For an inverting comparator circuit acting as a Schmitt Trigger, as shown in figure below, the expression of Hysteresis Voltage (Vny) is given by :
\(V_{ny}=\dfrac{R_{1}}{R_{1}+R_{2}}\left[+V_{sat}-\left(-V_{sat}\right)\right]\)
The hysteresis is the gap between the two trip points, and each trip point is the output fed back through a plain resistive divider — option 4.
Step 1 — find the trip points. The non-inverting terminal is driven from the output through the divider R1, R2, so whatever the output rail, the reference the comparator uses is
\(V_{ref}=\dfrac{R_{1}}{R_{1}+R_{2}}\,V_{o}\)
Since the output can only sit at one of the two saturation levels, there are exactly two such references:
\(V_{UT}=\dfrac{R_{1}}{R_{1}+R_{2}}\left(+V_{sat}\right),\qquad V_{LT}=\dfrac{R_{1}}{R_{1}+R_{2}}\left(-V_{sat}\right)\)
Step 2 — the hysteresis is their difference.
\(V_{ny}=V_{UT}-V_{LT}=\dfrac{R_{1}}{R_{1}+R_{2}}\left[+V_{sat}-\left(-V_{sat}\right)\right]=\dfrac{2R_{1}V_{sat}}{R_{1}+R_{2}}\)
A check that eliminates the other options at once. The divider ratio \(R_{1}/(R_{1}+R_{2})\) must lie between 0 and 1, since a passive divider cannot amplify. Option 1 multiplies a voltage by a resistance, which is dimensionally impossible. Option 2's ratio \(R_{1}/R_{2}\) exceeds 1 whenever \(R_{1}\gt R_{2}\), predicting more hysteresis than the supply rails allow. Option 3's denominator vanishes when \(R_{1}=R_{2}\), giving infinite hysteresis from a finite circuit — and turns negative beyond it.
Why the feedback must be positive. The divider returns to the non-inverting input, so any movement of the output reinforces itself. The moment the input crosses the threshold, the output starts to move, which moves the threshold away from the input — and the transition snaps through in nanoseconds instead of following the input's own slew rate.
What the hysteresis buys. A plain comparator with a slow, noisy input chatters: noise of a few millivolts around the threshold produces multiple output transitions. With hysteresis, once the output has switched, the input must travel the whole way back past the other threshold before it can switch again, so any noise smaller than \(V_{ny}\) is ignored entirely. Choosing \(R_{1}\) and \(R_{2}\) therefore amounts to choosing how much input noise the circuit should tolerate.
Hence, Vny = R1/(R1 + R2) [+Vsat − (−Vsat)].
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Important Questions from Op-Amp and Its Applications
In the circuit shown below, $R$ is $1\text{ k}\Omega$ and $R_f$ is $10\text{ k}\Omega$. If $V_{\text{in}}$ is $100\text{ mV}$ and the op-amps have supply voltages of $\pm 15\text{ V}$, then $V_{\text{out}}$ is _______ V.
For IC 741 circuit shown below, the maximum input frequency above which the output will be distorted is ____________.
Assume that an op-amp 741 connected as a unit gain inverting amplifier is applied with an input change of 10 V. Determine the time taken for the output to change by 10 V. [Slew rate for op-amp 741 = 0.5 V/µs]
What is the value of output voltage $V_o$ in the given circuit? Assume op-amp to be ideal and MOSFET operating in saturation region.
In an ideal op-amp, the common mode gain is _____________.
