- The current‒voltage relationship of a diode is given by $\text{i}=\text{i}_0\Big(\text{e}^{\frac{\text{eV}}{\text{kT}-1}}\Big)$
For a large value of voltage, 1 can be neglected.
$\text{i}\approx\text{i}_0\text{e}^{\frac{\text{eV}}{\text{KT}}}$
Again, we need to find the voltage at which
$\text{e}^{\frac{\text{eV}}{\text{KT}}}=100$
$\Rightarrow\frac{\text{eV}}{\text{kT}}=\text{In }100$
$\Rightarrow\text{V}=\frac{\text{In }100\times\text{kT}}{\text{e}}$
$\Rightarrow\text{V}=\frac{2.303\times\log\ 100\times8.62\times10^{-5}\times300}{\text{e}}$
$\Rightarrow\text{V}=0.12\text{V}$
- Given:
$\text{i}=\text{i}_0\Big(\text{e}^{\frac{\text{eV}}{\text{kT}-1}}\Big)\ ...(1)$
We know that the dynamic resistance of a diode is the rate of change of voltage w.r.t. current.
i.e. $\text{R}=\frac{\text{dV}}{\text{di}}$
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,
$\frac{\text{di}}{\text{dV}}=\text{i}_0\frac{\text{e}}{\text{kT}}\text{e}^{\frac{\text{eV}}{\text{kT}}}$
$\Rightarrow\frac{1}{\text{R}}=\frac{\text{ei}_0}{\text{kT}}\text{e}^{\frac{\text{eV}}{\text{kT}}}$
$\Rightarrow\text{R}=\frac{\text{kT}}{\text{ei}_0}\text{e}^{\frac{-\text{eV}}{\text{kT}}}\ ...(2)$
- Given,
$\text{R}=2\Omega$
On substituting this value in eq. (2), we get
$2=\frac{8.62\times10^{-5}\times300}{\text{e}\times10\times10^{-6}}\text{e}^{\frac{\text{-eV}}{8.62\times10^{-5}\times300}}$
$\Rightarrow\text{V}=0.25\text{V}.$
