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I was reading the following tutorial from Analog Devices:

https://www.analog.com/media/en/training-seminars/tutorials/MT-040.pdf

They include the following image and state:

In most op amp circuits, the inverting input impedance is reduced to a very low value by negative feedback, and only Zcm+ and Zdiff are of importance

I don't understand this point. How is Zcm- swamped by negative feedback and how can it possibly be reduced to a low value? Then the output of a follower would have to supply large current into the inverting input terminal, which is obviously not the case. To me it would make more sense to say that Zdiff is bootstapped by negative feedback and Zcm+ and Zcm- are therefore the only impedances that are significant.

What does Analog Devices mean by this quote?

enter image description here

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  • \$\begingroup\$ Bootstrap raises input impedance by positive feedback less than unity in BJT bias. neg FB uses resistors to drive Vin- to null the external input with respect to Vin+ the virtual null. That Rfb lowers the CM input impedance. But the virtual ground makes it near 0 Ohms differential impedance as long as the output is linear and can force the input Vdiff=0 \$\endgroup\$ – Sunnyskyguy EE75 Jun 29 at 1:51
  • \$\begingroup\$ @pr871 Did you recognize that the large common-mode resistances is in PARALLEL to the diff. resistance (which is drastically reduced due to the negative feedback effect)? \$\endgroup\$ – LvW Jun 29 at 8:48
  • \$\begingroup\$ @LvW No, I do not recognize that the large common-mode resistances are in parallel with the diff. resistance. If the non-inverting input is grounded (as in an inverting configuration), then I agree that Zcm- is in parallel with Zdiff and these impedances (combined with the Norton source impedance) are further scaled down by the amount of feedback, which makes the inverting amplifier have near 0 input impedance. But what about a voltage follower, where the feedback acts to increase the input impedance? Is Zcm- still reduced to a very low value as the article states? \$\endgroup\$ – pr871 Jul 2 at 21:01
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In addition to the other comments and answers you've received, take a look at Figure 5 on this webpage:

https://masteringelectronicsdesign.com/buildi-an-op-amp-spice-model-from-its-datasheet/

Figure 5 doesn't show the current sources IB+ and IB+ but its resistors Rin1 and Rin2 correspond to Zcm+ and Zcm-, respectively, and its resistor Rin corresponds to Zdiff. Redrawing Figure 5 with an emphasis on the op amp's amplifier+output stage (the dependent voltage-controlled voltage source VCVS) and negative feedback path yields the schematic below.

schematic

simulate this circuit – Schematic created using CircuitLab

Note that resistor \$R_2\$ is in parallel with the op amp's input impedances \$R_{\text{in2}}\$, \$R_{\text{in}}\$, \$R_{\text{in1}}\$, etc., and when \$R_2\$'s value is much less than the op amp's input impedance we have:

$$ R_2\;||\;(R_{\text{in2}}, R_{\text{in}}, R_{\text{in1}}, ...) \approx R_2 $$

Also note that VCVS and the op amp's "bias current" IB (not shown) both influence the voltage potential at the op amp's inverting input. When R2's resistance is small compared to the op amp's input impedance, IB's influence at the inverting input is "swamped out" by VCVS, R1, and R2.

$$ V_{\text{IN-}} \approx \text{VCVS} \frac {R_2\,||\,(R_{\text{in2}},R_{\text{in}},R_{\text{in1}},...)}{R_{\text{out}} + R_1 + (R_2\,||\,(R_{\text{in2}},R_{\text{in}},R_{\text{in1}},...))} \bigg\rvert_{R_2 \lll (R_{\text{in2}},R_{\text{in}},R_{\text{in1}},...)} \\ \Rightarrow V_{\text{IN-}} \approx \text{VCVS} \frac {R_2}{R_{\text{out}} + R_1 + R_2} $$

For what it's worth, that Analog Devices MT-040 data sheet doesn't clearly define or give typical values for the "bias currents" IB- and IB+, which only creates confusion. Assuming IB is the very tiny reverse leakage current (on the order of nanoamps) that flows through the reverse biased collector-base diodes of the two BJTs whose bases are connected to the inverting (IB-) and non-inverting (IB+) inputs, that IB current is so tiny that VCVS, R1, and R2 can easily dominate the voltage potential at the op amp's inverting input if R2's resistance value is much less than the op amp's input impedance. For example, add an independent current source at node IN- (the op amp's inverting input) whose amperage is, say, 10 nA, and see what effect it has on the voltage potential at IN-.

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That's because the negative input is fed by very low impedance network of feedback resistors. Even 1 kohm is low compared to the megaohms-teraohms range. Unity gain buffers may not even have resistors so it is driven by the output that could have 0.1 to 10 ohm range.

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  • \$\begingroup\$ This is not the correct explanation since you have considered the feedback NETWORK only - but not the feedback EFFECT. It this effect which reduced the input resistance by a very lage factor k=(1+loop gain). \$\endgroup\$ – LvW Jun 29 at 8:10
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The Vin- is labeled the "Virtual Ground" because with an Opamp having Avol (open loop) of 1,000,000 and a Vout of 1 volt, the Vin- will be at most 1 microvolt away from ZERO.

If the opamp has zero offset voltage, and other idealities.

That extremely low Vin (e.g. 1uV) is why the input impedance is so low.

Example: inverting-gain topology, Rin = 10Kohm, Rout = 100Kohm; with one volt in, and (-) 10 volts out, if Avol(DC) is 1,000,000, then the Vin will be 10volts/1,000,000 or 10 microVolts.

Yet the current flowing thru Rin is 1v/10Kohm = 100 microAmps. What is the apparent Rin?

Rin = Vin-/I_in = 10uV/100uA = 0.1 ohm

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