Ask & Answer Question

Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Physics Class 12th Semiconductor Electronics

The breakdown in a reverse biased p-n junction is more likely to occur due to
(a) Large velocity of the minority charge carriers if the doping concentration is small
(b) Large velocity of the minority charge carriers if the doping concentration is large
(c) Strongelectricfieldinadepletionregionifthedopingconcentrationissmall
(d) Strong electric field in the depletion region if the doping concentration is large

1 View | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

4 months ago

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, d)

Explanation- In reverse biasing, the minority charge carriers will be accelerated due to reverse biasing, which on striking with atoms cause ionization resulting secondary electrons and thus more number of charge carriers.

When doping concentration is large, there will be large number of ions in the depletion

region, which will give rise to a strong electric field.

0 Upvotes 0 Downvotes

Similar Questions for you

Two 10cm log, straight wires, each carrying a current of 5A are kept parallel to each other. If each wire experienced a force of 10^-5N, then separation between the wires is__________cm.

Vishal Baghel

$F = \frac{μ_{0} \times 2}{4 π} \frac{i_{1} i_{2}}{d} l$

$1 0^{- 5} = \frac{1 0^{- 7} \times 2 \times 5 \times 5}{d} = \frac{10}{100}$

$d = 5 \times 1 0^{- 2} m$

d = 5 cm

The positive feedback is required by an amplifier to act an oscillator. The feedback here means:

Vishal Baghel

A portion of the output power is returned back to the input.

The variation of acceleration due to gravity (g) with distance (r) from the centre of the earth is correctly represented by: (Given R = radius of earth)

Payal Gupta

$g \propto r 0 \leq r \leq R$

$g \propto \frac{1}{r^{2}} r > R$

A transistor is used in an amplifier circuit in common emitter mode. If the best current changes by 100 $μ A,$ it brings a change of 10 mA is collector current. If the load resistance is 2 $k Ω$ and input resistance is 1 $k Ω,$ the value of power gain is x × 10⁴. The value of x is_____________.

alok kumar singh

Power gain = ${(\frac{Δ i_{c}}{Δ i_{b}})}^{2} \times \frac{R_{0}}{R_{i}}$

= ${(\frac{10 \times 1 0^{- 3}}{100 \times 1 0^{- 6}})}^{2} \times \frac{2}{1}$

= 2 × 104 = x × 104

= 2

The typical transfer characteristics of a transistor in CE configuration is shown in figure. A load resistor of $2 k Ω$ is connected in the collector branch of the circuit used. The input resistance of the transistor is $0.50 k Ω .$ The voltage gain of the transistor is________.

alok kumar singh

Voltage gain = $\frac{Δ l_{C}}{Δ l_{B}} \times \frac{R_{C}}{R_{B}}$

$\frac{5 \times 1 0^{- 3}}{100 \times 1 0^{- 6}} \times \frac{2}{0.5}$

= 200

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
686k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question