Resistor value for LED

[RST Engineering (jw)]

Don ...

I've noticed this effect, also. Yet a hallowed source none other than HP
has a full chapter in AN-1005 describing how a pulsed drive increases the
optical output of the lamp. I either don't understand completely what HP is
telling me or my eyes are deceiving me, which is entirely possible.

Let's take a relatively trivial example. How about we generate a square
wave to drive our lamp. Let's not worry for the moment about how we
generate it, let's simply say that we can have either a square wave or a DC
signal for drive. Let's also make the math easy and say our LED has a
forward voltage of 2 volts at 20 mA and 2.5 volts at 40 mA (real data for
HLMP-1340, typical high brightness red LED).

So in the DC case, we choose a resistor value of 150 ohms, and in the square
wave case we choose a resistor value of 62 ohms. The battery doesn't care.
It is pumping out an average power of 100 mW in either case.

However, in the DC case, the (peak) power delivered to the LED is 40 mW and
in the square wave case the (peak) power delivered to the LED is 100 mW.
And yet, to my admittedly untrained eye, they are either the same brightness
or perhaps the square wave is just the littlest bit DIMMER than the DC
drive. What's happening? My understanding of optics is that the eye is
pretty much a peak detector when the PRF is greater than the flicker rate of
24 Hz. or so, but that isn't the case in the experimental lab. Comments?


Oh, and to the feller that wanted to use a PIC and an inductor at a cost of
$5 or so, you can achieve the same thing with a single CMOS quad NOR gate,
two sections as oscillator and one section as monostable multi plus a single
PNP transistor in saturation driving either the LED directly through a much
smaller resistor (the cheap method) or throught the inductor and a catch
diode (the expensive method) for less than two bits worth of parts (the
cheap method) or a buck's worth (the expensive method).

Jim

You need an inductor in addition, unless you were thinking of applying 5
volts directly to the LED at whatever duty cycle results in average
current of 20 mA. By and large, LEDs produce less light from this than
with 20 mA of steady DC - meaning pulsing at whatever duty cycle achieves
20 mA average current with 5V across the LED is less efficient than using
a resistor from a 5V power supply.

- Don Klipstein (don@misty.com)

 

[EN]

I played with some white LEDs awhile ago. I made a 10 LED
flashlite out of a VCR hand remote. 2 AA's, 1 hand made inductor
and 1 2222 or NTE123. I can dimly lite up the front of the house about
20 to 25 ft. away. I used 50/50 narrow and wide beam white LEDs,
you want a pic? I'll try to get it. I took a plastic straw wrapped it w/
fiberglass and epoxy. Cut this in 1/2. That's my 1/2 cylinder reflector.
I put aluminum foil in the straw as a better reflector.
I wasn't impressed with LED flashlites. I made one w/ a charge pump
from Micrel, using 6 LEDs and 2 cells. ahh. I had to make everything
SMD to fit in the back of the flashlite head.


"RST Engineering (jw)" <jim@rstengineering.com> wrote in message
news:117clqtflvemsbe@corp.supernews.com...

Don ...

I've noticed this effect, also. Yet a hallowed source none other than HP
has a full chapter in AN-1005 describing how a pulsed drive increases the
optical output of the lamp. I either don't understand completely what HP
is
telling me or my eyes are deceiving me, which is entirely possible.

Let's take a relatively trivial example. How about we generate a square
wave to drive our lamp. Let's not worry for the moment about how we
generate it, let's simply say that we can have either a square wave or a
DC
signal for drive. Let's also make the math easy and say our LED has a
forward voltage of 2 volts at 20 mA and 2.5 volts at 40 mA (real data for
HLMP-1340, typical high brightness red LED).

So in the DC case, we choose a resistor value of 150 ohms, and in the
square
wave case we choose a resistor value of 62 ohms. The battery doesn't
care.
It is pumping out an average power of 100 mW in either case.

