• A very popular connection of two bipolar junction transistors for operation as one “super beta “transistor is the darlington connection .This connection is as shown in the figure:

•The main feature of the darlington connection is that the composite transistor acts as a single unit with a current gain that is the product of the gains of individual transistors.If the connection is made using two separate transistors having current gains ofβ1 and β2 then the darlington pair provides a current gain of

βD= β1*β2

•If the two transistors are matched then β1=β2=β and darlington pair provides a current gain of βD = β^2

•Having very large current gain ,typically a few thousand.

An AC analysis of darlington pair is as shown in the figure:

In order to calculate the current gain of the configuration we assume that the input source is current source.

now we know that Ai =Io/Iin

from above figure it is seen that Vπ1 = Iin * rπ1........................1

therefore gm1 Vπ1 = gm1 Iin rπ1

but we know that gm1 rπ1 =β1

therefore gm1 Vπ1 = β1 Iin .........................2

now Vπ2 = (Iin + gm1 Vπ1) rπ2

substituting gm1 Vπ1 = β1 Iin in the above equation we get,

Vπ2 = (Iin + β1 Iin) rπ2

Vπ2 = (1 + β1 ) Iin rπ2.....................................3

Now the output current Io can be calculated from the output side of circuit ,

Io = (gm1 Vπ1 + gm 2 Vπ2)

Io = β1 Iin + ( 1 + β1 ) Iin gm 2 rπ2.....................................4

but gm 2 rπ2. = β2

Therefore Io = β1 Iin + ( 1 + β1 ) Iin β2

Io = Iin (β1 + ( 1 + β1 ) β2).................5

Ai = (Io/Iin ) = (β1 + ( 1 + β1 ) β2)

Ai = β1 + β2 + β1β2

But we know that β1β2 >> β1β2

Therefore Ai = β1β2.

As required.From above analysis it is seen that the small signal current gain of the darlington pair is the porduct of current gains of individual transistors. Hence the current gain is very high.