockroft-Walton multiplier . <A> The name 'charge pump' is the name given to devices that change voltages using capacitors as the main energy storage element, Boost/Buck regulators are the names given to devices that change voltages up or down using inductors as the main element respectively. <S> A charge pump has the advantage over a level shifter that it can generate a voltage either higher or of opposite polarity than the input, level shifters can only generate outputs that are within their supply rails, charge pumps do not have this limitation as they are full DC-DC converters as opposed to a glorified switch. <A> Is it possible to create a charge pump without using capacitors? <S> Yes you could use rechargable batteries as well <S> but it would be impractical as these are usually larger, they age and wear out. <S> Also most rechargable batteries do not like to be completely discharged so the circuit would have to be more complex to avoid that. <S> Also when the charge pump is off the rechargable batteries would still hold a voltage. <S> Maybe a circuit would be needed to disconnect the rechargable batteries from the load or from the circuit. <S> Rechargable batteries are more expensive than capacitors. <S> Capacitors do not mind being discharged completely. <S> They age but much less than rechargable batteries. <S> As far as I know there is no other way you can build a charge pump without using capacitors or rechargable batteries. <S> OK, rechargable batteries actually store the energy of the charge as chemical energy and not as an electrical charge as a capacitor does. <S> But for a chargepump like circuit that does not matter. <S> You could also make a increased voltage using an inductor but <S> that is not a charge pump but a boost converter. <A> Level-shifters don't store any energy. <S> This means they only work within the highest voltage available to them. <S> Typically they are connected to the two voltages they are shifting. <S> Level shifters act like a switch with some power converted to heat. <S> Some level shifters use 'transformers' to communicate a change of voltage between the two sides. <S> However, even these don't store energy to function, they are still acting as switches. <S> A DC-DC converter uses its power source to either produce a larger or smaller voltage. <S> A DC-DC converter stores energy in the capacitor. <S> When it's creating a larger voltage, a DC-DC converter can step well beyond its power supply voltage, unlike a level shifter. <S> A DC-DC converter, depending on its circuit structure, steps up a voltage, with reduced current (compared to its power source), or steps it down with increased current. <S> Of course, this process has losses. <S> However a step-down DC-DC converter can be much more efficient than a linear regulator. <S> An alternative short-term energy storage device would be an inductor.
| You need an element which can store charge which is what capacitors and rechargable batteries do.
|
Power draw from 120V AC Outlet According the the following link, a minimum of 4800W is drawn from an electric stove when all burners are turned on: http://www.ehow.com/how-does_5179441_many-watts-electric-stove-use_.html . I'm just wondering how a stove achieves this? Are they not limited to approximately 1800W like other power outlets are? <Q> That provides theoretical 9600 or 12,000 watt power, but due to "continuous load" restrictions, 80% of that, so 7680 or 9600 watts, respectively. <S> They don't make a receptacle for 40A, so they cut a special exception to use 50A receptacles, though they use a 40A circuit breaker if the wire is 40A. <S> You can count on the receptacle being NEMA 14-50, or in older homes NEMA 10-50. <S> These provide neutral, though the latter does not provide ground, which can create an electrocution hazard if there's a problem with the neutral. <S> Neutral is provided because US appliances often have bits which need 120V rather than 240V, notably the oven lamp. <S> Electric clothes dryers are provisioned with 240V 30A service, 7200W nominal, using a NEMA 14-30 or 10-30. <S> These are the only 240V heavy appliances where unngrounded service (NEMA 10) is commonly allowed Practices in other nations in the NEMA sphere of influence may vary. <A> In the US, an electric stove is typically powered from 220-240 V (two hot phases) rather than 120 V, and wired with 10 AWG wire or even 6 AWG, allowing something like 30 A or 50 A. <S> This would allow the stove to draw at least 6600 W. <A> Some electrical appliances draw large amounts like stove (cook-top), oven, water-heater, clothes-dryer, heater/furnace, etc. <S> In 120V territories, these high-current appliances are provided with special 240V, high-current branch circuits. <S> And furthermore these branch circuits are typically dedicated to the single appliance, and the power is not shared with other loads. <S> A 20 Amp, 240V circuit will supply 4800 Watts. <S> And 20A is probably the smallest size breaker and wire-size for a high-load branch circuit. <S> It is not unusual to find 30A or even 50A breakers and corresponding wire sizes.
| In the USA, electric stoves are supplied by a 240VAC, 40 or 50 amp circuit.
|
What is this suspicious stain under a VFD's glass? I wanted to give VFDs a try so I ordered a couple of 8x14 segments display modules. They arrived today, but I'm concerned because each has a big black stain in the corner. If it were on an LCD I'd suspect they're broken, but since I don't know this display technology at all, can someone tell me: Are those displays broken or is the stain a normal thing? If this thing has a name, what is it called? Here is a picture of the modules (the stain is on the top right corner): Note that after checking on the seller website and a few pictures on other online resources, this kind of stain is visible on some of them, so it may be a normal thing. <Q> It's getter, it should absorb any residual gases in the display. <S> Vacuum tubes have getters too (the black part on top). <S> If that spot turns white (though in some cases I have seen it become transparent/invisible or just fall off) <S> it means that the glass is broken and the VFD or tube is full of air and is no longer functional. <A> It is called a getter . <S> It's design to be a reactive deposit to absorb the remaining air in the vacuum. <S> The ones in the linked photo are probably fine, the Wikipedia article has photos of examples where the vacuum has been lost (they become completely whitened). <A> It is as designed. <S> No worries! <S> It'll get the remaining O2 that was left in there.
| It's called a getter.
|
Using LC filter instead of RC filter in mains powered circuits Above is a dimmer circuit. In some of these circuits I see a toroid inductor together with a capacitor forming a low pass LC filter which is called choke filter . My questions are: Why is RL used instead of RC and is that something to do with the character of mains freq. and voltage? Inductors are expensive or big thats why we use caps around opamps to obtain low pass filters right? What is being choked by the LC toroid cap couple? <Q> Ideal capacitors and inductors don't dissipate any power, whereas resistors do. <S> One advantage is therefore that L-C filters are more efficient than R-C or R-L filters. <S> Often it's not the actual waste of power that is the driving factor, but the mechanics required to get rid of the heat. <S> Lamp dimmers often need to fit into small spaces where keeping them cool would be difficult. <S> These are double-pole filters. <S> R-C and R-L filters are single pole. <S> Well into the stop band, a single pole filter attenuates by the frequency ratio to the rolloff point. <S> In log space, that's 20 dB per decade. <S> A two pole filter is like two poles applied in series. <S> They attenuate by the square of the frequency ratio to the rolloff point, which is 40 dB per decase in log space. <S> For low power ( <S> a few watts to 10s of watts) dimmers, the main reason to filter is to reduce the high frequencies so as not to interfere with radio communications. <S> These are many multiples of the 50 or 60 <S> Hz fundamental frequency. <S> A L-C filter will attenuate the higher frequency harmonics relatively more than the lower frequency ones. <S> This is useful since it's the higher frequency content that the regulators care about more. <A> LC noise filtering benefits were explained in other answers. <S> This is an addition specific to triacs or thyristors. <S> One reason to use an inductor in series with a triac is that a triac has maximum allowed rate of change of current (\$\frac{dI_T}{dt}\$ rating) when it is triggered into conduction. <S> This may or may not be an issue for a particular triac based circuit, but in general such problem does exist. <S> A quote from http://www.nxp.com/documents/application_note/AN_GOLDEN_RULES.pdf <S> (page 5): <S> When a triac or thyristor is triggered into conduction by the correct method via its gate, conduction begins in the die area immediately adjacent to the gate, then quickly spreads to cover the whole active area. <S> This time delay imposes a limit on the permissible rate of rise of load current. <S> A \$dI_T/dt\$ which is too high can cause localised burnout. <S> An MT1-MT2 short will be the result. <S> An inductor limits \$\frac{dI_T}{dt}\$, provided that the inductor does not saturate. <A> A dimmer typically works by using a triac to take only a portion of each 60Hz input power cycle. <S> When the cycle gets chopped off abruptly by the triac this creates a pulsed power draw at the input. <S> The pulsed power draw will have harmonics of the 60Hz waveform in it. <S> In order to comply with FCC regulations regarding conducted emissions a filter must be added isolate the mains power from those pulses. <S> So why use RC vs LC? <S> LC FILTER: <S> PRO: <S> Dissipates less power than RC filter (so less heat) <S> PRO: Second order filter, which means better noise rejection CON: <S> May resonate if not properly damped CON: <S> More expensive than RC due to higher inductor cost. <S> RC FILTER: <S> CON: <S> Dissipates more power than LC filter CON: First order, <S> worse noise rejection <S> PRO: <S> Inherently free of resonance <S> PRO: Cheaper than LC due to low cost resistor
| Another advantage is that L-C filters attenuate more in the stop band.
|
Can a Tesla coil's secondary wiring overlap? I am building a small tesla coil just as a proof-of-concept using a bug-zapper circuit. I unwound some enamel copper wire off a transformer and put it on my 12 cm length x 2 cm diameter PVC pipe and realized that after winding the first time, I still have lots of wire left over, so I just wound a second and third layer of secondary wire over the first layer. Is this okay to do? Every other Tesla coil I've seen doesn't do this, but why not? As I understand it, a Tesla coil is just a transformer and it shouldn't matter wether the coils overlap, and only the number of coils should matter, right? <Q> A Tesla coil is NOT a transformer. <S> It is a resonant structure that transfers energy via an coupled resonance. <S> Thinking of it as a transformer is incorrect. <S> So the answer to it being OK to have overlapped windings is no, due to the added capacitance of the overlapping windings and capacitive coupling. <S> A Tesla coil is better described as a quarter wavelength helical resonator. <S> It needs a high Q factor and the capacitive coupling of overlapping windings will dampen the resonance resulting in sub par performance. <S> Update (2 years later I got around to it...) <S> : I should make a diagram someday, while not technically 100% accurate, here is a good analogy. <S> Think of a tesla coil as a long wire that reflects waves at the top end back down. <S> When in resonance, every time the energy packet reaches the base after reflecting back from the top, a new amount of energy is added to the packet making it grow, and again sending it up the coil. <S> This will dampen the resonance via destructive interference etc. <S> You will get a loss of Q not only from over lapping windings, but also if the base material of the tube supporting the wire is lossy at RF, it will capacitivly couple energy out of the coil. <S> In some of my work I have found similar coils (dimensional and wire length etc.) <S> with cardboard vs. polypropylene tubes for the main body can have Q's of 25 vs 175. <S> That means the plastic tubed coil had an additional voltage gain of 7x, and it was very noticeable. <A> Honestly, MadHatter's answer is the closest to being correct, although Tesla coils ARE still a type of transformer. <S> Nobody else here seems to understand, however, that Tesla coils are resonant transformers, so they do not operate in the same way as common iron-core transformers. <S> The most important factor required for a Tesla coil to run correctly is that the secondary coil and topload LC circuit have the same resonant frequency as the primary coil/capacitor LC circuit. <S> This is how you get efficient energy transfer from the primary circuit to the secondary circuit. <S> Putting too many turns on the secondary would add too much inductance (and self-capacitance) that the secondary circuit will be significantly out-of-tune with the primary circuit. <S> You will get very little energy transfer between the two resonant circuits, causing there to be little to no output. <S> You will also run into the issues MadHatter suggested (the waveform will be discontinuous due to current being induced in the wrong portions of the coil). <S> Remove all but one layer of wire and just leave it as-is. <S> Then make sure that the secondary resonates at the same frequency as the primary. <S> You can use the following formula to calculate resonant frequency: where 'f' is the resonant frequency, ' <S> L' is the inductance of the coil, and 'C' is the capacitance of the system (the tank capacitor in the primary or the topload on the secondary plus the coil's self-capacitance). <S> Do the calculation for both the primary LC circuit and then the secondary LC circuit and make sure they match. <S> Otherwise your Tesla coil won't work at all. <S> If they don't match, you can "tune" the Tesla coil using different methods: If the primary resonant frequency is too low, do one or both of the following: <S> "Tap" the primary coil at different points to decrease <S> inductance of the primary coil (shorten the primary) <S> Decrease the capacitance of the primary tank capacitor <S> If the primary resonant frequency is too high, do the opposite. <S> If the secondary resonant frequency is too low, do one or both of the following: <S> Decrease the length of the secondary coil to reduce the inductance of the secondary coil <S> Reduce the size of the topload to decrease its capacitance If the secondary resonant frequency is too high, do the opposite. <S> You have to use math to determine which of the above to use, and how much to adjust each one. <A> Totally not, If the output voltage is like 100.000V <S> and you have 1000 turns that would mean there is a voltage of 100V/turn. <S> and when the turn are very close together it may happen that there will be an arc from one turn to another <S> and it will damage your tesla coil's secondary. <A> It all depends on the wire that you have removed from the transformer. <S> If it has any nicks or holes in the enamel insulation, it may short out from one layer to another. <S> Depending on the rating of the wire, it may short out even if there is not a defect in it. <S> You can use insulating tape between the layers, but given the voltages of the Tesla coil, the best approach is to use a single layer for the insulation. <A> Yes, one common technique for lowering the Tesla secondary's wire resistance and raising the Q-factor is to wind several parallel wires on the same coilform. <S> Solder your three windings together at each end <S> , so they behave like "Litz" wire. <S> The famous "Edmund Scientific" Tesla coil, the conical one, uses this technique. <S> I don't recall if it has three or four parallel conductors. <S> On the other hand, if the several conductors are wound atop each other, rather than lying in parallel against the coilform, there may be a circulating current between them (since their lengths differ.) <S> This would appear as a Q-factor lower than expected.
| If you have over lapping wires, as the energy wave travels through the wire some of it is lost via capacitive coupling to other winding layers, and those packets of lost energy continue on, but now out of phase, their reflection will not line up with the rest of the main energy packet.