However, in the DC case, the (peak) power delivered to the LED is 40 mW
and
in the square wave case the (peak) power delivered to the LED is 100 mW.
And yet, to my admittedly untrained eye, they are either the same
brightness
or perhaps the square wave is just the littlest bit DIMMER than the DC
drive. What's happening? My understanding of optics is that the eye is
pretty much a peak detector when the PRF is greater than the flicker rate
of
24 Hz. or so, but that isn't the case in the experimental lab. Comments?


Oh, and to the feller that wanted to use a PIC and an inductor at a cost
of
$5 or so, you can achieve the same thing with a single CMOS quad NOR gate,
two sections as oscillator and one section as monostable multi plus a
single
PNP transistor in saturation driving either the LED directly through a
much
smaller resistor (the cheap method) or throught the inductor and a catch
diode (the expensive method) for less than two bits worth of parts (the
cheap method) or a buck's worth (the expensive method).

Jim



You need an inductor in addition, unless you were thinking of applying
5
volts directly to the LED at whatever duty cycle results in average
current of 20 mA. By and large, LEDs produce less light from this than
with 20 mA of steady DC - meaning pulsing at whatever duty cycle
achieves
20 mA average current with 5V across the LED is less efficient than
using
a resistor from a 5V power supply.

- Don Klipstein (don@misty.com)

 

[ehsjr]

Anthony Fremont wrote:

<snip>

One final time, I wasn't wishing to debate this topic (as it's like
talking religion or politics to some people), I was looking for some
clever analog or digital based current source that would be more
efficient. Any ideas? ;-)

Sure. A single cell lead acid battery.

Ed

 

[Watson A.Name - \"Watt Su]

"mc" <mc_no_spam@uga.edu> wrote in message
news:4273b5e2@mustang.speedfactory.net...

"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> wrote in message
news:d07771dgjdgkjekceojour6c58dc3l1b9m@4ax.com...

Typical operation is with a current of 20 mA. I am surprised that
they
say
2.3 volts, because the normal forward voltage drop of a red LED is
1.8
volts, and that's what I'll assume.

Um, you're assuming the datasheet is *wrong*?

I just measured a 635nm red (that's less orange) of similar
brightness, and it came to 2.1V @ 20mA.

Generally, forward voltages seem to be creeping up (for the same
color
of emitted light) as efficiencies increase.

I wonder why. I thought the voltage was related to the wavelength by
a
basic law of physics. Is the internal resistance also increasing, I
wonder?

They're making LED chips with a whole new and different chemistry.

 

[Rikard Bosnjakovic]

Glenn Gundlach wrote:

Yeah, PIC or 555. They seem to be the solution to every problem.

As a person with software background since 19 years, that sentence
reminded me of a quote regarding regular expressions (regexps):

"Some people, when confronted with a problem, think 'I know, I'll use
regular expressions.' Now they have two problems."

- Jamie Zawinski


--
Rikard Bosnjakovic http://bos.hack.org/cv/

Anyone sending unwanted advertising e-mail to my address will be
charged $250 for network traffic and computing time. By extracting
address from this message or its header, you agree to these terms.

 

[mc]

They're making LED chips with a whole new and different chemistry.

But I thought the LED voltage corresponded to the energy level of the
photons... don't have the equation handy but can look it up. Hence my
question about whether the internal resistance is higher now. Maybe it is.

 

[Don Klipstein]

In article <4277fc44$1@mustang.speedfactory.net>, mc wrote:

They're making LED chips with a whole new and different chemistry.

But I thought the LED voltage corresponded to the energy level of the
photons... don't have the equation handy but can look it up. Hence my
question about whether the internal resistance is higher now. Maybe it is.

Voltage drop across the LED is often but not always close to that of the
"electron volts" of the emitted photons.

One thing that can happen is that when an electron "drops" from the
conduction band back to the valence band, it may have an intermediate stop
on the way down, causing the descent to have more than one step - one of
which radiates a photon that has less energy than that required to push an
electron from the valence band to the conduction band.
I suspect this is what happens in InGaN green LEDs, with voltage drop
well above the "electron volts" of their emitted photons even at 10% of
the current they were designed for.

