Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Phys-A5 — Semiconductor Devices & Circuits — National Exams, May 2015
3 hours duration. Closed book exam (useful constants, equations and device models are annexed to the exam 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. (semiconductor/diode physics Ch. 3–4, op-amp and active-filter circuits Ch. 2 & 12, MOSFET small-signal amplifiers Ch. 7, data converters Ch. 17, waveform-shaping/signal generators Ch. 13); M. M. Mano & M. D. Ciletti, Digital Design, 6th ed. (CMOS/NMOS logic-family gates Ch. 10); C. Kittel, Introduction to Solid State Physics, 8th ed. (semiconductor carrier transport and statistics Ch. 8).
Given. A precision square/triangle-wave generator (Figure P7) with the diode, Zener and op-amp ratings below.
Symbol
Value
$V_D$ (each of 4 diodes)
0.7 V
Zener current $I_Z$
1 mA
Max. current in $R_1$, $R_2$
0.2 mA
Op-amp saturation
±13 V
Integrator $C$
0.01 µF
$V_1$, $V_2$ amplitudes
±5 V
Frequency $f$
1 kHz
Figure P7 topology: op-amp 1 is an inverting integrator ($R$, $C$) driving $V_1$ (triangular) from $V_2$ (square). Op-amp 2's "$+$" input is a resistive summer fed by $V_1$ through $R_1$ and by $V_2$ (feedback) through $R_2$, with "$-$" grounded — a bistable (Schmitt) comparator. Its output drives $R_d$ into a 4-diode bridge with a Zener $Z$ across the bridge's other diagonal, which clamps node $V_2$ to $\pm(V_Z+2V_D)$ regardless of current direction.
Figure — the two-op-amp loop implied by Fig. P7: integrator produces the triangular $V_1$, which the bistable/Zener-bridge limiter compares and converts to the square-wave $V_2$, fed back to close the oscillation loop.
Approach. (a) the diode bridge always steers the Zener's breakdown current through exactly two forward-biased diodes in series with $Z$, regardless of which way current flows through $R_d$, so $V_2=V_Z+2V_D$. (b) the resistive-summer threshold and the current cap on $R_1$, $R_2$ fix their common value; symmetry ($V_1$ swinging $\pm5\,\text{V}$ like $V_2$) requires $R_1=R_2$. (c) size the integrator resistor so the triangular ramp swings its full $\pm5\,\text{V}$ over one half-period. (d) size $R_d$ from the op-amp's saturation voltage and the total current it must supply into the $V_2$ node: the Zener-bridge branch $I_D$, the summer branch $I_2$ and the integrator-input branch $I_1$, all three drawn in Figure P7.
Part (a) — Zener voltage rating. Whichever way current flows through $R_d$, the bridge routes it through exactly two forward diodes ($2V_D$) in series with the Zener ($V_Z$) to ground, clamping $|V_2|$ to that sum:
$$V_2=V_Z+2V_D\ \ \Rightarrow\ \ V_Z=5-2(0.7)=\boxed{3.6\ \text{V}}$$
Part (b) — $R_1$, $R_2$. With "$-$" grounded, op-amp 2's "$+$" node (no current into the op-amp) sits at the resistive average of $V_1$ (through $R_1$) and $V_2$ (through $R_2$); switching occurs when this node crosses zero: $V_1R_2+V_2R_1=0$. For the triangular wave's own $\pm5\,\text{V}$ peaks to be exactly the switching thresholds (matching the stated $\pm5\,\text{V}$ amplitude), symmetry requires $R_1=R_2$. At that threshold instant the current in each resistor is $|V_2|/R=|V_1|/R=5\,\text{V}/R$; sizing at the stated maximum:
$$\frac{5\,\text{V}}{R_1}\le0.2\,\text{mA}\ \ \Rightarrow\ \ R_1=R_2=\frac{5}{0.2\times10^{-3}}=\boxed{25\ \text{k}\Omega}$$
Part (c) — integrator resistor $R$. The integrator obeys $dV_1/dt=-V_2/(RC)$. With $V_2$ constant at $+5\,\text{V}$ (or $-5\,\text{V}$) for one half-period $T/2=1/(2f)=0.5\,\text{ms}$, $V_1$ must ramp through its full $10\,\text{V}$ peak-to-peak swing:
$$|\Delta V_1|=\frac{V_2}{RC}\cdot\frac{T}{2}=2V_2\ \ \Rightarrow\ \ R=\frac{T}{4C}=\frac{1/1000}{4(0.01\times10^{-6})}=\boxed{25\ \text{k}\Omega}$$
Part (d) — $R_d$. With the op-amp output saturated at $+13\,\text{V}$ and node $V_2$ clamped to $+5\,\text{V}$, Figure P7 shows the current $I_o$ through $R_d$ splitting three ways at node $V_2$, so it must supply the Zener-bridge current ($I_D=I_Z=1\,\text{mA}$), the current $I_2$ drawn by $R_2$ toward the summer node (at most $0.2\,\text{mA}$, from part (b), reached at the threshold instant when the “+” node sits at 0 V), and the current $I_1$ fed back along the top wire into the integrator resistor $R$, whose other end is held at virtual ground ($I_1=V_2/R=5/25\text{k}=0.2\,\text{mA}$, from part (c)):
$$I_o=I_Z+I_2+I_1=1.0+0.2+0.2=1.4\,\text{mA}$$
$$R_d=\frac{V_{sat}-V_2}{I_o}=\frac{13-5}{1.4\times10^{-3}}=\boxed{5.71\ \text{k}\Omega}$$
(a standard 5.6 kΩ resistor gives $I_o=1.43\,\text{mA}$ and keeps the Zener at or slightly above its 1 mA design current). Leaving out $I_1$ would give $8/1.2\,\text{mA}=6.67\,\text{k}\Omega$, which starves the Zener to $0.8\,\text{mA}$ when both side branches draw their full $0.2\,\text{mA}$.
