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When selecting wires or cables for a wiring harness, one of the first things to evaluate is their ability to handle the required current and voltage. Ensuring your wires and cables meet these requirements is critical for preventing hazards such as overheating or insulation breakdown.
The voltage rating of a wire specifies the maximum voltage its insulation can withstand before breaking down. At high enough voltages, arcing can occur even if the insulation remains intact, posing significant safety risks.
The current-carrying capability of a conductor, known as its ampacity, defines the maximum amount of electrical current it can safely handle without exceeding its temperature rating.
To select an appropriate wire, engineers often refer to standards like the American Wire Gauge (AWG), widely used in the U.S., or the IEC 60228, a metric-based standard used internationally. Both provide guidelines for wire sizes and their corresponding ampacity and voltage ratings. Understanding these standards helps you select the appropriate wires for your application.
Wire Ampacity: Guidelines for Current-Carrying Capacity
Table 1 provides general guidelines for the ampacity of copper wires. Actual ampacity depends on factors such as insulation properties (material type, thickness, and temperature rating) and environmental conditions (ambient temperature, airflow, and whether the wires are installed in open air or bundled together).
The Maximum Amps for Power Transmission (in table: Power) are based on conservative guidelines to minimize heat buildup and voltage drop, making them ideal for long cable runs. The Maximum Amps for Chassis Wiring (in table: Chassis) are also conservatively rated but are intended for wires installed in open air to allow better heat dissipation. For short wire lengths, such as those used in battery packs, wire selection often requires balancing resistance, current load, size, weight, and flexibility.
| AWG | Diameter (in) | Diameter (mm) | Resistance (ohms/1000ft) | Resistance (Ohs/ 1km) | Ampacity Chassis (A) | Ampacity Power (A) |
|---|---|---|---|---|---|---|
| 10 | 0.102 | 2.59 | 1.00 | 3.28 | 55.0 | 15.0 |
| 12 | 0.0808 | 2.05 | 1.59 | 5.21 | 41.0 | 9.30 |
| 14 | 0.0641 | 1.63 | 2.53 | 8.28 | 32.0 | 5.90 |
| 16 | 0.508 | 1.29 | 4.02 | 13.2 | 22.0 | 3.70 |
| 18 | 0.0403 | 1.02 | 6.39 | 20.9 | 16.0 | 2.30 |
| 20 | 0.0320 | 0.813 | 10.2 | 33.3 | 11.0 | 1.50 |
| 22 | 0.0254 | 0.645 | 16.1 | 52.9 | 7.00 | 0.920 |
| 24 | 0.0201 | 0.511 | 25.7 | 84.2 | 3.50 | 0.580 |
| 26 | 0.0159 | 0.404 | 40.8 | 134 | 2.20 | 0.360 |
| 28 | 0.0126 | 0.320 | 64.9 | 213 | 1.40 | 0.230 |
| 30 | 0.0100 | 0.254 | 103 | 339 | 0.860 | 0.140 |
| 32 | 0.00795 | 0.200 | 164 | 538 | 0.530 | 0.910 |
Table 1: American Wire Gauge Conductor Ratings Table for Copper Wire
Voltage Drop and Its Impact
Another critical factor is voltage drop—the reduction in voltage that occurs along the length of a wire due to its resistance. Over long distances, voltage drop can significantly affect the performance of sensitive electronics, leading to issues like brownouts or equipment malfunctions.
To calculate the voltage drop:
Identify your system’s maximum current draw and operating voltage.
Use the wire’s resistance per unit length to determine the total voltage drop over its length.
Voltage Drop Example: a Robotics Wiring Solution
Scenario:
An industrial robot arm operates on **48V DC **and draws 5.9A through a 14 AWG copper cable that's part of a larger wiring harness running 60 feet. The system requires the voltage drop to remain within 3% of the supply voltage. Can this wire meet the requirement?
Solution:
V_d = (2 × L × I × R) / (1000 ft)
Essentially this formula is Ohm's (V=IR) law modified for the length of the signal path.
Where:
2 accounts for the total length (supply + return wires).
L is the length of the wire in feet
I is the current in Amps
R is the resistance in Ohms/1000ft
1000ft normalizes resistance "per 1000 feet" to the actual wire length
Resistance of 14 AWG Copper: From the table, R=2.525 Ohms/1000 ft
Cable Length: L=60 ft
Current: I=5.9 A
Substitute into the formula:
V_d = (2 × 60ft × 5.9A × 2.525 Ohms/1000ft) / (1000ft) = (1784.7V) / (1000) = 1.78V
Conclusion:
The calculated voltage drop is 1.78V, which is well above the allowable limit of 1.44V (3% of 48V). The 14 AWG copper cable is suitable for this application.
**Solution: **Move to a lower gauge wire with less per unit resistance or shorten lead length to resolve the issue. (Question: Would 12 AWG wire solve this problem?)
Key Insights
**Importance of Correct Wire Specifications: **Using improperly specified wires can lead to overheating, excessive voltage drops, and potential system failure. Always ensure the wire gauge and material match your system’s requirements.
**How to Calculate Voltage Drop: **Use the modified Ohm’s law formula to determine the voltage drop. For system reliability, always consider the total signal path length (supply and return wires).
Choosing Wire Gauges:
For U.S. systems: Use the American Wire Gauge (AWG) chart.
For international systems: Refer to the IEC 60228 metric wire size standard.
Sources used and further reading:
American Wire Gauge (AWG) & Metric Gauge Wire Sizes, by Steven Errede, University of Illinois at Urbana-Champaign, 2015.
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