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.
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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...
$$\beta_D \approx \beta_1\beta_2$$