Ok. Assuming I up the voltage to 12 VDC, and use 36 series strings of 4
LEDs, what value/type of series resistor should I get for each string?
Vf = 3V (2.8-3.8V range)
Forward Current: 30 ma
I know this isn't the most efficient way, but its better than 144 resistors,
and I only have a choice of 5V, 9V, and 12V for power. I'm not familiar with
a regulator or DC-DC converter that would efficiently up 12V to something
like 24V or higher. I'm open to better configurations if anyone has any
ideas.
---
If you wanted to run it off the 120V (?) mains and you could assure that
no one could come in contact with any of the wirng you could do this
twice, which would get you down to two resistors:
ACIN>---[R]--+--[36LED>]---+
| |
+--[<36LED]---+
|
ACIN>----------------------+
36LED is a string of 36 LEDs wired in series, with two strings wired in
anti-parallel so that each string lights alternately as the polarity of
the input voltage reverses. Determining R analytically is tricky
because the LED array isn't ohmic, so I just scaled the array and ACIN
down by a factor of 9, built it, and adjusted R until I got about 30mA
RMS through it:
-->| |<---1.7VRMS
13.3AC>---[47R]---+--[4LED>]---+
| |
+--[<4LED]---+
|
13.3AC>------------------------+
<--28.5mA RMS-->
Scaled back up it should look like this:
120VAC>---[430R]---+--[36LED>]--+
| |
+--[<36LED]--+
|
120VAC>-------------------------+
With approximately 28mA flowing through the 430 ohm resistor it should
dissipate about 337mW, so I'd use a standard 5% 430 ohm 1/2 watt
resistor. There is one other problem to consider, and that's line
transients. While they'll be attenuated by the 430 ohm resistor, a sure
way to get rid of them would be to split the series R and use a couple
of Zeners in series opposition to soak it up, like this:
R1 R2
120VAC>---[220R]-+-[220R]--+--[36LED>]--+
| | |
| +--[<36LED]--+
[TVS] |
| |
| |
120VAC>----------+----------------------+
Since you'll have 36 LED's in each string they'll be dropping 126V peak,
and since 120VRMS ~ 170VP, the peak voltage at the junction of the
resistors will be 170 - ((170-126)/2) = 148VP with nominal (120V) mains.
For high line (132V) the voltage at the junction will be 187 -
((187-126)/2) ~ 163V, so you'll want to get something with a breakdown
voltage higher than 163V at the low end of its breakdown voltage
tolerance.
If you get, say, a 1000V spike across the mains and the TVS is rated for
180V, R1 will be dissipating (1000V-180V)²/220R ~ 3000 watts (YOW!!!!)
That's not as bad as it sounds, though... Since 1 watt is 1 joule per
second, if the spike lasts for 1 millisecond the resistor will only be
as hot (after 1 second) as if was dissipating 3 watts, _but_ it needs to
be something that can handle the pulse. Carbon composition or some sort
of bulk resitor (like carborundum) would work. Also, for a 1000V spike
the current which would be flowing through the TVS would be
(1000V-180V)/220R ~ 3.7A and it would be dissipating 670W during the
time of the pulse, so you'd need to get something that could handle
that.
If you're interested in following this approach but you need isolation
from the mains, you could use a little isolation transformer to do the
job.
http://www.premmagnetics.com/pdf/Page9.pdf