Effeciency loss from resistors

When I limit current to a LED by using a resistor, how much power is wasted by the resistor converting it to heat (or any other means)? I assume that using a resistor doest produce a circuit that consumes as much power per unit time as a "wide-open" circuit (I'm assuming this because EG a Surefire L1 runs for a LONG time on low mode), but I can't imagine that the total power delivered to the emitter over the life of a battery would be the same with and without the resistor, either.

You bring up a good point tyler.

I cant give any kind of #, but its alot of lose. /ubbthreads/images/graemlins/frown.gif

Even with circuits such as a BadBoy, we are seeing a typical 83% efficiencey.
Thats alot of lost power.

I would assume that a regular resistor woulf have alot more lose.

There are new converter circuits in the works, that could give us around 95% effeciency, but these are alittle ways off.

Makes you think.... if we stick a resistor in there, so that we limit our current down, in order to get some more runtime, well, we are wasting power converting it to heat, so, we might as well keep the light bright, and use the wasted power as brighter light.

ive wondered that my self.id figure direct drive is the most eficant ..but a course the light dims faster

When I limit current to a LED by using a resistor, how much power is wasted by the resistor converting it to heat...
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
It depends...
The resistor restricts the current flow to the led.
Also the farther the resistor must drop the voltage requires it to consume more power.

(VS-VL)/I=R (formula to calculate resistor value)
VS = supply voltage, VL = LED voltage, I = LED current

Three Alkaline cells: (Voltage times current is watts or power. V*I=W)
350mA Luxeon
(4.5-3.6)/.350 = .9/.350 = 2.57 ohms (20% wasted by the resistor)
Total watts = 1.57, Resistor = .31 watts, Luxeon = 1.26 watts

700mA Luxeon
(4.5-3.6)/.700 = .9/.700 = 1.28 ohms (20% wasted by the resistor)
Total watts = 3.15, Resistor = .63 watts, Luxeon = 2.52 watts

Direct Drive Luxeon (say 1.2 amps)
Total watts = 4.5*1.2 = 5.4 watts

Two 123 cells:
350mA Luxeon
(6-3.6)/.350 = 2.4/.350 = 6.86 ohms (40% wasted by the resistor)
Total watts = 2.1, Resistor = .84 watts, Luxeon = 1.26 watts

700mA Luxeon
(6-3.6)/.700 = 2.4/.700 = 3.43 ohms (40% wasted by the resistor)
Total watts = 4.2, Resistor = 1.68 watts, Luxeon = 2.52 watts

Direct Drive Luxeon (say ?.?? amps)
Total watts = 6*?.?? = 0 watts Because your Luxeon is toast.

The resistor is a good solution if you do not have to drop the voltage very far.

Another way to look at it is that whatever voltage you are dropping is being used by the resistor and not producing light.

3.6v/4.6v = .8 so 20% wasted by the resistor.
3.6v/6.0v = .6 so 40% wasted by the resistor.

yes, that's exactly what I wanted to see. Very illuminating.

Thanks for the calculations.

It should be remembered, though, that these figures refer to the power loss at first start. I wonder if anyone has ever integrated an entire power curve over time, and compared that to the total energy dissipated by the resistor. That would tell the whole story over the life of the batteries. Of course, with LEDs, the amount of runtime is a little subjective.

[ QUOTE ]
springnr said:
When I limit current to a LED by using a resistor, how much power is wasted by the resistor converting it to heat...
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
It depends...
The resistor restricts the current flow to the led.
Also the farther the resistor must drop the voltage requires it to consume more power.

(VS-VL)/I=R (formula to calculate resistor value)
VS = supply voltage, VL = LED voltage, I = LED current

Three Alkaline cells: (Voltage times current is watts or power. V*I=W)
350mA Luxeon
(4.5-3.6)/.350 = .9/.350 = 2.57 ohms (20% wasted by the resistor)
Total watts = 1.57, Resistor = .31 watts, Luxeon = 1.26 watts

700mA Luxeon
(4.5-3.6)/.700 = .9/.700 = 1.28 ohms (20% wasted by the resistor)
Total watts = 3.15, Resistor = .63 watts, Luxeon = 2.52 watts

Direct Drive Luxeon (say 1.2 amps)
Total watts = 4.5*1.2 = 5.4 watts

Two 123 cells:
350mA Luxeon
(6-3.6)/.350 = 2.4/.350 = 6.86 ohms (40% wasted by the resistor)
Total watts = 2.1, Resistor = .84 watts, Luxeon = 1.26 watts

700mA Luxeon
(6-3.6)/.700 = 2.4/.700 = 3.43 ohms (40% wasted by the resistor)
Total watts = 4.2, Resistor = 1.68 watts, Luxeon = 2.52 watts

Direct Drive Luxeon (say ?.?? amps)
Total watts = 6*?.?? = 0 watts Because your Luxeon is toast.

The resistor is a good solution if you do not have to drop the voltage very far.

Another way to look at it is that whatever voltage you are dropping is being used by the resistor and not producing light.

