0
\$\begingroup\$

I've got two differential driver circuit designs that take a single ended input and provide as output a buffered pair of +1 and -1 times the input and I'm struggling to understand the difference between the two.

In Circuit A, the inverting input to the bottom op-amp (U104B, configured to have gain +1) is attached to ground via R117. This is the "standard" topology as far as I can tell, and I understand how it works. (Note that the labels appear to be wrong, so the upper op-amp output labelled "Diff Out+" has gain -1 and the lower labelled "Diff Out-" has gain +1.)

enter image description here

In Circuit B on the other hand, the equivalent gain +1 op-amp, U12A (the upper one), has its inverting input connected to the input signal via R41. The non-inverting input is also connected to the input signal directly.

enter image description here

Why might someone connect the inverting input of a gain +1 op-amp to the input signal and not to ground like in the "standard" non-inverting op-amp topology? Lower noise perhaps? Better common mode rejection? (I'm not concerned about the op-amp types nor the extra resistors and capacitors at the output and feedback - just the differences between the inputs in the two circuits).

\$\endgroup\$
3
\$\begingroup\$

The upper opamp of circuit B works as a (positive) unity-gain amplifier - and R41 plays a special role !

In many cases, a unity gain configuration has a phase margin that is too small (step response with ringing effect or even instability). Here we have a circuit (unity gain) with a selectable loop gain which can be set to any value for ensuring a "good" and sufficient phase margin. Therefore, we even can use opamps which are not unity-gain- compensated.

In this circuit, the feedback factor is k=-R41/(R41+R40) and can be selected without touching the closed-loop gain of unity. More than that, both amplifier circuits (the upper and the lower one) can now have exactly the same loop gain for R40/R41=R43/R44 (same phase margin, same step response).

Without formal calculation we can see that the voltage Vp at the non-inv. input is Vp=Vin. For an ideal opamp with Vn=Vp=Vin there is no current through R41 and, hence, no current through R40. Therefore, wer have Vout=Vin and Vout/Vin=1.

\$\endgroup\$
1
  • \$\begingroup\$ So it's essentially a free way to match the loop gains of the two differential amplifiers exactly? That's smart! I guess this configuration is a good one to use in situations where the op-amp might be swapped out for a different model later - allowing adjustments to prevent gain peaking by suitable choice of R41. \$\endgroup\$
    – Sean
    Jul 24 '20 at 7:13
1
\$\begingroup\$

In the second circuit, it will work as shown — no current flows through R41 anyway — but it looks like an error. In order to balance the bias currents, R41 should be connected between Diff_In and the noninverting input, and the R40 should be the only thing connected to the inverting input.

But for the same reason, there should also be a 2.5k resistor between U12B pin 5 and ground.

It may be that the designer thought that by connecting R41 as shown, U12A would have the "same imbalance" in bias currents that U12B has, but it doesn't work that way.

\$\endgroup\$
1
  • \$\begingroup\$ No - it´s not a mistake...instead, a very versatile design for a unity-gain stage (see my answer) \$\endgroup\$
    – LvW
    Jul 23 '20 at 13:37

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.