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:
$$A_{voc} \approx -\frac{V_{CC}}{2(V_T + I_E\cdot R_4||R_7)} $$
Best Answer
In a common emitter amplifier with an emitter resistor the collector current and the emitter current are almost identical.
"Gain" is proportional to collector voltage / emitter voltage (by definition) so as Vc = Ic x Rc and as Ve = Ie x Rc and as Ic ~= Ie, Gain = Rc/Re.
Rc = collectpr resistor and Re = EFFECTIVE emitter resistance. This is the transistor internal Re_internal + theemitter resistor.
If there is no emitter capacitor across the emitter resistor then
Gain = Rc / (Re + Re_internal).
If the emitter resistor has a capacitor across it which allows AC to pass with no loss (or not much loss) then
Gain ~= Rc/Re_internal.
Re external is used to set the DC operating conditions and maximum AC gain occurs when the only emitter resistance = Re_internal.
For a silicon transistor Re_internal "just happens" to be 26/I ohms where I is the emmitter current in mA. This is a basic property of the silicon junction which I'll not go into further here. It makes sense when examined but is not intuitive.
ie if i=1 mA then Re_internal = 26 Ohms.
If Ie = 4 mA the Re_internal = 26/4 = 6.5 Ohms
If Ie = 10 mA then Re_internal = 2.6 Ohms
SO for a fully AC-bypassed Re (large enough Ce) :
Gain = Rc / Re_internal = Rc / (26/mA)
Following this through you arrive at the "fact" that
Gain = 38.4 x Vc
Not many people know that.
Few people believe it when told :-)