Total Current Draw Of LED And Resistor Combined

vk6kro

A resistor getting hot does use power but the LED current is the same.
You seem to be using the terms power and current as if they were the same thing.

LEDs are semiconductor devices which draw almost no current until they have a certain voltage across them and then they can draw enough current to destroy themselves.

So it is necessary to put a resistor in series with them to limit the current they can draw. In your case, this was 25 mA but we settled on a resistor that would let the LED draw a bit less than this.

Sometimes, we have a voltage we have to use and this determines the power we will use in the resistor.
The resistor is there to protect the LED, not to rob power from it.

You have some choices with your setup. I would have two wires going the full distance down your path and then take a resistor and a LED off at each point where you need a LED.

At each point you would have 20 mA flowing from the 12 volt line, so there would be 0.24 watts being used. (12volts * 0.02 amps = 0.24 watts)
0.07 watts of this would be going to the LED, so it isn't very efficient, but there isn't a lot of power being used so it probably doesn't matter.

One possible choice would be that you could put up to 3 LEDs in series for each (smaller) resistor.
It may be obvious how you would do this, but I will describe it if you like.
This would be more efficient, but the wiring would be more messy.

Another choice that may suit you is to get Christmas lights which are already designed for your mains voltage and then you just string them along your driveway. You would then just find some way to get mains voltage to the string of lights and do it safely.

John1397

I like watts and amps, but I was wondering how you came up with .07 watts going to the Led this seems what I want to know? Those numbers figure out to be 70% loss in inefficiency if that is correct?

John

vk6kro

3.5 volts * 0.02 amps = 0.07 watts.

Some of this inefficiency is necessary if you have to use a 12 volt supply.

If you had a 6 volt supply, the total power used would be 6 volts * 0.02 amps or 0.12 watts.
So the efficiency would be about 58 %. (0.07 watts / 0.12 watts * 100 = 58.333 % efficient.)

These losses are trivial, but larger LEDs can use currents of up to an amp and there are highly efficient regulators that can limit their current to safe levels without wasting a lot of power.

sophiecentaur

The power value was obtained by 3.5 volts * 0.02 amps, which was made clear, and the same formula (P = IV) was used to calculate input power.

The efficiency would be power out / power in. The phrase "loss in efficiency" is not a useful one as it tells you nothing about the actual efficiency. The 'retail' use of the percentage is creeping into descriptions of all sorts of things - along with "three times less" etc.. If you try to stick to the original definitions, you can communicate better on scientific matters, I think.

John1397

Least I can see how the formulas work. Now I have 13 Led's rated .25 ma at 3.3 volt connected to 12 volt supply with 13 - 1000 ohm resistor total sum of Led's should be 325 ma so when I connected an amp meter between 12 volt supply and Led's it reads .5 amp I just wonder if this is right or if I am leaking current somwhere, but 13 Led's at reduced current still must emit more light than one 6 watt 12 volt bulb so I should have gained something.

John

vk6kro

You have enough information to work this out for yourself now.

If you ignored the LEDs and just put 13 resistors, all 1000 ohms, in parallel across a 12 volt supply, each would draw 12 mA so the total current would be 156 mA. (12 mA * 13 = 156 mA)

Adding the LEDs would drop this current to 113 mA. ((12 volts - 3.3 volts) / 1000) * 13.

So, yes, your meter is telling you lies.

Your resistors could be as low as 390 ohms and this would have a big effect on the LED brightness.