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Linear amplifier:

In the following, we carry out a DC analysis of the circuit, to ascertain the range of operation over which it acts as a linear amplifier.

Supposing Q1 ≅ Q2 and β1,2 >> 1 so that we can assume IE1, 2 ≅ IC 1, 2, the application of KVL to the base loop gives

 VB1 = VBE1 - VBE 2  + VB 2

Under normal operating condition, according to the Ebers-Moll equations, we have

                                       Ic  ≅ I s  e (VBE /VT)       

where Is stands for the reverse saturation current and VT = kT/q. Therefore,

V BE      = V T ln (   IC  /IS)

From Eqs. (34) and (36), we have

VB1 - VB2 = VT ln [  ( IC1  /IS2)  /  ( IC1  /IS1)

For matched transistors IS1 = IS2 so that, we get

  690_Linear amplifier.png

on the other hand, one can write

IC1 / IC 2= e (VD /VT)                     


IC1 + IC 2  ≅ I E1 + I E 2  = I EE  = I0

Solving Eqs. (39) and (40) for IC1 and IC2 we obtain

I01 = IC1  = I0 /(1+  e (- VD  / VT )  )


I02 = IC 2 =  I0 /1 + e+ (VD/ VT  )

Figure illustrates a plot of these two equations. From the plot we can note that when VD

= 0, i.e. VB1 = VB2, IC1 = IC2 = IEE /2 which is an expected result due to the supposition of matched devices. Also, for VD /VT > 4, IC1 → IEE and IC2 → 0; likewise, for VD /VT → 4, IC1 → 0 and IC2 → IEE. Under either of these two conditions, one of the transistors is saturated and the other is cutoff.

1574_Linear amplifier1.png


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