Dual Input, Balanced Output Difference Amplifier:

The circuit is shown in fig. 1 v₁ and v₂ are the two inputs, applied to the bases of Q₁ and Q₂ transistors. The output voltage is measured between the two collectors C₁ and C₂, which are at same dc potentials.

Fig. 1

A.C. Analysis :

In previous lecture dc analysis has been done to obtain the operatiing point of the two transistors.

To find the voltage gain A_d and the input resistance R_i of the differential amplifier, the ac equivalent circuit is drawn using r-parameters as shown in fig. 2. The dc voltages are reduced to zero and the ac equivalent of CE configuration is used.

Fig. 2

Since the two dc emitter currents are equal. Therefore, resistance r'_e1 and r'_e2 are also equal and designated by r'_e . This voltage across each collector resistance is shown 180° out of phase with respect to the input voltages v₁ and v₂. This is same as in CE configuration. The polarity of the output voltage is shown in Figure. The collector C₂ is assumed to be more positive with respect to collector C₁ even though both are negative with respect to to ground.

Applying KVL in two loops 1 & 2.

Substituting current relations,

Again, assuming R_S1 / b and R_S2 / b are very small in comparison with R_E and r_e' and therefore neglecting these terms,

Solving these two equations, i_e1 and i_e2 can be calculated.

The output voltage V_O is given by

V_O = V_C2 - V_C1

      = -R_C i_C2 - (-R_C i_C1)

      = R_C (i_C1 - i_C2)

      = R_C (i_e1 - i_e2)

Substituting i_e1, & i_e2 in the above expression

Thus a differential amplifier amplifies the difference between two input signals. Defining the difference of input signals as v_d = v₁ – v₂ the voltage gain of the dual input balanced output differential amplifier can be given by

      (E-2)

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