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1. After cut-in voltage in piecewise linear model diode act as a
a) Resistor
b) Capacitor
c) Conductor
d) Insulator

Answer: a [Reason:] After cut –in voltage diode act as a resistor in piecewise linear mode. In normal operation diode current is exponentially related to voltage.

2. Reverse biased condition of a diode in piecewise linear model is equivalent to
a) Resistor
b) Capacitor
c) Conductor
d) Insulator

Answer: d [Reason:] For a diode in reverse bias mode current through the diode is in micro amperes or nano amperes. Hence we can assume it as zero. In piecewise linear model reverse current is assumed to zero. That is, as an insulator.

3. Voltage drop produced by a diode in piecewise linear mode is
a) Constant and equal to knee voltage
b) Varying linearly with voltage
c) Varies exponentially with voltage
d) Constant and equal to twice of knee voltage

Answer: b [Reason:] Voltage drop produced by diode in piecewise linear model is not constant. Since it contains effect of resistor, the diode voltage linearly increases as input voltage increases.

4. In the given circuit voltage V = 2V.cut-in voltage of diode is 0.7V. The current I through the circuit is
(Assume piecewise linear model for diode)

a) 0.235mA
b) 1.29mA
c) 1.63mA
d) 2.27mA

Answer: d [Reason:] Since diode is in forward bias mode it can replaced by the equivalent circuit I = (V-VD)/R1+RD = (3-0.7)/1010 = 2.27mA.

5. In the given circuit input voltage Vin is 3V and V2 is 1V. The resistance R1 is 1.5K. Cut-in voltage of diode is 0.5V. Forward bias resistance is 10Ω. The Voltage V will be

a) 2.03mA
b) 0.23mA
c) 1.58mA
d) 1.33mA

Answer: d [Reason:] Since both voltage sources are reverse bias to the diode, diode in the circuit disappears and equivalent circuit becomes as follows So current I = V1-V2/R = 3-1/1.5k = 1.33mA.

6. In the given circuit input voltage V1 is -3V and V2 is -1V. The resistance R1 is 1K. Cut-in voltage of diode is 0.5V. Forward bias resistance is 10Ω. The Voltage V will be
(Use piecewise linear model of diode)

a) -1mA
b) -2mA
c) -0.2mA
d) -0.1mA

Answer: b [Reason:] Since both voltage sources are in forward bias to diode, the equivalent circuit will be as follows Since voltage across diode is 1V. current I = -3+1/1k = -2mA.

7. In the given circuit input voltage V is 2V and VB is 1V. The resistance R1 and R2 is 1K. Cut-in voltage of diode is 0.5V. Forward bias resistance is 10Ω. The current I will be
(Use piecewise linear model of diode)

a) 0.29mA
b) 0.21mA
c) 0.36mA
d) 0.15mA

Answer: a [Reason:] Since V-VB = 1V forward biases the diode, we can use equivalent circuit of diode as follows Current through R1, I1 = 1V/R1 = 1mA. Current through R2, I2 = (1-VD )/(R2¬+RD) = (1-0.7)/1010 = 0.297mA.

8. In the given circuit input voltage V is 3V and VB is 1V. The resistance R1 and R2 is 1K. Cut-in voltage of diode is 0.5V. Forward bias resistance is 10Ω. The current I will be
(Use piecewise linear model of diode)

a) 0.96mA
b) 2.13mA
c) 1.56mA
d) 1.23ma

Answer: c [Reason:] Since diode is in forward bias we can assume equivalent circuit model and assume following circuit Let voltage across diode is V0 now voltage across branch is V-V0 Current I = (2-V0)/R1 +(2-V0-VB)/R2 =(2-V0)/1000+((1-V0 ))/1000 …………………….(1) V0 = VD+ IRD = 0.7+10I Put this value in eq(1) That is, I = (2-0.7-10I)/1000+((1-0.7-10I))/1000 => 1000I = 1.6 – 20I => 1020 I = 1.6 That is, I = 1.6/1020 = 1.56mA.

9. In the given circuit input voltage V is -3V and VB is 1V. The resistance R1 and R2 is 1K. Cut-in voltage of diode is 0.5V. Forward bias resistance is 10Ω. The current I will be
(Use piecewise linear model of diode)

a) 1.2mA
b) 0mA
c) 0.8mA
d) 1mA

Answer: b [Reason:] Since diode is in reverse bias mode there won’t be any current in the circuit. Both voltage sources reverse biases the diode.

10. In the given circuit input voltage Vin is 3V and VB is 1V. The resistance R is 1K. Cut-in voltage of diode is 0.5V. Forward bias resistance is 10Ω. The current I will be
(Use piecewise linear model of diode)

a) 1V
b) 3V
c) 2.3V
d) 1.3V