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)].
Similar Questions
Assertion (A) : An Op-Amp is a direct coupled high gain amplifier.
Reason (R) : It consists of one or more differential amplifiers and usually followed by a level translator and push pull stage.
The input to a differentiator is –5 V. Its output will be
Match the following :
List – I List – II a. h-parameters i. O/P voltage varies as the slope of i/p voltage b. differentiator ii. Noise division c. half-wave rectifier iii. Function of a Q point d. integrator iv. series diode clipper Codes :
Assertion (A) : Op-Amp is used for sensor circuit.
Reason (R) : A small signal amplifier amplify weak measured signals.
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(A) The output voltage of a summing amplifier (inverting configuration) with three inputs VA, VB and VC and input resistors RA, RB and RC is \(V_{o}=\left(1+\dfrac{R_{F}}{R_{A}R_{B}R_{C}}\right)\left[\dfrac{V_{A}}{R_{A}}+\dfrac{V_{B}}{R_{B}}+\dfrac{V_{C}}{R_{C}}\right]\)
(B) In a subtractor circuit, the output voltage is equal to voltage applied to non-inverting terminal minus voltage applied to inverting terminal
(C) The narrow band pass filter is called a Notch filter
(D) VCO is also called as frequency to voltage
(E) The all pass filter provides unity-gain with predictable phase shifts for different input frequenciesChoose the most appropriate answer from the options given below :
Consider the following circuit, the switch S1 allows the output to switch between two ranges of amplitudes from 0-0.1 V and 0-1 V. Arrange these values of R1, R2 and R3 in increasing order.
(A) Value of R1
(B) Value of R2
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Match List I with List II
LIST I LIST II A. Butterworth filter of order '2' I. Impedance matching B. Buffer II. CMRR = ∞ (infinity) C. Schmitt Trigger III. Positive feedback D. Ideal OPAMP IV. 40 dB/decade roll off Choose the correct answer from the options given below:
The given operational amplifier circuit corresponds to which electronic circuit application ?
Statements in connection to Op-Amp applications are :
A. If we use a square wave generator followed by integrator circuit we get a triangular wave at the output
B. The logarithmic amplifier called a log-amplifier or a logger, is basically a current to voltage converter.
C.
is a first order high pass filter with voltage followerD. If we use a square wave generator followed by a clipping circuit then we get a saw-tooth wave generator.
Choose the correct answer from the options given below:
Match List I with List II
LIST I LIST II A. Single polarity positive voltage clipper circuit I. 
B. Negative clamping circuit II. 
C. Differentiator circuit III. 
D. Logarithmic Amplifier IV. 
Choose the correct answer from the options given below:
Important Questions from Op-Amp and Its Applications
What is the typical value of open-loop voltage gain, AVOL, for a 741 op-amp?
An ideal Op-Amp is an ideal
- In an ideal OP-AMP, which one is correct?
Which of the following statements about the Op-Amp differential amplifiers is INCORRECT?
The total output offset voltage of an operational amplifier is a function of these effects.
