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Quick op amp question?

24Volts

Mar 21, 2010
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Hello,

I have very little experience with op amps. I know how to calculate the voltage outputs of a differential, non inverter and inverter op amp circuits.

But can someone explain to me in a simple sentence as to why it is important to be able to calculate the common mode gain and the differential mode gain?

Up to now, I never needed to use:

Vout = Acm x Vcm + Ad x Vd

to calculate vout!! Also can someone explain simply what is the common mode gain and the differential mode gain?

thanks
 

(*steve*)

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Common mode gain is what you would get if you tied both inputs together and applied a signal to them. The output (ideally) would not be affected.

Differential gain is where you apply a signal so that is appears across the two inputs. This is the normal way, and you expect *huge* variations in the output.

Incidentally, negative feedback can be thought of as providing a small amount of the output to the inverting input so that it is brought to the same voltage as the non-inverting input.

For many practical purposes, it is a good starting point to assume the differential gain is infinite and the common mode gain is zero.

For 99% of simple circuits, that's enough.

You won't generally find the common mode gain specified on datasheets, but you will find the CMRR (or Common Mode Rejection Ratio). This has a mathematical relationship with that gain.

See the Wikipedia page for the math.

Gain generally becomes important when it is substantially less than infinity.

Of course it's always less than infinity, but if the difference in performance calculated for infinite gain vs that actual gain is insignificantly different, we can ignore it.

It becomes important where the gain required is large, or where the op amp is operating in a region where its gain is reduces (typically at high frequencies). Most often in these cases you're looking at something more complex than a single figure though (you'll find graphs in the datasheets).

Although you might think that finite vs infinite gain is a large issue, often other things are more important. They might include things like slew rate (how fast the output can change), phase shift (affects stability), noise, input offset, and input/output characteristics.
 

24Volts

Mar 21, 2010
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Thanks (*steve)!

The reason I am asking is because I am a little confused on how to calculate my own ramps when it comes to op amps.

To keep subjects separated, I strat a new post on this.

thanks
24v
 

24Volts

Mar 21, 2010
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Hi (*Steve*),

I would start a new thread, but I figured since you and I started on this subject, it would be better if I continue it here. Sorry for bringing up an old subject!

Common mode gain is what you would get if you tied both inputs together and applied a signal to them. The output (ideally) would not be affected.

So I have been wondering what is the reason why a purely differential op amp that has all the same resistor values can get away by using the simple formula:

Vout = (vi-v2)(rf/ra)

Is it because in this special case, the common mode gain is very close to 0 and the voltage differential gain is 1 and therefore making the amplifier's gain have no effect to Vout right?

In reference to the following link:

http://www.ntu.edu.sg/home/aschvun/FAQ/DiffAmp.html

For example, if we take an op amp circuit with all its resistors equal to 10K. This would mean that:

Acm = [R3/(R1+R3)] [(R4 + R2)/R2 - R4/R2] = very close 0

Vd = 1/2[R3/(R1+R3)] [(R4 + R2)/R2 + R4/R2] = 1

and therefore we can simply use:

Vout = (vi-v2)(rf/ra)

right?

thanks
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Vout = (vi-v2)(rf/ra)

That simplification works because the resistors are the same value.

Is it because in this special case, the common mode gain is very close to 0 and the voltage differential gain is 1 and therefore making the amplifier's gain have no effect to Vout right?
Kinda.

It's more due to the fact that because the values of some resistors are equal, rf and ra can be substituted into the equation is such a way that most of the nastiness cancels out.

As is noted, the values of the resistors won't be exactly the same, so it's not quite accurate, but it's pretty close.

Acm = [R3/(R1+R3)] [(R4 + R2)/R2 - R4/R2] = very close 0
doesn't that work out to 0.5? (1/2)(2-1)

Vd = 1/2[R3/(R1+R3)] [(R4 + R2)/R2 + R4/R2] = 1

and therefore we can simply use:

Vout = (vi-v2)(rf/ra)

right?

I believe so.

However, see this page. It describes the circuit in another way which leads to equations that make more sense.
 
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24Volts

Mar 21, 2010
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Hi (*Steve*),

doesn't that work out to 0.5? (1/2)(2-1)
Yeah! ... your right.... I see what you mean.... I have a hunch
maybe there's a mistake with the brackets !!!! :confused:

When I have a chance I will try to do the math that involves
"Substitute eqns (5) & (6) into eqn (2)". :eek:

and see if its a misplaced bracket !!! I think (and i really am not
100% sure) those equations should perhaps be:

Acm = [((R3/(R1+R3)) ((R4 + R2)/R2)) - (R4/R2)]= 0
Ad = 1/2[((R3/(R1+R3)) ((R4 + R2)/R2)) + (R4/R2)] = 1

thanks for your help
 
Last edited:

Laplace

Apr 4, 2010
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Tried doing the substitutions but did not get anything like the NTU.EDU web page. Obviously, those guys are wrong.
 

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24Volts

Mar 21, 2010
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One more thing guys....

I figured after all this math I would be able to calculate an
op amp's gain in a box.

For example, jus by external measurements with our
VOM, can we figure out an amplifier's gain when
the op amp is wired up with some unknown resistor values
ra, rf, r1 and r2 (op amp in a black box)?

I am stumped .... gets me discouraged because I look at
all this maths and I understand what's going on but
for some reason my slow brain can't figure out how
to do this???

discouraged!
 
Last edited:

Laplace

Apr 4, 2010
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Suppose that your task was to design a curve tracer for an op-amp in a black box. How would you design it? {Oh, no pressure here, but your future employment at this firm depends upon how well you complete this task!}
 
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