Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Phys-A5 — Semiconductor Devices & Circuits — National Exams, May 2014
3 hours duration. Closed book exam (useful constants and equations annexed to the paper). Any FIVE (5) of the SEVEN (7) questions constitute a complete exam paper; all seven are answered here as a complete study resource.
Reference texts: A. S. Sedra & K. C. Smith, Microelectronic Circuits, 8th ed. (diode I–V characteristics and junction physics Ch. 3–4, op-amp circuits and active filters Ch. 2–12, BJT biasing and small-signal amplifiers Ch. 5–6, ADC architectures Ch. 17); C. Kittel, Introduction to Solid State Physics, 8th ed. (crystal structure and packing fraction Ch. 1, free-electron/semiconductor carrier statistics Ch. 8).
Standard voltage-divider-biased common-emitter stage: $R_1$/$R_2$ base divider, $R_C$ collector resistor, bypassed $R_E$ emitter resistor, AC-coupled $V_i$ at the base and $R_L$ at the collector.
Approach. (a) set the midband gain $-g_m(R_C\parallel R_L\parallel r_o)$ equal to $-150$ and solve for $R_C$; (b) close the DC collector-loop KVL for $R_E$; (c) size the base divider at the given boundary current $5I_{BQ}$; (d) combine the divider with $r_\pi$ for $R_i$, and $R_C\parallel r_o$ for $R_o$.
Part (a) — $R_C$ from the gain spec. With $R_E$ AC-bypassed, the midband gain (collector node loaded by $R_C\parallel R_L\parallel r_o$) is $A_v=-g_m(R_C\parallel R_L\parallel r_o)$. First combine the fixed loads: $R_L\parallel r_o=\dfrac{(1500)(80{,}000)}{81{,}500}=1474.8\,\Omega$. Setting $|A_v|=150$:
$$g_m\left(R_C\parallel 1474.8\right)=150\ \Rightarrow\ R_C\parallel1474.8=\frac{150}{0.250}=600\,\Omega\ \Rightarrow\ \boxed{R_C=1.013\ \text{k}\Omega}$$
Part (b) — $R_E$ from the DC bias point. $I_{EQ}=I_{CQ}\dfrac{\beta+1}{\beta}=6.5\,\text{mA}\times\dfrac{131}{130}=6.550\,\text{mA}$. KVL around $V_{CC}\to R_C\to$ collector-emitter$\to R_E\to$ ground:
$$V_{CC}=I_{CQ}R_C+V_{CEQ}+I_{EQ}R_E$$
$$25=(6.5\,\text{mA})(1.013\,\text{k}\Omega)+12+(6.550\,\text{mA})R_E\ \Rightarrow\ \boxed{R_E=980\ \Omega}$$
Part (c) — base divider at the boundary current. $I_{BQ}=I_{CQ}/\beta=6.5/130=0.05\,\text{mA}$. The base voltage needed is $V_B=I_{EQ}R_E+V_{BE(on)}=(6.550\,\text{mA})(980\,\Omega)+0.5=6.918\,\text{V}$. Sizing the divider at the stated limit $I_{R_2}=5I_{BQ}=0.25\,\text{mA}$ (the largest bias-stable resistors consistent with the constraint) gives $I_{R_1}=I_{R_2}+I_{BQ}=0.30\,\text{mA}$:
$$R_2=\frac{V_B}{I_{R_2}}=\frac{6.918}{0.25\,\text{mA}}=\boxed{27.7\ \text{k}\Omega}\qquad R_1=\frac{V_{CC}-V_B}{I_{R_1}}=\frac{25-6.918}{0.30\,\text{mA}}=\boxed{60.3\ \text{k}\Omega}$$
Part (d) — $R_i$ and $R_o$. The base divider loads the signal source in parallel with $r_\pi$:
$$R_i=R_1\parallel R_2\parallel r_\pi=60.3\,\text{k}\Omega\parallel27.7\,\text{k}\Omega\parallel520\,\Omega=\boxed{506\ \Omega}$$
Looking back into the collector, $R_o$ is set by $R_C\parallel r_o$ alone (the external $R_L$ is the load being driven, not part of the amplifier's own output resistance); using $R_C\parallel R_L\parallel r_o=(R_C\parallel r_o)\parallel R_L=600\,\Omega$ from part (a) and $R_L=1.5\,\text{k}\Omega$:
$$R_o\parallel1500=600\ \Rightarrow\ \boxed{R_o=1.00\ \text{k}\Omega}$$
