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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\$.

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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.

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  • \$\begingroup\$ The \$X_C\$ changes with frequency. All capacitors are frequency-dependent devices. You know how to compute it, yes? \$\endgroup\$ – jonk May 31 '18 at 19:12
  • \$\begingroup\$ @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. \$\endgroup\$ – Bill May 31 '18 at 19:20
  • \$\begingroup\$ 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. \$\endgroup\$ – jonk May 31 '18 at 19:27
  • \$\begingroup\$ Keep in mind that \$X_C\$ is out of phase. So it may not affect things a lot. But some. \$\endgroup\$ – jonk May 31 '18 at 19:37
  • \$\begingroup\$ So, I should expect a gain of -R_C/R_E=-4.7? I suppose not. I should expect something at the hundreds, right?! \$\endgroup\$ – Bill May 31 '18 at 19:39

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