Looks like a rather odd bias circuit there.
From what I gather the TR only turns on when
the AC signal is at a voltage that is 0.6V greater
than the voltage at the emitter.
Based on TR operating rules;
Rich (BB code):
++
/
+-{
\
-
or as is here;
Rich (BB code):
+
/
- -{
\
--
I'm stumped as to why there are two negative
DC supplies at -15V each.
Ie at 0.5mA ?
Ie includes Ib and Ic.
In this case Ic also goes through RL.
So Ie = Ib + Ic + Irl?
IC is ~ Ie, usually and thus Ib is neglected.
But Irl, that could be significant.
Perhaps we could cut Halim a break and assume their orientation of the supply connections is incorrectly drawn. Also forget about any missing coupling capacitors & so forth.
The emitter leg supply is -15V.
Presumably Halim just wants the steady state (quiescent) bias conditions - so we could assume the AC signal input is 0V for the moment.
The steady-state (quiescent) emitter current would be found from
\(15-Vbe=IeRe+Ie\frac{Rb}{(1+\beta)}\)
or
\(Ie=\frac{(15-Vbe)}{Re+\frac{Rb}{(1+\beta)}}\)
If Rb, Vbe, β are known and Ie=0.5mA one would re-arrange the equation to solve for Re. Which might be a useful exercise for Halim.