# Derivation of current gain and voltage gain for common base BJT

For a common base configuration BJT such as the one shown here: How can I derive the voltage gain and current gain?

Forgive me for using another schematic for common base configuration, but I found it more understandable in education purpose. First, voltage gain:$A_{v}=\frac{u_{iz}}{u_{ul}}$

Usual way to deal with this fraction is to derive the expression for both of the voltages separately and then evaluate.

Output voltage is equal to: $u_{iz}= -h_{fe}\cdot i_b \cdot R_c ||R_T$

Input voltage is equal to: $u_{ul}=i_{ul}\cdot R_E = -i_b \cdot r_{be}$. Notice the sign of the current and why we took $-i_b$(because the current flows in the negative terminal of the predefined voltage $u_{ul}$)

Therefore, voltage gain is equal to: $A_{v}=\frac{u_{iz}}{u_{ul}}=\frac{-h_{fe}\cdot i_b \cdot R_C||R_T}{-i_b \cdot R_E}=h_{fe}\frac{R_C||R_T}{R_E}$

Current gain is a bit difficult to derive, but don't give up :)

First, we need to see what is the expression for input resistance of the amplifier $R_{ul}$ (which is also an important parameter while designing an amplifier)

Input resistance is: $R_{ul}=\frac{u_{ul}}{i_{ul}}$

Currents are: $i_{ul}=i_{Re}+i_e; i_{Re} = \frac{u_{ul}}{R_E}; i_b = - \frac{u_{ul}}{r_{be}}$

If we apply KCL for node E we have: $i_e+i_b+h_{fe}i_b=0, i_e=-i_b(1+h_{fe})$

Expression for input current is: $i_{ul}=i_{Re}+i_e=i_{Re}-i_b(1+h_{fe})=\frac{u_{ul}}{R_E}-(1+h_{fe})=u_{ul}\cdot (\frac{1}{R_E}+\frac{1}{\frac{r_{be}}{1+h_{fe}}})$

Now when we take a step back and take a look at the input resistance expression: $R_{ul}=\frac{u_{ul}}{i_{ul}}=R_E || (\frac{r_{be}}{1+h_{fe}})$

From the last expression we can notice that the input resistance is equal to 2 resistor in parallel. First one is $R_E$ and the second one we can call $R_{ul}'$.

We are finally here, $i_{iz}=\frac{u_{iz}}{R_T}; i_{ul}=\frac{u_{ul}}{R_E || R_{ul}'}$

Current gain: $A_{i}=\frac{i_{iz}}{i_{ul}}=\frac{\frac{u_{iz}}{R_T}}{\frac{u_{ul}}{R_E || R_{ul}'}}=\frac{u_{iz}}{u_{ul}} \cdot \frac{R_E || R_{ul}'}{R_T} = A_V \cdot \frac{R_E ||R_{ul}'}{R_T}$

From here we can see that the current gain of these configuration can never be bigger than 1 and always positive.

I hope everything is clear now!