Experiment 16 - The Gain of the Darlington Pair

How do you calculate the gain of the Darlington pair \(\beta_D\)? Here you derive the formula.

The gain of the Darlington pair \(\beta_D\) is defined... $$\beta_D = {I_{CD} \over I_{BD}} \quad\mbox{(1)}$$ 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 Darlington pair.

You know (see illustration),  the base current of the Darlington pair is equal to the base current of the transistor T1... $$I_{BD} = I_{B1},$$ the emitter current of the transistor T1 flows into the base of the Transistor  T2... $$I_{B2} = I_{E1} = I_{C1} + I_{B1},$$ and the collector current of the Darlington pair splits to flow into both transistors... $$I_{CD} = I_{C1} + I_{C2}.$$ Set this into (1) to obtain... $$\beta_D = {I_{C1}+ I_{C2} \over I_{B1}}.\quad\mbox{(3)}$$ Rewrite also (2)... $$I_{B1} = {I_{C1} \over \beta_1}  \quad\mbox{und}\quad  I_{C2} = \beta_2(I_{C1}+I_{B1}),$$ and you obtain from (3)... $$\beta_D = {\beta_1 \over I_{C1}} (I_{C1} + \beta_2I_{B1} + \beta_2I_{C1})$$ $$= \beta_1\beta_2 + \beta_1 + \beta_2.$$ At sufficiently large gains \(\beta_1\) and \(\beta_2\) you obtain the approximation... $$\beta_D \approx \beta_1\beta_2$$