# resistors for leds



## msilhunter (Nov 5, 2012)

Im think of buying alot of resistors for my leds thats on the layout. If you have 2 leds at 14volts and forward voltage at 2.4 and the forward current at 20ma. It has the ohm at 470 and the 1/2 watt. If i use a 1/4watt resistor would it be x2 or how would you figure it out thanks Craig


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## rrgrassi (May 3, 2012)

NIMT can answer that one for you...


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## DonR (Oct 18, 2012)

Sean and Gunrunnerjohn can give you the math.

But the long and short of it is that I power my 3 LED 12 V strip
lights with a 1/4 watt 470 ohm resistor operating on a source of 9 vdc.
That same 470 ohm will light single 2 V dc LEDs also. The actual
resistance, in my experience, makes little difference in LED brightness
until you get resistance near 10 k. 

I'm having difficulty determining the resistance and power rating
of resistors for the 2.2 mil 1.5 volt, .04 amp incandescent bulbs. A 1/4 watt
200 ohm resistor with 9vdc source will make it glow the way I want it. Adding a
2nd 1.5 bulb in parallel reduces the current to just over 2 volts and
no detectable light from either. At that voltage they should
be bright. Why no light? Remove one of the bulbs
the other glows again. There is no short circuit.

I am going to try a half watt and a 1 watt resistor to see which will
do the job.

Don


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## gunrunnerjohn (Nov 10, 2010)

Basically, you figure 1K of resistance for one volt of drop drawing 1 milliamp. Extending that, if you want to power an LED at it's full rated 20ma, you need 50 ohms for each volt of desired drop from the supply.

Assume a 12 volt supply, and a white LED that has a 3V operating voltage, you'd need to drop 9 volts. 50 * 9 = 450, so that would be the lowest value of resistor you should use. I rarely operate any LED at the rated current, since they end up being too bright for modeling most of the time. So, if I wanted to operate this LED at 10ma, for instance, you need 100 ohms for each volt of drop, or 900 ohms for the resistor. I also simply round up the the next standard value.

For different supply voltages, LED operating voltages, series LED's, etc., simply do the same computation.

For instance, most red/yellow/green LED's operate at 1.5 volts. Many times, I'll wire several LED's in series to minimize component counts. If you wire four red LED's in series, you have a 6V operating voltage, using the same 12 volt supply, the resistor would be at least 300 ohms.


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## DonR (Oct 18, 2012)

John

I have a scratch built flood light on a pole.

There are 2 Model Power 1.5 volt 
.04 amp mini incandescent bulbs wired in
parallel. The source is 10 vdc. I want them
to glow less than full brightness for longevity.

What resistor in ohms and watts should I use
to accomplish this? I have been experimenting
using a 9 volt battery and failing to find a resistor
or combination in series to do the job. 

The bulbs work nicely powered by a 1.5 volt AA battery.

As I said in an earlier post, I can use various size resistors
(in ohms) and see little effect on brightness.

Appreciate your knowledge.

Don


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## gunrunnerjohn (Nov 10, 2010)

Well, you have .08A, or 80ma of power at 1.5 volts. So, at 10 volts DC, you need to drop 8.5 volts with whatever regulator you use, resistor or electronic. If I calculate the resistor, I get 106 ohms at .68 watts power dissipation.

For full brightness, I'd use a 2 watt resistor, the closest standard size is probably 120 ohms. You can move up from there. For instance, if you used a 220 ohm 1 watt resistor, you'd have roughly 1/2 the rated current, and the power would be .35 watts, so a 1 watt rated resistor would do the trick.

Here's a convenient Ohms Law Calculator that will allow you to do the *what-if* scenarios.

The obvious issue with incandescent lighting and using a resistor is the resistance of the lamp changes as it lights up! So, many of these calculations don't apply as directly, though they should generally work.


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## DonR (Oct 18, 2012)

Thank you John

I'll go on your calculations.

Something wrong with the ohmslawcalculator, tho. The 'calculate'
button has been deactivated. The reset works but not
the Calculate. Any idea?

Don


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## drabina (Mar 19, 2013)

I use this calculator for LED resistor values:

http://led.linear1.org/1led.wiz

There is a link for multiple LEDs scenarios as well.


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## gunrunnerjohn (Nov 10, 2010)

DonR said:


> Something wrong with the ohmslawcalculator, tho. The 'calculate'
> button has been deactivated. The reset works but not
> the Calculate. Any idea?
> 
> Don


Nothing wrong with it, you have to enter two of the values before it's possible to calculate, so the button is not active until there are two valid entries.


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## DonR (Oct 18, 2012)

John

Must have been the download. Tried your calculator again and
it worked this time. I had entered the 2 items of data and
calculator would not respond the other time.

Drabina's calculator works the same way. Both will sure come
in handy and answers the 'what resistor?' question other guys
on the Forum have also asked.

