the input resistance of a single opamp difference amplifier

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Ratchit said:
"Virtual ground" implies the same voltage as ground reference, but it is not obvious until explained.
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Hi, I have something to say from the point of view of a foreigner.

What "virtual ground" means is now clear for me, it is because I have taught what it means.

English technical terminology does sometimes cause confusion for beginners/learners, especially for someone who's not a native English speaker. While dictionaries are helpful, we (or at least me) sometimes have to guess or use our imagination.

I have encountered many similar situations, for example, "ideally a difference amplifier will reject completely the common-mode input." Why use "common-mode" input? What does it refer to anyway? Maybe it is easy to understand for English speakers, but not for everyone until it is clearly explained. Elegant or awkward doesn't seem very important to me as long as it is intuitive or straightforward.
 
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Heidi said:
Elegant or awkward doesn't seem very important to me as long as it is intuitive or straightforward.
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Well said Heidi. Terms like virtual ground, virtual short, common-mode etc. are ubiquitous in the language of electronics. Ratch would like to invent some new language that instantly teaches concepts with no effort, in addition to labeling. I'm sure his new language would be a wonderful thing, but the rest of us just get on with learning the concepts and the labels that are in place already.

I find some technical terms very poetic and interesting, even when they take a little extra effort to learn. I remember learning about "whispering gallery modes" in wave theory. Now there is a term that makes me want to investigate what the heck that is. I'm sure Ratch could invent a new name that is more apt, but I doubt I would like it as much, and I doubt his term would make me extra curious to learn about it.

Of course, now that I said that, I fully expect Ratch to come back with a response telling me that "whispering gallery mode" is perfectly descriptive and there is no better word possible. Oh, wait, now that I said that, he will say it's not a good term. Hmm, did i just trap him? No, he will still have a response that wastes our time. Oh wait, now that i said that, he will not respond. ... but now that I said that? ... and ad infinitem.
 
Certainly you have to learn what certain technical terms mean in English as even English speakers do, but I assume that would be true for technical terms in other languages as well.

Common-mode refers to voltages that are common (the same value) at both inputs of the difference amplifier. I'm not sure what name for that would convey it more clearly(?).
 
crutschow said:
Common-mode refers to voltages that are common (the same value) at both inputs of the difference amplifier. I'm not sure what name for that would convey it more clearly(?).
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In the early days of telephone engineering, they referred to longitudinal currents and transverse currents.
 
It is the word "mode". Does it have any special meaning here?

Why not just "common voltages"?
 
Heidi said:
It is the word "mode". Does it have any special meaning here?

Why not just "common voltages"?
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You can certainly say common-voltages and differential voltages, but common-mode and differential-mode is more general and highlights that a particular methodology is being used conceptually.

I think the word "mode" is confusing, even though English is my first (and only) language. The word itself has a few different meanings in general usage, and then in math and science you see it is so many places. After a while you just know what it means by examples. In one sense you can sometimes think of a "mode" as a particular natural solution to a problem. In this case, there is a mathematical symmetry to the circuit that makes the system behave in significantly different ways to the common-mode and differential-mode signals.
 
I view common-mode as basically one word which could be written as commonmode but this is not done in practice.
 
There is also pseudo-differential where the 2 inputs are measured with respect to ground and then subtracted. Just adds to single-ended, differential and pseudo-differential.
 
KeepItSimpleStupid said:
There is also pseudo-differential where the 2 inputs are measured with respect to ground and then subtracted.
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Why would you call that "pseudo-differential". Sounds like standard differential to me. It just depends upon how you do the subtraction.
 
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KeepItSimpleStupid said:
There is also pseudo-differential where the 2 inputs are measured with respect to ground and then subtracted.
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Very interesting to know there's a "pseudo-differential". I assume that the pseudo-differential = (V2-V1) in the following case:

fig.1

When I was learning about op-amp differential amplifier, I had been thinking "why are they introducing common-mode input signals?" "where did they come from?" "why bother splitting one single voltage into differential and common-mode components?". I was just focusing on the circuit in fig.1 above, where I arbitrarily set the differential input voltage as (V2-V1), which was the only way I could think of back then to creat a "voltage difference".
As a result, with R1=R3 and R2=R4 in fig.1, I could only see an amplified voltage difference, didn't know what role a common-mode input was playing.

Later I realized that I could arrange the output voltage expression in a different form, with arbitrary resistor values, to see how the so-called common-mode voltage affects the output voltage Vout in fig.1:

By superposition, Vout can be expressed as

Vout = -(R2/R1)*V1 + V2*[R4/(R3+R4)]*(1+R2/R1)

If we define the differential voltage Vd and the common-mode voltage Vc seperately as

Vd=V2-V1
Vc=(V1+V2)/2

and substitute

V1 = Vc - Vd/2
V2 = Vc + Vd/2

into Vout, we get

Vout = { R2/R1 + [R4/(R3+R4)]*(1+R2/R1) } * Vd/2 + [R4/(R3+R4)]*[ 1- (R2*R3)/(R1*R4)] * Vc

In the circuit in fig.1, because R1=R3 and R2=R4, the coefficient of Vc becomes zero, Vc is gone. The output voltage only has something to do with the differential voltage, Vout=Vd*R2/R1. That's why I didn't see the influence of a common-mode voltage on the output, and wondered why they were complicating things by introducing Vc and splitting V1 and V2 each into 2 terms.

