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Question: The current voltage relationship of a diode is \(I = {I_0}[\exp (\dfrac{{eV}}{{KT}}) - 1]\) where \(...

The current voltage relationship of a diode is I=I0[exp(eVKT)1]I = {I_0}[\exp (\dfrac{{eV}}{{KT}}) - 1] where ee the charge on electron is, VV is voltage applied, KK is Boltzmann constant, and TT is the temperature in Kelvin. Find the resistance of the diode if I0=1.5mA{I_0} = 1.5\,mA and T=398KT = 398\,K.

Explanation

Solution

In semiconductor devices, a PN junction diode is a very simple semiconductor device which permits the flow of current in only one direction by applying a reverse bias across negative side and hence depletion layer width increases.

Complete step by step answer:
As we have given the current and voltage relation of pn Junction diode as I=I0[exp(eVKT)1]I = {I_0}[\exp (\dfrac{{eV}}{{KT}}) - 1] and we will take the derivative of given current with voltage and since, I0{I_0} is the maximum current which is constant which is given to us
I0=1.5mA{I_0} = 1.5\,mA
Temperature is also a constant term which is given to us
T=398KT = 398\,K
And ee is the charge on electron which has a value of
e=1.6×1019Ce = 1.6 \times {10^{ - 19}}C
Voltage=2VVoltage = 2V Which is given in the circuit diagram.

Now, let us take the derivative current with respect to voltage which can be written as
I=I0[exp(eVKT)1]I = {I_0}[\exp (\dfrac{{eV}}{{KT}}) - 1]
Take derivative as dIdV\dfrac{{dI}}{{dV}}
dIdV=I0[(eKT)exp(eVKT)0]\dfrac{{dI}}{{dV}} = {I_0}[(\dfrac{e}{{KT}})\exp (\dfrac{{eV}}{{KT}}) - 0]
dIdV=I0(eKT)×exp(eVKT)\Rightarrow \dfrac{{dI}}{{dV}} = {I_0}(\dfrac{e}{{KT}}) \times \exp (\dfrac{{eV}}{{KT}})
First, let us find the magnitude of exp(eVKT)\exp (\dfrac{{eV}}{{KT}}) by putting values of given parameters
exp(eVKT)=exp(1.6×1019×21.38×1023×398)\exp (\dfrac{{eV}}{{KT}}) = \exp (\dfrac{{1.6 \times {{10}^{ - 19}} \times 2}}{{1.38 \times {{10}^{ - 23}} \times 398}})
exp(eVKT)=1.88 \Rightarrow \exp (\dfrac{{eV}}{{KT}}) = 1.88 \\\
And now, find the magnitude of I0(eKT){I_0}(\dfrac{e}{{KT}})
I0(eKT)=1.5×103×1.6×10191.38×1023×398{I_0}(\dfrac{e}{{KT}}) = \dfrac{{1.5 \times {{10}^{ - 3}} \times 1.6 \times {{10}^{ - 19}}}}{{1.38 \times {{10}^{ - 23}} \times 398}}
I0(eKT)=0.04\Rightarrow {I_0}(\dfrac{e}{{KT}}) = 0.04
Now, put these two parameters value in equation dIdV=I0(eKT)×exp(eVKT)\dfrac{{dI}}{{dV}} = {I_0}(\dfrac{e}{{KT}}) \times \exp (\dfrac{{eV}}{{KT}})
dIdV=0.04×1.88\dfrac{{dI}}{{dV}} = 0.04 \times 1.88
dIdV=15250\Rightarrow \dfrac{{dI}}{{dV}} = \dfrac{{15}}{{250}}
Taking reciprocal of above equation we get
dVdI=25015\dfrac{{dV}}{{dI}} = \dfrac{{250}}{{15}}
dVdI=503\therefore \dfrac{{dV}}{{dI}} = \dfrac{{50}}{3}
Since we know that R=VIR = \dfrac{V}{I}

Hence, dVdI=503\dfrac{{dV}}{{dI}} = \dfrac{{50}}{3} is the resistance of the diode.

Hence, resistance of the diode is 16.67Ω16.67\Omega .

Note: It should be remembered that, the basic conversions of units are as 1mA=103A1\,mA = {10^{ - 3}}\,A which a unit of current is and the relation between current, voltage and resistance R=VIR = \dfrac{V}{I} is known as Ohm’s law. Since, external resistance and diode are connected in series hence the same current follows through both the resistor and diode but the voltage across the series components in the electric circuit is never the same it gets distributed but incase of parallel components voltage is same in both components of the circuit.