The given operational amplifier circuit corresponds to which electronic circuit application ?
Half wave rectification
Recognise the topology. An inverting op-amp stage with two diodes in the feedback path — one returning the op-amp output directly to the summing node and the other passing through Rf to the load — is the standard precision (super-diode) half-wave rectifier.
How it works, half cycle by half cycle.
For one polarity of Vi, the op-amp output drives D2 into conduction while D1 is reverse biased. The feedback loop is then closed through Rf, and the circuit behaves as an ordinary inverting amplifier:
\(V_o = -\dfrac{R_f}{R_i}V_i\)
For the opposite polarity, D2 turns off and D1 conducts instead. D1 closes the loop around the op-amp so that it does not saturate, but the output node is disconnected from the amplifier, so
\(V_o = 0\)
One polarity is passed (amplified and inverted), the other is blocked — that is half-wave rectification.
Why the op-amp is there at all. A plain diode rectifier cannot handle inputs smaller than its 0.7 V forward drop, and it distorts small signals badly. Inside the feedback loop the diode drop is divided by the open-loop gain, so the effective cut-in voltage falls to microvolts:
\(V_{\gamma(effective)} \approx \dfrac{V_\gamma}{A_{OL}}\)
This is what makes precision rectifiers essential in a.c. millivoltmeters, AM detectors and signal-conditioning front ends. The role of D1 is equally important: without it the op-amp would slam into saturation on the blocked half cycle and its slew-rate recovery would distort the next half cycle.
Why the other options are wrong. Full-wave rectification needs a second stage (typically a precision half-wave stage summed with the original signal in an inverting adder), so a single op-amp with two feedback diodes cannot deliver it. A voltage doubler is a capacitor–diode charge-pump with no feedback resistor. A peak detector requires a hold capacitor across the output to store the maximum value, which this circuit does not have.
Hence, the circuit performs half wave rectification.
Similar Questions
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Reason (R) : It consists of one or more differential amplifiers and usually followed by a level translator and push pull stage.
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Match the following :
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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 \(V_{o}=\left(1+\dfrac{R_{F}}{R_{A}R_{B}R_{C}}\right)\left[\dfrac{V_{A}}{R_{A}}+\dfrac{V_{B}}{R_{B}}+\dfrac{V_{C}}{R_{C}}\right]\)
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(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
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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. 
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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.
