DIY LED driver for reef lighting

I know this won't help for efficiency, but may help with heat. If the voltage difference across string is enough to be producing a lot of heat form your dirver go get some power resistors and mount them to the heat sink with the LEDs. This will equalize the voltage across the drivers so less heat in the project box, and put the heat with the LEDs where there is a way to get rid of it.

If you used a several 0.5 ohm resistors even at 1 amp these could be half watt resistors. And would allow balancing of a string within a 1/2 volt of each other.
 
Is it worth it?

Is it worth it?

We started this to save money. By the time you order boards, pay shipping, and get all the parts what is the real cost. Does all the overhead stuff still make it cheaper.

I ordered for 5 boards (I only need 2, and I ordered for 2 boards for some one else, the third is a spare). I am guessing that by the time I am done I will have spent about $220 dollars for the five boards. That is about $5.50 per driver or $0.92 per LED. I had the caps a resistor already so I did save a little there, however I think they would not add more than $0.25 per driver.

So for those wondering yes it does pay to it yourself even on these small scales.
 
Indeed I feel that this project offers some very obvious cost savings!

For the sake of comparison, let's round this project's cost up to $1.25 to include project boxes, headers, and other misc. hardware you'd need to get a "finished" project.

By comparison, if you get a Meanwell driver and something to generate the required dimming signal for a bargain price ($30), you're spending exactly twice as much ($2.50) per LED.

But I also started this thread and DIY driver effort to create something that was easy to assemble (check!), easy to dim (check!), efficient (check!), flexible (check!), and fun (check!). So, as I sit here next to 18 LEDs fading up and down randomly on my test rig, I can say that yes, I think we've hit all the original goals. :)

I've thought about looking for a more efficient DC power supply as a next direction of effort but I'm not QUITE convinced it would be worthwhile unless the cost didn't go up much. These $15 power supplies are maybe 80% efficient. A 10% increase to 90% would net maybe $20/year savings in electricity for a rig the size mine will be (50w savings for 10h/day at 12 cents per kwh). It strikes me that LEDs are just so bloody efficient as is, that we're at a point of diminishing returns. Saving 10% on a 1200w MH rig would be significant. Saving 10% on a 400w LED rig isn't such a big deal.
 
Confuses

Confuses

I am rereading the data sheet. The CAT4101 data sheet says
Up to 25 V operation on LED pin
Click to expand...
I am wondering if we read this wrong. We read this as the supply voltage. I think that the pin can be at 25 volts - not that I would want to do it at our currents. A little later in the description
The LED pin is compatible with high voltage up to 25 V, allowing the driving of long strings of LEDs.
Click to expand...
Well six LEDs (at least in my mind) is not a long string (look at the meanwell). So my question is could we supply 50 volts and as long as the string drops at least 25 the chip would function.

DWZM, you refer to a minimum .5 volts + output voltage for the input. Basically the LED pin needs to be at .6 volts. Sorry I am not finding this in the data sheet, but as you see I am rereading. Could you point me to it. So rephrasing the question if we keep the LED pin at .6 volts what is limiting the LED string length and hence the supply voltage.

I figure I must be missing something, but I am not finding other than the "typical Application Circuit" on page 1.

Also page 4 has a LED current Change vs. Temperature Chart that goes to 120 Celsius so they must be able to run pretty hot.
 
TheFishMan65 said:
I am rereading the data sheet. The CAT4101 data sheet says
I am wondering if we read this wrong. We read this as the supply voltage. I think that the pin can be at 25 volts - not that I would want to do it at our currents.
Click to expand...

I agree, we may be misinterpreting that. I suppose there's one way to find out. . . :lol: The caps I'm using are 50v caps so it shouldn't be a problem to have a longer LED string. At some point though it becomes a tradeoff of spending money on a higher-voltage DC power supply vs. spending money on more drivers. In terms of watts per dollar, the 24v 6.5a supplies from MPJA are dirt cheap.

If I had a higher-voltage DC supply handy, I'd try a longer string. I think you're right because later in the datasheet there's reference to the voltage on the LED pin and it's CLEAR that they're talking about drop across the chip, not supply voltage for the LED string.

DWZM, you refer to a minimum .5 volts
Click to expand...

First sentence in the datasheet, and backed up in the electrical specs table:

The CAT4101 is a constant−current sink driving a string of
high−brightness LEDs up to 1 A with very low dropout of 0.5 V at full
load.
Click to expand...
 
