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Adjustable AC voltage

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by fezder, May 14, 2012.

  1. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,

    Oh ok that's very good.

    What you should do next is measure the resistance of the coil you are testing right now. We should see what that is. It may not be right for this initial set up.

    There are a couple things we will check over, and this may take a little while to get everything right. But once we get it right it will be easier to switch coils and such for more testing of the other coils and whatnot.

    So we need to know the resistance of your coil next, and you can measure that with a regular ohm meter. Doesnt have to be super accurate get the measurement as good as you can.
     
  2. fezder

    fezder Well-Known Member

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    ok, my dmm shows 2.8ohms, strange, i wouldh've guessed smaller resistance :D
     
  3. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Well a lot of the smaller chokes have higher resistance like that. The bigger ones usually lower.
    With 3 ohms there you'll probably have to go to a higher load resistance. Maybe 50 ohms, maybe 100.
    The idea here is we want to get a square wave across the choke. If we cant get that then something might be wrong, like the drive to the transistor.

    I noticed other resistors on your breadboard. Like in series with the power buss, what are they for?
     
  4. dave

    Dave New Member

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  5. fezder

    fezder Well-Known Member

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    http://i46.tinypic.com/u7bdh.jpg
    you mean at this picture? 22k resistors are for freguency, and 330ohm resistor before fet is r3 on your given schematic
    http://www.electro-tech-online.com/attachments/
     
  6. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    R3 is there to help the power supply handle the peak current, but if the power supply can already handle the peak current then this isnt needed, at least for now. If you do use R3 however, then you also need the 1000uf capacitor or something like that.
    Also, 330 ohm is much too high. 0.1 ohms should probably be the highest value used there. For low inductor currents might get away with 1 ohm.
    But that 1000uf cap is really a necessity if the resistor is used or else the resistor starts to play too much with the other circuit components.

    So far so good, we just have to iron out a few things.
     
    Last edited: Jun 9, 2012
  7. fezder

    fezder Well-Known Member

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  8. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,

    Ok that looks better so now we have something to work with.
    You'll note that the current waveform is somewhat like a ramp up and ramp down, with noise in between, and the voltage waveform is a square wave.
    The only strange thing is that the voltage waveform seems to be inverted from what it would look like if we used two separate scopes. Perhaps something is still not right there yet, but we can see that the duty cycle is not 50 percent so using that we'll estimate the inductance using the positive voltage pulse and the low going current ramp until we clear up the possible sync problem.

    Now keep in mind that normally you would use the positive voltage pulse and the positive going ramp for example.

    I've attached the two scope pictures with the quantities labeled on the picture so you can see how there measurements are made. Once we take the measurements we then go ahead and estimate the inductance based on the well known inductance equation:
    v=L*di/dt

    So looking at the first attachment which is the voltage, we see the voltage goes above zero by 5.5 volts. I assume you have the zero adjustment of the scope done properly. Looking at the pulse width shown there also, we see the pulse width is 8us.
    Now looking at the second attachment which is the current, we see the voltage fall from about 0.0063 to 0.0003 which means we have a differential of 0.006 volts, and 0.006 divided by 0.1 ohms (i assume you are using 0.1 ohms for the current measurement) we get 0.060 amps.

    Solving the inductance equation for L we get:
    v=L*di/dt
    L=v*dt/di

    Now we know v is 5.5, dt is 0.000008, and di is 0.060, so we plug these values in and we get:
    L=v*dt/di
    L=5.5*0.000008/0.060
    so
    L=733uH approximately.

    So that's the way to get the inductance value. Note again that we used the positive voltage pulse and the low going current ramp, but normally we would use the positive voltage pulse and the high going current ramp. See if you can get both the voltage and the current on the scope face at the same time using the dual channel feature. It might be better NOT to invert any waveshapes for now.
     
    Last edited: Jun 10, 2012
  9. fezder

    fezder Well-Known Member

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  10. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Ok that looks good. We'll now estimate the inductance again this time using your resistor value of 1 ohm...

    As shown in the attachment, the voltage is measured from the zero line to the *negative* peak of the square wave because the voltage is inverted. The current is measured as the rise of *only* the ramp itself, not the peak above zero like we did the voltage. This gives us the differential current di. The differential time period is about 9us so dt is 0.000009.

    Starting again with the inductor equation:
    v=L*di/dt

    solving for L;
    L=v*dt/di

    we get di from the current ramp rise differential which is the voltage rise divided by the resistance used to do the measurement:
    di=0.005/1=0.005 amps

    and dt is the time period where the voltage square wave is negative (it's really positive because it is inverted):
    dt=0.000009

    and the square wave amplitude average is:
    v=7

    So now we just plug in the values and do the calculation:

    L=7*0.000009/0.005
    L=12.6mH

    So the inductor is about 12mH.

