Q: LED Margin?

[Mark Jones]

Curious, can two identical LEDs with Vf of 3V (30mA) be powered
reliably in series from a 5V regulated supply without using a dropping
resistor? Or is this not enough margin for die differences, thermal
runaway, etc. This is at room temperature. Thanks.

 

[Reinardt Behm]

Curious, can two identical LEDs with Vf of 3V (30mA) be powered
reliably in series from a 5V regulated supply without using a dropping
resistor? Or is this not enough margin for die differences, thermal
runaway, etc. This is at room temperature. Thanks.

You know that 2*3V is greater that 5V ?

 

[Mark Jones]

Mark Jones wrote:




You know that 2*3V is greater that 5V ?

Obviously. The question isn't wether it will work, but rather *is it
reliable?*

 

[Reinardt Behm]

Obviously. The question isn't wether it will work, but rather *is it
reliable?*

Forget it. You might have luck and the LEDs will light up. But do not leave
out a resistor. I would expect any heat up will kill your LEDs.
In my experience with LEDs at 30mA you must expect heat up.

 

[Boris Mohar]

Obviously. The question isn't wether it will work, but rather *is it
reliable?*

It will work reliably for about a nanosecond. LEDs are not resistors.

 

[TCS]

Obviously. The question isn't wether it will work, but rather *is it
reliable?*

It'll be rock solid relible if zero light output is what you're trying to
achieve.

Put 1.68V to a 1.7V LED and it won't light up.
Put 1.72V to a 1.7V LED and it'll go up in smoke.

LED's aren't voltage operated devices. They're current operated devices.
Subtract the voltage drop from the supply voltage and then compute a series
resistor to get the correct current. That, or use a current regulated supply.

 

[Mac]

Curious, can two identical LEDs with Vf of 3V (30mA) be powered
reliably in series from a 5V regulated supply without using a dropping
resistor? Or is this not enough margin for die differences, thermal
runaway, etc. This is at room temperature. Thanks.

From what you've said, I question whether they would turn on at all at 5V.
Is there a minimum specified for Vf?

The LED's whose datasheets I've examined show wide spread in the forward
voltage.

With LED's, you really need to use a current source to drive them. The
typical thing, as you seem to know, is to just put a resistor in series
with the LED. If there is very little headroom, as when driving a blue LED
from 5 Volts, you may need a better current source than a resistor to VCC.

One circuit I've seen is to put the LED in series with a voltage follower
driving a resistor. Like this: (Use courier or another constant-width font)



VCC
|
+---------------+
| |
/ ____
\ R1 \ / LED
/ ---
\ |
| |
| /
| /
+-------------| Q1 (NPN)
| |
/ \
\ R2 \_>
/ |
\ |
| /
| \ R3
GND /
\
|
|
GND

The idea is that Q1 is a follower. You pick the voltage you want across
R3, and you choose R1 and R2 to set that voltage, keeping in mind that
there will be a drop of around 0.6 V from the base to emitter of Q1. You
also want to keep in mind that Q1 will have some base current, so R1 and
R2 should be chosen so that the base current won't throw off the voltage
divider by much. If possible, you should look over the datasheet for Q1
and see if you can estimate the Vbe drop more accurately than just 0.6 V.

So if Beta is 100, and you want 30 mA in the LED, then the base current
will be 300 uA. So you should choose R1 and R2 to satisfy this inequality:

VCC/(R1+R2) >= 300 uA * 10

In other words, the voltage divider current should be at least 10x the
base current.

Oh, R3 will probably be pretty small.

HTH

--Mac

 

[John Larkin]

It'll be rock solid relible if zero light output is what you're trying to
achieve.

Put 1.68V to a 1.7V LED and it won't light up.
Put 1.72V to a 1.7V LED and it'll go up in smoke.

LED's aren't voltage operated devices. They're current operated devices.
Subtract the voltage drop from the supply voltage and then compute a series
resistor to get the correct current. That, or use a current regulated supply.

Actually, lots of people run led's with constant-voltage drive; done
properly, it seems to work fine. LEDs don't have a brick-wall forward
conduction curve.

