Problem

A transistor is connected in a CE configuration. The voltage drop across the load resistance ($R_c$) is 6 V, and the load resistance is $3 \, \mathrm{k} \Omega$. Find the base current ($I_B$) if the current gain ($\beta$) of the transistor is 0.97.

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Data

• Load resistance: $R_c = 3 \, \mathrm{k} \Omega = 3 \times 10^3 \, \Omega$
• Voltage drop across load: $V_c = 6 \, \mathrm{V}$
• Current gain: $\beta = 0.97$

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Prerequisite Concepts

1. The current gain ($\beta$) is defined as:

$$\beta = \frac{I_C}{I_B}$$

Rearranging for $I_B$:

$$I_B = \frac{I_C}{\beta}$$

2. The collector current ($I_C$) can be calculated using Ohm’s law:

$$I_C = \frac{V_c}{R_c}$$

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Solution

Step 1: Calculate the Collector Current ($I_C$)

Using Ohm’s law:

$$I_C = \frac{V_c}{R_c}$$

Substitute the values:

$$I_C = \frac{6 \, \mathrm{V}}{3 \times 10^3 \, \Omega}$$ $$I_C = 2 \times 10^{-3} \, \mathrm{A} = 2 \, \mathrm{mA}$$

Step 2: Calculate the Base Current ($I_B$)

Using the formula for current gain:

$$I_B = \frac{I_C}{\beta}$$

Substitute the values:

$$I_B = \frac{2 \times 10^{-3} \, \mathrm{A}}{0.97}$$ $$I_B \approx 2.06 \times 10^{-3} \, \mathrm{A} = 2.06 \, \mathrm{mA}$$

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Answer

The base current ($I_B$) is approximately $2.06 \, \mathrm{mA}$.