Question 3.6: A BJT has α = 0.99, iB = IB = 25 μA, and ICBO = 200 nA. Find......

(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 \%