Dear Sir, please answer
I tested circuit (please, see attach, "picture 2"). I set 3A initial current for charging 1-cell lead acid battery (2.3V). This circuit works. VT1 transistor and VD1 diode must be placed on the heatsink.
I would like to remove the VD1 diode. The 2.3 V battery is constantly connected to the device, and the 5.5 V power supply is periodically turned off.
If to remove VD1 diode in my circuit, then collector-base junction of the VT1 will be shifted in the forward direction, when power supply 5.5 V is turn off (as the battery remains connected to the device). Therefore, I can’t remove VD1 diode in my circuit.
If I use external Quasi-Darlington pair or NPN Emitter-Follower (please, see page 14 of the BQ24450 datasheet), then, if to turn off 5.5 V supply, then emitter-base junction of the Q2 (or Qext) will be closed.
QUESTIONS:
1) Can I remove VD1, if I use this external Quasi-Darlington pair ? Since this pair has a large current gain, is it necessary to provide measure to prevent oscillation ?
2) Since the input voltage of 5.5 V is enough for the NPN Emitter-Follower circuit to work, can it be better for me to use this NPN Emitter-Follower circuit (without VD1 diode) ?
3) What max current I can set through Rp resistor (NPN Emitter-Follower circuit )? Is it possible to set 50 mA current?
TIA
Sincerely,
Vladimir Naumenkov
www.agat.by