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Variable Bipolar Power Supply

Discussion in 'Electronic Projects' started by Hero999, Sep 27, 2006.

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  1. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,

    Yes, i find that simulations dont always equal real life, but what else
    i find is that the simulations often tell me a lot about the theory of
    the circuit, in that once it is working in theory there is often a way to
    get it working in real life with some minor modification unless the
    circuit is too far from real life to begin with.
    For my simulation i used op amps that had good models that model
    their frequency characteristics quite well. I did have to simulate
    the two ICs though, with op amps and transistors.

    Just in case you are interested, there might be a way to replace the
    op amp with transistors and that may greatly improve the performance
    in the real life circuit.
     
  2. Hero999

    Hero999 Banned

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    Try it without the capacitor first and add it if you have any problems.

    If you change to 120R, you'll need to adjust the value of the pot or it'll give 27.4V out.

    For 15V the pot will need to be about 1k3 so try finding a pot in this value.

    Change the 120R to 82R and use a 1k pot, this will give 16.5V out or use 91R instead of 82R for exactly 15V.

    For adjustable current limiting use a a separate transistor current limiter in series with each regulator. Of course the current limiting for each side will be separate so use a dual ganged pot if this is a problem.

    Apart from building your own op-amp I can't think of anything so post your idea.
     
    Last edited: Jun 23, 2009
  3. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again Hero,


    Ok, i'll post something soon.

    In the mean time, isnt it true that that 10uf cap screws up the transient
    response? In other words, did you try the real life circuit without that cap
    or rather with that cap significantly reduced to say 1uf instead?
    Yes, the ripple isnt as good, but that's life :) use more filter caps.

    ADDED LATER:
    I found out that the two transistor op amp replacement works for tracking
    with a single set point (like plus and minus 10v for example) but as the
    voltage is adjusted the two voltages (plus and minus) go out of sync
    a little. Thus, reducing that 10uf cap and using a high speed op amp might
    be the best way to go, or possibly even simply lowering that 10uf cap.
     
    Last edited: Jun 24, 2009
  4. dave

    Dave New Member

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  5. Hero999

    Hero999 Banned

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    I don't see how removing C3 would improve the transient response; how could it slow down the speed of U1's output transistor?

    I'm pretty certain that a high speed op-amp won't make any difference because the voltage at U2's ADJ pin should remain constant during transients. The op-amp's output only needs to change when the power supply is adjusted which happens very slowly.
     
  6. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,


    No, C3 only slows the turn on transient.

    When do you consider the (steady state) transient to be bad?
    Did you try it with one LM317 alone and compare to the dual circuit?
     
  7. Hero999

    Hero999 Banned

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    I don't care about the turn-on transient, the fact that it might take a couple of hundred ms to turn on is not a problem.

    What do you mean by steady state transient? That sounds meaningless to me.

    The problem is the transient response to an increase and decrease in loading isn't as fast as I feel it should be.
     
  8. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,

    I meant the transient response after the circuit has had time to reach steady state.
    That would be what you were testing for i believe.

    What i was wondering now is exactly how good do you want the transient
    response to be? That is, how many microseconds before the output gets
    back to within 1 percent of the original output?
    In other words, say we have a 10v output, then an extra load is switched onto
    the output and it causes a dip down to 9v, and it takes 100us to get back
    to 9.9v. Is 100us ok or do you need faster?

    What i am doing here is trying to get an idea how good you need this circuit to
    be. What i was proposing is that you test one LM317 without any extras (just
    one 10v output for example) and see if the transient response is good enough
    for your purpose. If not, then certainly a dual circuit will not be good enough.

    If you let me know what kind of transient response you need i can try a few
    different things in order to try to attain that specification, but first i would
    myself test the LM317 by itself to see if it can even do it when it is working
    all by itself (no extra transistors or controlling op amps).

    The transistor replacement circuit is pretty cool, but unfortunately when the
    output is adjusted for a different level of 2x the first level the voltages move
    apart by as much as 0.1v, which is too much for a tracker. For that one
    set voltage though it works pretty nice.
     
  9. Hero999

    Hero999 Banned

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    To be honest, I haven't thought much about how good I want it to be.

    I suppose, even if it's near-perfect i.e. <1:mu:s for the output to be within 5% of the set value at full current, then the inductance of the cables will probably make the transient response worse.

    Please post you're circuit, I'm still interested, even if it isn't perfect.
     
  10. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,


    Ok sure, i'll have to draw it up.
     
  11. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Here's the circuit...

    ADDED:
    I almost forgot to mention that the 661 resistor is changed to adjust output voltages,
    but some adjustment in the 1.18k resistor is also needed to keep the tracking perfect.
     

