GATE EC || ANALOG ELECTRONICS|| PYQs (2000-2025)

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Question 1

For the circuit in the below figure, the voltage V0 is (ASSUME IDEAL DIODE)

Screenshot-2025-09-02-113026


(GATE 2000 || EC || MCQ || 2 MARKS)

  • 2V

  • 1V

  • -1V

  • None of the above

Question 2

For the circuit shown in figure. D1 and D2 are identical diodes with utility factor of unity. The thermal voltage VT = 25 mV.

         (a) Calculate V1 and V2.

         (b) If the reverse saturation current of D1 and D2 are 1 pA then compute the current I through the circuit.

Screenshot-2025-08-20-120430


(GATE 2001 || EC || PYQ || NAT || 5 MARK)

  • V1 = 14.15 mV,  V2 = 35.85 mV

     I = 0.76pA

Question 3

The circuit shown in the figure is best described as a

Screenshot-2025-09-02-113457

(GATE 2003 || EC || PYQ || MCQ || 1 MARK)

  • bridge rectifier

  • ring modulator

  • frequency discriminatory

  • Voltage doubler

Question 4

In a full-wave rectifier using two ideal diodes, Vdc and Vm are the dc and peak values of the voltage respectively across a resistive load. If PIV is the peak inverse voltage of the diode, then the appropriate relationships for this rectifier are



(GATE 2004 || EC || PYQ || MCQ || 2 MARK )

  • [Tex]V_{dc} = \frac{V_m}{\pi}, \quad \text{PIV} = 2V_m[/Tex]

  • [Tex]V_{dc} = \frac{2V_m}{\pi}, \quad \text{PIV} = 2V_m[/Tex]

  • [Tex]V_{dc} = \frac{2V_m}{\pi}, \quad \text{PIV} = V_m[/Tex]

  • [Tex]V_{dc} = \frac{V_m}{\pi}, \quad \text{PIV} = V_m[/Tex]

Question 5

The correct full wave rectifier circuit is



(GATE 2007 || EC || PYQ || MCQ || 1 MARK)

  • Screenshot-2025-08-20-142121


  • Screenshot-2025-08-20-142203


  • Screenshot-2025-08-20-142238


  • Screenshot-2025-08-20-142316


Question 6

In the circuit below, the diode is ideal. The voltage V is given by

Screenshot-2025-09-02-114113

(GATE 2009 || EC || PYQ || MCQ || 2 MARK)

  • min (Vi, 1)

  • max (Vi, 1)

  • min (–Vi, 1)

  • max (–Vi, 1)

Question 7

In the circuit shown below, assume that the voltage drop across a forward bias diode is 0.7 V. The thermal voltage [Tex]V_t = \frac{KT}{q} = 25\ \text{mV}[/Tex]. The small signal input [Tex]V_i = V_p \cos(\omega t)[/Tex] where [Tex]V_p = 100\ \text{mV}[/Tex].

gate_

The bias current IDC through the diodes is


( GATE 2011 || EC || PYQ || MCQ || 2 MARK)

  • 1 mA

  • 1.28 mA

  • 1.5 mA

  • 2 mA

Question 8

In the circuit shown below, assume that the voltage drop across a forward bias diode is 0.7 V. The thermal voltage [Tex]V_t = \frac{KT}{q} = 25\ \text{mV}[/Tex]. The small signal input [Tex]V_i = V_p \cos(\omega t)[/Tex] where [Tex]V_p = 100\ \text{mV}[/Tex]. The ac output voltage Vac is

gate_

(GATE 2011 || EC || PYQ || MCQ || 2 MARK)

  • 0.25 cos(ωt) mV              

  • 1 cos(ωt) mV

  • 2 cos(ωt) mV

  • 22 cos(ωt) mV

Question 9


The diode and capacitors in the circuit shown are ideal. The voltage v(t) across the diode D1 is

Screenshot-2025-11-19-045131

(GATE 2012 || EC || PYQ || MCQ || 1 MARK)

  • cos(ωt) – 1

  • sin(ωt)

  • 1 - cos(ωt)

  • 1 - sin(ωt)

Question 10

A voltage 1000sinωt volts is applied across YZ. Assuming ideal diodes, the voltage measured across WX in volts, is

Screenshot-2025-11-19-045316

(GATE 2013 || EC || PYQ || MCQ || 2 MARK)

  • sinωt

  • [Tex]\frac{\sin \omega t + |\sin \omega t|}{2}[/Tex]

  • [Tex]\frac{\sin \omega t - |\sin \omega t|}{2}[/Tex]

  • 0 for all t

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