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Author Topic: Opamp CMRR  (Read 799 times)
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promach
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« on: August 31, 2016, 06:50:41 06:50 »

I do not understand how CMRR are affected as in http://i.imgur.com/VRKgsSM.png and http://i.imgur.com/v6A9Zdq.png

I am reading Razavi book now and trying to understand.

Any help ? Thanks!
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Old_but_Alive
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« Reply #1 on: August 31, 2016, 08:38:40 08:38 »

Taking the simple circuit, assume that V1=V2

If  the resistors are ideally matched so R1's are the same, and R2's are the same ( or their ratios are the same), the op amp will force its output to be 0V no matter what the voltage level of V1 and V2 as long as they are the same..

imagine current flowing down the V2 line down to ground.

In the ideal situation if V1=V2 ie no input voltage difference, then the current flowing in the V1 line must be the same, so Vout must also be at 0V (ground).

So, if V1=V2, but at a level of say 10volts,( the common mode voltage), then any deviation from the ideal resistors, will give a difference in the currents, and thus a difference in Vout, which would be an error.

I have assumed of course an ideal op amp, with infinite input impedance, and zero input offset voltage
 
« Last Edit: August 31, 2016, 08:47:58 08:47 by Old_but_Alive » Logged
Signal
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« Reply #2 on: August 31, 2016, 07:39:49 19:39 »

Looking is not reading. Getting an explanation is not understanding. Such endless hopeless despair...

promach, did you try to derive a formula from first slide for basic op-amp based differential amplifier by yourself? (With resistors designated as R1, R2, R3, R4, of course.) Can you share this formula for general case: Vout=F(V1, V2, R1, R2, R3, R4)?

Try to answer following questions:
If V1 = V2 = Vcm, can you derive Vout as function of following arguments: Vcm, R1, R2, R3, R4?
If V1 and V2 are different what is the value of "common mode" input Vcm then?
If V1 and V2 are different how can you express V1 in terms of Vcm and Vdiff?
What is Vout for given Vcm, Vdiff, R1, R2, R3, R4 :  Vout=F(Vdiff, Vcm, R1, R2, R3, R4)?
What is the expression for differential gain Vout/Vdiff?
How CMRR measured in dB is connected to "common mode" gain Vout/Vcm?

I'd, personally, appreciate if you post answers here in this thread.
« Last Edit: August 31, 2016, 07:43:04 19:43 by Signal » Logged

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vern
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« Reply #3 on: August 31, 2016, 10:19:51 22:19 »

Promach,
for the simple circuit:
both R1/R2 are voltage dividers, one goes to ground, the other to the op-amp output, which should be regulated to ground level by the op-amp.
If you apply the same voltage to both dividers, they will present the same voltage at the op-amps input.
Any mismatch at the resistors will result in a small voltage difference at the inputs, which will of course be amplified by the op-amp and result in an error voltage at the output.
If you change the input voltage from 0 to 10V on both inputs and the output changes from 0V to 0.01V then you have a CMRRof 60dB (20log (10V/0.01V)
As old_but_alive pointed out that is only for ideal op-amps, in reality the op-amp itself has a limited CMRR, input offset and a limited input resistance.
The second circuit is basically the same, there are just two amplifiers added to get a high input impdance. The two amplifiers will of course add some errors to the overall CMRR of the circuit. This is why R5 is variable, to correct for any errors.
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Vineyards
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« Reply #4 on: September 01, 2016, 06:16:22 18:16 »

As far as I can tell, you are trying to figure out why you do not get enough CMRR performance where on paper everything must work. These circuits would work flawlessly on a simulator and since all calculations are based on ideal components they can happily do so forever. However, when you decide to build this circuit problems begin to emerge. First of all, there is no such thing like a 0 tolerance component. The best you can get without burning your pocket would be 1% components which are still not good enough considering how tight tolerance requirements must be for a good enough CMRR. Active components are also far from being perfect and in many cases far from reflecting the performance stated in the datasheets for one thing, datasheets tend to remain on the sunny side of the street by presenting you with mostly those specs that look handsome and encouraging. For example, CMRR is often indicated for a certain frequency (mostly DC) and these specs do not actually apply to higher frequencies.

