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NEET2019PhysicsSemiconductorsActual

In an n-p-n transistor (in common emitter mode) 10¹⁰ electrons enter the emitter in 10⁻ ^6 s. Only 2 % of the electrons are lost in the base. Calculate the current amplification factor. (Charge on electron is 1.6 10⁻¹⁹ C ).

Options

  1. A20
  2. B72
  3. C22.5
  4. D49

Correct answer

D. 49

Step-by-step solution

i_e= q t = 10¹⁰ 1.6 10⁻¹⁹ 10⁻⁶ =1.6 ~mA Base current, i_b=2 % of i_e= ( 2 100 ) 1.6 ~mA =0.032 ~mA i_c=i_e-i_b=1.568 ~mA Therefore current amplification factor, = i_c i_b = 1.568 0.032 =49

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