Q: LED Margin?

[Mark Jones]

Reinardt Behm wrote:

Mark Jones wrote:


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 ?

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

 

[Reinardt Behm]

Mark Jones wrote:

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 ?

--
Reinhardt Behm, Nauheim, Germany, reinhardt.behm@t-online.de

 

[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]

Mark Jones wrote:

Reinardt Behm wrote:
Mark Jones wrote:


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 ?


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.

--
Reinhardt Behm, Nauheim, Germany, reinhardt.behm@t-online.de

 

[Boris Mohar]

On Sat, 25 Dec 2004 12:00:35 -0500, Mark Jones <abuse@127.0.0.1> wrote:

Reinardt Behm wrote:
Mark Jones wrote:


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 ?


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.

--

Boris Mohar

 

[Ben Bradley]

On Sat, 25 Dec 2004 10:42:09 -0800, John Larkin <john@spamless.usa>
wrote:

On Sat, 25 Dec 2004 11:18:13 -0600, TCS
The-Central-Scrutinizer@p.o.b.o.x.com> wrote:

On Sat, 25 Dec 2004 12:00:35 -0500, Mark Jones <abuse@127.0.0.1> wrote:
Reinardt Behm wrote:
Mark Jones wrote:


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.

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

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


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.

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

-----
http://mindspring.com/~benbradley

 

[Active8]

On Sat, 25 Dec 2004 11:29:25 -0500, Mark Jones wrote:

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

--
Best Regards,
Mike

 

[Joop]

Mark Jones <abuse@127.0.0.1> wrote:

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

 

[Apostrophe Police]

On Sun, 26 Dec 2004 00:22:32 -0500, Active8 wrote:

On Sat, 25 Dec 2004 11:29:25 -0500, Mark Jones wrote:

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:

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.
--
Rich Grise, Self-Appointed Chief,
Apostrophe Police