Question

Find the value of emitter current for a transistor with \[ \alpha_{dc}=0.98 \] and \[ I_{CBO}=5~\mu A \] when the base current is measured as \[ 100~\mu A. \]

• A. \(5.25~mA\)
• B. \(5.15~mA\)
• C. \(4.9~mA\)
• D. \(0.25~mA\)

Hint:

A useful transistor relation is \[ I_E=\frac{I_B+I_{CBO}}{1-\alpha}. \] When \(\alpha\) is close to unity, a small base current can produce a much larger emitter current.

Answer 1

The Correct Option is A

Concept: For a transistor including leakage current, \[ I_C=\alpha I_E+I_{CBO} \] and \[ I_E=I_C+I_B. \] Combining these equations gives \[ I_E=\frac{I_B+I_{CBO}}{1-\alpha}. \]

Step 1: Write the given values.
\[ \alpha=0.98 \] \[ I_B=100\mu A \] \[ I_{CBO}=5\mu A \]

Step 2: Substitute into the formula.
\[ I_E = \frac{I_B+I_{CBO}}{1-\alpha} \] \[ = \frac{100+5}{1-0.98} \mu A \] \[ = \frac{105}{0.02} \mu A \] \[ = 5250\mu A \]

Step 3: Convert into milliampere.
\[ I_E = 5.25mA \] Hence, \[ \boxed{I_E=5.25mA} \]