Common mode signal enters the circuit (non-inverting and differential inputs are especially vulnerable) usually through the input leads. Actually, we would prefer noise to be in the form of common mode noise. For example, in signal transmission, twisted pairs are used for this purpose. When the two wires in a pair are exposed to the same noise due their near identical geometric patterns in relation to each other, they essentially transmit the smae noise which is filtered out due to CMRR. Problems begin when the balance is disturbed. For one thing, the difference will leak through the inputs. The imbalance may also form because of the small differences between the two inputs. It is quite difficult to develop analog filters for differential inputs. You need to balance everything using caps which are ten times larger than those used in either input etc. Another common problem with these amps is that even when they have huge CMRR values, they usually have very small common mode ranges and in some cases this can be as small as a few hundred millivolts. So there are lots of things to keep in mind when working with this stuff. Most people don't use either configuration except for undemanding applications. They buy in-amps which support laser etched resistors trimmed to very tight tolerance levels. They solve most of above problems but many of them still suffer from the tight common mode range I mentioned earlier. Furthermore, it is still problematic to develop filters for in-amps.

Posted on: September 01, 2016, 06:12:50 18:12 - Automerged

Since they tend to be fragile where the common mode range is exceeded, they are usually protected with diodes and the diodes are generally very leaky things and the addition of each components makes it a little bir more difficult to keep everything in balance.
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bigtoy
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« Reply #5 on: September 17, 2016, 06:45:42 06:45 »

At my work we use the first circuit you provided (the single op-amp circuit) quite often. So we have a lot of experience with getting it to work, as well as its limitations.

You can see the op-amp is quite "balanced". Each of the 2 op-amp inputs (+ & -) sees an impedance of R1//R2.  If that holds true then the 2 inputs see the same thing and the circuit balances nicely. However, if the 2 inputs see slightly different things, then the circuit is no longer perfectly balanced and the output of the op-amp will change slightly, reflecting this input imbalance.

As vineyards points out, there are no perfect components. 1% tolerance resistors are common. We use 0.5% tolerance resistors in this circuit. Their price is reasonable. You can buy 0.1% tolerance resistors but they're very expensive. Resistors also vary their resistance with temperature, so a "perfect" circuit at room temperature probably won't be perfect at a hotter or colder temperature, due to the resistors changing value.

Ultimately you need to decide what's an acceptable level of error, or accuracy, for your application. We use LTSpice, which is free, to model these types of circuits. Draw up your circuit, start with perfect resistors, and run the simulation. Then change resistor values. For example a 10k resistor can be changed to a 10.1k to simulate a 1% tolerance. Run the sim again and see what happened. It's an easy way to get a feel for how the circuit behaves, without having to go through the effort of soldering one together.

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promach
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« Reply #6 on: November 27, 2016, 12:00:23 12:00 »

Promach,
for the simple circuit:
both R1/R2 are voltage dividers, one goes to ground, the other to the op-amp output, which should be regulated to ground level by the op-amp.
If you apply the same voltage to both dividers, they will present the same voltage at the op-amps input.
Any mismatch at the resistors will result in a small voltage difference at the inputs, which will of course be amplified by the op-amp and result in an error voltage at the output.
If you change the input voltage from 0 to 10V on both inputs and the output changes from 0V to 0.01V then you have a CMRRof 60dB (20log (10V/0.01V)
As old_but_alive pointed out that is only for ideal op-amps, in reality the op-amp itself has a limited CMRR, input offset and a limited input resistance.
The second circuit is basically the same, there are just two amplifiers added to get a high input impdance. The two amplifiers will of course add some errors to the overall CMRR of the circuit. This is why R5 is variable, to correct for any errors.

Hi, Vern.
a) why would variable resistor R5 alone be able to maximize overall CMRR as in http://i.imgur.com/v6A9Zdq.png?
b) what is Vout for given Vcm, Vdiff, R1, R2, R3, R4 :  Vout=F(Vdiff, Vcm, R1, R2, R3, R4) as in http://i.imgur.com/VRKgsSM.png ?
« Last Edit: November 27, 2016, 10:53:40 22:53 by promach » Logged
vern
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« Reply #7 on: November 28, 2016, 09:25:48 09:25 »

