LED light bar question

[jjake]

I asked this on another website but thought it might be practical to ask here as well. I have a 36v EZ GO golf cart running on 6-6v batteries (when fully charged the battery pack puts out about 38v). I will be adding a light bar but don't want to add weight or more "stuff" to the cart. There are a couple of ways to power the light bar at 12v. To avoid potentially messing up the battery pack, some folks purchase a voltage reducer or they purchase a separate 12v battery both of which I would like to avoid. Others run the 12v from two of the batteries which I understand may damage the battery pack due to the unbalanced use.

Most of the LED light bars that I found having a working voltage of 9v-33v. Is there such a thing as an LED light bar with up to 38v working voltage? I could then use the entire battery pack to avoid unbalanced discharge, it wont reduce the required amperage needed and would minimize adding the weight of a 12v battery or more wiring for a reducer.

Alternatively could I run the light bar across 4 -6v batteries at 24v and maybe lessen the risk of unbalanced discharge damaging the battery pack?

Thoughts? Has anyone already done this? If so, which light bar did you use for 36v-38v?

Thanks

[BobBoyce]

Originally Posted by jjake

I asked this on another website but thought it might be practical to ask here as well. I have a 36v EZ GO golf cart running on 6-6v batteries (when fully charged the battery pack puts out about 38v). I will be adding a light bar but don't want to add weight or more "stuff" to the cart. There are a couple of ways to power the light bar at 12v. To avoid potentially messing up the battery pack, some folks purchase a voltage reducer or they purchase a separate 12v battery both of which I would like to avoid. Others run the 12v from two of the batteries which I understand may damage the battery pack due to the unbalanced use.

Most of the LED light bars that I found having a working voltage of 9v-33v. Is there such a thing as an LED light bar with up to 38v working voltage? I could then use the entire battery pack to avoid unbalanced discharge, it wont reduce the required amperage needed and would minimize adding the weight of a 12v battery or more wiring for a reducer.

Alternatively could I run the light bar across 4 -6v batteries at 24v and maybe lessen the risk of unbalanced discharge damaging the battery pack?

Thoughts? Has anyone already done this? If so, which light bar did you use for 36v-38v?

Thanks

Whether you unblance 2 batteries or 4 batteries, the result is the same. What is the current draw of the LED light bar? There are 10A and lower voltage reducers out there rather than the 30A typically used.

Bob

[rabbitreborn]

If the lightbar is capable of 33v and the highest charge of a 36v cart is just below 43v (while on charge), why not wire it to the whole pack with a few resistors in line to reduce the maximum pack output to be below 33v?

I'm not a math genius, but we have some GREAT electrical people here that could probably tell you exactly what resistors you would need to accomplish this......

This way you would be equally drawing from the entire pack while powering this lightbar....

Not to mention how bright it would be......

[jjake]

Bob - okay. I was thinking the draw across the 24v would be low and minimal risk to the battery pack but I tend to agree. Most of the 20" light bars seem to be around 120w.

Rabbit - My thoughts exactly! Ohm's Law runs through the veins of many on this site.

[cgtech]

I would need to know the actual amp draw of the light bar at its highest practical voltage (maybe hook to 5 batteries and read it). The closer the input voltage is to the output voltage is, the better (less energy will need to be burned off in the resistor). And the resistors couldn't be the common little 1/4 watt resistors, they would have to be beefy ones.

[rabbitreborn]

Ok this is what I found by doing a quick search.......

excuse the crude drawing but I am tired and it was the first app I could find to try to draw it out.

Picture the black rectangle as your battery pack. The red line is the 36v + going to the brown switch and then the output of the switch (also red) goes to a Blue 5 Ohm resistor. The other side of the Blue resistor goes to a Green 10 Ohm resistor and an Orange output wire. The Orange output wire goes to the positive lead for the yellow LED light. The other side of the Green resistor goes to the black ground wire. The Black ground wire goes from the battery pack to the ground on the LED. They call this a Voltage Divider circuit apparently.

I ran an online calculator and it says :

pack voltage of 43 (peak charge) light would get 28.667v
nominal voltage of 38 light would get 25.333v
lowest pack voltage of 36 (time to recharge) light would get 24v


Hopefully someone can come along and confirm that but I think it would work.......

[jjake]

This is the description for one of the 20" LED light bars that I am looking at. I believe the 126W at 36v would be about 3.5amps.

Power: 108W~198W
42pcs*3W high intensity Cree LEDs
Working Voltage: DC 9-32V
Lumen: Approx 8820lm~12600lm
Beam: Flood Beam(60 degree) & Spot Beam(30 degree) combo
Color Temperature: Pure White 6000K
Working Temperature:-40~85 degree Celsius
Dimension: 505*73*107mm/20inch*2.9inch*4.2inch approx
Protection Rate: IP67
Single LED power: 3W

If this gets too complicated I may just carry a flashlight. :)

[Sergio]

You really should use a 36vDC to 12vDC converter for this application.

It is not practical to reduce that much current using resistors.

If You just use your 5Ω current limit resistor (voltage divider would not work) and the LED bar light draws 5 amps, the power dissipated on the resistor would be:

P=RxI^2 = 5Ω x 5amps^2 = 125watts

Since the resistor selected should have double that capacity, You are looking at a 250watt resistor.

A voltage reducer would be way more efficient.

[cgtech]

I agree, plus the posted "specs" are all hear-say to me, China lies all the time. Unless you measured it yourself, it ain't real. If all this jargon is too much like work, and you like the idea of not simply "burning-off" the extra power to suit your needs, get a convertor. There is no good reason not to get one (they make them plenty compact, so room is not a good one, absolute rock-bottom cheapness 'efficiency-be-damned' is all that's left) . A decent DC convertor can run all your incandescent lights on less than 5 amps (your LEDs will draw far less). Still wanna make the resistor setup? I'll still help, but you have to measure the actual load of your lights, not the "ebay rating".

Originally Posted by Sergio

You really should use a 36vDC to 12vDC converter for this application.

It is not practical to reduce that much current using resistors.

If You just use your 5Ω current limit resistor (voltage divider would not work) and the LED bar light draws 5 amps, the power dissipated on the resistor would be:

P=RxI^2 = 5Ω x 5amps^2 = 125watts

Since the resistor selected should have double that capacity, You are looking at a 250watt resistor.

A voltage reducer would be way more efficient.

[rabbitreborn]

Originally Posted by Sergio

You really should use a 36vDC to 12vDC converter for this application.

It is not practical to reduce that much current using resistors.

If You just use your 5Ω current limit resistor (voltage divider would not work) and the LED bar light draws 5 amps, the power dissipated on the resistor would be:

P=RxI^2 = 5Ω x 5amps^2 = 125watts

Since the resistor selected should have double that capacity, You are looking at a 250watt resistor.

A voltage reducer would be way more efficient.

OK, I'm an idiot when it comes to the science of this. Can you break this down to the level of a 15 year old for me? I'm not trying to be a smart ***, just trying to understand why what I proposed would not work.
You said a voltage divider circuit won't work.... Why?

Where did you get the 5 amp figure for your calculation?