[UPDATED] How should I power a 12V-DC, 12.3A LED circuit?

Thread Starter

-Ty-

Joined Feb 5, 2017
83
Hello everyone!

I currently have a thread running that seems to have reached its natural conclusion for the questions I originally posed in it. However, I have a different set of questions now, and I felt it was more appropriate to give it its own thread, rather than try to merge it with the first, and change the title. Thank you to everyone who helped. This is going to be a long read though, so apologies in advance.

So, I'm building a set of photography lights that make use of 5050 LED strips. There are four panels in total, two large, and two small. The two large panels use 10m of 5050LED strips, which yields a (multimeter-tested) current draw of 12.3A at 12V-DC EACH. The two small panels are half of this, each. In the spoiler below is a picture of one of the small panels, just for illustrative purposes.

20190124_220243.jpg

LED Panels Circuit.jpg

Now, the problem I'm having is that when purchasing all of my parts for this project, I bought some brandless, nameless, waterproof LED drivers from Ebay, at the well-meaning suggestions of some people on the forums.

These power supplies arrived, and ended up being ridiculously heavy. Over 5 lbs each for the 200W variant.

My original plan was to mount the power supplies to the panels directly and feed into them with just a few inches of wire, but that simply isn't possible with that kind of weight, so now, the power supplies have to sit on the ground and feed the power to the panels through a long wire. The wire has to be AT LEAST 11 feet long, to reach the top of the tripods these lights get mounted to. Using a typical 18AWG cable, I'd lose something stupid like 4V from the 12V supply, since it's being driven at 12.3A.

Suffice it to say, in the previous thread, people helped me settle on the idea of buying some thick-a-- cables, like 8 or 10 AWG four-conductor, to minimize the voltage losses due to wire resistance. Fine and dandy, except that between the wire, and other components I'd need to make that work, I'm looking at $140. I'd really like to avoid spending that much given that the entire project was only around 520.

SO, after some dealings with EBAY, it seems like I MIGHT be able to return the shitty power supplies after all! That's where this thread comes in. I'm hoping to get some help picking out PROPER power supplies to get this system working. I STILL need to feed the power through a long cable, though. As i started putting these panels together, I realized that I simply can't add any more weight to them. So we're still looking at a 11 foot cable minimum, but 20ft preferably.

The Short Story:

So, if everything with Ebay goes through and I'm able to return the crap power supplies, my two ideas are:

1) To buy a LRS Series power supply outputting at around 14V. Then get a standard 16AWG cable at 20 feet, hook it up, and set the supply's output voltage to whatever it needs to be in order to get 12V at the end of the wire. Now, this would be done without a load hooked up, so there would be essentially no voltage drop through the wire. However, I will rely on the online calculations which show that I can expect a 2V drop across the wire at 12.37A, so i will set the power supply to its lowest output, around 13.5V, and work up from there once it's hooked up to the panels, until I'm reading 12.0V while the whole system is running.

or 2) to find and buy a power supply with a voltage that's much higher than what I need, like around 24V, then feed it through the long thin wire, and then use a DC-DC step down regulator to bring whatever the voltage at the end of the line is, back down to 12V for the load.

The only complication is this:

20190124_220200.jpg

What you see in that spoiler is the control box for the panel, with the three 25kHZ PWM dimmers inside. And you might notice at the bottom... a C13 connector.

Yes, yes, I know, don't use AC plugs for DC projects. I get it, I learned my lesson. But, there's no going back now, they can't be swapped out. Those aluminum panels were cut to fit the plugs and I can't swap it for something different unless it has the exact same shape and size.

SO, whatever power supply I get, I'll wire the output up to a C13 Computer power cord.

Anyways, yeah, any help, suggestions, or ideas I can get would be greatly appreciated. Thank you for your time!
-Ty
 
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oz93666

Joined Sep 7, 2010
737
Nice work .... first how long do you need to operate these lights for ??? It maybe simpler and much cheaper to use rechargeable batteries , very light , Half a kilo ($20) should run the lights for 10 or 15 mins and you would be free from mains wires. On low light setting the battery should last for hours .

for dimming , since these lights are not in constant use , it would be easier to put resisters in series ... lengths of wire ... have 10 or so switchable settings for light intensity.

Ah ... so that black box is the dimmer ??? ..12V in and three variable outputs ?
 
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Thread Starter

-Ty-

Joined Feb 5, 2017
83
Nice work .... first how long do you need to operate these lights for ??? It maybe simpler and much cheaper to use rechargeable batteries , very light , Half a kilo ($20) should run the lights for 10 or 15 mins and you would be free from mains wires. On low light setting the battery should last for hours .

for dimming , since these lights are not in constant use , it would be easier to put resisters in series ... lengths of wire ... have 10 or so switchable settings for light intensity.

Ah ... so that black box is the dimmer ??? ..12V in and three variable outputs ?
Unfortunately, Im going to need hours worth of working time with the lights near full brightness/current draw, so batteries are out of the question for now. As for dimming, I want complete control, not incremental steps. The three dimmers in the one shown control each of the R, G, and B channels in those strips, which happen to be RGB LEDs. By being able to dial in a specific brightness, i can mix and match for any colour under the rainbow.

