When A pn junction Is Reverse Biased, The Flow Of Current Across The Junction Is Mainly Due To

Why Kaysons ?

Video lectures

Access over 500+ hours of video lectures 24*7, covering complete syllabus for JEE preparation.

Online Support

Practice over 30000+ questions starting from basic level to JEE advance level.

Live Doubt Clearing Session

Ask your doubts live everyday Join our live doubt clearing session conducted by our experts.

National Mock Tests

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

Organized Learning

Proper planning to complete syllabus is the key to get a decent rank in JEE.

Test Series/Daily assignments

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

SPEAK TO COUNSELLOR ? CLICK HERE

Question

When a pn junction is reverse biased, the flow of current across the junction is mainly due to

Solution

Correct option is

Diffusion of charges

The flow of current across the junction is mainly due to the diffusion of the majority charge carries across the junction.

SIMILAR QUESTIONS

Q1

On increasing the reverse bias to a large value in a pn junction diode, the current  

Q2

When arsenic is added as an impurity to silicon, the resulting material is

Q3

What is the voltage gain in a common emitter amplifier, where the input resistance is  and load resistance ? Given .

Q4

In a pn junction dipole at a high value of reverse bias, the current rises sharply. The value of reverse bias is known as

Q5

The total current in the block is 

Q6

In a pnp transistor, the p-type crystal acts as

Q7

The current gain in a common emitter transistor is

Q8

If the input and output resistances in a common-base amplifier circuit are  respectively, what is the voltage amplification when the emitter current is 2 mA and current gain ?

Q9

When npn transistor is used as an amplifier

Q10

The V-i characteristic of a silicon diode is given in the figure. Calculate the diode resistance in forward bias at V = +2 volt.