The Transistor — Question 6

Question 6

Emitter degeneration trades voltage gain for linearity

For a low-frequency NPN amplifier use $i_c=g_mv_{be}$, $i_b=i_c/\beta$, $v_e=(i_c+i_b)R_E$, $v_{in}=v_{be}+v_e$, and $v_o=-i_cR_C$. Let $g_m=.0400\,\mathrm S$, $\beta=100$, $R_E=200\,\Omega$, and $R_C=2.00\,\mathrm{k\Omega}$. Neglect output resistance.

1. Derive the gain from base input voltage to collector output voltage.

2. Evaluate the gain and compare with the gain when the emitter resistor is bypassed for ac signals.

Question 6 - Solution

Emitter degeneration trades voltage gain for linearity

(a) Method and calculation

$$v_{in}=v_{be}[1+g_m(1+1/\beta)R_E].$$$$\boxed{A_v=\frac{v_o}{v_{in}}=-\frac{g_mR_C}{1+g_m(1+1/\beta)R_E}}.$$

(b) Interpretation and further calculation

$$\boxed{A_v=-80/[1+.0400(1.01)(200)]=-8.811}.$$$$R_{E,\mathrm{ac}}=0\Rightarrow\boxed{A_{v,\mathrm{bypass}}=-80.0,\quad |A_v|/80=1/9.08=.1101}.$$

Physical check

The emitter voltage follows changes at the base, reducing the change across the base-emitter junction. The negative sign denotes collector output inversion.