In a bipolar junction transistor, the collector current is
\(\dfrac{\alpha_F I_B}{1-\alpha_F}+\dfrac{I_{CO}}{1-\alpha_F}\)
Start from the common-base relation, which is where αF is defined:
\(I_C=\alpha_F I_E + I_{CO}\)
where ICO is the collector-base reverse saturation (leakage) current.
Bring in Kirchhoff's current law for the three terminals:
\(I_E=I_B+I_C\)
Substitute and collect the IC terms.
\(I_C=\alpha_F(I_B+I_C)+I_{CO}\)
\(I_C-\alpha_F I_C=\alpha_F I_B+I_{CO}\)
\(I_C(1-\alpha_F)=\alpha_F I_B+I_{CO}\)
Divide through.
\(I_C=\dfrac{\alpha_F I_B}{1-\alpha_F}+\dfrac{I_{CO}}{1-\alpha_F}\)
which is option 2.
Recognise the familiar form hidden inside it. Since
\(\beta=\dfrac{\alpha_F}{1-\alpha_F} \qquad\text{and}\qquad 1+\beta=\dfrac{1}{1-\alpha_F}\)
the same equation reads
\(I_C=\beta I_B+(1+\beta)I_{CO}=\beta I_B + I_{CEO}\)
the standard common-emitter result. That is also why leakage is so much more troublesome in the CE configuration: the small ICO is multiplied by (1 + β), which for αF = 0.99 means a factor of 100.
Eliminating the other options. Option 1 drops the αF from the base-current term, so it would give the wrong gain. Option 3 is dimensionally impossible — \(\alpha_F/(1-\alpha_F)\) is a pure number being added to a current. Option 4 has (1 + αF) in the denominator, which would make the current gain less than 1 instead of large.
Hence, \(I_C=\dfrac{\alpha_F I_B}{1-\alpha_F}+\dfrac{I_{CO}}{1-\alpha_F}\).
Similar Questions
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Important Questions from Bipolar Junction Transistors
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- In a PNP transistor, the base is of________ material.
In a BJT the collector current in common emitter configuration is