Now I see once there's an imbalance in the resistor values, the common-mode voltage begins to play a role. In addition to amplifying the voltage between the amplifier's two input terminals, it also amplifies another kind of voltage - the common-mode voltage. If the common-mode voltage is too large, the voltage at the output terminal might saturate.
 

I don't think that "pseudo-differential" applies to the analog circuit that Heidi asked about. According to the paper you linked, it applies to the track and hold switching that the ADC sampling circuit uses to receive input. That is not what Heidi is doing. She is using a straight analog circuit.

Ratch
 

Just in case, Jim, go the Escorial monastery (Spain). Near one of the entrances, there is place where you could experiment it. Incredible.

And then... in the Academy where I spent four of my youngest years wehre every night (or morning) we could enjoy it. Real.

/Edit

 
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Hola Heidi,

I hope I am not helping to derail this thread. If you ask, I start a new one. Sure.

In the Steve's notes I run across the definition of "common-mode" voltages as the half-sum of both. I am now perplex because I always believed that the common-mode value was the smallest of both, so the difference was what mattered.

I've been searching but could not find a reason why it is so defined. What it actually represents?

I know it is simply the average so, if V1= 7V and V2= 2V, where do I measure 4,5V?
 
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Of the three different types averages, the half the sum of two values is the mean average. 4.5 volts is 2.5 volts from 7 volts and 2.5 volts from 2 volts. Common-mode means the same voltage on both the positive and negative inputs. It has nothing to do with the relative voltages because there is only one voltage to apply. An ideal differential amplifier would show no output at any voltage applied equally to its input pins. Non-ideal amplifiers show an offset output voltage which is related to the common-mode rejection ratio (CMRR).

Ratch
 
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This is not an obvious thing, and it really needs to be studied vigorously, with consideration of the mathematics and the symmetry (and anti-symmetry) of the system. It turns out that many systems have symmetries that make the sum and difference of variables as more interesting/useful than the actual ones we measure. You should not worry so much whether you can directly measure the common-mode and differential mode. In many cases you can measure them directly, and you can devise ways to measure them in other cases, but this is not the main point. Just to give an example, you might deliberately inject a differential mode signal as the input, but you might forget that there is common-mode noise injected by nature.

So some systems can be studied by using the sum and differences of standard variables. The common mode is a summation, but we divide by two to get the average value, as you said. But, it's still a summation of the variables. The differential-mode is a direct difference of the variables. So they do represent the average and the difference, as you said, but what is more relevant is that we use these variable only for circuits that have the symmetry needed for these signals to have usefulness. So, many amplifiers will try to reject the common-mode, which can be good because noise is often common-mode. Also, as we see with Heidi's example, the circuit has different gains and input resistances for the two types of signals.

Anyway, one of the reasons why I presented those notes is that, even though they will be hard to follow at first, they provide a basic outline of how these systems need to be analyzed in order to understand them and to understand why the CM and DM signals are defined and are useful.

This CM/DM trick is very handy, by the way. Whenever you study a new system, you should see if it has a symmetry that can allow this technique to simplify the problem. I have used it many times in my career to simplify the analysis and modeling of systems. Just to give one example, the SEPIC converter is tricky to analyze and model. If you make a simulation directly using SPICE or SIM-Power in Matlab, their can be inaccuracies and crashes. But, if the system is rewritten using CM and DM variables, the simulations become very robust.

Basically, this is very important stuff to understand, and it shows up all over the place in engineering.
 
atferrari said:
I've been searching but could not find a reason why it is so defined. What it actually represents?
Click to expand...
Hola atferrari,

Please refer to the differential amplifier in post #33, where the op amp was assumed to be ideal with infinite open-loop gain and CMRR. If R1/R3 ≠ R2/R4, then the amplifier will be considered as non-ideal and has a finite CMRR. Under this condition, the output voltage can be expressed as

Vout = { R2/R1 + [R4/(R3+R4)]*(1+R2/R1) } * (v2-V1)/2 + [R4/(R3+R4)]*[ 1- (R2*R3)/(R1*R4)] * (V1+V2)/2

If we give the term (V1+V2)/2 a special symbol, say Vc, Vcm or whatever, then the error

[R4/(R3+R4)]*[ 1- (R2*R3)/(R1*R4)] * (V1+V2)/2

can be expressed in a more concise way

[R4/(R3+R4)]*[ 1- (R2*R3)/(R1*R4)] * Vc

I think that's why they defined the common-mode voltage that way.

However, it's only my guess, because, then they should have defined the differential-mode voltage as (V2-V1)/2 rather than (V2-V1).
 
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