## Q. 3.6

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.

## Verified Solution

(a)  With $α = 0.99$, (3.2) gives
$β(≡ h_{FE}) ≡ \frac{α}{1 – α} ≡ \frac{I_C – I_{CEO}}{I_B}$        (3.2)

$β = \frac{α}{1 – α} = 99$

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