Question 2 of 7: Fixed-Bias BJT Amplifier — Design and the $R_b=0$ Failure Case
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Comp-A1, Electronics — National Exams, undated sitting (page 1 of the paper reads "May 2018" while every question page reads "May 2019", so the exam period is uncertain). Open-book, 3 hours; the paper's own instructions read "FIVE (5) questions constitute a complete exam paper" while seven distinct 20-mark questions are printed — all seven are answered below as a complete study resource. Where a diode is used without a stated drop, $V_D=0.7\text{V}$.
Reference texts: Sedra & Smith, Microelectronic Circuits (diode rectifiers/limiters, BJT and MOSFET biasing and small-signal amplifiers, op-amp limiting circuits, CMOS logic gate sizing, R-2R ladder D/A converters) — the single reference text covering every question on this paper.
Question 2: Fixed-Bias BJT Amplifier — Design and the $R_b=0$ Failure Case (20 marks)
Given. Fixed-bias NPN stage: $V_{CC}=10\text{V}$ through $R_b$ to the base, $V_{CC}$ through $R_c$ to the collector, emitter grounded (no $R_E$).
Given data
Quantity
Value
$V_{CC}$
10V
$V_{BE}$ (at $I_C=1\text{mA}$)
0.7V
$\beta,\ V_T,\ V_A$
100, 25mV, 100V
Target $I_B$, $V_{CE}$
$100\ \mu\text{A}$, 5V
Find. $R_b,R_c$ for the stated targets; small-signal output resistance $R_o$; repeat both with $R_b=0$.
Approach. DC KVL around the base loop and the collector loop, using $V_{BE}$ corrected to the actual operating current via the diode law (since $0.7\text{V}$ is only stated at $I_C=1\text{mA}$); then $R_o=R_c\|r_o$ with $r_o=V_A/I_C$ for the small-signal part.
Part (a) — bias point and resistors. $I_C=\beta I_B=100(100\ \mu\text{A})=10\text{mA}$. Because $V_{BE}$ was only specified at $1\text{mA}$, correct it to the actual $10\text{mA}$ operating point:
$$V_{BE}=0.7+V_T\ln\!\frac{I_C}{1\text{mA}}=0.7+0.025\ln(10)=\boxed{0.758\text{ V}}$$
Base loop: $V_{CC}=I_B R_b+V_{BE}\ \Rightarrow\ R_b=\dfrac{10-0.758}{100\ \mu\text{A}}=\boxed{92.4\text{ k}\Omega}$. Collector loop: $V_{CC}=I_C R_c+V_{CE}\ \Rightarrow\ R_c=\dfrac{10-5}{10\text{mA}}=\boxed{500\ \Omega}$.
Part (b) — output resistance. $r_o=V_A/I_C=100\text{V}/10\text{mA}=10\text{k}\Omega$. Looking into the collector with $v_i=0$:
$$R_o=R_c\|r_o=500\|10{,}000=\boxed{476\ \Omega}$$
Check — engineering assumption
Part (c) sets $R_b=0$, tying the base directly to the $10\text{V}$ rail with no current-limiting resistor at all — the very mechanism that sets $I_B=100\ \mu\text{A}$ in part (a) is gone. This is solved as the standard textbook demonstration of why fixed-bias needs a well-defined $R_b$: the base is forced hard on, driving the transistor into saturation rather than leaving it at the part-(a) operating point. $R_c$ is kept at its part-(a) value ($500\,\Omega$, the only value given for the collector loop), and a typical $V_{CE,sat}\approx0.2\text{V}$ is assumed (not stated in the source, a standard silicon value) to get a numeric saturation current.
Part (c) — $R_b=0$, repeat (a). With $R_b=0$, $V_B=V_{CC}=10\text{V}$ directly — far beyond any normal $V_{BE}$ — so $I_B$ is no longer a controlled $100\ \mu\text{A}$; it becomes whatever the base-emitter junction and its (unmodelled) bulk resistance allow, which the $I_C=\beta I_B$ relation shows would demand a collector current far beyond what $R_c$ and $V_{CC}$ can supply while staying in the active region. The transistor is driven hard into saturation: $V_{CE}\approx V_{CE,sat}$, and $I_C$ is now set by $R_c$ and $V_{CC}$ alone (NOT by $\beta I_B$):
$$I_{C,\text{sat}}=\frac{V_{CC}-V_{CE,sat}}{R_c}=\frac{10-0.2}{500}=\boxed{19.6\text{ mA}}$$
nearly double the part-(a) design current, with $V_{CE}\approx0.2\text{V}$ instead of the designed 5V.
Part (c) — repeat (b). In saturation the Early-effect small-signal model ($r_o=V_A/I_C$) no longer applies, because $Q_1$ has left the active region entirely. The incremental output resistance collapses to the transistor's own small saturation resistance $r_{CE,sat}$ (typically tens of ohms, dominated by the collector-emitter path of a saturated device, not by $R_c\|r_o$) — the stage stops behaving as an amplifier with a current-source-like output and instead behaves like a nearly-closed switch, a resistance far below the $476\,\Omega$ found in part (a).
