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Question: In PN- junction diodes the reverse saturation current is \[{10^{ - 5}}\] amp at \[{27^0}C\]. The for...

In PN- junction diodes the reverse saturation current is 10−5{10^{ - 5}} amp at 270C{27^0}C. The forward current for a voltage of 0.20.2 volt is
(A) 2037.6×10−3\left( {\text{A}} \right){\text{ }}2037.6 \times {10^{ - 3}} amp
(B) 203.76×10−3\left( {\text{B}} \right){\text{ }}203.76 \times {10^{ - 3}} amp
(C) 20.376×10−3\left( {\text{C}} \right){\text{ }}20.376 \times {10^{ - 3}} amp
(D) 2.0376×103\left( {\text{D}} \right){\text{ }}2.0376 \times {10^3} amp

Explanation

Solution

Diode: It is a two-terminal electronic device that produces current primarily in one direction; it has low resistance in one direction and high resistance in other direction.
P-type semiconductor: Trivalent impurity (Boron, Gallium, Indium, and Aluminum) is added to an intrinsic or pure semiconductor (Silicon, Germanium).
N-type semiconductor: Pentavalent impurity (Arsenic, Antimony, and Phosphorus) is added to an intrinsic or pure semiconductor (Silicon, Germanium).
PN-junction: It is formed when a p-type semiconductor is fused to an n-type semiconductor.

Formula used:
i = is(eev/kT−1){\text{i = }}{{\text{i}}_{\text{s}}}({{\text{e}}^{{\text{ev/kT}}}} - 1),
Here i=i = forward current in the PN-junction, is={i_s} = saturation current in the PN-junction,
e=e = Charge of the electron,  v=\;v = voltage source, k=k = Boltzmann’s constant, T=T = absolute temperature

Complete step by step answer:
It is given that the question,is= 10−5amp{i_s} = {\text{ }}{10^{ - 5}}amp, T= 270CT = {\text{ }}{27^0}C= (273+27)(273 + 27) k=300Kk = 300K, v=0.2Vv = 0.2V
Now using the values and formula we get that i = is(eev/kT−1){\text{i = }}{{\text{i}}_{\text{s}}}({{\text{e}}^{{\text{ev/kT}}}} - 1)
⇒i=10−5[e1.6×10−19×0.21.4×10−23×300−1]\Rightarrow i= {10^{ - 5}}\left[ {{{\text{e}}^{\dfrac{{1.6 \times {\text{1}}{{\text{0}}^{ - 19}} \times {\text{0}}{\text{.2}}}}{{1.4 \times {\text{1}}{{\text{0}}^{ - 23}} \times {\text{300}}}} - 1}}} \right]
⇒i=10−5[e0.32×10−19420×10−23−1]\Rightarrow i= {10^{ - 5}}\left[ {{{\text{e}}^{\dfrac{{0.32 \times {\text{1}}{{\text{0}}^{ - 19}}}}{{420 \times {\text{1}}{{\text{0}}^{ - 23}}}} - 1}}} \right]
⇒i=10−5[2038.6−1]\Rightarrow i= {10^{ - 5}}[2038.6 - 1]
On some simplification we get,
⇒i=20.376×10−3\Rightarrow i = 20.376 \times {\text{1}}{{\text{0}}^{ - 3}}A

Hence the correct option is (C)\left( {\text{C}} \right).

Note:
Forward current: If the electrical current flows in the direction of lower resistance in a diode is called forward current.
A certain amount of positive voltage to be applied across the PN-junction to perform forward current.
A steady or equilibrium stage will be reached whereby electrons are diffusing out of the N-region and drifting (pulling by the huge positive charge left behind) back in at the same rate.
PN-junction is used in solar cells, a forward-biased diode is used in LED, and it is also used in rectifiers in many electric circuits.
In a PN-junction diode, the reverse saturation current is due to the diffusive flow of minority electrons from the p-side to the n-side and the holes move from the n-side to the p-side.