The diode used in the circuit shown in the fi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The maximum current $I$ that the diode can handle is determined by its maximum power rating $P$ and its constant voltage drop $V_d$.Given $P = 100

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The maximum current $I$ that the diode can handle is determined by its maximum power rating $P$ and its constant voltage drop $V_d$.Given $P = 100\; mW = 100 	imes 10^{-3}\; W$ and $V_d = 0.5\; V$.The maximum current is $I = \frac{P}{V_d} = \frac{100 	imes 10^{-3}}{0.5} = 0.2\; A$.Applying Kirchhoff's Voltage Law to the circuit,the voltage across the resistor $R$ is $V_R = V_{source} - V_d = 1.5\; V - 0.5\; V = 1.0\; V$.Using Ohm's law,$V_R = I 	imes R$,so $R = \frac{V_R}{I} = \frac{1.0}{0.2} = 5\; \Omega$.

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 $\Omega$)

Similar Questions

In a $p-n$ junction diode,the direction of diffusion current is from

Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is: (in $A$)

The circuit shown in the figure contains two ideal diodes $D_1$ and $D_2$. If a cell of emf $3 \text{ V}$ and negligible internal resistance is connected as shown,then the current through the $70 \Omega$ resistance (in ampere) is:

The potential barrier of a $P-N$ junction is $0.5 \ V$. The thickness of the depletion layer is $5.0 \times 10^{-7} \ m$. Calculate the electric field.

$A$ potential barrier of $0.50 \ V$ exists across a $P-N$ junction. If the depletion region is $5.0 \times 10^{-7} \ m$ wide,the intensity of the electric field in this region is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module