Your DPDT switch won't work with the resistor your proposing because the LED resistors are in parallel. So if you tried it you wouldn't get 10mA but 20mA spread across the LEDs. Your resistor value will need to be much lower.How are you powering the board?At 2 amps+ wouldn't it be easier to have a regulator produce 3.3v and either switch +5v / +3.3v.
The LED's get 20ma now. DPDT would switch from full power, to double resistance on each line, dropping to 10ma. 150 ohm LED's have an extra 150 ohm in the main power line, 90 ohm LED's have an extra 90ohm resistor in power line. But I don't know whether to get 3 watt resistors (1/4A @ 5V) or 10 watt resistors (1.12A @ 5V).
Quote from: magnethead794 on Dec 21, 2011, 07:15 pmThe LED's get 20ma now. DPDT would switch from full power, to double resistance on each line, dropping to 10ma. 150 ohm LED's have an extra 150 ohm in the main power line, 90 ohm LED's have an extra 90ohm resistor in power line. But I don't know whether to get 3 watt resistors (1/4A @ 5V) or 10 watt resistors (1.12A @ 5V).How many DPDT switches are you fitting? One per LED? If not then this "new" resistor will need to be a lot less than you think.If I just stick to the 39 LEDs you're connecting directly to +5v, you're proposing to connect all the 150 ohm resistors to the centre terminal of a DPDT switch, with one terminal of the switch connected to +5v and the other terminal to +5v via a single 150 ohm resistor?If so, the 150 ohm LED resistors are all in parallel but the new resistor (if it's to only drop 1.5 volts) will need to be 150 / 39 = ~ 3.9 ohms. Since you're dropping 1.5 volts across the resistor it will need to be 0.6 watts (but you could go with a 1 watt to be safe).Iain
Close. IE the 39 all in parrallel to top center postThe other 56 all in parrallel to bottom center post.Both posts on one side have clean +5VOn other side, top post has +5V with 150 ohm resistorBottom post has 90 ohm resistor.Tthe segment needs a daylight (full power) and night (half brightness). The 56 LEDs are 5V -> 90 ohm resistor -> LED -> Darlington array --> groundThe LEDs are rated 2V 20mA. The darlington saturation voltage is 1.2.
How do you figure that? I've not seen the expression ohm/(load quantity) before..?
Quote from: magnethead794 on Dec 22, 2011, 06:19 amHow do you figure that? I've not seen the expression ohm/(load quantity) before..?If you have identical resistors in parallel, the equivalent resistance is divided by the number of resistors.So for two identical resistors in parallel the equivalent resistance is half. For three identical resistors it's a third, for four identical resistors it's a quarter and so on. Your LED resistors are in parallel with this one resistor. If you want to halve the current you need to use the equivalent resistance to all these parallel resistors. This means you need:150 ohm / 39 (identical resistors) = ~ 3.9 ohm90 ohm / 56 (2 chains of 28 identical resistors, all in parallel) = ~ 1.5 ohmFor the power calculation I used P = I * I * R. Using your value of 10mA per LED I got the following:Current through 3.9 ohm resistor is 10mA * 39 = 0.39A.Therefore P = 0.39 * 0.39 * 3.9 = 0.6 watts. I'd get a 1 watt resistor to be safe. Current may actually be higher than 10mA.e.g. Current is actually 11mA per LED and the power dissipated in the 3.9 ohm resistor is 0.72 watts.Current through 1.5 ohm resistor is 10mA * 56 = 0.56A.Therefore P = 0.56 * 0.56 * 1.5 = 0.47 watts. Again I'd get a 1 watt resistor.If you can't get a 3.9 ohm or 1.5 ohm 1 watt resistor you can make them by combining higher value resistors in parallel, but you may not need 1 watt resistors any more. It will depend on the actual values you can buy. Worst case you could wire 39 150 ohm resistors in parallel to make your 3.9 ohm resistor. Each resistor would only dissipate 0.01 * 0.01 * 150 = 0.015 watts.Iain
Quote from: sixeyes on Dec 22, 2011, 10:45 am Your LED resistors are in parallel with this one resistor.No, it's in series....
Your LED resistors are in parallel with this one resistor.
Quote from: magnethead794 on Dec 22, 2011, 11:32 amQuote from: sixeyes on Dec 22, 2011, 10:45 am Your LED resistors are in parallel with this one resistor.No, it's in series....If you don't like my terminology, think about the current. You need 10mA per LED. That's 390mA through a single resistor that needs to drop the same voltage as a 150 ohm resistor drops at 10mA. The voltage dropped by the 150 ohm @ 10 10mA is 1.5 volts. Using R = V/I you get 1.5 / 0.39 which yields 3.9 ohms. You get exactly the same result, so you can consider the LED resistors to be in parallel (from an analysis point of view).The end result is that the resistor values I've quoted are correct and if you're still having a hard time believing it, use a 150 ohm resistor instead. You won't burn it out but each LED will get about 0.5mA each. They might glow dimly but it won't be 10mA per LED that you're hoping for.Iain
Then I may need less than half current...I just put a 150 on the power line, and it's the exact brightness I need.
So if it's dropping 2.9 volts, I is 19.3 mA ,
Passing 2.1 volts to the LED resistor, which is then dropping current to 14mA. So how many volts are being dropped by the second resistor/how many volts are the LED's seeing?