Voltage Gain

From Fig. 2.12 we see that the input is into the base (just as with the CE amp), but the output is at the emitter. We can thus use our knowledge of BJT's to immediately see that the voltage gain of an EF amp is approximately 1, Since the emitter voltage is always approximately 0.7V below the base voltage, if a small signal is appled to the base, then you will see that same signal at the emitter, just offset by 0.7V. Since the voltage gain actually reflects the change of voltage, and since the change at the base and emitter are approximately the same, the voltage gain is approximately one. To see this analytically, substitute the BJT with its equivalent circuit as shown above. First perform the base-emitter loop analysis.

V_b=V_be+I_e R_E

V_e=I_e R_E

Providing a small signal input leads to the following:

$$v_{in}= \Delta V_b = \Delta V_{be} +\Delta I_e R_E$$ (50)

$$v_{out} = \Delta V_e = \Delta I_e R_E$$ (51)

Taking the ratio of v_out to v_in, and recalling that $\Delta V_{be}=\frac{\Delta I_e}{g_m}$, we obtain for the gain

$$A_V= \frac{v_{out}}{v_{in}}=\frac{R_E}{\frac{1}{g_m}+R_E}\approx 1$$ (52)

Finally, since $\Delta V_{be}$, is usually much smaller than $\Delta I_e R_E$,the voltage gain of an EF amp can often be approximated as one.