Overcurrent Detection circuit for PSU

Discussion in 'The Projects Forum' started by witssq, May 13, 2010.

  1. witssq

    Thread Starter Active Member

    Mar 29, 2009
    48
    0
    I would like to connect to Regulator EN pin, when excess current is drawn to shutdown regulator.
    Some schematics is found but seems to be not suitable.
    The flow of Power supply is Bridge Rectifier -> Conditioning Capacitor -> (7V/7.5A) -> Linear LDO Regulator. I'd like to insert detection circuit between capacitor and Regulator,if possible, in parellel for low voltage drop.
    Then 5V logic overcurrent detection line would be connected to EN pin of regulator.

    Could you advise me the above overcurrent protection circuit?

    Thanking you in advance

    SunSung Hwang
     
  2. oidium45

    Active Member

    Apr 24, 2010
    130
    8
    This chip is internally limited and protected. It will automatically shut down when overloaded. Read page 8 of the data sheet that you supplied.
     
  3. witssq

    Thread Starter Active Member

    Mar 29, 2009
    48
    0
    Thanks for comment.


    Whether IC is defined with overcurrent protection, I'd like to protect for low current value than noted because I should use this chip.

    Could you help me?

    SunSung Hwang
     
  4. oidium45

    Active Member

    Apr 24, 2010
    130
    8
    So what you are looking for is a simple circuit that will connect to the PIN1 (EN) of the MIC29712 and trigger the IC to turn off when a specific amount of current is reached or exceeded?
    Would it be acceptable to just limit the output current of the IC?

    Anyway, I don't think that I can help you. But, you may want to re-post this thread and reword it a bit differently and you may get more assistance from other people.
     
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  5. oidium45

    Active Member

    Apr 24, 2010
    130
    8
    Actually you should do a search online and look into the following. Maybe one of these will suit your needs?
    Current sensing resistor
    SHUNT (ammeter shunt)
    Overload protection relay
    Over current switch

    Just a thought...
     
    Last edited: May 13, 2010
    witssq likes this.
  6. k7elp60

    Senior Member

    Nov 4, 2008
    478
    69
    You could conect a low value resistor between the filter capacitor and the Vin terminal. Monitor the voltage drop with a comparator and when the current reaches the desired value(IR drop on resistor) the comparator would shut off the regulator. You no doubt would have to add a latch otherwise the circuit will turn on and off all the time when the specified current is reached.
     
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  7. sage.radachowsky

    Member

    May 11, 2010
    241
    38
    I think to do what you describe, you need to have a current-sense resistor at the bottom of the load.

    Say it is 0.1 Ohms... this will read 50 mV at 500 mA. I don't know what your current limit is, but that's good for 500 mA range, without too much droop on your regulated voltage. Lower it to 0.05 or 0.02 if you need more current. You could ground the regulator before the sense reisstor, and there will be *no* voltage droop.

    Then you need an accurate op-amp to do an inverting amplification with gain, like say 20X gain... then 50 mV will become 1V.

    Then you need a voltage reference and a comparator. You can use a voltage divider off the 5V supply. Anyway, compare the reference to the amplified current sense signal, and use the output of that to drive the Enable pin of the regulator you're controlling.

    That should do it... it should be a voltage source with an upper current limit, and it will lower its voltage when the current limit is reached.




    I assume you have another power supply in the circuit... let's say it's another linear 5V regulator.
     
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  8. witssq

    Thread Starter Active Member

    Mar 29, 2009
    48
    0
    Current sense resistor seems to be too expensive for this application.
    If we are using 0.5Ohm resistor for current sense, resistor power rating would be 0.5 x 15V x 15V = 450 Watt resistor as I know. It is not easy to find, and seems too expensive for PSU.

    As your reference, my PSU rating is +/-15Vx3A, +5Vx7.5A.

    Please comment about this matter?

    SunSung Hwang
     
    Last edited: May 14, 2010
  9. Bychon

    Member

    Mar 12, 2010
    469
    41
    mouser part number 588-MCS3264R005FER .005 ohms 1% 40 cents
     
  10. sage.radachowsky

    Member

    May 11, 2010
    241
    38
    You make a mistake in the calculations -- the sense resistor will never see 15V across it. It will see 50 mV across it maximum.

    What do you want to limit your power supply to?

    If you say 5A then you can use a 0.010 Ohm sense resistor, and it will come to 50 mV when there is 5A flowing across it. Or use the one Bynchon mentioned (0.005) and it will come to 25 mV at 5A... then set your reference into the comparator accordingly.

    Hope this helps.
     
  11. witssq

    Thread Starter Active Member

    Mar 29, 2009
    48
    0
    Could you upload sample schematics under consideration so that I can read your comment?

    Thanking you in advance.

    yours
     
  12. witssq

    Thread Starter Active Member

    Mar 29, 2009
    48
    0
    Under condition that the PSU rating is 5V 7.5A.

    1) At the above, you may mean that power dissipation is 7.5Ax7.5Ax0.01Ohm = 562mW. Is this right?

    2) When sense resistor is connect in series with load, if load is shorted to
    ground(zero resistance) at the worst case, power would be
    5V*5V/0.01. Is this not right?
    If it is correct, please advise how I could avoid this problem.
    for example, whether I should use parellel resistor.

    3) I found current sense IC as attached, SC310(positive only) and
    LT6105 (dual polarity). At this case. the above 2) question remains
    too.

    4) If available, please advise me a link for your schematics sample

    Thanks,
     
    Last edited: May 15, 2010
  13. Bychon

    Member

    Mar 12, 2010
    469
    41
    1) 562 milliwatts is correct.
    2) when load is shorted, 5 volts does not exist anymore. Even if it did, the resistor would dissipate 250 milliwatts. Why is this a problem that you need to avoid?
     
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