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Wednesday, May 11, 2011

Operation with a Common-Mode Input voltage

To see how the differential pair works, consider first the case of the two gate terminals 
interjoined together and connected to a voltage vCM called the common-mode voltage. That is as shown in Fig (2.2)                
Since Ql and Q2 are matched, it follows from symmetry that the current I will divide equally between the two transistors. Thus
and the voltage at the sources, vs will be

where VGS is the gate-to-source voltage corresponding to a drain current of I/2. Neglecting channel-length modulation, VGs and I/2 are related by

or in terms of the overdrive voltage   
The voltage at each drain will be

Thus, the difference in voltage between the two drains will be zero.

The MOS differential pair with a common-mode input voltage
Fig. 2.2      The MOS differential pair with a common-mode input voltage

Now, let us vary the value of the common-mode voltage vCM Obviously, as long as Q1 and Q2 remain in the saturation region, the current I will divide equally between Q1 and Q2 and the voltages at the drains will not change. Thus the differential pair does not respond to (i.e., it rejects) common-mode input signals.
       An important specification of a differential amplifier is its input common-mode range.This is the range of vCM over which the differential pair operates properly. The highest value of vCM is limited by the requirement that Q1 and Q2 remain in saturation, thus

The lowest value of vCM is determined by the need to allow for a sufficient  voltage across current source I for it to operate properly. If a voltage Vcs is needed across

the current source, then

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