For a control or instrument device to function properly, it’s important that cables contain the appropriate conductor size. The standard ampacity (ampacity is a measure of the current required to raise the conductor to the rated temperature of the insulation) tables found in electrical handbooks were created for 60 cycle power cables. The standard ampacity tables are fine for power circuits, but they’re not valid for most control and instrumentation cables.
To calculate the conductor size you will need to know the maximum voltage drop that is acceptable for your application.
You must know the following:
•Acceptable voltage drop
•Current the wire will carry
•Length of the wire
Using Ohm’s law where: V=IR
I =Current (Amps)
The maximum allowable voltage drop Vdrop = Vsource - V minimum load
= 84V – 80V = 4V
The current required is 3.7A
The total length of wire in this case is the distance from the source to the load and back again. In this case it’s 25 ft twice giving a total of 50 ft. Given a current of 3.7A the maximum resistance allowable:
V / I = R 4 / 3.7 = 1.08 Ohms
Find a conductor that will have a resistance equal to or lower than 1.08 Ohms per 50 ft. Wire tables always give the conductor resistance in Ohms per 1000 feet. We need to convert the 1.08 Ohms per 50 ft to the value of Ohms per 1000 feet.
1.08 Ohms/50 ft where 1.08/50 = .0216 Ohms per ft
.0216 Ohms/ft x 1,000 ft = 21.6 Ohms / 1,000 ft
From the Wire Table, select a conductor with a resistance lower than 21.6 Ohms/ 1,000 FT.
|AWG||Stranding||DC Resistance(Ohms/M Ft@ 20?C)|
Note: M FT = 1,000 FT
Looking at the conductor tables we find that 22 AWG or larger will suffice. The stranding will be determined by the application. For some, it’s easier to understand the concepts of voltage, current, and resistance in term of hydraulics.
Voltage, Current and Resistance can be thought of in terms of fluids where:
Voltage (V) = Pressure
Current (I) = Flow Rate
Resistance (R) = Constriction
We have a device to be attached at the end of a cable. It requires at 4.6 to 5.2 VDC supply at 120 mA. The source supplies 5.0V exactly. We need 65 ft of cable between the load and the source. What gage of wire should we be using?
- Acceptable voltage drop across the cable is 5.0V – 4.6V = 0.4V
- Current 0.120A
- Wire length is 65 feet times 2 (for both ways)
- R=V/I=0.4/.120= 3.3 Ohms
- Total Resistance to device and back is 3.3 Ohms per 2 x 65′, 3.3 Ohms/130 ft
- 3.3 Ohms/130ft= .02564 Ohms/ft This is 25.64 Ohms/1,000′
- From Wire Table find the conductor with a resistance of 25.64 or less.
- Use 24AWG or Larger.
For assistance with custom wire & cable design, contact a design expert at
Calmont Wire & Cable, Inc.
420 East Alton Ave.
Santa Ana, CA 92707
Phone: 800-905-7161 ext. 135