Yes, that is exactly right. The formula for the maximum current is 0.6V/R7 which in the circuit of post #76 works out to be 0.6V/4.7R = 0.128 Amps
No problem BM.
From the formula, Imax =0.6V/R7 you can see that the higher resistance R7 is the less current will flow into the battery.
How does this work:
(1) With no current flowing into and out of the battery on charge there is no voltage drop across R7, so Q2 is turned off (Q2 will only turn on when its base/emitter voltage reaches 0.6V).
(2) As a result Q2 has no affect on the LM317 regulator output voltage. Thus the LM317 acts like a normal voltage regulator.
(3) But if you connect a discharged battery across the charger output terminals, the battery will try to take a large current.
(4) But as soon as the current flowing through R7 reaches 0.6V, Q2 turns on and the current from Q2 collector drags the LM317 ADJ terminal down with the result that the charger voltage will drop and thus reduce the charging current.
(5) After a while of being charged with this constant current, the battery voltage will start to rise until it reaches the charger output voltage that you have set.
(6) The battery will continue to charge, but the current will gradually taper off as the battery reaches full charge.
I hope that helps to explain how the circuit operates.
spec