Question 4 of 7: BJT Common-Emitter Amplifier with Dual Supply

Question 4: BJT Common-Emitter Amplifier with Dual Supply (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. $V_{CC}=5\text{V}$, $V_{EE}=-5\text{V}$, $R_C=1\text{k}\Omega$, $R_B=1\text{k}\Omega$, $R_E=3.3\text{k}\Omega$ (unbypassed), $V_{BE}=0.7\text{V}$, $\beta=100$, $V_i=0\text{V}$ DC.

Find. DC bias point ($I_B,I_C,I_E,V_{CE}$); small-signal circuit and gain; $I_C$-$V_{CE}$ sketch with Q-point; a bias change for max swing.

Approach. Write one KVL loop from $V_i$ through $R_B$, the B-E junction, and $R_E$ to $V_{EE}$ (with $I_E=(\beta+1)I_B$) to get the DC point; linearise with the hybrid-$\pi$ model (RE unbypassed, so it stays in the AC loop) for the gain; read swing limits off the load line.

1. Part (a) — DC bias point. KVL: $V_i-I_BR_B-V_{BE}-I_ER_E=V_{EE}$, with $I_E=(\beta+1)I_B$: $$I_B=\frac{V_i-V_{BE}-V_{EE}}{R_B+(\beta+1)R_E}=\frac{0-0.7-(-5)}{1000+101(3300)}=\boxed{12.9\ \mu\text{A}}$$ $$I_E=(\beta+1)I_B=\boxed{1.30\text{ mA}}\qquad I_C=\beta I_B=\boxed{1.29\text{ mA}}$$ $$V_C=V_{CC}-I_CR_C=5-1.29=3.71\text{V}\qquad V_E=V_B-V_{BE}=(V_i-I_BR_B)-0.7=-0.713\text{V}$$ $$V_{CE}=V_C-V_E=\boxed{4.43\text{ V}}\quad(\gg0.2\text{V sat}\Rightarrow\text{ confirmed active})$$
2. Part (b) — small-signal gain. $r_e=V_T/I_E=25\text{mV}/1.30\text{mA}=19.2\,\Omega$; $r_\pi=\beta V_T/I_C=1.94\text{k}\Omega$. With $R_E$ unbypassed it stays in the emitter AC loop:

3. Solving the small-signal loop ($v_i=i_bR_B+i_br_\pi+(\beta+1)i_bR_E$, $v_o=-\beta i_bR_C$): $$A_v=\frac{v_o}{v_i}=\frac{-\beta R_C}{r_\pi+(\beta+1)R_E}=\frac{-100(1000)}{1943+101(3300)}=\boxed{-0.298\ \text{V/V}}$$ (equivalently the familiar approximation $A_v\approx-R_C/(r_e+R_E)=-1000/3319=-0.301$, matching within rounding).
4. Part (c) — $I_C$-$V_{CE}$ load line. DC load line: $V_{CC}-I_C R_C-V_{CE}-I_ER_E=V_{EE}$ with $I_E\approx I_C$: $V_{CE}=(V_{CC}-V_{EE})-I_C(R_C+R_E)=10-I_C(4300)$.

5. Part (d) — maximizing signal swing. The Q-point sits at $V_{CE}=4.43\text{V}$, much closer to the upper limit ($V_{CC}$ end, cutoff) than the lower (saturation) end — available room is only $\approx1.3\text{V}$ before $V_C$ hits $V_{CC}$ (cutoff) but $\approx4.2\text{V}$ before $V_{CE}$ collapses to $0.2\text{V}$ (saturation): a very asymmetric, non-optimal bias. To maximize symmetric swing, INCREASE the bias current (e.g. raise $V_i$, or reduce $R_B$, or add a base voltage-divider supplying a higher DC $V_B$) so the operating point moves down the load line until $V_{CE}$ sits near the midpoint between $V_{CC}$ and the saturation edge, $V_{CE}\approx\big[(V_{CC}-V_{EE})+V_{CE,\text{sat}}\big]/2\approx5.1\text{V}$ measured from $V_{EE}$ — i.e. roughly doubling $I_C$ from its present $1.29\text{mA}$ would centre the point and maximize the undistorted output swing in both directions.