Question 2 of 6: Common-Emitter Amplifier Design

Question 2: Common-Emitter Amplifier Design (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. $R_1$, $R_2$, $R_C$ for the stated bias and gain; the output resistance $R_O$; and the maximum undistorted peak-to-peak swing.

[Figure not reproduced: Q2 circuit (redrawn): PNP common-emitter amplifier. RE (bypassed by C3) sets the DC emitter current from VCC; R1/R2 bias the base; RC is the collector load to ground; the signal couples in through C1 and out through C2 to RL. See the official exam paper.]

Approach. Set the small-signal transconductance from the bias current, size $R_C$ for the required gain (emitter fully bypassed by $C_3$), then place the DC operating point with the $R_1/R_2$ divider and check the swing against the AC load line.

1. Bias-point currents. With $\beta=100$, $I_C=\alpha I_E=\tfrac{\beta}{\beta+1}I_E=\tfrac{100}{101}(2\ \text{mA})\approx1.98\ \text{mA}\approx2\ \text{mA}$, and $I_B=I_C/\beta\approx20\ \mu\text{A}$. The transconductance is $$g_m=\frac{I_C}{V_T}=\frac{1.98\ \text{mA}}{25\ \text{mV}}\approx79\ \text{mS}.$$
2. Size $R_C$ for the gain. $C_3$ bypasses $R_E$, so the mid-band gain is $v_{out}/v_s=-g_m\,(R_C\parallel R_L)$ (the source drives the base directly). Setting $|{-}g_m(R_C\parallel R_L)|=100$: $$R_C\parallel R_L=\frac{100}{g_m}=1.26\ \text{k}\Omega \;\Rightarrow\; R_C=\left(\frac{1}{1.26\text{k}}-\frac{1}{10\text{k}}\right)^{-1}=\boxed{1.45\ \text{k}\Omega}.$$
3. Place the DC operating point. The emitter sits at $V_E=V_{CC}-I_E R_E=10-2\ \text{mA}\times1\ \text{k}\Omega=8\ \text{V}$, so the base is $V_B=V_E-V_{EB(on)}=8-0.7=7.3\ \text{V}$.
4. Bias divider $R_1,R_2$. Choose a stiff bleeder current $I_{R_1}=10\,I_B=0.20\ \text{mA}\gg I_B$ for bias stability. Then $$R_1=\frac{V_B}{I_{R_1}}=\frac{7.3}{0.20\ \text{mA}}\approx\boxed{36.9\ \text{k}\Omega},\qquad R_2=\frac{V_{CC}-V_B}{I_{R_1}+I_B}=\frac{2.7}{0.22\ \text{mA}}\approx\boxed{12.4\ \text{k}\Omega}.$$
5. (b) Output resistance. Looking back into the collector with $V_A=\infty$ (so $r_o=\infty$), $$R_O=R_C\parallel r_o=R_C=\boxed{1.45\ \text{k}\Omega}.$$
6. (c) Maximum undistorted swing. The quiescent collector voltage is $V_C=I_C R_C=1.98\ \text{mA}\times1.45\ \text{k}\Omega\approx2.86\ \text{V}$, so $V_{ECQ}=V_E-V_C=8-2.86=5.14\ \text{V}$. On the AC load line $R_{ac}=R_C\parallel R_L=1.25\ \text{k}\Omega$, the two clipping limits are the cut-off peak $I_C R_{ac}=1.98\ \text{mA}\times1.25\ \text{k}\Omega=2.50\ \text{V}$ and the saturation head-room $V_{ECQ}-V_{EC(sat)}=5.14-0.3=4.84\ \text{V}$. The smaller governs: $\hat v_o=2.50\ \text{V}$, so $$V_{o(pp)}=2\hat v_o=\boxed{5.0\ \text{V}_{pp}}.$$