amplifieranalysisdifferential
Let's consider this circuit:
Razavi makes the large signal analysis in the ideal case, that is with an ideal current generator. He shows that Ic1, Ic2, Vout1, Vout2 and (Vout1 – Vout2) are function of only the difference (Vin1 – Vin2), as shown in the following image:
Question: in a more realistic circuit (that is, with a real current source with a resistor in parallel), how would large signal analysis modify? How would the previous graphs change?
Thanks
If Vin1 and Vin2 rise together (within reasonable boundaries of the upper and lower power supplies), Vout and Vout2 don't change at all.
You have a current source and that is contant and the current is spread equally between M1 and M2. It's still spread equally between M1 and M2 when Vin1 and Vin2 change together. The current through M1 and hence through R1 remains the same and therefore Vout1 remains fixed.
So when you say a text book says Vout1 and Vout2 both rise this text book is incorrect.
An ideal current source has infinite output impedance (i.e. \$dV/dI = \infty\$). A real current source is equivalent to an ideal current source in parallel with some finite output impedance Ro.
A current source in parallel with a resistor is equivalent to a voltage source in series with a resistor (source transformation).
simulate this circuit – Schematic created using CircuitLab
So the long-tail resistor in your circuit is just a non-ideal current source.
In the traditional vacuum tube version, V- would be a very large voltage and Re would be a very large resistance, so that the resulting current was correct but it was closer to the ideal of infinite resistance.
Best Answer
Qualitative/intuitive answer (without exact calculation):
If the current source in the common emitter path is non-ideal (parallel resistor RE), there are two effects:
the common mode gain Acm=-gmRc/(1+2gmRE) has a finite value,
the maximum possible current through the amplifying transistors is reduced.
As a consequence, in the transfer curves the maximum possible current is reduced and the minimum possible output voltage is enlarged. More than that, due to the finite common mode gain (negative sign) the crossing of the curves at Vin1=Vin2 (Vd=0) is somewhat shifted. Hence, the curves are not fully symmetrical anymore.