|
auto-recalibration of a power supply through something stable over time? (thought experiment) Highly accurate instruments with a calibration are quite expensive, and anything electronic will tend to change over time. The effects of heat and oxidation, and perhaps changes in the material due to current directly. Perhaps there are other reasons circuits change over time. So I was wondering about some kind of reference that could be built into equipment so it could recalibrate itself. Suppose you wanted to have a fixed power supply that would stay stable at a precise voltage or at least be capable of re-calibration without sending it for expensive recalibration at the factory. Thinking about what kind of components could be stable, I think a wire-wound resistor with thick connectors and soldered connections. Zener diodes would seem less stable. So can anyone suggest a way to build a circuit that would provide an extremely stable voltage reference? Given such a voltage reference, I think it would be possible to build an extremely stable power supply. This question is mostly theoretical out of interest. I'm not asking for a practical design at this point though I wouldn't say no! <Q> How far you have to go <S> depends on how accurate you need. <S> For many applications, a zener diode is sufficient. <S> The next level might be a zener at a specific temperature, held there by a heater. <S> There are also "reference" cells, which are really batteries that as little as possible current is drawn from. <S> These need to be at the right temperature too for the best accuracy. <S> If you keep going, you have to worry about the temperature sensor drift used to keep whatever you are using as a reference at the right temperature. <S> Then you have to worry about how exactly you are comparing the reference voltage to something else. <S> Every means of comparison will have some error, which will drift over time, be dependent on temperature, etc. <S> Even more extreme, you have to think about thermocouple offsets due to different parts of the circuit being of different temperatures and made of different materials. <S> Without a number of how accurate "good enough" is, it's impossible to say how far you have to go <S> , how much you have to spend to get there, and whether it's even possible with current technology. <S> Even if you have a perfect power supply, what about the voltage drop in the wires between it and the load? <S> You have to look at the whole picture and think carefully about what problem you are really trying to solve. <A> It requires a lot of voltage (more than 7V) and significant power (especially if ovenized). <S> There are more precise references such as the Josephson junction standard , but that's out of the budget of even most calibration labs, as well as being somewhat inconvenient, what with the cryostat and liquid helium. <S> There are other technologies with various advantages and disadvantages- <S> you can read the datasheets- <S> if it's a good reference they will give drift per 1000 hours or similar, typically the drift rate decreases over time as in the classic 1/f noise spectrum. <S> Also the stability over time will be at a certain temperature and operating conditions. <S> If you ovenize the reference, for example, if you allow the oven to go out of control there will be more drift in the reference, and there will typically be small changes from thermal hysteresis that are not time related. <S> Stable resistors are all well and good (as well as necessary) but they are required in addition to the voltage reference, not instead of, if it's a power supply. <S> A stable reference resistor might be useful for an ohmmeter. <S> I question your premise- if you have an extremely stable reference built-in then why would you re-calibrate using the reference? <S> Would it not be better use the reference directly to determine the output voltage rather than having a worse reference that you would calibrate to the good one... unless there are power constraints etc. <S> Consider the auto-zero dual slope ADC- it removes errors due to long term drift in the integrator capacitor, integrator resistor, amplifier offset voltage and clock frequency- <S> each of those parameters only needs to be stable enough not to change much during each measurement cycle (milliseconds). <S> As a huge advantage, this also removes errors due to initial tolerances and due to temperature change (provided the temperature does not change much during the measurement cycle). <A> There are stable references such as MAX6194AESA+ that drift by as little as 50ppm per 1000 hours of operation. <S> But no matter how stable your reference, you can always beat that accuracy by putting in an adjustment to periodically calibrate the supply from some external NIST traceable reference.
| Probably the most stable voltage reference commonly available is a buried zener diode. There are many circuits that continuously 'recalibrate' such as ratiometric measurements to a reference resistor or voltage, or chopper amplifiers.
|
Hot and neutral have similar voltage to ground - finding the problem I have an APC battery backup and surge protector for my computer. It has a light, labeled "Building wiring fault." This light is on, and my computer will not start. I am unemployed and my landlord is useless, so I bought a multi-meter (Etekcity MSR-C600). Checking voltages at the outlet, I found the following: Hot to neutral: 117.5 Hot to ground: 60.8 Neutral to ground: 55.9 I then checked the voltage several other outlets. All of them have the same hot to neutral voltage, but the hot to ground and neutral to ground vary considerably. For example, another socket had: Hot to ground: 77.2 Neutral to ground: 36.5 Is this sufficient to tell me the most likely cause of the problem? I'm guessing that hot and neutral are cross-wired or shorted in some outlet. Is that probably correct? This circuit is rather overloaded, but it does not seem to me like that should cause the low hot to ground voltage and the high neutral to ground voltage that I observe, unless there is a partial short in some appliance. All the outlets in my apartment are on a single circuit. How do I go about locating and identifying the actual wiring problem? <Q> Yes, you have a problem. <A> The normal (if a bit optimistic) values you'd be looking for are 120V Hot->Neutral, 120V Hot->GND, and 0V Neutral->GND. <S> It sounds like you pretty much get that already. <S> I'd be strongly inclined to believe the fault is not in the socket itself, but rather either in a connected appliance or in the wiring itself. <S> The wiring TO the socket could be at fault, especially in a socket that had been habitually overloaded (and heated up as a result). <S> You might start by unplugging everything in your apartment and testing again. <S> If that clears the fault, plug half of them back in, see if it reoccurs, then step-wise in that binary fashion narrow it down to a single appliance. <S> If that doesn't clear the fault, I would head to the breaker box and take measurements with your circuit's breaker in the OFF position. <S> But this is probably where you should start looking for someone like an electrician, because this gets you into the (more) dangerous zone, likely having to remove the cover plate from the breaker box. <S> But a measurement here would show whether the fault is inside or outside your apartment. <A> You have a SERIOUS PROBLEM. <S> Potentially life-threatening and structure-endangering. <S> As @scanny suggested unplug EVERYTHING in your apartment to make the differential diagnosis whether it is the building wiring or some appliance of yours that is causing the problem. <S> Else their property is at risk of fire. <S> If the landlord does not respond appropriately, report the problem to local authorities as this is a violation of safety codes and laws. <S> Your backup gadget has possibly saved your life and property.
| If the problem still exists with everything unplugged, then the building owner must be notified ASAP. This suggests a failed connection, and this is potentially life-threatening. Basically, your "ground" is floating with respect to neutral.
|
Is dual power supply's ground coupled? I am using a dual power supply model CPX200 from TTI in my school lab, and I am trying use it to supply 1.8V and 3.3V to a PCB, so I figured it was a good idea to check whether the ground of the dual port is coupled. However, when I use multimeter DC voltage mode to measure the voltage difference between two ground terminal, it shows that the voltage difference is 0.3V and it starts decreasing slowly. I am having difficulties understanding this phenomenon. So is the ground of dual power supply coupled? Why would my measurement decrease slowly? Thanks a lot. <Q> The spec sheet for that supply says it has two isolated outputs. <S> Since the two supplies are isolated, they could also be used to provide positive and negative supplies (as often needed for audio amplifiers and other analog circuits) by connecting the negative terminal of one supply to the positive terminal of the other, and calling that connection "ground". <A> That sounds like they are not connected, and what you are reading is slowly varying leakage into their capacitance. <S> You must connect the grounds together if you want to be sure they are both system ground. <S> I would fully expect that the two supplies are independent, for best flexibility. <S> But sometimes, grounds are connected with diodes so that they can float a little to cope with small differences, but are still safe and able to blow fuses if live shorts to ground. <S> As an experiement, you could connect the ground of one to the output of the other. <S> With a current limited bench power supply like that, the worst that could happen if they are connected with diodes internally is that the current limit comes on. <S> Then measure the voltages at all terminals. <A> The CPX200 has two completely independent and isolated outputs. <S> This means that the negative terminal of the output is not connected to ground. <S> If you want it grounded, you must provide ground connection externally. <S> Anyways it is very probable that there is a capacitor between the negative output and the ground, and this capacitor can be charged for different reasons, so when you measure the voltage between the terminals you are discharging this capacitor via the input resistance of your voltmeter. <S> You can verify that the output are isolated with an ohmeter.
| If you want to use them as two positive supplies, with the same ground, you MUST connect the two negative terminals together, and to the ground of your circuit. The fact that you measured any significant voltage difference says that they are not hard coupled together.
|
Protection of BeagleBone Black from eMMC corruption on power loss I'm considering an application where a BeagleBone Black, preferably running one of the recommended Debian images loaded in eMMC, is powered by a 5V source that can be removed at any moment. I want to prevent catastrophic data loss in this event, including if a write to file is underway. At least, I want the BBB to still be able to reboot from eMMC on return of power, repair the (probably ext4) file system, and run my application. If I can have the insurance that any file that was properly flushed is consistent, that's ideal. The problem I fear is that, when power is lost while a write in underway in the eMMC, there could be data loss in an unpredictable sector, because a recent sector write handled by the eMMC triggered a Flash page erase, and the other sectors in that page are lost, and (because of wear leveling and sector reallocation internal to the eMMC) they belong to files or volumes that have not been written recently, including perhaps in read-only file systems or files. That problem is mentioned here : a power cycle at the wrong time can destroy data anywhere on the (SD or eMMC) card, no matter where you THINK you're writing and that could justify the warning in the BBB manual and here that [powering down using the power button] will also help prevent contamination of the SD card or the eMMC. as well at the terse warning made by what seems to be the primary eMMC source for the BBB Avoid power-down during WRITE and ERASE operations. To what degree can such data loss occur? What is the duration of the vulnerability window, if any? Are there standard solutions? Recommendations? Is there any built-in software mechanism against this, and how does it work? In particular, is the TPS65217C PMIC programmed to generate an NMI on loss of VDD_5V, aka AC, as the hardware allows, and is it handled in a way that stops ongoing eMMC write activity? If not, how can I at least add that? Is stopping any eMMC-related activity good enough for the eMMC to enter a safe state by itself after some delay? Which delay? What about a capacitor or Supercap (like Murata DMT3N4R2U224M3DTA0: 0.22F, 4.2V, 0.3Ω) on the battery connector to provide some power reserve? Are there modes/settings in the eMMC to configure a write performance/safety compromise? What are the defaults, are they adjustable? Important update: I have located authoritative information in the March 2010 eMMC specification about reliability when power is lost during write: In general, an interruption to a write process should not cause corruption in existing data at any other address. However, the risk of power being removed during a write operation is different in different applications. Also, for some technologies used to implement eMMC, there is a tradeoff between protecting existing data (e.g., data written by the previous completed write operations), during a power failure, and write performance. Lots of valuable information follows in that source, including description of how the eMMC declares that it promises (or not) to protect previously written data from haphazard erasure in case of power loss, and what seems to be truly write-protected areas. Addition: a comment mentions that mobile phones use read-only file systems in eMMC, and are not known to experience corruption of that. However, mobile phones have a large battery, and that practically ensures they do not experience abrupt power loss in normal operation, thus are not concerned by the eMMC corruption mechanism that I fear. <Q> All layers of the stack need to cooperate here. <S> The eMMC storage device has an integrated load balancer that needs to handle power loss situations gracefully. <S> I'd expect it to do that, by pretending that incompletely written blocks were not written at all. <S> The file system needs to be able to handle power loss between block writes. <S> The ext3 and ext4 file systems fulfill this, as they are journaled and enforce the ordering of writes so the journal accesses are kept separate from actual file system modifications. <S> The weak point in your setup is the package manager, which doesn't offer a good solution if the power fails during a package upgrade. <S> Usually you should still be able to boot unless the error happened precisely during renaming files after an installation, and the files being renamed were important system files. <S> For almost all purposes, that window is sufficiently small that you can ignore the problem. <S> You can shrink it more by running dpkg --configure <S> -a and dpkg -iGROEB /var <S> /cache/apt/archives during each boot -- these should get you back into a sane state from an interrupted update, at the expense of a bit of time, and they cannot handle cases where dpkg itself or the init system were damaged as a result. <S> Last but not least, you can set up a rescue system in an initrd that will check and try to repair and/or reinstall the system if needed, and anchor this in the bootloader. <A> simulate this circuit – Schematic created using CircuitLab <S> To compare intupt voltage and to do some action until cap has still charge <A> When power fails, you have sufficient time to call an fsync or perhaps even a kernel halt, or if nothing else works, a kernel panic. <S> Anything to prevent cutting power during eMMC write/erase cycles; any pure software interruption is dealt with by the file system journal. <S> I'm looking into this method for a project I'm working on.
| If you are powering your Beagle Bone from an AC supply you can control, you can connect the non-smoothed, rectified AC with a small timer cap to one of the GPIOs.
|
Torque in relationship to Voltage Regarding a DC motor designed to operate at a voltage up to 22 volts: If this motor is powered by a 18.5 volt lithium battery, will it not produce more torque than if the same motor is powered by a 14.8 volt lithium battery? I know more volts create more speed, but the increased current available in the 18.5 volt pack will provide the additional torque, right? If this is true, will the same motor stall at different amperage depending on how much voltage is being supplied? Thank you for your replies. <Q> More voltage means more current which means more torque output from the motor. <S> Different amperage implies you have a current limiter, and can adjust (limit) <S> the drive current separate from the drive voltage. <S> If that is the case the motor speed will reduce under the same load but with less current. <S> Drop the current low enough and the motor will stall for a given load. <S> Do not exceed the voltage or amp rating for the motor and it should be ok. <S> If it stalls under full current (the maximum allowed for the motor) <S> then the load is too much or the motor is too small. <S> Make sure the load is well lubricated and free of rust. <A> When you increase the voltage, the stall torque is higher due to higher current. <S> However the rated torque(current) remains equal and is related to motor construction. <S> In the picture you can see what happens if the voltage is reduced, it can be also viewed opposite what should happen if the voltage is increased. <S> The max speed will be increased, also the torque characteristics is offset, meaning that the same load will find the ballance at higher speed, higher torque -> higher torque means higher current. <S> Therfore your motor might became overloaded, this is not true if measure the current and you determine that motor is running with <= rated current <A> A useful first-order model of a motor is a resistor in series with a voltage source. <S> That voltage source is a function of the motor speed, and apposes the applied voltage. <S> Torque is proportional to the current. <S> This simple model tells you a lot. <S> When the speed is 0, the current is just the applied voltage divided by the resistance. <S> That's the stall current, and is when you get maximum torque. <S> If you let the motor go (no load on it), then it will accelerate due to the torque. <S> That makes it go faster <S> , so now the internal voltage source starts offsetting some of the applied voltage. <S> Eventually the motor will get to the steady state speed for that voltage. <S> That is also called the unloaded speed of the motor at that applied voltage. <S> This is where the internal voltage source has offset most of the applied voltage. <S> The little that's left across the resistance allows just enough current to flow to produce just enough torque to overcome the friction and other losses to keep the motor spinning. <S> Now you should hopefully be able to understand why the first sentence is true. <S> One gotcha of this is higher power dissipation. <S> The motor's dissipation comes from the resistance, and is therefore proportional to the square of the current.
| More voltage means more torque at the same speed .
