Question

Consider a p-n junction diode having the characteristic $i-i_0(e^{eV/kT}-1)\mathrm{where}{i}_0=20\mathrm{\mu A}$. 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
$$\begin{gathered} i=i_0\left(e^{\frac{eV}{KT}}-1\right) \\ \Rightarrow i=20\times {10}^{-6}\left(e^{\frac{0.3}{8.62\times 300\times {10}^{-5}}}-1\right) \\ \Rightarrow i=20\times {10}^{-5}\left(e^{\frac{100}{8.61}}-1\right) \\ \Rightarrow i=2.18\mathrm{A}\approx 2\mathrm{A} \end{gathered}$$

(b) We need to find the voltage at which the current doubles so that the new value of the current becomes 4 A.
$$\begin{gathered} \Rightarrow 4=20\times {10}^{-6}\left(e^{\frac{eV}{8.62\times 3\times {10}^{-3}}}-1\right) \\ \Rightarrow {\mathrm{e}}^{\frac{V\times {10}^3}{8.62\times 3}}-1=\frac{4\times {10}^6}{20} \\ \Rightarrow {\mathrm{e}}^{\frac{V\times {10}^3}{8.62\times 3}}=200001 \\ \Rightarrow \frac{V\times {10}^3}{8.62\times 3}=12.2060 \\ \Rightarrow V=\frac{12.206\times 8.63\times 3}{{10}^3} \\ \Rightarrow V=318\mathrm{mV} \end{gathered}$$