The circuit is shown in figure belowV1 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. Because of the equal dc potential at the two collectors with respect to ground, the output is referred to as a balanced output.


D.C. Analysis

To obtain the operating point (ICC and VCEQ) for differential amplifier dc equivalent circuit is drawn by reducing the input voltages V1 and V2 to zero as shown in figure below.


The internal resistances of the input signals are denoted by RS because RS1 RS2. Since both emitter biased sections of the different amplifier are symmetrical in all respects, therefore, the operating point for only one section need to be determined. The same values of ICQ and VCEQ can be used for second transistor Q2.

Applying KVL to the base emitter loop of the transistor Q1.


The value of RE sets up the emitter current in transistors Q1 and Q2 for a given value of VEE. The emitter current in Q1 and Q2 are independent of collector resistance RC.

The voltage at the emitter of Q1 is approximately equal to -VBE if the voltage drop across R is negligible. Knowing the value of IC the voltage at the collector Vis given by

       VC = VCC - ICRC
and VCE = VC - VE
              = VCC - ICRC + VBE
       VCE = VCC + VBE - ICRC      _______  (E-2)

From the two equations VCEQ and ICQ can be determined. This dc analysis applicable for all types of differential amplifier.

Differential Input Resistance

Differential input resistance is defined as the equivalent resistance that would be measured at either input terminal with the other terminal grounded. This means that the input resistance Ri1 seen from the input signal source V1 is determined with the signal source V2 set at zero. Similarly, the input signal V1 is set at zero to determine the input resistance Ri2 seen from the input signal source V2. Resistance RS1 and RS2 are ignored because they are very small.
Substituting ie1,
Similarly,
The factor of 2 arises because the re' of each transistor is in series.

To get very high input impedance with differential amplifier is to use Darlington transistors. Another ways is to use FET.

Output Resistance

Output resistance is defined as the equivalent resistance that would be measured at output terminal with respect to ground. Therefore, the output resistance RO1 measured between collector C1 and ground is equal to that of the collector resistance RC. Similarly the output resistance RO2 measured at C2 with respect to ground is equal to that of the collector resistor RC.

RO1 = RO2 = RC     _______   (E-5)

The current gain of the differential amplifier is undefined. Like CE amplifier the differential amplifier is a small signal amplifier. It is generally used as a voltage amplifier and not as current or power amplifier.



The circuit is shown in Figure 1 below 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.

Figure 1

A.C. Analysis

In previous article dc analysis has been done to obtain the operating point of the two transistors.

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

Figure 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 RS1 β and RS2 β are very small in comparison with RE and r'e and therefore neglecting these terms,


Solving these two equations, ie1 and ie2 can be calculated.


The output voltage Vo is given by

Vo = VC2 - VC1
     = - RC iC2 - (- RC iC1)
     = RC (iC1 - iC2)
     = RC (ie1 - ie2)

Substituting ie1, & ie2 in the above expression


Thus a differential amplifier amplifies the difference between two input signals. Defining the difference of input signals as Vd = V1 – V2 the voltage gain of the dual input balanced output differential amplifier can be given by


Differential Input Resistance

Differential input resistance is defined as the equivalent resistance that would be measured at either input terminal with the other terminal grounded. This means that the input resistance Ri1 seen from the input signal source V1 is determined with the signal source V2 set at zero. Similarly, the input signal V1 is set at zero to determine the input resistance Ri2 seen from the input signal source V2. Resistance RS1 and RS2 are ignored because they are very small.


Substituting ie1,


Similarly,

The factor of 2 arises because the r'e of each transistor is in series.

To get very high input impedance with differential amplifier is to use Darlington transistors. Another ways is to use FET.

Output Resistance

Output resistance is defined as the equivalent resistance that would be measured at output terminal with respect to ground. Therefore, the output resistance RO1 measured between collector C1 and ground is equal to that of the collector resistance RC. Similarly the output resistance RO2 measured at C2 with respect to ground is equal to that of the collector resistor RC.

RO1 = RO2 = RC     ________   (E-5)

The current gain of the differential amplifier is undefined. Like CE amplifier the differential amplifier is a small signal amplifier. It is generally used as a voltage amplifier and not as current or power amplifier.

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