Newbie LED question

The LEDs will want a nominal 20 mA of current and will drop a nominal

3.2V at that drive. You could put two LEDs in series with a 330 ohm resistor in series with each pair. That would give you about 17 mA through the LEDs (should be plenty) and you'll drop about 100 mW in the resistor, so a common 1/4 W 5% 330 ohm resistor (available at Radio Shack) will be fine.

.------o------------. | | | | .-. .-. o | | 330 | | 330 .-----. | | | | | | '-' '-' | + | | | | | | | |12 V | | | | | V -> V ->

| - | - - | | | | | | | | '-----' | | o V -> V ->

| - - | | | '------o------------' (created by AACircuit v1.28.6 beta 04/19/05

If the LEDs are separated enough that you'll want them wired individually, then use each LED in series with a 470 ohm resistor and drive the four sets in parallel from the battery.

--
Rich Webb     Norfolk, VA

--
No.

Thanks for your reply.

Not really sure how to read your diagram. I think it is saying: +12 to 300ohm to LED1 to LED2 to gnd +12 to 300ohm to LED3 to LED4 to gnd (basically group1 in series through resistor to, group2 in parallel through second resistor) the vertical lines (2nd, 4th row) are confusing me.

And yes, I would love to know why?

Thanks again.

Assuming the transformerless 12 volt adapter is actually 12 volts, then your solution is good. Common 12 volt transformer types will almost certainly be under loaded, in that case the actual voltage will be 14 volts plus. That permits all 4 LEDs in series and a single 100 ohm resistor will likely work fine. For the OP, each parallel LED needs one resistor to limit the current. 12 volts would cause too much current for the LED, and it would burn out. So you add the dropping resistor. Since the LED drops 3.2 volts, each resistor will drop 8.8 volts. (12 minus 3.2) Since voltage divided by current equals the desired resistance, then 8.8 divided by 10 milliamps equals 880 ohms. 10 milliamps will give good brightness without risking the LED. Each LED can be made a different brightness by varying the resistance to each LED. You could try 1000 ohm resistors (or higher) to see how changes in resistance change the brightness. Don't go below 680 ohms if you use one resistor for each LED.

Thank you

If you mean the ones directly under the W and the T in "wall wart," they're just the sides of the box that the 12V supply is in. All the rest are wires.

LEDs need current limiting, because their voltage/current curve is ridiculously exponential - a very small change in voltage can cause a dramatic change in current. If the current is regulated (by, in this case, the 300 ohm resistors), then the voltage will be whatever it needs to be at that forward current.

The reason he's got two LEDs in series is so that there's less power dissipated in the series resistors.

Hope This Helps! Rich

--- My pleasure. :-)

sorry about the garbled diagram; here it is corrected:

. WALL-WART . +-------+ .MAINS>--|~ +12|---+-[300R]--[LED1>]--[LED2>]--+ . | | | | . | | +-[300R]--[LED3>]--[LED4>]--+ . | | | .MAINS>--|~ 0V|-------------------------------+ . +-------+

The "vertical" lines in the second and fourth row are braces/brackets and I like to use them to bound components.

The circuit is connected the way you read it, and the cathode ends of the LEDs are identified by the ">" inside the brackets.

The tech stuff:

LEDs are basically just diodes and, once they're forced into forward conduction in the region of the knee, a very small change in voltage across the diode can/will result in a very large change in current through it.

All LEDs have negative temperature coefficients of resistance, so if they're driven with a stiff enough voltage source, as they warm up their resistance will drop, allowing more more current through them, which will heat them up more and more and allow more and more current through them, and on and on, until the LED gives up the ghost and releases the magic smoke.

In order to keep that from happening, what's been done is that the voltage band across an LED is specified with a certain current through it, as specified on the LED manufacturer's data sheet.

Knowing the width of that band, we can then approximate a constant-current source which will keep the LED from committing suicide.

From the data sheet you provided, the manufacturer hasn't specified a band but, we do have that, with 20mA through the LED, the voltage dropped across it will be 3.2 volts so, two in series will drop 6.4 volts.

Hence, if we have a stiff 12V supply and we want it to supply 20mA to a load which will drop 6.2V with 20mA through it, we must supply the means to limit the current into the load to 20mA while dropping the unwanted 6.4V from the supply.

We can do that by subtracting the load's forward voltage from the supply voltage and then dividing by the current into the load:

Vs - Vl 12V - 6.4V R = --------- = ------------ = 280 ohms Il 0.02A

and interposing that resistance between the 12V source and the 2 LEDs in series load.

300 ohms is a standard 5% value, and will be fine in this application.

-- JF

Got it (some anyway)

--
I probably clouded the issue with too much irrelevant data; sorry
about that.

What is it that's not clear to you?

AACircuit... interesting... Thanks!

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