Yes I finally found it in several place. I reread while I had a chat session with them. The first answer was figure 1 shows only 25 volts. I referred to the power dissipation which is a function of voltage and current on the LED pin. I got him to admit it is confusing. So he is forwarding the question to a product engineer. So I will post when I get more info.
 
Waiting to hear!

In the meantime I realized I forgot a BOM for the EAGLE project I put up for the CAT driver last night so just added it to the google site, and here it is in image form:

cat4101_3_BOM.gif


It feels like cheating to have a BOM with so few lines. :lol:
 
TheFishMan65,

"Well six LEDs (at least in my mind) is not a long string (look at the meanwell). So my question is could we supply 50 volts and as long as the string drops at least 25 the chip would function."

Here is the part you are missing: You could string together 100 LEDs & drive them at 300V and the driver ( on the low end of the string) should be fine.

However, the minute you pulse that PWM pin to turn off the string, the FET inside now has to tolerate the entire 300V !!!

As long as you had a relay to engage/disengage the PS and fault protection circuitry to protect the driver from these high voltages ( and NEVER tried to dim it ) , you might be able to make it work.


I have stated before on another thread that an IRF510 MOSFET can give us all we need for an LED driver up to 100VDC but the circuitry required to make sure you never blow up the string is non-trivial.

You CANNOT use software to regulate current in these applications as there are too many fault cases or boot issues to contend with.

We could make these drivers bullet-proof if there was something like "infallible software that boots instantly" And I'll sell you a block of un-obtanium....;-)

Stu
 
Gee stu you put that so clearly. Why couldn't their tech support say that. Thanks for making it simple. I just never took dimming into account!
 
DER or anyone else, on the dual driver version I notice that the D1-D4 have a line on one side of the silkscreen diagram. Is there a direction that these need to go:

D1 36v zener
D2 1A 40V schottky
D3 36v zener
D4 1A 40V schottky

Here is the board again:

LB1.jpg


Also how were you guys connecting the leads from the LED array and PS to this board. Seems a bit tight.

Down to the last bit of soldering and should be going live tonight or tomorrow. Hopefully tonight as a third of tank as been in the dark for a bit.
 
Last edited:
Shack, yes it matter, but I can't remember which way is which. Try looking online, but don't solder till you get it figured out. Sorry I am not more help, but I hope I am at least keeping you from having to take them out :)
 
From what I read online:

"A Zener Diode is a special kind of diode which permits current to flow in the forward direction as normal, but will also allow it to flow in the reverse direction when the voltage is above a certain value - the breakdown voltage known as the Zener voltage."

So I am holding off until I can either figure it out or someone chimes in. Hint Hint :)
 
After doing a bit more searching would the black band being the Cathode end be correct?

"It can be seen that the band around the package corresponds to the line on the diode circuit symbol and this can be an easy way of remembering which end is which. For a Zener diode operating in its reverse bias condition the band is the more positive terminal in the circuit. "

zener-diode.gif


0_5w_Series_Glass_sealed_Zener_Diode.jpg
 
The band on the physical part should go towards the line on the silkscreen - that's the cathode end.

FWIW, In this application, the schottky diode is to act as a one way gate basically where the output of the IC's switch and the inductor meet. The zener diode is a clamp on the output in case the LEDs aren't connected or fail open - without that there for protection, the IC would get no current feedback and boost voltage until it killed itself.
 
Ok, so here is the dual driver board already to be wired to my arrays:

Thanks DMZ for giving me a hand with the last of it.

ldriver.jpg


Also so that I am reading this corrrectly:

+L2 refers to the positive connection of the 2nd string
-L2 refers to the negative connection of the 2nd string
P2 refers to the PMW control from the Arduino

+L1 refers to the positive connection of the 1st string
-L1 refers to the negative connection of the 1st string
P1 refers to the PMW control from the Arduino

VIN refers to + Voltage in
GND refers to - Negative in

If I don't have this hooked up to my Arduino, will it simply power the LEDS or do I need to connect it?

Exciting to say the least.

Can't wait to get it connected.
 
You have the connections all correct. If you leave the PWM pins floating (not connected to anything) it'll be a "zero" signal and the drivers will just pass the input voltage (minus about half a volt of drop across the circuit). It's a fine way to test if you made any glaring errors but since the chip is "off" the circuit won't actually be running.

If you want to test without a PWM signal you can just connect a positive voltage to the PWM pins. Check the datasheet as I think it'll take a pretty large range.
 
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