# Analog Circuits Questions and Answers – Characteristic Equation of Diode-1

This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Characteristic Equation of Diode 1”.

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

Explanation: 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 = 0.0252V, η = 1, V = 0.4V
Therefore, I = 1.17×10-9xe0.4/0.025 -1 = 9.156mA.

2. Calculate the thermal voltage when the temperature is 25°C.
a) 0V
b) 0V
c) 0.026V
d) 0.25V

Explanation: 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 = 298.15/11600 = 0.0257V.

3. Calculate the reverse saturation current of a diode if the current at 0.2V forward bias is 0.1mA at a temperature of 25°C and the ideality factor is 1.5.
a) 5.5x 10-9 A
b) 5.5x 10-8 A
c) 5.5x 10-7 A
d) 5.6x 10-10 A

Explanation: 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 = 0.0256V.
Therefore, reverse saturation current
IO=0.00055mA = 5.5×10-7A.

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

Explanation: 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 = 0.0256V, η = 1.5, I = 1mA, I0 = 10-10A

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

Explanation: 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

We know thermal voltage VT = TK/11600. Therefore, TK = VTx11600 = 0x11600 = 342.65K = 69.65°C.
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6. Consider a silicon diode with η=1.2. Find the change in voltage if the current changes from 0.1mA to 10mA.
a) 0.154V
b) 0.143V
c) 0.123V
d) 0.165V

Explanation: 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

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) 0.0718V
d) 0.00728V

Explanation: η = 1.2, I2 = 10mA, I1 = 1mA and take VT = 0.026V
.

8. What will be the ratio of final current to initial current of a diode if the 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

Explanation: η = 1.5, ΔV = 0.8725V and take VT = 0.026V

9. What will be the current I in the circuit diagram below. Take terminal voltage of diode as 0.7V and I0as 10-12A.

a) 2.4mA
b) 0.9mA
c) 1mA
d) 4mA

Explanation: 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)) = 0.5864V.
Now the current becomes (3-0.5864)/1000 = 2.41mA.

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.

a) 0.84mA
b) 1.84mA
c) 2.35mA
d) 3.01mA

Explanation: 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 = 1.8mA.
The VD = VT ln((I/IO)+1) = 0.58V
Hence current is (3-(2×0.58))/1k = 1.84mA.

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