3.6v/4.6v = .8 so 20% wasted by the resistor.
3.6v/6.0v = .6 so 40% wasted by the resistor.

[/ QUOTE ]

Not quite though. You've forgotten that the Vf of the Luxeon will vary with current. You get a Vf of 3.6V at 350mA, or at 700mA, but not both.

I've done simulations, using real modeling of battery voltage over usage, and luxeon Vf variations over current levels.

As the luxeon current drops, the percentage of power consumed by the resistor also drops.

Below is for a 1W luxeon, Vf of 3.3V @ 350mA, run off 3 alkaline cells, 1.3 ohms of resistance.

Vf is the forward voltage of the luxeon. If is the current. P is the total power delivered by the batteries. Rpwr is the power dissipated by the resistor.

<font class="small">Code:</font><hr /><pre>
Vin Vf If (A) P(W) Rpwr(W) Eff
4.5 3.58 0.71 3.20 0.66 21% 79%
4.2 3.47 0.56 2.35 0.41 17% 83%
3.9 3.36 0.42 1.64 0.23 14% 86%
3.6 3.23 0.28 1.01 0.10 10% 90%
3.3 3.09 0.16 0.53 0.03 6% 94%
3.0 2.91 0.07 0.21 0.01 3% 97%
2.7 2.68 0.02 0.05 0.00 1% 99%
</pre><hr />

As you can see, as the batteries deplete, the efficiency of a simple resistor circuit goes up.

[ QUOTE ]
Lux Luthor said:
Thanks for the calculations.

It should be remembered, though, that these figures refer to the power loss at first start. I wonder if anyone has ever integrated an entire power curve over time, and compared that to the total energy dissipated by the resistor. That would tell the whole story over the life of the batteries. Of course, with LEDs, the amount of runtime is a little subjective.

[/ QUOTE ]

I have done exactly that.


Assuming, at high currents, that an alkaline AA will deliver 1800mAh, then for the above example (1.3 ohm, 3 cells, luxeon Vf of 3.3V @ 350mA):

Total energy delivered by the batteries = 6.2 W-h
Total energy dissipated by the luxeon = 5.6 W-h
Total energy dissipated by the resistr = 0.6 W-h

Efficiency = 90%


This was modeled assuming that the total energy in the cell is equally distributed over the voltage range of 1.5-0.8V/cell (according to manufacturer's specifications).


I've done simulations for the above in a variety of other configurations - from 4 to 10 cells, all with the same luxeon, and initial condition of I=700mA.

Snap shots are useful for conveying concepts.

"Another way to look at it is that whatever voltage you are dropping is being used by the resistor and not producing light."

As you show in your chart the LED is an active component which can play beautifully off a passive resistor.


______________
The Art of Electronics is in the curves.

Please excuse me if this question is off topic, however...

I have a 9 volt 5 watt direct drive LED light that I want to evaluate with Jon Burly's rechargeable 123's (not Pila's). The initial voltage is reported (I don't have them yet) to be 11 volts.

If I put a 2 ohm resister in the circuit (I think that will work), how does that effect the run time of the batteries?

I would like to give a proper evaluation of the batteries in this light, but am not sure how to handle the loss through the resistor. I suppose I could just report my set up and go from there.

Tom

The short answer is: the more resistance you put in, the longer your batteries will last (at the expense of brightness, of course)


And the long answer:
It's hard to guess how the light will perform without knowing a few things - internal resistance of the batteries, the voltage curve over the life of the batteries, and the Vf of your particular 5W at various current levels.

Assuming no internal resistance in your batteries (this sounds like a physics problem /ubbthreads/images/graemlins/wink.gif ), and a Vf of 6.8V @ 700mA, 7V @ 1A, 7.3V @ 1.5A, and 7.5V @ 2A; then we can make some guesses based on those numbers as to what the current will be at 11V with 2 ohms...<grumble...mumble> about 1.8A.

Given that the battery voltage will sag under load, it should be somewhat lower than 1.8A, so like 1.5A (guessing).

Now, with the same 2 ohms, and assuming that the nominal voltage is 9V (after the batteries have calmed down), then your current (neglecting cell resistance) will be 1A, so with the batterys' internal resistance, probably less.

Jon was worried about cell reversal with a three cell setup.

Hello Evan9162,

Thanks, that gives me a place to start from.

Springnr,

I am well aware of the risks.

I plan to take protective measures to control the risks. I also have inquired as to cost of replacement if the LED is damaged during the experiment. I also plan to run the light on 1 and 2 cells and see how that works.

Tom

[ QUOTE ]
Lux Luthor said:
It should be remembered, though, that these figures refer to the power loss at first start. I wonder if anyone has ever integrated an entire power curve over time, and compared that to the total energy dissipated by the resistor. That would tell the whole story over the life of the batteries. Of course, with LEDs, the amount of runtime is a little subjective.

[/ QUOTE ]

You want this thread. /ubbthreads/images/graemlins/grin.gif Be sure to read the whole thing, since there are some revisions and corrections.

Evan and Andrew,

Thanks for the information. Nice work!!!