amplifierbjtbypassgain
I am trying to design a high-gain common emitter circuit. I know that I need to use a bypass capacitor with emitter resistor to increase the gain. My problem is the formula
A = Rc / re
makes things so much different than designing same circuit without it. I already designed one with gain of 5 in less than a hour but I am stuck in this one.
Any suggestions?
There are several gains associated with voltage amplifiers. Consider the following model
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 \$
The gain of a true common-emitter (emitter at AC common or "ground") is approximately:
$$A_v = -g_mR_C||R_L = -g_m 7.5 \Omega$$
for your resistor values. For a (magnitude) gain of 7, the required transconductance is:
$$g_m = \frac{7}{7.5} = \frac{I_C}{V_T} \Rightarrow I_C \approx 23mA$$
But, according to your schematic:
$$I_C \approx 8mA$$
Back to the drawing board!
Also, do realize that the above calculation is for small-signal gain. Since you have the emitter at AC ground, this amplifier is quite non-linear for large signals and the distortion is quite evidently large in both transient simulations you provide.
Best Answer
For a genuine common emitter configuration (emitter at AC ground), the small-signal open-circuit voltage gain is approximately
$$A_{voc} = -g_m \cdot R_C||r_o \approx-g_mR_C = -\frac{I_CR_C}{V_T} = -\frac{V_{CC} - V_C}{V_T}$$
Thus, the AC gain is fixed by your choice of supply voltage \$V_{CC}\$ and DC collector voltage \$V_C\$
If, like many, you choose (or require that) \$V_C = \frac{V_{CC}}{2}\$, the gain of the (genuine) common emitter amplifier is just
$$A_{voc} \approx -\frac{V_{CC}}{2V_T}$$
Which is to say that you don't have the necessary degree of freedom to choose the AC gain independent of the supply voltage.
Now, by adding a resistor \$R_4\$ in series \$C_4\$, you gain the degree of freedom for the AC gain:
$$A_{voc} \approx -\frac{\alpha R_C}{r_e + R_4||R_7} $$
where
$$r_e = \frac{V_T}{I_E}$$
Assuming once again that you choose \$I_CR_C = \frac{V_{CC}}{2}\$, the AC gain is:
$$A_{voc} \approx -\frac{V_{CC}}{2(V_T + I_E\cdot R_4||R_7)} $$