Re: voltage divider or series current resistor
- From: John Popelish <jpopelish@xxxxxxxx>
- Date: Mon, 05 Mar 2007 16:17:31 -0500
NJM wrote:
I am just a beginner so forgive me if this question is stupid.
On a simple circuit, like 2 LEDs with a 12v supply, which would be better, a voltage divider to lower the voltage or a series resistor to restrict the current? Or am I completly wrong in my understanding of a voltage divider?
Please correct me if I am wrong, but it seems the divider would actually waste some energy because it is being sent to ground.
What is the determining factor in deciding which to use?
Is the reason a series resistor with LEDs because the LED has a resistance of its own therefore creating a sort of divider?
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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.
.
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