In a p-n junction diode, the current I can be expressed as
$\boxed{\mathrm{I} = {\mathrm{I}}_0\exp \left(\frac{\mathrm{eV}}{2{\mathrm{k}}_{\mathrm{g}}\mathrm{T}}-1\right)}$
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, k_B is the Boltzmann constant (8.6 x 10^–5 eV/K) and T is the absolute temperature. If for a given diode I₀ = 5 x 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?

Given,
Reverse saturation current, ${\mathrm{I}}_0 = 5 \times {10}^{-12}\mathrm{A}$
Absolute temperature, T = 300 K 
$$\begin{gathered} \mathrm{Boltzman} \mathrm{constant}, {\mathrm{k}}_{\mathrm{B}} = 8.6 \times {10}^{-5}{\mathrm{eVk}}^{-1} \\ = 8.6 \times {10}^{-5}\times 1.6 \times {10}^{-19} {\mathrm{JK}}^{-1} \end{gathered}$$ 
(a) Forward voltage, V = 0.6 V, then 

$\frac{\mathrm{eV}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}} = \frac{1.6 \times {10}^{-19}\times 0.6}{8.6\times {10}^{-5}\times 1.6\times {10}^{-19}\times 300} = 23.26$

and

      $$\begin{gathered} \mathrm{I} = {\mathrm{I}}_0\left[\exp \left(\frac{\mathrm{eV}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}-1\right)\right] \\ = 5\times {10}^{-12} [{\mathrm{e}}^{23.26}-1] \\ = 5\times {10}^{-12}[1.259 \times {10}^{10}-1] \\ = 5\times {10}^{-12}\times 1.259\times {10}^{10} \\ = 0.063 \mathrm{A} \mathrm{is} \mathrm{the} \mathrm{required} \mathrm{forward} \mathrm{current}. \end{gathered}$$ 

(b) If the forward voltage is increased to V = 0.7 V,  then 
           $$\begin{gathered} \frac{\mathrm{eV}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}=\frac{1.6\times {10}^{-19}\times 0.7}{8.6\times {10}^{-5}\times 1.6\times {10}^{-19}\times 300} \\ =27.14 \end{gathered}$$

Therefore, 
$$\begin{gathered} \mathrm{Forward} \mathrm{current}, \mathrm{I} = {\mathrm{I}}_0\left[\exp \left(\frac{\mathrm{eV}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}-1\right)\right] \\ = 5\times {10}^{-12}[{\mathrm{e}}^{27.14}-1] \\ =5\times {10}^{-12}[6.07\times {10}^{11} - 1] \\ =5\times {10}^{-12}\times 6.07\times {10}^{11} \\ = 3.035 \mathrm{A} \end{gathered}$$

Thus, increase in current is,
         $∆\mathrm{I} = (3.035 - 0.063) = 2.972 \mathrm{A}$ 

(c) As, $∆\mathrm{I} = 2.972 \mathrm{A},$  $∆\mathrm{V} = 0.7 - 0.6 = 0.1 \mathrm{V}$

Therefore, '
Dynamic resistance =$\frac{∆\mathrm{V}}{∆\mathrm{I}} = \frac{0.1 \mathrm{V}}{2.972\hspace{0.17em}\mathrm{A}} =0.0336 \mathrm{\Omega }$ 

(d) For both the voltages, the current I will be almost equal to I_0. This implies that the circuit offers infinite dynamic resistance in the reverse bias.