Besides, as Kevin keeps assuring us, all junction semiconductors are
voltage-operated devices.

But they're cheap enough that the OP should just try it and see what
happens.

John

 

[Anthony Fremont]

Actually, lots of people run led's with constant-voltage drive; done
properly, it seems to work fine. LEDs don't have a brick-wall forward
conduction curve.

Besides, as Kevin keeps assuring us, all junction semiconductors are
voltage-operated devices.

But they're cheap enough that the OP should just try it and see what
happens.

I agree, experimentation is the best teacher. People can guess all day,
but the real truth can be found out in seconds. I have some LEDs (grab
bag special) that I use when tinkering with PIC chips on a breadboard.
I often hook them straight to an output pin and they usually work fine,
though they get a tad bright and tend to load the output pins kinda
hard. ;-) I have one in particular that turns orange and then
yellowish after about 5 seconds of continuous on time. It gets quite
warm but somehow survives the abuse. I've had others (usually small
ones) blink out like a flash bulb too.

My advice: If you want LEDs (and the parts that drive them) to work
reliably, provide plenty of voltage and use a resistor.

 

[Ben Bradley]

It's only at room temperature for the first nanosecond you power it
up...

Actually, lots of people run led's with constant-voltage drive; done
properly, it seems to work fine. LEDs don't have a brick-wall forward
conduction curve.

No more than any other diode.

Besides, as Kevin keeps assuring us, all junction semiconductors are
voltage-operated devices.

It "can" be done, if you use a photodiode to measure the LED's
light output and have that drive an op-amp which powers the LED. Using
this, the LED's light output will stay constant over a long period (as
opposed to dropping substantially, as I've read that LED's do in there
first few months of operation).

But they're cheap enough that the OP should just try it and see what
happens.

He should try a few on the bench, not in production. If the OP's
idea is to somehow save the cost or space of a resistor, I'd say don't
do it. Resistors are ridiculously cheap, compared to any other
possible way to current-control a LED.

 

[John Woodgate]

I read in sci.electronics.design that Anthony Fremont

I agree, experimentation is the best teacher. People can guess all day,
but the real truth can be found out in seconds. I have some LEDs (grab
bag special) that I use when tinkering with PIC chips on a breadboard. I
often hook them straight to an output pin and they usually work fine,
though they get a tad bright and tend to load the output pins kinda
hard. ;-) I have one in particular that turns orange and then
yellowish after about 5 seconds of continuous on time. It gets quite
warm but somehow survives the abuse. I've had others (usually small
ones) blink out like a flash bulb too.

Just hope that getting an LED to act as a laser by passing a large
current through it is just an urban myth.

 

[Rich Grise]

I read in sci.electronics.design that Anthony Fremont


Just hope that getting an LED to act as a laser by passing a large
current through it is just an urban myth.

Wouldn't that cause an EMP on the order of an attoton?

;-)
Rich

 

[Rich Grise]

Forget it. You might have luck and the LEDs will light up. But do not leave
out a resistor. I would expect any heat up will kill your LEDs.
In my experience with LEDs at 30mA you must expect heat up.

People sure do overdrive their LEDs.

There was a thread a couple of years ago, essentially, "Can I make
an LED fade out with a Capacitor?"

There was some disagreement, imagine my surprise. ;-)

So I put an LED, a resistor, a voltmeter, and another voltmeter on
the bench with some supply, and a BMF cap. I watched it as the cap
discharged.

From about 20 mA down to about 10 mA, I didn't see any difference
in brightness. "Diminuition of illumination was indiscernable."
From 10 mA on down, it was as linear as my eye could distinguish.
i.e. at 5 mA, it was half as bright as at 10 mA, and so on, all
the way down to zero.

It didn't "wink out" ;-) .

So, limit the current, and you can probably get away with half
the spec. current, or even less - when you get under 10 mA, you
might want to eyeball it.

Cheers!
Rich

 

[Active8]

Curious, can two identical LEDs with Vf of 3V (30mA) be powered
reliably in series from a 5V regulated supply without using a dropping
resistor? Or is this not enough margin for die differences, thermal
runaway, etc. This is at room temperature. Thanks.

