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Chapter 19: Problem 2640

Semi-conductor has phosphorus as impurity then it will have (A) \(\mathrm{n}_{\mathrm{e}}>\mathrm{n}_{\mathrm{h}}\) (B) \(\mathrm{n}_{\mathrm{e}}<<\mathrm{n}_{\mathrm{h}}\) (C) \(\mathrm{n}_{\mathrm{e}}=\mathrm{n}_{\mathrm{h}}\) (D) \(n_{e}=n_{h}=n_{i}\)

Short Answer

Expert verified
In an n-type semiconductor, doped with phosphorus as an impurity, the number of free electrons will be greater than the number of holes. So, the relationship between electron concentration (\(n_e\)) and hole concentration (\(n_h\)) is given by \(n_e > n_h\). Therefore, the correct answer is (A).

Step by step solution

01

Understand n-type semiconductors

When a semiconductor is doped with phosphorus, it becomes an n-type semiconductor. This is because phosphorus has 5 valence electrons, which means it provides an extra free electron to the semiconductor. These extra electrons become mobile and contribute to the conductivity of the material. As a result, the electron concentration in the semiconductor (\(n_e\)) will increase while the hole concentration (\(n_h\)) will decrease.
02

Write the relation for an n-type semiconductor

In n-type semiconductors, the number of free electrons is greater than the number of holes. Therefore, we have the following relationship: \(n_e > n_h\)
03

Choose the correct answer

Based on the relationship we derived above, we can now choose the correct answer from the given options: (A) \(n_{\mathrm{e}} > n_{\mathrm{h}}\) So, the correct answer is (A).

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Most popular questions from this chapter

In a N-P-N transistor circuit, the emitter, collector and base current are respectively \(\mathrm{I}_{E}, I_{C}\) and \(I_{B} .\) The relation between them is (A) \(\mathrm{I}_{\mathrm{C}}<\mathrm{I}_{\mathrm{E}}<\mathrm{I}_{\mathrm{B}}\) (B) \(\mathrm{I}_{\mathrm{B}}<\mathrm{I}_{\mathrm{C}}<\mathrm{I}_{\mathrm{E}}\) (C) \(\mathrm{I}_{\mathrm{B}}>\mathrm{I}_{\mathrm{C}}<\mathrm{I}_{\mathrm{E}}\) (D) \(\mathrm{I}_{\mathrm{B}}>\mathrm{I}_{\mathrm{C}}>\mathrm{I}_{\mathrm{E}}\)

A n-p-n transistor is used in common emitter made in an amplifier it. A change of \(40 \mu \mathrm{A}\) in the base current changes the output current by $2 \mathrm{~mA}\( and \)0.04 \mathrm{~V}$ in input voltage. An amplifier has voltage gain \(A_{V}=1000\). The voltage gain in \(\mathrm{dB}\) is (A) \(20 \mathrm{~dB}\) (B) \(30 \mathrm{~dB}\) (C) \(3 \mathrm{~dB}\) (D) \(60 \mathrm{~dB}\)

A n-p-n transistor is used in common emitter made in an amplifier it. A change of \(40 \mu \mathrm{A}\) in the base current changes the output current by $2 \mathrm{~mA}\( and \)0.04 \mathrm{~V}$ in input voltage. The current amplification factor is (A) 20 (B) 30 (C) 50 (D) 400

A N-P-N transistor conducts when collector is and emitter is with respect to base. (A) positive, negative (B) positive, positive (C) negative, negative (D) negative, positive

In forward bias made, the PN junction diode resistance will (A) infinity (B) zero (C) less (D) more
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