Question

The current−voltage characteristic of an ideal p-n junction diode is given by
$i=i_0(e^{eV/KT}-1)$
where, the drift current i₀ equals 10 µA. Take the temperature T to be 300 K. (a) Find the voltage V₀ for which $e^{\mathrm{e}\mathrm{V}/\mathrm{k}\mathrm{T}}=100.$ One can neglect the term 1 for voltages greater than this value. (b) Find an expression for the dynamic resistance of the diode as a function of V for V > V₀. (c) Find the voltage for which the dynamic resistance is 0.2 Ω.

Answer 1

(a) The current‒voltage relationship of a diode is given by
$i=i_0\left(e^{eV/kT-1}\right)$
For a large value of voltage, 1 can be neglected.
$i\approx i_0{e}^{eV/kT}$
Again, we need to find the voltage at which
$e^{eV/k\mathrm{T}}=100$
$$\begin{gathered} \Rightarrow \frac{eV}{kT}=\ln 100 \\ \Rightarrow V=\frac{\ln 100\times kT}{e} \\ \Rightarrow V=\frac{2.303\times \log 100\times 8.62\times {10}^{-5}\times 300}{e} \\ \Rightarrow V=0.12\mathrm{V} \end{gathered}$$

(b) Given:
$i=i_0\left(e^{eV/kT-1}\right)$ ...(1)
We know that the dynamic resistance of a diode is the rate of change of voltage w.r.t. current.
i.e. $R=\mathit{}\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{i}}$
As the exponential factor dominates the factor of 1, we can neglect this factor.
Now, on differentiating eq. (1) w.r.t. V, we get
$$\begin{gathered} \frac{di}{dV}=i_0\frac{e}{kT}{\mathrm{e}}^{eV/kT} \\ \Rightarrow \frac{1}{R}=\frac{e{i}_0}{kT}{\mathrm{e}}^{eV/kT} \\ \Rightarrow R=\frac{kT}{e{i}_0}{\mathrm{e}}^{-e\mathrm{V}/k\mathrm{T}}...\left(2\right) \end{gathered}$$

(c) Given:
R = 2 Ω
On substituting this value in eq. (2), we get
$$\begin{gathered} 2=\frac{8.62\times {10}^{-5}\times 300}{e\times 10\times {10}^{-6}}{\mathrm{e}}^{-eV/8.62\times {10}^{-5}\times 300} \\ \Rightarrow V=0.25\mathrm{V} \end{gathered}$$