Wheatstone Bridge & Strain Gauge 3

[UsmanLula]

Hi All

I have a technical question relating to the Wheatstone bridge for a medical kit.

Background

I am using an S250 Loadcell (strain gauge) which incorporates an inbuilt 4-wire closed Wheatstone bridge. I am applying an excitation voltage of 5V (+/- 0.1V) across the bridge (as supply rails are 5V). The Rated Output (R.O.) of the S250 is 2mV/V.


The Big Question:

If the S250 incorporates a 3kOhm bridge i.e. each arm has a resistance of 750 Ohms, then how do you calculate the maximum Vout from the bridge? This should also show the tolerance of the output voltage.


My Own Analysis:

Example: If the resistance in one arm changes to 745 Ohms, then does this mean:

Vout = R x Vexc

where R = Change in Resistance calculated using the Wheatstone bridge equation i.e. calculated as 0.998 Ohms or is it 0.002 Ohms?

Taking both cases:

Vout = 0.998 x 5 = 4.99V (calculated using Bridge Eq)

or is it....

Vout = 0.002 x 5 = 0.01V (realistic) i.e R=1-0.998

I think if R.O = 2mV/V, then Vout (max) = 10 mV (if using a 5V supply rail). This seems ok but how do I calculate the value of 10 mV using the Wheatstone Bridge Equation and apply the tolerance of +/- 0.1V of Vexc??? i.e. What is the error in the Wheatstone bridge Output???? Are my calculations correct?

Hope this helps.

Regards

Usman
Medical Physicist

 

[VEBill]

Ref:

http://www.smdsensors.com/detail_pgs/s250.htm

Maybe I'm missing something, but I don't see anything on the ref. webpage about output voltage (or resistance) per unit of applied force (kg). Is that information listed anywhere?

Are you sure that the '3000 ohm' bridge is really 750 ohms per leg? Is that an assumption, or a standard way to define Wheatstone bridges' resistor values?

Once these questions are clarified, then it would seem to be a very simple analysis (two voltage dividers).

PS: Lots of blank <CR>s on your posting...

 

[UsmanLula]

Hi VE1BLL

Thanx for that. I have assumed the value of the resistance arm to be 750 Ohms. I may have to contact SMD to find out what they mean by 3KOhm (+/- 2%) bridge.

Also, they have given the Rated Output (R.O) as 2mV/V under the Electrical Characteristics.

I have done the calculations again as follows:

Assuming each arm resistance is 750 Ohms and there is a change in one arm of 7 Ohms, then with a 5 volt excitation I get 10 mV (Vout). What if the resistance is greater than 7 Ohms? The Vout is >10mV. But the R.O is 2mV/V i.e. for 5V, it is 10mV(max). So i assume that the maximum change in resistance can only be 7 ohms. Is this correct???

Now going with this theory, if I account for the Vexc tolerance of +/-0.1V, then the tolerance on Vout will be +/-2mV (calculated by using 5.1V (Vexc) and 4.9V (Vexc)) in the bridge equation.

Am I going wrong somewhere, or is this what I should be getting???

thanx

Usman

 

[UsmanLula]

Sorry VE1BLL - Correction to my last posting...

The output tolerance on Vout will be 0.2 mV (not 2 mV!) if Vexc has a tolerance of +/- 0.1V.

Also, the resistance can only change by 6 Ohms max (one leg changed to 744 Ohms and the rest remained at 750 Ohms)

Usman

 

[VEBill]

If by 'Rated Output' they mean that is what the bridge will produce at full scale (like 10kg for one example), then the maximum output signal is only 10mV for a 5 volt excitation. Obviously it would be 40mV for 20 volt in. (Fairly low signal levels...) I didn't know what they meant by Rated Output, but it makes sense now.

Since the output is obviously in direct proportion to the excitation, there is no power supply rejection ratio. Therefore the error from the excitation variability is simply 'percentages in' equals 'percentages out' - no need for any complicated analysis.

Ideally, you should be able to hold the 5 volt much tighter than 0.1 volts.

You could add an opamp (or software) to remove the excitation error by also measuring the excitation voltage. But this is somewhat circular because you could just use your ultimate voltage reference as the excitation voltage (perhaps buffered if necessary).

