Current flows from collector to base of PNP transistor

(1) Your batteries are drawn backwards. The long line is positive.

(2) See the arrow on the emitter? That shows that the base-emitter junction is a diode, with the current flow in the direction of the arrow. Thing is, the base-collector junction is a diode as well, pointing the opposite way. (For a PNP transistor, the base is the cathode of both the BE and BC junctions when they're running as diodes.)

When you remove the supply voltage, there's nothing to pull the base above the collector, so the base voltage is determined by the BC diode. Since the collector is at +4.2V and the base is pulled towards ground by the LM334 current source, the BC junction starts conducting, as you're seeing.

A simple solution is to put a diode between the collector of the transistor and the battery.

Your rationale for transistor behaviour is right, but only applies in normal bias (i.e. when the BC junction is not forward-biased enough to start conducting like a diode, as in your case).

Cheers

Phil Hobbs

Err.. OK, I believe you. Don't worry, they are the right way round in the real circuit.

Right. Thanks. I thought it would be something like that...

OK, but that has the (minor?) disadvantage of a small additional voltage drop and reduced efficiency in the case where I'm charing from a solar panel. Can anyone suggest a fix that doesn't have this problem?

How about putting a resistor between the base and the LM334, calculated such that the maximum current that can flow through the base is the current limit divided by the transistor gain? In this case that would be 250 mA / ~100, which is 10X better than the present discharge (and may also fix the other problem that I described where too much current flows through the LM334 if the supply is weak).

Thanks, Phil.

The right answer for an efficient charger is a switch-mode power supply, but that's well beyond beginner level. If you use a Schottky diode you can save most of the voltage drop.

Cheers

Phil Hobbs