|
Is it possible to de-cap a ceramic capacitor Would it be possible to remove the case from a ceramic capacitor ? Like the ones surrounding the IC in this picture? <Q> They consist of fine conducting membranes between layers of ceramic; the same ceramic body also forms the outer shell. <S> The electrodes at the ends connect the plates to each other and the outside world. <S> If you want to see how they work, I'd suggest splitting them in half along the long axis and using a microscope. <A> What Simon said . <S> I would only like to add pictures. <S> ( source ) <S> ( source ) <A> No. <S> There is no "cap" in a monolithic surface-mount ceramic capacitor like those. <S> That is what "monolithic" means: "one-piece". <S> If you had mentioned WHY you think you want to do this, more helpful responses may be possible.
| These capacitors do not have an outer case, so there is nothing to remove.
|
Is it possible to induce RF Emissions from a RF enabled Device Would it be possible to take a device anything such as a wireless router or a wireless video and induce unintentional RF emissions while no power is supplied to the device ? Can you send it some range of RF waves and attempt to measure the RF emissions due to incidental waves (re-radiation) ? <Q> A significant part of emissions problems relates to non-linearity (which will depend on the bias point of the junctions). <S> You will also not manage to induce the same level of RF signal. <S> This can tell you about where in a circuit you need to focus your efforts to resolve problems, maybe if you can't determine which direction an interferer is propagating. <S> Assuming you have a CPU and an RF section in your design, there are probably three different energised states, CPU only , CPU plus receive , CPU plus Tx . <S> The worst state from an emission point of view is when the transmit circuit is active, and a CPU circuit is physically close. <S> From a testing point of view, test with the CPU running and not powered (to find the worst case). <S> If you have a radio which meets emissions on its own, but it fails with harmonics once you connect to the CPU, no amount of work on the radio will fix the problem - you need to decouple the CPU input pins. <A> Would it be possible to take a device anything such as a wireless router or a wireless video and induce unintentional RF emissions while no power is supplied to the device ? <S> No, if it has no power then it can't emit. <S> Can you send it some range of RF waves and attempt to measure the RF emissions due to incidental waves (re-radiation) ? <S> Emissions come from the originating source (what you "send it") and what you appear to be describing are reflections and yes, you can attempt (and actually) measure these reflections. <A> Mostly any RF device will emit RF signals if it is exposed to an suitable EM field. <S> Nearly every RF circuit contains a resonator or at least somehow resonant parts. <S> If you expose this circuit and particularly its antenna to a field that can excite the resonant parts then these parts will store energy harvested from the EM field. <S> If the EM field vanishes, the resonant parts will continue their oscillation while radiating the stored energy. <S> In most cases this won't take long as a long post-pulse oscillation is not a desired behaviour <S> but it will happen and with considerable delicate equipment it can be measured. <S> If the resonator of a RF circuit contains nonlinear parts (e.g. Diodes and Transistors) the circuit can emit frequencies different from the ones contained in the incident EM field.
| What you can do is (assuming you have a suitable RF test and development licence, screened room, etc) connect a signal generator to part of a circuit and analyse the harmonic and radiation characteristics of that part. No, you can't perform any useful RF emissions analysis of an active device when it is not powered.
|
Mosfet motor problems with bluetooth module I'm playing around with a bluetooth module and want to control a motor using the module and a phone. I manage to control a small motor (±30 mA), but as soon as I'm connecting a bigger motor I'm getting some inconsistent problems. I'm powering the whole circuit using a 3v/ 1.5A voltage regulator (which is powered by a benchtop power supply). The motor is normally drawing about 700mA if connected directly to a 3v source (benchtop power supply) I'm using a N-Channel Mosfet to control the motor ( mosfet datasheet ). I've hooked it up according to the following schematic The problems I'm experiencing are: When I set IO5 HIGH the motor starts spinning, but slower and only draws around 250mA (which slowly increases over time to ±270mA). It's not letting the motor draw all the current. (I've tried different Mosfets, all with a Vgs < 3v) Sometimes after switching it ON, or when the motor is ON for a longer time the module resets (restarts) itself. Seems like something is happening the module doesn't like (power peak or something). It definitely should not do this and is clearly connected to the motor actions. I've been putting in some extra decoupling capacitors, but this doesn't solve anything. I put a resistor between IO5 and the Mosfet gate, but also didn't solve anything. Am I missing something fundamentally, or doing something else wrong? Any help is appreciated. Cheers Ruben <Q> The regulator has an internal current limiter which is close to your average motor current. <S> Look at the motor voltage with a scope and see if it is being pulled low periodically with the brushes. <S> The occasional shutdown might occur either from the regulator dropping too low and pulling down your controller or possibly the thermal limit being reached. <S> Something to try might be to run your controller from the regulator but run the motor/FET circuit directly from your benchtop supply. <S> This might be better practice so that motor back EMF will not create high voltage on your controller. <A> ALL inputs should be connected to ground or Vcc or erratic behavior can occur. <S> If the pin can also be an output it is best to connect it to ground or Vcc through a 1K ohm 1/8th watt resistor. <A> I managed to find the causes of the problems: <S> First of all the motor caused too much noise on the power line. <S> This made the voltage drop below 2.8v at some moments, switching off/ resetting the module. <S> Indeed ,like @Bruce Abbott suggested, the gate voltage on the Mosfet was not high enough. <S> I tested with a higher voltage and is was passing more current. <S> I did the following (see schematic below): <S> I used a 3.3v voltage regulator instead of 3.0v. <S> This gives it a bit more freedom to fluctuate without causing too much problems. <S> This also increased the logic voltage tot 3.3v which was high enough for the Mosfet. <S> (I want to power eventually with a Lipo <S> so will wonder what happens as soon as that voltage will drop below 3.3v, but that's a worry for later) <S> I put a resistor (R13) in series with the Motor, limiting the current, but also the noise on the powerline. <S> It rotates a bit less strong, but still enough. <S> As @Sparky256 suggested I pulled the Reset and IO0 pins up. <S> Don't think it made a difference in this case, but still good to do. <S> I will use IO6,PWM 1 and PWM3 for other purposes later so no need there. <S> And also used a 10 ohm resistor (R12) in line with IO5.
| A motor does not draw a constant current but has a "current ripple", so it is possible that the current ripple is hitting its peak and dropping the output voltage accordingly. I noticed you have a few input pins 'floating', including your 'reset' pin.
|
Powering a microcontroller inductively from mains wire I am in the process of making a mains power monitor using a SCT-013-000 CT clamp and an arduino mini, running on batteries. Even though I should be able to get many months of operation from two small AA batteries, would it be possible to use the clamp to harvest power to run the unit forever? If I'm doing the maths correctly, two 10F 2.7v capacitors in series would keep a unit drawing an average of ~0.1mA, 5.4v->2.7v running for a couple of days even if there is no power usage on the mains wire? <Q> If you want to monitor mains power precisely, you have to measure not only mains current but also mains voltage. <S> When you measure mains voltage you can use a small transformer to power the microcontroller inductively because you already have a connection to both wires of mains. <S> No energy harvesting necessary. <A> Uwe already pointed out the obvious, I am posting this answer for something else. <S> Please do some research before jumping on in this project and later on end up scratching your head why this power meter doesn't work. <S> I have implemented an energy meter using BL0921 IC. <S> There are various metering ICs available in the market and there is a good reason you should prefer those ICs over your arduino. <S> First, you are going for ac power measurement which is not as straight forward as P = <S> V <S> x <S> I. So don't assume you can do analog reads every few milliseconds and multiply it to get result. <S> You will have to take those measurements at a very high frequency and keep on integrating over time. <S> After that power factor will come in to ruin your day. <S> Those metering ICs take care of all that stuff for you and give you a nice output. <S> They also give you simple methods for calibration of your energy meter. <S> You can use arduino to read the output and use it the way you like. <S> You can use an ESP instead of arduino to directly send it to a server over wifi. <S> Possibilities are endless. <A> Using your current clamp for both power and signal would result in a significant degradation in the accuracy of your measurements. <S> The high turn-count of the secondary on the CT requires thin wire, producing a relatively high output resistance. <S> Applying the dynamic load of the Arduino would change the load on the CT output, causing inaccuracy. <S> While inductive energy harvesting is definitely possible, you wouldn't want it to be from your primary sensing device. <S> As for using electrolytic supercapacitors in your application, be aware that safely charging/discharging them is best done with a dedicated IC for low-power backup applications. <S> Such controllers include the LTC3225 , the LTC4425 , and <S> the LTC3226 . <S> However, these controllers would require having a low-voltage supply available normally such as a 5V USB power brick/"wall wart". <S> For higher voltage inputs, you could use a Li-Ion battery and the LTC4079 . <S> You will need to ensure that the CT output voltage will be appropriate to drive the energy harvester at both low and high measured currents. <S> I am uncertain how the combined energy harvesting and supercapacitor implementations would interact and they might not be easily combined.
| Implementing inductive energy harvesting could be done with the LTC3588-1 and a second CT.
|
Using 220v transformer on 110v input I've been looking for 115v primary 9v secondary 3.2va power rating PCB transformer and couldn't find any to replace the one that was fried. All I find is dual one, instead for example 4 pins it has 8 pins, and 220v once. Could I use 220v primary 9v seconday 3.2va even the input is 110v?It's for coffee machine Saeco. Can I use the dual output and just use one pin? Here is a picture of a transformer I need (size and shape) Thank you <Q> A 220:9 transformer with 110 input will only be capable of about 1/2 <S> the 220V rated VA. <S> The 220V, 8-pin transformer needs to have the two primaries connected in parallel to get the full VA rating. <S> Using just 2-pins will give 1/2 VA. <A> However, since you did not actually identify it, we can't confirm whether is is really suitable. <S> There are many such transformers with "split windings". <S> So, for example, you can connect the two primary windings in parallel for 110V or in series for 220V. <S> And in a similar manner the two secondary windings can be used either in parallel (for half voltage, double current) or in series (for double voltage, half current), etc. <S> However, it sounds like your replacement transformer won't be a PHYSICAL replacement because it doesn't have the same "pinout" or even the same number of pins. <S> And of course the exact size and spacing of the pins is also likely to be a mis-match. <S> In that case, you could hot-glue the transformer upside-down and run wires from the pins down to the board pads, etc. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Double-primary, double-secondary wiring options. <S> The twin 110 V primary type transformer can be configured as shown in Figure 1 (a) and (b) to work on either 220 V or 110 V wiring. <S> Just ensure that the transformer VA rating matches or exceeds that which is replacing - and given that the original burnt out "exceeds" would be better. <S> You may have to stand it off the board and connect with flying leads.
| It is quite possible that you have identified a suitable FUNCTIONAL replacement transformer.
|
DC motor causing noise on the power line I've been building a circuit where I'm controlling a motor using a bluetooth module + phone. I've been having some problems, which mostly are fixed (see topic: previous topic ). The main problem I'm facing at the moment is that the DC motor is causing quite some noise on the power line (even if I power the motor directly from V+). I've managed to reduce this by limiting the motor current (using resistor R13) and put in some capacitors + a diode (D3) over the motor. However there is still siginificant noise. Therefore I would like to know if there is a way to reduce the noise on the power line even more beside having these capacitors and the diode(D3), when the motor switched on (or have additional capacitors)? And what capacitors are best in reducing noise? See schematic below: <Q> Solder a 100nf capacitor as close to the motor brushes as you can, and if the motor has a metal case, also 100nf from each brush to the case. <S> Similar to what is shown at the bottom of this page on the Pi2Go build instructions. <S> Keep the capacitor leads as short as possible, even a large SMT ceramic capacitor would be a good option. <S> This is assuming the noise is RF, interfering with the bluetooth. <S> If its interference with an MCU, you could increase the capacitor value maybe 10x. <A> You could put a pie filter in front of the motor. <S> This could reduce your overall load spikes on the system. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> I've been testing with bigger capacitors, and a 1000uF (C5) solved all the issues. <S> Unfortunately I won't have enough space in my casing for such a big capacitor. <S> But even a 100uF made it a lot better. <S> I will see what I can squeeze in. @ <S> Sean Houlihane, I'm not able to solder directly to the motor or very close to the brushes. <S> The motor is placed in a very tight space in the casing and a couple cm away from the print. <S> Luckily replacing C5 did the trick well enough. <S> @JWL, unfortunately the pie filter won't be an option, since the motor is also powered from the 3.3v <S> so they share the same power source. <S> I don't want to power it directly from the LIPO battery since the current the motor draws at 4.2 volt is too high. <S> At 3.3 volt the current stay nice and even. <A> Unfortunately the motor is alternatively pulling energy from the power line and dumping it back in. <S> Then put an electrolytic across the processor with a resistor in series to give you some isolation. <S> But most importantly, make sure the battery or power source is connected to the motor terminal and MOSFET source such that the current from the power source going to the motor does not pass through or under the processor circuit. <S> Use wide traces when you can, but it is most important to keep the motor current separate from the low power stuff, including processor power and signals.
| I would might try the following approach: put a low ESR cap (ceramic is best) as big as you can across the motor to knock out the high frequency.
|
Simple soft start for high power mains device I have seen, but cannot find again, a really simple implementation of a mains soft start circuit.The thing I'm thinking of involves a power resister inline of the mains. When the device is started, current is restricted until a relay is energised, which shorts out the resistor allowing full power The thing is, I can't remember (and so can't decide on) a good way to energise the relay after a second or so. Does anyone know of a link to a reference implementation for such a circuit? Cheers <Q> [ This diagram is close but has too many parts: K1 and K2 are terminal block - or large wire nuts. <S> K1 goes to what should be a 20 to 30 amp 250 volt slow-blow fuse. <S> Assuming Re1 has a 120VDC coil, <S> R1=R2=R3 = <S> 2.2K ohm 3 watts each. <S> C1 can be omitted as can C2. <S> B1 can be a bridge rectifier with a 1 amp 600 volt rating, or make one from 4 1N4007 diodes. <S> C3 can be 47uF 150VDC, enough so Re1 does not chatter. <S> Increase C3 value if longer delay time is needed. <S> Re1 should be a relay with a 120VDC coil (common) and contacts rated for 20 to 30 amps 250VAC. <S> The high current rating will ensure a long life. <S> R4, R5, R6, R7 should be 2 ohm 300 watt <S> ceramic resistors mounted to a panel or on a fiberglass threaded rod. <S> They take some of the punishment if the motor stalls, giving time for the fuse to blow. <S> I would suggest 600 watt resistors but they get expensive (in case the motor is jammed and cannot rotate). <S> These are all bulky brute-force parts designed to last even if the motor stalls. <S> I suggest adding a thermal cut-off next to R4-R7 unless it is built into the motor. ] <A> I have seen, but cannot find again, a really simple implementation of a mains soft start circuit. <S> The thing I'm thinking of involves a power resister inline of the mains. <S> When the device is started, current is restricted until a relay is energised, which shorts out the resistor allowing full power <S> Something like this: - Or <S> maybe like this: - <S> Or just use one of these: <S> - It's a thermistor that when cold has a resistance of tens of ohms and rapidly warms down to a low on resistance <S> but it's slow enough to take the edge of the inrush current. <A> Back when high-powered tube-based amplifiers were commonly used by amateur radio operators, it was common to control the main contactor (the one that bypasses the current-limiting resistor) using a thermal time-delay control relay. <A> There are several ways to do the delayed energize of the resistor bypass relay coil. <S> One method is that there is a separate isolated power supply off the mains that is designed with its PWR_GOOD indicator status to come valid at the end of the 1 to 2 second delay period. <S> This indicator signal can turn on a transistor that energizes the bypass relay coil using the voltage rail of this separate power supply to give power to the relay. <S> Some products that need a soft start run on a off mains switch mode power supply. <S> These products often have rectifier diodes right off the line and <S> then the resistor to the main SMPS switcher curcuit operates as a DC rail. <S> On these cases it is possible to use a FET to bypass the soft start resistor.