Hi promach,
long time since I looked at the schematics.
to a: you could also use R4 or one of the R3, but since R5 goes to GND it is the best resistor to chose. And R5 should of course not be the same value as R4, because the resolution would be bad. If for example R4 is 10k, R3 should be a fixed 9.9k and a variable resistor of 200R.
Then you have a range of 9.9 - 10.1k
In real life for a precision amp you find the right value and put in a precision resistor, because variable resistors add noise and are not really stable.
to b: you have the formula right there.
Example: R1 10k, R2 100k and Vdiff 0.1V  will give you 1.000V output, plus or minus the CMRR, if this is for example 60dB the error would be 1/1000, the output could be 0.999 to 1.001 V.
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promach
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« Reply #8 on: November 28, 2016, 03:24:59 15:24 »

Hi Vern,

a) I understand that if R5 is not matched with R4, then there will be CMRR issue due to "unbalanced" inputs ? is it possible to relate this "unbalanced" issue with CMRR mathematically ? and What do you mean by bad resolution ?
b) I am a bit confused with your error calculation concept. Why is it 1/1000 ? and Why is it 1V plus or minus CMRR ? my understanding is that CMRR = Ad/Acm
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vern
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« Reply #9 on: November 28, 2016, 05:31:34 17:31 »

Hi promach,
resolution in this case means how much your voltage changes if you turn the trimmer.
Bad resolution: you turn the trimmer a little bit and the voltage makes a big jump.
In my description above the voltage change for a full turn is only about +/- 1%, very easy to make a very fine adjustment.
And my error calculation is 1/1000 because as an example I assumed a CMRR of 60dB, which comes to 1/1000.  (https://en.wikipedia.org/wiki/Decibel)
In the case above I have an gain of 10, if your input voltage changes from 0 to 0.1 V your output will cange form 0 to 1V  +/- the CMRR error (0.999 / 1.001V)
If your input changes to from 0 to 0.5V, your output will change to 5V    +/- CMRR Error, 4.995 / 5.005V
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promach
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« Reply #10 on: November 29, 2016, 12:42:49 00:42 »

Thanks Vern.

a) For the case of 60dB, CMRR = Ad/Acm = 1000.  You meant Acm/Ad = 1/1000 = Vcm/Vd ?
b) Why is it 1V plus or minus (Vcm/Vd) ?
c) How is R5=R4 related to bad resolution ?
d) Is there an established equation relating CMRR to the unbalanced resistors matching for http://imgur.com/VRKgsSM ?
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vern
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« Reply #11 on: November 29, 2016, 09:51:50 09:51 »

Hi promach,
a: yes
b: because the CMRR error can be positive or negative
c: because you want to change R5 only a tiny bit
d: no, because CMRR is not just a function of the resistors, even if you match them perfectly you still have the OP Amp errors. (offset, bias voltage, bias current, phase differences etc.)
In fact, to correct for the Op Amp errors you might end up with a different value for R4 and R5 to get a good overall CMRR.
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promach
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« Reply #12 on: November 29, 2016, 03:02:17 15:02 »

Thanks Vern.

a) I still could not get what you mean exactly by:  R5=R4 will give bad resolution.

b) We get back the same equation as in http://imgur.com/VRKgsSM because the two R1 and two R2 are the same, however, if they deviate by a little, i.e., we have R1a, R1b, R2a and R2b, we will see that we do not get (R2/R1)*Vd, the Vcm will also appear in the equation. Is there a way to model this statement in an equation with Vcm terms ?
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vern
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« Reply #13 on: November 30, 2016, 09:37:16 09:37 »

Hi promach,
R5=R4 will give bad resolution. This is not what I said.
If you try to make R5 equal to R4, you want to have a setup where you can make very fine adjustments. This what I mean with "resolution"
You want to change the resistor only by 1 or 2% with a turn of the trimmer, not by 100%.
That is why you take a fixed resistor and add an adjustable resistor to it.
Example: a standard trimmer has about 270 degrees to turn, if it has 10k then 1 degree turning means a resistor change of 37 Ohms (10k / 270)
If you take a fixed resistor of 9.9k and a trimmer of 200 Ohms, a turn of one degree means a change of 0.74 Ohms, it has therefore a much better resolution.
If you want to have a high CMRR, for example 80dB, which means 1/10.000, you have to make very fine adjustments. In this case you have to match the resistors better than 1 Ohm to archive 80dB. You see it is very important to have a fine resolution of the resistor / trimmer combination.
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