CMR Help

[UsmanLula]

Hi guys (correct me if im wrong anywhere in my method)

I am looking into the minimum CMR that is required in an Instrumentation Amplifier (IA) used to convert a differential signal (from the S250 strain gauge) to a single ended signal (ready to be ADC'd).

If the excitation voltage (Vexc) changes by 2% (using TC55 regulator), then the CM voltage at the bridge output will change by 1% (half of Vexc). If the effects of the CM shift is limited to an accuracy specification of 1/4 (i.e. 11.5 bit accuracy at best), then what must be the minimum CMR?

Note: Vexc = 5V;
ADC = 12 bit
S250 Output = 2mV per volt of excitation (i.e.10mV)

For a gain of 1 on the IA, I calculate the minimum required CMR to be 98.3 dB. However, am I right to say that if I have a gain of 427 say, the requirement of a higher CMR (when G=1) will be reduced to something like 46 dB????

This means that the CMR will improve with an increase in gain i.e. I will only require a CMR of 46dB to reject the 2% variation in the regulator if i use a gain of 427. However, I will require a higher CMR of 98dB if I only use a gain of 1 on the IA.

I would be grateful if some could verify my calculations and comments.

Cheers

Usman

 

[UsmanLula]

A further query - about strain gauge resistance

We all know the equation to the change in resistance in a wire -

dR = pl/A; dR = change in resistance; p=resitivity; A=area

This means that as the length of a wire increases (in tension), the resistance also increases; and vice versa for the wire in compression (resistance decreases).

Also using the formula for the gauge resistance i.e.

dR/R = GdL/l; R=initial resistance; G=gauge factor (2);
dL=change in length; l=oroginal length

If I substitute numbers into this equation, I find that as the length of the wire increases (tension), the resistance increases.

So, is the following statement incorrect??

"...the loadcell of interest is one that is capable of measuring the force both in tension (WHERE THE RESISTANCE FALLS) and in compression (WHERE THE RESISTANCE INCREASES)..."

I think its incorrect (when looking at the analysis above) but in case i've missed something.

Any light? Cheers

Usman

 

[UsmanLula]

Well its definitely not correct. Somehow I had this in the portfolio I gave in. I must have been sleeping when I write that statement!

Any light on the CMR question????

Usman

 

[UsmanLula]

Guys

I have completed the CMR analysis. I found that with my homebrew (classic 3-opamp IA), my CMR was 119dB (with a 5% resistor matching). With a common mode voltage of 2.25V, the IA would attenuate this CM voltage down to 2.5uV. However, if I matched my resistors well (using metal film 1%), this attenuation would be down to 0.5uV (as the IA would have a CMR of ~130dB).

I am happy with these calculations. However, there are a few important points for discussion - namely if I have a voltage offset of say 200uV (which is small compared to the CM voltage of 2.25), then will this offset voltage be rejected by the classic IA?

Question: Will the amplifier offset voltage of 200uV (and the zero balance voltage of 1.35mV of the strain gauge) be rejected by the Instrumentation Amplifier??? Or will these appear at the amplifier output??? Since the main purpose of the amplifier is to amplify a differential voltage and to reject a common mode voltage. If it does get rejected, then why would anyone have to tweak the offset voltages using a POT??? I am sure im mistaken somewhere here.

I know the amplifier offset voltages are inherent mainly because it is attempting to cause a zero voltage at the output of the IA. The only way to achieve a zero voltage at the amplifier output of say AD620 is to create an offset of say 200uV at the inputs. The real question here is then is this offset is common to both inputs (i.e. CM voltage)? If so, it will get rejected wont it?

I am perfoming an error budget analysis and attempting to calculate the accuracy of the whole circuit. I know i am using a 12-bit ADC, but by looking at these offset voltages, etc the accuracy of the whole circuit is then only around 6-bits!!! I am down to an accuracy of 11.8bits just by having a CM voltage (the bridge excitation voltage).

For 12 bits: 1LSB = 2.2uV
11.8 bits: 1LSB = 2.5 uV
10 bits: 1LSB = 8.8uV **** This is minimum accuracy

However, I have already used up 28% of my error budget (for 10-bit accuracy) - that is due to the 2.5uV error due to the amplifier rejection of the bridge excitation CM voltage of 2.25V!

Any comments appreciated.

Usman

 

[UsmanLula]

Addendum

I didnt adjust the offset voltage (of 200uV) on any one of the amplifier inputs using a POT to create a zero output voltage (which I was supposed to do!).

What I did instead, was level shift the signal (using a POT) so that it rides between the 0-4.5V supply rail i.e. at around 2.25V.

