In a p-n junction diode, the current I can be expressed as

Solution:

In a p-n junction diode, the expression for current is given as:

$I=I_{0} \exp \left(\frac{\mathrm{eV}}{2 k_{\mathrm{B}} T}-1\right)$

Where,

I0 = Reverse saturation current = 5 × 10−12 A

T = Absolute temperature = 300 K

kB = Boltzmann constant = 8.6 × 10−5 eV/K = 1.376 × 10−23 J K−1

V = Voltage across the diode

(a) Forward voltage, V = 0.6 V

$\therefore$ Current, $I=5 \times 10^{-12}\left[\exp \left(\frac{1.6 \times 10^{-19} \times 0.6}{1.376 \times 10^{-23} \times 300}\right)-1\right]$

$=5 \times 10^{-12} \times \exp [22.36]=0.0256 \mathrm{~A}$

Therefore, the forward current is about 0.0256 A.

(b) For forward voltage, V’ = 0.7 V, we can write:

$I^{\prime}=5 \times 10^{-12}\left[\exp \left(\frac{1.6 \times 10^{-19} \times 0.7}{1.376 \times 10^{-23} \times 300}-1\right)\right]$

$=5 \times 10^{-12} \times \exp [26.25]=1.257 \mathrm{~A}$

Hence, the increase in current, ΔI = I' − I

= 1.257 − 0.0256 = 1.23 A

(c) Dynamic resistance $=\frac{\text { Change in voltage }}{\text { Change in current }}$

$=\frac{0.7-0.6}{1.23}=\frac{0.1}{1.23}=0.081 \Omega$

(d) If the reverse bias voltage changes from 1 V to 2 V, then the current (I) will almost remain equal to I0 in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.