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91Ó°ÊÓ

A n-p-n transistor circuit has \(\alpha=0.985 .\) If \(\mathrm{I}_{\mathrm{c}}=9 \mathrm{~mA}\) then the value of \(\mathrm{I}_{\mathrm{b}}\) is (A) \(0.003 \mathrm{~mA}\) (B) \(0.66 \mathrm{~mA}\) (C) \(0.015 \mathrm{~mA}\) (D) \(0.13 \mathrm{~mA}\)

Short Answer

Expert verified
The short answer is: \( I_{b} \approx 0.13 \mathrm{~mA} \). The correct option is (D).

Step by step solution

01

Rearranging the equation to solve for Ib

First, we will rearrange the equation to isolate Ib: α = (Ic - Ib) / Ic Multiplying both sides by Ic, we get: α * Ic = Ic - Ib Adding Ib to both sides and subtracting α * Ic from both sides, we get: Ib = Ic - α * Ic
02

Plug in the given values

Now, we can substitute the given values of α and Ic: Ib = 9 mA - 0.985 * 9 mA
03

Calculate Ib

Performing the arithmetic, we find the value of Ib: Ib = 9 mA - 8.865 mA Ib ≈ 0.135 mA
04

Choose the correct answer

Comparing the calculated value of Ib with the given options, we find that the closest match is: (D) 0.13 mA So, the correct answer is option (D).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bipolar Junction Transistor (BJT)
A Bipolar Junction Transistor (BJT) is a type of transistor that uses both electron and hole charge carriers. In electronics, BJTs are widely used to amplify current in various circuits. There are two types of BJTs: n-p-n and p-n-p. The n-p-n version has one p-type layer sandwiched between two n-type layers. This configuration allows for efficient electron flow, making the n-p-n transistor very effective in amplifying current. When it comes to operation, BJTs have three main regions: the emitter, which emits electrons; the base, which controls the flow; and the collector, where the electrons are collected. Understanding these parts is crucial when working with BJTs in circuit applications.
Current Amplification
Current Amplification in BJTs is characterized by the parameter alpha (α), which is the ratio of the collector current (Ic) to the emitter current (Ie). It can be expressed as:
  • \( \alpha = \frac{I_c}{I_e} \)
In practical terms, α represents the efficiency of the BJT in transferring the emitter current to the collector. The closer α is to 1, the more efficient the current amplification is. This is essential for applications where increasing the strength of an electrical signal is needed. Typically, in n-p-n transistors, the base current (Ib) controls the larger amount of current flow from the emitter to the collector, which is central to the transistor's function as a current amplifier.
Base Current Calculation
Calculating the base current (Ib) involves understanding the relationship between Ib, Ic, and α in BJTs. Given the formula:
  • \( \alpha = \frac{I_c - I_b}{I_c} \)
By rearranging, the base current can be found:
  • \( I_b = I_c - \alpha \cdot I_c \)
This equation highlights how the base current is a small fraction of the collector current. For instance, given α = 0.985 and Ic = 9 mA, we substitute these values into the equation:
  • \( I_b = 9 \, \text{mA} - 0.985 \times 9 \, \text{mA} \)
  • \( I_b \approx 0.135 \, \text{mA} \)
This detailed approach ensures accuracy, especially in selecting the correct multiple-choice answer, such as option (D) in our exercise. Understanding this calculation process is crucial for anyone working with BJTs in electronic circuits.

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Most popular questions from this chapter

A n-p-n transistor is used in common emitter mode in an amplifier it. A change of \(40 \mu \mathrm{A}\) in the base current changes the output current by \(2 \mathrm{~m} \mathrm{~A}\) and \(0.04 \mathrm{~V}\) in input voltage. The input resistance is (A) \(1 \mathrm{k} \Omega\) (B) \(10 \Omega\) (C) \(10 \mathrm{k} \Omega\) (D) \(100 \Omega\)

In a common emitter amplifier, output resistance is \(5000 \Omega\) and input resistance is \(1000 \Omega .\) If peak value of signal voltage is \(1 \mathrm{mV}\) and \(\beta=100\), then the peak value of output voltage is (A) \(0.1 \mathrm{~V}\) (B) \(0.3 \mathrm{~V}\) (C) \(0.2 \mathrm{~V}\) (D) \(0.5 \mathrm{~V}\)

When NPN transistor is used as an amplifier then (A) electron moves from base to collector (B) hole travels from emitter to base (C) hole goes to emitter from base (D) electron goes to base from collector

Reverse bias applied on a junction diode: (A) raises the potential barrier (B) increases majority charge carrier current (C) lowers the potential barrier (D) increases the temperature of junction

Direction for Assertion - Reason type questions (A) If both Assertion and Reason are true and reason is the correct explanation of assertion. (B) If both Assertion and Reason are true but Reason is not the correct explanation of assertion. (C) If Assertion is true but Reason is false (D) If both assertion and reason are false A: Photodiode are operated in reverse bias \(\mathrm{R}\) : In reverse bias fractional change in minority charge carrier is more (A) a (B) \(b\) (C) c (D) \(\mathrm{d}\)

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