Also, when I supplied a calibrated voltage of upto 2.5V to the ADC channel 0 of LTC1298 (12bit), my output signal had a 1.3mV offset (i.e. at the intercept). What is this offset due to??? (due to the ADC itself?? how can this be??). Puzzling

Usman

 

[UsmanLula]

Just another addition.

I have attempted to claculate the CMRR of a 3-opamp (home-brew) Instrumentation Amplifier.

However, all my resistances were equal (except the gain resistor). This means that my common mode gain was equal to zero!!! And my diiferential mode gain was 427. So how on earth do I calculate the CMRR????

Is the CMRR above 100 dB??? or below?? is it 119dB (as ive calculated using a cmrr equation considering resistor mismatch) or is it equal to 52dB????

I have used the LT1014 to build the homebrew IA. But somehow, if i plug the resistances into the CMRR equation, i get an error on the calculator. That is because my common mode gain is zero. So whats teh fix for this??? any light??

regards

Usman

 

[felixc]

USman,
what kind of precision are you expecting from your whole system? It looks like you are placing a lot of energy on these calculus.
The output of a strain gauge bridge is ratiometric, that is the variation for a specific load is directly proportional to the power supply. So yes a variation of 1% in the power supply will result in a variation of 1% of the output of the strain gauge.
Take a look at a load cell specification. Iis output is stated in mV/V. Ratiometric!
Try to get some litterature on load cells, or maybe from the strain gauge manufacturers. You will find theory on this subject, better than anything I could recap in a reply here.
Obviously you are using the A/D of your PIC to read the output of the IA. You can then use the same A/D to read the supply voltage at the power side of the bridge.
Felixc

 

[itsmoked]

Actually you reference the PIC's A/D to the same VCC you excite the bridge with, then the voltage aught to drop out.

 

[UsmanLula]

Hey guys

Thanks for post.

By ratiometric - as smokey said - I used the same reference voltage on the ADC (LT1298) to excite the S250 strain gauge. So, the voltage variation/drift is not a problem (ADC and bridge move together). However, its a full bridge strain gauge (active) and thus even if the ratiometric principle bears some benefits over power supply variations - the strain gauge result is directly dependant on two aspects - the change in resistance and the excitation voltage (also TCR of gages). So I was accounting for the variation in the Vexc and its errors in the bridge output. Its a 2% regulator variation. If 1 LSB = 8.8 uV, then the error of power variation is 0.2uV. i.e. I have to ensure my error budget is at most 8.6uV (for 10bit accuracy). I am ok with the error aspect.
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However, I built a classic IA using 3-opamps. But how do i calculate the CMRR and does anyone know what the typical CMRR is for a classic IA??? For your information, I have used 10kOhm resistors in the IA except the gain resistor which is 47Ohms (G=426.5).

I did use the formula but the common mode gain is zero!! so i get an error on the calculator. Is there another method to compute the CMRR just by knowing the resistance values and gain???

Usman

 

[itsmoked]

National had an ancient data book that explained how to calculate the CMRR on a 3 O.A. I wonder if the info is on there plush site? BTW, when building a 3 O.A. you are supposed to ONLY use resistor networks. This is because the CMRR is very dependant on thermal matching of the 10k resistors which networks inherently provide. The values are actually less important.

 

[UsmanLula]

Smokey thanks

I have looked into matching resistance - the difference between a carbon film resistor (5%) and a metal film resistor (1%) can be 10dB (in terms of CMRR).

However, the problem I am finding is in calculating the CMRR for a 3-opamp. I have used the CMRR equation:

CMRR = 20log (Gd/Gc) where Gd = 426.5
Gc = 0!!!

However, if I assume Gc=1, then the CMRR is merely 53dBs!

Just the LT1014CN quad op-amp (used to build the IA) gives me a CMRR of 114dB! So perhaps the CMRR is above 100dB??? Confusing. lol

Usman

 

[cbarn24050]

Hi, you neeed to excite your bridge with an ac voltage if you want an accurate system. As someone pointed out your bridge supply needs to be accurate as it produces a differential output as well as a common mode output.

 

[UsmanLula]

Hi

Its a portable system and running on a PP3 9V. An AC excitation can produce a more accurate system-true.

In terms of the CMRR, i will stick to the values of CMRR stated on the LT1014 of 114dB (in the low frequency range). Also, this will be worst case cmrr as the classic type will improve on this. perhaps the 119 db is a realistic value of a classic 3-opamp. will have to liase with collegue on this.

ta

Usman