Another thing that can happen is that factors besides bandgap energy
affect the wavelength of the emitted light. For example, the chip may
have layers of such thickness that have interference effects that
reinforce a particular wavelength. Of course, efficiency may be reduced
if this is a wavelength other than that at which the chip would otherwise
emit, especially if the wavelength has photon energy greater than the
bandgap energy.

Still another thing that could happen is that an electron has more than
one mode of descending from the conduction band to the valence band - a
radiating mode and a non-radiating one.
If the radiating one has average photon energy greater than the bandgap
energy, what happens is that most photons take the non-radiating path.
Thermal agitation gives a few electrons the above-average energy needed to
take the radiating route, but thermal agitation is actually detremental as
a net by causing what I believe is "thermal quenching" - kicking electrons
onto the non-radiating route. At lower temperature, fewer electrons take
the non-radiating route, and the voltage drop is higher.
Example: GaP and InGaP green LEDs. A really efficient one with
wavelength in the upper 560's or around 570 nm - the usual "chartreuse"
color - is about 1.5-2% efficient. The "pure green" varieties of these
chemistries (wavelength around 550-555 nm, still "lime green" as in less
yellowish but still yellowish) are much less efficient - and have average
photon energy around 2.25 eV and voltage drop maybe usually 2.1 V.

- Don Klipstein (don@misty.com)

 

[Don Klipstein]

In article <4273ee67$1@mustang.speedfactory.net>, mc wrote in part:

I know what the standard values. Some people have to shop at Radio Shack,
which has 220 but not 160 or 150 available singly.

<SNIP>

Actually, Radio Shack has 150 ohm resistors all by themselves in
5-packs, both in 1/4 watt (271-1312) and 1/2 watt (271-1109). Probably
because, as you said, this is one of the values that 20% resistors came
in.

They also have 180 ohms (a value that 10% resistors came in) in 1/2
watt (271-1110).

Assuming none of these were discontinued since they printed the catalog
that I just pulled out (the 2002 one).

- Don Klipstein (don@misty.com)

 

[Don Klipstein]

In art. <117clqtflvemsbe@corp.supernews.com>, RST Engineering \(jw\) wrote:

Don ...

I've noticed this effect, also. Yet a hallowed source none other than HP
has a full chapter in AN-1005 describing how a pulsed drive increases the
optical output of the lamp. I either don't understand completely what HP is
telling me or my eyes are deceiving me, which is entirely possible.

Let's take a relatively trivial example. How about we generate a square
wave to drive our lamp. Let's not worry for the moment about how we
generate it, let's simply say that we can have either a square wave or a DC
signal for drive. Let's also make the math easy and say our LED has a
forward voltage of 2 volts at 20 mA and 2.5 volts at 40 mA (real data for
HLMP-1340, typical high brightness red LED).

So in the DC case, we choose a resistor value of 150 ohms, and in the square
wave case we choose a resistor value of 62 ohms. The battery doesn't care.
It is pumping out an average power of 100 mW in either case.

However, in the DC case, the (peak) power delivered to the LED is 40 mW and
in the square wave case the (peak) power delivered to the LED is 100 mW.
And yet, to my admittedly untrained eye, they are either the same brightness
or perhaps the square wave is just the littlest bit DIMMER than the DC
drive. What's happening? My understanding of optics is that the eye is
pretty much a peak detector when the PRF is greater than the flicker rate of
24 Hz. or so, but that isn't the case in the experimental lab. Comments?

This business of the eye being a peak detector even when the pulse rate
is above that necessary to avoid visible flicker (more like 50-60 Hz
rather than 24 Hz if duty cycle is near or under 50%) is largely not true.
This is a myth (at least mostly) that started due to an LED chemistry
commonly used in digital displays appearing to produce more light at high
instantaneous current (10's of mA per chip) than at low instantaneous
current (a couple to a few mA per chip) even with the same average
current.
The truth is that the reason they looked brighter when pulsed was
because average light output actually increased from pulsing them
even with the same average current, because the LED chips were nonlinear,
with higher efficiency at higher current.
This was fortunate, since the circuitry got easier with multiplexing
that had only one segment in each digit (or sometimes even in the whole
display) on at a time.