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    Last edited: Jun 26, 2009
  12. Hero999

    Hero999 Banned

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    That's a cleaver little circuit. I can see why it doesn't track perfectly: it's because of the base voltage isnt' it? Perhaps an old germanium transistor might help.

    It turns out the transient response isn't that bad after all. The negative is slightly worse, it rings a little bit when the load is disconnected.

    Here are some screen shots of the positive supply. For some reason my camera doesn't like photographing the 'scope so sorry about some of the shots being a bit blurred.

    Test conditions:

    The power supply was set for 13.2V. Two 22R thin film resistors were connected in parallel, these which switched on and off by an IRL540 MOSFET driven from a signal generator. The frequency was varied so I could clearly see the load connect and disconnect transients.

    All the waveforms shown below are from the positive side of the supply:

    1 & 2 The whole waveform
    1) 5µs 200mV
    2) 2µs 200mV

    The transient occurring when the load is disconnected.
    3) 1µs 100mV
     

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  13. Hero999

    Hero999 Banned

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    Positive side of the supply again.

    The transient occuring when the load is connected.

    1µs 200mV
     

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  14. Hero999

    Hero999 Banned

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    Here are the waveforms on the negative side of the supply.

    1) 10µs 200mV
    2) 5µs 200mV
    3) 2µs 200mV

    I think it's weird how the ringing only occurs when the load is disconnected, maybe a compensation capacitor might help after all.
     

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  15. Hero999

    Hero999 Banned

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    1 The nasty ringing transient when the load is disconnected.
    1) 2µs 100mV

    2) A the load connected from the + to the - with it set at 8.25V or ±16.5V for a load current of 1.5A.

    Look at how large the turn off transient is 1.7V!

    2) 5µs 500mV

    3) Picture of test device.

    A MOSFET plus thin film resistors were used for this test.

    Note the overly thick gauge wire used for minimal inductance. The thin film resistors also have a very low inductance and are rated to 500MHz.


    EIDT:
    The transient response seems to be better than it was when I tested it previously. The only things I can think of is that this time I've made an effort to minimise the impedance between the load and the PSU and the 'scope was connected directly to the supply terminals rather than across the resistor.

    I think I'll test my load connected to a 12V SLA (which should have damn near perfect transient response) as a control.
     

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    Last edited: Jun 27, 2009
  16. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,

    Oh yeah, i didnt think about exactly why it doesnt track well yet,
    i have only been fooling around with this circuit for a day or two now.
    That's a good idea though...try to find out exactly what is causing
    the tracking problem and maybe find a way to fix it. It very well
    could be the voltage as you are thinking.

    So you are saying that the transient response looks ok now?
    That's good to hear. Maybe this circuit (with the op amp)
    will work out after all which would be very nice.

    BTW thanks for the scope shots. It's nice to see what is
    actually happening for a change.
     
  17. kchriste

    kchriste New Member Forum Supporter

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    It could be due to the fact that the NPN is working with a current source and the PNP with a 5K resistor. So the current through the PNP varies as the output voltage is adjusted but the current through the NPN does not. Try tying the top of the 5K resistor to a separate 5V source such as a 78L05 powered from the +20V instead to see what happens.
     
    Last edited: Jun 26, 2009
  18. Hero999

    Hero999 Banned

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    Here's a picture of the test being conducted on a 7Ah 12V SLA with a 1000µF capacitor in parallel.

    2µs 20mV

    It's still odd how the disconnection and connection transients differ. The connection transient is longer with a lower voltage than the disconnection transient.

    I suppose it's dude to the switching characteristics of the MOSFET. Still the transients are much smaller than with the regulator which proves that my little load transient tester is working properly.
     

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  19. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,


    kchriste:
    But doesnt that 5k resistor have to be connected to the output so it sees
    the feedback? I could try it anyway though.

    Hero:
    That's interesting, maybe there is still some inductance in there some where.
    That spike looks like inductive kick back. Some battery models have a small
    inductance in series with the rest of the model (capacitors and all). I have
    to wonder if it could be that perhaps.
     
  20. kchriste

    kchriste New Member Forum Supporter

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    No, the feedback path is via the 1.18K and 1.2K divider. The transistors, combined with the LM317, try to keep the voltage at the junction of the divider at zero. Thus when the negative voltage goes lower it reduces the voltage on the base of the PNP and thus also the base voltage of the NPN which causes it's collector voltage to rise. That causes the output of the LM317 to rise until the positive voltage equals the negative voltage output by the LM337. The 5K resistor is simply a bias resistor which supplies base current to the NPN so it can turn on. If you keep the bias current through the 5K resistor constant, the regulators may track better.
     
  21. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again kchriste,

    Ok, i havent quite looked at it as thoroughly as you have so i will take your
    word for it, and try it a little later today. I'll get back here with the results.
     
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