I have a problem of calculating the voltage gain of cascade common emitter BJT amplifier using small-signal model

Assume that the reactance of the capacitor is negligible. The circuit is bias at \$I_{C1}=0.1mA\$ and \$I_{C2}=1mA\$ and \$\beta=100\$ for both BJT



simulate this circuit – Schematic created using CircuitLab

$$A_v=\frac{V_{out}}{V_{in}}$$ $$V_{out}=100i_{b2}*R_2=100(\frac{100i_{b1}\frac{R_1r_{be2}}{R_1+r_{be2}}}{r_{be2}})R_2=100(\frac{100\frac{V_{in}}{r_{be1}}\frac{R_1r_{be2}}{R_1+r_{be2}}}{r_{be2}})R_2$$

using \$r_{be}=\frac{25\beta}{I_c}\$, \$r_{be1}=25k\Omega\$ and \$r_{be2}=2.5k\Omega\$

by then I get the result \$A_v=1785\$

However, the result of my text book says that it's 1765 and \$r_{in}=25.25k\Omega\$ which I think that \$r_{in}=r_{be1}\$

Can anybody explain what I missed in calculating or what I did wrong.

  • \$\begingroup\$ The voltage gain will be somewhat lower than calculated by you because the input resistance of the second stage will be smaller than 2.5kOhms. This is because the resistors biasing the second stage have been forgotten. Without this biasing the circuit will not work. \$\endgroup\$
    – LvW
    Commented Oct 14, 2015 at 8:04

1 Answer 1


You should use \$25(\beta+1)/I_C\$ for the input impedance; the base current also flows through the emitter in addition to the collector current.

Also don't forget that the 25mV figure is an approximation for the thermal voltage; the textbook seems to be using 25mV, the same as you, but a more accurate figure is 25.85mV.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.