| Another method is to design a direct off mains power circuit that is not isolated but can provide power to a 1 second delay circuit and then drive the relay coil.
|
Please explain why this simple circuit is latching 'on', but only when a fan is attached My son and I made a circuit using an electronics kit. It's supposed to work so that when the IR receiver receives a signal from a remote control, it turns on the motor. It does that, but I noticed that if you hold the remote control button long enough for the motor to reach top speed, then the circuit latches and the motor continues to spin even after you release the remote's button. It only does this if the green fan is attached to the motor, otherwise the motor stops when you release the remote's button. My first thought was something environmental, like heat from the motor (though I doubt that), light from outside - but it happens even if the IR is covered. So is it something about the load on the motor affecting the transistor? I'm a novice at electronics. simulate this circuit – Schematic created using CircuitLab <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Your circuit. <S> A search for LFN shows up the PIC-1018SCL via this site . <S> It's an infra-red receiver with built-in logic to pull its output low when IR is received. <S> In normal operation it will draw current from the base of Q1 via R1 when an IR signal is received. <S> This will pass current to the motor and start it up. <S> The fact that it keeps going is a slight mystery as the LFN should be either fully off or on. <S> One possibility is that C1 is leaking a little current. <S> If so, this will turn on the Q1 a little but not enough to supply the stall current and start up the motor. <S> Try removing C1 and see if you get the same effect. <A> I think this circuit needs a pull-up resistor of about 2.2k 1/4 watt from the base to the emitter of Q1. <S> This is from the spec sheet for this part: Voh = Vcc-0.5 Volts. <S> Vol = 0.2 to 0.4 volts max <S> Even though the output is not OC type, <S> the LFN out rail is low enough to induce a trickle current in Q1, so it may NOT shut off even if LFN out = high. <S> A pull-up resistor is needed or the state of Q1 can be unpredictable. <A> The problem is probably because the battery voltage collapses when the motor is loaded. <S> That runs the IR receiver out of spec, and it pulls its output low regardless of what it is receiving. <S> That keeps the fan on, which keeps the power voltage low, etc. <S> This is just a guess on my part from looking at the schmatic and the symptoms, but it seems quite plausible. <S> There are two ways to verify this. <S> First, put a meter on the power voltage. <S> If I'm right, you will see significantly more sag when the motor is loaded (had the propeller on it) than when unloaded. <S> A scope would be better since maybe there are short low-going spikes. <S> One obvious problem with the circuit is a lack of capacitance across the supply. <S> Second, use a separate battery to power the IR receiver and the motor. <S> If the problem is as I described, then this should work even when the battery running the motor drops in voltage. <S> A datasheet for the IR receiver would be useful, of course.
| If all is well removing the IR source will result in the transistor switching off and the motor slowing to a stop. If, however, LFN pulls low, Q1 turns fully on and the motor spins up to speed, the current caused by leakage may be enough to keep the motor running.
|
Electronically, how to control this 5v arduino-style relay board from a PC serial port I have the following relay board (pictured below), and to activate any of the relays, you simply connect one of the IN1-IN8 pins to ground. Easy. The current going between gnd and INx is negligible, because of the transistors already on the board, so the current going through the serial port of the PC is not an issue. How can I electronically use the serial port to to activate a relay on this board? I'm not concerned about the software aspect of it. The standard serial port has DTR and RTS (pin 4 and 7) that can be set high or low through software. I believe that "high" on a serial port is -3 to -25 volts and "low" on a serial port is +3 to +25 volts. Is there some easy way to use that to set IN1 to GND? Look at the schematic below, and see if there is an easy way to accomplish this: By the way, VCC and JD-VCC are connected together with the jumper, I'm not doing high voltage or concerned about isolation. <Q> If by Electronically you mean by Software, yes you can Programmatically Toggle a Specific Serial Line. <S> In LabVIEW this would look as follows: Toggle a Specific Serial Line in LabVIEW <S> In Python you can use the pySerial Library, where the code would look like this: __ <S> init__(port=None, baudrate=9600, bytesize=EIGHTBITS, parity= <S> PARITY_NONE, stopbits=STOPBITS_ONE, timeout= <S> None, xonxoff=False, rtscts= <S> False, write_timeout= <S> None, dsrdtr= <S> False, inter_byte_timeout=None) , where <S> rtscts= <S> False sets RTS to Low and drsdtr= <S> False sets DTR to low. <S> And you are right <S> , the Serial Port Voltages can reach from -25 to +25 Volts, but in most cases it will be around -10 to 10 Volts. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Serial to relay-board interface. <S> This circuit should work. <S> When serial-out line goes positive Q1 is turned on via R1. <S> This will turn on the opto-LED on the relay board. <S> When serial-out line goes negative Q1 will be turned off and D1 will prevent the base of Q1 being pulled more than 0.7 V negative. <S> Q1 can be any small signal NPN transistor with a Vce greater than your relay board supply voltage subject to base current below. <S> +/-25 <S> V seems extreme for RS232-type serial port. <S> I always thought it was +/-12 <S> V. <S> Most devices use something like the MAX232 chips to double the 5 V supply to give 8 or 9 V out for reliability. <S> Working on +12V on Ser-Out you would get 12 mA into the base of Q1 which should be fine for something like the 2N2222. D1 will easily handle the reverse current. <S> If you really expect Ser-Out > <S> 12 V increase the value of R1 accordingly. <A> This is an 8 realy board. <S> A standard (9 pin) serial port doesn't have 8 outputs. <S> Are you going to use (at least) <S> 2 serial ports? <S> The quickest solution would be to put an Arduino (Uno or Nano) between your PC and the relay board. <S> Or, for the more adventurous: an FT245RL usb-to-parallel chip. <S> If you realy want to use serial ports: use the pins that are outputs. <S> My gut feeling is that they can drive your optocouplers directly (disconnect VCC-JDvcc, connect VCC to the serial ground). <S> If this doesn't activate the relays, use a transistor per input as 'transistor' suggests (how could he suggest anything else!), or an ULN2803 for 8 lines.
| Relay board GND needs to be connected to serial port GND - pin 5 on the 9-pin PC port or pin 7 on the 25-pin port.
|
How do you tell if a Li-ion battery has a protection circuit? I bought a li-ion battery off of ebay. The listing claims that the battery has a protection circuit, but a week later, the battery didn't cut out or anything. With a multimeter, it is reading 10V when the battery is listed at 10.8-12.6V. Does this mean the battery doesn't have protection or should the cutoff be lower? <Q> The battery is most likely a 3S Li-ion pack, i.e. 3 cells/packs in series. <S> Protection circuits for single cell Li-ion normally have overdischarge protection set somewhere in the range 2.5V-3.2V per cell, which translates to 7.5V-9.6V for a 3S pack. <S> So this is the range that you should test to ensure that the undervoltage protection correctly triggers. <S> The listed output voltage 10.8V-12.6V is probably meant to be a conservative estimate of the usable voltage range, since there will be little accessible capacity left below 10.8V <S> (except at very small currents). <S> Edit <S> To respond to a comment on the question, if you do a search on the model number "DC 12680 protection" you will find other listings (e.g. on Amazon) where it claims to include protection circuitry. <S> You will also find pages that unmask the typical overinflated capacity claims on these packs. <S> For example, here it is discharged at 1A down to 8.5V and tests at only 25Wh. <S> Notice that this is 3.7V <S> * 6.8Ah = 25Wh, not 11.1 <S> * 6.8Ah <S> = 75Wh. <S> The industry standard is to quote capacity at the cell nominal voltage 3.7V not the pack voltage (here 11.1V). <S> Many unscrupulous sellers exploit this ambiguity to mislead the buyer into thinking that the pack has much higher capacity, here 75Wh vs. 25Wh. <S> Caveat emptor. <A> Without actually looking in the pack to determine if there is protection circuitry, it would be very difficult to test for the circuitry. <S> The main reason being that in order to do a voltage or current test, you would need to test the extremes( Highest voltage, lowest voltage, max current draw, temp, etc.) <S> and if the battery is not protected you will most likely end up destroying the battery. <S> Ex: <S> You decide to test if it has under voltage protection, so you start to drain the battery and observe the voltage. <S> outcome a) <S> The battery has protective circuitry, so as the voltage reaches a low level, around 2.5V per cell(can vary quite a bit, determined by the manufacturer), the output shuts off and you suddenly read 0V. outcome b) <S> There is no protective circuitry, so the battery keeps draining and the voltage reaches a range where you can determine that there is no protective circuitry and the voltage is still high enough that the cells have not completely died, around 2V-2.5V, depends on chemistry. <S> outcome c) <S> There is no protective circuitry, and the voltage on the cells becomes low enough that they start plating lithium and become useless. <S> It then becomes a game of trying to achieve outcome b while avoiding outcome c. <S> If testing is the only option to determine if it has protective circuitry, I would recommend draining the battery to about 2.4V per cell or so, and if it gets that low with out shutting off <S> then there is no circuitry, at least none of any value. <S> You need to be very careful not to go much lower, because all it takes is one cell dropping below 2V to make the entire battery useless. <S> I would not recommend test over voltage or current, as these can easily lead to the battery "rapidly disassembling". <S> TLDR; It can be done by very carefully lowering the voltage, however it should only be done as a last resort as cell damage can easily occur if not careful. <A> re: "... <S> battery has a protection circuit... <S> "What <S> kind of protection? <S> Over charge protection (max current &/or above the cutoff V), over discharge protection (max current &/or below the cutoff V), max current protection or all three? <S> If you want to know for certain which type/s of protections your battery provides, your battery will need to be monitored (V & A) while charging & discharging it. <S> If the li-ion cells are 3.7 Vnom , then the max charge should be 4.2 vdc per cell an 12.6 vdc for 3 cells that are connected in series. <S> If the charging current cuts-off before the in circuit charging V exceeds 12.6V, then your battery has over charge protection.
| If the discharge current cuts-off at or before the battery reaches typical cut-off voltage levels (~3v per cell), then your battery has over discharge protection.
|
How to make multiple GND pours in Altium? I have a PCB design that requires two ground pours on the bottom layer, one is AGND and the other is DGND. Most of the board is AGND ('L' shape) and the remaining should be DGND. I have tried to pour polygon with AGND over the entire board, draw another polygon for cutting a square out of the AGND pour and tried drawing yet another polygon for DGND. I think this is a naive approach and it did not work. If anyone can enlighten me it would be super! <Q> Option 1: Just bite the bullet and make the AGND pour in an L shape. <S> Option 2: In the polygon manager, adjust the pour order so the DGND pour happens before the AGND pour. <S> I'm not clear what you mean about "draw another polygon for cutting a square out of the AGND". <S> If DGND is poured first, and your design rules are set correctly, Altium will respect the different nets clearance rule and produce a clearance between the DGND polygon and the AGND being poured around it. <S> However I actually prefer the first option because it is less likely to cause trouble if you edit the design later, shelve some of the pours, need to add a new polygon that overlaps the first two, etc. <A> Draw the AGND polygon pour over the entire board. <S> Draw the DGND polygon pour where you want it. <S> Set the DGND pour priority to something higher than the AGND pour ( this is somewhere in the pour settings <S> IIRC <S> *see below). <S> Altium should repour the AGND pour. <S> If something's not right, check your clearance settings. <S> Basically, a poly pour with a lower priority should "wrap" around a higher priority pour with clearance set by DRC settings. <S> *edit: <S> rats, I think I'm misremembering Altium with Kicad. <S> In Kicad, you can easily set pour priority, but it's not so easy in Altium. <A> You can also create the shape in a CAD program, which is a lot more flexible than trying to use Altium's primitive drawing tools, and then export the shape in DXF and then import it onto a mechanical layer and use "convert selected to polygon fill" to create an arbitrary polygon. <S> Unfortunately it cannot properly handle shapes with internal cutouts.
| However, you should be able to edit pour order in the Polygon Manager, which has a list of polygons and lets you adjust the order.
|
Converting a 120V universal motor to run on 240V Is it as simple as rewinding the field coils with twice the number of turns and wire half the cross-sectional area of the original?or would the armature coils also need to be rewound as well? The motor is rated 120V, 200W. I live in SE Asia, so wall voltages are 240V. I do have a voltage transformer, I just wanted to know what is involved in converting the motor to run on a different voltage. <Q> You do not tamper with the motor . <S> You buy a step-down transformer rated for the current the motor needs. <S> The transformer needs to handle the start-up current as well (it may be listed on the motor), so it maybe rated for twice the motor run current - or more. <S> They are plentiful and have many sources and should not cost much. <S> They are a quick and safe solution. <A> Re-winding the rotor and stator coils is not a reasonable option. <S> The step-down transformer (or auto-transformer) is the PROPER method. <S> Since it is a "universal" motor it runs on DC, so you could try simply rectifying your 220V mains using a half-wave rectifier. <S> But that would yield only half-power, and we don't know your application to know whether that is a practical solution or not. <A> You'd have to re-wind all the windings in the motor. <S> Usually this is not worth doing, especially on such a small motor, use a transformer or thyristor speed speed control.
| Get the voltage and current ratings of the motor, then just search the web for 'step-down' transformers that match your 230 to 240 vac line voltage on the primary side, and 120vac on the secondary side.
