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AC to 12V to multiple Buckpucks?

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thepaan

Member
I'm hoping this is a basic question for someone.

I'm planning to build a light using an AC adapter as my DC source. I want to hook it up to a buckpuck (led driver to supply the correct current) which will then drive the LEDs. I can run them all at 700mA forward current but the problem I have is there are multiple forward voltages. Some are 2.35 and some are 3.5. I had the idea to use multiple Buckpucks but then I wasn't sure if the AC adapter was powerful enough.

If I hook up 1 LED then it draws 700mA and 3.5V. So my source has to also provide at least 700mA? If I hook 3 of them in series I know that the forward current is still 700mA for all 3 but the voltage is multiplied; 3.5*3=10.5V. If I add a second string of 3 I know I still only need 10.5V to drive all 6 but how much current is needed? is it still 700mA. Take this to the next step and drive two buckpucks with a 12V AC adapter. Is 12V/750mA/15W enough to drive say 2 series of 3 leds on each Buckpuck? What is the math you use to figure all that out?
 

Chippie

Member
I'm hoping this is a basic question for someone.


If I hook up 1 LED then it draws 700mA and 3.5V. So my source has to also provide at least 700mA? If I hook 3 of them in series I know that the forward current is still 700mA for all 3 but the voltage is multiplied; 3.5*3=10.5V. If I add a second string of 3 I know I still only need 10.5V to drive all 6 but how much current is needed? is it still 700mA. ?
No the load current is doubled as you have added a second set of leds...
 

Hero999

Banned
How much current is drawn by the buck pucks?

Because a buck puck is really a SMPs, the input current will be less than the output current. With a single 3.5V LED connected the input current will probably be about 245mA, with two 3.5V LEDs in series the current will probably be 490mA.

What's the maximum output current from the adaptor? It might 15W but are you sure that isn't the input power taken from the mains?
 

thepaan

Member
The input voltage I was looking at is either 12 or 24 since I'm not sure how many LEDs I can hook up yet. This specific 12V model I was looking at is 18W so 750mA. The LEDs all pull 700mA. I did some more reading since I posted this and found out what Chippie said that you have to add the mA for each series to get the total draw.

I'm more concerned with the two pucks (output whatever the voltage needed at 700mA. I didn't see anywhere the input current specification so I'm assuming it will draw as much as it needs up to the max the AC adapter will put out... and that is what I'm worried about. If I hook the AC adapter to two pucks, how do I calculate the draw since there are two different voltages? I can add all the amps together to see if that is fine but what about the voltage? Do I add the voltage 3.5 and 2.35 together?
 

Hero999

Banned
Sorry I find your verbal description confusing, could you please upload a schematic.

Yes, a buck puck will draw as much current as it needs. I'm not sure if you understood my point about it being a switched mode power supply and that the input current will actually be less than the output current if there are few LEDs connected to it. My advice is measure the current drawn with different numbers of LEDs connected.

The total voltage drop is equal to the voltages of all the LEDs connected to it. There will be a limit to the maximum output voltage of the buck puck, probably a couple of V below 12V so you might not be able to power three 3.5V LEDs with one.
 

thepaan

Member
I don't mean to be confusing so I attached the schematic and thank you for your patience.

I understand how switched mode power supplies provide power only in bursts measured in maybe nanoseconds instead of 100% of the time but I can't find any documentation on that feature for the Buckpuck. I'm still planning so I cannot test any of the equipment yet as I have not yet purchased it. The data sheet says there is only a limit on the input voltage (5-32V) which in turn must limit the output voltage so I think a single unit can easily power more than 3 LEDs.

Due to the nature of the buck regulator, the input voltage must always be higher than the total forward voltage drop of the LED junction(s) connected in series (2.0V for DC models ...). Thus, for a series string of six junctions having an average forward drop of 3.5V, the required input voltage will be 21V DC.
The above quote from the data sheet makes it seem to me its like a basic circuit using inline resistors only it needs 2 more volts for it's own functions. The input voltage range is 5-24 and after reading that again I might have to rearrange my series strings. If the output current is fixed no matter how many LEDs are connected then a 700mA puck will split that current between 2 strings for 350mA (I want at least 700 per string not 700 total).

But back to the question. In the diagram, how would I figure the minimum voltage/current needed by all pucks?
 

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Hero999

Banned
A buck regulator is a kind of switched mode power supply.

Easy, just add 2V onto the total LED voltage drops, so for the 14V and 9.4V you need a minimum of 16V and 11.4V respectively.

The LEDs will probably not completely matched so they won't share the current equally. Although connecting them as strings of many helps, you may need current balancing resistors to help split the current.

What's the maximum current rating of the LEDs?

You shouldn't run them continuously at their absolute maximum rating, especially in this case where poor current sharing could result in one group exceeding the current rating and burning out. If they're rated for 420mA or more then I wouldn't worry about it.
 

thepaan

Member
In writing my response I just figured this out. For the example I would use at least a 16V AC adapter. The two pucks could be considered a parallel circuit and therefore the voltage is the same for both branches. Since each puck produces 700mA I will assume it also draws that much (I agree that this is probably not the case, but I will be generous just to be sure) which means the AC adapter will have to output at least 1400mA because the current rating for each branch of a parallel circuit are added. Using the rule Watts=Amps.Volts I would require a 16V 22.4W AC adapter.


Thanks for the help :p
 

Hero999

Banned
I would recommend an old laptop PSU, usally 18V >3A and can be purchased cheeply off ebay.
 
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