Appendix d: supply voltage requirements, Dual conductor supply – Laser beta lasermike LS8000-3 User Manual
Page 202
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LaserSpeed 8000-3 Instruction Handbook
Appendix D: Supply Voltage Requirements
Part No. 93463 / Drawing No. 0921-01561
Page 202 of 221
Revision A (Sep 2007)
Appendix D: Supply Voltage Requirements
The LS8000-3 gauge requires a supply voltage of 20 to 28 Volts DC at the
gauge. Because the gauge can draw over 2A of current, Voltage drop across
long cable distances may be a problem. The LS8000-3 normally only
demands this high current just after power on, or when the ambient
temperature is near the upper limit. This is when the gauge is cooling the
internal optical components with the most power. If you are using cables not
purchased from Beta LaserMike, please review this section before designing
the cable interface to the gauge.
The LS8000-3 gauge has two power input pins (pins 24 and 25) and two
power ground pins (pins 12 and 13). Cabling wires to all four pins (Quad
Conductor Supply) will allow you to use smaller wire diameters while avoiding
voltage drop problems. Cabling wires to only two pins (one power, one
ground—called Dual Conductor Supply) requires larger wire diameters. This
manual section will discuss how to size your supply wires correctly to eliminate
voltage drop problems.
Dual Conductor Supply
If you are using two conductors (one supply wire, one return wire), the
following table will help you determine your total voltage drop across the cable.
Wire Size
(AWG)
Wire Size
(mm
2
)
Voltage Drop per Foot
(@2.0A)
Voltage Drop per Meter
(@2.0A)
26 0.129
0.16
0.54
24 0.205
0.10
0.34
22 0.326
0.065
0.21
20 0.518
0.041
0.13
18 0.823
0.026
0.083
16 1.31
0.016
0.053
14 2.08
0.010
0.033
12 3.31
0.0064
0.021
Voltage Drop using Single Supply/Single Return
Example: Your power supply is 200 feet (61 meters) away from the LS8000-3.
You are using a single 20 AWG wire to supply power to pin 24, and a single 20
AWG wire as a ground (connected to pin 12). Your voltage drop in the cable
will be:
Voltage Drop = #Feet * Voltage Drop/Foot = 200 * 0.041 = 8.2 volts