There is some good info at the bottom of this page:

Link:

http://tinyurl.com/cg9hk

Learning from the above, the Full Bridge means that all four of Wheatstone's resistors are active strain sensors.

 

[UsmanLula]

Hi VE1BLL

Thanx for that once again. I have already developed the instrumentation. The tolerance on the power supply thus remains +/-0.1V (using the TC55 Series regulator from 9V to 5V). If you look at the bridge resistance, it also has a tolerance of +/-2%.

Without accounting for the tolerance of teh bridge resistance, the largest tolerance on Vout seems to be 0.2mV. However, the largest change in the leg resistance then seems to be 6 Ohms. I sent in a query to the managing director at SMD. He said the leg resistance is 3KOhm. I will have to confirm this and repeat the calculations.

I dont know whether my calculations of errors is correct or not?? Need some light in this.

Usman

 

[VEBill]

Aren't all the errors simply percentages? If so, then you simply need to combine the percentages. If you're converting all the error factors back to ohms, why? Isn't everything simply percentages?

Some of the errors would be fixed offsets and could thus be trimmed out (one time calibration). You might require two calibration trimmers - one for the zero offset and one for the scale slope (the old Y=aX+b thing). Once that was zeroed and calibrated at full scale, then you're left with just various drifts and non-linearity.

From what I have seen, setting the zero point seems to be the most common (most important?) daily calibration. All my strain guage scales have 'zero' buttons...

You probably already know most (all?) of this.

If you need a more complicated analysis, it is simply two voltage dividers (perhaps with non-infinite R loads).

 

[UsmanLula]

Hi VE1BLL

I can zero the offset voltage but the 5V excitation voltage also has a tolerance +/-2%. Moreover, the Full 3kOhm bridge has a tolerance of +/-2%.

If Vout = R x Vexc

...then the tolerance on Vout will be +/-4%? i.e. if the Full R.O is 10 mV for my application then the error on that will be 0.4 mV?? Is this what you are trying to say?

Are you also saying that if I zero balance the SMD250, I remove the error in tolerance caused by the gauge resistance? i.e. +/-2% resistance tolerance no longer exists on Vout tolerance?

Usman

 

[VEBill]

The link that I provided earlier shows some options for calibrating out the offset.

It'll still drift of course.

Most similar products on the market have a 'zero' button - there must be a reason for this.

 

[Skogsgurra]

The reason is very often the need to zero out the tare, as well as zero out offset inherent in bridge and amplifiers.

Gunnar Englund

http://www.gke.org

 

[UsmanLula]

Hi VE1BLL/skogsgurra

Thankyou so much for your advice. I have completed the data analysis. Without the zero offset error (which is almost +/-1.5mV (SMD Datasheet) if using 5V excitation), the overall bridge output tolerance is +/-0.4mV.

In the instrumentation I developed, I forgot to add a POT to 'ZERO' the SMD250 strain gauge. Indeed, I did add a POT to the amplifier to bias signal output level. I was using a single supply classic instrumentation amplifier and had to shift the signal level from 0V to 2.5V - as I wanted to measure in tension and in compression and the supply rail was at 5V.

This data analysis has helped me understand why I had this peculiar offset in my results. It most probably was the voltage offset.

Anyway, a lesson learnt. Thanks once again.

Best regards

Usman

 

[felixc]

Your loadcell has two ratings, one is the rated load, in kg or pounds, the other is the sensitivity, expressed in mV/V.

So if 2mV/V is for a 100Kg loadcell, for example, then if you feed the bridge with 10 volts you will see a variation of 20mV when 100kG is applied to the loadcell.

Few loadcells have their zero balanced, because just setup that keeps it in place puts some strain on it and does generate a voltage output even if none of our intended measurements are taking place. This is why you need a means to cancel this voltage offset to "zero" your measurement. This can be done analogically or digitally if you are using an A/D to read the voltage from the load cell.

Variations of the resistance of the load cell elements are part of the calibration done on the load cell to reach its sensitivity value and the linearity specs, you do not need to take these into account.