Disconnecting load from battery while charging the battery

Discussion in 'General Electronics Chat' started by Johnny1010, Mar 23, 2016.

  1. Johnny1010

    Thread Starter Member

    Jul 13, 2014
    83
    2
    I am designing a switching circuit that disconnects the load from battery when it is charging and connects battery back to load after charging is removed. I am doing this with the help of PNP and I am using TP4056 battery charging module.In the figure attached, IN+ (5V input from USB is applied at base of PNP), bat+ (output voltage for battery charging module and battery's +ve terminal is connected at emitter), Collector is connected to voltage regulator (3.3V regulator). Battery is of 3.7V.

    When 0V (in case charging is disconnected) applied at IN+ (base of PNP) then PNP should operate in active mode and 3.7V should appear on emitter as well as on collector but don't know why collector (connected to voltage regulator's input) shows 2.5V?. I think voltage regulator is loading the PNP but can't understand?

    Mini USB Lithium Battery Charging Board Charger Module 5V 1A-1500x1500.jpg Circuit.JPG
     
  2. crutschow

    Expert

    Mar 14, 2008
    13,001
    3,229
    If you apply 0V (ground) to the transistor base without a series resistor you will likely zap the transistor, since there's nothing to limit the base-emitter current.
    A base resistor is need to limit the current to about 10% of the maximum collector current (for good saturation of the transistor when on).
     
  3. Johnny1010

    Thread Starter Member

    Jul 13, 2014
    83
    2
    Now collector voltage is dropped to 1.5V (rather it has to be 3.7V)?
     
  4. crutschow

    Expert

    Mar 14, 2008
    13,001
    3,229
    Are you referring to a simulation or a real circuit?
    What is the collector current?

    Where did you get a value of 10kΩ for R2? :confused: That's way too large.
    I stated in post #2 that the base current should be 10% of the collector current, so for 100mA collector current, Ib shoud be 10ma, giving a value for R2 of (3.7v-0.7V) / 10ma = 300Ω.
     
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