Question 5 of 5: Common-Collector (Emitter-Follower) Design

Question 5: Common-Collector (Emitter-Follower) 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. An emitter follower on ±5 V rails: base fed from v_s through R_S–C₁ with R_B to ground; collector to R_C and AC-bypassed to the +5 V rail by C_C; output taken at the emitter through C_E, with R_E to −5 V.

Find. (a) R_E and R_C; (b) output resistance R_o at the emitter; (c) maximum undistorted peak-to-peak swing.

Approach. Fix the DC operating point from I_E = 2 mA to size R_E; choose R_C to keep Q₁ active with symmetric headroom. Then find the output resistance looking into the emitter, and finally the clipping limits that bound the undistorted swing.

1. (a) DC bias — R_E. The base current $I_B=I_E/(\beta+1)=2/101=19.8\ \mu$A drops only $I_B R_B\approx2$ mV across R_B, so $V_B\approx0$ and $V_E=V_B-V_{BE}=-0.7$ V. With the emitter returned to −5 V,
   $$R_E=\frac{V_E-(-5)}{I_E}=\frac{-0.7+5}{2\,\text{mA}}=\boxed{2.15\ \text{k}\Omega.}$$
2. (a) DC bias — R_C. The collector carries $I_C=\alpha I_E=\tfrac{100}{101}(2)=1.98$ mA. Since C_C makes the collector an AC ground, R_C only sets the DC collector voltage; choosing $V_C=+3.9$ V gives symmetric headroom (see part c) and keeps $V_{CE}=V_C-V_E=4.6$ V well above saturation:
   $$R_C=\frac{5-V_C}{I_C}=\frac{5-3.9}{1.98\,\text{mA}}\approx\boxed{560\ \Omega.}$$
3. (b) Output resistance. With $V_T=25$ mV, $r_e=V_T/I_E=25/2=12.5\ \Omega$. Looking back from the emitter, the base network reflects down by $\beta+1$: the base sees $R_S\|R_B=100\|100=50\ \Omega$ (input source zeroed), so
   $$R_o=R_E\ \Big\|\ \Big[r_e+\frac{R_S\|R_B}{\beta+1}\Big]=2150\ \Big\|\ \Big[12.5+\frac{50}{101}\Big]=2150\|13.0\ \Omega=\boxed{13\ \Omega.}$$
   (V_A=∞ so r_o is infinite and drops out.)
4. (c) Maximum undistorted swing. The emitter sits at −0.7 V. Two limits bound its excursion: on the positive swing the transistor saturates when $V_{CE}\!\to\!V_{CE(sat)}$, i.e. at $V_E=V_C-0.3=+3.6$ V — a headroom of $3.6-(-0.7)=4.3$ V; on the negative swing the transistor cuts off as the emitter approaches the −5 V rail, a headroom of $-0.7-(-5)=4.3$ V. Choosing V_C=3.9 V made these equal, so the maximum symmetric swing is
   $$\pm4.3\ \text{V}\ \Rightarrow\ \boxed{v_{o,\text{pp}}\approx8.6\ \text{V peak-to-peak.}}$$