Linear regulator with a pass transistor protection.

Discussion in 'General Electronics Chat' started by illusive, Jul 10, 2015.

  1. illusive

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    Jul 9, 2015
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    Hi everyone,
    I want to make a variable power supply with lm317/338 and a PNP pass transistor. I need a max output of 2.5-3A and 1.25-28V. I will use the transistor for its better power dissipation. The problem is that in this way the short circuit protection of the regulator is useless.
    Here i found a schematic (2nd one) that deals with this problem by using a resistor in the emitter, but i didn't quite get the principle of operation so i can change the values for my requirements.
    http://www.bowdenshobbycircuits.info/page12.htm
    I also want to change the 0.7ohm resistor to a bigger value, so that the regulator won't be needing a cooling heat sink, at least not very big one.
    Any ideas?

    edit: I didn't get the title quote right, but i think you get the idea :D
     
  2. DickCappels

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  3. Jony130

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    Simple use LM350/LM333
     
  4. dl324

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    What part don't you understand?

    The referenced schematic is a questionable design. The Author designed it so the external pass transistor doesn't start conducting until the regulator is passing about 1A and he is counting on the LM317T to handle more than the guaranteed 1.5A.

    A better approach would be to set up a current divider (more correctly, a different one). The ratio should be based on the current your pass transistor can handle. To be conservative, I'd assume that the regulator will pass more than typical current before it starts limiting (say 2A).
    I suggest that you reconsider this. The LM317 also has thermal protection. If you mount the regulator and pass transistor on the same heat sink, the regulator will also protect your pass transistor if it starts throttling due to over temperature.
     
    Last edited: Jul 10, 2015
  5. illusive

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    Jul 9, 2015
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    Is there any info on theory of operation in this configuration with pass transistor? I can only find schematics but no real explanation. I want to know how the current is determined, how much of it will flow through the regulator and how much through the transistor and so on...
     
  6. dl324

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    It's just a current divider. This is from the 1980 Nat Semi Voltage Regulator Handbook:
    regCurDiv.jpg
    The key is finding a diode (D) that matches the forward voltage drop of the transistor BE junction, but a perfect match isn't required so don't waste a power transistor wired as a diode.
     
  7. Jony130

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  8. illusive

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    Jul 9, 2015
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    Thanks!
    Still, all of these calculations need an emitter resistor, but most of the circuits don't have it.
    I mean, for example the formula from the Handbook:
    I1=(R2/R1)*Ireg
    now R1 is 0, or if we count wire resistance lets say very close to 0. So in the moment when the transistor starts conducing the amperage jumps really high. How this works? Or this formula applies only with the diode in the circuit?

    My idea here is to let the transistor do all the hard work and built protection circuit (thermal and short) for it. I have damaged several voltage regulators when relying on their protections. More specifically, when i'm drawing the max current form the power supply (2.5A) i want the regulator to be still cool or warm and not hot. I guess this means that at worst case scenario - 34 volts input to the reg, about 5V output and i draw 2.5A from the whole power supply, the current through the reg should be about 100mA so the power dissipation will be close to 3W and the rest goes through the transistor. Of course here i'm assuming the transistor (or two of them) can handle that but let's focus on the reg.
    This whole thing is just to be safe, i will rarely put the power supply at that high stress.
    At the end, its cheaper and easier to change a damaged transistor rather then a regulator.
     
  9. dl324

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    Post a schematic of an example you want help analyzing.
    The simple resistance ratio requires the diode that approximates the BE junction of the pass transistor. Without it, the current division won't be "constant".
    IMO, that would be more work than it's worth. I recommend letting the regulator provide thermal and over current protection...
    In 40 years, I've only had one regulator die. It was in the first supply I built and I didn't provide sufficient heat sinking.
    I don't think this is a reasonable expectation. The regulator can safely operate to 125C. That will be *very* hot to the touch for most...
    With that kind of power being dissipated in the regulator+pass transistor, a switching regulator might be more appropriate. If it's only for a short time, let the regulator and pass transistor dissipate the power...

    Or maybe you should consider building a triple supply with a switching supply for low voltages at high current.
    It has been my experience that power transistors cost more than LM317T.
     
    Last edited: Jul 11, 2015
  10. Jony130

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    A lot of heat you want to dissipate.
    29V*2.4A ≈ 70W
    And during "short circuit " you get 34V*2.4A = 81.6W. So you need a very big heat-sink.
     
  11. illusive

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    Jul 9, 2015
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    Ok, thanks for the advice.

    For example in this circuit how can i calculate how much current will flow through the transistor and the regulator for a given output voltage and current? Is there some kind of ratio set by the resistor ? I know the transistor wont conduct until there is 0.7V across the resistor, so until that point all the current will flow through the regulator.
    [​IMG]
     
  12. dl324

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    The amount of current through the pass transistor is based solely on the amount of current through the regulator.
    The disadvantage of doing this is that the regulator won't provide thermal and overcurrent protection for the pass transistor. But this is the most typical application of an external pass transistor because it's in many datasheets.
     
  13. dl324

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    Regarding the capacitors across the rectifier diodes... In general I think they're unnecessary at line frequencies. The diodes in this schematic are ultra fast rectifiers with reverse recovery times of 60nS, so even in high frequency applications (e.g. switching regulator) paralleling with caps is probably unnecessary.
     
  14. ian field

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    In the classic current limiting supply; you'd have the B/E junction of a current sense transistor across the current sensing resistor with its emitter to the load, its base to the pass transistor emitter and its collector to the pass transistor base. As current increases the voltage developed on the sense resistor rises, when it gets to 0.7V it biases the sense transistor on and its collector starts to shunt current away from the pass transistor base.

    So far - I've never seen that arrangement used with a pass transistor boosted regulator chip.
     
  15. dl324

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    It isn't required because an appropriately designed regulator with an external pass transistor will take advantage of current limiting and over temperature protection afforded by the regulator.
     
  16. illusive

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    Jul 9, 2015
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  17. dl324

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    The arithmetic given is correct. The assumption regarding when the LM317T will begin current limiting isn't. Do the arithmetic solving for 3A total.

    I wouldn't do it that way. I'd use the current divider from the Nat Semi handbook and make sure the ratio would protect your pass transistor regardless of the actual current that the LM317 actually started limiting.
     
  18. dl324

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    FYI, here are the current limit values based on the datasheet current limiting specs:
    icalc.jpg

    2A is what the Author used, 1.5, 2.2, and 3.4 are the min, typical, and max current limit specs from the datasheet. That means the pass transistor needs to be able to handle 5.6A.
     
  19. illusive

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    Jul 9, 2015
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    Well in order to do the arithmetic i need to know how to find the current through the emitter resistor. For example in the link i have given, the
    Author says:
    " Current limiting occurs at about 2 amps for the LM317 which will drop about 1.4 volts across the 0.7 ohm resistor and produce a 700 millivolt drop across the 0.3 ohm emitter resistor."
    How did he find the current so he can calculate the drop of 700mV across the 0.3 resistor?
     
  20. dl324

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    It's simple arithmetic and he went through the calculation step by step:
    Itot = ((2A * 0.7 ohms - 0.7V) / 0.3 ohms) + 2A = 4.3A
     
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