Experiment 17 - The Gain of the Sziklai Pair
How do you calculate the gain of the Sziklai pair \(\beta_S\)?
Here you derive the formula.
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.
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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...
$$\beta_S \approx \beta_1\beta_2$$
