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Question: Find static and dynamic resistances of a PN junction germanium diode for an applied forward bias of ...

Find static and dynamic resistances of a PN junction germanium diode for an applied forward bias of 0.2volts0.2volts, if the temperature is 300oK{300^o}K and reverse saturation current of 1A1A:
(A) 2.85Ω,22.5Ω2.85\Omega, 22.5\Omega
(B) 22.85Ω,2.5Ω22.85\Omega, 2.5\Omega
(C) 5.85Ω,26.5Ω5.85\Omega, 26.5\Omega
(D) 5.63Ω,55Ω5.63\Omega, 55\Omega

Explanation

Solution

Hint In these kinds of situations the first thing important is to notice the given problem statement. Specially this is a question regarding little manipulation. First we just have to sort what we got and then we just have to interpret the missing quantities for the equation. In this question we will get the answer simply by putting the values in the equation for current and then resistances.

Complete step by step answer:
Here, first we have to find sort out the given quantities:
T=300K V=0.2volts k=1.38×1023  T = 300K \\\ V = 0.2volts \\\ k = 1.38 \times {10^{ - 23}} \\\
And the Is{I_s} for the Germanium diode is 106A{10^{ - 6}}A where the Is{I_s} is the saturation current.
Now we have to define the formula for the forward current through the diode, which is:
I=Is[exp(eVkT)1]I = {I_s}[\exp (\dfrac{{eV}}{{kT}}) - 1]
Now the final step is just to put up the formula and obtain the value:
I=Is[exp(eVkT)1] I=106[exp(1.610190.21.381023300)1] I=2.27mA  I = {I_s}[\exp (\dfrac{{eV}}{{kT}}) - 1] \\\ \Rightarrow I = {10^{ - 6}}[\exp (\dfrac{{1.6*{{10}^{ - 19}}*0.2}}{{1.38*{{10}^{ - 23}}*300}}) - 1] \\\ \Rightarrow I = 2.27mA \\\
Therefore the static resistance would be
rs=VI rs=0.22.27×103 rs=88Ω  {r_s} = \dfrac{V}{I} \\\ \Rightarrow {r_s} = \dfrac{{0.2}}{{2.27\times{{10}^{ - 3}}}} \\\ \Rightarrow {r_s} = 88\Omega \\\
And, the dynamic resistance would be:

rd=VTI=kTeI rd=26×1032.27×103 rd=11.4Ω  {r_d} = \dfrac{{VT}}{I} = \dfrac{{kT}}{{eI}} \\\ \Rightarrow {r_d} = \dfrac{{26\times{{10}^{ - 3}}}}{{2.27\times{{10}^{ - 3}}}} \\\ \Rightarrow {r_d} = 11.4\Omega \\\

Note Most important thing here is to notice the given values, then we can easily derive the missing quantities. Apart from that it is necessary to observe what process is straight and easy to find the rest values and then we can execute the formulas. In these types of questions interpreting the correct formula to be used is necessary unless it will get twisted.