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NCERT Class XII Chapter
Semiconductor Electronics
Questions With Solutions

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Question : 14
Total: 19
In a p-n junction diode, the current I can be expressed as
I = I0[exp(
eV
kBT
)−1] where I0 is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, kB is the Boltzmann constant (8.6 × 10–5 eV/K) and T is the absolute temperature. If for a given diode I0 = 5 × 10–12 A and T = 300 K, then
(a) What will be the forward current at a forward voltage of 0.6 V?
(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V ?
Solution:  
The current I through a junction diode is given as
I0[exp(
eV
kBT
)−1]
where I0 = 5 × 10−12 A , T = 300 K,
kB = 8.6 × 10−5 eV k−1 = 8.6 × 10−5 × 1.6 × 10−19 J K−1
(a) When V = 0.6 V
eV
kBT
 =
1.6×10−19×0.6
8.6×1.6×10−24×300
 =
600
8.6×3
 = 23.26
∴ I = I0[exp(
eV
kBT
)−1] = 5 × 10−12 [exp (23.26) - 1] A
= 5 × 10−12 [1.2586 × 1010 - 1] A = 5 × 10−12 × 1.2586 × 1010 A = 0.06293 A
(b) When V = 0.7 V,
eV
kBT
 =
1.6×10−19×0.7
8.6×1.6×10−24
 = 27.13
∴ I = I0[exp(
eV
kBT
)−1]
= 5 × 10−12 [exp (27.13) - 1] A = 5 × 10−12 × [6.07 × 10−11 - 1] A
= 5 × 10−12 × 6.07 × 1011 A = 3.035 A
Increase in current ,
ΔI = 3.035 – 0.06293 = 2.972 A.
(c) For ΔV = 0.7 – 0.6 = 0.1 V, ΔI = 2.972 A
Dynamic resistance, rd =
ΔV
ΔI
 =
0.1
2.972
 = 0.0336 Ω
(d) For both the voltages, the current I will be almost equal to I0, showing almost infinite dynamic resistance in the reverse bias.
I = - I0 = - 5 × 10−12 A
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