LM317 as current limiter question?

Discussion in 'General Electronics Chat' started by Dyslexicbloke, Jun 7, 2011.

  1. Dyslexicbloke

    Thread Starter Active Member

    Sep 4, 2010
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    Hi folks,

    I know that an LM317, 317L in this case, will work well as a current limit but in order to do so it needs to drop circa 1.2V across the sense resistor.
    See circuit A

    What I would like to know is can I do something like circuit B to reduce the overall voltage drop whilst limiting is not active.

    [​IMG]

    I have a small monolithic DC/DC converter with a maximum output of circa 160mA and want to protect it with but with the minimum voltage drop when the current is below Imax.

    Any and all suggestions welcome …
    Al
     
  2. tom66

    Senior Member

    May 9, 2009
    2,613
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    A better way to do this would be to use a series pass transistor and an op-amp, in a current regulator configuration. See screenshot.
    [​IMG]
    The 1 ohm pass resistor drops 1V per amp. If you replace it with a 0.1 ohm or 0.01 ohm resistor it will drop 100mV/A and 10mV/A respectively; adjust the control input respectively to account for this.
     
  3. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    From what I simmed it looks like it will work, but beware both A and B are current SOURCEs, not current limiters.
     
  4. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Unlike those, this IS an actual current limiter: http://forum.allaboutcircuits.com/attachment.php?attachmentid=5070&d=1225117974

    I designed it for high voltage operation, thus the need for very low gain in X2 and fixing that in X4 (to keep the measured voltage inside the rails). You could merge those two into one opamp with 1:1 gain if you want to current-limit a rail inside the opamp´s range.
     
  5. Dyslexicbloke

    Thread Starter Active Member

    Sep 4, 2010
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    Thanks both ... good stuff.

    Kubeek ... I dont understand the distinction .. curent source / current limmit.
    What am I missing?

    Given that my input voltage is controlled at 12V surly the only effect of A or B would be to reduce its output below that when the current threshold was reached.
    Obviously both will drop voltage in any condition acros the sens resistor and the reg.

    My intention with B was to minimise the voltage drop when current limmiting wasnt in play.

    Have I got something fundamentally wrong with my reasoning.
    I like the opamp solution by the way if a darlington pare dosnt work I will try that.

    Lastly, am I correct in assuming that a base current limmiting resistor isnt necessarry because the reg will begin to turn off when the transistor starts to conduct?

    Thanks
    Al
     
  6. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    The difference is that the current source mainly tries to force the set current into the load, but since it doesn´t have enough voltage to do it when the current demand is low, you are left with the 1.2 or 0.65V drop in A or B.

    Real current limit leaves the mosfet fully open all the time, so the drop is only on the Rds of the fet and the sense resistor, which both can be very low. When the current reaches the limit, the mosfet starts closing to keep the current below the threshold.

    So if you want to minimize the drop the opamp limit is the better way, even though it will take a little more space on the board.

    PS darlington will keep the drop on the resistors on 2x0.65, so youre back to square one. The least drop will be with the single PNP, but you will still lose 0.2W at max current.

    PSS i don´t think you need the base resistor, but my model of voltage regulator is flawed so I can´t be sure.
     
    Last edited: Jun 7, 2011
  7. Dyslexicbloke

    Thread Starter Active Member

    Sep 4, 2010
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    OooPs,
    Spoke too soon RE the opamp solution ....
    My load circuit is charging FET gates and is floating so cant measure current with a sense below the load.

    I guess I could still use an opamp to detect diferential voltage acros a sense resistor elsewhere though but I would need one that was capable of handling inputs right up to its supply rail wouldnt I.
    Common mode including supply I mean.

    Al
     
  8. #12

    Expert

    Nov 30, 2010
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    Leave the resistor in or the opamp will try to send the whole 200ma through the base emitter junction if something goes wrong with the collector or the +12 connection.

    That circuit in post #2 is a very good way to do this.

    I would like to mention that, while repairing a garage door opener, I discovered a circuit that intentionally used a 317L simply for its current limit, same as your circuit A in the first post. I had never thought of that use. I just stumbled across it one day. Point is, it's a valid use of the 317L chip.
     
  9. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    The limiter I posted was good for about up to 100V rail, and the current sense is in the high side. IIRC the circuit need +/-15V for the opamps, but it should be able to work from single 15V supply with some minor changes.
     
  10. Dyslexicbloke

    Thread Starter Active Member

    Sep 4, 2010
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    Good stuff folks, thyanks ... I think I have something to work with now.

    I will strat with B, test performance, given my stock parts, and then look at improvinng it by using an opamp rather than a transistor to see how low I can get the value of the sense resistor.

    I think I have some single rail JFET opanps somewher ... I'l give them a go and see what happens.

    Thanks
    Al
     
  11. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Here is a 160mA current limiter with about 180mV dropout voltage. You can make it lower by using a lower resistor value for R1, and changing the input reference divider accordingly. The limiting factor is mostly the input offset voltage of the op amp. This one is not great.
    The op amp needs to have rail-to-rail input and output. I picked this one because it's cheap. It cannot run on a 12V supply (5.5V max), so I had to pick a logic-level p-channel MOSFET. This one is available at Mouser.com.
     
  12. Dyslexicbloke

    Thread Starter Active Member

    Sep 4, 2010
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    Thanks so much for the time and effort .... Some interesting concepts in that circuit which I had not considdered and probably wouldnt have thought about to be honest.

    Sorry I didnt reply earlier, I was lashing up a testbed which I need for this weekend but the final installed unit is going to have to include far better solutions than I have cobbled together thus far and I expect I will be using a derivertive of your ideas.

    I have realised that I have a significant problem with a power supply for my circuit if you are up for a conversation RE that .... New thread though I think.

    Summerising the problem as simple question isnt going to be easy to start with ..

    Thanks
    Al
     
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