# Cabling and Current-Resistance Loss Considerations for Programmable DC Power Supplies and Precision DC Sources

## Aperçu

This tutorial is part of the National Instruments Measurement Fundamentals series. Each tutorial in this series, will teach you a specific topic of common measurement applications, by explaining the theory and giving practical examples. This tutorial contains information you may find useful as you connect specific types of loads to a power supply.

For additional power supply only concepts, refer to the Power Supply Fundamentals main page.
For the complete list of tutorials, return to the NI Measurement Fundamentals Main page.

## Current-Resistance Loss

Current-resistance loss is introduced by the cabling wires that connect the power supply to the load terminals. The amount of current-resistance loss is determined by the resistance of the cabling wire (a property of the wire gauge and length) and the amount of current flowing through the wire. Current-resistance loss results in a voltage drop between the power supply and the load.
To minimize voltage drop caused by cabling, keep each wire pair as short as possible and use the thickest wire gauge appropriate for your application. The lower the American Wire Guage (AWG) rating, the thicker the wire. NI recommends 18 AWG or lower.

To reduce noise, twist each wire pair. Refer to the table below to determine the wire gauge appropriate for your application.

Caution: Use wire that is thick enough to avoid overheating if the output current from the power supply were to short circuit.

 AWG Rating mΩ/m (mΩ/ft) 10 3.3 (1.0) 12 5.2 (1.6) 14 8.3 (2.5) 16 13.2 (4.0) 18 21.0 (6.4) 20 33.5 (10.2) 22 52.8 (16.1) 24 84.3 (25.7) 26 133.9 (40.8) 28 212.9 (64.9)

Table 1. Wire gauge comparison

## Calculating Maximum Voltage Drop

When cabling a power supply to a constant load, be sure to account for voltage drop in your application, and if necessary, adjust the output voltage of the power supply. Use the amount of current flowing through the cabling wires and the resistance of the wires to calculate the total voltage drop for each load, as shown in the following example.

Example: Operating within the recommended current rating, determine the maximum voltage drop across a 1 m, 16 AWG wire carrying 1 A:

V = I × R

V = 1 A × (13.2 mΩ/m × 1 m)

V = 13.2 mV

As illustrated in the preceding example, a 1 m, 16 AWG wire carrying 1 A results in a voltage drop of 13.2 mV.

Note: When calculating voltage drop for a pair of wires, multiply the voltage drop by two. Thus, the total voltage drop for a pair of wires is 26.4 mV. To compensate for the voltage drop across the wire pair and ensure the correct power is supplied to the load, increase the output voltage of the power supply by 26.4 mV, or, if available, use 4-wire remote sensing.

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For the complete list of tutorials, return to the NI Measurement Fundamentals Main page