    From here you would raise the current a little and see if you can get it to saturate. This choke has quite a high resistance however so it might take a higher voltage supply to see it saturate. Once you see the current start to peak up sharply you would have an idea what the sat current is as that would be the peak of the current waveform.
    You might also see the inductance go down as the current is increased, before it actually starts to peak very sharply. This is normal too as the inductance could change quite a bit before it actually saturates deeply.

    Since you already know that the DC resistance is 3 ohms, that means this inductor would not be too good in a power supply that has to produce too much power. Maybe to drive a couple small LEDs or something like that in a boost converter.
     
    Last edited: Jun 10, 2012
  11. fezder

    fezder Well-Known Member

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    ok, how do i increase the current to the coil? by dropping resistance of the load?
     
  12. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Yes, but because this inductor has such a high resistance you cant go too low. See what happens.
     
  13. fezder

    fezder Well-Known Member

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    hi,
    i started dropping load resistance, its now 5ohms (not right away, at small steps), the coil is super hot, cant keep finger there second longer :D is this sign of upcoming saturation?
    http://i47.tinypic.com/2h7pjco.jpg
     
  14. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    If you look at the current wave shape you'll see that it curves up slightly now, rather than a straight line. That means the inductance is changing as the current is increasing. That's not too bad though so far so it may not be near saturation, but if the coil is also overheating then that means the DC current is too high for the coil.
    See if you can find a level of current where the coil just gets warm but not hot. That's probably a safe operating level for this coil.
     
    Last edited: Jun 10, 2012
  15. fezder

    fezder Well-Known Member

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    hmm, ok its doesnt get too hot now, should i now use bigger voltage? oh and thanks so much for your patience helping me out with this :)
     
  16. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,

    You're welcome. I guess you are going to have to experiment a little too and get the feel for how this works. I hope you are starting to get the idea how to do this now.

    As far as this coil is concerned, it looks like what happens is we reach the DC operating current limit before we reach saturation. That means that the current limit is going to be the one where it gets slightly warm but not hot. The wire inside is probably too thin so this coil is made for light duty applications.
     
  17. fezder

    fezder Well-Known Member

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    ok, well i think too that youve helped me enought, to think about it, its pretty hard to destroy these components, at least if im careful :). thanks for your help again, if id ask this from teacher, he wouldnt have the knowledge to help :D
     
  18. fezder

    fezder Well-Known Member

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    hey, i would appreciate if you could help me once again. i started testing power inductors, but the problem is that my power resistors are getting hot, i could add more resistors that the energy would settle well between all resitors. but how could i hook transistor as variable load to inductor? i wonder this because my potentiometers are 1/2w linear 100k and 10k, so they cant take much current on their own. i was planning to use 2n3055 or tip122, any suggestions?
     
    Last edited: Jun 19, 2012
  19. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,


    If i understand you correctly you just want to vary the output load of the tester/powersupply so you can increase/decrease current on demand more easily and not have to buy a bunch of power resistors.

    The simplest transistor DC load tester is simply a potentiometer driving the base of a bipolar transistor. That's super simple. It's not the best load but it works. I've actually used this in a pinch. The drawback is that if the DC voltage to the pot changes it changes the current through the transistor and thus the equivalent (power) resistance changes. It still works though, as you can just readjust the pot if the voltage drops (or increases) to get the required current level for the load.
    There is also some variation with temperature. As the transistor heats up the Beta changes and so the collector current changes and thus changes the equivalent load resistance again. This too however can be made up for by readjusting the pot for the correct current level.

    The idea is simple: take a NPN transistor, connect its emitter to ground, connect its collector to the positive voltage output you wish to load. Connect the potentiometer to the same output and also connect the 'arm' of the pot to the base of the transistor. When the pot resistance is varied the base current changes and thus changes the conductance of the transistor collector to emitter and that changes the equivalent load on the output.

    There are variations to this you might want to try, such as changing the pot resistance, adding some resistance in series with the pot to limit the adjustment range, etc. There is also a limit to how low you can go on the pot because it will drive the base too hard and/or burn out the pot. For example a 50 ohm 0.5 watt pot connected to the output when it is around 10 volts would burn out the pot due to overheating of the pot. So you'll have to deal with that using Ohm's Law and the variant Power Law.

    You also need a nice size heat sink for the transistor if you intend to use it for significant load current tests.

    There are also more advanced load testers such as a constant current load. This would be made using an op amp in addition to a couple transistors and that would set the load current to some constant value.
     
    Last edited: Jun 19, 2012
  20. fezder

    fezder Well-Known Member

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  21. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Well not exactly but that's close. That circuit supplies a current to a load, not being a load tester per se, although you might be able to use it for that too somehow.

    The circuit i was talking about uses an NPN transistor and has the emitter resistor going to ground. The rest is just about the same. If you need a schematic i can draw one up.

    See attachment.
     
    Last edited: Jun 20, 2012

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