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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(A) Value of R1
(B) Value of R2
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A. If we use a square wave generator followed by integrator circuit we get a triangular wave at the output
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C.
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Choose the correct answer from the options given below :
Match the following :
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i) 
b) 
ii) 
c) 
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\(\dfrac{d^{2}y}{dt^{2}}+y=0\) can be solved using an Operational Amplifier. Which one is preferred ?
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 _____________.
