Willen
Well-Known Member
Here's a simple 12V battery charger. I simulated first to understand it but became MORE confuse than before!
I used to think that when battery gets full power then the transistor starts conduct and will short the ADJ pin of the Reg IC to the Gnd then the charging will stop. But how transistor conducts when bettery gets around 13.8V. Because when bettery starts to get 13.8V, the 1 ohm current limiting resistor has just few mV (voltage across it) or, upper side of the resistor is just few mV. It means the base of transistor gets just few mV too. Which means the voltage difference between emitter and base of the transistor is just few mV. The main question is how the transistor is conducting and stopping the IC, with just few mV base voltage? (base needs at least 600mV to conduct)
Another again, when bettery has around 12V (or less), voltage across 1 ohms resistor is around 650mV. It means 650mV is in the base of the transistor. Now why transistor is not conducting when it is getting 650mV base voltage? Which thing making me silly?
I used to think that when battery gets full power then the transistor starts conduct and will short the ADJ pin of the Reg IC to the Gnd then the charging will stop. But how transistor conducts when bettery gets around 13.8V. Because when bettery starts to get 13.8V, the 1 ohm current limiting resistor has just few mV (voltage across it) or, upper side of the resistor is just few mV. It means the base of transistor gets just few mV too. Which means the voltage difference between emitter and base of the transistor is just few mV. The main question is how the transistor is conducting and stopping the IC, with just few mV base voltage? (base needs at least 600mV to conduct)
Another again, when bettery has around 12V (or less), voltage across 1 ohms resistor is around 650mV. It means 650mV is in the base of the transistor. Now why transistor is not conducting when it is getting 650mV base voltage? Which thing making me silly?
Attachments
Last edited: