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In article <3834@phri.UUCP> roy@phri.UUCP (Roy Smith) writes:
=  First, the DC model:
=
=
=                  o +6V
=                  |                          [key to ascii art: < and > are
=                  >                          resistors, V are ground points,
=                  < 10k       C              and the transistor is NPN]
=                  >         /              
=                  |     B |/
=                  +-------|
=                  |       |\
=                  >         \ E
>                  < 2.2k     |
>                  >          <
=                  |          > 1k
=                  V          <
=                             |
=                             V
=
=	There are, of course, connections to the collecter, but all of them
=use a capacitor as a series element, so there is no DC collector connection
=at all.  How does one go about analysing a DC circuit like this?
=
=	After taking the thevinin equivalent of the 10k/22k voltage divider
=you get 1.1V through 1.8 k.  Assuming Q is active, you use KVL to write
=
=	1.8V = 1.8k*Ib + 0.7V + 1k*Beta*Ib
=and solve for Ib.  This gives you an open-circuited current source of
=(Beta+1)*Ib at the collector.  Obviously, with Ic=0, Q is cutoff, but
=that's not very useful.  Am I missing something?

Ic=0, therefore Ib=Ie  (Ie to be taken out of the emitter).
Voltage division at the base gives 6*2.2/(10+2.2)=1.08V. Therefore,

        1.08V = 1.8K*Ib + 0.7 + 1K*Ib

You have nothing to do with Beta in the DC model, because the transistor
is not in the active region. Base acts like a forward biased diode.

-=-
Bulent Abali
Ohio State Univ., Dept.of Electrical Eng.
2015 Neil Av. Columbus, Ohio 43210
abali@baloo.eng.ohio-state.edu