Why is CMRR infinite for coupled differential amplifiers when Re increases ?? I cant seem to understand what happens on making Re infinite.
Electrical – Common mode Rejection ratio
diff-amp
Related Solutions
That parameter says nothing about the CMRR. All it is telling you is the maximum common mode voltage (\$V_{comm}\$) the amplifier can handle while still meeting the other specifications.
simulate this circuit – Schematic created using CircuitLab
$$ V_{comm} = \frac{V_+ + V_-}{2}$$
Provided that \$V_{comm} < 1.2V\$ (and really, they probably mean \$-1.2V < V_{comm} < 1.2V\$) then the amplifier will function according to specifications. Ostensibly, this common mode signal is attenuated, but if the CMRR is unspecified, then we don't know by how much.
I think the 100 ohm resistor should be 100K. The differential gain is thus very close to 10. The CMRR of the op-amp is 10^5 and the open-loop gain of the op-amp is 10^6.
You then calculate the CMRR of the differential amplifier, assuming perfectly accurate resistors, and I get the suggested answer.
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Best Answer
For a single common-emitter transistor amplifier, voltage gain boils down to collector resistor divided by emitter resistor. The bigger the emitter resistor the smaller the gain.
When applied to a differential amplifier (aka long-tailed pair) the common mode gain is in fact the gain of the single transistor so, if the emitter resistor is very high compared to Rc then common mode gain is very small.
Given that the common mode rejection for a perfect long-tailed pair with the output taken differentially across the collectors is theoretically 100% for any size emitter resistor (within reason), any slight imbalance in the transistors doesn't contribute a massive common-mode signal at the output but, as explained above, whatever imbalance there is will be eradicated by a high value of Re.
In fact, most op-amps (if not all?) will use a constant current source for Re and, if you did some research, you'd find that a perfect constant current source has infinite impedance.