Question:

The base current of CE transistor in active region is 15 μA, find \( V_{CE} \), assuming \( \beta = 150 \), \( R_C = 2~k\Omega \) and \( V_{CC} = 12~V \).

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Use \( I_C = \beta I_B \) and \( V_{CE} = V_{CC} - I_C R_C \) for CE transistor in active mode.
Updated On: May 23, 2025
  • 8.5 V
  • 9.0 V
  • 9.5 V
  • 7.5 V
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The Correct Option is D

Solution and Explanation

Given: \( I_B = 15~\mu A \), \( \beta = 150 \Rightarrow I_C = \beta I_B = 150 \times 15~\mu A = 2.25~mA \) \[ V_{CE} = V_{CC} - I_C R_C = 12~V - (2.25 \times 10^{-3}~A)(2 \times 10^3~\Omega) = 12~V - 4.5~V = 7.5~V \]
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