Q159 / 160 · TS EAMCET 2018 (05 May Shift 2) · Physics

The voltage-current characteristic of a diode during forward bias is given by $I=7.8 \times 10^{-5} e^{6.5 V_D}$, where $I$ is the current in $\mathrm{mA}$ and $V_D$ is the diode voltage in $\mathrm{V}$. Find the dynamic resistance of the diode in $\Omega$, when the current is $4 \mathrm{~mA}$

  • A
    18.6
  • B
    21.7
  • C
    28.2
  • Correct answer
    D
    36.2

Solution

Differentiating $I_D$ w.r.t. $V_D$, we get $$ \begin{gathered} \frac{d}{d V_D}\left(I_D\right)=\frac{d}{d V_D}\left(7.8 \times 10^{-5} e^{6.9 V_D}\right) \\ =7.8 \times 10^{-5} \times 6.9 \times e^{6.9 V_D} \\ =I_D \times 6.9 \end{gathered} $$ Dynamic resistance is $\frac{d}{d I_D}\left(V_D\right)$ $$ \Rightarrow \quad \frac{d V_D}{d I_D}=\frac{1}{6.9 \times I_D} $$ When $I_D=4 \mathrm{~mA}=4 \times 10^{-3} \mathrm{~A}$ Then, $$ \begin{aligned} \frac{d V_D}{d I_D} & =\frac{1}{6.9 \times 4 \times 10^{-3}} \\ & =36.23 \Omega \end{aligned} $$ $\therefore$ Dynamic resistance $=36.23 \Omega$

Hence, the correct answer is option: 'D'.

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