Quick answer: Add up the LED module wattage, add roughly 10 percent for wiring loss, then divide by 0.8 so the power supply never runs above 80 percent of its rating. For 1,000 modules at 0.72 W: 720 W of LED load, about 790 W with wiring loss, so a supply capacity near 990 W — two 600 W units or three 400 W units.
Sizing a transformer for illuminated letters is simple arithmetic with one safety rule attached. The arithmetic is LED load plus cable loss; the safety rule is that a power supply should run at no more than about 80 percent of its rated output. Get either one wrong and the sign works on the bench and fails on a hot afternoon.
The three numbers you need
- Module wattage — from the LED datasheet, for example 0.72 W per module
- Module count — the total across every letter, including spares inside large letters
- Cable loss — typically 5 to 10 percent, depending on run length and conductor size
Step 1: total LED load
Multiply the single-module wattage by the number of modules:
LED load = 0.72 W × 1,000 = 720 W
Count every module, not every letter. A large letter may hold 30 modules and a small one five; the power supply sees the total. If the sign mixes module types, calculate each group separately and add the results.
Step 2: add wiring loss
Cable resistance turns a small part of the output into heat. The longer the run and the thinner the conductor, the larger the loss. On a normal channel letter set, allow 5 to 10 percent:
System load = 720 W × 1.10 ≈ 790 W
Voltage drop matters more on 12 V systems than on 24 V, because the same wattage draws twice the current. Long runs, thin cable and daisy-chained letters are the usual reason a supply that looked adequate trips on load. The 12V vs 24V LED modules comparison explains when to step up to 24 V.
Step 3: apply the 80 percent rule
A power supply running at 100 percent of its rating produces full heat and has no margin for a hot day, a slightly higher LED bin or an added module. Divide the system load by 0.8:
Required capacity = 720 W ÷ 0.8 = 900 W
Using the example without cable loss gives 900 W. Add the 10 percent wiring allowance and the number is about 990 W, which is why the practical answer is two 600 W supplies or three 400 W supplies. The selection method is set out in more detail in the channel letter power supply guide.
Why split into two or three supplies
One 1,000 W unit is rarely the best answer:
- Heat — a single large supply concentrates heat in one enclosure, often inside a raceway with little airflow
- Service — if one unit fails, the whole sign goes dark; with two or three, the rest keeps working
- Wiring — shorter low-voltage runs from several points reduce voltage drop
- Availability — 400 W and 600 W units are standard stock; 1,000 W units are less common and harder to replace
Balance the load across the supplies rather than filling one and leaving the last nearly empty. Two 600 W units should share the 790 W roughly evenly; three 400 W units should carry about 265 W each.
Worked example
| Item | Calculation | Result |
|---|---|---|
| LED modules | 1,000 × 0.72 W | 720 W |
| With 10% cable loss | 720 × 1.10 | ≈ 790 W |
| 80% rule | 720 ÷ 0.8 | 900 W |
| Practical supply | 2 × 600 W or 3 × 400 W | 1,200 W or 1,200 W rated |
Common mistakes
- Counting letters instead of modules
- Forgetting the modules hidden behind a diffuser or in a return
- Using the LED’s nominal wattage at 12 V when the sign runs at 24 V, or the reverse
- Filling a supply to 95–100 percent because “it is rated for it”
- Ignoring cable length on a large roof sign
FAQ
Is 80 percent a rule or a regulation?
It is an engineering practice, not a legal limit. It keeps the supply cooler, extends its life and leaves room for small load increases. Some electrical codes separately require circuits to be loaded to no more than 80 percent of rating.
Can I use one 1,000 W supply instead of two 600 W?
Electrically yes, if it is rated for the load and the enclosure is ventilated. In practice two smaller units are easier to service, easier to fit and less of a single point of failure.
How do I know the module wattage?
It is on the LED module datasheet, usually stated per module at a given voltage. If only current is given, multiply voltage by current: 12 V × 0.06 A = 0.72 W.
Next step
Send the module datasheet, the total module count and the distance from the supply to the letters. We will return the load calculation, the recommended supply split and the cable size. For layout detail, read the channel letter LED layout guide, or start with the complete channel letter guide.
Reference: voltage drop, Ohm’s law, US Department of Energy on LED lighting.