The transistor is BC 108. Transistor gain is 165
Where did you get that information from?
Is it the Hfe? If so, that's irrelevant to the voltage gain of the circuit, because the Hfe refers to the current gain, which is totally different to the voltage gain.
The open loop gain of this circuit is equal to R
C in parallel with the R
L multiplied by the transistor's transconductance, i.e. the change in I
C vs V
BE.
At room temperature the transconductance of a BJT is equal to:
g
m = I
CmA/25.6mV
Let's look at the top circuit.
Calculate the approximate bias voltages and therefore the collector current:
V
BE = 0.6V the usual diode drop.
V
B = 12*2.3/(8.2+2.3) = 2.6V
V
E = V
B - V
BE = 2.6V - 0.6V = 2V
I
E = V
E/R
E = 2/(386+235) = 3.22mA
I
C is near enough equal to I
E, as the Hfe is high.
R
L in parallel with R
CR
C|L = (1800*80*10
3) / (1800+80*10
3) = 1760
Now the open loop gain:
g
m = 3.22/25.6 = 0.126 = 126mS
A
V_OL = R
C|L*g
m = 0.126*1760 = 222
Finally the closed loop gain:
A
V_CL = A
V_OL/(1+A
V_OL*β)
β = R
E/R
C|L = 386/1760 = 0.219
A
V_CL = 226/(1+222*0.219) = 4.48
That's only a little less than R
C/R
E, which is 4.66