Electrical Circuit Question

[AndrewBird]

I've got a question for those of you who are electronics smart. I'm trying to figure out what size resistor I would need to dim an LED array. Now I know lowering the forward voltage to an LED isn't the most ideal way to dim one, but it should work ok for the application I'm using it in.

I have an array of 28 LEDs. Normal operating voltage is 3.5V @ 10mA. Supply voltage is 12v, so I'm going with an 820ohm resistor per LED. This supplies 3.8v to each LED, which is within its max limits.

Now, since each LED has it's own resistor, they are wired in parallel. I want to dim this whole array with one resistor instead of having a whole other set (one for each LED) that is a different resistance,

The reason I want a separate resistor is because I want to be able to switch between full power at 3.8v and then dim down to ~3v.

I calculated the total resistance of the resistors in the array, which came to 30.4ohms. I then also calculated what resistor per LED I would need to produce 3V to each LED and came up with 900ohms. Working off of this, the total resistance of that circuit is 32.1ohms, a difference of 1.7 ohms. IS this the size of resistor I need t make this work?

The array would draw 280mA, which at 12v, works out to be 3.36watts. So I would need at least a 4 watt ceramic resistor, correct?

Sorry for the long post and thanks in advance.

[andyman2487]

Maybe it's early or something, but I keep getting ~88 ohms for the resistor in series that you want. I started by finding the equivalent resistance of all those resistors in parallel. Got R parallel = ( 820/28 ) = 29.29 ohms. Then i put that as an equivalent resistor in a simple series circuit composed of a 12V source, the 29.29 ohm resistor, and another resistor of unknown value. You know that the voltage drop across the 29.29 ohm resistor will be 3V and that the current is constant in a series circuit, so the current will be I = V/R = 3/29.29 = .1024A. Since the resistor of unknown value has to drop 9V (9+3=12V) and the current is .1024 A, the resistance would be R = V/I = 9/.1024 = 87.87 ohms. I could be wrong though, I'm only a student.

Edit: And for your initial calculation, you would have to put in a 63.2 ohm resistor instead of an 87.87 ohm resistor to deliver 3.8V to the LEDs.

[benDASHDASH]

Hey Andys,

According to my calculations yer all wrong!: :p

R1 = LED's Dim
R2 = LED's Bright

R1 = V/I = (12-3.0V)/.28 ohms = 32.14 ohms
R2 = V/I = (12-3.8V)/.28 ohms = 29.29 ohms

To simplify the circuit and reduce resistors, I would definitely wire 1 resistor of each value in series with the LED's all in parallel. The LED's are in parallel so they all obtain the same voltage.

[andyman2487]

Where are you guys getting .28 Amps from? RustlerDriver's calculations are fucked so the current value will be different.

[benDASHDASH]

Each LED is drawing .010 A, multiplied by 28 LEDs

[AndrewBird]

I realized yesterday that I screwed up on the current rating. It's 20mA. 10mA was reverse current.

I was talking to a friend, who is an electrical engineering student, and he says since LED's don't dim linearly with voltage, it would be better to calculate the resistor I need based off a given current, not voltage. So, I'm just going to wait until I get the LED's (I haven't ordered them yet) and test them before I calculate anything. I'm not sure how accurate the specs I''m looking at are anyway (eBay ad), so...

Thanks for the help.

[mrsleeve]

I may dumb here but why not put a rehoistat (sp) on the supply line before the resistors to the LED's.