White LED's: get 20% more efficiency


Flashlight Enthusiast
Sep 9, 2001
New Jersey
White LED\'s: get 20% more efficiency

Energy conscious apps might wish to take special note.

Looking at the curve for the relative
intensity of a typical white LED vs
its forward current, pulsing doesnt increase
efficiency because the LED has to operate at
a higher current even though its for a
shorter time period. The funny thing is
though, operating the LED at a LOWER current
does give an increase in efficiency over an
LED operating at 20ma because the light
intensity vs current goes up slightly for
lower currents.

Looking at this short chart of current
vs relative intensity (20ma):
we can see that for the higher current
of 50ma the eff goes down by 20%, while
at 10ma the eff goes up by 20%.

This means using two LED's each running at
10ma will produce 20% more light then one
LED running at 20ma, while both
configurations consume exactly same amount
of energy!
Also, running two LED's each running at
8ma will produce the same amount of light
as a single LED running at 20ma, while the
8ma LED's will consume 20% less energy then
the single 20ma LED.

So far, this is the only way to increase
efficiency that i have found to be valid.



Aug 30, 2001
Colorado Springs, CO
Re: White LED\'s: get 20% more efficiency

Yup. This would work. But then, you have to have a lot more LEDs for the same amount of light. This impacts size and cost.


Re: White LED\'s: get 20% more efficiency

al: Great information. Thanks. Got a question for ya: you limit the current by the resistor right? not lowering the voltage across the LEDs.



Re: White LED\'s: get 20% more efficiency

The best way is to simply lower the voltage until you reach the desired current through the LED.

Of course, that's a lot easier said than done. Batteries tend to have fixed voltages, more or less.

You can use a resistor to reduce the current. The advantage to this is that it's simple and easy. However, it wastes energy. For example, if an LED draws 10mA at 3 volts, and you use a 6-volt battery, the necessary 300-ohm resistor will consume as much energy as the LED.

Since only half the battery power is going toward lighting the LED, the circuit's electrical efficiency is only 50%.

A linear voltage regulator is just as inefficient.

A more efficient approach would be to use a pulse circuit. If you pulse 40mA every millisecond then rest a millisecond, and repeat this cycle, the LED "sees" 20mA average current through it.

However, the 40mA pulses don't cause the LED to produce twice as much light, because LEDs aren't as efficient at 40mA as they are at 20mA. One way to improve this is to put a large capacitor in parallel with the LED. The capacitor smoothes the pulses, perhaps between 19-21mA, and makes the LED much happier.


Flashlight Enthusiast
Sep 9, 2001
New Jersey
Re: White LED\'s: get 20% more efficiency

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by yahoosier:
al: Great information. Thanks. Got a question for ya: you limit the current by the resistor right? not lowering the voltage across the LEDs.


Well that pertains to an LED alone, without
much series resistance, as in a down converter
or boost converter where you can adjust
the output voltage across the LED that
is in series with a fairly small resistor.

One thing that 20% doesnt take into account
is the drop in voltage itself, which would
also contribute to the increase in efficiency by increasing the electrical
efficiency, which would improve overall
eff by about 10% more.

I can see why this is a little more
complex then other aspects of LED driving,
and that's because the overall eff depends
not only on the LED alone, but also the
circuit used to drive it. This is probably
why there is so much difference in opinion
about how to get more eff and such.

I havent found one pulsing circuit though
that didnt contain at least one inductor
that improved eff at all (except for the
switched capacitor voltage doubler with
syncronous rectification).
I checked out pulsing an LED with 50% duty
cycle rectangular waveform (square wave)
driving the LED through a series resistor to
10ma average, which requires a peak LED
current of 20ma. Because the resistor can
be lower than driving the LED with constant
direct current (which also raises the LED
voltage), the electrical eff comes up
by about 5% over the simple series resistor
circuit, but the luminous eff goes down
by about 20%, so you end up with an overall
eff of about -15% relative to an LED
driven at constant current with a series
resistor, meaning overall the simple series
resistor with LED driven to 10ma is more
efficient then the LED driven to 10ma avg
with pulsing 50% duty cycle square wave
Apparently, whenever you pulse the LED you
lose efficiency. This doesnt apply for
all CIRCUITS that pulse though, because
some circuits pulse the current through
an inductor to achieve voltage conversion,
but dont pulse the LED itself.
These kinds of circuits typically are used
to provide a voltage match between LED
and power source without consuming tons of
energy in the process. A good example is
running a single LED from a 100v dc power source,
where a series resistor would consume
quite a lot of overhead energy that doesnt
contribute to the total light output.

But anyway, back to the main topic:

Two LED's in parallel each drawing 10ma
will put out more light then a single
LED drawing 20ma.
Of course the overall cost will increase
because you have to buy two LED's instead of
one :) but energy consious app's dont
usually worry about cost as much.

One thing to remember when comparing the
efficiency of one method to another, is
that there are three efficiency's to
consider when dealing with the LED and
the circuit used to drive it:

1. electrical efficiency
2. relative luminous efficiency
3. overall efficiency

The electrical efficiency is usually just
the output power divided by the input power.

The luminous efficiency is the light output
that is obtained while running the LED at
various currents. Sometimes this is very
difficult to calculate by hand, because
for some wave shapes (such as a triangle
wave) you have to take the LED through
each point on the wave, determine the
relative efficiency for each of these points,
and then sum the results. I did this once
with a sawtooth wave, and the data for the
efficiency i got off the web somewhere in
the form of a curve, and did a simple curve
fitting procedure to come up with a model
of the intensity vs forward current, and
applied the sawtooth wave to the curve. The
results quickly showed that driving an LED
with a triangle wave is the worst method on
earth :) because in order to get some decent
average current though the LED, the triangle
current wave has to be twice as high as a
rectangular shaped wave, and the higher the
current wave peak, the lower the luminous
efficiency, hence the lower the overall

The overall efficiency is the combined effect
of both electrical and luminous efficiency.

Lastly, the practical application of the
nature of efficiency has to be taken into
account. In other words, if you have a
circuit/LED combination that isnt great on
efficiency, you can reduce LED current by
maybe 10% or so, and have the batteries last
just as long as with a higher efficiency
circuit/LED combo, and the only difference
will be the lower eff circuit will put out
a little less light, which will probably
be unnoticable except in the most demanding