A p - n junction diode has the characteristic i = i_0\! (e^ eV/kT -...

The diode current is given by $i = i_0 \left(e^{eV/kT} - 1\right)$.

At $T = 300 \text{ K}$, the thermal voltage is \dfrac{kT}{e} = \dfrac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}} \approx$25.87 \text{ mV}$.

For $V_1 = 300 \text{ mV}$, the term $e^{eV_1/kT} = e^{300/25.87} \approx e^{11.6} \gg 1$. Thus, we can approximate the current as $i \approx i_0 e^{eV/kT}$.

Let the new voltage be $V_2$ where the current becomes $2i_1$.

$2i_1 = i_0 e^{eV_2/kT}$

Substituting $i_1 \approx i_0 e^{eV_1/kT}$:

$2 i_0 e^{eV_1/kT} = i_0 e^{eV_2/kT}$

$e^{e(V_2 - V_1)/kT} = 2$

Taking natural logarithm on both sides:

\dfrac{e(V_2 - V_1)}{kT} = \ln$2$

$V_2 - V_1 =$\dfrac{kT}{e} \ln$2$

$V_2 - V_1 = 25.87 \times 0.693 \approx 17.9 \text{ mV}$

$V_2 = 300 + 17.9 = 317.9 \text{ mV} \approx 318 \text{ mV}$