Voltage or Current Excitation for Wheatstone Bridge

[yashrk]

Hey guys,

I am working with a pressure sensor with a strain gauge arranged in a Wheatstone Bridge. I have a two-parter question.

1. Why does the sensor performs better with voltage excitation vs current excitation, for a temperature test over -40°C to 80°C?
    With this question, I want to get some understanding of our findings.
    We are getting approximately 1.166 %error/°C when we are using 0.5mA current source across the sensor. There is a 3V voltage drop across the sensor.
    Whereas if we use say 2.5V voltage excitation we get 0.233 %error/°C. Of course, the absolute error of the current/voltage source doesn't account for the difference in error
    Intuitively I would think a higher voltage across the sensor will yield better stable readings, irrespective of how it is generated (voltage/current excited). We also tested the sensor with a higher voltage of 4V but there were diminishing returns, we just got 0.2166 %error/°C.

2. The Second question I have is about the configuration shown in the image below
    Here the main advantage is that this configuration doesn't need a stable reference source. What I want to know is apart from not needing a precision voltage reference is there any added advantage for such a configuration?

[mikerj]

If you drive a Wheatstone bridge with a constant current then a change in impedance on one side of the bridge will change the current flowing in the other side of the bridge (and therefore the voltage in the middle of the divider).

[yashrk]

If you drive a Wheatstone bridge with a constant current then a change in impedance on one side of the bridge will change the current flowing in the other side of the bridge (and therefore the voltage in the middle of the divider).

Isn't that what we want a voltage difference between two points. And I am not saying it doesn't work on current excitation it's just less temperature stable with the current. 

[bob91343]

Perhaps a careful analysis is in order.  Figure the Thevenin equivalent of the bridge arms and so on.

[mikerj]

If you drive a Wheatstone bridge with a constant current then a change in impedance on one side of the bridge will change the current flowing in the other side of the bridge (and therefore the voltage in the middle of the divider).

Isn't that what we want a voltage difference between two points. And I am not saying it doesn't work on current excitation it's just less temperature stable with the current.

If the resistance on one side of your bridge has not changed, why would you want it's voltage to change?

[fcb]

Driving a bridge with a CC (constant current) will result in a change in gain (non-linearity) UNLESS it is perfectly balanced (i.e. both sides keep an identical resistance regardless of strain/pressure) - unlikely.

In your 2^nd part - the results will be better if the ADC reference voltage and the bridge drive voltage are the same, or at least ratiometrically linked - having the bridge driven by a 5V regulator and then the ADC reference being perhaps a precision 4.096V will lead to changes in the gain (non-linearity) if either deviates - in that case you either want the reference to be the 5V rail or perhaps the drive the bridge from the 4.096V.

The higher voltage leads to more dissapation in the strain-gauges and you will probably see greater thermal effects.

[Wolfram]

The output voltage of the bridge depends on the ratio of the resistors when you feed it from a voltage source, but also on the absolute value of the resistors if you feed it from a current source. With current source feed you basically get an extra error term from the temperature coefficient of the resistors.

[Doctorandus_P]

You're measuring an output voltage differential, and therefore must also excite the thing with a voltage.
It's as simple as that.

A part for putting elements in a bridge configuration is for temperature compensation for the elements.
Let's say that  strain gauges can deviate 2% with temperature. If you excite the bridge with a current, then that would directly result in a 2% error when you measure the output differential voltage.

[RandallMcRee]

The answer to your question 2 is that the bridge becomes ratiometric. (Italicized because this is a good google search term. Hint).

In a proper ratiometric bridge the error from the voltage reference drops out, since it goes both to the bridge and the ADC.

This can be combined with bridge reversal, e.g.
https://www.analog.com/media/en/technical-documentation/application-notes/an96fa.pdf
and also
https://www.analog.com/en/analog-dialogue/articles/transducer-sensor-excitation-and-measurement-techniques.html

Randall.
P.S. post mod'ed to add links.

[David Hess]

The output voltage of the bridge depends on the ratio of the resistors when you feed it from a voltage source, but also on the absolute value of the resistors if you feed it from a current source. With current source feed you basically get an extra error term from the temperature coefficient of the resistors.

Resistive bridges usually have very tight absolute tolerances however there are exceptions like semiconductor bridges which a pressure sensor may very well be.

1. Why does the sensor performs better with voltage excitation vs current excitation, for a temperature test over -40°C to 80°C?
    With this question, I want to get some understanding of our findings.

Maybe it is a semiconductor bridge?  Externally they operate like resistive bridges but with higher sensitivity and lower accuracy.

Intuitively I would think a higher voltage across the sensor will yield better stable readings, irrespective of how it is generated (voltage/current excited).

That is generally true but only up to the point where self heating effects manifest.  Sometimes pulsed high voltage excitation is used to get the advantage of high voltage without self heating.

2. The Second question I have is about the configuration shown in the image below
    Here the main advantage is that this configuration doesn't need a stable reference source. What I want to know is apart from not needing a precision voltage reference is there any added advantage for such a configuration?

Ratiometric operation removes both excitation drift *and* excitation noise.