# What's the relevance of differential (and common-mode) input impedance?

Short version: as stated in the subject: what's the relevance of these parameters when designing a "normal" op-amp circuit, for example for audio: a Sallen-Keys active filter, or an RIAA-equalized phono amplifier?

Detailed version:
In this Analog Devices MT-040 Tutorial, they state that the differential impedance is the impedance between the two inputs. The common-mode impedances are the impedances between each input to ground. An application note from T.I. (SLOA011B) is more specific: it defines differential input impedance as "small-signal resistance between two ungrounded input terminals".

The A.D. tutorial further states that "In most op amp circuits, the inverting input impedance (Zcm‒) is reduced to a very low value by negative feedback, and only Zcm+ and Zdiff are of importance".

EDIT:
After thinking about it, I think they misspoke and instead of "reduced to a very low value", they meant "increased to a very high value" (am I right on this?). But then, it looks to me like that's also (or mainly, or instead?) the case for the differential impedance. Through negative feedback, the differential voltage across ‒ and + inputs is brought to zero (assuming the ideal op-amp model except for these input impedances); so, even though there is a real impedance connecting ‒ and +, there is no current, that impedance has no effect on the circuit. An LTSpice simulation agrees with this, where I connect an external resistor||capacitor between the two input terminals. The frequency response does not change at all varying that resistor from 1k to 100k, or the capacitor from 1pF to 1nF.

From this, it would seem like the answer to the paragraph below is "it is not relevant for any circuit where negative feedback maintains linear operation" (within reasonable range, excluding stability/compensation issues, perhaps?)
END EDIT

In addition to that: how is that relevant to a circuit design of, for example, an active filter such as a Sallen-Keys configuration, or an RIAA equalization curve? Intuition would seem to suggest that the resistors used in the feedback network should be much larger than the differential input impedance so that the circuit is unaffected by this low(ish) impedance. However, running LTSpice simulations (with the downloaded SPICE model of an actual op-amp with diff impedance of 20kΩ) shows no difference in frequency response when switching from a feedback resistor in the order of 1kΩ to a resistor in the order of 1MΩ — the response is essentially identical to the theoretical response under the ideal op-amp model and all ideal components (at least in the range below 100kHz).

Quote: "But then, if anything, that would mean that the differential impedance is the one that is brought to near-zero by the negative feedback."

No - that is a misunderstanding. The (internal) impedances of the opamp do not change when feedback is applied. You must distinguish between the properties of the (naked) opamp and the whole circuit with feedback.

For the classical inverting gain block (non-inv. input node grounded, gain=-R2/R1) the input impedance at the signal input is considered to be identical to R1 (because the inv. input is assumed to have also ground potential due to the extremely high open-loop voltage gain). This resistance is, of course, much smaller than the diff. input impedance of the opamp unit.

As far as (positive-gain) Sallen-Key filter circuits are concerned, the external frequency-determing network (and so the filter input node) is connected to the non-inv. opamp input terminal. The inverting input node is connected to the gain determining resistive voltage divider.

• LvW: I do understand that the internal parameters of the op-amp do not change; but the effective parameters do change; for example, I've always perfectly understood that despite any (significant) output impedance in an op-amp, when applying negative feedback, the output impedance of the whole circuit is brought to near-zero (op-amp's output capabilities permitting — slew rate, maximum output swing, ability to drive capacitive loads, etc.) May 12, 2022 at 12:18
• Regarding your comment "This resistance is much smaller than the diff. input impedance of the opamp unit": no, not really. The OPA1611 specifies a 20kΩ || 8pF differential input impedance. However, I assume that a simple/classical inverting configuration would always have input impedance = R1 and gain = ‒R2/R1, regardless of whether I choose R1 = 1k or R1 = 100k. Assuming that this previous statement is correct, that brings me back to my initial question: what is the relevance of this parameter? May 12, 2022 at 12:25
• OK - of course, there are special-purpose opamps (like the 1611) which intentionally have not an input resistance in the MegOhm range. The relevance of this parameter is as follows: The input currents (DC and ac) can - in most cases - be neglected during calculation of the feedback loop (closed-loop gain). Such a simplification drastically reduces the complexity of the calculations (provided the feedback loop does not contain resistors larger than lets say 100kOhms).
– LvW
May 12, 2022 at 12:38