12V into the box, 25kHZ PWM dimming out.
 

MisterBill2

Joined Jan 23, 2018
8,505
Thanks for creating a new thread directly related to the actual challenge.
Now here is the question: What voltage comes out of those dimmers? I am asking because if those dimmers are intended to be supplied by a 12 volts DC supply and deliver a 12 volt PWM controlled output, you may be all set. The other question is about the LED assemblies, do they contain any electronics or are they only LEDs , possibly with resistors on board? It may be that the solution is simple, depending on the actual hardware that you have on hand.
 

Thread Starter

-Ty-

Joined Feb 5, 2017
83
Thanks for creating a new thread directly related to the actual challenge.
Now here is the question: What voltage comes out of those dimmers? I am asking because if those dimmers are intended to be supplied by a 12 volts DC supply and deliver a 12 volt PWM controlled output, you may be all set. The other question is about the LED assemblies, do they contain any electronics or are they only LEDs , possibly with resistors on board? It may be that the solution is simple, depending on the actual hardware that you have on hand.
I was actually worried I would be breaking some forum rule I wasn't aware of by creating a new thread like this.

So for your first question, the PWM dimmers take a 12-volt input directly from the power supply, and modulate it at 25kHz, yielding what looks to my multimeter like a lowered voltage. So yes, like you said, it's still just a 12 volt PWM-controlled output. They're actually advertised as motor speed controllers, if that's any help. Here's a description of the dimmers from their product page:

Input voltage: DC10-60V
Output voltage type: Load Linear
Output Current: 0-20A
Continuous power: 1200W
Speed Type: adjusting flow
Speed mode: potentiometer (linear)
Speed range: 0 - 100%
Control frequency: 25KHZ


As for the LEDs, unfortunately there's no datasheet available for them, but then again, they function like pretty much every other LED strip in the world: three diodes with some resistors, per section. I haven't added any other electronics to the circuit, its just:

Power supply--------= Switch-----------= PWM Input-----------=PWM Output---------------=LED +/-.
 

Reloadron

Joined Jan 15, 2015
5,917
yielding what looks to my multimeter like a lowered voltage.
When measuring a PWM signal with a simple DMM. Just as an example with a 25 KHz PWM signal the period is the reciprocal of the frequency so 1/25000 = 40 uSec. So with a 50% duty cycle the 12 volt PWM will be on for 20 uSec and off for 20 uSec a simple DMM will read about 6 Volts or the average or 50% of 12 = 6.0. So what the DMM reads will be a function of the PWM. Something you may wish to consider when using a DMM to measure a PWM signal.

Ron
 

Thread Starter

-Ty-

Joined Feb 5, 2017
83
When measuring a PWM signal with a simple DMM. Just as an example with a 25 KHz PWM signal the period is the reciprocal of the frequency so 1/25000 = 40 uSec. So with a 50% duty cycle the 12 volt PWM will be on for 20 uSec and off for 20 uSec a simple DMM will read about 6 Volts or the average or 50% of 12 = 6.0. So what the DMM reads will be a function of the PWM. Something you may wish to consider when using a DMM to measure a PWM signal.

Ron
Yeah I realize that, which i why i mentioned that's what the voltage "looks like" to my multimeter. I dont own an Oscilloscope unfortunately so I have no other means of testing available to me.
 

Thread Starter

-Ty-

Joined Feb 5, 2017
83
So, i've run into another problem here I don't know how to fix...

So, using online calculators, It shows that for 18AWG wire, at 20' long, I'll lose over 3V off of my supply voltage, when the current draw is the full 12.3A. So, if i pick up a 15V supply with a 3% adjustable output, i can get up to 15.45 volts from the supply, which after resistive losses, i can dial in so that it would come to 12V at the load.

HOWEVER, once i dim the lights below maximum, the current draw will drop as well, which means less voltage is lost to resistance, which means that where the panels were once getting 12V, they'll start to get 12.5, 13, 13.5... etc., and will immediately burn out. Basically, if i dim the circuit, it'll explode. What can i do to work around this?
 

Danko

Joined Nov 22, 2017
1,073
HOWEVER, once i dim the lights below maximum, the current draw will drop as well, which means less voltage is lost to resistance, which means that where the panels were once getting 12V, they'll start to get 12.5, 13, 13.5... etc., and will immediately burn out. Basically, if i dim the circuit, it'll explode. What can i do to work around this?
When you dim panels, using PWM, every pulse has the same parameters (current, voltage), except duration.
You change only averaged in time current through panel.
So nothing bad will happen.
EDIT:
No spikes.
See simulation:
upload_2019-1-28_0-55-6.png
 
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Norfindel

Joined Mar 6, 2008
326
This voltage drop calculator gives a 1.73v voltage drop with 18 AWG wire. That's 14.4% of the 12v supply.

https://www.calculator.net/voltage-...e=11&distanceunit=feet&amperes=12.3&x=74&y=15

If you use 16 AWG, the drop is 1.09v, or around 9%.
What's the tolerance on the led strips? Most equipment will run if the difference in PSU voltage is within a certain tolerance.
Those wire sizes are as thin as they go for mains cabling here, and we use 220v AC!

