Lead resistance of 3 wire RTD100

I am using a three wire RTD100 as a temperature sensor.

I know the advantage of the 3 wire RTD100 is that it gives accurate temperature output by canceling the lead resistance. First the resistance between line RED (R1) and BLACK (R2) is measured the total resistance is R1 + R2 + RE and then the resistance between line BLACK (R2) and BLACK(R3) is measured. All three wires have the same resistance i.e R1=R2=R3 therefore we will have to subtract (R2+R3) from (R1 + R2 + RE) to get an accurate temperature value.

I researched about the circuits that I can use with 3 wire RTD100. A Wheatstone bridge is used to find the resistance change in the RTD100.

Circuit 1:

In the above circuit, the wire resistance of the RTD100 is labelled as RW1, RW2, RW3. I understand that the change in resistance will unbalance the Wheatstone bridge. The output nodes of the Wheatstone bridge are given to the difference amplifier where we get the voltage output corresponding to the change in temperature, but I am not understanding how the Wheatstone bridge and the difference amplifier is canceling the lead resistance.

Important Note- the above circuit is only applicable when all the resistances are nominally at the same value i.e. the same magnitude balance current flows into both R1 and R2. I'm mentioning this because if the left half of the bridge has a different impedance to the right half, 3-wire connections to an external device don't produce wire-resistance cancellation correctly. In other words, at balance, R3 must equal R4.

So, I've added a label in red (inside a red circle) called "C". This is the "new lower power feed point" and you can now ignore the old node in black called "C": -

Can you see that R3 is in series with RW2 and equally, R4 is in series with RW1 and hence the balance is maintained? No? Read below: -

Consider that in balance, R3 equals R4 and R1 equals R2. This produces 0 volts output between "a" and "b". If R3 and R4 both increased by 10%, the bridge would still be in balance. And increasing both by 10% is the same as adding series resistance to them both. That series resistance comes from the interconnecting wires (RW1 and RW2) and, it can be presumed to be identical for both. So the bridge remains in balance no matter what value RW1 and RW2 is providing that they are both the same value.

RW3 is a red-herring because it doesn't imbalance anything when its value changes; it just affects the "gain" of the circuit.

• but how lead resistance is cancelled ?@Andy aka – power machines Sep 17 '20 at 9:21
• I've added some more information. – Andy aka Sep 17 '20 at 9:27
• Okay i will think about it, sorry for confusion – power machines Sep 17 '20 at 10:30
• @Andyaka i realized my mistake in interpretation of circuit ,thanks for help – power machines Sep 17 '20 at 10:44
• @Andyaka is there any thing i can do to minimize the effects of RW3 on gain – power machines Sep 17 '20 at 10:55

Wheatstone won't help here.

Two matched current sources, then $$\ V_{dif}=I_1\cdot R_W + V_{rtd} - I_2\cdot R_W \$$

I1=I2, Rw equal then

Vdif=Vrtd

A reference voltage for ADC is then Vref=(I1+I2)/Rref;

simulate this circuit – Schematic created using CircuitLab

• thanks for help @MarkoBuršič – power machines Sep 17 '20 at 11:49