# Voltage gain of common emitter amplifier

I'm posting this question just to be sure I'm understanding this correctly.

I want to estimate the voltage gain of the following common emitter amplifier either as $A_v=-4.7$ or $A_v=-145$.

It's pretty obvious that $A_v=-\frac{R_C}{r_e}<<-4.7$ due to the shunt capacitor. However, I am asked to estimate the voltage gain at frequencies 1kHz-10kHz. Does anything change at these frequencies? I think I have HPF filters due to the coupling capacitors but I'm not too sure which resistance goes with which capacitor. I'd appreciate any help.

• The $X_C$ changes with frequency. All capacitors are frequency-dependent devices. You know how to compute it, yes? – jonk May 31 '18 at 19:12
• @jonk Of course. I just don't know if I will have different voltage gain from the A_v=-R_C/r_e at low frequencies due to HPF filters. – Bill May 31 '18 at 19:20
• Yes, $r_e$ must be added to $X_C$ as must any Ohmic emitter resistance. You also have two other capacitors that interact. Your output capacitor is low enough that it will affect things added later on as a load. Your input capacitor has a high enough impedance, and your bypass capacitor has a low enough impedance, reflected back to the base, that the gain will be measurably affected, I suspect. – jonk May 31 '18 at 19:27
• Keep in mind that $X_C$ is out of phase. So it may not affect things a lot. But some. – jonk May 31 '18 at 19:37
• So, I should expect a gain of -R_C/R_E=-4.7? I suppose not. I should expect something at the hundreds, right?! – Bill May 31 '18 at 19:39