# Common mode impedance in operational amplifier

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.

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?

• 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... – Circuit fantasist Apr 9 at 6:50
• @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. – Sayan Apr 9 at 8:49
• 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... – Circuit fantasist Apr 9 at 11:16
• @Circuitfantasist from a DC point of view, I think the ac source is nothing but a dc resistor, correct me if i am wrong – Sayan Apr 9 at 11:23
• Interesting... For starters it is good. And what is the function of this "resistor"? What does it do? How is it connected? – Circuit fantasist Apr 9 at 12:18

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.

• 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... – Circuit fantasist Apr 9 at 7:36

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.

• 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... – Circuit fantasist Apr 9 at 6:59
• @Circuitfantasist Looking forward to your contribution here! – jonk Apr 9 at 18:26
• 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:) – Circuit fantasist Apr 9 at 20:16
• jonk, I finished my "fairy tale" about the famous pair drawing inspiration from your story. Do you mind participating in our imaginary experiment:)? – Circuit fantasist Apr 11 at 18:50

Answering your question directly is not that easy. So I decided to tell you how I imagine the invention of this famous circuit happened. I was inspired by the @jonk's great story... and I decided to enrich it with more observations and thoughts accumulated over the years.

To really understand the circuit(s), several things are needed - full current paths, local voltage drops across the elements, discerning well-known circuit building blocks and circuit concepts, mimicking the more sophisticated electronic elements with humble electrical elements, etc.

# Inside the input source

First of all, we have to know how the AC input voltage source is made... what is inside this circle with a small sine wave.

Regardless of the specific implementation, the output stage of the input voltage source is something like a controlled voltage divider (potentiometer) which is "stretched" between the supply rails. Its two elements can be a transistor and resistor (simple stage) or two transistors (complementary stage) where at least one of the elements is controlled. The general property of these elements is resistance (in a large sense of the word - linear, nonlinear, etc.). By varying the one, other or both "resistances" in opposite directions, we make the two voltage drops across them vary… and we take one of them as an output (the two voltages are complementary).

Note that only one variable resistor is not enough to regulate the voltage when there is no load (open circuit); a second resistor is needed to allow a current to flow through the first resistor and create a voltage drop across it. This is the idea of ​​the 19th century potentiometer. Then why not mimic the voltage source with such a simple potentiometer P1 (see Fig. 1 below)? I will slowly move its wiper up and down and, if you wish, I will do it in a sinusoidal manner.

The split supply consists of two voltage sources (V+ and V-) in series. Their midpoint is used as a reference (ground). Thus the potentiometer is supplied by a common voltage which is a sum of the two voltages V+ and V-. If its output (the wiper) is referenced to the midpoint (the black probe of the voltmeter is connected to this point), it will produce bipolar voltage between V- and V+. But with the same success, its output voltage can be referenced to V-; then it will produce only positive voltage from 0 to 2V... and such a point of view may be more convenient to you...

Note that the current always exits the potentiometer wiper because the emitter resistor is connected to the point with the most negative voltage (the negative terminal of the lower source). This is because the transistor input (base-emitter) junction must always be forward biased. So, the emitter voltage must always be lower than the base voltage… and this is achieved by connecting the emitter resistor to V-.

Thus there are two currents in the input part of the circuit. Both exit the positive terminal of the upper (positive) voltage source V+. Ip flows through the potentiometer and creates two complementary voltage drops - Vr1 and Vr2. Ib1 is the base current. It exits the potentiometer output (wiper), flows through the base-emitter junction, then through the emitter resistor Re, and finally enters the negative terminal of the lower source V-.

This was a story about how transistor stages control their output voltage. Now let's unravel the secrets of the differential pair...

# Half differential pair

When they had to invent this "tailed" circuit, they had to solve a seemingly strange problem - how to make the common-emitter stage... not amplify:) Hmm… a really interesting device - an amplifier that does not amplify...

## … with emitter static resistor…

They solved this problem by inserting a resistor between the emitter and ground (the so-called "emitter degeneration"). We can see this trick in a half of the pair - Fig. 1, for example the left one, because the right half does the same and in fact, it does not affect the left half. That is why I only hinted at the right half by drawing it in pale gray.

Fig. 1. Half differential pair

This topology is known as an amplifier stage with emitter degeneration or as a common-emitter amplifier stage with current-type negative feedback. Here, the input source tries to create a copy of its voltage across the emitter resistor Re by passing a current Ib through the base-emitter junction and the resistor. The transistor senses its "desire" and passes its collector current Ic through the emitter resistor, in parallel to Ib (see Ic path in Fig. 1)… and so it "helps" Ib by doing its "job". Thus it creates the most of the voltage drop across Re and "raises" the emitter voltage close to the input base voltage. At the same time, the collector current creates another voltage drop across Rc that is proportional (Rc/Re) to the emitter (input) voltage... and is used as an output voltage. Figuratively speaking, the voltage drops VRe and VRc are connected by the common collector current like an "electrical transmission"... or Re and Rc can be considered as two cascaded (voltage-to-current and current-to-voltage) converters.