- Don Klipstein (don@misty.com)

 

[Watson A.Name - \"Watt Su]

"Don Klipstein" <don@manx.misty.com> wrote in message
news:slrnd7r3co.b2e.don@manx.misty.com...

In article <4277fc44$1@mustang.speedfactory.net>, mc wrote:


They're making LED chips with a whole new and different chemistry.

But I thought the LED voltage corresponded to the energy level of the
photons... don't have the equation handy but can look it up. Hence
my
question about whether the internal resistance is higher now. Maybe
it is.

Voltage drop across the LED is often but not always close to that of
the
"electron volts" of the emitted photons.

One thing that can happen is that when an electron "drops" from the
conduction band back to the valence band, it may have an intermediate
stop
on the way down, causing the descent to have more than one step - one
of
which radiates a photon that has less energy than that required to
push an
electron from the valence band to the conduction band.
I suspect this is what happens in InGaN green LEDs, with voltage
drop
well above the "electron volts" of their emitted photons even at 10%
of
the current they were designed for.

Another thing that can happen is that factors besides bandgap energy
affect the wavelength of the emitted light. For example, the chip may
have layers of such thickness that have interference effects that
reinforce a particular wavelength. Of course, efficiency may be
reduced
if this is a wavelength other than that at which the chip would
otherwise
emit, especially if the wavelength has photon energy greater than the
bandgap energy.

Still another thing that could happen is that an electron has more
than
one mode of descending from the conduction band to the valence band -
a
radiating mode and a non-radiating one.
If the radiating one has average photon energy greater than the
bandgap
energy, what happens is that most photons take the non-radiating path.
Thermal agitation gives a few electrons the above-average energy
needed to
take the radiating route, but thermal agitation is actually
detremental as
a net by causing what I believe is "thermal quenching" - kicking
electrons
onto the non-radiating route. At lower temperature, fewer electrons
take
the non-radiating route, and the voltage drop is higher.
Example: GaP and InGaP green LEDs. A really efficient one with
wavelength in the upper 560's or around 570 nm - the usual
"chartreuse"
color - is about 1.5-2% efficient. The "pure green" varieties of
these
chemistries (wavelength around 550-555 nm, still "lime green" as in
less
yellowish but still yellowish) are much less efficient - and have
average
photon energy around 2.25 eV and voltage drop maybe usually 2.1 V.

- Don Klipstein (don@misty.com)

Thanks for an explanation to a noticeable difference between older and
newer LEDs. I measured some yellow ultrabright LEDs yesterday, and they
were 2.3V at 20mA. That's less than I expected, since some of the red
and orange high brightness LEDs are almost that high.

 

[tlbs]

Notice in the data sheet, that the forward current (If) listed for
those maximum voltages is 20 mA. You need to drop half of your voltage
across the resistor (~2.5 V), so use Ohms law.

R = V / I
R = 2.5 / 0.02
R = 125 Ohm

The resistor power is calculated thus:
P = V^2 / R
P = 2.5^2 / 125
P = 0.05 W

so a 1/8 W resistor or larger will do just fine.

Since the LED will handle up to 50 mA continuously (although I would
not recommend operating it like that) the resistor value can go as low
as 50 Ohms. The LED will PROBABLY light up dimly with as little as 10
mA, so the resistor value can go as high as 250 Ohms.

 

[mc]

"Tom" <no@address.invalid> wrote in message
news:42712E86.9A60113F@address.invalid...

I have a led that needs 2.3 to 2.6 V max, I want to run it off a 5V
supply at close to its maximum voltage.

The exact led is here:

http://tinyurl.com/cmleb

Could someone tell me what value resistor I need to use?

Typical operation is with a current of 20 mA. I am surprised that they say
2.3 volts, because the normal forward voltage drop of a red LED is 1.8
volts, and that's what I'll assume.

You need a voltage drop of 3.2 volts (to take 5.0 down to 1.8 volts) at 20
mA = 0.020 A. Ohm's Law:

R = E / I = 3.2 / 0.020 = 160 ohms.

