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I am planning to design a LED lamp using a 6V power supply from a lead-acid battery with a rating of 4.5Ah. I want to connect 20 LEDs (LED's rating is 20mA 2V) in parallel to achieve maximum brightness. I calculated the resistor I will use is about 100 ohms to 120 ohms.
Using Ohms law. Since we know the current, source voltage, and LED forward voltage drop, we must calculate for the series Resistor. R = (V Source - V Forward) / I.
(6V - 2V) / 0.02A = 4V / 0.02A = 200Ω
220Ω is the next resistor up.
Now since we have the resistance, we can calculate Wattage of the Resistor. P = V (of Resistor) * I
4v * 0.02A = 0.08W or less than 1/8th (0.125) Watts. A 1/8W Resistor would work.
A better solution is two leds in series, sharing the current.
(6 - 2 - 2) / 0.02 = 100Ω
2v * 0.02A = 0.04W. A 100Ω 1/8W would work.
The ohm's law solution is correct, but using dropping resistors in the first place wastes power, as I^2R, or I*V.
So the first iteration is to put pairs of LEDs in series. Now, we have 6V-4V=2V, to be dropped when 20mA is flowing, or you need a 100 ohm resistor. Power wise, the 100 ohm resistor has 2V * 20mA = 40mW being dissipated in it...you will need to work hard to find a resistor with that small of a power rating, and get out the tweezers to pick it up and a magnifier to see it.
You can also try this with triplets of LEDs, trying perhaps 10 ohm resistors to balance the currents between the triplets. This means you are putting very little power into the resistors. However, you might not get the best results as the battery weakens.
A more sophisticated solution, especially if the triplets of LEDs vary too much in brightness, is to put all the LEDs in a string and use a current-mode DC-DC converter circuit to get whatever battery voltage is needed to drive the single string, or perhaps just a few strings with balancing resistors.
