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
- A20
- B72
- C22.5
- 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