BJT Gain Stage: 22-Elec-A5 Electronics, May 2014, Q4

Question 4: BJT Gain Stage (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.

Check / topology. The emitter is bypassed to ground by $C_E$ and the output is taken from the collector node $v_{C1}$ through $C_C$ into $R_L$. Electrically this is a common-emitter amplifier (a true common-collector follower cannot have a voltage gain of 100). The design below treats it as common-emitter, which is the only reading consistent with the stated gain of 100 V/V.

Given. $\beta=100$, $V_{BE}=0.7\ \text{V}$, $V_A=\infty\;(r_o=\infty)$, $R_S=100\ \Omega$, $I_E=2\ \text{mA}$, open-circuit $|A_v|=100$, $R_L=1.25\ \text{k}\Omega$; thermal voltage $V_T=25\ \text{mV}$.

Find. $R_E,R_C$; output resistance $R_O$; the two output waveforms.

Q4 - The emitter is bypassed by C_E (a.c. ground) and the output is taken from the COLLECTOR via C_C into R_L: electrically a common-emitter amplifier.

Approach. With $\beta$ large take $I_C\approx I_E=2\ \text{mA}$; the transconductance sets $R_C$ from the gain spec, and the emitter d.c. drop sets $R_E$; then load the gain with $R_L$ for part (c).

Transconductance and $R_C$ (part a). $g_m=\dfrac{I_C}{V_T}=\dfrac{2\ \text{mA}}{25\ \text{mV}}=80\ \text{mA/V}$. The bypassed-emitter open-circuit gain is $|A_v|=g_mR_C=100$, hence $$R_C=\frac{100}{g_m}=\frac{100}{0.08\ \text{S}}=\boxed{1.25\ \text{k}\Omega}.$$
Emitter resistor from the d.c. bias. The base d.c. path is through $R_S$ to ground, so $V_B\approx-I_BR_S\approx0$ ($I_B=I_C/\beta=20\ \mu\text{A}\Rightarrow V_B\approx-2\ \text{mV}$). Then $V_E=V_B-V_{BE}=-0.7\ \text{V}$, and $R_E$ carries $I_E$ to the $-5\ \text{V}$ rail: $$R_E=\frac{V_E-(-5)}{I_E}=\frac{-0.7+5}{2\ \text{mA}}=\boxed{2.15\ \text{k}\Omega}.$$
Output resistance (part b). The output looks back into the collector, where $r_o=\infty$, so $$R_O=R_C\parallel r_o=R_C=\boxed{1.25\ \text{k}\Omega}.$$
Loaded gain and waveforms (part c). With $R_L=1.25\ \text{k}\Omega$ coupled to the collector, the a.c. load is $R_C\parallel R_L=625\ \Omega$ and $$A_v=-g_m(R_C\parallel R_L)=-(0.08)(625)=-50\ \text{V/V}.$$ A 1 mV$_{pp}$ input therefore gives $v_o=50\ \text{mV}_{pp}$, inverted. The collector node carries the same 50 mV$_{pp}$ sine riding on its d.c. level $V_{C1}=+5-I_CR_C=+5-2.5=2.5\ \text{V}$; $C_C$ removes the d.c. so $v_o$ is centred on 0 V. Both are small, undistorted sinusoids.

Q4(c) - Both nodes carry the same 50 mV pp sine (inverted vs input); v_C1 rides on the 2.5 V collector d.c. level, v_o is centred on 0 V (amplitudes exaggerated for clarity).

Q4 results
Quantity	Value
$g_m$	80 mA/V
$R_C$ (for $\|A_v\|_{oc}=100$)	1.25 kΩ
$R_E$	2.15 kΩ
$R_O=R_C$	1.25 kΩ
Loaded gain	−50 V/V
$v_o$ amplitude	50 mV$_{pp}$ (0 V d.c.)
$v_{C1}$	50 mV$_{pp}$ on +2.5 V d.c.

Topic: Common-emitter gain design with a bypassed emitter and the effect of loading.

Key relations: $g_m=I_C/V_T$; open-circuit $A_v=-g_mR_C$; loaded $A_v=-g_m(R_C\parallel R_L)$; $R_O=R_C\parallel r_o$; $V_{C1}=V_{CC}-I_CR_C$.

Why this works: The bypass capacitor grounds the emitter for signals, so the full $g_m$ appears and the collector resistor converts signal current to voltage. Coupling $R_L$ through $C_C$ parallels it with $R_C$, halving the gain here; the collector keeps its d.c. bias while the coupled output is a.c.-only.

Common pitfall: Believing the “common-collector” label and reporting a gain near 1, or forgetting that $R_L$ loads the collector (via $C_C$) and reduces the gain from 100 to 50.

Source: Sedra & Smith 7th ed., §6.5–6.6 (BJT biasing and the common-emitter amplifier).

Quantity	Value
\(g_m\)	80 mA/V
\(R_C\) (for \(\|A_v\|_{oc}=100\))	1.25 kΩ
\(R_E\)	2.15 kΩ
\(R_O=R_C\)	1.25 kΩ
Loaded gain	−50 V/V
\(v_o\) amplitude	50 mV\(_{pp}\) (0 V d.c.)
\(v_{C1}\)	50 mV\(_{pp}\) on +2.5 V d.c.

Question 4 of 5: BJT Gain Stage

Question 4: BJT Gain Stage (20 marks)