The higher the resistance Re is, the smaller the current through it and the gain will be. As though, Re "softens" the emitter voltage and makes it "movable". Thus I (input source) "move" the base but the transistor "moves" its emitter in the same direction with (almost) the same rate… and we do not do anything. We can say that a "pull-up" network of three elements in series (the collector-emitter part of the transistor, the collector resistor Rc and the voltage source V+) is connected to the emitter. The input source "sees'' extremely high resistance. This topology resembles a balanced bridge and the phenomenon of such an artificial resistance increasing is figuratively named "bootstrapping".

To understand how this trick enormously increases the input resistance, let's do a simple experiment by disconnecting the collector. Now the transistor does not act as a transistor; it is just a base-emitter diode. There is no "pull-up" network. The input voltage source (potentiometer) is loaded only by a "pull-down" network of three elements in series - the base-emitter junction (with zero AC resistance), the emitter resistor (with Re) and the voltage source V- (with zero AC resistance). This network is connected between the wiper and the midpoint (ground). Significant base current Ib is consumed from the input source (see Ib path in Fig. 1) and it "sees'' a total resistance Re (the genuine resistance of the emitter resistor).

## … with emitter dynamic resistor

But we cannot increase the resistance very much because, in the other (differential) mode, we will need more current for a higher gain. Also, when the input voltage approaches V-, the voltage drop across Re and the current through it will be too small. And here they came up with the ingenious idea to replace the "static" (constant) resistor with a "dynamic" (varying) resistor.

Then let's do it here - we just replace Re with a variable resistor ("rheostat") and ask someone to control it (I think that, in the name of revealing this ingenious idea, @jonk will not refuse us:) Then, when I increase/decrease Vin, he will increase/decrease Re with the same rate. As a result, the emitter current Ie = Vin/Re will stay constant. The input voltage source varies its voltage but the base current Ib stays constant… and it has the illusion that the input resistance of this stage is infinite! And what is important here, the gain is zero since the output voltage drop across Rc does not change. What a wonderful trick! By the way, this element is implemented by another transistor and is known as an "emitter current source".

So, let's summarize the three types of input resistance of the common-emitter stage:

1. Extremely low resistance rbe only of the base-emitter junction when the emitter is grounded (no matter if the collector is connected)

2. Moderately high input resistance Re when a static emitter resistor is inserted and the collector is not connected

3. High input resistance Re when a static emitter resistor is inserted and the collector is connected

4. Extremely high input resistance when a dynamic emitter resistor is inserted and the collector is connected

# Full differential pair

Now let's assemble the complete differential pair by adding the right half - Fig. 2. Nothing changes except that the collector current is divided into two equal parts - Ic1 and Ic2. The second input voltage is produced by another potentiometer P2. The whole arrangement resembles a balanced bridge where the input of the differential pair is connected to the bridge output.

Fig. 2. Full differential pair

## Common mode

Now I move both wipers so that the two input voltages simultaneously change in the same direction and rate… and the whole circuit is equivalent to the half circuit above. It can be considered as two emitter followers working on a common load with high resistance. If it is high enough or it is dynamic, the collector currents do not change… the voltage drops across Rc1 and Rc2 do not change as well (there is no gain). So the output voltages and their difference do not change.

The emitter voltage is "soft" and can be changed without any effort from the input voltage sources. Each of them "sees" extremely high input resistance.

## Differential mode

Now I move both wipers in opposite directions so that the two input voltages simultaneously change in opposite directions. Each of the followers acts as a load with extremely low resistance for the other follower. The emitter voltage is fixed and the gain is maximum. The current is steered between the collectors; the voltage drops across Rc1 and Rc2 crossfade. So the output voltages and their difference vigorously change. Each of the input sources "sees'' extremely low input resistance.

## Generalization

Here is the clever idea behind the long-tailed pair - the emitter resistance (voltage, gain) depends on the mode:

• At common mode, there is a resistance in the emitter; so the emitter voltage is "soft", "movable"... and there is no gain.
• At differential mode, there is no resistance in the emitter (it is shunted by the low output resistance of the other emitter follower); so the emitter voltage is "stiff", "immovable"... and there is a maximum gain.

# As a conclusion

I rudely sketched out my "magic tale" about the long-tailed pair and hurried to bring it to your attention; it only remains to refine it with some details. Indeed, it turned out to be quite long... but as I often say, this is the price of understanding; the other is just knowledge. I do not expect everything to be clear to you but you will still find answers to some of your questions - now or in the future when they arise. Even just as an illustration of a different way of explaining, this story can be useful to you...