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Chapter 20: Problem 25

In order to forward bias a \(P-N\) junction, the negative terminal of battery is connected to (A) \(P\)-side (B) Either \(P\)-side or \(N\)-side (C) \(N\)-side (D) None of these

Short Answer

Expert verified
In order to forward bias a P-N junction diode, the negative terminal of the battery must be connected to the N-side of the diode, which counteracts the built-in electric field and allows for the flow of charges through the junction. Hence, the correct answer is (C) N-side.

Step by step solution

01

Recall the Structure of P-N Junction Diode

A P-N junction diode is formed by bringing a P-type semiconductor and an N-type semiconductor in contact with each other. The P-type side has an excess of positive charge carriers (holes), while the N-type side has an excess of negative charge carriers (electrons). When a P-N junction is formed, an electric field, called the built-in electric field, is established across the junction, preventing charges from flowing through the diode.
02

Understand Forward Biasing

To forward-bias the P-N junction diode, an external voltage must be applied across the diode to counteract the built-in electric field, so that electrons from the N-side can recombine with holes from the P-side, enabling current flow through the junction. This requires connecting the positive terminal of the battery to the P-side and the negative terminal of the battery to the N-side of the diode.
03

Choose the Correct Option

Based on the understanding of forward biasing of a P-N junction diode, we can conclude that the negative terminal of the battery must be connected to the N-side of the diode. Therefore, the correct answer is: (C) N-side

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Most popular questions from this chapter

The manifestation of band structure in solids is due to \([2004]\) (A) Heisenberg's uncertainty principle. (B) Pauli's exclusion principle. (C) Bohr's correspondence principle. (D) Boltzmann's law.

In NPN transistor the collector current is \(10 \mathrm{~mA}\). If \(90 \%\) of electrons emitted reach the collector, then (A) emitter current will be \(9 \mathrm{~mA}\). (B) emitter current will be \(11.1 \mathrm{~mA}\). (C) base current will be \(0.1 \mathrm{~mA}\). (D) base current will be \(0.01 \mathrm{~mA}\).

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When the \(P\) end of \(P-N\) junction is connected to the negative terminal of the battery and the \(N\) end to the positive terminal of the battery, then the \(P-N\) junction behaves like (A) A conductor (B) An insulator (C) A super-conductor (D) A semi-conductor

The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit? (A) \(1.33 \mathrm{~A}\) (B) \(1.71 \mathrm{~A}\) (C) \(2.00 \mathrm{~A}\) (D) \(2.31 \mathrm{~A}\)
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