Two Quadrant Power Supply MOSFET Help

Discussion in 'The Projects Forum' started by Heisler, Apr 4, 2013.

  1. Heisler

    Thread Starter New Member

    Apr 4, 2013
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    Greetings,

    I'm working on a 2 quadrant power supply based on the LT1970. It is a pretty neat part with configurable source and sink current limits. I've based my design on Linear's example circuit but am having a few issues. I'm using external FETs since I need more current than the LT1970 can thermally handle and the FETs are easy to put big heatsinks on. I'm currently using +/-8V supply rails and everything mostly works. The issue I'm having is I can't output greater than about 4.7V and I need to get to at least 5.75V. The attached diagram shows the basic topology and FETs I'm using (except I'm using +/-8V). I've increased R2 and R3 to 200 Ohms which does help some but not enough. If I increase much higher it starts to oscillate. Any suggestions would be appreciated..... my analog skills are very rusty.:)

    Many thanks.
     
  2. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
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    Are you using all of the caps as shown in the schematic?
    How are your grounds connected?
    Is the circuit built on a bread board?
     
  3. panic mode

    Senior Member

    Oct 10, 2011
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    increase your supply voltage, +/-8V is too low, those mosfets need considerable Vgs to turn on.
    if you have to use low supply voltage, consider replacing MOSFETs with BJTs.
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
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    How about logic level MOSFETs?
     
  5. Heisler

    Thread Starter New Member

    Apr 4, 2013
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    The circuit is on a custom PCB.

    I don't want to raise the rails since the parts will then have to dissipate even more power at low output voltages. I actually want to reduce the rails eventually.

    I only need to handle just over 500mA. Any suggestions for a different FET or BJTs that would fit the application.

    Thanks.
     
  6. panic mode

    Senior Member

    Oct 10, 2011
    1,320
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    logic level mosftes are fully on with 3-5V.
    BJTs are fully on with 0.7-1.4V (in case one uses Darlingtons).
    if the goal is to reduce supply voltage, the best bet is going with BJTs.
    perhaps this thread gives you some ideas:
    http://forum.allaboutcircuits.com/showthread.php?p=84026
    the circuit from origianl post will work for BJTs too with slight adjustments (values for R2 and R3 for example).
     
  7. panic mode

    Senior Member

    Oct 10, 2011
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  8. Heisler

    Thread Starter New Member

    Apr 4, 2013
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    I'm looking at using On Semi D44H11/D45H11 Complementary power transistors. How do I best determine the resistor values?

    Thanks for all the ideas.
     
  9. panic mode

    Senior Member

    Oct 10, 2011
    1,320
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    resistor value is R=V/I

    where V is Vbe of transistor (0.7V)
    and I= max output current you want from IC alone (if there was no transistors).

    suppose your IC can handle 0.5A but you want to be conservative and choose 0.4A as max current that IC can output.

    that current has to come from somewhere and that is power pin of the IC. if the output is positive current (flowing out of output), then it gets into IC from the positive rail and - through R2.

    when voltage across R2 reaches Vbe, transistor turns on and provides additional current. this way output current is combination of currents coming through IC output AND transistor Q1.

    this high current raises voltage drop across load and IC detects that through feedback Rf which means that IC will compensate and the output voltage/current stabilize at some equilibrium point.

    if Vbe=0.7V and Ir2=0.35A then R2=0.7/0.35 = 2 Ohm.

    of course if you choose darlington transistors, you get different Vbe but principle is the same.
    Now that you figured out R2, repeat same computation for R3 - just kidding, it's the same as R2.
    here is practically identical circuit
    http://www.circuitstoday.com/40w-audio-amplifier

    The next thing to keep in mind is transistor ratings. There are 3 numbers you need to become familiar with: V, I and P

    Example: your transistor is rated for absolute max. 10A, 80V and therefore it should handle

    P=VI=800W

    cool eh?

    nope, that was a trick question, 800W would kill it instantly.
    you cannot have both high voltage and high current at the same time. ;-)
    max rated power is 50W.
    50W=50V*1A
    50W=25V*2A
    50W=10V*5A
    50W=5V*10A
    50W=8V*6.xxx A

    if your supply is +/-8V, then Q1 will see up to 16V (if output is shorted to neg. rail and that means some 3Amax through transistor(plus current from IC).
     
  10. Heisler

    Thread Starter New Member

    Apr 4, 2013
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    0
    OK.... thanks for the ideas so far. I've replaced the FETs with the On Semi D44H11/D45H11 BJTs. I've changed R2/R3 to 10 Ohms and seem to have it mostly working.... however.....

    1. I'd like to pass as little current through the LT1970 as possible. It is easier for me to dissipate heat from the TO-220 package BJTs with nice heatsinks. If I increase R2/R3 back to 100 Ohm things I don't get the proper output. I find this odd since I should still be pulling the same current from the base of the BJT(PNP on the + side and NPN on the negative). It seems like, theoretically, I should be able to remove R2/R3 so all current goes through the BJT. I'm wondering what part R4 and R5 play in all this. It looks to me like R4 will limit the current through the base of the BJTs and R5 will reduce the current as the output voltage increases.

    2. I'd like the circuit to handle a short circuit and basically source current at the limit specified by the VCsrc signal. Regular loads work fine but a short causes oscillation. Any ideas on how to tweak the feedback to prevent this?

    Many thanks.
     
  11. panic mode

    Senior Member

    Oct 10, 2011
    1,320
    304
    read the previous post. if you want less current through IC and more through transistors, you need to turn on transistors sooner. this means increasing values for R2 and R3

    R=V/I,

    you know that V=0.7V then just change desired IC output current
    instead of I=0.5A use I=0.1A

    R=0.7/0.1 = 7Ohm

    want even less?

    try I=0.02A
    R=0.7/0.02=35Ohm


    w have seen sample schematic but how about posting YOUR schematic (actual build)? don't try to talk about it ("i just changed blah blah..), post your circuit.
     
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