Common-Emitter & Common-Source Amplifier — Worked Example

CE microphone pre-amp design

Target spec: $A_v = -50\,\text{V/V}$, $V_{CC} = 12\,\text{V}$, $I_C = 2\,\text{mA}$, $\beta = 100$.

Find: $R_C$, $R_E$, bias divider values, and bypass capacitor $C_E$.

1. Target spec.
   $$A_v = -50\,\text{V/V}, \quad V_{CC} = 12\,\text{V}, \quad I_C = 2\,\text{mA}$$
   The negative sign confirms inverting operation. With $I_C = 2\,\text{mA}$ we can leave $V_{CE} \approx 6\,\text{V} = V_{CC}/2$ for maximum symmetric output swing.
2. Pick $R_C$ from gain target.
   $$g_m = \frac{I_C}{V_T} = \frac{2\,\text{mA}}{25\,\text{mV}} = 80\,\text{mA/V}$$
   With the bypass cap installed (full gain), $A_v \approx -g_m R_C$:
   $$R_C = \frac{|A_v|}{g_m} = \frac{50}{0.080\,\text{S}} = 625\,\Omega$$
   Pick the nearest E24 standard value: 620 Ω. This gives $A_v = -0.080 \times 620 = -49.6\,\text{V/V}$ — within 1% of target.
3. Pick $R_E$ for bias stability.Rule of thumb: $V_E \geq 1\,\text{V}$ for temperature stability. Choose $V_E = 2\,\text{V}$:
   $$R_E = \frac{V_E}{I_C} = \frac{2\,\text{V}}{2\,\text{mA}} = 1\,\text{k}\Omega$$
   Check $V_{CE}$: $V_{CE} = 12 - 2\,\text{mA}(0.62 + 1) = 12 - 3.24 = 8.76\,\text{V}$. This is above $V_{CC}/2$ — we can live with it for this single-stage demo. A lower $R_C$ would re-centre the Q-point.
4. Voltage divider: choose $R_1$ and $R_2$.
   $$V_B = V_E + V_{BE} = 2\,\text{V} + 0.7\,\text{V} = 2.7\,\text{V}$$
   Use the rule $I_{div} \approx 10 \times I_B = 10 \times I_C/\beta = 200\,\mu\text{A}$:
   $$R_1 + R_2 = \frac{V_{CC}}{I_{div}} = \frac{12}{200\,\mu\text{A}} = 60\,\text{k}\Omega$$
   $$R_2 = \frac{V_B}{V_{CC}}(R_1 + R_2) = \frac{2.7}{12} \times 60\,\text{k}\Omega = 13.5\,\text{k}\Omega$$
   Pick standard values: $R_1 = 33\,\text{k}\Omega$, $R_2 = 10\,\text{k}\Omega$. Check: $V_B = 12 \times 10 / 43 = 2.79\,\text{V}$ — close enough. The Q-point barely shifts.
5. Bypass capacitor $C_E$.Size so that $X_{C_E} \ll R_E$ at the lowest signal frequency (100 Hz). Rule: $X_{C_E} \leq R_E / 10$ at $f_{low}$:
   $$C_E \geq \frac{1}{2\pi \cdot f_{low} \cdot (R_E/10)} = \frac{10}{2\pi \times 100 \times 1000} \approx 16\,\mu\text{F}$$
   Pick 100 µF for generous headroom. This ensures $C_E$ is effectively a short above 1.6 Hz, well below the audio band.