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
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.
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.
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
You know that 2*3V is greater that 5V ?
-- Reinhardt Behm, Nauheim, Germany, reinhardt.behm@t-online.de
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
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
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.
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. 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).
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.
-----
I read in sci.electronics.design that Anthony Fremont wrote (in ) about 'Q: LED Margin?', on Sat, 25 Dec 2004:
Just hope that getting an LED to act as a laser by passing a large current through it is just an urban myth.
-- Regards, John Woodgate, OOO - Own Opinions Only. The good news is that nothing is compulsory. The bad news is that everything is prohibited. http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
If you wan't them to emit:
5 V | +--+---+ | | .-. .-. | | | | | | | | '-' '-' | | | | | | V V - - | | | | +--+---+ | 0 V created by Andy´s ASCII-Circuit v1.22.310103 Beta-- Best Regards, Mike
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
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 ( snipped-for-privacy@misty.com)
dropping
to
devices.
series
supply.
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.
(as
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.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.