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While looking at this simulation, as they say no current flows at the non inverting terminal of the op-amp, but looking at the differential stage of the operational amplifier shouldn't there be some current from ground to -15V point THROUGH THE NON INVERTING INPUT?

Secondly, they define common mode impedance as the impedance between the input terminal and ground point.

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Now the current producing by the action of the AC voltage source has to travel from -VEE to ground. So the impedance should be of this path, then how impedance between ground and terminal is defined as the common mode impedance?

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  • \$\begingroup\$ Sayan, You have two very good answers (concise and detailed) to your question from two very good professionals. Do you want to see another detailed answer from a (creative thinking) teacher? My idea is to explain the meaning of all this to you. This will require some extra efforts... but they will reward you with a true understanding of the phenomenon... \$\endgroup\$ – Circuit fantasist 2 days ago
  • \$\begingroup\$ @Circuitfantasist I am satisfied with the answer already given, If i could accept both of these , I would. but since no such option here so i have upvoted them. anyway, please show what do you want to explain, it has been always great for me to learn new things. \$\endgroup\$ – Sayan 2 days ago
  • \$\begingroup\$ Sayan, it is good that you are still responding to my comment (here OPs often ask intriguing questions... pretend to be very interested in the answers... and then "disappear"). It would be good to respond to the answers here in some way (even if it is wrong)... as they say, "to give a feedback". This would result in something like a discussion in the course of which the truth emerges. By the way, I want to ask you, "Can you imagine what is inside the AC input voltage source (the circle with a small sine wave inside)?" This is necessary to understand where the current produced by it travels... \$\endgroup\$ – Circuit fantasist 2 days ago
  • \$\begingroup\$ @Circuitfantasist from a DC point of view, I think the ac source is nothing but a dc resistor, correct me if i am wrong \$\endgroup\$ – Sayan 2 days ago
  • \$\begingroup\$ Interesting... For starters it is good. And what is the function of this "resistor"? What does it do? How is it connected? \$\endgroup\$ – Circuit fantasist 2 days ago
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There is a small bias current that flows into the input, less than 1% of half of Iem at balance, for transistors with hFE > 100.

Common mode input impedance will be very high because that bias current does not change much with small changes in input CM voltage.

In many cases you can ignore both input bias current and input CM impedance when modern op-amps are used with resistors in the few K ohm range, but it doesn’t hurt to run the numbers and establish that for a fact.

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  • \$\begingroup\$ A concise and accurate answer... but OP obviously needs more detailed explanations. The questions below the OP's figure are quite interesting; they give rise to thought... \$\endgroup\$ – Circuit fantasist 2 days ago
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Common mode means that both inputs "move" equally up or down.

To keep this simple, start out by imagining both inputs to be the exact same voltage (same source, even) and midway between the rails. In this case, both BJTs will share equally the current generated in \$R_\text{EM}\$. The split currents will generate their respective voltage drops across \$R_{\text{COL}_1}\$ and \$R_{\text{COL}_2}\$ and therefore the differential output between the two collectors will be \$0\:\text{V}\$.

Now, imagine that the two inputs (both moving identically, still) move slightly upward. When that happens, both emitters are also dragged upwards, too, and equally as much. This means more current is induced in \$R_\text{EM}\$. This current splits equally, again, between the two BJTs and, again, there is no difference in the output taken between the collectors. That's good.

But what change in current is incurred at each base? Well, each base will require a little more recombination current because the collector currents increased (by about half the current change in \$R_\text{EM}\$.) So it's pretty easy to see that the sources will "see" an impedance (the slightly changed voltage divided by the slightly changed base current.) Roughly speaking, and ignoring the Early Effect, this will be on the order of \$2\,\beta\,R_\text{EM}\$ for each input and is therefore considered "high impedance." (The differential impedance is, conversely, quite low. But that's another story.)

It's not quite so often that the common mode impedance is spoken of as more than "it's high," though. (These are usually driven by significantly lower impedance sources.) Instead, the CMRR is usually more important to know. In theory, this would be infinite as both collectors would move equally for any common mode change (assuming also that the resistors were ideal and exactly the same value, of course.) But the reality is that the BJTs themselves are never perfectly identical to each other and the resistors won't be perfectly matched, either. So there will always be some non-zero output difference to start out as well as some further change in that non-zero output difference when the inputs move identically together up or down. And that kind of thing is often more important to know about.

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    \$\begingroup\$ jonk, It was a real pleasure for me to read your brilliant "philosophical" explanation cleared of unnecessary (in this case) math. I like the style of your explanation the most, which makes me reread it many times to feel the pleasure of it again abd again... \$\endgroup\$ – Circuit fantasist 2 days ago
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    \$\begingroup\$ @Circuitfantasist Looking forward to your contribution here! \$\endgroup\$ – jonk yesterday
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    \$\begingroup\$ Thanks, jonk; it is great that there are people like you here. I intend to fabricate another "fairy tale" about the famous pair. I drew diagrams; it remains to decorate them with text. In short, I will not be bored at the weekend. By the way, I feel like a father from the good old days who bought a little train for his son... but he plays with it:) \$\endgroup\$ – Circuit fantasist yesterday

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