Q.CE amplifier circuit using germanium transistor is shown in figure given below. Calculate i) $V_2$, ii) $I_E$, iii) $r_e'$, iv) $A_v$, Given $r_e' = 52\,\text{mV}/I_E$.

For this voltage-divider biased CE amplifier the base voltage sets the emitter current, from which the ac emitter resistance $r_e'$ and the voltage gain $A_v = R_{c(ac)}/r_e'$ follow, where the ac collector load is $R_C \parallel R_L$. …

[!TLDR]

The divider gives $V_2=3.66\,\text{V}$, so $I_E=3.34\,\text{mA}$, $r_e'=15.57\,\Omega$; with an ac collector load of $R_C\parallel R_L=10\,\text{k}\parallel5\,\text{k}=3.33\,\text{k}\Omega$ the voltage gain is $A_v=R_{c(ac)}/r_e'\approx214.1$.

In a common-emitter amplifier this Karnataka II PUC Electronics problem tests the small-signal model. The dc bias fixes the operating point; the ac gain then depends only on the ac collector load and the transistor's internal emitter resistance $r_e'$.

Step 1 - Base voltage (voltage divider):

Step 2 - Emitter current (germanium, $V_{BE}=0.32\,\text{V}$):

Step 3 - ac emitter resistance (germanium uses $52\,\text{mV}$):

Step 4 - Voltage gain. The ac load seen at the collector is $R_C$ in parallel with the coupled load $R_L$:

$R_{c(ac)} = R_C \parallel R_L = 10\text{k}\parallel5\text{k} = 3.33\,\text{k}\Omega,$ …