|
For 3.3V microprocessors, what's the best way to power the gate of a MOSFET? The new processors are all 3.3V and dropping. So while I am used to 5V arduinos being able to power the gate of a MOSFET, at least most of the way, this is not going to be true any more. And for an IRF630, someone pointed out that I really need to drive the gate to 10V to get the rated on-resistance. So what's the canonical way to do this? Do I have a 10V power supply, and drop the voltage for the processor, have a charge pump that generates the 10V from 3.3V? The current will be very small, because the gate has massive impedance. Finally, assuming the 10V power supply, what's a good way to switch that voltage to the gate? Do I have to use a small junction transistor because I don't have the voltage to switch a MOSFET? The junction transistor is shown in this solution: Multiplying the voltage of an output pin on an Arduino board I'm just asking if there is another way. <Q> Three options. <S> Typically known as a logic level mosfet. <S> Use a simple npn transistor as a switch to drive the mosfet at a higher voltage. <S> Logic would be inverted. <S> Use a dedicated mosfet driver IC. <A> Assuming you only have 3.3V available, that your MCU can supply at least 1mA thru the gpios, and that your application does not require more than 10A, I would use a N mosfet compatible with 3.3V. <S> For example the PMV16XNR has an On-Resistance of only 20 mOhm when powering the gate at only 3V, and can source a bit more than 6A. <S> There are many other compatible MOSFETS. <S> Just be careful to add a resistor from your gpio to the gate, so <S> the current spikes when switching the mosfet are not too large. <S> For the PMV16XNR I use 500 Ohm resistors before the gate so the spikes are 6mA and the maximum switching frequency is somewhere near 300 KHz. <S> If using this option remember also to put a large resistor from the gate to ground, so it is powered off if gpio is floating. <S> If you need a much bigger MOSFET, then a mosfet driver or a charge pump could be necessary. <S> If you tell us your application and more details, maybe we can help you better. <A> For P-channel FETs, another option is to use a combination N/P MOSFET device. <S> These have <S> both P and N FETs built into the same pacakge. <S> The P can be used as the high-side switch and the N can be driven at 3.3V to turn the P on/off. <S> For example, SI4559ADY-T1-E3 . <A> I've been using the NPN MOSFET AO3400 with a Rds(at Vgs = 2.5V) <S> < 52mΩ with good results. <S> They can be bought cheaply <S> (100 pieces for less than 3 dollars) online from aliexpress or ebay. <S> There is also the PNP MOSFET AO3401 with Rds(at Vgs <S> =-2.5V) < 85mΩ for high side switching. <S> The AO3401 is also useful for reverse polarity protection.
| Use a mosfet with a VGS compatible with 3.3V.
|
Tap into 4 wire RTD Is there any way to tap into an existing 4 wire RTD without changing the resistance, and therefore changing the temp that the existing device will read? Here's the scenario. The device is a medical refrigerator. It doesn't have an API I can use to get the temperature it's reading from a built in 4 wire RTD. I'd like to tap into the RTD with something like an Arduino to be able to read the temperature that way, but it seems like doing that will alter the resistance which will cause the temps to be off. Is there something I'm missing, or is this just not going to work? <Q> You could tap into the sense lines using a decent instrumentation amplifier . <S> The ground would have to be common between the two circuits so that the RTD stays within the common-mode range of the instrumentation amplifier. <S> You would be at the mercy of the energizing current supplied by the measuring instrument. <S> If that current is constant, you'll get a reading proportional to resistance. <S> If it's variable to linearize the RTD you'll get a reading more-or-less linear with temperature (with offset). <S> In particular, if the sensor you're trying to tap into is used with a controller, you should be careful not to cause any unforeseen issues with safety. <S> Eg. <S> INA826 : <S> Typically Pt100 RTDs have >200uA of energization current <S> so the 35nA bias current will have little effect on the reading. <A> It may be possible to tap into the sense circuit, but this is clearly much more invasive than your question is asking for. <S> Potentially you can tap off add inputs somewhere, with a low enough load to not affect the original circuit. <A> It is unlikely that tapping into the existing RTD is going to allow retaining the existing device temperature accuracy. <S> There are several possible reasons for this: <S> The existing sensor circuit may have a particular input impedance that acts in parallel with the RTD resistance. <S> When you tap in you potentially add an additional parallel impedance that changes the net sensor impedance. <S> The existing sensor may very well have a linearization table built into its controller that converts the RTD reading to actual temperature reading. <S> Simple tapping into the RTD bypasses this on the way to your parallel readout device. <S> A tapped in connection could upset the bridge or compensation circuitry. <S> I think you should really think about adding your own additional temperature probe device inside the unit and interface that to your logging device. <S> You can get RTDs, PtResistance probes or even thermocouples. <A> Something like this should work. <S> The ground is isolated from RTD ground.
| The RTD device may be connected into some type of bridge circuit or temperature compensation circuit in the existing controller.
|
How to remove noise from power adapter I am making an audio amplifier using an LM2879 . I am using the schematic provided in the data sheet. For a starting point, this design works pretty well, although the base cutoff point could be a bit lower. When I power this circuit on, I am greeted by quiet, high pitched static regardless of the volume of my device. When I play music, the quality of the music is good. I believe the static is coming from my power adapter because someone else in the reviews for this product said it was electrically noisy. Is there a way to filter out this noise at all? If you have any other ideas of where the static is coming from, I would appreciate the help. Thank you! <Q> What you describe looks almost sure as an interference from your switching power supply to your amplifier. <S> The design from previous reply does not help at all, sorry. <S> The origin of the problem is: <S> the 24 V output of your PS (both wires!) is effectively "jumping" with respect to ground potential, most likely, at switching frequency, its harmonics and sub-harmonics. <S> The bad news are that it is very difficult to solve this problem. <S> One (may be - easiest) solution is to use another 24 V DC source, based on traditional iron core 220 / 24 V transformer and rectifier. <S> Use 2 common mode chokes on these wires and a pair of capacitors between them: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> In the second case, the mid-point of capacitors is connected to "true ground". <S> You can try using the third line (0 or GND) on your Mains circuit, or some ground connection, if it is available. <S> Capacitors must be larger then 10 uF and rated for 50 V at least. <S> Chokes must be rated for 2 A DC current and their inductance must be no smaller than 1 mH, preferably much more. <S> And the last circuit to try is to add a differential mode LC filter. <S> Hope it is not needed. <A> As noted in the comments you will be able to reduce this issue by placing a decoupling capacitor across the power supply. <S> The wikipedia article on the subject provides a very good explanation of the technique. <S> Place the capacitor as close as possible to the IC, and if this does not work you can use multiple capacitors of different types (e.g. electrolytic, tantalum) to gain the benefits of each type (e.g. high frequency performance). <S> If you are using a dual-rail design then ensure that you are using capacitors between each rail and ground, and also between both of the rails: <A> Having recently taken apart a 20V laptop power supply of a kind which is rather noisy (basically, as soon as you plug it in you get a terrible ground loop that sounds like frying bacon) <S> I concluded that the solution is to find a different power supply. <S> Consider a linear supply perhaps (transformer, bridge rectifier and big electrolytic caps) which although larger do not suffer the noise problem inherent in switched mode power supplies.
| If you wish to use this kind of switching supply, you can try: Install a common mode choke on both DC 24 V wires
|
what is the maximum output voltage from an opamp in general case? In general, what is the output voltage that an amplifier (not rail-to-rail output type) can provide,if it is supplied for example Vdd=+24V and Vee=-24V which is the maximum value I can get "for sure" from the output, can it reach +24V and -24V? in datasheets they always mention the minimum and the typical output voltage in some cases but never the maximum one. http://www.ti.com/lit/ds/symlink/opa2604.pdf <Q> Firstly note that the opamp given has an unusually large supply voltage range, most opamps at +/-24V will only give out smoke. <S> Secondly. <S> the spec you're looking for is in Table 6.5, page 5, conveniently labeled "Voltage Output". <S> All figures in that table are for supply of +-/15V, and it guarantees an output of +/-11V <S> (typically +/-12V) under the specified conditions. <S> So at <S> +/-24V you could expect +/-20V or slightly better. <S> (Whether it can still drive 600 ohms at these voltages is another matter, I wouldn't trust it to drive better than 1 kilohm thanks to power dissipation) <S> Most datasheets show the output voltage over a wider range of conditions, different load impedances, etc. <S> but Burr-Brown (now TI) don't for this one. <A> The data sheets don't give the maximum output voltage because a) <S> it's poorly defined under amplifier output driving conditions and b) very few people want to know the maximum, they are interested in the minimum it is guarranteed to drive, and the typical voltage it can drive to. <S> The only time you would want to know the maximum is if the amplifier was driving something that needed protection from extreme voltages. <S> In this case, the safe and sensible thing to do is to take the rail voltages, these will not be exceeded if the amplifier is doing the driving, and into a resistive load. <S> Warning. <S> Typically the ESD protection on an IC will clamp any pin to a diode drop beyond the rails (a few pins on specialist ICs may be specified to be able to go beyond the rails) , but that only works up to some current, after which you can damage the diode. <S> Check the ABS max ratings on your data sheet for the largest current you can send into a pin. <A> The important thing to realize is that sometimes the min and max seems reversed, but it is actually correct. <S> You are looking for the maximum voltage. <S> What the manufacturer is giving you that they guarantee that maximum voltage to be 11V MINIMUM (when powered with +/- <S> 15V). <S> They also tell you that under the test conditions most of the chips will be able to reach 12V, but... <S> no guarantees! <S> Loading matters a lot. <S> Because they specify a bunch of parameters with 600 Ohm load, they apparently find that a normal load. <S> Usually when you have less load, say a 10k load resistor (or circuit that acts as such), the highest achievable voltage will be higher. <S> The problem is that the datasheet gives you no guarantees in this field. <S> Depending on the internal structure, there is probably a fixed offset (XX Volts) and a current dependent part. <S> The rail-to-rail opamps often have the fixed part as zero. <S> The OPA2604 is a weird opamp. <S> Even though they clearly allow you to power it with +/- <S> 24V, most of the specs are given with +/- <S> 15V powersupply. <S> This could be taken as a hint that they are not really intending it to be used under such conditions, but they do promise the device will still work more-or-less. <S> Suppose you design a circuit with +/- <S> 20V power rails. <S> And you've studied the datasheet and it says: max input offset voltage: <S> +/- <S> 5mV. <S> So you design your circuit to allow for +/- <S> 5mV of offset. <S> Now, when you test your circuit... it doesn't work. <S> Turns out the offset is 10mV! <S> Does the chip conform to specifications? <S> Sure it does! <S> They only guarantee that <S> +/- <S> 5mV at the +/- <S> 15V power supply voltages. <S> I think this is a shortcoming of the way modern datasheets are written. <S> There really are very little guarantees when you might want to use a chip slightly outside the tested conditions. <S> In this case almost nothing is specified for power supply voltages <S> +/- <S> 5V or +/- <S> 24V.
| If the load is inductive, or sourcing a current into the amplifier, then the output may well go beyond the rails.
|
VCC pour and decoupling capacitors on a dual layer board I am designing a simple dual layer board which has some ICs that I have to put decoupling capacitors near them. Initially I wanted to make both top and bottom layer as ground plane and route signals and power with traces. But then I found this answer here and decided to make my top layer a power plane (but honestly did not understand why!) So my question is, should I forget the VCC plane and make both top and bottom planes as ground? My board has only one 5V supply input called VCC_IN which then goes into a EMI filter (which I am not sure if it is going to be any good) and after R11 (R11 and R12 are going to be 1206 ferrite beads) and C14 it will be called VCC and turns into a power pour. So, I wonder what kind of effects I am going to see on C1 and C2 as decoupling capacitor compared to if the top pour was also ground and I just routed VCC using a normal track? The ground plane is ripped off in the image for better visibility of top layer <Q> There are pros and cons. <S> The advantage of ground on both sides is that you can tie the two ground planes together directly with vias. <S> Whatever you do for high speed signals you need to think about your high frequency return paths. <S> The return path should generally be as close as possible to the signal path to minimise inductance. <S> If part of the return path flows through a power plane and part through a ground plane you can tie them together with a capacitor. <S> A 2 layer board will always be a compromise for high speed because your planes inevitablly end up quite "cut up". <S> As far as providing power decoupling for the ICs themselves the general rule is that it should be as close to the IC as practical. <S> Your board seems to have a lot of unnessacery space between the capacitors and the IC. <A> Keep the VCC plane. <S> Keep in mind current travels in loops. <S> What happens is that current travels from the power input to the capacitor, through it and back to the ground connection through the ground plane. <S> The capacitor acts as open circuit at DC. <S> If you get some high frequency interference, it will go through the capacitor and will not interfere with your power supply. <S> What will happen if you have a track instead of a VCC plane is that the track will introduce higher impedance in series with the capacitor. <S> This will reduce the capacitor's ability to provide low impedance connection to ground filtering the interference. <S> Another problem is the current loops. <S> Power planes will help you minimize them reducing your emissions. <A> It is good practice to have power and ground planes, as the post you linked to states that having your layer stack-up in this manner creates a fairly nice bypass capacitor in the fab, and can reduce EMC/noise/common-mode and differential mode emissions from the PCB. <S> Keep the traces as short as possible to IC1 and your power/ground planes. <S> If there is any ripple/noise on your power, these capacitors will ensure your power to the IC remains smooth and stable. <A> Keep the Vcc plane over as much ground plane as possible. <S> This creates a board-sized capacitor to keep RF noise to a minimum. <S> But you need vias to connect the gnd layers often, especially near decoupling caps ground side. <S> You can keep any power traces about .100" wide for a local feed (coming off of the Vcc plane), then slim it down to .020 to .030" up at the smd parts. <S> Any ground plane on top should only be used to isolate and shield HF / RF signal traces on both sides of the board. <S> No issues for C1 or C2. <S> The only issue that would change the board layout is if you were using frequencies over 200MHZ, then you would use custom board layout software for routing the traces and gnd and Vcc coverage. <S> In such a case the gnd plane would be in a polygon pattern to break up standing waves
| I would keep your VCC plane, and try to place the decoupling capacitors as close to their pins on IC1 as you can. The advantage ofone side power and the other side ground is that it can provide lower DC voltage drop.
