# Whether Internal source resistance will have effect on the circuit

Calculate the voltage gain for the network of Fig. 3.12 if the source has an internal resistance of 100 in series with Vi.Vi=500mV,R= 1K ?

For this problem I had tried a solution but I am not sure whether it is correct or not? Please Help. Below is the solution that i have found

Ii = Vi/Ri = 25mA For this ckt, Ie=Ii=25mA

Since, Ic is approx equal to Ie, Ic=25mA VL=Ic*R=25V Hence, Av=Vl/Vi=50.

Is my solution correct? I am assuming that the internal resistance of source will not have any effect on the circuit. Correct me if I am wrong.

• The 'key' point is, how the gain is defined? That is, which is the 'input' and which is the 'output'? – diverger Nov 9 '14 at 12:07

From the given input resistance $r_{i}=20\;Ohms$ I deduce that the transconductance is $g_{m}=\frac{1}{r_{i}}=0.05A/V$. Because of
$$I_{c}=g_{m} \times V_{t} \quad(V_{t}=temeperature\: voltage, V_{t}=25mV)$$
we can calculate the DC current $I_{c}=I_{e}=0.05 \times 25 mA=1.25 mA$ (neglecting $I_{b}$).
The voltage gain referred to the base node is $A_{1}=g_{m} \times (r_{o}||R_{L})$ and the gain referred to the signal source (with internal $R_{s}=100\;ohms$) is $A_{0}=A_{1} \times r_{i}/(r_{i}+R_{s})$.