Question 4 of 5: Zener‑Limited Astable Function Generator — Output Waveform

Question 4: Zener‑Limited Astable Function Generator — Output Waveform (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. $A_1$ has its inverting input grounded and its non‑inverting input fed from the limited node $v_1$ through $R_2$ and from the output $v_{OUT}$ through $R_1$; $R_3$ feeds the back‑to‑back zener pair that clamps $v_1$; $A_2$ is an inverting integrator built from $R$ and $C$, and its output is $v_{OUT}$.

Find. The waveform at $v_{OUT}$ — its shape, its peak levels, its slew rate and its period — drawn to scale with levels and timings marked.

Approach. Identify the two blocks, fix the clamped square‑wave level, find the bistable’s switching threshold referred to $v_{OUT}$, then divide the peak‑to‑peak swing by the integrator’s slew rate to get the half period.

1. Recognise the two blocks. $A_1$ has its feedback returned to the non‑inverting input, so it is a bistable (Schmitt trigger), not a linear amplifier: its output rests at one saturation rail until the voltage at that input crosses zero. $A_2$ has $C$ from output to inverting input and its non‑inverting input grounded, so it is an ideal inverting integrator. A bistable driving an integrator whose output drives the bistable back is the standard astable function generator, and it needs no input.
2. Fix the level at $v_1$. $A_1$ saturates at $\pm15\ \text{V}$, but $R_3$ feeds the two zeners connected cathode‑to‑cathode from $v_1$ to ground. Whichever way $v_1$ tries to go, one diode conducts forward ($0\ \text{V}$) and the other breaks down ($10\ \text{V}$), so $$v_1=\pm\left(V_Z+V_F\right)=\pm10\ \text{V}$$ and $R_3$ absorbs the difference, carrying $(15-10)/10\ \text{k}\Omega=0.5\ \text{mA}$.
3. Find the switching threshold. Superposition at the non‑inverting node of $A_1$, fed by $R_1$ from $v_{OUT}$ and by $R_2$ from $v_1$, gives $v_+=(v_{OUT}R_2+v_1R_1)/(R_1+R_2)$. The bistable flips when $v_+=0$, i.e. when $$v_{OUT}=-v_1\frac{R_1}{R_2}=\mp10\times\frac{10}{10}=\boxed{\mp10\ \text{V}}$$ so the triangle turns around at exactly $+10\ \text{V}$ and $-10\ \text{V}$, a 20 V peak‑to‑peak swing.
4. Find the ramp rate. The integrator gives $v_{OUT}(t)=-\dfrac{1}{RC}\displaystyle\int v_1\,dt$, and $v_1$ is constant within each half cycle, so $v_{OUT}$ is a straight line of slope $$\frac{dv_{OUT}}{dt}=-\frac{v_1}{RC}=\mp\frac{10}{(10\times10^{3})(100\times10^{-12})}=\mp10\ \text{V}/\mu\text{s}$$ since $RC=1\ \mu\text{s}$. A positive $v_1$ ramps the output down and vice versa, which is exactly the sign that sustains oscillation.
5. Convert the swing into a half period. Each half cycle takes the output across the full 20 V between thresholds at 10 V/μs: $$\frac{T}{2}=\frac{2\left|v_{OUT,pk}\right|}{\left|dv_{OUT}/dt\right|}=\frac{20}{10}=2\ \mu\text{s}$$
6. State the period and frequency. The two halves are symmetric, so $$T=4RC\frac{R_1}{R_2}=4(10\times10^{3})(100\times10^{-12})(1)=\boxed{4\ \mu\text{s}},\qquad f=\frac{1}{T}=\boxed{250\ \text{kHz}}$$ Note that the closed form depends only on $R$, $C$ and the resistor ratio — the zener voltage sets the amplitude but cancels out of the frequency, which is what makes this topology a practical function generator.
7. Describe the sketch. $v_{OUT}$ is a symmetric triangle wave running between $-10\ \text{V}$ and $+10\ \text{V}$ with straight $\pm10\ \text{V}/\mu\text{s}$ flanks, each flank 2 μs long, period 4 μs, mean value zero. It is accompanied at node $v_1$ by a square wave of the same period switching between $\pm10\ \text{V}$, whose edges coincide exactly with the triangle’s peaks; $A_1$’s own output is the same square wave at $\pm15\ \text{V}$.