When using the 8508A in 2-wire resistance mode, finding the total uncertainty can be a complex task.

To start, we need to consider the base specification. I am using Normal Mode and I am making the assumption that I am not going outside of ±1 °C from the temperature at which the 8508A was last zeroed. I am using the 1 Year specification and using the absolutely uncertainties to include the effect of the last calibration.

Next I'll review the applicable footnotes for this table:

[1] Specifications apply for max resolution in each function, normal mode

[2] Assumes 4 hour warm-up period

[3] Input zero or offset null required whenever the temperature moves more than ±1 °C from the temperature at which the previous null/zero was performed.

[10] Tru Ohms mode available on 2 Ω to 20 kΩ ranges. Read Rate reduced in Tru Ohms Mode. Specification for Tru Ohms same as corresponding Normal or Lo Current range

[15] The maximum display value for the Analog to Digital converter is 199 990 000 counts. This sets the maximum value measurable on each range to be a one followed by four nines. For example, the maximum measured values on the 2 V range on DC Voltage are ±1.999 900 00 V. However, the 1000 V ranges are limited to a maximum 1050 V.

None of these footnotes apply to my example, so no additional uncertainties need to be added from them. Since I have already stated that my temperature has not varied more than 1 °C, footnote 3 does not require further action.

Next I will look at the secondary specifications for Normal mode.

I am not doing transfer mode or adding anything for temp in this example so I can disregard it. I *do *need to make note of my measurement current because that contributes to my 2-wire adder formula.

Footnotes for the secondary specs are the same as above and like above I can ignore it all together.

Thus far I have 9% of reading plus 0.7 parts per million (ppm) of range.

Now we can look at the 2-Wire Adder formula.

I can ignore the ±3 mΩ/°C as I've stipulated that my temperature is still within 1 °C.

My Ir is 10 mA (taken from the secondary specifications table).

Here is my formula for the adder:

±(10 pA/10 mA) * 1E6 ppm of reading ± 50 mΩ

We can calculate this to be:

((10E-12 A / 10E-3 A) * 1E6 ppm) * 10 Ω + 50 mΩ, or

(0.01 ppm) * 10 Ω + 0.05 Ω = 50 mΩ

In the case of this example, the 10 pA/10 mA * 10E6 PPM of 10 Ω becomes such a small number (10E-9 Ω) that it is several orders of magnitude smaller than the fixed contribution. So assuming a reading of 10 Ω, the only additional adder you need is the 50 mΩ.

My initial spec was:

9% of reading plus 0.7 ppm of range

My 2-wire adder is 50 mΩ.

So my total specification will be:

(9% * 10 Ω) + (0.7E-6 * 20 Ω) + 0.05 Ω = (0.09*10 Ω) + (0.000007 * 20 Ω) + 0.05 Ω, or

±0.950014 Ω.

Douglas KuikHi Krista,

An error in the calculation. 9% should be 9 ppm of 10 Ohms.

Krista E.Thanks Doug. Even my reviewers didn't catch it. ;) I'll get it corrected.