Minimizing Voltage Drop Loss when Cabling
- Updated2023-09-29
- 2 minute(s) read
Minimizing Voltage Drop Loss when Cabling
Voltage drop loss is introduced by the cabling wires that connect the DUT to the electronic load terminals.
The voltage drop due to 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. Instruments with remote sense capabilities can compensate for voltage drop by measuring the voltage across the load terminals with a second set of leads that do not carry a significant current.
To minimize voltage drop caused by cabling:
- Keep each wire pair as short as possible.
- Use the thickest wire gauge appropriate for your application. NI recommends 18 AWG or lower.
To reduce noise picked up by the cables that connect the instrument to a load, twist each wire pair. Refer to the following table to determine the wire gauge appropriate for your application.
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) |
Calculating Voltage Drop
When cabling a power supply or SMU to a constant load, be sure to account for voltage drop in your application. If necessary, adjust the input voltage of the device or, if available, use remote sensing.
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:
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.