# Examples of Calculating Accuracy

Version:

Note

Specifications listed in examples are for demonstration purposes only and do not necessarily reflect specifications for this device.

## Example 1: Calculating 5 °C Accuracy

Calculate the accuracy of 900 nA output in the 1 µA range under the following conditions:

 ambient temperature 28 °C internal device temperature within Tcal ± 5 °C [] self-calibration within the last 24 hours.

Solution

Since the device internal temperature is within Tcal ± 5 °C and the ambient temperature is within 23 °C ± 5 °C, the appropriate accuracy specification is:

0.03% + 100 pA

Calculate the accuracy using the following equation:

$\mathrm{Accuracy}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}900\text{\hspace{0.17em}}\mathrm{nA}*\text{\hspace{0.17em}}0.03%\text{\hspace{0.17em}}+\text{\hspace{0.17em}}100\mathrm{pA}$ $\text{\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}}270\mathrm{pA}+\text{\hspace{0.17em}}100\mathrm{pA}$

$=\text{\hspace{0.17em}}370\mathrm{pA}$

Therefore, the actual output will be within 370 pA of 900 nA.

## Example 2: Calculating 1 °C Accuracy

Calculate the accuracy of 900 nA output in the 1 µA range. Assume the same conditions as in Example 1, with the following differences:

 internal device temperature within Tcal ± 1 °C[]

Solution

Since the device internal temperature is within Tcal ± 1 °C and the ambient temperature is within 23 °C ± 5 °C, the appropriate accuracy specification is:

0.022% + 40 pA

Calculate the accuracy using the following equation:

$\mathrm{Accuracy}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}900\text{\hspace{0.17em}}\mathrm{nA}*\text{\hspace{0.17em}}0.022%\text{\hspace{0.17em}}+\text{\hspace{0.17em}}40\mathrm{pA}$

$=\text{\hspace{0.17em}}238\mathrm{pA}$

Therefore, the actual output will be within 238 pA of 900 nA.

## Example 3: Calculating Remote Sense Accuracy

Calculate the remote sense accuracy of 500 mV output in the 600 mV range. Assume the same conditions as in Example 2, with the following differences:

 HI path lead drop 3 V HI sense lead resistance 2 Ω LO path lead drop 2.5 V LO sense lead resistance 1.5 Ω

Solution

Since the device internal temperature is within Tcal ± 1 °C and the ambient temperature is within 23 °C ± 5 °C, the appropriate accuracy specification is:

0.016% + 30 μV

Since the device is using remote sense, use the remote sense accuracy specification:

Add (3 ppm of voltage range + 11 µV) per volt of HI lead drop plus 1 µV per volt of lead drop per Ω of corresponding sense lead resistance to voltage accuracy specifications.

Calculate the remote sense accuracy using the following equation:

$\mathrm{Accuracy}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left(500\text{\hspace{0.17em}}\mathrm{mV}*\text{\hspace{0.17em}}0.016%\text{\hspace{0.17em}}+\text{\hspace{0.17em}}30\mathrm{\mu V}\right)+\text{\hspace{0.17em}}\frac{600\text{\hspace{0.17em}}\mathrm{mV}*\text{\hspace{0.17em}}3\mathrm{ppm}\text{\hspace{0.17em}}+11\mathrm{\mu V}}{1V\mathrm{of}\text{\hspace{0.17em}}\mathrm{lead}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{drop}}*\text{\hspace{0.17em}}3V+\text{\hspace{0.17em}}\frac{1\mu V}{V*\text{\hspace{0.17em}}\mathrm{\Omega }}*\text{\hspace{0.17em}}3V\text{\hspace{0.17em}}*\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\mathrm{\Omega }+\text{\hspace{0.17em}}\frac{1\mu V}{V*\text{\hspace{0.17em}}\mathrm{\Omega }}\text{\hspace{0.17em}}\text{\hspace{0.17em}}*\text{\hspace{0.17em}}2.5V*\text{\hspace{0.17em}}1.5\mathrm{\Omega }\text{\hspace{0.17em}}$

$\text{\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}}80\mathrm{\mu V}+\text{\hspace{0.17em}}30\mathrm{\mu V}+\text{\hspace{0.17em}}12.8\mathrm{\mu V}*3+\text{\hspace{0.17em}}6\mathrm{\mu V}+\text{\hspace{0.17em}}3.8\mathrm{\mu V}$ $=\text{\hspace{0.17em}}158.2\mathrm{\mu V}$

Therefore, the actual output will be within 158.2 µV of 500 mV.

## Example 4: Calculating Accuracy with Temperature Coefficient

Calculate the accuracy of 900 nA output in the 1 µA range. Assume the same conditions as in Example 2, with the following differences:

 ambient temperature 15 °C

Solution

Since the device internal temperature is within Tcal ± 1 °C, the appropriate accuracy specification is:

0.022% + 40 pA

Since the ambient temperature falls outside of 23 °C ± 5 °C, use the following temperature coefficient per degree Celsius outside the 23 °C ± 5 °C range:

0.0006% + 4 pA

Calculate the accuracy using the following equation:

$\mathrm{Temperature}\mathrm{Variation}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left(23°C-\text{\hspace{0.17em}}5\text{\hspace{0.17em}}\text{\hspace{0.17em}}°C\right)\text{\hspace{0.17em}}-15°C=\text{\hspace{0.17em}}3°C\text{\hspace{0.17em}}$

$\mathrm{Accuracy}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left(900\text{\hspace{0.17em}}\mathrm{nA}\text{\hspace{0.17em}}*\text{\hspace{0.17em}}0.022%\text{\hspace{0.17em}}+\text{\hspace{0.17em}}40\mathrm{pA}\right)+\text{\hspace{0.17em}}\frac{900\text{\hspace{0.17em}}\mathrm{nA}*\text{\hspace{0.17em}}0.0006%\text{\hspace{0.17em}}+4pA}{1°C\text{\hspace{0.17em}}}*\text{\hspace{0.17em}}3°C$

$\text{\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}}238\mathrm{pA}+\text{\hspace{0.17em}}28.2\mathrm{pA}$

$=\text{\hspace{0.17em}}266.2\mathrm{pA}$

Therefore, the actual output will be within 266.2 pA of 900 nA.

• 1 Tcal is the internal device temperature recorded by the PXIe-4139 at the completion of the last self-calibration.