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Super cheap Dimmable LED Driver (cost under $3)

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    Posted: September 13 2016 at 7:21pm
Another thought is: Why not go for the cheapest option? Even if it burns up prematurely, if it cost you half or less, you can spend the difference on the next light 2+ years later. And by then there is new technology, so you get even more for your money. I've had great success so far with cheaper options and it's fun to experiment.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote MadReefer Quote  Post ReplyReply Direct Link To This Post Posted: September 06 2016 at 7:20pm
You are completely correct, David. But the voltage has to remain steady. I found a power supply for a sensitive electronic that provides 9v up to 1A and I use it for a 10w LED directly hooked up. If the voltage is too high, the current jumps up really fast. What started as a simple test (the 10w light above) has become a useful light. It is the only light in my house that I can read tiny numbers from tiny capacitors and other electronics.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Krazie4Acans Quote  Post ReplyReply Direct Link To This Post Posted: July 15 2015 at 7:35am
^^^^ Agreed. Plus the LDD's are much more efficient and will automatically regulate upto 56v and will hold the rated output current rock steady. Not much use for 317's in the LED driver world.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Fatman Quote  Post ReplyReply Direct Link To This Post Posted: July 15 2015 at 7:21am
This is a long dead thread, this works but it's not very efficient. There are better ways to build a driver, but at this point I'm not seeing a need to build one when you can get Meanwell LDDs for $7.00. Building a more efficient one than the one based on the 317 IC is not too expensive, but will still be very close to the cost of a Meanwell.

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Post Options Post Options   Thanks (0) Thanks(0)   Quote jonas Quote  Post ReplyReply Direct Link To This Post Posted: July 09 2015 at 12:03pm
The original schematic, posted by Seti007 is not correct.
In order to build a constant current driver, the resistor must be inserted between the Vout pin and the load.

The one proposed by Davidwillis is correct.

Hope this helps.


Edited by jonas - July 09 2015 at 12:10pm
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Post Options Post Options   Thanks (1) Thanks(1)   Quote Krazie4Acans Quote  Post ReplyReply Direct Link To This Post Posted: April 20 2013 at 8:47pm
That's what I thought makes sense
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Post Options Post Options   Thanks (0) Thanks(0)   Quote chris.rogers Quote  Post ReplyReply Direct Link To This Post Posted: April 20 2013 at 7:30pm
Yeah, as best as I can tell, this is what's going on.  Let me know if any of that doesn't make sense.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote chris.rogers Quote  Post ReplyReply Direct Link To This Post Posted: April 19 2013 at 10:01am
But from what I can tell, essentially what you are saying is correct.

R1 is a fixed resistance, R2 is your pot of varying resistance.

The circuit will see R1 and R2 as one resistance.

When R2 is set to minimum resistance, the R value for use in Ohm's law to calculate the current is going to be essentially R1 plus whatever real-life minimum R2 value produces.  (In theory a pot goes from 0 to max, but in practice it's not always the case.)

When R2 is set to maximum resistance, the R value for use in Ohm's law would be R1 + R2(max).

So if you're curious what would happen, you'd measure the resistance across the pot in both min and max, substitute in 0 for R1 (if you were to remove it) and do the math.  Odds are the current you'd produce according to Ohm's law would saturate the regulator and burn it out, as you'd essentially be shorting the two pins.

So you provide the regulator with a minimum resistance, R1, which would be your max brightness.  So you pick your max current you want to provide, set it to I, you have your supply voltage and then you'd solve for your R1. 

Then you select your pot.  Say it's 5Ohm.  So now you plug in R1 + 5, keep your reference voltage, and solve for a new I.  That will be your "dimmest" current.

Make sense?

(I'll google and verify whether or not I'm full of it.  I'm full of it a lot.  Watch thread for my future posts containing errata.  )
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Post Options Post Options   Thanks (0) Thanks(0)   Quote chris.rogers Quote  Post ReplyReply Direct Link To This Post Posted: April 19 2013 at 9:55am
Krazie,

It's been a long time since I had my analog circuits class and outside of dinky lab stuff I've never done what you're asking in person.

I've been thinking about it and when I have a minute this weekend I'll google and see what I can find, unless, of course, someone else can handle it.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Krazie4Acans Quote  Post ReplyReply Direct Link To This Post Posted: April 12 2013 at 7:49am
Seti007,
       So I have one question that I don't see answered in your thread. In your picture for the dim-able configuration, you show two resistors (R1 and R2), I am presuming that the value of R1 is the calculated value to set the maximum constant current output of the LM317 and that R2 is additional resistance to reduce the current output to dim the lights.

     Is that correct? Without R1 would the output runaway to the maximum current of the power supply if the POT was turned down below ~2Ohms? Just curious what the result would be?

Sorry to throw a new question on an old thread. lol

   Krazie
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Post Options Post Options   Thanks (0) Thanks(0)   Quote chris.rogers Quote  Post ReplyReply Direct Link To This Post Posted: March 19 2012 at 10:19am
The longterm savings come from fine tuning the LEDs to max efficiency of power consumption and efficiency between source and load.

You're probably not over-driving your LEDs, but if your power supply begins to wear out your LEDs will take the hit, where if you had a constant current circuit, it would protect the load.

