2
\$\begingroup\$

If you think of a better title, please do edit or let me know to edit, but this is how I thought I could word the question.

I am finding difficulty in understanding how a differential pair exactly works. I commonly find that textbooks first go through the 'tail' load being a perfect current source before using real world components.

Say, the tail load was a current source and that the entire circuit looked like this.

schematic

simulate this circuit – Schematic created using CircuitLab

Say the current through the current source I1 is Iss. If Vin1=Vin2 (i.e. common mode input) rises, then, textbooks say that the Vout1 and Vout2 both rise by equal amounts and that this is how common mode signals are removed (if we take Vout1-Vout2).

But if Iss is fixed by the current source and M1 and M2 are at identical states (Vin1=Vin2, R1=R2), then I believe I can assume that Id1=Id2=Iss/2. Since Iss is fixed, Id1 and Id2 cannot possibly change for varying (common mode) values of Vin1 and Vin2. So then, how could Vout1 or Vout2, ever possibly rise?

In other words, if Id1 is fixed as Iss/2, then Vout1 is fixed at Vout1 = (Iss/2)*R1 for as long as Id1=Id2=Iss/2, irrespective of what value Vin1 or Vin2 are at (as long as Vin1=Vin2).

This is inconsistent with what textbooks say and I don't understand why? Is there some mistake in my thinking?

\$\endgroup\$
5
\$\begingroup\$

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.

\$\endgroup\$
1
\$\begingroup\$

@midnightBlue,

"What textbooks say" above refers to a differential pair operating at a common-mode input (Vin1 = Vin2) and having a differential output (Vout1 - Vout2). First at all, let's just to fix that when "Vout1 and Vout2 both rise by equal amounts", Vout1 and Vout2 both fall (not rise) by equal amounts since the common-emitter stage is inverting.

This "textbook explanation" would do some work in the case of an imperfect differential amplifier with emitter resistor and a differential output; but the latter is rarely used in practice. In most cases, we put a current source in the “tail” and take a single-ended (referred to ground) output signal from only one of the collectors. In this case, the second (your) textbook explanation is valid. It does a good job for teachers and writers... but unfortunately, it is formal... and even misleading for readers and students. Let me explain why...

In fact, there is no current source (or any other source) in the “tail”... because if there was a source there, we would supply the circuit with it... and we would not need a power supply:) What is there is a current-stabilizing element… a kind of dynamic resistor. It is implemented by the output (collector-emitter or drain-source) part of a transistor with a constant input voltage (at the base or gate). Now I will explain its operation in so simple way that you will remember it for a lifetime...

Let's reproduce the famous Ohm’s experiment but in a little unusual and funny way. Connect a variable voltage source to a variable resistor (rheostat)... also, an ammeter in series and a voltmeter in parallel. Then set the rheostat’s slider somewhere in the middle and vary the voltage. Look at the ammeter - you will see the current proportionally varies obeying Ohm's law - I = V/R.

Now I decide to join the "game" and start moving the slider so that the resistance changes in the same direction and rate as your voltage. The result is surprising - the current does not change at all! This is because both the numerator and denominator in Ohm’s law change in the same way; so I = Vvar/Rvar = const.

Looking at the ammeter and voltmeter, you have the feeling the resistance of this dynamic resistor is infinite. Actually only its differential resistance dR = dV/dI is infinite and constant while its static resistance R is finite and varying.

See also my answer to the question How does the resistor at the tail act as a current source in a differential pair?

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.