22-Elec-A5 Electronics · December 2013

Question 6 of 6: Op-Amp + Diode Wave-Shaping Circuits

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams — 07-Elec-A5 Electronics, December 2013. Closed book; non-communicating calculator permitted. Six questions of 20 marks; any five constitute a complete paper — all six are solved here as a study resource. Op-amps ideal with ±15 V supplies unless stated; diode forward drop as given per question.

Reference texts. A. Sedra & K. Smith, Microelectronic Circuits, 7th ed. (Oxford) — MOS/BJT amplifiers (ch. 7–9), diode wave-shaping (ch. 4), op-amp circuits (ch. 2), CMOS logic (ch. 14). Cross-checks from C. Sadiku, Fundamentals of Electric Circuits.

Check / source notes. All 16 schematics follow the printed drawings; diode orientations are as drawn (Q5 a/b/d, Q6 b). Q2's data block is internally inconsistent (it labels the BJT NPN yet lists V_EB(on), names an R_L that is not in the drawing, and gives R_E = 1 kΩ while part (a) asks you to size R_E); assumptions used to resolve these are stated in Q2 per the exam's "state any assumptions" instruction.

Question 6: Op-Amp + Diode Wave-Shaping Circuits (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. Inverting op-amp stages with diodes in the feedback or input path; rails ±15 V. Find. the transfer relation and one-cycle output sketch for each of (a)–(e), including any output clipping at the rails.

Approach. Hold the − input at the virtual ground (0 V) while feedback is intact. Determine, for each input polarity, whether the diode conducts; solve the resulting linear network. Where a conducting diode clamps a feedback node and removes the op-amp's control of its own input, the loop is broken and the output saturates to a rail — a graded feature of these circuits, confirmed here by an independent nodal (Newton) solve.

(a) T-feedback (5 kΩ–5 kΩ) with a diode from the mid-node to ground

Fig. 6a — input 5 kΩ; feedback is two 5 kΩ in series with a shunt diode (anode up) at the centre tap.

Diode off (v_IN ≥ −0.7 V). The centre node sits at $-v_{IN}$; with it below +0.7 V the diode is off and the two 5 kΩ act as one 10 kΩ feedback: $v_{OUT}=-2\,v_{IN}$ (clipped at −15 V once $v_{IN}\gt7.5$ V).
Diode on (v_IN < −0.7 V). The centre node is pinned at +0.7 V, so the op-amp output can no longer control its own inverting input — feedback is lost and the output saturates positive: $$\boxed{v_{OUT}=-2v_{IN}\ (v_{IN}\ge-0.7),\qquad v_{OUT}=+15\text{ V}\ (v_{IN}\lt-0.7)}.$$

Fig. 6a output: an inverted ×2 sine clipped at −15 V on the positive-input half, and a +15 V rail block on the negative-input half.

(b) Two anti-parallel input branches (diode+1 kΩ and diode+2 kΩ), 5 kΩ feedback

Fig. 6b — the 1 kΩ branch conducts on negative input, the 2 kΩ branch on positive input.

Dead zone. For $|v_{IN}|\lt0.7$ V neither branch conducts, so $v_{OUT}=0$.
Each half. $$\boxed{v_{OUT}=\begin{cases}-2.5\,(v_{IN}-0.7)&v_{IN}\gt0.7\\[2pt]-5\,(v_{IN}+0.7)&v_{IN}\lt-0.7\end{cases}}$$ The positive half (gain −2.5) reaches the −15 V rail once $v_{IN}\gt6.7$ V; the negative half (gain −5) reaches the +15 V rail once $v_{IN}\lt-3.7$ V. A full-wave-rectifier-like output with unequal gains and a small dead band at the zero crossings.

Fig. 6b output: dead band near zero, then hard clipping to −15 V (positive input) and +15 V (negative input).

