Tech Notes

--------------------------------------------------------------------------------------------------------------------------------

Title: Calculating Readout Accuracies

Created: 11 Jan 2013

Last Revised:

--------------------------------------------------------------------------------------------------------------------------------

Applies To: Various Readouts

Problem Description: How does 8 ppm translate to ⁰C accuracy?

Resolution/Work Around:

To help us determine the accuracy, here is an example of how to calculate the accuracy of a readout device, like the 1594. We will use the specification of the 1594 and the thermistor calibration report to help us.

Example: Measuring a 10 kΩ Thermistor

In this example, a 1594 is used to measure a 10 kΩ thermistor probe at 0 °C. The 1594 accuracy, when measuring the thermistor probe is based on the following:

●● Resistance accuracy of the 1594

●● Measurement noise

Resistance Accuracy

First, calculate the 1594 absolute resistance accuracy at 26839.94 (the resistance of the 10-k Ω thermistor probe at 0 °C) when using the 10 k Ω internal reference resistor. The one-year absolute resistance accuracy is 8 ppm of reading (k = 1). This is converted to temperature by dividing 8 ppm by 1.0 × 10⁶ and multiplying by 26839.94 Ω. The result is then divided by dR/dt of the thermistor probe at 0 °C which, in this example, is 1244.9 Ω/°C. The final result is 0.000172 °C.

8 / 1000000 = 0.000008

0.000008 * 26839.94 = 0.21471952

0.21471952 / 1244.9 = 0.000172

So, 0.000172°C is the accuracy of the 1594 for a thermistor at 0°C.

A few more Examples:

First, calculate the 1595A absolute resistance accuracy at 257 Ω (the resistance of the 100 Ω PRT at 420 °C). The one-year absolute resistance standard uncertainty (k = 1) of the 1594A, using the internal 100 Ω resistor, is 2.0 ppm. To convert this to an uncertainty in temperature, multiply 2.0 ppm by 1.0 × 10⁶ then multiply by 257 Ω. Divide the result by the sensitivity of the PRT (dR/dT) at 420 °C. This can be found in the PRT calibration report. For this example, 0.4 Ω/°C is used. The resulting uncertainty is 0.0013°C.

2 / 1000000 = 0.000002

0.000002 * 257 = 0.000514

0.000514 / 0.4 = 0.001285

So, 0.0013°C is the accuracy of the 1594 for a 100 Ω PRT at 420°C.

The resistance of the SPRT at 157 °C is 41.1 Ω. Using the 25 Ω internal reference resistor, the 1595A one year resistance standard uncertainty is 2.5 ppm. This uncertainty, in terms of temperature, is calculated by first dividing 2.5 ppm by 1.0 × 10⁶ then multiplying by 41.1 Ω. The result is then divided by the sensitivity (dR/dT) of the SPRT at 157 °C which is 0.1 Ω/°C. This results in a standard temperature uncertainty of 0.001028 °C.

2.5 / 1000000 = 0.0000025

0.0000025 * 41.1 = 0.00010275

0.00010275 / 0.1 = 0.0010275

So, 0.001028°C is the accuracy of the 1594 for a 25 Ω SPRT at 157°C.

## Comments

Please sign in to leave a comment.