Battery Powered Lighted Marquee

Thread Starter

AIRSPD

Joined May 27, 2015
2
image.jpg Hello everyone. I am exploring the possibility of lighting a home made marquee of the word "CHICAGO" for a dance stage performance with BATTERY POWER. I would greatly appreciate any advice or solutions for this project if anyone has any good ones. My current thought is trying to light a store bought string of 75 to 100 bulbs of the G40 look using a car battery or similar as simply and cheaply as possible. I only need 5 minutes of light at most.
 

elec_mech

Joined Nov 12, 2008
1,500
Welcome to AAC.

Assuming the marquee you're using is the same as the one pictured, you'll need 76 bulbs. However, the first C in the sign above has 12 bulbs while the second C has 11. If you can rearrange the bulbs on the first C so you only need 11, then you'll need a total of 75 bulbs making for a less expensive solution.

Based on your description, you'll need three 25-bulb G40 packs, one inverter, and one car or large lead-acid battery. Length of run time will depend on the battery size, but you'll want a minimum rating of 6Ah from a 12V battery. I don't know what kind of efficiency the inverter has, so the larger the battery, the better. The inverter has a switch, so you can use that to turn the lights and inverter off when not in use to save battery power. You don't have to use the inverter above, but don't get one with less than a 500W rating. Each bulb uses 5W bulbs: 5 x 75 = 375 watts. While there are 375W inverters, you don't want to push the inverter to its limit and when a bulb needs replacing, someone might replace it with a 7W bulb so you want some overhead.

Whatever battery you use, you'll want to be darn careful to make sure no one touches the leads. These have the potential to harm or kill. I'd strongly covering the battery such that users won't ever come in contact with the leads.

Also make sure the battery is charged properly and regularly, otherwise you'll have a heavy and expensive paperweight in a short time. This is from personal experience. Good luck.
 

Thread Starter

AIRSPD

Joined May 27, 2015
2
Welcome to AAC.

Assuming the marquee you're using is the same as the one pictured, you'll need 76 bulbs. However, the first C in the sign above has 12 bulbs while the second C has 11. If you can rearrange the bulbs on the first C so you only need 11, then you'll need a total of 75 bulbs making for a less expensive solution.

Based on your description, you'll need three 25-bulb G40 packs, one inverter, and one car or large lead-acid battery. Length of run time will depend on the battery size, but you'll want a minimum rating of 6Ah from a 12V battery. I don't know what kind of efficiency the inverter has, so the larger the battery, the better. The inverter has a switch, so you can use that to turn the lights and inverter off when not in use to save battery power. You don't have to use the inverter above, but don't get one with less than a 500W rating. Each bulb uses 5W bulbs: 5 x 75 = 375 watts. While there are 375W inverters, you don't want to push the inverter to its limit and when a bulb needs replacing, someone might replace it with a 7W bulb so you want some overhead.

Whatever battery you use, you'll want to be darn careful to make sure no one touches the leads. These have the potential to harm or kill. I'd strongly covering the battery such that users won't ever come in contact with the leads.

Also make sure the battery is charged properly and regularly, otherwise you'll have a heavy and expensive paperweight in a short time. This is from personal experience. Good luck.
Elec_Mech thank you for the thorough solution! You're right, I'll need to limit the final bulb count to 75. The picture above is just a Photoshop visualization I made quickly.
 

Externet

Joined Nov 29, 2005
2,202
Also search for "light rope" and LED strips; made with LEDs, work on 12VDC, can be surprisingly bright at certain angles, and an automotive battery does the job
 
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