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1. Calculate the forward bias current of a Si diode when forward bias voltage of 0.4V is applied, using characteristic equation of diode. The reverse saturation current is 1.17 x 10-9 A. Thermal voltage is 25.2 mV.
a) 9.156mA
b) 8.23mA
c) 1.256mA
d) 5.689mA

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Answer: a [Reason:] Equation for diode current I=I0×(e(V/ηVT)-1) where I0 = reverse saturation current η = ideality factor VT = thermal voltage V = applied voltage Since in this question ideality factor is not mentioned it can be taken as one. I0 = 1.17 x 10-9A, VT = 0V, η = 1, V = 0.4V Therefore, I= 1.17×10-9xe0.4/0.025 -1 = 9.156mA

2. Calculate thermal voltage for a temperature of 25oC.
a) 0V
b) 0V
c) 0.026V
d) 0.25V

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Answer: b [Reason:] Thermal voltage VT is given by k T/q Where k is the boltzman constant and q is the charge of electron. This can be reduced to VT = TK/11600 Therefore, VT at T= 25+273=298 is 298/11600 = 0V

3. Calculate the reverse saturation current of a diode if the current at 0.2V forward bias is 0.1mA at a temperature 25 oC and ideality factor is 1.5
a) 5.06x 10-9 A
b) 5.06x 10-8 A
c) 5.06x 10-7 A
d) 5.06x 10-10 A

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Answer: c [Reason:] Equation for diode current I=I0×(e(V/ηVT ) -1) where I0 = reverse saturation current η = ideality factor VT = thermal voltage V = applied voltage Here, I = 0.1mA, η = 1.5, V= 0.2V, VT = TK/11600 Therefore, VT at T= 25+273=298 is 298/11600 = 0V. Therefore, reverse saturation currentanalog-circuits-questions-answers-characteristic-equation-diode-1-q3

4. Find the applied voltage on a forward biased diode if the current is 1mA and reverse saturation current is 10-10A. Temperature is 25 oC and take ideality factor as 1.5
a) 0.658V
b) 0.726V
c) 0.526V
d) 0.618V

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Answer: d [Reason:] Equation for diode current I=I0×(e(V/ηVT)-1) where I0 = reverse saturation current η = ideality factor VT = thermal voltage V = applied voltage VT at T= 25+273=298 is 298/11600 = 0V, η = 1.5, I = 1mA, I0 = 10-10A analog-circuits-questions-answers-characteristic-equation-diode-1-q4

5. Find the temperature at which a diode current is 2mA for a diode which has reverse saturation current of 10-9 A and ideality factor is 1.4, the applied voltage is 0.6V forward bias.
a) 69.65oC
b) 52.26oC
c) 25.23oC
d) 70.23oC

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Answer: a [Reason:] Equation for diode current I=I0×(e(V/ηVT)-1) where I0 = reverse saturation current η = ideality factor VT = thermal voltage V = applied voltage I0 = 10-9A, η = 1.4, V =0.6V, I = 2mA analog-circuits-questions-answers-characteristic-equation-diode-1-q5 We know thermal voltage VT = TK/11600 .Therefore, TK = VTx11600 = 0x11600 = 342.65K = 69.65oC.

6. Consider a silicon diode with η=1.2. Find change in voltage if the current changes from 0.1mA to 10mA.
a) 0.154V
b) 0.143V
c) 0.123V
d) 0.165V

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Answer: b [Reason:] Equation for diode current I=I0×(e(V/ηVT)-1) where I0 = reverse saturation current η = ideality factor VT = thermal voltage V = applied voltage η = 1.2, I2 = 10mA, I1 = 0.1mA and take VT = 0.026V analog-circuits-questions-answers-characteristic-equation-diode-1-q6

7. If current of a diode changes from 1mA to 10mA what will be the change in voltage across the diode. The ideality factor of diode is 1.2
a) 0.718V
b) 7.18V
c) 0V
d) 0.00728V

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Answer: c [Reason:] η = 1.2, I2 = 10mA, I1 = 1mA and take VT = 0.026V analog-circuits-questions-answers-characteristic-equation-diode-1-q7.

8. What will be the ratio of final current to initial current of a diode if voltage of a diode changes from 0.7V to 872.5mV. Take ideality factor as 1.5
a) 90.26
b) 52.36
c) 80.23
d) 83.35

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Answer: d [Reason:] η = 1.5, ΔV = 0.8725V and take VT = 0.026V analog-circuits-questions-answers-characteristic-equation-diode-1-q8

9. What will be the current I in the circuit diagram below. Take terminal voltage as 0.7V and I0 as 10-12A.
analog-circuits-questions-answers-characteristic-equation-diode-1-q9
a) 2.4mA
b) 0.9mA
c) 1mA
d) 4mA

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Answer: a [Reason:] Let VD be the voltage of diode, then by Kirchoff’s loop rule 3V = VD + IR1 This method of assumption contains small error but it is the simplest method. Let VD be 0.7V. Now the current I = (3-0.7)/1k = 2.3mA. Now the diode voltage for 2.3mA VD = VT ln⁡(I/I0 ) = 0.026 x ln((2.3×10(-3))/10(-12) ) = 0.560. Now the current becomes (3-0.560)/1000 = 2.44mA.

10. Find current I through the circuit using characteristic equation of diode. The terminal voltage of each diode is 0.6V. Reverse saturation current is 10-12A.
analog-circuits-questions-answers-characteristic-equation-diode-1-q10
a) 0.845mA
b) 1.892mA
c) 2.359mA
d) 3.012mA

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Answer: b [Reason:] Let VD be the voltage of diode, then by Kirchoff’s loop rule 3V = 2VD + IR1 This method of assumption contains small error but it is the simplest method. Let VD be 0.6V. Now the current I = (3-1.2)/1k = 2.8mA. The VD for 2.8mA is 0.554V. Hence current is (3-(2×0.554))/1k =1.892mA.