I'm not saying you're headed for ruin.  Keep us posted.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Davidwillis Quote  Post ReplyReply Direct Link To This Post Posted: March 18 2012 at 5:58pm
That is an interesting read.  However the data does not match very closely to my LED's.  Possibly cheaper LED's vary more (or maybe I control my temperature better, or that I run my LED's on the low end of the maximum power).

Anyway, from my readings, my current increases with temperature from cold to hot (reading after 8 hours on).  My increase in current is about 15%.  I start out with my LED's running at about 50%, and after fully heated, it is running at about 65%.  So I don't feel there is any risk of running too much current.  Also, since I have a trimpot on the power supply, I can still have complete control over the color output.  Sure it varies a little with the increase of temperature, but it is not noticeable.

I don't see how avoiding the constant current supply negates both long term savings or controllability.  Sure, I can't push my LED's to the limit with out the risk of running too much current through them, and shortening their life, but if you don't plan on maxing out your LED's then I don't see how there is a need.

Again, I may be completely wrong and my LED's may not last how I am running them.  So I am not advising anything, just sharing my experience so far (which isn't much).

Thanks for the article.


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Post Options Post Options   Thanks (0) Thanks(0)   Quote chris.rogers Quote  Post ReplyReply Direct Link To This Post Posted: March 18 2012 at 4:09pm
David,

I've been thinking a lot about the setup you just tried, LEDs with just a power supply and then matching the Vf * I drops across your LEDs.


There are two important takeaways:
1.) Avoiding max current rating of the LED
2.) Making sure you control color output and luminous intensity

I don't know just how much it'll vary without the constant current supply, I imagine a lot of that depends on your supply.  But if we're buying LEDs because of the controllability of light, and longterm savings from efficient use, avoiding the constant current supply seems to negate both of those benefits LEDs bring.  
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Post Options Post Options   Thanks (0) Thanks(0)   Quote chris.rogers Quote  Post ReplyReply Direct Link To This Post Posted: March 10 2012 at 11:54am
I don't think they always assume you're using a regulated power supply.  If the power supply says it's made for LEDs there may already be some current regulation included in there.

At the very least, something *must* limit current or you will burn out the LED.  The LEDs you get at Radio Shack are rated from 20mA to 60mA, and the ones in this thread between 500mA to 1A.  So if you strung up enough LEDs to have the voltage drop match the output voltage of the power supply, and something internal is regulating the current to 2A, and you have two separate 1A strings, and there's no fluctuations in visible light and nothing is burning out, then yes, your power supply is doing something to regulate the current.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Davidwillis Quote  Post ReplyReply Direct Link To This Post Posted: March 10 2012 at 11:02am
I wonder if anyone has done any testing.  There must be some reason why they say to run a constant current driver.  But mine works just fine (as far as I can tell) without one.  The power supply does say it is made for LED's.




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Post Options Post Options   Thanks (0) Thanks(0)   Quote chris.rogers Quote  Post ReplyReply Direct Link To This Post Posted: March 09 2012 at 4:36pm
David,

All you're doing is running the LEDs without a constant current driver.  I'm not sure what effect they'll have longterm on the LEDs.

Just speculating here, but I imagine that if your power supply dies or slowly fades the fluctuations in the light will be noticeable, or you might overdrive the LEDs and burn the LEDs out rather than the driver.  Another possibility is that the LED life is shortened, even if there are no fluctuations right now, none noticeable to your eyes (and we can only perceive about 50 Hz), that is.

I've noticed a lot of LED sources are selling 1W LED solutions.  Anyone have reasoning as to why 3W are better, assuming SPS/Clams/Anemones are wanted in the tank?
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Post Options Post Options   Thanks (1) Thanks(1)   Quote Davidwillis Quote  Post ReplyReply Direct Link To This Post Posted: February 25 2012 at 10:46am
I am trying an even easier and cheaper method, and it seems to work fine so far.  Maybe those of you that know more than me can let me know if this will cause problems down the road.  I just hooked two strings into a 2a 24v power supply (just like the one linked to in my post above), no chip, or potentiometer, just directly into the power supply.  You have to get the correct number of LED's, but then you can use the trim-pot to dim.  Works great, more efficient (no heat loss in the potentiometer or LM317.   Also you could use any voltage (adjust your string length to match or get the voltage to match the string length you need).
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Post Options Post Options   Thanks (0) Thanks(0)   Quote rbtron Quote  Post ReplyReply Direct Link To This Post Posted: February 25 2012 at 10:25am

The LM317HV will work up to 57V.

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Post Options Post Options   Thanks (0) Thanks(0)   Quote easydo Quote  Post ReplyReply Direct Link To This Post Posted: February 21 2012 at 9:14pm
Hey guys.  Great thread, lots of useful info. I have a bit of a problem though. I have a 48v power supply, but it seems that the max output is for the LM317 is 37v. Is there a similar chip to the LM317 that I can use that would allow me to use the 45 or so volts output I would have, or perhaps a different workaround for this problem? Thanks.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote wick246 Quote  Post ReplyReply Direct Link To This Post Posted: November 21 2011 at 10:40pm
Thanks David, that helps. I need to read through the thread again about dimming - if I use the dimming version of the drivers does each driver need a seperate   variable resistor or can all the drivers for each color led use one variable resistor? Also when shopping for leds some are mounted to smaller boards than the 21mm stars and are a little cheaper - would these still be suitable for reef lights if properly mounted for heat transfer?
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