Welcome: LED pixle string-LED module-LED rigid strip-Point light source. Taiwan Quality. Since 1991
Language: Chinese line  English

News

Volt drop and calculator

LED light color and brightness are best when 5/12/48-volt power feed wires from the power supply to the LED strip,

or array of strips, is delivering as close to 5/12/48 volts as possible.


Voltage drop is a natural occurrence in low-voltage lighting systems. It is the gradual decrease in voltage that

occurs along the length of the 5/12/48-volt power feed wires to the lighting, and varies depending on the type

and size of theLED tape light installation. It is a function of wire length, wire thickness and the energy or total

watts used by the lighting.


Voltage drop only becomes undesirable if you notice the brightness in one area of your lighting is objectionably different

than in another area. With longer lengths of LED tape lighting, voltage drop also occurs along the tape light. As a practical approach, test your lighting prior to final installation.


If voltage drop appears to be a concern, shorten your 5/12/48-volt power feed wires or switch to a heavier gauge wire

(lower AWG number),or shorten the length of your LED tape lighting. You can also consider using an additional power

supply to create a second, separate installation.

-------------------------------------------------------------------------

Voltage drop calculations

DC / single phase calculation

The voltage drop V in volts (V) is equal to the wire current I in amps (A) times 2 times one way wire length L

in feet (ft) times the wire resistance per 1000 feet R in ohms (Ω/kft) divided by 1000:

Vdrop (V) = Iwire (A) × Rwire(Ω)

= Iwire (A) × (2 × L(ft) × Rwire(Ω/kft) / 1000(ft/kft))

The voltage drop V in volts (V) is equal to the wire current I in amps (A) times 2 times one way wire

 length L in meters (m) times the wire resistance per 1000 meters R in ohms (Ω/km) divided by 1000:

Vdrop (V) = Iwire (A) × Rwire(Ω)

= Iwire (A) × (2 × L(m) × Rwire (Ω/km) / 1000(m/km))

3 phase calculation

The line to line voltage drop V in volts (V) is equal to square root of 3 times the wire current I in amps

(A) times one way wire length L in feet (ft) times the wire resistance per 1000 feet R in

ohms (Ω/kft) divided by 1000:

Vdrop (V) = √3 × Iwire (A) × Rwire (Ω)

= 1.732 × Iwire (A) × (L(ft) × Rwire (Ω/kft) / 1000(ft/kft))

The line to line voltage drop V in volts (V) is equal to square root of 3 times the wire current I i

 amps (A) times one way wire length L in meters (m) times the wire resistance per 1000 meters

R in ohms (Ω/km) divided by 1000:

Vdrop (V) = √3 × Iwire (A) × Rwire (Ω)

= 1.732 × Iwire (A) × (L(m) × Rwire (Ω/km) / 1000(m/km))

Wire diameter calculations

The n gauge wire diameter dn in inches (in) is equal to 0.005in times 92 raised to the power

of 36 minus gauge number n, divided by 39:

dn (in) = 0.005 in × 92(36-n)/39

The n gauge wire diameter dn in millimeters (mm) is equal to 0.127mm times 92 raised to the

power of 36 minus gauge number n, divided by 39:

dn (mm) = 0.127 mm × 92(36-n)/39

Wire cross sectional area calculations

The n gauge wire's cross sercional area An in kilo-circular mils (kcmil) is equal to 1000 times

 the square wire diameter d in inches (in):

An (kcmil) = 1000×dn2 = 0.025 in2 × 92(36-n)/19.5

The n gauge wire's cross sercional area An in square inches (in2) is equal to pi divided by

4 times the square wire diameter d in inches (in):

An (in2) = (π/4)×dn2 = 0.000019635 in2 × 92(36-n)/19.5

The n gauge wire's cross sercional area An in square millimeters (mm2) is equal to pi

 divided by 4 times the square wire diameter d in millimeters (mm):

An (mm2) = (π/4)×dn2 = 0.012668 mm2 × 92(36-n)/19.5

Wire resistance calculations

The n gauge wire resistance R in ohms per kilofeet (Ω/kft) is equal to 0.3048×1000000000

times the wire's resistivity ρ in ohm-meters (Ω·m) divided by 25.42 times the cross

sectional area An in square inches (in2):

Rn (Ω/kft) = 0.3048 × 109 × ρ(Ω·m) / (25.42 × An (in2))

The n gauge wire resistance R in ohms per kilometer (Ω/km) is equal to 1000000000

 times the wire's resistivity ρ in ohm-meters (Ω·m) divided by the cross

sectional area An in square millimeters (mm2):

Rn (Ω/km) = 109 × ρ(Ω·m) / An (mm2)

AWG chart

AWG # Diameter
(inch)
Diameter
(mm)
Area
(kcmil)
Area
(mm2)
0000 (4/0) 0.4600 11.6840 211.6000 107.2193
000 (3/0) 0.4096 10.4049 167.8064 85.0288
00 (2/0) 0.3648 9.2658 133.0765 67.4309
0 (1/0) 0.3249 8.2515 105.5345 53.4751
1 0.2893 7.3481 83.6927 42.4077
2 0.2576 6.5437 66.3713 33.6308
3 0.2294 5.8273 52.6348 26.6705
4 0.2043 5.1894 41.7413 21.1506
5 0.1819 4.6213 33.1024 16.7732
6 0.1620 4.1154 26.2514 13.3018
7 0.1443 3.6649 20.8183 10.5488
8 0.1285 3.2636 16.5097 8.3656
9 0.1144 2.9064 13.0927 6.6342
10 0.1019 2.5882 10.3830 5.2612
11 0.0907 2.3048 8.2341 4.1723
12 0.0808 2.0525 6.5299 3.3088
13 0.0720 1.8278 5.1785 2.6240
14 0.0641 1.6277 4.1067 2.0809
15 0.0571 1.4495 3.2568 1.6502
16 0.0508 1.2908 2.5827 1.3087
17 0.0453 1.1495 2.0482 1.0378
18 0.0403 1.0237 1.6243 0.8230
19 0.0359 0.9116 1.2881 0.6527
20 0.0320 0.8118 1.0215 0.5176
21 0.0285 0.7229 0.8101 0.4105
22 0.0253 0.6438 0.6424 0.3255
23 0.0226 0.5733 0.5095 0.2582
24 0.0201 0.5106 0.4040 0.2047
25 0.0179 0.4547 0.3204 0.1624
26 0.0159 0.4049 0.2541 0.1288
27 0.0142 0.3606 0.2015 0.1021
28 0.0126 0.3211 0.1598 0.0810
29 0.0113 0.2859 0.1267 0.0642
30 0.0100 0.2546 0.1005 0.0509
31 0.0089 0.2268 0.0797 0.0404
32 0.0080 0.2019 0.0632 0.0320
33 0.0071 0.1798 0.0501 0.0254
34 0.0063 0.1601 0.0398 0.0201

CATEGORIES

CONTACT US

Contact: Kyle Yung (Senior Manager)

Phone: +86-18148918807(大陆)

Tel: +886-3-459-1828(台湾) +86-769-82300929(大陆)

Email: info@hsuyu.com

Add: NO.7, Daren Str, Pinzhen Xian, Taoyuan City, Taiwan.