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From: DaveH, 8-22-00
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>For parallel hookup the ohm calculation works out this way,
(voltage source - LED voltage )/ LED current. White LED's are usually rated at 20 ma, but some products their driven at 50 ma, so lets set the LED current at 30 ma.
So substituting in, (6 - 1.7)/.03 = 143 ohms more or less. You can probably get 150 ohm resisters at radio shack in a pack of 6 for $1.<HR></BLOCKQUOTE>
From: Doug P. 7-17-01
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>((Voltage supplied)-(Voltage required)) / Current = Ohms resistance
(Vs-Vr) / I = Ohms
so to apply to the above problem
if Vs = 4.5
and Vr = 3.6 (single LED or Paralled LEDs)
and I = .020 Amps
then you would need a (4.5 - 3.6)/.020 resistor = 45 ohms if you play "by the rules" and keep it in the factory specs for Nichias (if that's what you're using)
The LED will dim when the battery gets weak no matter what you do. The resistor just "uses up" the extra voltage that would destroy the LED. Essentially the resistor (in the example above) becomes a little tiny .9V heater that you run whenever the LED is on, wasting the .9V as heat but protecting the LED. If you want, you could build a switch into the circuit that jumps the current around the resistor when the LED starts to get dim.<HR></BLOCKQUOTE>
Thanks to both!
From: DaveH, 8-22-00
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>For parallel hookup the ohm calculation works out this way,
(voltage source - LED voltage )/ LED current. White LED's are usually rated at 20 ma, but some products their driven at 50 ma, so lets set the LED current at 30 ma.
So substituting in, (6 - 1.7)/.03 = 143 ohms more or less. You can probably get 150 ohm resisters at radio shack in a pack of 6 for $1.<HR></BLOCKQUOTE>
From: Doug P. 7-17-01
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>((Voltage supplied)-(Voltage required)) / Current = Ohms resistance
(Vs-Vr) / I = Ohms
so to apply to the above problem
if Vs = 4.5
and Vr = 3.6 (single LED or Paralled LEDs)
and I = .020 Amps
then you would need a (4.5 - 3.6)/.020 resistor = 45 ohms if you play "by the rules" and keep it in the factory specs for Nichias (if that's what you're using)
The LED will dim when the battery gets weak no matter what you do. The resistor just "uses up" the extra voltage that would destroy the LED. Essentially the resistor (in the example above) becomes a little tiny .9V heater that you run whenever the LED is on, wasting the .9V as heat but protecting the LED. If you want, you could build a switch into the circuit that jumps the current around the resistor when the LED starts to get dim.<HR></BLOCKQUOTE>
Thanks to both!