voltage divider or series current resistor

An LED MUST have a series resistor to limit the current or it will burn out in a fee milliseconds. It's not a question of lowering the voltage, it's one of limiting the current to a safe value. LED's need a current of 10-20mA through them to glow fairly brightly, so select a resistor for each LED to supply this.

Correct

When the load uses a constant current a series resistor is enough.

Sort of.

For calculating led circuits, try this program:

It can also do a lot of other electric calculations:

Basic principles:

LEDs produce light in rough proportion to the current passing through them.

LEDs, like most junction diodes drop a voltage that is not proportional to the current through them, the way resistors do. Very small changes in voltage produce large changes in current. So voltage drive is not practical.

Different colored LEDs drop different voltages, with shorter wavelengths generally dropping more. For example, most red LEDs will light with less than 2 volts across them, while blue ones often drop more than 3.

Your method should at least roughly control the current passing through the fairy fixed voltage drop of the LEDs.

The series resistor does that a lot better than the voltage divider. Besides, the voltage divider wastes current that does not pass through the LEDs.

So lets say that you want to light 3 green LEDs that drop about 3 volts each, when their operating current is in the normal range. You select an operating current of about 10 mA (because, say, it is safely below their maximum current rating and you judge, from some experimentation, that this produces a reasonable amount of light). So the LEDs, in series, will drop about 6 of the 12 volt supply. The resistor must be selected to drop the remaining 6 volts while passing about 10 mA. 6V/.01A=600 ohms. The next higher standard 5% resistor would be 620 ohms and the next lower one would be 560 ohms. Take your pick. The resistor will produce some heat, so you might want to check that you have one large enough to cook off that power. You can calculate the power dumped into the resistor either with P=amps times volts, volts squared divided by ohms, or amps squared times ohms. Since we are trying to achieve 6 volts drop while passing .01 amp, the power is 6V*.01A=0.06 watts, so even a 1/8th watt resistor would be large enough.

This is a newsgroup for beginners. Welcome.

While LEDs are voltage controlled devices, the critical parameter is current.

I think you may have gaps in your understanding of LEDs.

True. That's how they work.

The specifications of the LED. Specifically the maximum continuous allowable current.

Interesting question. The answer is no. The LED provides minimal resistance once sufficient voltage to light it is present.

The answer is that an LED (like all diodes) will conduct all available current once they turn on. The problem is because of construction, if too much current goes across, then the diode burns up.

So the purpose of the resistor is to limit the amount of current in the circuit. Off the top of my head Kirchoff's law states that the amount of current across every component in a circuit is equal. A resistor impeads current based on its resistance. The LED is a virtual dead short once it turns on. So when you combine the two together, the amount of current that flows across the LED is the same as the current across the resistor. Hence the name current limiting resistor.

Hope this helps. You don't need a voltage divider. The LED will consume whatever voltage required to light it (its forward voltage drop) leaving the rest of the voltage to the rest of the circuit (which is the resistor).

A quick example. I once modified a cheap motion detecting floodlight for some security testing by connecting the 120VAC output to the LED of an optoisolator. Now since that circuit powered a 150W incandecent floodlight, there was well over an amp of power @ 120VAC available. Definitely needed a current limiting resistor.

The 2V that the LED actually comsumes is negligable. So to simplify my calculation, I used a 130VAC (added some slop) voltage and wanted 15mA across the resistor. So:

V=IR

I think for good measure I used a 10K resistor giving a final current of 13mA. Finally I needed to compute the power

P=VI P=130V*0.013A P=1.69W

So I pulled a 2W 10K resistor for the project.

Finally LEDs (and other diodes) don't care for high reverse voltages either. So I put a second visible LED in reverse parallel to the opto LED. So one was on for half the AC cycle, and the other on for the other half. Plus it gave me a visual indicator of when the motion detector activated.

Note that all of this occured with a 120 VAC input. No need to divide the voltage as long as the current is kept under control.

Hope this helps.

BAJ

A little perhaps. When you put the LED into the split off of the the divider, you're basically going to be shorting the bottom resistor to ground. Not quite because there is a small voltage drop across the diode, and thus a small one across the the bottom resistor. Most of the voltage will be dropped across the top resistor, so in effect you have a current limiting resistor with another resistor that bleeds a little bit of current off to ground.

The voltage divider process depends on the fact that the resistance of whatever is being driven by the divider is much greater than the Thevenin resistance of the divider (in this case, the upper and lower resistances in parallel). If it is, then the output Voltage of the divider given by the Voltage Divider equation. If it isn't, then the output voltage sags due to the voltage dropped across the upper resistor.

On Mon, 5 Mar 2007 14:16:57 -0500, "tempus fugit" Gave us:

Fairly good answer aside from you being a top posting Usenet retard.

You seem to forget that there are 1 amp LEDs now.

Yes, OP (original poster), they MUST be current limited, and that limit is defined by their declared specs from their manufacturer.