The $V-I$ characteristic of a silicon diode is shown in the Figure. Calculate the resistance of the diode at $(a) \; I_{D} = 15 \, mA$ and $(b) \; V_{D} = -10 \, V$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) To calculate the resistance of the diode,we use Ohm's law,$R = V / I$ or $r = \Delta V / \Delta I$.
$(a)$ For the forward bias region,we calculate the dynamic resistance at $I_{D} = 15 \, mA$. From the graph,we can take the slope between $I = 10 \, mA$ $(V = 0.7 \, V)$ and $I = 20 \, mA$ $(V = 0.8 \, V)$.
$r_{f} = \Delta V / \Delta I = (0.8 \, V - 0.7 \, V) / (20 \, mA - 10 \, mA) = 0.1 \, V / 10 \, mA = 10 \, \Omega$.
$(b)$ For the reverse bias region at $V_{D} = -10 \, V$,the current is $I = -1 \, \mu A$.
The static resistance is $r_{r} = |V| / |I| = 10 \, V / 1 \, \mu A = 10 \, V / (1 \times 10^{-6} \, A) = 1.0 \times 10^{7} \, \Omega$.

Explore More

Similar Questions

Two $PN$-junctions can be connected in series by three different methods as shown in the figure. If the potential difference across the junctions is the same,then the correct connections will be:

The region near the junction of an unbiased $p-n$ junction diode is known as the depletion layer. This layer is depleted of

The diode used in the circuit shown in the figure has a constant voltage drop of $0.5\; V$ at all currents and a maximum power rating of $100\; mW$. What should be the value of the resistor $R$,connected in series with the diode,to obtain the maximum current? (in $\Omega$)

In the given circuit,the current $(I)$ through the battery will be $..........\,A$.

The ratio of the resistance of a $P-N$ junction diode in forward bias to its resistance in reverse bias is approximately:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo