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Semiconductor Electronic: Material, Devices And Simple Circuits — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsSemiconductor Electronic: Material, Devices And Simple Circuits5 Marks
Question
In a p-n junction diode, the current I can be expressed as,
$I=I_0\ \text{exp}\Big(\frac{\text{eV}}{2\text{k}_{\text{B}}\text{T}}-1\Big)$
where $I_0$ 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, $k_g$ is the Boltzmann constant $(8.6\times 10^{-5} eV/K)$ and T is the absolute temperature. If for a given diode $I_0 = 5\times 10^{-12} A$ and $T = 300 K$, then,
  1. What will be the forward current at a forward voltage of 0.6V?
  2. What will be the increase in the current if the voltage across the diode is increased to 0.7V?
  3. What is the dynamic resistance?
  4. What will be the current if reverse bias voltage changes from 1V to 2V?
✓

Answer

In a p-n junction diode, the expression for current is given as:
$I=I_0\ \text{exp}\Big(\frac{\text{eV}}{2\text{k}_{\text{B}}\text{T}}-1\Big)$
Where,
$I_0$ = Reverse saturation current $= 5 \times 10^{-12} A$
$T =$ Absolute temperature $= 300K$
$k_B= $ Boltzmann constant $= 8.6 \times 10^{-5} eV/K = 1.376 \times 10^{-23} J K ^{-1}$
V = Voltage across the diode
  1. Forward voltage, V = 0.6 V
$\therefore\text{Current,}\ I=5\times10^{-12}\Bigg[\text{exp}\Bigg(\frac{1.6\times10^{-19}\times0.6}{1.376\times10^{-23}\times300}\Bigg)-1\Bigg]$
$= 5 \times 10^{-12} \times$ exp $[22.36] = 0.0256 A$
Therefore, the forward current is about 0.0256 A.
  1. For forward voltage, V = 0.7 V, we can write:
$I=5\times10^{-12}\Bigg[\text{exp}\Bigg(\frac{1.6\times10^{-19}\times0.7}{1.376\times10^{-23}\times300}-1\Bigg)\Bigg]$
$= 5 \times 10^{-12} \times$ exp $[26.25] = 1.257 A$
Hence, the increase in current, $\Delta{I}=I'-I$
= 1.257 - 0.0256 = 1.23 A
$\text{Dynamic resistance }=\frac{\text{Change in voltage}}{\text{Change in current}}$
  1. Dynamic resistance
$=\frac{0.7-0.6}{1.23}=\frac{0.1}{1.23}=0.081\Omega$
  1. If the reverse bias voltage changes from 1V to 2V, then the current $I$ will almost remain equal to $I_0$ in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.

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