|
Why is the regulation of my 5 volt power supply so poor and how to address it? I purchased a 5 volt power supply off Amazon. It runs from 120 VAC and produces DC power. It's a switch mode power supply. If I put a multimeter on the output it always shows approximately 5.0 VDC. I ran into extreme difficultly in using this power supply for any sort of actual project. The output is extremely noisy. I connected some small capacitors with a value of 10 pF and 10000 pF to the output of the power supply. I would think the power supply would have small capacitors of this type in any case, but apparently not. These eliminate lots of HF noise that is coming out of the power supply. Unfortunately this noise is not really the issue. Here is what the power supply looks like on my oscilloscope with no load. Adjusting the time scale and voltage scale I saw this The channel in blue is the output of the power supply. The channel in yellow is the output of a filter network I built. I used the low voltage side of a mains transformer and a large electrolytic capacitor. Here are pictures of those, although I doubt it matters at all. The inductor is wired in series with my load(if any), and the capacitor is parallel with the power supply. I decided to test just the power supply with a resistive load. I selected a 10 ohm resistor. This should provide a load of approximately 500 milliamperes. The filter network deals with some of the oscillations, but there is still an almost 1 volt spike on the output of the filter network. I tried moving the capacitor around, but it makes little difference. In fact, even with the capacitor disconnected the output does not change much. Here is a small transformer removed from a switching power supply. I connected the 5 volts in series with the primary of this transformer. And the view from my oscilloscope: It seems that almost any inductor filters out the oscillation with the period of approximately 3.5 microseconds. But that huge spike the precedes the oscillation remains. In this case the power supply jumps around by over 2 volts. 2 volts on a power supply meant for 5 volts is 40%. The interesting thing about this is the capacitor seems to make no difference. It is old, but I've tried several and gotten the same result. They all have some capacitance, although it may be slightly diminished over time. Given the fact the voltage still swings all over the place with the capacitor, my only theory is that the circuit inside the power supply is actually shorting its own output. If the regulator on the output of the power supply just turned off, the voltage would just taper downwards because the capacitor would slowly discharge. It's almost as if the power supply is internally shorting for a brief period of time, then the regulator goes a little nuts and "rings" as it tries to find 5 volts again. Why is the regulation of my 5 volt power supply so poor and how to address it? Although I cannot imagine it would help, here is a picture of the power supply with its case off Update: I performed an additional test with the mains transformer as a filter wired in series with 4 resistors. One of the resistors was the 10 ohm resistor, the other three were 6 ohm. This should give a resistance of 1.66 ohms for approximately 3.125 amps of current. This does not change anything significantly in the observed output. I reversed my probes in this test, so the colors in this screenshot are reversed as well. Here is a closeup shot of the "spike" as I called it. I also tried connecting a 1 microfarad capacitor across the power supply while it was driving the 10 ohm load. Here is what that looked like <Q> A 39000uF cap and a transformer winding is way too much for a filter. <S> I suspect your power supply is either defective, poorly-designed, or being operated way out of spec. <S> The spikes are happening at a rate of ~85-90 kHz, which could be the switching frequency. <S> The higher-frequency ringing afterward is clearly due to the spikes. <S> If you can zoom in on the spikes with your scope, it might tell you (and us) more. <S> A link to the Amazon page or a datasheet would also be helpful. <S> Return or replace the supply. <S> The Photon's suggestion of trying a larger load is also a good one. <A> Eric Urban,With respect it sems to me <S> thay <S> you do not fully understand that useful abomination the switcher? <S> In particular do yo know how dangerous it can be to go inside.? <S> The best saolution for you is to purchase a switcher with (say) more than 10-volts output and follow it wibh a voltage-regulator chip that will given you 5-V at very-low o/p impedance. <S> That uis what is inside the larger and more expensive items. <A> This looks like a typical AC to DC converter. <S> 350W maybe and could be any two switch topology given two switches shown, hard to say for sure. <S> The transformer provides isolation from primary to secondary in any case. <S> I recommend using a differential probe when looking at the secondary side waveforms or cautiously float the scope through an isolation transformer or cheater plug. <S> What you are seeing might be a ground loop issue (?). <S> A agree with others that supplies typically need a load to regulate to else they may operate in bust mode or some other mode to try to generate feedback to regulate to the set voltage.
| They do by nature have a very poor regulation because every rule is bnroken in the inrerest of small size. Regardless, your options are: Try a more reasonable filter -- maybe a few microfarads ceramic capacitor.
|
What is the main source of variation between thermocouples of the same type Thermocouples have stated tolerances [ wikipedia ]. However, the characteristic behaviour is defined by the materials in contact, so there aren't size or shape considerations as for RTDs. What are the main sources of the variation/tolerances in thermocouples? A comment to this answer blames variation in metallurgical properties, which seems plausible, but this is unsourced. <Q> In my experience the difference between thermocouples of the same type typically comes from the the manufacturing process. <S> I have been making a number of K-type thermocouples using electrical arc welding, with "exactly" the same wire lengths, at the same temperature and with the same amplifier they display different temperatures in the range of ±0.5 Celsius from each other. <S> My only explanations for this is that during the fusing different amounts of impurities are introduced into the junction, which is mostly based on the quality of the conductors, and that the phenomenon is related to the temperature of the junction at the moment of fusing. <S> Unsurprisingly Wikipedia has some useful info on this as well. <S> Basically, I'd say the guy in the thread you linked is right. <A> Thermocouples are nothing but two (or more) metals or alloys in contact, so the inherent variations must be metallurgical in nature. <S> Metallurgical effects include the alloys, how evenly distributed they are during the drawing of the wire, and effects such as annealing and cold-working. <S> Of course the construction of a thermocouple can lead to differing measurements depending on way the thermocouple is made or mounted and how heat flows from the object being measured down (or up) <S> the leads. <S> Extension wire is not always made from the same type of metals, or made to the same specifications, so the entire thermocouple assembly may show errors if the transition to extension leadwire is far from room temperature. <S> The main reason for using different materials is to save money and allow the use of robust conductors, especially with precious metal thermocouple types (most commonly types S, R, B but gold has been used as well). <A> To understand thermocouples there are three things you have to keep in mind: <S> Thermocouples measure their own temperature. <S> They do not measure air temperature, water temperature or anything else; they measure the temperature of the wire. <S> Thermocouples do not measure absolute temperature, they measure differences in temperature or relative temperature. <S> If you have a thermocouple based temperature gauge that gives an absolute reading another temperature reference is used also. <S> (This is called cold junction compensation.) <S> Thermocouple junctions are important and insignificant. <S> The thermoelectric effect is a wire effect, the junctions are just where you change wires. <S> There are people who are trying to make the junction as small as possible because they think that the thermocouple measures the temperature at the junction. <S> thermocouples measure the temperature across the wire. <S> It is not the thermal mass of the junction, but the thermal mass of the wire that is important. <S> but many times significant source of differences in thermocouple readings is not the thermocouple, but the rest of the system. <S> Common mode noise can be a problem as can rounding error. <S> Wire length is almost never a source of error (assuming you are using thermocouple wire), but shielding, connectors and oxidation can all be problems, specifically hot junction oxidation (even on welded junctions) can lead to errors and is the most common reason to replace old thermocouples (Heat, dissimilar metals and current is a recipe for unwanted reactions.).
| The most common source of variation in thermocouples is the wire (which results in changes of thermal mass and electrical resistance and in millivolt systems like thermocouples resistance is very significant).
|
Is this a power resistor? I'm trying to repair an electrophysiology amplifier (AXOPATCH 200A). It has this power resistor which I desoldered because I suspected it was broken. The color code (red, green, black, gold) gives 25 Ohms / 5%, but it measures 16.6 Ohms. It is positioned directly between +5V and GND, so at 25 Ohms it is disipating 1W. My questions are: Is it really a power resistor? Is it in a bad shape or am I confused and it is another type of component? Which is it approximate power rating? (there are some components on the image that can give an idea of its size) Which could be the purpose of a resistor not doing anything other than disipating power?? <Q> I am pretty sure that is a 2-Watt carbon resistor (about 18 mm long), based on comparing it to the ICs. <S> Here is a picture of 1/2, 1 and 2-Watt resistors: <A> To answer your questions : 1-The resistor you show is not a power resistor, however it drains more current than the small ones on your picture. <S> 2-Here is a link that might help you to determine the dissipation power : http://learn.mikroe.com/ebooks/componentsofelectronicdevices/chapter/introduction-to-resistors/ <S> 3-There are various reasons to use a resistor in a circuit, if it seems useless it could be simply a load to make sure the circuit as a minimum current to operate. <S> A good example is for relays, they need a minimal current to make sure the contact do not oxide. <S> If you need more help, you should provide the schematics. <A> Yes, you can consider that a "power" resistor relative to common resistors of today and other resistors on the board. <S> Since it's clearly bigger than other resistors, it was chosen in part for its power rating. <S> From the size I'm guessing it's in the 1-2 W range. <S> The much smaller resistors in the upper right section are probably 1/4 W. <S> The value seems to be 15 Ω (BRN, GRN, BLK). <S> Keep in mind that "power resistor" isn't a definative spec. <S> All resistors have a power rating. <S> Nowadays, anything over 1/4 for thru hole is probably a power resistor because if the extra power dissipation wasn't required, a 1/4 W resistor would have been used. <A> Are you sure it is " RED(2) green(5) <S> black(10^ <S> 0=1) gold(+-5%)"? <S> Color quality of your photograph seems very good, and to my vision, the first band is BROWN, and NOT RED. . <S> So it seems the resistor is "BROWN(1) green(5) <S> black(0) <S> black(10^ <S> 0=1) <S> " i.e. 15 ohms. <S> that is close to 16.6 ohms. <S> Indeed color codes are often troublesome. <S> Read more about it on <S> Why do resistors still use color coding? <S> Such banded Color coding is common in 3 parts: Resistors, Non-electrolytic color-codes (some old-type), and some coils (maybe rare one , I've never seen one, just read @Wikipedia). <S> If you've check Ohms in proper-method with a multimeter; Then it is surely a resistor; and if it was a non-electrolytic capacitor, it would show infinite resistance (non-conductor) <S> (in case of digital-multimeter, 1 (High Value) or OL(Over Limit or Open Line)).If <S> it was a small coil, it would show Zero-ohms or hardly 1 ohms (just like conductive wire). <S> From an intermediate resistance value very close to color-coded printed-value, it seems to be Resistor, nothing else. <A> No, it's not a power resistor as most colloquially know a power resistor to be, a fairly arbitrary term really. <S> Those would be large white ceramic resistors <S> Of course, 2 Watt resistors are fairly large, and not typically used for signaling or other purposes. <S> It's clearly there to use as a power resistor. <S> As to why it is there, if it actually connects directly between 5V and GND may be one of two reasons. <S> Heating the board or enclosure to make sure it is within operating temperature range. <S> Not likely here. <S> Providing a significant load to help ensure a stable voltage rail. <S> Especially for an Amp that likely has a transformer based power supply, not having a sufficient load may result in noise or higher than safe voltages.
| "Power resistor" just means that this is a large point of that resistor, which varies by usage.
|
What determines voltage levels in RS-232 What determines the peak voltage levels for RS-232 lines? Is it determined by the software? The hardware? Something else? <Q> The RS-232 standard specifies the range as: 0 (space) <S> Asserted <S> +3v <S> to +15v1 <S> (mark) <S> Deasserted −15v to −3v <S> In addition, inputs must tolerate voltages up to ±25v, and outputs must tolerate indefinite shorts to ground. <S> In the past, a lot of equipment used ±12v since it was available from a minicomputer power supply for example. <S> Once personal computers became popular, most of them switched to ±5 since it is commonly available and still within the spec. <S> (Well, the +5 is commonly available, additional circuitry is necessary to get the -5V <S> if the main power supply does not provide it.) <S> As an example, the MAXIM series of UART RS-232 interface chips such as the MAX220-MAX249 series use a voltage doubler and voltage converter to generate ±10v.) <A> Voltage levels are determined by the HW drivers. <S> A typical driver has a switched capacitor power supply to transform from logic levels to RS232 bipolar levels <A> The software neither knows nor cares what voltages represent mark or space (0 or 1). <S> As long as the receiving circuit can discern the difference. <S> In some cases it is not unusual to find simple TTL levels (0 and 5V) in modern digital equipment decades removed from the vintage technology.
| Specifically the voltage levels for mark and space are determined by the power supply rails of the transmitting circuit.
|
Calculate Back EMF from Torque Constant I have a DC motor with a torque constant of 1.05 mNm/A. How do I calculate back emf constant from torque constant (mNm/A)? I have information that is in the maxon catalog.(DCX 10 S metal brush motors) <Q> OK I'm going to turn the comments into an answer since the question raises a surprisingly subtle point, and one that is most easily grasped if you consistently use the SI system rather than traditional units. <S> Because a motor translates electrical power into mechanical power (and vice-versa in generator mode) it must obey the conservation of energy. <S> So (ignoring friction, resistive and other losses) power in = power out. <S> Or, voltage * current = rotational speed * torque. <S> Rearranging, Voltage/Speed = Torque/Current. <S> Torque/Current (Nm/A) is known as the torque constant Kt. <S> Speed/Voltage (rad/s/volt) is known as the speed constant Kv (commonly seen is RPM/V but here expressed in SI units. <S> So, given the torque constant for a motor, the speed constant is also known, and presumably its inverse is known as the back EMF constant in some circles (though I haven't personally ever seen that). <S> EDIT : <S> Following Gregory Kornblum's comment : who says it's the same power? <S> The principle of conservation of energy. <S> Now clearly this is the simplest, most ideal situation - as I said above, ignoring all losses. <S> You can define anything any way you like, but the most generally useful approach is to start with the simplest ideal situation, then separately account for energy losses until you have a satisfactory model for your purpose. <A> you cannot make the assumption that \$K_t \equiv K_e\$ for a couple of reasons. <S> the definition of Kt & Ke where Kt and Ke are defined now #1 is easy enough to manage. <S> Kt is Nm/A. Peak of the AC (or quasi if BLDC) will give the torque product. <S> The equivalent for Ke is V/\$\omega\$ peak line-line for mechanical velocity. <S> An ideal motor, with a stator-pack that DOES NOT saturate. <S> Ke and Kt (for the above statement) are interchangeable & if you wanted rms phase voltage a simple factor is all that is needed. <S> However, there is no such thing as an ideal motor & this is where the main difference comes into play. <S> \$K_t\$ is determined at PEAK current. <S> \$K_e\$ is defined as the OPEN-CIRCUIT voltage. <S> If you happen to have a "lazy motor" that is inefficiently using the stator pack and ONLY operating in the linear region of the B-H curve then <S> yes... <S> \$K_t <S> \approx K_e\$ <S> but that is a very, very poorly design motor. <S> The optimal point of a motor design is around hte knee and as such \$K_t != <S> K_e\$. <S> Its close but not 1:1. <S> There is no magic "frig factor" to convert between \$K_t\$ and \$K_e\$ because it is dependent on the magnetic design & operating point. <S> If you do not have access to the magnetic design work & are not provided with \$K_e\$, the only guaranteed way is to backdrive the motor and determine the open-circuit voltage for a given rotor velocity... Or accept the deviation <S> The majority of the time \$K_t\$ is more useful, not only for torque calculation but also for operational BackEMF as when the motor is loaded, at speed, the core is becoming saturated and naturally \$K_e\$ drifts towards \$K_t\$. <A> First of all you should have done a bit of research before asking this question, you also havent mentioned what sites did you go to before asking and also why other similar questions could not answer your problem. <S> Now as mkeith said the wikipedia page on Motor constants answers your question. <S> Wikipedia <S> Also you haven't specified which kind of motor it is. <S> If its a brushless motor then $$Kemf=Kt$$ or the back emf is equal to the torque. <S> The different types of motor will always have a specific ratio between EMF and the Torque. <S> This EE.SE answer might also help you : Stack Exchange answer <S> This is also a good website for learning about such calculations <S> Micromo.com <S> If you still have a doubt comment and I'll try to clear it. <A> These two constants <S> Kt and Kv are not related, knowing torque constant won't help you to calulate the generated voltage. <S> EDIT, with enlightenment of Brian Drummond: <S> \$P= <S> M\omega\$ <S> \$V\cdot <S> I= <S> I\cdot <S> K_T[\dfrac{Nm}{A}]\cdot V\cdot\frac{1}{K_V[\dfrac{V}{rpm}]} \cdot{\frac{2\pi[rad]}{60[s]}} <S> \$ <S> \$K_V[\dfrac{V}{rpm}]=\dfrac{2\pi}{60}\cdot K_T[\dfrac{Nm}{A}]\$ <S> \$K_V[\dfrac{V}{krpm}]=\dfrac{2000\pi}{60}\cdot K_T[\dfrac{Nm}{A}]\$ \$K_V[\dfrac{rpm}{V}]=\dfrac{60}{2\pi}\cdot <S> \dfrac{1}{K_T[\dfrac{Nm}{A}]}\$ <S> \$K_V[\dfrac{krpm}{V}]=\dfrac{60000}{2\pi}\cdot \dfrac{1}{K_T[\dfrac{Nm}{A}]}\$ <S> Also it can be written that 3 phase PMSM is little different: <S> \$P= <S> M\omega = \sqrt{3}\cdot <S> V <S> \cdot <S> I\$ <S> so: <S> \$K_V[\dfrac{V}{krpm}]=\dfrac{2000\pi}{\sqrt{3}\cdot 60}\cdot <S> K_T[\dfrac{Nm}{A}]\$ <S> These are data of 3phase PMSM: <S> \$K_T=0.835\$ <S> calculating <S> \$K_E\$ for 3ph PMSM gives result 50.5 V/krpm, which is approximately the declared 53.0 V/krpm.
