0
\$\begingroup\$

Why doesn't the tail resistor feature in the denominator for differential gain (whereas it does for common mode gain) in the differential amplifier below? (Fig 5N.28 from Learning the Art of Electronics)

enter image description here enter image description here enter image description here

LAoE explains this by pointing out that in pure differential mode with an equal but opposite difference at each input, the voltage across the tail resistor won't change whereas it does change with a signal common to both inputs. That much is clear but I still don't understand why the tail resistor can be ignored for calcualtion of differential gain.

And what about a differential signal that isn't equal and opposite?

\$\endgroup\$
  • \$\begingroup\$ Show the formulas you refer to. \$\endgroup\$ – Andy aka Feb 28 at 8:43
  • 2
    \$\begingroup\$ Where the Gcm can increase significantly is when external resistor ratio tolerances for Vin+ and Vin- are mismatched by x% ( eg.differential 4 resistor config) which is often greater than the internal differences on each side for \$r_e+R_E\$ that result in degrading huge CMRR in some IC specs. \$\endgroup\$ – Sunnyskyguy EE75 Feb 28 at 9:54
2
\$\begingroup\$

The usual explanation is one of symmetry.

schematic

simulate this circuit – Schematic created using CircuitLab

Applying a differential signal means that the middle will not change its voltage because the differential pair is constructed in a symmetrical fashion. If the left pulls it up with a certain voltage, then the right will pull it down an equal amount. This means that the middle point can be considered an AC ground (0V) for differential signals.

For common mode signals, you don't have this canceling out, because both left and right pull in the same direction.

Of course, the actual circuit isn't perfectly symmetrical (it would be if the left also had a load resistor), but a BJT transistor will "shield" effects happening at its collector from the emitter (in the active region, when \$v_{CE}\$ changes, \$i_{C}\$ doesn't change much so it doesn't affect the emitter current). So the approximation is still very much valid. However, if you look really, reaaally closely, you will find a dependency on the tail resistor because of that asymmetry.

And what about a differential signal that isn't equal and opposite?

Such a signal can be broken down again in a non-zero common mode signal and a differential signal that is equal and opposite. For example:

$$\begin{align} v_+ &= 1V \\ v_- &= -2V \\ v_d &= v_+ - v_- = 3V \\ v_{cm} &= \frac{v_+ + v_-}{2} = -0.5V \end{align}$$

\$\endgroup\$
  • \$\begingroup\$ A ground at AC... ...right, I like that idea. I'm guessing that looking at gains in terms of voltages produced by currents through resistances, the equations above can be derived where there's no change in current through the tail resistor, and where there is. I'll try that at the weekend. Breaking down the signal looks good too. I'll do some work on that. Cheers! \$\endgroup\$ – David1 Feb 28 at 20:07

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.