LM317 reverse voltage/current

Discussion in 'General Electronics Chat' started by Kay_niv, Jan 17, 2015.

  1. Kay_niv

    Thread Starter New Member

    Jan 17, 2015
    I'm currently doing a project where I need to check the battery protection circuit of a LiPo .
    My current setup is a 12v supply from which by making use of LM317,Arduino(i vary the Vadj) to get a voltage range of 2V ~ 5V output.

    when i connect the charger it would produce 4.2V at the 3.7V (output of the LM317).
    I'm not clear as to how LM317 handles reverse current and how much it can handle ?

    My aim is to check the outputs Discharging,Charging over the different voltage range ( 2V ~ 5V).
    How do i protect my test setup so that varying the bat_voltage(LM317 outputs) does not affect/burn the setup when i connect the charger.

  2. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
    The LM317 will be damaged if the Vout exceeds Vin by more than a volt or so.
    A common method is to connect, say a 1Amp diode in reverse parallel across the LM317. ie: diode Cathode to Vin side of Lm317 and Anode to Vout.
    So the maximum reverse voltage across the LM317 will be clamped at ~0.7V
  3. GopherT

    AAC Fanatic!

    Nov 23, 2012
    If you are measuring voltage of the battery, you can put a pair of 100k resistors in series from your battery (+) to ground and measure the mid-point and double the measured voltage instead of measuring at the battery (+). Battery charging is fairly slow so you won't suffer too much if you give your ADC a bit of time to charge through the 100k resistors (that is, you may need more than a few microseconds for the ADC to stabilize since you are measuring though a 100k resistor).
  4. bertus


    Apr 5, 2008

    Here is the part about the protection diodes from the datasheet:


  5. ian field

    AAC Fanatic!

    Oct 27, 2012
    Don't pass reverse current through any 3-terminal regulator. Bypass the input and output terminals with a diode oriented so its reverse biased in normal operation.
  6. ronv

    AAC Fanatic!

    Nov 12, 2008
    The way I understand the circuit he has 12 volts on the input of the regulator and has set the output to 3.7 volts. Then attaches the charger to the output (and the battery). In this case no reverse current. I think the 317 will just turn off???
  7. ian field

    AAC Fanatic!

    Oct 27, 2012
    With a battery connected to the output - if the input isn't supplied with anything, you have the potential for reverse current in the regulator.

    Charging the input capacitor via the regulator's output is enough current to damage the regulator.

    It needs a protection diode - either bridging the input & output and oriented to be reverse biased in normal operation, or a forward conducting diode in series with the output.