| There is just one way, spin the motor and measure voltage and speed, make a table and write a linear function.
|
If negative input ==> 0 V Otherwise, input voltage I am struggling, and I would like to know how I can make a circuit that holds the following conditions: If the input voltage is negative, then the output should be zero volts. If the input voltage is zero or positive, then the output voltage should be the same as the input voltage. Thanks in advance. <Q> simulate this circuit – <S> Schematic created using CircuitLab <S> But, if you want to compensate for the voltage loss over the diode you'll need an Opamp . <S> simulate this circuit <A> The simplest circuit: simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Single diode rectifier. <S> The diode will conduct when Vin <S> > <S> Vout. <S> The diode will block when Vin < Vout. <S> Whether this is suitable for your application depends on whether you can tolerate a 0.6 or 0.7 V drop across the diode as well as any factors you haven't told us. <S> Sorry, (1) <S> it is a DC voltage, (2) <S> It comes from the output of an operational amplifier which compares two voltages. <S> (3) It will be sent to a microcontroller pin. <S> Use @dim's circuit: simulate this circuit Figure 2. <S> GPIO interface for bipolar signal. <S> Use (a) when Vin ≤ Vcc. <S> Use (b) if Vin ≥ Vcc. <S> Let's say that Vin is <S> 15V. Isn't it dangerous to send that voltage to the GPIO? <S> Yes. <S> We wouldn't connect 15 V directly to the input. <S> If Vin ≥ Vcc then use Figure 2(b). <S> If Vin goes above V+ then D3 shunts the current to V+ and R2 limits the current to around a milliamp. <S> Between 0 <S> and V+ the diodes don't affect the circuit. <S> Below 0 V D2 conducts and prevents the GPIO <S> seeing any less than -0.6 to -0.7 V. <A> There is a very simple solution, but whether this works will depend on the input impedance of the subsequent circuit the output impedance of the previous circuit <S> the fact you can tolerate a small amount of remaining negative voltage (~0.4v). <S> You can put a resistor of about 1-10k between the input signal and the cathode of a diode (preferably schottky, like BAT54) with the anode of the diode tied to ground. <S> The output signal will be taken between the resistor and the diode. <S> If the impedances/remaining negative voltage is a problem, a more complex solution would be to use an active device like an opamp, but I'll let someone else suggest this.
| What you are looking for is a rectifier , if the voltage drop of the diode is of no importance ,then this circuit will do the trick.
|
Designing a PCB layout without having a schematic diagram (or netlist)? Could you please let me know if it's possible to design a PCB layout without a schematic diagram? I outsourced the development and manufacturing of the ARM board which has 512MB DDR3 RAM and 4GB eMMC. The outsourcing company promised to make it in 45 days, but they didn't give me the product even it's already more than 100 days from beginning of the project. When they said they were manufacturing its PCB last month, I asked them to send its schematic diagram. But the answer was that they didn't finalise the schematic diagram and need 2~3 days to make it. As far as I know, you can't make PCB without importing a netlist from a schematic diagram file, especially when you have two DDR3 RAMs. It has been 1 month from the time and still I don't have the product. I need to decide how to react and would like to know if the company lied. I spent about $6,000 for the outsourcing and now I have no idea what to do. <Q> It is technically possible to design a PCB layout without first completing a schematic, but extremely unlikely. <S> It just makes it unnecessarily complicated. <S> I think whoever you hired is not being fully honest with you. <S> I think a likely alternate explanation is that the company you hired turned around and outsourced the project to yet another company! <S> But they may want you to believe they are doing the work in house, so they're keeping you in the dark and coming up with weird excuses. <S> It's also possible that they are doing the work in house, but underestimated the skill level necessary to complete a board of that complexity (it turns out that high-speed signals are hard to design for). <S> So they're stringing you along until they figure it out. <S> Either way, your prospects don't look good. <S> I would think very seriously about trying to get your money back and go with a company that can keep their promises. <A> Is very possible that the pcb files you have do not correspond to the circuit you want. <S> it "smells" like they just took a similar board and just gave you the finish board. <S> Not supplying the schematic is a bad sign. <S> my humble oppinion is they are playing with you. <S> Be carefull, never pay for dev in advance, or bad stuff can happen to you. <A> I would consider anyone who would do it for a non-trivial design (and this design is far from trivial) to be incompetant. <S> It is possible that there is a schematic <S> but it's not in a state where they consider it fit for release. <S> For example component values may be missing or it may just be very messy (on complex board <S> you inevitablly end up modifying the schematic to swap pins arround while you are doing the layout, so it can make sense to leave tidying up until after layout). <S> It's also very much possible that they lied about having the boards made. <A> As an addition to the absolute correctness in Dan's answer the only possibility I see for there to be the smallest hint of truth is them using a set of pre-made blocks from a library as smart companies would and them having to re-do some stuff to not expose the entire library. <S> Apart from that the total run time on this is about a factor 5 over what I would need based on the limited details, so I would abandon these people permanently whether or not they lied. <S> But then, as @Daniel says in a comment: My fee is nowhere near as low as $6000. <S> It might be just about enough as a down-payment given a proper contract with deliverables and payment obligations was made along with it. <S> I just wanted to add this (and it didn't fit in a comment) for balance and clarity that not always not wanting to give the schematic right away when the boards were already ordered <S> means something untoward is happening. <A> I routinely design complex PCB layouts before I create the schematic. <S> However, this is not a practice that I recommend for anyone else. <S> For me, it's easy. <S> I have an extremely detailed picture in my head of what I want <S> and I simply do the layout. <S> When I'm done the layout, I extract the netlist from the layout and use my CAD package's back-annotation process to create the schematic. <S> I can then check the schematic for errors and omissions. <S> I can do this because I started laying out complex PC boards long before there was inexpensive CAD systems available. <S> I'm talking about the Bishop Graphics Puppets era here - and even the time period before Puppets were available. <S> Not having a computer to manage your netlist means that you need to keep your schematic in your head. <S> Do this for a long enough period and it becomes second nature. <S> This process is good (for me) for up to several hundred nets. <S> Once the number of nets get into the thousands, I do things the way everyone else does: create the schematic, then use the netlist to create the PCB layout. <S> Where I'm going with this is that it IS possible that your subcontractor has done things this way. <S> Unlikely, but possible.
| It's possible to lay out a PCB without a schematic but the chances of error are extremely high.
|
Sharp and Simple Analog Filter suggestions I'm looking for schematics of some simple opamp-based analog filter, which has a sharp band-pass functionality. I want it to control small robot with sounds of different pitch. The matter is like this - to amuse my pupils I already made a robot which is controlled by digital filters. Perhaps this short demo explains better . However some of them asked if they can build similar thing. My design is rather simple, but makes usage of a small MCU and about hundred lines of code. They complain they are not advanced enough in programming so I'm thinking of proposing "more analog" solution to them. That is why I'm looking for substituting digital filter (and MCU at all) with analog schematics. I do know how to build first-order low-pass and high-pass with RC chain. And I can use several of them with opamps... But filtering of 20 or 40 dB per decade seems to be not too sharp for my goal. The main problem is that controlling pitches should not be too far (e.g. I do not want to use 261 and 440 Hz) to ensure that sounds have roughly equal loudness. With simple plastic recorder tones of 784, 880 and 988 are ones of most easily produced and quite loud (indoors) - that is why I used them. So now I'm looking for schematic consisting of preferably opamps, resistors and capacitors (preferably no coils) which will allow to extract very narrow band. Currently it gives about 0.3-0.4 of amplitude on, say, 784 or 988 Hz filter output when 880 Hz tone is played (and amplitude on 880 Hz filter is about 1.0). Probably I just do not know proper keywords for googling. What such kind of filter could be called? Thanks in advance for any suggestions! <Q> This might sound silly, but it could work and result in a simple circuit with large margins for errors: demodulate the signal to a lower frequency. <S> By multiplying the signal with a ~750 tone, you will have a ~30Hz and ~250Hz signals, that can easily be separated with a first order filter. <S> You can easily achieve this by “chopping” the signal with the output of a 750Hz astable and a transistor or two, or even just with some diodes. <S> You also need to low-pass it at ~400Hz to remove the images. <S> A simpler alternative, because it would reuse circuits and concepts, would be to have two oscillators/demodulator/filter chains. <S> One at ~750Hz the other at ~950Hz. <S> You just have to use stable capacitors for the oscillators, the rest can have very large tolerances. <S> This design can easily be moved to any desired frequency, just by retuning the oscillators. <A> The simple answer is that you are out of luck. <S> With "tones of 784, 880 and 988 <S> " Hz, a second-order filter simply won't work. <S> While, in theory, you could make the Q high enough to get the selectivity you need, component sensitivity will eat you alive. <S> Instead, you'll need to take another approach. <S> The early days of DTMF touch-tone detection used PLLs, with the LM567 being the classic chip. <S> For Googling purposes, try "touch tone detection" and "touch tone detection pll". <S> The problem with the first string is that for quite some time now the digital approach (such as you originally used) has been the norm, and most recent articles assume a digital approach. <S> In some respects, while a PLL will work well, it does not address your students' objections that they don't know enough to properly use the recommended circuit. <S> While understanding code can be a problem, PLL theory is not a beginner's field, either. <S> Frankly, I don't see an easy path for you. <S> Your tones are very close to each other, and simple circuits are not going to work. <A> If you want to do things <S> Analog you could consider a BiQuad Active filter that is a modification of the Classic State Variable Filter .Your <S> Frequency is low so you do not need special Opamps so this will be cheap .You do not need too much Q so component tolerances will not be a problem .Your <S> Cap values will not be to big because frequenvy is not too low. <S> When these approaches were commonly taught in colleges opamps were expensive but it can be shown that the effect of component tolerances are not as bad as single opamp designs.
| So another possible way to do it is to use a synchronous demodulator, which will give excellent results, but this will require enough digital skills to produce the clock frequencies you'll need, and this too may be too much for your students.
|
Trying to make an AC 3 phase motor run as smooth as possible at low speeds. Is there a benefit in smoothness of rotation by increasing the outer rotor magnetic poles? Is there a rule of thumb for a 24 slot core vs number of magnetic poles? <Q> Now that I know that your building a turntable, I have some tips, although not necessarily an answer. <S> Instead, I will expand on Charles answer above. <S> To minimize wow and flutter (turntable terms), you need to minimize torque ripple (a motor term). <S> Besides the motor, the drive electronics and platter play a big roll. <S> What you want is a motor that is magnetically optimized for a sine-wave drive. <S> Many (maybe even most) brushless motors are optimized for a trapazoidal drive, and they will likely have too much torque ripple. <S> Then, your electronics need to provide a sine-wave drive. <S> That means that you can't depend on hall sensors for feedback, but need an encoder for high enough resolution. <S> But I assume you have the encoder already, since you need it to phase-lock to. <S> Another way to produce the sine-wave drive instead of the encoder is to do what Sansui did in their high-end direct-drive turntables (look at the SR-929 and SR-838). <S> They added windings to the motor to sense the rotor position in an analog fashion, and using the sine-wave sensed from those windings to drive the motor windings. <S> It had very simple drive electronics, but the motor was custom. <S> As you know, the heavier the platter, the smoother the ride. <A> Synchronous speed of induction machine:\$N[rpm]=\dfrac{60[s]\cdot <S> f[Hz]}{p}\quad p..\text{number of pole pairs} \$ <S> Increasing the poles number, the nominal speed is decreased. <A> In my estimation, for smooth operation, you would like as many slots per pole as possible and slots as narrow as possible. <S> The phase coils should be distributed rather than limited to one pair of slots per phase. <S> The slots will cause the reluctance of the magnetic flux to vary with rotor angle. <S> I suspect that it would be preferable to have an interior permanent magnet rotor, one in which the permanent magnets are in the interior of the rotor iron rather than affixed to the surface. <S> For the VFD, I suspect that some form of multilevel inverter topology would be preferable. <S> Is there a rule of thumb for a 24 slot core vs number of magnetic poles? <S> Excerpt from John H. Kuhlmann, Design of Electrical Apparatus, Second Edition John Wiley & Sons, New York, 1940 More than one coil can be wound through a given slot. <S> I believe that coils can overlap to some extent, but I don't have an explanation of that.
| Increasing the number of poles raises the frequency of any torque ripple that you may have left, and that makes the rotational inertia of your platter more effective at damping any speed variations caused by that ripple.
