Question: Draw the darlington pair and explain its working.
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Subject: Electronic Devices and Circuits II

Topic: Introduction of Multistage amplifier

Difficulty: Low

edc2(34) • 689 views
 modified 12 months ago  • written 14 months ago by Amruta Padhye • 20
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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 darlingtone pair provides a current gain of .βD = β^2 Darlington pair Have very large current gain ,typically a few thousand .

*Working *

-In darlington pair, it has emitter follower configuration has a pair of transistors.The emitter current of first transistor is act as input current i.e base current of transistor second transistor.

Ie1 =Ib2......................1

Input voltage is applied at the base of first transistor and output voltage is obtained at collector of second transistor. Total collector current IC is calculated by Ic = Ic1+Ic2.......................2 But we know that Ic=β* Ib Let substitute it in above equation

IC =β1* Ib1+β2* Ib2..................3

Substituting the value of Ib2 in above equation from equation 1

We get Ib2 = Ie1=Ic1+ Ib1

=β1.Ib1 +Ib1

= Ib1(β1 + 1)......................4

Substitute this Ib2 value in the above equation 2

we get,Ic= β1.Ib1+ β2. Ib1(β1 + 1)....................5

If Ib1 = Ib2= Ib

IC= β1.Ib + β2. Ibβ1 + β2. Ib

= (β1+ (β2.β1) + β2). Ib

In the above equation, β1 and β2 are gains of individual transistor.

 written 12 months ago by Amruta Padhye • 20