They don't make 160-ohm resistors, so use either 150 or 220.

Now... Will an ordinary 1/8-watt resistor handle the current? Let's see.

P = E I = 3.2 * 0.020 = 0.06 watt

Yes.

 

[Spehro Pefhany]

On Thu, 28 Apr 2005 16:25:05 -0400, the renowned "mc"
<mc_no_spam@uga.edu> wrote:

"Tom" <no@address.invalid> wrote in message
news:42712E86.9A60113F@address.invalid...
I have a led that needs 2.3 to 2.6 V max, I want to run it off a 5V
supply at close to its maximum voltage.

The exact led is here:

http://tinyurl.com/cmleb

Could someone tell me what value resistor I need to use?

Typical operation is with a current of 20 mA. I am surprised that they say
2.3 volts, because the normal forward voltage drop of a red LED is 1.8
volts, and that's what I'll assume.

Um, you're assuming the datasheet is *wrong*?

I just measured a 635nm red (that's less orange) of similar
brightness, and it came to 2.1V @ 20mA.

Generally, forward voltages seem to be creeping up (for the same color
of emitted light) as efficiencies increase.


Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

 

[Glenn Gundlach]

Yeah, PIC or 555. They seem to be the solution to every problem.
GG

 

[Glenn Gundlach]

Yeah, PIC or 555. They seem to be the solution to every problem.
GG

 

[Watson A.Name - \"Watt Su]

"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> wrote in message
news:d07771dgjdgkjekceojour6c58dc3l1b9m@4ax.com...

On Thu, 28 Apr 2005 16:25:05 -0400, the renowned "mc"
mc_no_spam@uga.edu> wrote:


"Tom" <no@address.invalid> wrote in message
news:42712E86.9A60113F@address.invalid...
I have a led that needs 2.3 to 2.6 V max, I want to run it off a 5V
supply at close to its maximum voltage.

The exact led is here:

http://tinyurl.com/cmleb

Could someone tell me what value resistor I need to use?

Typical operation is with a current of 20 mA. I am surprised that
they say
2.3 volts, because the normal forward voltage drop of a red LED is
1.8
volts, and that's what I'll assume.

Um, you're assuming the datasheet is *wrong*?

I just measured a 635nm red (that's less orange) of similar
brightness, and it came to 2.1V @ 20mA.

Generally, forward voltages seem to be creeping up (for the same color
of emitted light) as efficiencies increase.

Yeah, and as the number of elements increase: gallium, indium, antimony,
aluminum, arsenic, rat poison...

And that doesn't even get into the various phosphors they use to get
white LEDs.

Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the
reward"
speff@interlog.com Info for manufacturers:
http://www.trexon.com
Embedded software/hardware/analog Info for designers:
http://www.speff.com

 

[Don Klipstein]

In article <R58de.35651$h6.10828@tornado.texas.rr.com>, Anthony Fremont wrote:

"RST Engineering (jw)" <jim@rstengineering.com> wrote in message
news:1177ap112ioto34@corp.supernews.com...
I'm astonished, nay FLABBERGASTED, that none of the regulars on this
newsgroup told this fellow to use a microcontroller to do the job.


{;-)

Make fun if you like, but it's way more power efficient than a resistor,
even if it's not cheaper.....yet. ;-) AFAIK, even an exceptional SMPS
couldn't do it any more efficiently. IME a nano-watt PIC running at
32kHz should be ~99% efficient driving the LED with at least 10mA
average current.

How difficult would it be to beat that using another method?

You need an inductor in addition, unless you were thinking of applying 5
volts directly to the LED at whatever duty cycle results in average
current of 20 mA. By and large, LEDs produce less light from this than
with 20 mA of steady DC - meaning pulsing at whatever duty cycle achieves
20 mA average current with 5V across the LED is less efficient than using
a resistor from a 5V power supply.

- Don Klipstein (don@misty.com)

 

[Watson A.Name - \"Watt Su]

"Don Klipstein" <don@manx.misty.com> wrote in message
news:slrnd7b818.7s.don@manx.misty.com...