|
What cause extreme heat when electricity connect to it? I want to know is there a item/chemical etc, that cause extreme heat if it comes into contact with even a small bit of electric currents? I ask cause I am doing a school project and I need something to generation heat using electricity as it would be very covenant. Thanks Update: Sorry this question was unclear. I have done a bit of research since posting and hopeful I can make this more clear. I am trying to find away to amplify heat created by joules. ohmic heating is kind of what I'm talking about. How can I make a large volume of heat using that law? Do I need a large amount of electricity flowing or could I use a small bit of electricity and some how amplify it. <Q> The overarching principle is that conservation of energy applies in a closed system. <S> See https://en.wikipedia.org/wiki/Conservation_of_energy <S> A very common example of this <S> the combustion of fuel in a spark-ignition engine. <S> See https://en.wikipedia.org/wiki/Spark-ignition_engine , https://en.wikipedia.org/wiki/Otto_cycle <S> An extreme example would be triggering a nuclear explosive. <S> In these these examples the potential (stored) energy in the device is liberated as heat, mechanical action, etc; after the reaction has been initiated by a bit of heat provided by electrical system. <A> It seems you want more heat energy out than electrical energy you put in. <S> Conservation of energy means that this therefore requires some other form of energy to be put in, which is then controlled electrically. <S> The question then comes down to what other form should this energy take. <S> This depends on your application. <S> One obvious way to get a lot of heat is via chemical reaction. <S> It's easy enough to initiate a exothermic chemical reaction with electricity, but controlling it roughly proportional to some electric signal is more difficult. <S> One possibility may be electrically controlling a valve that lets more or less fuel into a combustion chamber. <S> It gets more tricky when the total output needs to be able to go to 0, since then you need a way to start up the chemical reaction, which can be quite different from sustaining it. <S> Traditional solutions include pilot lights or spark ignitors. <S> Or, you can chose chemicals that react exothermically on contact. <S> These will be harder to procure and may attract unwanted attention from the authorities. <S> This controls the magnitude of the chain reaction. <S> The advantage is that this method can produce large amounts of heat for small amounts of material, but getting the material will be difficult. <S> The system can also go into exponential runaway if carbon rods are removed too far. <S> This will have bad consequences, so overall I don't recommend this method. <A> The only way you can 'amplify' the amount of heat you getc i.e. output more energy as heat <S> then the amount of electrical energy you put in is if you consume some sort of fuel e.g. use electricity to spark a fire. <S> Energy, including heat must come from somewhere, if it's not from the electricity it has to be from something else like the fuel.
| If you put enough uranium or plutonium close enough with some carbon rods separating things, you can have the electric signal control how much the carbon rods inserted into the pile or not.
|
Do consumer high temperature RF ASICS exist at 200C? So at lunch I was browsing Indiegogo projects and I saw this one here It's a wireless meat thermometer, I've seen it in some permutation or another a few times on sites like this. Apparently no one can handle wires coming out of their oven or BBQ. I'm skeptical that this could be built cheaply, do ASIC and RF components even exists at 200C temperature ranges? I'd never thought about it so I did a little searching and saw the main market is oil drilling and probably military. The one little MCU I found from TI is $359 on digikey :) This project claims their thermometer's communicate via bluetooth! Based on my experience I find it hard to believe such a chip exists, but maybe I just don't know any better. So anyway just trying to expand my knowledge if something exists or fuel my skeptacism of this project Oh I also read this SE article which more or less says what I expect, but maybe things have changed. <Q> They do (At least for some ASIC components) -- some of them leverage Silicon-On-Insulator technology to help achieve those high temperature ratings. <S> X-Rel for example manufactures many "common" analog / etc components that can hit 230C+ temperatures. <S> They don't make anything RF at the moment, but I'd be curious to see if Peregrine's SOI process is similarly tolerant of extremely high temperatures. <A> The electronics can be in the bottom and cooled by the phase change material (moisture in the meat) so that ordinary components can be used, as @Jay says. <S> 100 <S> °C or 125°C is (relatively) easy with ordinary parts, especially if they don't have to last all that long at high temperature. <S> Most ordinary parts, even commercial parts, will work to something approaching the absolute maximum junction temperature, for a while, if the specs are relaxed (reduce clock speed, keep supply voltage moderate etc). <S> The temperature in a car parked under the sun in Phoenix in summer can probably approach 150°C. <S> Honeywell is one supplier of such parts. <S> In this case, (meat thermometer) 160 <S> °F (71°C) is enough for turkey. <S> 63 <S> °C for a medium steak, so <S> the 'tip of the spear' will stay relatively cool even in a hot oven. <S> Ruth's Chris (a chain of steakhouses) claims their cooking temperature is 1800°F (982°C) <S> but, of course, the center of the meat never gets hotter than 60-70 <S> °C (and much less if the customer wants it very rare). <S> Of course if the consumer leaves it in the oven long enough to char the meat to a crisp they might have to buy another electronic meat thermometer. <A> I think an animation from seller is self explanatory. <A> The directions for the product say to insert the probe into the meat and then place it in the oven. <S> Once the probe reaches the desired temperature the wireless probe sends a notification to the users' smart phone. <S> Most meat only needs to be cooked to around 160F (71°C) to be safe. <S> The probe will not get any hotter than the meat it is tuck into. <S> So it is possible to construct it using industrial temp parts rated in the 100C to 125C range. <S> Those types of parts are a little more expensive than 0 to 70C parts, but not unreasonably so. <S> If the user decides to ignore the notification or doesn't notice it; then it is possible that if the meats' internal temperature passed 100C or 125C (depending on the selected parts) <S> the probe could be destroyed. <S> For example Atmel (now Microchip) makes a chip with an ARM processor and Bluetooth transceiver that operates from -40C to 125C and sells for about $14.36 at 1pc or $9.50 at 1000pcs on Digikey. <S> ATSAMR21B18-MZ210PA <S> http://www.atmel.com/images/atmel-42486-atsamr21b18-mz210pa_datasheet.pdf <S> So the product is viable if the user pays attention.
| There are components designed for down-hole instrumentation that will operate at 200°C+ with degraded specs and extremely high prices.
|
Does wire diameter affect the electricity bill? Does thickness of wire affect the electricity bill, for same load? Thicker Wire has lower resistive losses ,does this reduce power bill? Does a longer length of the same diameter wire increase bill by the same reasoning? <Q> You are REQUIRED by code to use specific wire sizes for specific loads and in no case can you use a wire that is rated for less than 125% of a continuous load. <S> So if you follow the law, your wire heating is minimal. <S> so there is almost never a return on that investment. <S> The energy savings may be only a few dollars per year, whereas the cost of the larger wire for an entire house may result in hundreds of dollars in increased cost. <S> You would not likely live long enough for the energy savings to pay for itself. <A> "Power consumption drops when line resistance increases" is true when you have a resistive load, but not true when you have an active (regulating) load. <S> And it is not interesting to do so. <S> Let's assume that you have an apparatus that needs 100W to work. <S> For example your fridge that requires 100W to cool down to the set temperature level, or your computer that uses and adaptor and consumes 100W to do its work. <S> Suppose that your fridge or adaptor can run at 100V and also at 10V. <S> So at 100V the current would be 1A, and at 10V it would be 10A.If the line voltage is 100V, then at 0Ohm of line resistance, the apparatus gets 100V and all delivered power is used by the apparatus. <S> If the line resistance is changed to 9 Ohms then there would be a 90V loss in the lines at 10A and the apparatus would get 10V and consume the 100W it needs. <S> However the loss in the lines would be 900W - 9 times the usefull power. <S> In practice you will not have such a big difference, but it will always work the same way. <S> For a constant "usefull" power, the lower the line resistance, the lower the consumption and the lower the bill. <S> If you accept the principle of getting less usefull power, the bill will be lower with higher line resistance, but you will also get a lower amount of light from light bulbs and the cost per useful <S> Wh will be higher. <S> With (partially) inductive and capacitive loads, the computation is more complex but the line resistance will consume a lot with respect to the power actually consumed in the load. <S> This is why there is a lower limit on what is called the "Power factor", but that is another (related) story. <A> Power consumption drops when line resistance increases <S> - it's a case of more resistance in series with the load takes less current and hence power is less. <S> However, if you are boiling a kettle it will take longer to boil so, you maybe be taking less power but you are taking it for a longer time <S> and so the power loss in the cable (due to \$I^2R\$) will add a bit on your bill. <S> Remember you are not billed on power but power x time i.e. energy. <A> Theoretically the thinner the wire and longer it is <S> , the higher it's resistance and the overall load seen from the outside (power plant?). <S> Some of the energy entering your house will just dissipate as heat on the wires and will not produce any useful work, while you are still billed for it. <S> But practically it should be unnoticeable on your bill. <A> Thinner wire will have higher resistance and will reduce current and voltage to the load. <S> This will reduce power consumed but a higher percentage of that will be lost in the wires. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1 (a) Low-resistance wire and (b) high resistance wire. <S> Figure 1 shows a 10 Ω load fed from 100 V supply. <S> With zero wire resistance \$ <S> I = \frac { <S> V}{R} = \frac {100}{10} = <S> 10~A <S> \$. Power is given by \$ P = <S> I^2R = <S> 10 <S> ^ <S> 2 <S> \cdot 10 <S> = 1000~W \$. <S> (a) has 0.2Ω series resistance. <S> \$ <S> I = \frac { <S> V}{R} = \frac {100}{10.2} = <S> 9.8~A <S> \$. Power in load is given by \$ P = <S> I^2R = <S> 9.8 <S> ^ <S> 2 <S> \cdot 10 <S> = 961~W \$. Power lost in wire is given by \$ P = <S> I^2R = <S> 9.8 <S> ^ <S> 2 <S> \cdot 0.2 = <S> 19.2~W <S> \$. <S> Overall, we've saved a little on our bill and lost 20 W to the wire. <S> (b) has 2Ω series resistance. <S> \$ <S> I = \frac { <S> V}{R} = \frac {100}{12} = <S> 8.33~A \$. Power in load is given by \$ P = <S> I^2R = <S> 8.3 <S> ^2 <S> \cdot 10 <S> = 694~W \$. Power lost in wire is given by \$ P = <S> I^2R = <S> 8.3 <S> ^2 <S> \cdot 2 <S> = 132~W \$. <S> Overall, we've reduced our bill by 174 W (which means that the equipment won't run as intended) and lost 132 W to the wire. <S> Summary tableR <S> Load power Loss in wire Total <S> (billed)0 <S> Ω <S> 1000 W 0 W 1000 <S> W0.2 Ω 961 W 19 W 982 <S> W2.0 Ω 694 W <S> 132 W 826 W <S> For AC (Alternative Current)? <S> DC or AC. <S> Same rules apply.
| If you increase the wire size to be larger than necessary, it will technically reduce the losses, but at a much lower rate of increase in benefit
|
What is the purpose of MOSFETs in this schematic? On the main power line there are two MOSFETs but for what reason? image http://www.ti.com/ods/images/SLUSA78C/sys_sch_A_lusa78.gif The MOSFETs in question are Q1 and Q2. <Q> Q1, Q2 and Q5 are under host control and determine if battery is charging, and if 'System' is on or not. <S> Q5 works with Q2 and both are 'ON' when the system is battery powered. <S> U1 can be pulsed 'ON' (Q2 is 'ON') to maintain charge under load. <S> This circuit can both charge the battery and supply system power if required. <S> Todays smart phones can both charge and play videos or surf the net at the same time. <S> Older cell phones had to charge first before you could use the phone as a phone, or send messages, etc. <S> The newer charger IC's (U1) can handle the extra load of doing both task at the same time. <S> If the phone is just charging, Q2 and Q5 will be 'OFF'. <S> The user turning on the phone will cause Q5 and Q2 to come on in the proper order to supply system power. <S> If the adapter is not plugged in or the battery is at full charge <S> then Q1 is 'OFF' <A> A power MOSFET has got a parasitic body diode . <S> Often, body diodes are implied and not drawn on circuit diagrams for brevity. <S> If we draw a P-channel MOSFET with a body diode, it looks like this. <S> Due to the body diode, a power MOSFET can block current only in one direction. <S> When you put a pair of MOSFETs back-to back, they can block current in either direction. <S> (image from here ) <A> Most likely this is for "Break before Make" logic to control the source of your system. <S> In the schematic shown above, there is a SMBus power management IC or controller of some sort that will select the system source. <S> Typically the two mosfets on the top left will be lifted before a connection to the battery is made with Q3 to prevent damage to it.
| Q1 and Q2 select whether the battery or the adapter is sending power to U1.
|
Polygon pour over keep-out layer In Altium Designer, I'm using Keep-out layer lines to enforce the manufacturer's board-edge clearance restrictions. The problem is, that the top layer polygon will pour around these lines, which is not necessary. How can I make the polygon pour act like the keep-out layer was a silk-screen layer, i.e. just ignore it. If I set the clearance rule to 0 or matching "NOT OnLayer('Keep-Out Layer') ", it will have no clearance but still have no copper directly below the line. <Q> Personally, I use the built-in Board Outline Clearance rule located within the "Manufacturing" category: <S> This should produce a DRC error if components or traces are placed too close to the outline (if it lets you place them too close in the first place), and will also prevent polygon pours from pouring too close or outside of the outline. <A> Personally, I like Derstrom's answer for its simplicity about using Altium's built-in solution. <S> However, if you need keepout clearances not on your board edge, it won't help you there. <S> Then I create a similar rule called BoardInnerLayer_Clearance, make it lower in priority, to catch the other layers, and this second rule is exactly the same as the first except second object matches "All. <S> " <S> This lets me set the pullback for inner layers to something higher, say, 15mils, since often the board houses recommend this. <S> Then I just need draw <S> a boundary around my board and any pours or online routing rules will respect this. <S> Here you can actually see I put my KOR already pulled back at the inset, so a bit of overkill here on this board. <S> But in this image, the polygon pour goes all the way to the board edge, so the pour is respecting the rule: <S> Of course, the real beauty of this method is that it will work everywhere. <S> So here I have a footprint for a screw hole, with silkscreen to indicate to the designer <S> but there is also a circle drawn in keepout layer there that will keep any screw heads from eating through and biting into copper. <S> The rule handles this for me and all I have to do is draw keepout wherever I need it once I've set up the rule. <A> DerStrom8's and Joel's solutions do not work for older versions of Altium. <S> The easiest solution is adding a new constraint for the clearance definition. <S> To make the keep out layer solid ground, define the following constraint. <S> " <S> All And Not IsKeepOut"
| What I do is create a rule for my clearance on my outer layers, called say BoardOuterLayer_Clearance:
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.