# Resistor value for LED

T

#### Tom

Jan 1, 1970
0
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?

Thanks
Tom

D

#### Dave Platt

Jan 1, 1970
0
Tom said:
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.

What the data sheet tells us is that the LED's absolute forward
current level is 50 mA continuous, or 100 mA if fed a low-duty-cycle
high-speed pulse train. More current than that will damage it.

Running an LED near to its absolute-maximum rating is, I believe,
likely to shorten its lifetime significantly. I usually prefer to
keep them at half-maximum or a bit less. This LED's brightness (which
is considerable) is rated by the manufacturer at a forward current of
20 mA, so that's what I'd suggest using.

At that current level, the LED is going to have a forward voltage of
2.3 to 2.6 volts. Let's use 2.5 volts as a working figure (it won't
matter very much if the particular LED you use is a bit higher or
lower).

So... a 5-volt supply, with 2.5 volts appearing across the diode.
This means that we must choose a resistor which will have (5.0 - 2.5)
= 2.5 volts at a forward current level of 20 mA.

You can solve for this by using Ohm's Law, which states that E=I*R, or
(by solving for R) R=E/I.

R = 2.5 / .02 (if we do it in volts and amps) or
R = 2500 / 20 (if we do it millivolts and milliamps)

In either case, R works out to be 125 ohms.

Let's figure out the power dissipation. P = E*I. We know that E
(across the resistor) is 2.5, and I is .02, so P = .05 watts of heat
dissipated in the resistor. It's generally a good idea to "de-rate"
resistors by 2:1 (that is, don't ask them to dissipate more than half
of their rated capability) to ensure long life. In this particular
case that's no problem, since the smallest standard leaded resistors
are 1/8 watt.

So - choose a resistor value in the neighborhood of 125 ohms (whatever
you have that's closest, plus or minus 10-20%), in whatever size is
convenient (1/8 watt or larger), and you should be fine.

T

#### tlbs

Jan 1, 1970
0
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.

M

#### mc

Jan 1, 1970
0
Tom said:
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.

T

#### Tom

Jan 1, 1970
0
tlbs said:
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.

Thanks for the excellent replies!

W

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

Jan 1, 1970
0
Tom said:
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?

Looks like 68 ohms would be a good choice. Use at least a quarter watt
resistor.

S

#### Spehro Pefhany

Jan 1, 1970
0
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

R

#### RST Engineering $$jw$$

Jan 1, 1970
0
I'm astonished, nay FLABBERGASTED, that none of the regulars on this
newsgroup told this fellow to use a microcontroller to do the job.

{;-)

Jim

M

#### mc

Jan 1, 1970
0
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?

M

#### mc

Jan 1, 1970
0
RST Engineering (jw) said:
I'm astonished, nay FLABBERGASTED, that none of the regulars on this
newsgroup told this fellow to use a microcontroller to do the job.

AVR or PIC ?

G

#### Glenn Gundlach

Jan 1, 1970
0
Yeah, PIC or 555. They seem to be the solution to every problem.
GG

G

#### Glenn Gundlach

Jan 1, 1970
0
Yeah, PIC or 555. They seem to be the solution to every problem.
GG

W

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

Jan 1, 1970
0
mc said:
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.

The older GaAs LEDs were lower, maybe below 2VDC. But I took a bunch of
indicator grade red LEDs, maybe just a few mCd, and measured them at
20mA. They were all 2VDC or even more. Last night I took a single high
brightness red LED and put 25mA thru it, and it dropped 2.3VDC. All the
newer GaInAs whatever LEDs seem to be well over 2V drop.

I put this high brightness red LED into a '50s vintage 'jewel'
indicator. These have a red glass multifaceted 'jewel' in a threaded
brass base a half inch across, which was mounted on the equipment panel
with a #47 lamp behind it. The red LED looks really good behind the
jewel. Maybe a bit too bright at 20mA, tho. I looked into it and had
spots in my vision for a minute afterwards.
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.

Again, you've missed the boat. The 160, 1.6k, etc., 5% resistor values
are no harder to get than 1k or 2.2k. Standard resistor values can be
http://www.rfcafe.com/references/electrical/resistor_values.htm

W

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

Jan 1, 1970
0
Spehro Pefhany said:
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
http://www.speff.com

M

#### mc

Jan 1, 1970
0
Again, you've missed the boat. The 160, 1.6k, etc., 5% resistor values
are no harder to get than 1k or 2.2k. Standard resistor values can be
http://www.rfcafe.com/references/electrical/resistor_values.htm

I know what the standard values. Some people have to shop at Radio Shack,
which has 220 but not 160 or 150 available singly. Radio Shack has
assortments that have 150 but not 160. Admittedly, "they don't make" was an
erroneous oversimplification.

I have actually not seen 160 (or 1.6, 16, 1600, etc.) very often at all.
Not all of the 5% standard series is equally widely available.
Long-established good practice, even with 5% resistors, is to prefer the
values in the 20% series first, then the 10% series, then the 5% series.

The 20% series goes: 1.0, 1.5, 2.2, 3.3, 4.7, 6.8, 10.

I spent a long time writing construction project articles for magazines and
developed a fairly strong aversion to hard-to-get parts. That, and I
started designing things back when resistors were 10% carbon composition!

A

#### Anthony Fremont

Jan 1, 1970
0
RST Engineering (jw) said:
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?

S

#### Spehro Pefhany

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.

By your definition a resistor is 100% efficient, then?

Best regards,
Spehro Pefhany

A

#### Anthony Fremont

Jan 1, 1970
0
Spehro Pefhany said:
By your definition a resistor is 100% efficient, then?

I don't know what you mean. I was thinking the dropping resistor
approach was much less than 50% efficient since the resistor ends up
dissipating more power than does the LED. Is that not correct?

I was thinking that by PWM'ng the LED and using a bit of inductance to
limit the current surges thru the i/o pin, would be much more efficient.
A nano-watt 16f88 claims to dissipate a max of 40uA at 5V running at
32kHz. That's only 200uW of power used by the PIC. The LED dissipates
many times (at least 100?) as much power. What am I doing wrong?

S

#### Spehro Pefhany

Jan 1, 1970
0
I don't know what you mean. I was thinking the dropping resistor
approach was much less than 50% efficient since the resistor ends up
dissipating more power than does the LED. Is that not correct?
Yes.

I was thinking that by PWM'ng the LED and using a bit of inductance to
limit the current surges thru the i/o pin, would be much more efficient.
A nano-watt 16f88 claims to dissipate a max of 40uA at 5V running at
32kHz. That's only 200uW of power used by the PIC. The LED dissipates
many times (at least 100?) as much power. What am I doing wrong?

The increased efficiency would be because of the inductor and
switch(s) or switch + diode, not the control circuitry. I didn't see
either taken into account in your efficiency calculation. They'll not
likely be very close to being ideal.

Best regards,
Spehro Pefhany

E

#### EN

Jan 1, 1970
0
Anthony Fremont said:
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?

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.

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