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Chapter: Electronic Circuits - BJT Amplifiers

Transistorised Differential Amplifier

Let us study the circuit operation in the two modes namely • Differential mode operation • Common mode operation

Transistorised Differential Amplifier

 

The transistorised differential amplifier basically uses the emitter biased circuits which are identical in characteristics. Such two identical emitter biased circuits are

 


 

The magnitudes of + Vcc and – V EE are also same. The differential amplifier can be obtained by using such two emitter biased circuits. This is achieved by connecting emitter E1 of Q1 to the emitter E2 of Q2. Due to this, R E1 appears in parallel with R E2 and the combination can be replaced by a single resistance denoted as R E. The base B1 of Q1 is connected to the input 1 which is V S1 while the base B 2 of Q2 is connected to the input 2 which is Vs2. The supply voltages are measured with respect to ground. The balanced output is taken between the collector C1 of Q1 and the collector C2 of Q 2. Such an amplifier is called emitter coupled differential amplifier. The two collector resistances are same hence can be denoted as R C..

 

The output can be taken between two collectors or in between one of the two collectors and the ground. When the output is taken between the two collectors, none of them is grounded then it is called balanced output, double ended output or floating output. When the output is taken between any of the collectors and the ground, it is called unbalanced output or single ended output. The complete circuit diagram of such a basic dual input, balanced output differential amplifier is shown in the Fig.

 


As the output is taken between two output terminals, none of them is grounded, it is called balanced output differential amplifier.

Let us study the circuit operation in the two modes namely

 

      Differential mode operation

 

      Common mode operation

 

1. Differential Mode Operation

 

In the differential mode, the two input signals are different from each other. Consider the two input signals which are same in magnitude but 180" out of phase. These signals, with opposite phase can be obtained from the center tap transformer. The circuit used in differential mode operation is shown in the Fig


Assume that the sine wave on the base of Q 1is positive going while on the base of Q 2 is negative going. With a positive going signal on the base of Q 1 , m amplified negative going signal develops on the collector of Q1. Due to positive going signal, current through R E also increases and hence a positive going wave is developed across R E. Due to negative going signal on the base of Q2, an amplified positive going signal develops on the collector of Q 2 . And a negative going signal develops across R E, because of emitter follower action of Q 2. So signal voltages across R E, due to the effect of Q1 and Q2 are equal in magnitude and 180o out of phase, due to matched pair of transistors. Hence these two signals cancel each other and there is no signal across the emitter resistance. Hence there is no a.c. signal current flowing through the emitter resistance. Hence R E in this case does not introduce negative feedback. While Vo is the output taken across collector of Q1 and collector of Q 2. The two outputs on collector L and 2 are equal in magnitude but opposite in polarity. And Vo is the difference between these two signals, e.g. +10 - (-10) = + 20.

 

Hence the difference output Vo is twice as large as the signal voltage from either collector to ground


 

2. Common Mode Operation

 

In this mode, the signals applied to the base of Q1 and Q2 are derived from the same source. So the two signals are equal in magnitude as well as in phase. The circuit diagram is shown in the Fig.

 

In phase signal voltages at the bases of Q1 and Q2 causes in phase signal voltages to appear across R E, which add together. Hence R E carries a signal current and provides a negative feedback. This feedback reduces the common mode gain of differential amplifier.


While the two signals causes in phase signal voltages of equal magnitude to appear across the two collectors of Q 1 and Q2. Now the output voltage is the difference between the two collector voltages, which are equal and also same in phase, Eg. (20) - (20) = 0. Thus the difference output Vo is almost zero, negligibly small. ideally it should be zero.


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