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 \[e^{eV/kT}  = 100 .\]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.

(Use Planck constant h = 4.14 × 10^-15 eV-s, Boltzmann constant k = 8·62 × 10^-5 eV/K.)

Answer 1

 (a) The current‒voltage relationship of a diode is given by 

\[i =  i_0 ( e^{eV/kT - 1} )\]

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

\[e^{eV/kT}  = 100\]

\[\Rightarrow \frac{eV}{kT} = \text{ln }100\] 

\[ \Rightarrow V = \frac{\text{ ln  }100 \times \text{ kT }}{e}\] 

\[ \Rightarrow V = \frac{2 . 303 \times \log  100 \times 8 . 62 \times {10}^{- 5} \times 300}{e}\] 

\[ \Rightarrow V=0.12\] V

(b) Given:-

\[i =  i_0 ( e^{eV/kT - 1} )  ...........(1)\]

We know that the dynamic resistance of a diode is the rate of change of voltage w.r.t. current.
i.e. 

\[R   =   \frac{d V}{\text{d i}}\]

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

\[\frac{\text{di}}{\text{dV}} =  i_0 \frac{e}{kT} e^{eV/kT} \] 

\[ \Rightarrow \frac{1}{R} = \frac{e i_0}{kT} e^{eV/kT} \] 

\[ \Rightarrow R = \frac{kT}{e i_0} e^{- eV/kT} ............(2)\]

(c) Given:-

R = 2 Ω

On substituting this value in eq. (2), we get

\[2 = \frac{8 . 62 \times {10}^{- 5} \times 300}{e \times 10 \times {10}^{- 6}} e^{- eV/8 . 62 \times {10}^{- 5} \times 300} \]

\[ \Rightarrow V = 0 . 25 \] V