On Mon, 05 Mar 2012 15:39:48 -0800, Winston
put finger to keyboard and composed:

Franc Zabkar wrote:

(...)

As for R22, I don't understand how an open circuit could reduce the
voltage on the base of the transistor.


Emitter current through R24 is adjusted by the setting of
R22, per the note to the right of R24. As R22 is adjusted
higher in resistance, we can expect less current to flow
from the base to emitter, causing AF126 to tend towards
cutoff. (The base becomes less negative *in relation to
the emitter*).

AISI, the voltage readings on the circuit diagram are using the
positive terminal of the 9V battery as the 0V reference. Therefore all
the voltage measurements are negative numbers, which means that the
transistor is being turned on harder during the fault condition. The
voltage on circuit ground would be -9V.

Or am I having a brain fart? (Quite likely)

- Franc Zabkar
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