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Chapter 40: Problem 2928

A transistor is connected in common emitter configuration. The collector supply is \(8 \mathrm{~V}\) and the voltage drop across a resistor of 800 in the collector circuit is \(0.5 \mathrm{~V}\). If the current gain factor \(\alpha\) is \(0.96\), then base current will be (A) \(27 \mu \mathrm{A}\) (B) \(26 \mu \mathrm{A}\) (C) \(25 \mu \mathrm{A}\) (D) \(24 \mu \mathrm{A}\)

Short Answer

Expert verified
The short answer is (B) \(26 \mu \mathrm{A}\).

Step by step solution

01

Find the current flowing through the resistor

We are given the voltage drop across the resistor as 0.5 V, and the resistance as 800 Ω. We can use Ohm's Law to find the current flowing through the resistor: \(I = \frac{V}{R}\) Where I is the current, V is the voltage drop, and R is the resistance. Plug in the given values: \(I = \frac{0.5}{800} = 0.000625 \mathrm{~A}\)
02

Calculate the collector current

In a common emitter configuration, the current flowing through the resistor is the same as the collector current (Ic). Therefore, \(I_c = 0.000625 \mathrm{~A}\)
03

Use the current gain factor to find the base current

We are given the current gain factor α as 0.96. The relationship between α, base current (Ib), and collector current (Ic) can be expressed as: \(\alpha = \frac{I_c}{I_c + I_b}\) We can rearrange the equation to find the base current (Ib): \(I_b = \frac{I_c}{\alpha} - I_c\) Now, plug in the given values: \(I_b = \frac{0.000625}{0.96} - 0.000625 = 2.604 \times 10^{-5} \mathrm{~A}\)
04

Convert base current to microamperes and check the options

To convert the base current from amperes to microamperes, multiply by 10^6: \(I_b = 2.604 \times 10^{-5} \mathrm{~A} \times 10^6 = 26.04 \mu \mathrm{A}\) Checking the given options, we find that the closest answer to our calculation is: (B) \(26 \mu \mathrm{A}\)

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Most popular questions from this chapter

In a transistor amplifier, \(\beta=62, \mathrm{R}_{\mathrm{L}}=5000 \Omega\) and internal resistance of the transistor is \(500 \Omega\). What will be the ratio of power amplification to voltage amplification? (A) 61 (B) 62 (C) 60 (D) 100

What is the relation of the transistor current? (A) \(\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{E}}\) (B) \(\mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{E}}+\mathrm{I}_{\mathrm{B}}\) (C) \(\mathrm{I}_{\mathrm{E}}=\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{B}}\) (D) \(\mathrm{I}_{\mathrm{C}}=\mathrm{I}_{\mathrm{E}}+\mathrm{I}_{\mathrm{B}}\)

The current gain of a transistor in common base mode is 0.99. To change the emitter current by \(5 \mathrm{~mA}\). What will be the necessary change in collector current ? (A) \(5.1 \mathrm{~mA}\) (B) \(4.95 \mathrm{~mA}\) (C) \(2.45 \mathrm{~mA}\) (D) \(0.196 \mathrm{~mA}\)

In a common base amplifier circuit, calculate the change in base current if that in the collector current is \(2 \mathrm{~mA}\) and \(\alpha=0.98\). (A) \(980 \mathrm{~mA}\) (B) \(2 \mathrm{~mA}\) (C) \(0.04 \mathrm{~mA}\) (D) \(1.96 \mathrm{~mA}\)

In a common emitter amplifier, using output resistance of \(5000 \Omega\) and input resistance of \(2000 \Omega\), if the peak value of input signal voltage is \(10 \mathrm{mV}\) and \(\beta=50\) then what is the peak value of output voltage? (A) \(2.5 \times 10^{-4}\) Volt (B) \(5 \times 10^{-6}\) Volt (C) \(1.25\) Volt (D) 125 Volt
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