Shouldn't be that much of an expense to buy some mains wire, of the ones used to make extensions.
 
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Thread Starter

-Ty-

Joined Feb 5, 2017
83
When you dim panels, using PWM, every pulse has the same parameters (current, voltage), except duration.
You change only averaged in time current through panel.
So nothing bad will happen.
That makes sense, but its hard to wrap my head around how the voltage from the supply would react in response...

Pulse on, first thing that happens is a voltage of 15v (direct from the supply) presents itself to the LEDs (since there's currently no current draw, and so no loss of voltage through the wire), this voltage then motivates the flow of electrons, at a rate set by the internal resistance of the LED circuit, so, 12.3A. This current draw then sets the amount of resistance through the supply wires, which creates the feedback that lowers the presented voltage from the supplied 15V to 12. At least, that's how im picturing the chain of events in my head.. but its that beginning part of every pulse, where the system is off and then the pulse hits, where there isn't any current draw, so the LED's would be getting hit with a 15V spike at the beginning of every PWM wave, wouldnt they?

I mean I realize that I'm getting down into nanosecond timings here, I wouldn't be surprised if quantum mechanics reared its head, but still.

What im wondering now is... if i can't really use my multimeter to determine what the actual current draw is at less-than-maximum brightness, and I'm to assume its just the same 12.3A, then the best way to set this up would be... what? Start off with a length of 20' of wire, and with the 15V power supply turned all the way down to its minimum of 14.55V, and just hope and pray that when i connect it to the panel, everything will work itself out and the voltage will drop as I've calculated it should... and then use my multimeter to see what the voltage is in steady-state at maximum brightness, and adjust the output upwards until it reads 12.0V, and assume it will stay steady at that when the panels are dimmed?

This voltage drop calculator gives a 1.73v voltage drop with 18 AWG wire. That's 14.4% of the 12v supply.
If you use 16 AWG, the drop is 1.09v, or around 9%.
What's the tolerance on the led strips? Most equipment will run if the difference in PSU voltage is within a certain tolerance.
Those wire sizes are as as thin as they go for mains cabling here, and we use 220v AC!
Using THIS calculator (which shows the same results as two others) shows what you said. for a 12.3A load at 12.30V, across a distance of 11 feet, I lose 1.73V and am left with 10.57V, in 18AWG. Yes, I believe that's still within a usable range for the LED strips, but it will be dim, very dim. I'd be losing over half of the potential brightness of these lamps, and given that the whole point of them is to use them for photography.... well. I need to get the voltage at the load to be as close to 12.0V as possible... so im thinking get an adjustable 15V supply, dial it up to 15.14V, and then I can get away with using even 20 feet of wire, and the voltage at the load will be 12.0, after a drop of 3.14V.
 

Thread Starter

-Ty-

Joined Feb 5, 2017
83
I did not read all the posts, I am tired.

Here is your power supply:
Free Shipping 1pcs LM2596S DC-DC 4.5-40V adjustable step-down power Supply module NEW High Quality LM2596S

The losses in long wires are described by the telegraphic equation. You have to reduce the wire resistance, a battery in the middle will do.
Now i aint no electamatronic enginur or nuthin', but I dun reckon the ol' LM2596 board there with its "output current: rated current 1A" aint gonna fare too good with my 12.3A load ;)
 

sghioto

Joined Dec 31, 2017
2,336
Why can't you just change the wire size? The cost is not that significant.
I've seen 25 ft of 2 conductor #12 wire for 10.00 on ebay.
With #12 wire you will only lose about .75 volts, I doubt if you will even notice the difference using a 12 volt supply
Another point, #18 wire is nominally rated at 10 amps max. and at 12.3 amps the wire will dissipate 38.5 watts or almost 2 watts per ft. Use #12 wire.
SG
 
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BobTPH

Joined Jun 5, 2013
3,132
Pulse on, first thing that happens is a voltage of 15v (direct from the supply) presents itself to the LEDs (since there's currently no current draw, and so no loss of voltage
No, it does not work that way. The strip never sees the voltage before the current starts flowing. It is the movement of the electrons through the strip that creates the voltage across it.

Bob
 

ArakelTheDragon

Joined Nov 18, 2016
1,353
Now i aint no electamatronic enginur or nuthin', but I dun reckon the ol' LM2596 board there with its "output current: rated current 1A" aint gonna fare too good with my 12.3A load ;)
Sorry about that, "12.3A" means "12V, 3A" when you are tired :D.

EDIT:
You will have enourmous losses trying to transfer DC current. Electricity is only transferred at high voltages with low current and with VAC. An DC-AC/AC-DC option is better.

EDIT_1:
A power supply of 450W or more will be needed.

Maybe this.
 
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Thread Starter

-Ty-

Joined Feb 5, 2017
83
When you dim panels, using PWM, every pulse has the same parameters (current, voltage), except duration.
You change only averaged in time current through panel.
So nothing bad will happen.
EDIT:
No spikes.
See simulation:
View attachment 168879
Interesting... Thank you for taking the time to create the graphic. I learned something today!
 
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