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Mr. Vogan,
Many thanks for the link to www.superbrightleds.com !
I think this is a great idea. For those who want to make their own ganged
LED clusters, but aren't sure how to size the current limiting resistors for LED
circuits, perhaps these few thoughts and comments may be useful:
1) LEDs are indeed a diode. If they
go into the circuit backwards (reverse biased) they won't light up. To
install them properly, the anode (+ terminal) should be connected to the + side
of the battery pack. The kathode (- terminal) should be attached to the
current-limiting resistor, and the other end of the resistor goes to the
negative side of the battery pack. Typically, the Anode is the long leg of
the LED, and the kathode is the short leg.
2) LEDs (like all diodes) have a fixed forward
voltage drop, regardless of the total voltage of the circuit they're installed
in. To find the exact forward voltage drop of the LED (Vf) you need to
read the specs on the package. However, it's almost always somewhere
in a range from 1.2V to 2.0V.
3) LEDs also have a definite current they
are rated at. At lower currents they will still light up, just not
as brightly. At higher currents they will also light up, and then
burn out (or blow up) depending on how high the current is. Typical
LEDs are rated for about 20mA (.02A) of current. The purpose of the
current-limiting resistor is (as the name implies) to limit the current to 20
mA.
4) LEDs can be connected in series, anode to
kathode, anode to kathode, etc. When series connected, their foward
voltage drops are added together, and the string is treated like a single
component needing only a single current-limiting resistor.
5) LEDs can also be connected in parallel by
tying all the anodes together. In this case, each LED needs it own
current-limiting resistor.
6) To find the value of the current limiting
resistor, first subtract the forward voltage drop of the LED from the total
circuit voltage. The difference will be the voltage drop on the
resistor. Then, keeping in mind that you want this resistor to limit
current flow to 20 mA (.02A), just use Ohm's law to solve for the resistor
value...R = E/I.
Example: A single LED in a 36V
circuit...36V(source) - 2V(LED drop) = 34V (resistor drop)...R = E/I = 34V/.02A
= 1,700 ohms (or as close to that as you can get.
Example: Six LEDs connected in parallel in a 36V
circuit...this is exactly the same problem as the first example, except all six
LEDs will need their own 1,700 ohm resistor (or, again, as close as you can get
to that).
Example: Six LEDs connected in series in a 36V
circuit...LED drop = 6 x 2V = 12V...36V (source) - 12V (LED drop) = 24V
(resistor drop)...R = E/I = 24V/.02A = 1,200 ohms (1.2Kohms)...a single 1.2Kohm
resistor is all that's needed.
Thanks again, Ralph. I'm
definitely putting LEDs in my E-16.
Tim Wilhelm
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