Wheatstone bridge with four active strain gauges - extension cable length compensation formula

[MarsBravo]

Suppose a Wheatstone bridge with four equal active strain sensors (350 ohm each).
The output nodes are connected to a high impedance instrumentation amplifier, so no current draw from there. The bridge is placed at consiberable distance from the exitation voltage source (and the amplifier), so the length of the extension cable (with all four wires) must be accounted for in such a way that it adds to the total resistance of the bridge but not to the sensitive part (the bridge). It will reduce the sensivity of the bridge, but in a small way.
Can the bridge and the extension cable resistors considered as three resistors in series (the bridge resistance measured from the exitation source is still 350 ohm!), or has one take into account the split of resistors within the bridge (dividing the exitation current) as the extension cable resistors 'serves' both branches?
In the attachment, the difference in the formula is in red. CF is the calibration factor, CF' is the adjusted CF. Which formula is right?

 

[Baluncore]

The resistance of the wires, from the bridge to the amplifier, play no part in the sensitivity. The voltage across the bridge is reduced by the lead resistance, which reduces the excitation current.

The 4 resistor bridge looks like one resistor to the excitation current, Ie.
The sensitivity of the bridge is proportional to Ie. Which is reduced by the series voltage drop in the wires.

Without wires, Ie = Ve / Rb;
With two wires of resistance, Rw;
Ie = Ve / ( Rb + Rw + Rw );
Ie and the sensitivity will be reduced to; Rb / ( Rb + Rw + Rw );
350 / ( 350 + 4 + 4 ) = 0.9777 = -2.23 %
So your first solution is correct.

 

[MarsBravo]

(Quoting seems not to work...)
"The voltage across the bridge is reduced by the lead resistance, which reduces the excitation current."
Both are true, but not related as such. The higher overall resistance (Rb + 2*Rw) reduces Ie.

"The 4 resistor bridge looks like one resistor to the excitation current, Ie."
That's the opinion of one camp, the other side claim the splitting of Ie in the bridge makes the difference.
The bridge is not one resistor, but a network.

 

[Baluncore]

Both are true, but not related as such. The higher overall resistance (Rb + 2*Rw) reduces Ie.

I apologise for not welcoming you top PF, and for providing too much information.

"The 4 resistor bridge looks like one resistor to the excitation current, Ie."
That's the opinion of one camp, the other side claim the splitting of Ie in the bridge makes the difference.
The bridge is not one resistor, but a network.

It is a 4 terminal network, but only two terminals are used to excite the bridge. The sensitivity of the bridge is proportional to the bridge excitation current.

 

[MarsBravo]

The supply on the input terminals do not really 'exite' the bridge, but the strain on the sensors brings the bridge out of balance and can be measured on the output terminals. The supply is a passive voltage source.
But the question remains: is the Wheatstone bridge network exactly the same as an equivalent resistor with small cable resistors in series with the shared voltage supply?
Looking only from the outside supply world, the bridge can be considered as an eqivalent resistor (first option), but inside the bridge this supply current Ie is split in two, so 'seen' from the strain sensors these cable resistors can be considered as of double value (Kirchhof, second option).
This dispute is amongst several well seasoned engineers (MSc, BSc) at our work, ranging from 20 to 40 years of experience in electronics. The formula of the first option is in use for over twenty years, but network analysis seems to suggest the obvious understanding as wrong. From the traditional outside exitation viewpoint there is no doubt, but from another angle, the inside bridge viewpoint, each strain sensor 'sees' a double cable resistor value. Hence the second option. The key here is to understand what the splitting of the exitation current will mean really. Hammering down on the tradition only sounds like a one hand clapping machine.
What's 'top PF' btw?

 

[Baluncore]

Welcome top PF.

If you view the bridge as two branches, you do not change the differential output, but you must remember it is there.

The supply on the input terminals do not really 'exite' the bridge, but the strain on the sensors brings the bridge out of balance and can be measured on the output terminals. The supply is a passive voltage source.

You invented the word "exite", I used the word "excite".

To quote, highlight the text, then a box appears, select either 'add to quotes', or 'reply with quote'.

I refer to the electrical excitation current, that results in a differential output voltage when the four gauges are strained. Your passive voltage source is not at the bridge. Perform an experiment, insert a 330 ohm resistor into each line from the voltage source. How much does that change the sensitivity?

Nowhere do you actually define the orientation polarity of the gauges used to form the bridge. You must verify that the assumed arrangement is actually being used in practice. If one gauge is rotated, half the signal will be lost, and the experiment will be wasted.

Your argumentative nature will keep this going, so long as you can find people silly enough to argue against the results of a practical experiment, or the results of a SPICE simulation. Do you also argue that the world is flat?

 

[Tom.G]

Hi @MarsBravo and
..

Probably the easiest way to settle your problem is to visit the website of the strain gauge manufacturer and look for some application notes that explain the usage details with nice drawings.

Cheers,
Tom

 

[Baluncore]

Here is an LTspice model of the 350Ω bridge with lead resistance of 4Ω.
Resistance change is ±1%.

The model shows that it does not matter how you conceptually model the currents in the wires.

For the same resistive change of the bridge;
Without lead resistance, the output is 9.95000 mV;
With lead resistance, the output is 9.72767 mV;
The ratio is 9.72767 / 9.95000 = 0.977655
1 – 0.977655 = 0.022345 = 2.2345%

That agrees well with my original calculation in post #2;
1 - (350 / ( 350 + 4 + 4 ) ) = 0.022346 = 2.2346%

 

[Dullard]

Post #2 answers your question. Post #8 drives a stake through it's heart.

I would (humbly) add:

Because the instrumentation amplifier is 'high impedance,' the resistance of those wires may be ignored.

People who actually do this (strain gauge instrumentation) tend to use 'shunt cals' to characterize the error due to cable resistance. If you read up on the details of that, the answer provided here may make more sense to you.