Question

Consider a p-n junction diode having the characteristic \[i -  i_0 ( e^{eV/kT}  - 1) \text{ where }  i_0  = 20\mu A\] . The diode is operated at T = 300 K . (a) Find the current through the diode when a voltage of 300 mV is applied across it in forward bias. (b) At what voltage does the current double?

Answer 1

(a) Given:-

Drift current, i₀ = 20 × 10⁻⁶ A

Temperature, T = 300 K

Applied voltage, V = 300 mV

The variation in the current with respect to the voltage is given by

\[i =  i_0 \left( e^\frac{eV}{KT} - 1 \right)\]

\[ \Rightarrow i = 20 \times  {10}^{- 6} \left( e^\frac{0 . 3}{8 . 62 \times 300 \times {10}^{- 5}} - 1 \right)\]

\[ \Rightarrow i = 20 \times  {10}^{- 5} \left( e^\frac{100}{8 . 61} - 1 \right)\]

\[ \Rightarrow i = 2 . 18  A \approx 2  \] A

(b) We need to find the voltage at which the current doubles so that the new value of the current becomes 4 A.

\[\Rightarrow 4 = 20 \times  {10}^{- 6} \left( e^\frac{eV}{8 . 62 \times 3 \times {10}^{- 3}} - 1 \right)\]

\[ \Rightarrow  e^\frac{V \times {10}^3}{8 . 62 \times 3}  - 1 = \frac{4 \times {10}^6}{20}\]

\[ \Rightarrow  e^\frac{V \times {10}^3}{8 . 62 \times 3}  = 200001\]

\[ \Rightarrow \frac{V \times {10}^3}{8 . 62 \times 3} = 12 . 2060\]

\[ \Rightarrow V = \frac{12 . 206 \times 8 . 63 \times 3}{{10}^3}\]

\[ \Rightarrow V = 318  \] mV