In article <1177ni69odjseba@corp.supernews.com>, Watson A.Name -
\"Watt
Sun, the Dark Remover\" wrote:

"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> wrote in message
news:d07771dgjdgkjekceojour6c58dc3l1b9m@4ax.com...
On Thu, 28 Apr 2005 16:25:05 -0400, the renowned "mc"
mc_no_spam@uga.edu> wrote:


"Tom" <no@address.invalid> wrote in message
news:42712E86.9A60113F@address.invalid...
I have a led that needs 2.3 to 2.6 V max, I want to run it off a
5V
supply at close to its maximum voltage.

The exact led is here:

http://tinyurl.com/cmleb

Could someone tell me what value resistor I need to use?

Typical operation is with a current of 20 mA. I am surprised that
they say
2.3 volts, because the normal forward voltage drop of a red LED is
1.8
volts, and that's what I'll assume.

Um, you're assuming the datasheet is *wrong*?

I just measured a 635nm red (that's less orange) of similar
brightness, and it came to 2.1V @ 20mA.

Generally, forward voltages seem to be creeping up (for the same
color
of emitted light) as efficiencies increase.

Also consider that voltage varies surprisingly little with
wavelength/color within a given chip chemistry. For example,
less-yellowish green LEDs and red ones of the same chemistry (InGaAlP,
or GaP) have surprisingly similar voltage drop.
One thing that does usually happen is that efficiency falls bigtime
if
the wavelength gets shorter to an extent that has the photon energy in
electron volts exceeds the voltage drop in volts.

Yeah, and as the number of elements increase: gallium, indium,
antimony,
aluminum, arsenic, rat poison...

And that doesn't even get into the various phosphors they use to get
white LEDs.

Phosphors don't affect the voltage drop...

Right, sorry if I may have caused any confusion.

But most white LEDs have blue LED chips and phosphor. So white LEDs
will have the same voltage drop as related blue ones.

I noticed that some white LEDs are now using wavelengths even shorter
than blue.

> - Don Klipstein (don@misty.com)

 

[Watson A.Name - \"Watt Su]

"Anthony Fremont" <spam@anywhere.com> wrote in message
news:6fmde.35922$h6.8439@tornado.texas.rr.com...

"Watson A.Name - "Watt Sun, the Dark Remover"" <NOSPAM@dslextreme.com
wrote in message news:117bk88qlcead6f@corp.supernews.com...
[snip]

I have nothing against using the right tool for the job. But it's
deceptive and unfair to tell someone to use something without
telling
them what they will have to pay and what they will be going thru to
complete their project.

One might also consider it deceptive and unfair for someone to suggest
that micros are as difficult and unreachable as you describe.

I didn't describe "how unreachable" they are.

I did say that it's deceptive to portray them as Plug and Play when they
simply are not. When a person asks for a simple answer to a simple
problem that is low cost, giving him an answer that isn't (and saying
that your answer is the best) isn't telling the whole truth.

The primary difference being that your anology is absurd. Learning
micros or purchasing the needed equipment is no where near the
difficulty you describe. Ten or twenty hours of self study and $75
for
a programmer is not quite in the same league as ramping up to make
your
own LASER tubes.

Maybe my analogy flew over the top of your head. What I'm trying to say
is that "Ten or twenty hours of self study and $75.." is *not* the
appropriate answer to a question that should cost $5 and take a half
hour using skills that are already learned.

But what's more to the point, telling the asker that it is the answer,
and can do all these wonderful things, and leaving out the "Ten or
twenty hours of self study and $75.." part is deceptive.

And there is *nothing* Godwinesque about it. Deception is
deception.

If there is nothing Godwinesque about it, then why is the same old
tired
argument used to shout down the micro crowd every time?

It has nothing to do with micros. It has to do with deception.

One final time, I wasn't wishing to debate this topic (as it's like
talking religion or politics to some people), I was looking for some

Exactly. Some people consider Pics to be a religion, and give that
answer to those problems that could be solved with a simple 555 timer.
And I (and others) are only trying to dispel the myth that a Pic is the
answer to all your problems, even the smallest.

[snip]