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

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?

• What's with the '------------------------------------>'? – Transistor May 10 '18 at 20:35
• Something wrong is happened – Kono May 10 '18 at 20:36

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.

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.

• That's the same as: R1/R2=Rg/R3. I have written all equatations 100 times :( – Kono May 10 '18 at 20:57
• I think your mistake is thinking that a Wheatstone bridge measures Rg by measuring Vo. It doesn't. It adjust the other bridge R to balance the bridge. – Henry Crun May 11 '18 at 1:30
• I know. Thats a way. But almost in every case they measure vo to find unknown resistance. Including literature and wikipedia. – Kono May 11 '18 at 8:12
• No. Thats THE way. What you are trying to do is just an offset voltage measurement. Nothing in common with a bridge technique. "They" you refer to, being people who complain that RL is not compensated, and it doesn't work properly when I use it wrongly, and can you please dig up Charles Wheatstone so I can give him a proper talking to. – Henry Crun May 11 '18 at 8:19
• Well if you have it then it's definitely the way. Thanks for asking the question. I had no idea about the RL cancelling - a useful trick to know. – Henry Crun May 11 '18 at 9:27

$R_g + R_L = R_3 + R_L$ 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.

• Who says that Rg and R3 is the same? – Kono May 11 '18 at 15:53
• If they are not the same but the bridge is balanced initially due to R1 and R2 being proportionately the same as Rg and R3 then RL does not factor into the equations for balance. To balance the bridge R1/R2 does not need to equal (Rg + RL)/(R3 + RL) - is that what concerns you? – Andy aka May 11 '18 at 15:56
• If i balance the bridge with Rg and R3 proportional to R1 and R2; but not identical to each other. The bridge balances; and adding RL to Rg and R3 disrupts the bridge; the bridge will unbalance. So RL does not get compensated – Kono May 11 '18 at 16:29
• R1/R2 needs to be same (Rg + RL)/(R3 + RL) to balance yje bridge – Kono May 11 '18 at 16:32
• OK I see what you mean. I get where you are coming from. Yes, if the ratio of R1 to R2 is not 1:1 then a three wire measurement will not work because RLa and RLb would need to be different. – Andy aka May 11 '18 at 16:32

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.

• So Rg/R3 is not same as Rg+RL/R3+RL? (only if all resistances are the same). Why is this circuit used? – Kono May 10 '18 at 21:17
• As I said, it's not much used. If you pick numbers like 0-100 degrees C (balanced at 0'C) and 1 ohm line resistance it reduces the error by more than 10:1 so it's better than nothing. – Spehro Pefhany May 10 '18 at 21:23
• Does this principle also count for this issue with strain gauges? – Kono May 10 '18 at 21:27
• electronics.stackexchange.com/questions/373627/… – Kono May 10 '18 at 21:27
• This works much better for strain gauges, where full scale might only be 1%. However, unfortunately, a single ended strain gauge would be a bit of a nightmare, as it is still a nickel RTD, with perhaps 0.4%/degC tempco i.e. it would be a better temperature sensor than strain sensor. – Henry Crun May 11 '18 at 4:52

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.

• Welcome to EE.SE. Proper sentence structure is easier to read and less likely to be misinterpreted than ... stream of consciousness-style ... punctuation. Note that judging by the OP's comments to Andy aka over seven years ago that s/he was well aware of that no current was flowing in the third wire. The point requiring clarification was that R1:R2 must be 1:1 for this compensation scheme to work. – Transistor Nov 21 '18 at 21:17