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I trying to build a UV exposure box so I can do my own PCB at home. The problem is I don't have a power brick that can output 32V @ 1A. the matrix consists of 10 columns \$\times\$ 10 rows.

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The power brick am using is 20V @ 4A. Be much easier to buy one but I thought this might be good exercise to do.

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Where are your current-limiting (and sharing) resistors? – Connor Wolf May 28 '12 at 9:40
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@Fake - they were out of stock. – stevenvh May 28 '12 at 12:57
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Why not just build a Constant current driver. That is the way led are driven these days. – user10030 May 29 '12 at 18:51
@Richard - I see this was an answer first, and you're lucky, because I would have downvoted it :-). A constant current driver doesn't answer OP's question. You'll still have to deal with a supply voltage, and the 20V still isn't sufficient. You also don't explain what to do with the 10 parallel branches. The upvote for your comment is ridiculous. – stevenvh May 30 '12 at 4:11
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@kevlar1818: What's hostile about a downvote if the answer is no good? With a constant current driver alone the circuit won't work! 20 V isn't enough. Also just applying 1A from a constant current source won't guarantee that all branches get the same current. This would have been a bad answer, and now is a bad comment. – Federico Russo Jun 13 '12 at 5:23
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1 Answer

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The simple solution is to rearrange your LED matrix into 20 x 5 LEDs. In theory strings of 6 LEDs would also do for a 20V supply, but it would only give you 0.8V to control the current and that's too little, a small variation in LED or supply voltage would give a large change in current. So calculate the series resistor (one for each string, it's not on your schematic) as R = (20V - 5 * 3.2V)/(LED current).
If the current is 100mA (you talk about 1A for 10 strings) R would be 40\$\Omega\$. Make that 1W resistors. Total power is 20V \$\times\$ 20 LED-strings \$\times\$ 100mA = 40W.

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