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f-alpha.net » Electronics » Basic Circuits » Darlington Pair » Know more... » Experiment 17 - Calculation Gain II

Experiment 17 - The Gain of the Sziklai Pair

The gain of the Sziklai pair \(\beta_S\) is defined... $$\beta_S = {I_{CS} \over I_{BS}} \quad\mbox{(1)}$$ and the gain of the individual transistors \(\beta_1\) and \(\beta_2\)... $$\beta_1 = {I_{C1} \over I_{B1}} \quad\mbox{and}\quad \beta_2 = {I_{C2} \over I_{B2}}. \quad\mbox{(2)}$$

Illustration currents in the Sziklai pair.

You know (see illustration), the base current of the Sziklai pair is equal to the base current of the transistor T1... $$I_{BS} = I_{B1},$$ the base current of the transistor T2 flows into the collector of the transistor T1... $$I_{B2} = I_{C1}$$ and the collector current into the Sziklai pair splits up in the transistor T2... $$I_{CS} = I_{C2} + I_{B2} = I_{C2} + I_{C1}.$$ Set this into (1) to obtain... $$\beta_S = {I_{C2}+I_{C1} \over I_{B1}}.\quad\mbox{(3)}$$ Rewrite also (2)... $$I_{B1} = {I_{C1} \over \beta_1} \quad\mbox{and}\quad I_{C2} = \beta_2I_{C1},$$ and you obtain from (3)... $$\beta_S = {\beta_1 \over I_{C1}} (I_{C2} + I_{C1})$$ $$= \beta_1\beta_2 + \beta_1.$$ At sufficiently large gains \(\beta_1\) and \(\beta_2\) you obtain the approximation... $$\beta_S \approx \beta_1\beta_2$$

 
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