Question 3 of 5: Op-Amp Comparator with Positive Feedback (Schmitt Trigger)

Question 3: Op-Amp Comparator with Positive Feedback (Schmitt Trigger) (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. Ideal op amp, output clamped at $\pm10$ V. The input reaches the inverting terminal ($v_1$) through the $R_3$–$R_4$ divider; $R_2$ feeds the output back to the non-inverting terminal ($v_2$), which also has $R_1$ to ground. Input: $\pm10$ V triangle, 1 kHz.

Find. The time waveforms of $v_1$ and $v_2$.

Approach. Feedback to the $+$ input is positive, so the stage is a bistable comparator (Schmitt trigger): the output sits at a rail and the input node voltages set the two switching thresholds.

1. Inverting-node voltage. No current enters the op-amp input, so $v_1$ is just the divided input: $$v_1=v_{IN}\frac{R_4}{R_3+R_4}=\tfrac12 v_{IN},$$ a triangle of amplitude $\pm5$ V that follows the input directly.
2. Non-inverting-node voltage (the thresholds). $v_2$ is the output fed back through the $R_1$–$R_2$ divider: $$v_2=v_{OUT}\frac{R_1}{R_1+R_2}=\frac{v_{OUT}}{11}=\pm0.909\ \text{V}$$ for $v_{OUT}=\pm10$ V. This is a square wave that jumps with the output.
3. Switching condition. The comparator changes state when $v_1=v_2$. Starting at $v_{OUT}=+10$ V ($v_2=+0.909$ V) the output holds until $v_1$ rises past $+0.909$ V, i.e. $v_{IN}=+1.82$ V, then snaps to $-10$ V. Starting at $v_{OUT}=-10$ V it holds until $v_1$ falls below $-0.909$ V, i.e. $v_{IN}=-1.82$ V. $$\boxed{\text{trip points } v_{IN}=\pm1.82\ \text{V},\quad v_{OUT}=\pm10\ \text{V}}$$
4. Resulting waveforms. $v_1$ is the $\pm5$ V triangle; $v_2$ is a $\pm0.909$ V square wave in phase with the output; the output itself is a $\pm10$ V square wave whose edges are inverted relative to the input (input at the $-$ terminal) and offset by the hysteresis.