# Resistor value for LED

W

#### Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
EN said:
The learning curve. When someone asks how to turn on a LED.
There was a space constraint to the project box as I recall.
Most of everyone thinks of a DC source,1 resistor and the LED.
Simple enough. Someone posts the actual calculations, E=IR and
W=EI . Without knowing about E=IR or W=EI, you expect to jump
into Microcontrollers? Writing uC programs ?
How long will it be before you research about PICs, build or buy a pic
programmer, learn to program a PIC. There's a big jump between
a transistor battery and a PIC project, don't you think?
To run, you 1st got to learn to crawl & stand up.

My exact sentiments. However there are a lotta pic programmers who
don't have a clue about analog electronics. They could connect point A
to point B and make a digital circuit (and with the limited number of
pins on a pic, it's really easy), but wouldn't have a clue as to how to
calculate the resistor for the LED.

I'm not suggesting that they _get_ a clue. They do just fine without
one.

A

#### Anthony Fremont

Jan 1, 1970
0
EN said:
The learning curve. When someone asks how to turn on a LED.
There was a space constraint to the project box as I recall.
Most of everyone thinks of a DC source,1 resistor and the LED.
Simple enough. Someone posts the actual calculations, E=IR and
W=EI . Without knowing about E=IR or W=EI, you expect to jump
into Microcontrollers? Writing uC programs ?

Whoa nellie, I didn't suggest the OP do any such thing. ;-) I was only
following up on Jim's comment about PICs.
How long will it be before you research about PICs, build or buy a pic
programmer, learn to program a PIC. There's a big jump between
a transistor battery and a PIC project, don't you think?
To run, you 1st got to learn to crawl & stand up.

I really wish that every mention of a PIC didn't have to result in this
statement. It's sorta becoming the Godwin's law of electronics. It's
almost like saying, "everything about electronics is worth learning, no
matter how difficult or expensive, as long as it doesn't involve micros
because they are too hard and too expensive". At any rate, I didn't
suggest using one for the OP's problem, I only mentioned that it could
be quite efficient, as a response to Jims comment.

I wasn't looking to beat that dead horse again, I was only hoping that
someone could/would (if possible) show me something more efficient using
"conventional" components.

D

#### Don Klipstein

Jan 1, 1970
0
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...

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

- Don Klipstein ([email protected])

D

#### Don Klipstein

Jan 1, 1970
0
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 ([email protected])

W

#### Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Anthony Fremont said:
Whoa nellie, I didn't suggest the OP do any such thing. ;-) I was only
following up on Jim's comment about PICs.

I really wish that every mention of a PIC didn't have to result in this
statement. It's sorta becoming the Godwin's law of electronics. It's
almost like saying, "everything about electronics is worth learning, no
matter how difficult or expensive, as long as it doesn't involve micros
because they are too hard and too expensive". At any rate, I didn't
suggest using one for the OP's problem, I only mentioned that it could
be quite efficient, as a response to Jims comment.

I wasn't looking to beat that dead horse again, I was only hoping that
someone could/would (if possible) show me something more efficient using
"conventional" components.

It has _nothing_ to do with "..micros because they are too hard and too
expensive". What it has to do with is deception.

Deception. If there were money being exchanged, it could be called
something illegal, like fraud. But there's no money being exchanged.
The deceiver is advocating the use of something because it's his
fervent, almost religious, belief that it's _the_ solution to all of the
OP's problems. Which may be true.

But the deceptive part is that the deceiver has not told the OP (who is
usually a neophyte and probably knows little about electronics OR pics)
that there is a price he will have to pay in up front costs for the
hardware and software to program the pic. And there is a learning curve
that the OP must go thru to learn how to program and debug the software.
It is simply NOT Plug and Play!

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.

It just so happened that microcontrollers were the subject in past
discussions. Four decades ago it could have been using a laser as a
pointer for presentations. We all know nowadays that the little
hand-held ones do an excellent job. Back then, 40 years ago, you would
have had to blow your own glass, fill it with helium and neon, and use a
high voltage power supply to excite it. If a presenter could do all
that, he could then pull out his gatling gun sized pointer and look like
the termninator as he pointed to his presentation. But it would be
deceptive to not tell him that he would have to jump thru all those
hoops to get this cool looking presentation tool.

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

W

#### Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Don Klipstein said:
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.

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.

A

#### Anthony Fremont

Jan 1, 1970
0
Watson A.Name - "Watt Sun said:
It has _nothing_ to do with "..micros because they are too hard and too
expensive". What it has to do with is deception.

Here we go again. :-(
Deception. If there were money being exchanged, it could be called
something illegal, like fraud. But there's no money being exchanged.
The deceiver is advocating the use of something because it's his
fervent, almost religious, belief that it's _the_ solution to all of the
OP's problems. Which may be true.

So, suggesting the use of a micro is now tantamount to fraud? The
zealot thing is cute too. Not very realistic, but cute.
But the deceptive part is that the deceiver has not told the OP (who is
usually a neophyte and probably knows little about electronics OR pics)
that there is a price he will have to pay in up front costs for the
hardware and software to program the pic. And there is a learning curve
that the OP must go thru to learn how to program and debug the software.
It is simply NOT Plug and Play!

I don't recall anyone (and especially not me) saying that it was PnP.
Of course there is a learning curve, and I believe it is simply another
step in someone's education in electronics. As I've said before, if I
was recommending one to a newbie, I'd likely be giving him the code and
offering to flash the chip for them.

To be clear, I never suggested the OP use a micro. I just wanted to
respond to Jim's comment and hopefully spawn some conversation about
more efficient ways to drive the LED. It looks like that was simply a
pipe dream on my part.
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.
It just so happened that microcontrollers were the subject in past
discussions. Four decades ago it could have been using a laser as a
pointer for presentations. We all know nowadays that the little
hand-held ones do an excellent job. Back then, 40 years ago, you would
have had to blow your own glass, fill it with helium and neon, and use a
high voltage power supply to excite it. If a presenter could do all
that, he could then pull out his gatling gun sized pointer and look like
the termninator as he pointed to his presentation. But it would be
deceptive to not tell him that he would have to jump thru all those
hoops to get this cool looking presentation tool.

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. 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? 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? ;-) W #### Watson A.Name - \Watt Sun, the Dark Remover\ Jan 1, 1970 0 Anthony Fremont said: "Watson A.Name - "Watt Sun, the Dark Remover"" <[email protected]> wrote in message news:[email protected]... [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. 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] A #### Anthony Fremont Jan 1, 1970 0 "Anthony Fremont" <[email protected]> wrote <Snipped pointless rhetoric about using microcontrollers> As I have now said multiple times, I do not wish to have this "religious" debate. Very well, I take it that you wish not to contribute anything useful to the subthread about driving an LED with extreme power efficiency. R #### RST Engineering $$jw$$ Jan 1, 1970 0 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

E

#### EN

Jan 1, 1970
0
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.

E

#### ehsjr

Jan 1, 1970
0
Anthony Fremont wrote:

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

W

#### Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
mc said:
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.

M

#### mc

Jan 1, 1970
0
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.

R

#### Rikard Bosnjakovic

Jan 1, 1970
0
Glenn said:
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/

charged \$250 for network traffic and computing time. By extracting
address from this message or its header, you agree to these terms.

D

#### Don Klipstein

Jan 1, 1970
0
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
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 ([email protected])

D

#### Don Klipstein

Jan 1, 1970
0
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 ([email protected])

D

#### Don Klipstein

Jan 1, 1970
0
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 ([email protected])

W

#### Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Don Klipstein said:
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
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 ([email protected])

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.

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