A BJT has α = 0.99, i_B = I_B = 25 μ\text{A}, and I_{CBO} = 200 \text{nA}. Find (*a*) the dc collector current, (*b*) the dc emitter current, and (*c*) the percentage error in emitter current when leakage current is neglected.

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(*a*) With α = 0.99, (*3.2*) gives

β(≡ h_{FE}) ≡ \frac{α}{1 – α} ≡ \frac{I_C – I_{CEO}}{I_B} (*3.2*)

Using (*3.3*) in (*3.2*) then gives

I_{CEO} = (β + 1)I_{CBO} (*3.3*)

I_{C} = βI_B + (β + 1)I_{CBO} = 99(25 × 10^{-6}) + (99 + 1)(200 × 10^{-9}) = 2.495 \text{mA}

(*b*) The dc emitter current follows from (*3.1*):

α(≡ h_{FB}) ≡ \frac{I_C – I_{CBO}}{I_E} (*3.1*)

I_E = \frac{I_C – I_{CBO}}{α} = \frac{2.475 × 10^{-3} – 200 × 10^{-9}}{0.99} = 2.518 \text{mA}

(*c*) Neglecting the leakage current, we have

I_{C} = βI_B = 99(25 × 10^{-6}) = 2.475 \text{mA} \quad \text{so} \quad I_E = \frac{I_{C}}{α} = \frac{2.475}{0.99} = 2.5 \text{mA}

giving an emitter-current error of

\frac{2.518 – 2.5}{2.518}(100 \%) = 0.71 \%

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