How does 3-wire RTD or resistance measurement works in bridge of wheatstone?

Question

I cannot understand how 3-wire measurement will improve the resistance measurement.

They say that IF lead resistance \$ R_L \$ is the same, it will compensate. How can this happen?

If bridge is balanced (without \$ R_L \$): \$ \frac {R_1}{R_2} = \frac {R_g}{R_3} \$

If \$ R_L \$ is added: \$ \frac {R1}{R2}= \frac {R_g+R_L}{R_3+R_L} \$

\$ \frac {R_g}{R_3} \$ is not same as \$ \frac {R_g+R_L}{R_3+R_L}\$; is this true only if \$ R_G=R_3 \$ in the beginning? So this means that \$R_L\$ does not get compensated?

Answer 1

Start from the voltage divider equation (since it is Vo you are balancing), and see where you get to...

If bridge is balanced (without \$ R_L \$): \$ \frac {R_2}{R_1+R_2} = \frac {R_3}{R_g+R_3} \$

In the specific case that R3=RG (and R1=R2), then RL cancels out.

A Wheatstone bridge does not work by measuring Vo. It works by adjusting the known components, to balance the bridge (Vo=0).

In this case we will adjust R3 (whilst keeping R1=R2 constant), until Vo=0. The bridge will be balanced. Rg will = R3, and we know R3, as it is calibrated. For all values of RL we can measure Rg.

It is a common misuse to call a differential arrangement where we measure the differential voltage and use that to calculate the unknown a "bridge", "Wheatstone" or other. The true bridge arrangements have identities that are true at balance, and not true away from balance. Cancellation was able to eliminate computation, and often electrical precision.

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But even if you do use the Vo measurement, this arrangement still works. If you had a two wire sensor, the RL causes a direct and immediate offset error. With the 3 wire arrangement, there is 0 error at balance, and the eror term grows ratiometrically with the change in Rg. For small changes in Rg (strain gauges, small temperature range RTD), this error term will be small.

A modern measurement system would also measure VoR3, and thus could calculate and remove RL, by brute arithmetic.

Answer 2

\$R_g + R_L = R_3 + R_L\$ and so the balance does not change.

But the "gain" due to the introduction of \$R_L\$ will change. Life isn't always perfect and quarter-bridge sensing has other problems that mask this less-serious problem. The point is that the offset of the bridge will not change and that a small amount of increased resistance in the cable may not always be a big problem in many systems.

Answer 3

Yes, they are assuming the resistances R=R1=R2=R3~=Rg are all nominally equal. There will be an error term that increases as Rg varies from R.

Nobody in modern times does 3-wire RTD circuits like that.

Answer 4

I think you are missing the point. You only get a voltage drop across a resistor when current FLOWS. So... the middle leg, which 'feels' the change in temperature at the Platinum Resistor, terminates at the + ...there is no current flowing in this leg.... there never was, even when it was a part of the two wire... this was not a cause of the (two wire) error (there is not current and hence no voltage drop regardless of resistance). It was the other lead and ITS resistance. The addition of the the third wire, counteracts the error caused by this other lead.... the resistances are the same and balance the bridge.

V=IR.... Voltage drop across a resistor (the lead wire going to centre of bridge to measure voltage) by function of their resistance is a function of current flowing: If no current flows... Voltage drop= ZERO times R = ZERO; 0=0*R.

Imagine, if you will, a hosepipe: If water flows, then pressure at the end is a function of resistance of the pipe. YET; put your finger over the end and stop the water coming out. You will soon feel full line pressure (Voltage). No current flowing means no voltage drop.