Op-Amp Limiter with a Zener in the Feedback Path: 22-Elec-A5 Electronics, December 2015, Q2

Question 2: Op-Amp Limiter with a Zener in the Feedback Path (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. An inverting configuration whose only feedback element is the zener D₁ (anode at the output, cathode at the summing node). The input v_s is a ±7.5 V triangle.

Given data
Quantity	Value
R₁	10 kΩ
D₁	V_Z = 5 V, V_γ = 0.7 V
Slew rate	0.5 V/μs
Supplies	±15 V
v_s	±7.5 V triangle, T = 100 μs (slope 0.3 V/μs)

Find. The output waveform v_o(t) with its two levels, transition timing, and slew-limited edges.

Figure 2.1 — v_s drives the inverting node through R₁; the non-inverting input is grounded; D₁ is the sole feedback element from v_o back to v₋.

Approach. The op-amp holds its inverting input at virtual ground, so the input current $i=v_s/R_1$ must return through D₁. The sign of that current decides whether the zener is forward-conducting or in reverse breakdown, which sets v_o to one of two fixed levels; the slew rate then rounds the edges between them.

Virtual ground and feedback current. With $v_-=0$, the resistor delivers $i=\dfrac{v_s}{R_1}$ into the summing node, and this same current must flow through D₁ between $v_-$ and v_o.
Negative input — diode forward. For $v_s\lt 0$ the current flows from v_o toward $v_-$, the forward direction (anode at v_o). The output sits one forward drop above the virtual ground: $v_o=v_-+V_\gamma=+0.7\ \text{V}.$
Positive input — zener breakdown. For $v_s\gt 0$ the current must flow from $v_-$ toward v_o, the reverse direction for the diode. The zener holds this off until it enters reverse breakdown at V_Z, clamping $\boxed{v_o=-V_Z=-5\ \text{V}\ \ (v_s\gt0),\qquad v_o=+V_\gamma=+0.7\ \text{V}\ \ (v_s\lt0).}$ Note the two clamp levels occur on opposite half-cycles — they are not added.
Timing and slew-limited edges. The output switches state each time v_s crosses zero. For the ±7.5 V, 100‑μs triangle the zero crossings fall at t = 25, 75, 125, 175 μs. The full step is $\Delta v_o=0.7-(-5)=5.7$ V, so each edge takes $t_\text{slew}=\frac{\Delta v_o}{SR}=\frac{5.7\ \text{V}}{0.5\ \text{V}/\mu s}=11.4\ \mu s,$ comfortably shorter than the 50‑μs half-period. The result is a rectangular wave between −5 V and +0.7 V with 11.4‑μs ramped transitions centred on the input zero crossings.

Figure 2.2 — v_o (blue): −5 V while v_s>0, +0.7 V while v_s<0, with 11.4‑μs slew edges at each input zero crossing (grey = v_s).

Question 2 — results
Interval	v_o
v_s > 0 (zener breakdown)	−5.0 V
v_s < 0 (diode forward)	+0.7 V
Output step	5.7 V
Slew time per edge	11.4 μs

Topic: a single zener as the feedback element makes a two-level op-amp limiter; finite slew rate rounds the edges.

Key relations: virtual ground $i=v_s/R_1$; forward clamp $v_o=+V_\gamma$; breakdown clamp $v_o=-V_Z$; slew time $t=\Delta v_o/SR$.

Why this works: because the diode is the only path around the amplifier, the output can never settle to a linear gain — it moves until the feedback element can carry the demanded current, which happens only at the diode’s forward drop or at its reverse-breakdown voltage. The input’s sign selects which, so the output is a clean two-level wave. The op-amp still tries to change v_o instantly at each crossing, but the slew rate caps the rate of change, giving finite-slope edges whose duration is the total swing divided by SR.

Common pitfall: adding the 0.7 V and 5 V drops, or forgetting that they land on opposite half-cycles. Also, the drop that appears when v_s>0 is the reverse-breakdown V_Z (5 V), not the forward 0.7 V.

Source: Sedra & Smith, Microelectronic Circuits 8th ed., Ch. 4 (limiters & clamps) and Ch. 2 (slew rate).

Question 2 of 5: Op-Amp Limiter with a Zener in the Feedback Path

Question 2: Op-Amp Limiter with a Zener in the Feedback Path (20 marks)