Discussion
In two transistor analogy of a thyristor, if '\(\alpha\)' is current gain and ICBO is the collector leakage current, the expression for anode current 'IA' is given by :
Answer: D
The anode current is \(I_A=\frac{\alpha_2I_G+I_{CBO1}+I_{CBO2}}{1-(\alpha_1+\alpha_2)}\).
In the two-transistor model, the thyristor is a p-n-p transistor Q1 and an n-p-n transistor Q2, with \(I_{K} = I_{A} + I_{G}\). The collector currents are \(I_{C1} = \alpha_{1}I_{A} + I_{CBO1}\) and \(I_{C2} = \alpha_{2}I_{K} + I_{CBO2}\), and the anode current is their sum:
\[ \begin{aligned} I_{A} &= \alpha_{1}I_{A} + \alpha_{2}(I_{A} + I_{G}) \\ &\quad + I_{CBO1} + I_{CBO2} \end{aligned} \]
Solving for \(I_{A}\) gives the keyed form. As \(\alpha_{1} + \alpha_{2}\) approaches 1 the denominator approaches zero and the anode current grows without limit: the device latches on. The other forms have the wrong sign for a leakage current, a plus in the denominator, \(\alpha_{1} - \alpha_{2}\), or \(\alpha_{1}\) on the gate current.
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