Electrical – voltage gain and current gain in a common emitter

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 \$

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:

$$A_{voc} \approx -\frac{V_{CC}}{2(V_T + I_E\cdot R_4||R_7)} $$