Hysteresis width of a non-inverting op-amp Schmitt trigger
BlinkNBuild practice problem · GATE standard, authored and verified in-house
In the non-inverting Schmitt trigger shown, the op-amp saturates at $\pm 12\ \text{V}$, the inverting input is grounded, $R_1 = 10\ \text{k}\Omega$ (input to the non-inverting node) and $R_2 = 20\ \text{k}\Omega$ (output back to the non-inverting node).
The hysteresis width $V_{TH} - V_{TL}$, in volts, is ________.
Show the step-by-step derivation
Step-by-step derivation
- The op-amp is in positive feedback, so it never settles in a linear region - the output is always at one saturation rail. The switching condition is the moment the two inputs are equal.
- The inverting input is grounded, so $V_- = 0$. Switching therefore occurs when $V_+ = 0$.
- Find $V_+$ by superposition at the non-inverting node, which is driven by $V_{in}$ through $R_1$ and by $V_{out}$ through $R_2$: $$V_+ = V_{in}\frac{R_2}{R_1+R_2} + V_{out}\frac{R_1}{R_1+R_2}.$$
- Set $V_+ = 0$ and solve for the input threshold: $$V_{in}\frac{R_2}{R_1+R_2} = -V_{out}\frac{R_1}{R_1+R_2} \;\Longrightarrow\; V_{in} = -V_{out}\,\frac{R_1}{R_2}.$$ The $(R_1+R_2)$ terms cancel.
- Upper threshold. While the output sits at $V_{out} = -12$ V, the input must rise to $$V_{TH} = -(-12)\frac{10\text{k}}{20\text{k}} = +12 \times 0.5 = +6\ \text{V}$$ to trip the comparator.
- Lower threshold. While the output sits at $V_{out} = +12$ V, the input must fall to $$V_{TL} = -(+12)\frac{10\text{k}}{20\text{k}} = -6\ \text{V}.$$
- Hysteresis width: $$V_{TH} - V_{TL} = 6 - (-6) = \mathbf{12\ V}.$$
- Compact form. For this topology the width is always $$\Delta V = 2\,V_{sat}\frac{R_1}{R_2} = 2(12)\left(\tfrac{1}{2}\right) = 12\ \text{V},$$ which is why $R_1/R_2$ is the single knob for noise immunity.
The idea behind this question
A Schmitt trigger is a comparator with positive feedback, so it has two thresholds - one for a rising input and one for a falling input. The gap between them is the hysteresis, which stops noise on a slowly changing input from making the output chatter. The thresholds depend on whether the input drives the inverting or the non-inverting side.
Try a variation
With $R_2 = 30\ \text{k}\Omega$ instead, what is the hysteresis width?
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Answer: $8$ V
The thresholds are $\pm 12 \times 10/30 = \pm 4$ V, so the width is $8$ V.
Other mistakes to avoid
- Treating it as a negative-feedback amplifier and assuming both inputs are at the same voltage.
- Giving one threshold when the question asks for the width between them.