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Question Bank (5 marks) · Q6

Q.Obtain an expression for anode current IAI_A when gate is applied with IGI_G.

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[!TLDR]

With IK=IA+IGI_K=I_A+I_G, the two-transistor model gives IA=α2IG+(Ico1+Ico2)1−(α1+α2)I_A=\dfrac{\alpha_2 I_G+(I_{co1}+I_{co2})}{1-(\alpha_1+\alpha_2)}.

Using the two-transistor equivalent of the SCR (pnp transistor T1T_1 with gain α1\alpha_1 and npn transistor T2T_2 with gain α2\alpha_2, leakage currents Ico1I_{co1}, Ico2I_{co2}), now a gate current IGI_G is fed into the base of the npn transistor T2T_2 (the p-base). The cathode current is then

IK=IA+IGI_K=I_A+I_G

The collector currents are:

IC1=α1IA+Ico1I_{C1}=\alpha_1 I_A+I_{co1}

IC2=α2IK+Ico2=α2(IA+IG)+Ico2I_{C2}=\alpha_2 I_K+I_{co2}=\alpha_2 (I_A+I_G)+I_{co2}

The anode current is the sum of the two collector currents:

IA=IC1+IC2=α1IA+Ico1+α2(IA+IG)+Ico2I_A=I_{C1}+I_{C2}=\alpha_1 I_A+I_{co1}+\alpha_2 (I_A+I_G)+I_{co2}

Collecting the IAI_A terms:

IA−(α1+α2)IA=α2IG+Ico1+Ico2I_A-(\alpha_1+\alpha_2)I_A=\alpha_2 I_G+I_{co1}+I_{co2}

IA [1−(α1+α2)]=α2IG+(Ico1+Ico2)I_A\,[1-(\alpha_1+\alpha_2)]=\alpha_2 I_G+(I_{co1}+I_{co2})

Therefore

IA=α2IG+(Ico1+Ico2)1−(α1+α2)I_A=\frac{\alpha_2 I_G+(I_{co1}+I_{co2})}{1-(\alpha_1+\alpha_2)} …

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