Quote:
Originally Posted by agutt /img/forum/go_quote.gif
I have found a White LED at radioshack that has a 20mA rating, which is the same as the LED on the BOM. So I should have no problems with this one correct?
The MCD is rated at 1100, and the "Typical Voltage" is 3.6V
I would like to avoid having to swap out my resistors since I have soldered them in already...
Clearly my understanding of this in terms of electronics is poor...any explanation would be a great help!
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The LED positions on the MAX are fed with V+ voltage, whatever you've set for that. Most use the recommended 27VDC. The LED and the LED resistor are in series, so they will share that full voltage. Meanwhile, whatever current flows through them will be the same for both. Ohm's Law establishes this relationship as V = I*R. So, if V is 27VDC, then we can establish "I" by choosing the correct value for R. If we were sizing it for the full current rating of the LED you mention above (it's typical), then we'd use 0.02A (20ma) for "I". Then R = 27VDC/0.02, or 1350 ohms.
Now, we don't really want to set a resistor for exactly that amount, because we can't be too sure of the rest of the components. If that 20ma is an absolute rating and our resistor is 5% low, then more current will flow and the LED might burn out pretty quickly.
That said, there is another factor to consider and that is the power handling ability of the resistor. Power = I^2 * R. For the values we just arrived at, Power = (0.02^2) * 1350, or 0.54W. Unfortunately, resistor power ratings go in fractional increments like 1/8, 1/4, 1/2, and 1W. So, for 0.54W, we'd need to use a 1W resistor. Even then, if anything was the slightest bit off, the LED would still burn out.
Luckily, the brightness curve for an LED is not really proportional. I don't know what it is exactly, but experience seems to indicate you can probably get at least 3/4 of the maximum brightness rating at 1/2 current. If we use 10ma up there as the current - so as to assume a decent safety factor on burning out the LED - then R = 27VDC/0.01A = 2700 ohms.
That's an odd resistor size, so let's pick a 2K resistor and see how that works. I = V/R, so I = 27VDC/2000 = 13.5ma. Hmm not too bad - still a decent safety factor, but with a standard-issue, common resistor size. If we check the power equation, then P = (0.0135^2) * 2000, or P = 0.365W. That's a bit more than 1/4W, but plenty below 1/2W to allow for a decent safety factor.
Hence, we chose 2K - 1/2W resistors for the LED positions.
You can follow the same logic by setting a higher resistor for a super-bright LED in order to knock down the brightness to something reasonable for a panel LED. Let's say we used the 10K resistor. Then we have I = 27/10000, or I = 2.7ma. Checking for power, we'd have P = (0.0027^2) * 10,000, or P = 0.073W. Power should have a much higher safety factor (extreme failure could result in flames), so even though a 1/10W resistor might work, a 1/4W would probably be better.
Hope that clears up some of your question.
