Common Emitter Impedance

Please find for the Common Emitter, the values of $R_1$ and $R_2$ the voltage gain as and its input and output impedance. Values are $I_E=1.5 mA$ and $beta=100$.

$I_c=beta I_b$ and $I_e=I_c+I_b=(beta+1)I_b$

$beta =I_e/(beta+1)=(1.5 m)/101=14.85 mu A$

Current through divider $R_1+R_2$ need to be at least 10 times bigger than base current for the base current not to modify the polarization of the base itself. take $I_{divider} = 1 mA$ (about 70 times bigger than $I_b$). On the lower loop

$V_b =V_{be}+V_e=0.7+I_e*R_e=0.7+1.5*10^{-3}*500=1.45 V$

$V_b =V(R_1)=1.45 V$ so $R_1=V(R_1)/I_{divider}=1.45/0.001=1.45 KOmega$

$V(R_2)=12-V(R_1)=12-1.45=10.55 V$ and $R_2=10.55/0.001=10.55 KOmega$

(Standard value for resistors are $R_1=1.5 KOmega$ and $R_2=10 KOmega$

Of course one can take $I_{divider} =150 mu A$ (10 times bigger) and re-compute the values of $R_1$ and $R_2$. Gain is

$G=frac{Delta V_{out}}{Delta V_{in}}=frac{Delta V_c}{Delta V_b} =frac {-g_m R_c}{g_m(R_e+r_e)}=-frac{R_c}{R_e+r_e}$

where $r_e$ is the emitter intrinsic resistance. Usually

$r_e=V_{thermic}/{I_e}=(0.026 V)/(0.0015 A)=17.3 Omega$

Without $C_e$ the gain is $G=-R_c/R_e=-4.7/0.5=9.4$

With $C_e$ the gain is $G=-R_c/r_e=-4700/17.3=272$

In reality not all the value of $r_e$ is decoupled by $C_e$ so the gain is somewhere the minimum and the maximum values above.

For the input impedance the total emitter resistance $(R_e+r_e)$ is transmitted into the base circuit multiplied with $beta(R_e+r_e)$ so that $R_{in}=R_1||R_2||beta(R_e+r_e)$.

The voltage sources are “in short” like when computing the Thevenin equivalent resistance. In DC

$R_{in}=10.55 K||1.45 K||100(500+17.3)=1.244 KOmega$

In AC, $R_e$ is missing (being decoupled by $C_e$):

$R_{in}=10.55 K||1.45 K||100*17.3=734 Omega$

The output impedance is equal to the collector impedance

$R_{out}=R_c=4.7 KOmega$