Question

For a CE transistor amplifier, the current amplification factor is 59 and the emitter current is 6.6 mA. Then the base current is

• A. \( 0.11 \) mA
• B. \( 1.1 \) mA
• C. \( 11 \) \( \mu \)A
• D. \( 0.11 \) A

Hint:

In a CE transistor, remember the relationships between the currents: \( I_E = I_B + I_C \) and the current amplification factor \( \beta = I_C / I_B \). Use these equations and the given values to solve for the required current. Pay attention to the units (in this case, mA).

Answer 1

The Correct Option is A

For a Common Emitter (CE) transistor amplifier, the current amplification factor \( \beta \) (also denoted as \( h_{fe} \)) is defined as the ratio of the collector current \( I_C \) to the base current \( I_B \): $$ \beta = \frac{I_C}{I_B} $$ We are given \( \beta = 59 \).
The emitter current \( I_E \) is the sum of the base current \( I_B \) and the collector current \( I_C \): $$ I_E = I_B + I_C $$ We are given \( I_E = 6.
6 \) mA.
We can express \( I_C \) in terms of \( I_B \) and \( \beta \): $$ I_C = \beta I_B = 59 I_B $$ Substitute this into the equation for \( I_E \): $$ I_E = I_B + 59 I_B = 60 I_B $$ Now, we can solve for the base current \( I_B \): $$ I_B = \frac{I_E}{60} = \frac{6.
6 \text{ mA}}{60} $$ $$ I_B = \frac{6.
6}{60} \text{ mA} = 0.
11 \text{ mA} $$ The base current is \( 0.
11 \) mA.