The cap is discharging while the mosfet is in the conducting state
(shorting its power source). Q one's bias is "on," with its base
close to zero and emitter positive. Resistor is also discharging the
gate of the mosfet's capacitance.
The signal diode is there to remove the discharge path through the
shorted power source - the objective isn't to discharge the cap
rapidly - that's what is holding the gate of the mosfet up and keeping
the signal lamp on.
It seemed that the diode could have been added only on purpose to
remove that path, but I didn't understand why you'd care to. The
discharge path then must only be through the emitter of the bjt and
via the base and the 220k and also from the emitter of the bjt and via
the collector and the two resistors there. Okay. So you wanted to
remove the 100k from being in parallel with all that. Got it.
I kind of wonder, without thinking too closely right now about the
relative value of the currents via the bjt vs the 100k, whether that's
all that important to add the diode. The charge rate is through the
100k. Having a large part of the discharge also through it seems to
suggest roughly equal timing at first blush. Did you find the diode
really helping here?
(It's cheap to add, I just wonder more if you did do the analysis
first and inserted it on the basis of theory instead of practice and
if you did, then I'm going to sit down and work through the details on
paper to see the same reasoning. I might ask for help, if I have
trouble there.)
Jon