(c) Feedback diode (−in→M) with M pulled to +10 V (3 kΩ) and to output (1 kΩ)

Fig. 6c — input 1 kΩ; a diode conducts from the virtual ground to node M, which is biased from +10 V.

v_IN > 0 (diode conducts). Node M is held at −0.7 V; KCL at M (with the +10 V through 3 kΩ and the output through 1 kΩ) gives $$v_{OUT}=-v_{IN}-\tfrac{10.7}{3}-0.7=-v_{IN}-4.27\text{ V}.$$ At the +10 V input peak, $v_{OUT}\approx-14.3$ V (just short of the rail).
v_IN < 0 (diode blocks). The only feedback path is through the diode, which is now reverse-biased, so the loop opens and $$\boxed{v_{OUT}=+15\text{ V}\ (v_{IN}\lt0),\qquad v_{OUT}=-v_{IN}-4.27\ (v_{IN}\gt0)}.$$

Fig. 6c output: a downward ramp from −4.3 to −14.3 V on the positive-input half, and +15 V on the negative-input half.

(d) Two anti-parallel feedback branches (diode+3 kΩ and diode+5 kΩ)

Fig. 6d — input 1 kΩ; the 5 kΩ branch feeds back on positive input, the 3 kΩ branch on negative input.

Each half (feedback intact both ways). $$\boxed{v_{OUT}=\begin{cases}-(5\,v_{IN}+0.7)&v_{IN}\gt0\\[2pt]-3\,v_{IN}+0.7&v_{IN}\lt0\end{cases}}$$ An inverting amplifier with a larger magnitude on the positive half (−5) than the negative (−3), with the usual ±0.7 V crossover offset.
Rail limits. The positive half hits −15 V for $v_{IN}\gt2.86$ V; the negative half hits +15 V for $v_{IN}\lt-4.77$ V. Both halves therefore clip on a ±10 V input.

Fig. 6d output: inverted, asymmetric gains, clipping at both rails.

(e) Feedback 3 kΩ in parallel with a 3 V source (no diode)

Fig. 6e — an ideal 3 V source sits directly in the feedback path (+ at the virtual ground).

Battery clamps the output. The ideal source forces $v_{(-)}-v_{OUT}=3$ V. With the virtual ground $v_{(-)}=0$, $$\boxed{v_{OUT}=-3\text{ V (constant)}}.$$ The parallel 3 kΩ merely carries the input current $v_{IN}/1\text{k}$; the output is a flat DC line independent of the input.

Fig. 6e output: a constant −3 V (the source pins it, regardless of the input sine).

Q6 — transfer relations (±15 V rails)
Part	Positive input	Negative input
(a)	v_o = −2v_IN (clip −15)	+15 V (rail)
(b)	−2.5(v_IN−0.7)	−5(v_IN+0.7); dead band \|v\|<0.7
(c)	−v_IN−4.27	+15 V (rail)
(d)	−(5v_IN+0.7) (clip −15)	−3v_IN+0.7 (clip +15)
(e)	−3 V constant (battery clamp)

Topic. Diodes in op-amp feedback/input paths — precision rectifiers, limiters and the loss of feedback.

Key relations. Virtual ground $v_-=0$; inverting gain $-R_f/R_{in}$; a diode that clamps a feedback node so the output cannot influence $v_-$ breaks the loop ⇒ the op-amp saturates to a rail; an ideal source in feedback fixes the output directly.

Why this works. When the feedback diode conducts, it either steers current to set a defined gain (precision-rectifier action) or pins an intermediate node and disconnects the output's control of the summing junction — in the latter case the ideal op-amp, unable to null its input error, drives fully to a rail. The direction of the rail follows the sign of the residual input error.

Common pitfall. Assuming feedback is always maintained: in (a) and (c) one input polarity breaks the loop and the output rails rather than following a finite gain. Also, forgetting the ±15 V clip when the computed gain would exceed the supply on a ±10 V input.

Source. Sedra & Smith, Microelectronic Circuits, 7th ed., §4.5 (precision rectifiers) and §2 (ideal op-amp, saturation).

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