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# GATE-2012 ECE Q13 (circuits)

by on January 22, 2013

Question 13 on analog electronics from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

## Solution

The first half of the circuit is a negative clamper circuit and the second half is a peak detector circuit as shown in the figure below.

(Discussed in Section 3.8 of MicroElectronic Circuits Sedra/Smith (from Amazon.com, from Flipkart.com) or in  Chapter 6.17 of Millman’s Electronic Devices and Circuits (from Amazon.comfrom Flipkart.com)

The negative clamper circuit works as follows :

The diode D1 will be initially conducting till the voltage across the capacitor C1 is charged to the peak voltage of 1 volts. In the following cyclesDuring the first half of the +ve cycle, the diode D1 will be ON and the capacitor is charged to peak voltage. The diode D1 will remain OFF during the further cycles and the voltage ${V(t)}$ is given by,

${V(t)=-1+\cos(\omega t)}$

## .

The second half of the circuit i.e the peak detector circuit, provides a negative voltage of -2 volts at the output of capacitor C2. Anyhow, given that we are only interested in the output of the negative clamper, the  voltage ${V(t)}$ is

${V(t)=-1+\cos(\omega t)}$

Based on the above, the right choice is (A) 1

## References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

[3] Millman’s Electronic Devices and Circuits ( Buy from Amazon.comBuy from Flipkart.com)

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Pravin Kumar January 23, 2013 at 1:28 pm

For C2 to charge, D2 must be ON. This happens only during the negative cycles of the input. During the positive cycles. D2 is OFF only during the +ve half cycle.

Krishna Sankar January 24, 2013 at 5:36 am

@Pravin: Hope you agree that C2 gets charged to -2volts. Once that happens chances of D2 becoming ON is very less

Pravin Kumar January 22, 2013 at 8:59 pm

During the complete positive cycle of the input, D1 is ON making the output 0 for this half-cycle.
In the positive half-cycle, D1 is OFF (can be removed). Now D2 is ON and the voltage V(t) would be the voltage across C2, which is the capacitive division of C1/(C1+C2)*cos(wt) assuming that D2 is ideal. This is definitely not cos(wt) -1.