If you wan't them to emit:

5 V
|
+--+---+
| |
.-. .-.
| | | |
| | | |
'-' '-'
| |
| |
| |
V V
- -
| |
| |
+--+---+
|
0 V
created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de

 

[Joop]

Curious, can two identical LEDs with Vf of 3V (30mA) be powered
reliably in series from a 5V regulated supply without using a dropping
resistor? Or is this not enough margin for die differences, thermal
runaway, etc. This is at room temperature. Thanks.

I would think so. Current is the same though both LEDs.
If you are sure that they can handle 30mA and at this current they
both have 3V drop (or at least >2.5V each), then the actual current
level will be less then 30mA.
How much less is hard to tell. Perhaps even so small you will not see
them light up at all.

Joop

 

[Don Klipstein]

I would think so. Current is the same though both LEDs.
If you are sure that they can handle 30mA and at this current they
both have 3V drop (or at least >2.5V each), then the actual current
level will be less then 30mA.
How much less is hard to tell. Perhaps even so small you will not see
them light up at all.

3V is not a common typical forward voltage drop at 30 mA for the usual
LEDs that have maximum current around 30 mA and characteristics specified
at 20 mA. I suspect this may be a maximum value for the forward voltage
drop for an orangish red, orange, yellow or yellow-greeen LED, and the
typical voltage drop at 30 mA could be 2.2 to 2.4 volts.

If this series pair of LEDs is connected directly to a 5V supply, I
imagine it will probably draw more than 30 mA. More likely than not, I
imagine it will survive - but I don't see adequate reliability for putting
this in any sort of production units.

If there is a diode drop from a non-CMOS IC output in the way, then you
get about 4.4V rather than 5V. A series apir of LEDs of color orange-red
to yellow-green as well as high brightness red ones will probably glow and
probably run safely, but the current may vary excessively with
temperature. Current will probably also vary excessively from one unit to
another if you make a lot of them. If you make a production run like
this, I consider it possible that a significant percentage opf the LEDs
may fail in the long term.

Most CMOS IC outputs normally have enough internal resistance
for driving from 5V a series pair of LEDs that are high brightness red,
high efficiency red, or orange-red to yellow-green.

You can get good reliability of safe operation if you add in series with
this LED pair a resistor of value 33 or 47 ohms.

- Don Klipstein ([email protected])

 

[R.Lewis]

It's only at room temperature for the first nanosecond you power it
up...



No more than any other diode.


It "can" be done, if you use a photodiode to measure the LED's
light output and have that drive an op-amp which powers the LED.

Then you will get a led that changes its output with ambient temperature
which, depending upon the led type (as its colour chages with temperature
and hence the response of the photodiode) , can be made to go up, down, but
not, as yet, sideways.

 

[Joop]

3V is not a common typical forward voltage drop at 30 mA for the usual
LEDs that have maximum current around 30 mA and characteristics specified
at 20 mA. I suspect this may be a maximum value for the forward voltage
drop for an orangish red, orange, yellow or yellow-greeen LED, and the
typical voltage drop at 30 mA could be 2.2 to 2.4 volts.

If this series pair of LEDs is connected directly to a 5V supply, I
imagine it will probably draw more than 30 mA. More likely than not, I
imagine it will survive - but I don't see adequate reliability for putting
this in any sort of production units.

<snip>
Sure, for production units this is bad design. Therefore I mentioned
if he was sure the forward voltage was 3V.
But from a theoretical point of view it might work. If the OPs remark
to 'reliably' would mean will components fail, then they will not if
he can guarantee the rating he has given.

But if he means can it be used in production volumes, give a
predictable amount of light, using random LEDs with Vf of 2.2 to 3V,
then the answer is indeed no. Production units were however not
mentioned in his post.

 

[Apostrophe Police]

If you wan't them to emit:

Five Demerits for the Most Egregious Misplaced Apostrophe In The
English Language.

"Want" is one single word, and that's all there is to it.