Electronic – Gain of common-emitter BJT amplifier with bypass capacitor

amplifierbjtgain

I have this common-emitter BJT amplifier:

schematic

simulate this circuit – Schematic created using CircuitLab

The text book says that with a bypass capacitor, the gain is -Rc/re, where Rc is resistor R6 and re is the internal resistance of the BJT. The internal resistance of the BJT is found by computing VE = VB – VBE (VB is found by voltage division, VBE is 0.7V), then IE = VE/RE = VE/R7, then re (internal resistance) = 25mV/Ie

Now I want to simulate this circuit and use simulated values to directly compute the gain by dividing the output by the input voltage, so I can compare with the theory, but I do not know exactly what is the input and what is the output. Are they taken right before and after the BJT, or right before and right after the caps, or…?

I have tried to take different combinations of output/input but the result does not even come close to the theoretical -Rc/re.

Any guidance will be appreciated. Thank you.

Best Answer

There are several gains associated with voltage amplifiers. Consider the following model

enter image description here

In this model, the gain \$A_{VO} \$ is the open circuit voltage gain of the amplifier which, in your circuit, is given by \$R_C/r_e\$

But note that the output resistance of the amplifier (which is about \$R_C\$ in your circuit) forms a voltage divider with the load \$R_L\$.

So, the loaded voltage gain is:

\$A_{V} = A_{VO} \dfrac{R_L}{R_{out}+ R_L}\$

But, note that the input voltage \$V\$ is less than the source voltage due to the voltage division between the source resistance and the input resistance of your amplifier. Thus, the loaded gain with respect to the source is:

\$A_{VS} = A_V \dfrac{R_{in}}{R_S + R_{in}} = A_{VO} \dfrac{R_L}{R_{out}+ R_L} \dfrac{R_{in}}{R_S + R_{in}}\$

So, you cannot expect to measure anything close to the open circuit gain \$R_C/r_e \$