# Series Resister or not

Discussion in 'Electronic Basics' started by GrimReaper, Oct 21, 2003.

1. ### GrimReaperGuest

Would I be right in assuming that if the supply voltage was the same as the
forward voltage of the LED then there is no need to fit a series resister or
do you have to have a higher supply voltage and always fit a series resister
to drop the supply voltage back to the forward voltage of the LED to limit
the current?

GrimReaper

2. ### GrimReaperGuest

Thank you my friend, I now know what to do

Regards
GrimReaper

3. ### BaphometGuest

You should always use a higher voltage supply and limit the current to the
led with a series resistor, unless you're using a current limited power
supply.

4. ### John GGuest

Very simplicticly.
You could imagine a LED as a constant voltage device and given that, the
current will rise to infinity or until the source limits it or the diode
turns to smoking. The resistor on the other hand, drops more volts as the
current rises and soon there is equilibrium.

5. ### dBGuest

Yes, use a higher voltage and a series resistor.

http://pub40.ezboard.com/fbasicelectronicsfrm5.showMessage?topicID=16.topic

6. ### Robert MonsenGuest

While I agree with the sentiment, diodes do actually have a 'resistance' of
sorts; its simply nonlinear, ie, it doesn't obey ohms 'law', so its a bit
more difficult to figure out. The forward voltage is related to the natural
log of the current through it, so the more current, the more voltage across
it, just like a resistor. IIRC, one rule of thumb is that if you increase
the current by 10 times (a decade), the forward voltage will go up something
like 60mV, at room temperature (at least this works for bipolar transistor
diode junctions).

I think this means that the OP can omit the resistor without explosions...

Regards
Bob Monsen

7. ### Dimitrij KlingbeilGuest

As a general rule, you should use a series resistor and a slightly higher
(+0.5-1V) voltage. However a LED is not a constant voltage device and most
LEDs have noticeable resistances that allows them to operate quite well
ballanced at a constant voltage. Note nevertheless that the difference in
the forward voltage rises very slowly on substantial current increase due to
a LED's nonlinearity. This means that your power supply voltage should be
_really_ constant, small changes may become lethal for the LED. Never rely
on a 'trafo, diode and cap' style of power supply for a voltage stability.
Note also that an LED is a temperature-sensitive device. When operating at a
constant voltage it should be kept away from high temperatures and mounted
in a case that can provide slight heatsink capabilities. If you plan to
operate the LED with batteries, their internal resistance can help making a
series resistor unnecessary, but batteries change condition easier than one
can sometimes think, so I would classify a device of this sort as 'not
reliable'.

8. ### GrimReaperGuest

Thanks for all your info. I want to connect 21 Blue LED's as part of a Coral
Tank lighting system. The LED's have a forward voltage of 3.4v. A current of
supply and put 3 LED's in series with a 91 ohm resistor and repeat this
series configuration 7 times. Does that make sense?
No good at drawing that here. I am a marine biologist and what I know about
electronics I have gained from here and Google searches - that's it.

Regards
GrimReaper

9. ### Lord GarthGuest

That'll do nicely GR... now using P=I²R, calulate the power rating of the
resistor you
need....Oh, use the nearest standard value resistor, it isn't at all
critical. Don't forget
to add a little extra power rating for slop margin.

You have 11.2 volts across the LEDs, leaving .8 volts across the resistor,
when supplied
with 12 volts.

10. ### John FieldsGuest

---
Yes.

The difference between the supply voltage and the sum of the LED forward
voltages divided by the LED current is 90 ohms and you have chosen the
nearest standard 5% value, 91 ohms. Well done.

All that's left to do is determine the wattage required for the resistor
to dissipate, and that'll be the difference between the supply voltage
and the sum of the LED forward voltages _multiplied_ by the LED
current,

(12V-10.2V)*0.02A = 0.036 watts

so a 1/4 watt resistor will be fine.
---

11. ### Lord GarthGuest

Oops John caught me !

Seems I flubbed that 3.4 times three ....back to second grade I go!!!

12. ### GrimReaperGuest

John / Lord Garth
Thank you for that. I had not thought about the power dissipated in the
resistor

Kind Regards
GrimReaper

13. ### BaphometGuest

That's certainly one of the things that makes this such a powerful medium. A
final solution becomes a collaborative effort, each one seeing a possible
solution from a slightly different perspective.

14. ### Watson A.Name - Watt SunGuest

"Just like a resistor"? You're saying that the resistor, which obeys
ohm's law, obeys the natural log relatonship? I don't think so.

And there is a resistive component to the current/voltage
relationship, just that it's not a predominant component. There is
the bond wire resistance, the internal resistance of the chip itself,
etc. But as the LED gets hotter, the V drop gets lower, so putting
the LED on a constant voltage supply means that it will draw excessive
current as it warms up. This could let the smoke out.

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15. ### Robert MonsenGuest

No, I'm not saying that. You apparently misread my post...

Regards,
Bob Monsen

16. ### Watson A.Name - Watt SunGuest

It looks like you have never measured the voltage across a LED while
it's operating. If you had, you would know that it has a negative
tempco. As the LED gets warmer, it drops less voltage. This is
really pronounced on big LEDs such as the Luxeon Star. If you connect
a LED to a constant voltage, it draws more current as it gets hotter,
which causes it to draw more current and get even hotter, in a vicious
circle which will destroy the LED if it doesn't shorten its life

[snip]

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17. ### Watson A.Name - Watt SunGuest

Eh? 3.4V times 3 is 10.2V. That leaves 1.8V across the res. There
will be some variation in the voltage drop, but it shouldn't be too
much. That can be measured with a DMM. If it's too much, change the
LED.

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18. ### Watson A.Name - Watt SunGuest

If you can't get that relatively hard-to-find value, use two 47 ohms
(much more common) in series, or 100 ohms will work okay with no
perceptible difference in brightness.
I bought a couple dozen blue LEDs and I've found that every one of
them that I've taken out of the package is intermittent. And it's
really strange, too. I crank the PS up to 1.5V and the LED starts to
draw current, maybe a few mA, but the current is really erratic, I
can't get a stable reading. If I go up to 3.5V at 25 mA, sometimes
the LED will light steady, sometimes it will blink, maybe a few times
a minute, maye less, once every few minutes I've checked at least 6,
probably more, and they're all like that. Bad news. Sucks.

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Don't be ripped off by the big book dealers. Go to the URL
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http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
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19. ### Don KlipsteinGuest

Maybe it's static damage. Most blue LEDs are static-sensitive.

- Don Klipstein ()

20. ### Dan DunphyGuest

In an LED VI Curve, the independant variable is current, and the
dependant variable is voltage.
Use the resistor. You can series a number of leds with a higher
voltage supply and one resistor, or use a current scource, if you want
to get sophisticated.
Dan