Question about LVD dropping out with quick inrush current.

Discussion in 'Power Electronics' started by wd8cxb, Apr 17, 2017.

  1. wd8cxb

    Thread Starter Member

    Jul 8, 2009
    10
    0
    I am controlling a motor off of a 3S1P 10.8V Li-Ion Battery. It is simple yet it has more than enough design problems for what I thought would be a simple design problem. When turning on the motor under load the battery voltage dips down and triggers the LVD. This condition is even worse at low temperatures. The motor drive a screw drive, at times the instantaneous current can be about 34A. Would not normally be a problem but you can see this happen by viewing an LED that is powered off the battery through a LM340 regulator. If I remove the LVD everything works fine so the LDO is not dropping out because of the inrush current.

    I placed a capacitor on the output of the OP-Amp. I don't believe I need the 1K current limiting resistor but thought it would be a good idea just to be on the safe side as the 1M resistor will limit the current very effectively.

    The capacitor does it's job and limits it dropping out to some extent. However it takes just as long to charge as it does to discharge so there is a delay for drop out but also a delay to turn back on if it dropped out.

    I had hysteresis in the circuit but removed it out because it would cause the motor to stop and shut off until the battery voltage recovered. The motor is controlled by a small microcontroller to provide a few extras features such as current detection, LED delay off, etc but it is full on or off on the motor. The micro also provides a low battery indicator under no load voltages.

    The is no way at the moment to combine the microcontroller with the LVD to ignor the voltage dips under high current conditions.

    Any ideas how to prevent the LVD from dropping out and taking the same amount of time to disconnect and connect?

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  2. crutschow

    Expert

    Mar 14, 2008
    22,244
    6,491
    Add a capacitor from the junction of R2 and R3 to the (-) side of the circuit.
    This will delay the tripping of the LVD circuit.
    The delay will be in the neighborhood of 2.7*C where C is the capacitance is microfarads.
     
  3. ronv

    AAC Fanatic!

    Nov 12, 2008
    3,735
    2,852
    Seems like you could put a signal diode across the 10 meg with the anode going to the output of the op amp. That would make it recover quickly.
     
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