Question

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 $mA$ and $V_{D}$ is the diode voltage in $V$. Find the dynamic resistance of the diode in $\Omega$, when the current is $4\, mA$.

• A. 18.6
• B. 21.7
• C. 28.2
• D. 36.2

Answer 1

Answer: (D)

Differentiating $I_{D}$ w.r.t. $V_{D}$, we get
$\frac{d}{d V_{D}}\left(I_{D}\right) =\frac{d}{d V_{D}}\left(7.8 \times 10^{-5} e^{6.9 V_{D}}\right)$
$=7.8 \times 10^{-5} \times 6.9 \times e^{6.9 V_{D}}$
$=I_{D} \times 6.9$
Dynamic resistance is $\frac{d}{d I_{D}}\left(V_{D}\right)$
$\Rightarrow \frac{d V_{D}}{d I_{D}}=\frac{1}{6.9 \times I_{D}}$
When $I_{D}=4\, mA =4 \times 10^{-3}\, A$
Then,
$\frac{d V_{D}}{d I_{D}} =\frac{1}{6.9 \times 4 \times 10^{-3}}$
$=36.23\, \Omega$
$\therefore $ Dynamic resistance $=36.23\, \Omega$