The nature of the Kelvin measurement is that you have a separate current path and measurement path so that no current (except leakage and bias currents) flow through the measurement conductors.

Thus, the PCB layout (unless the **design** itself is faulty) is remarkably easy, and I don't expect you'll have any troubles, since series resistance hardly matters.

^{simulate this circuit – Schematic created using CircuitLab}

In the above schematic, the exact value of the resistors R1, R2, R3, and R4 hardly matter, provided they are reasonably low. R4 affects the common-mode voltage the instrumentation amp sees, R1 and R2 affect errors due to input offset current (and noise) a bit, but really the PCB layout is not very important until currents and voltage drops start to become significant wrt the common mode range of U1. So you want to keep the voltage drop across R4 reasonably low (not a problem typically unless you use very thin traces and/or very high currents).

Regard it as a transmission line problem and from that, we know that the characteristic impedance is: -

\$Z_0 = \sqrt{\dfrac{R + j\omega L}{G + j\omega C}}\$

It's not too tricky to prove this - see my answer **here**

So for DC this reduces to \$Z_0 = \sqrt{\dfrac{R}{G}}\$

R is series resistance per unit length and G is parallel conductivity per unit length. For example if R is 1 ohm per metre and G is 1 micro siemen per metre then the characteristic impedance is 1000 ohms.

Looks like I'll have to do the proof. Imagine one section of the line having R series ohms and 1/G parallel Mohms. If this section is repeated to infinity the impedance looking into the 1st section is the same impedance as if the 1st section is discarded and you looked into the 2nd section.

From this simple and straightforward observation you can say: -

\$Z_{IN} = R + \dfrac{1}{G} || Z_{IN}\$. In other words this: -

^{simulate this circuit – Schematic created using CircuitLab}

So, \$Z_{IN} = R + \dfrac{\frac{Z_{IN}}{G}}{Z_{IN} + \frac{1}{G}} \$

\$Z_{IN} = R + \dfrac{Z_{IN}}{1+Z_{IN}\cdot G}\$

\$Z_{IN} + Z_{IN}^2\cdot G = R + Z_{IN}\cdot G\cdot R + Z_{IN} = R(1 + Z_{IN}\cdot G) + Z_{IN}\$

As the sections of cable are made infinitely small, \$Z_{IN}\cdot G\$ becomes an insignificant term (right hand side of the equation) hence we are left with: -

\$Z_{IN}^2\cdot G = R\$ or \$Z_{IN} = \sqrt{\dfrac{R}{G}}\$

## Best Answer

Yes, it should be possible. No, it doesn't sound like a good idea.

Regular residential air conditioners are designed with some assumptions, including a lower limit on the temperature to cool too. The unit might work very inefficiently, or not at all, at a temperature much below what it is intended for. Icing will also quite likely be a problem, unless maybe you are in a very dry environment.

It sounds like you really want a refridgeration unit, not a air conditioner.