Answer
1— When
both transistors are turned on, the base voltage of Q1
is at 3.7 V (VE + VBE) and its collector
is around 3.4 V (VE + VCE(sat)).
The
base of Q2 is at 4.3 V (5.0 V – VBE), so the
voltage across R4 is 0.9 V and the current through it
is 208 µA.
Treating
R1 and R2 as a Thevenin source with impedance 66.7 K (and
ignoring the effects of R3), the source voltage is 8 V
at the lowest nominal input voltage of 12 V. The minimum
base current of Q1 is therefore 64 µA, so this transistor
is operating with a collector current that’s only about
3.3 times its base current.
Assuming
that the optoisolator’s LED has a forward voltage drop
of 2 V, the collector current of Q2 is 2.6 mA through
the LED, plus 0.46 mA through R5, for a total of 3.06
mA. This is about 18.9 times its base drive.
Contributor: David Tweed
Published
January 2004
