Boost converter or turn series to parallel in an LED work light.

[HR19]

I have a Coast LED work light, 2,500 lumens, plug-in, and I'm trying to convert it to use my Milwaukee tool batteries, 21V max, 18V nominal. It runs at 72V currently, using 3 parallel sets of 24 LEDS for 72V nominal. I bought a boost converter for $5, but I quickly fried it by powering it backwards (blew the fuse, but even after replacing the fuse it seems to have a short and the MOSFET gets SUPER HOT). Should I buy a new one? Would making a boost converter be easy enough I could DIY it (I've seen a few videos, I just don't know the calculations)? Or should I scrape part of the solder pad and make it into 12 parallel sets of 6? (I'll upload pictures in a few minutes.)

 

[Pommie]

Do you know the current or watts used? Is there a plate with the wattage on it? Or, do you have a link to it?

Mike.

 

[HR19]

Do you know the current or watts used? Is there a plate with the wattage on it? Or, do you have a link to it?

Mike.

24W is the rating.

 

[Pommie]

If you divide it into 4 (or 6?) 12 LED strings they will require 36V at ~660mA. What do they have limiting the current - resistor etc. Or is it done with a constant current supply?

Mike.

 

[HR19]

If you divide it into 4 (or 6?) 12 LED strings they will require 36V at ~660mA. What do they have limiting the current - resistor etc. Or is it done with a constant current supply?

Mike.

My bad, edited OP, meant 12 sets of 6. I was thinking of using a rheostat for it as the load should be under 2 amps, but a rheostat would give me variable brightness levels, and at such low current there would be very little wasted energy.

 

[augustinetez]

It runs at 72V currently, using 2 parallel sets of 24 LEDS

Depending on how you interpret the above, the sums don't add up.

I count 72 LEDs in the pic - have you physically measured the voltage coming out of the power supply - red and blue wires.

 

[HR19]

Depending on how you interpret the above, the sums don't add up.

I count 72 LEDs in the pic - have you physically measured the voltage coming out of the power supply - red and blue wires.

Yes, 72V, and my DMM says they have about 2.49Vf

 

[Pommie]

It still doesn't add up. How many LEDs are there? What is the supply voltage? How many strings are there?

Mike.

 

[HR19]

It still doesn't add up. How many LEDs are there? What is the supply voltage? How many strings are there?

Mike.

72 LEDs, OP was typo, it's 3 parallel sets of 24 series, at 3V makes 72V total

 

[Pommie]

You said Vf was 2.5V so 24 in series is 60V. Is there a series resistor?

Mike.

 

[HR19]

I don't know, there's some sort of power supply to convert 120VAC to 72VDC (71.5 actual measured during operation).

But I went the easy way and scraped up the contacts, and opted for 73.33333333333333333333333333333333333333333333 ohm resistor (3x220 in parallel, because that's what I had) for now, but I'll probably add a rheostat later, we'll see. For now, seems to work well despite my initial mess ups.

 

[Pommie]

Assuming 60V for the LEDs (24x2.5) gives 12V across your resistor which is a current of 166mA.
Three strings = 500mA times 72V = 35W.
But you said they're rated at 20W. Something is still wrong.

Mike.

 

[HR19]

Yeah, I'm using an 18V power supply now, running 12 parallel sets of 6 in series, which I said was the goal in the original post.

 

[Nigel Goodwin]

My bad, edited OP, meant 12 sets of 6. I was thinking of using a rheostat for it as the load should be under 2 amps, but a rheostat would give me variable brightness levels, and at such low current there would be very little wasted energy.

2A isn't 'low' currant - and you're going to waste as much power as you use, and possibly more - this isn't the 1950's, we don't use rheostats for light dimming.

The obvious, simple, and efficient, option would be to use a constant current boost converter, and simply have adjustable constant current if you want variable brightness.

 

[HR19]

A boost converter would be a big mistake as I'm now using 15V of LED in series with an 18V source. A buck converter might do OK, but at the moment I've no way to justify the cost and complexity of adding another circuit. I measured 4.41V across the resistor (73.3ohms), so that gives about 60mA. 60mA, across 12 parallel sets, means 5mA per LED right now. 5mA times 2.5V makes 12.5mW, times 72 LEDs makes for less than 1W. Unless I screwed up my math somewhere, I'm using less than 1W for the LEDs, less than 1/4W for the "waste," and it's still really bright.

 

[Nigel Goodwin]

A boost converter would be a big mistake as I'm now using 15V of LED in series with an 18V source. A buck converter might do OK, but at the moment I've no way to justify the cost and complexity of adding another circuit. I measured 4.41V across the resistor (73.3ohms), so that gives about 60mA. 60mA, across 12 parallel sets, means 5mA per LED right now. 5mA times 2.5V makes 12.5mW, times 72 LEDs makes for less than 1W. Unless I screwed up my math somewhere, I'm using less than 1W for the LEDs, less than 1/4W for the "waste," and it's still really bright.

You mentioned above that it's a 24W light? - and 1W shared between 72 LED's doesn't sound 'really bright', and there would be no point using 72 of them, a great many less would easily be just as bright.

 

[HR19]

I'm modifying an existing light I already have, that's why it is the way it is.

 

[Pommie]

I measured 4.41V across the resistor (73.3ohms), so that gives about 60mA. 60mA, across 12 parallel sets, means 5mA per LED right now.

This suggests you have one resistor for all the LEDs.
You need a resistor for each string.
As you have 6 strings, they should consume ~3.3W per sting (20/6).
Current per sting should therefore be 3.3/18 = 183mA.
The LED strings appear to have a Vf of 13.5V leaving 4.5V to be dropped by the resistor.
Resistance required = 4.5/0.18 = ~25Ω. 22Ω - 1W will probably suffice.
You're wasting 4.5V of the 18V so 4.5/18 = about 25%. Or about 5W - nearly 1W per resistor.

Mike.

 

[HR19]

This suggests you have one resistor for all the LEDs.
You need a resistor for each string.
As you have 6 strings, they should consume ~3.3W per sting (20/6).
Current per sting should therefore be 3.3/18 = 183mA.
The LED strings appear to have a Vf of 13.5V leaving 4.5V to be dropped by the resistor.
Resistance required = 4.5/0.18 = ~25Ω. 22Ω - 1W will probably suffice.
You're wasting 4.5V of the 18V so 4.5/18 = about 25%. Or about 5W - nearly 1W per resistor.

Mike.

At the moment yes, one resistor (3 in parallel, but effectively one), and then the 12 strands of 6 series are in parallel. I brought the resistance down to 8.8ohms for a test and it's WAY brighter but it burns through the 10ohm, 1/4W resistor (running about 10x the rated current through it).

Each LED is appx. 0.33W rated (24W/72LEDs). My measuring is about 2.5Vf/LED, 4.41V across the resistors was because the battery is fully charged, it's over 19V right now.

 

[Pommie]

If you use one resistor to limit the current through multiple LED string then the string with the lowest Vf will get ALL (well most of) the current and maybe burn out.

If we take your value of 0.33W per LED with a Vf of 2.5 then that is a current of 0.33/2.5 = 130mA. This does of course ignore the losses in the resistor.

Just pick a resistor, 22Ω 1W seems to fit the bill, try it and see.

Mike.