# Why is 2-terminal sensing less accurate than 4-terminal sensing if the internal resistance of leads is known?

I have a pretty good understanding of 2-terminal vs. 4-terminal (kelvin) sensing measurements. I understand that in the 4-terminal measurement, current is supplied by 2 terminals that are separate from the voltage sensing terminals. This eliminates the need for (an appreciable amount) of current to run through the leads of the voltmeter, which would cause a slight voltage drop due to the internal resistance of the leads.

My question is - can't that internal resistance be measured quite easily? If it is known, how is 2-terminal sensing any less accurate than 4-terminal sensing if the internal resistance fact is used to compensate?

can't that internal resistance be measured quite easily?

No, it cannot always be measured quite easily as it can be very small. Now, if it's so small, why do we care? Well, we normally use a 4-terminal measurement precisely when the resistance we're trying to measure is itself rather small. If we have some uncertainty in the value of the probe resistance, than the subtraction you suggest would result in a large uncertainty in the extracted resistance value. To be specific, consider a case where we're trying to measure a device with resistance $R$ using a probe with resistance $R_\text{probe}$. We put a current $I$ through the series combination of the device and probe and measure the voltage

$$V = I (R + R_\text{probe}) \, .$$

We then compute the resistance of the device as

$$R = (V/I) - R_\text{probe} \, .$$

If there's an uncertainty $\delta R_\text{probe}$ in the resistance of the probe, then the uncertainty in $R$ is also $\delta R_\text{probe}$. Therefore, the two-terminal measurement is bad in the case that the uncertainty in the probe resistance is an appreciable fraction of the resistance you're trying to measure. In those cases, it's better to use a procedure that eliminates this uncertainty entirely.

There is a very common case where the uncertainty in the probe resistance is huge: contact resistance. Suppose I'm trying to measure the resistance of some structure on a chip by touching that structure with probes connected to the readout electronics. The contact resistance between the probes and the metal is not reproducible from trial to trial, so there's a big $\delta R_\text{probe}$.

• In addition, any variation in the probe impedance with current does not affect the Kelvin pair. – Carl Witthoft Mar 8 '16 at 14:21