But, alas, I got so excited about knowing what to do, got in a hurry,
and, would you believe, I connected those 2 1.5 volt bulbs directly to the 10 volts
Source. May they rest in peace. 

Not bad enuf, I got replacements but they are different. 1.5 v but
draw .0125. Wisely I tested with only one bulb...thru the 1 watt 220 ohm
resistor, poof. Another basket case.  Still, a 1.5 v .04 amp bulb
only glowed warmly in that circuit. Obviously the current draw must
be a part of the equasion. The calculator will solve that. 

I am still shaky on one point. Does the resistor power factor (1/4, 1/2, 1 or 2 watts)
affect the equasion other than to permit heavier draws? In other words,
If a 1.5 v .04 amp bulb glows right with a 220 ohm 1/2 watt resistor
would it have a different brightness (voltage) if the resistor were 220 ohms but only 1/4
watt.

Thank y'all

Don


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## gunrunnerjohn (Nov 10, 2010)

The current draw obviously enters into the computation of voltage drop, that's what causes voltage drop. 

The wattage of the resistor is simply how much power it will have to dissipate. I like the power in the resistor to be no more than 1/2 of the rated power of the resistor.


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## DonR (Oct 18, 2012)

Thanks again John

I had hoped that the power rating of a resistor was not
pertinent to the equation. So for these small bulbs I have
a good deal of freedom of resistor capability. I have, 1/4,
1/2 and 1 watt resistors available.

Don


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## gunrunnerjohn (Nov 10, 2010)

When you calculate the value needed, note the power and double it, that's the minimum size resistor I'd use. Normally, for LEDs, 1/4W is all that's required. Incandescent bulbs can draw much more.


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## Patrick1544 (Apr 27, 2013)

msilhunter

Here's a neat little calculator for getting the resistance values correct. It even lets you swing an array of more than one LED. Hope this helps. I use it a lot, myself. You just need the voltages from the LED package to compute the resistance required.

http://led.linear1.org/led.wiz


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## Hutch (Dec 19, 2012)

I like it. Don't know if it will be useful but it's cool and I bookmarked it. Thanks



Patrick1544 said:


> msilhunter
> 
> Here's a neat little calculator for getting the resistance values correct. It even lets you swing an array of more than one LED. Hope this helps. I use it a lot, myself. You just need the voltages from the LED package to compute the resistance required.
> 
> http://led.linear1.org/led.wiz


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## Southern (Nov 17, 2008)

DonR said:


> John
> 
> I have a scratch built flood light on a pole.
> 
> ...


wire them in series. see if you like that better. I had six street lights that are two bright. I rewired them so that now I have 3 sets of two.


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## golfermd (Apr 19, 2013)

How many in series?. Would think parallel would be better since voltage drop is equal for each circuit. In series you will get bright at head and drop in brightness each succeeding light.

Dan


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## gunrunnerjohn (Nov 10, 2010)

Actually, series is better. You will NOT get brighter and dimmer, ALL the current has to go through every item in a series string. In truth, there's more chance that a parallel connection will have varying brightness. In addition, series wiring will drop less voltage in the resistor or other current limiting device and use more of it to actually light the lamps.


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## DonR (Oct 18, 2012)

I did use a series circuit for my street lights. Turned out the number
of lights in series equaled the source voltage so no
resistance was required. Good old fashioned Christmas 
tree light style. Just hoping one doesn't burn out. Can't
unscrew em like the tree lights. 

These tiny incandescent bulbs are perfect for street lights.
I use a 'chrome' SEQUIN for the 'shade'. They look good
as a ceiling light for a porch, or again with a sequin shade
as an outdoor light over a door or to light a freight platform.

I am debating with myself tho about that very series vs parallel 
method for my twin bulb flood light.

I ran into a 'bump' when I went to use the
resistor calculator in a series circuit. I was unsure
how to enter the current. Each 1.5 v bulb draws .0125 amps.
Would the 2 in series = .0250 amps which seems 
logical to me but sometimes electrical rules are
different. 

Don


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## Hutch (Dec 19, 2012)

Nope. Same current flowing through both. You'll have a voltage drop across each one.


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## gunrunnerjohn (Nov 10, 2010)

Take a look at Series and Parallel Circuits, saves me from typing a lot of stuff.


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## Hutch (Dec 19, 2012)

If you have a spreadsheet program like excel on you computer, you can enter all those equations into it and play with the values as needed and you don't have to remember them or break out your calculator.


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## gunrunnerjohn (Nov 10, 2010)

Lots of simple calculators on the Internet for any of this stuff.


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## DonR (Oct 18, 2012)

John

After answering so many electronic questions from me it may
shock you to learn that during high school I studied
electronics...alas..that was long ago. 

The wiki statement regarding series circuits is:
"In a series circuit the current is the same for all elements."

That is murky at best. So, let me offer, as example, my flood light with
2 -1.5 volt incandescent bulbs, each drawing .0125 AMPS.

I agree that in series the total voltage
is 3 volts and THINK that the total current draw is .0125.

Is that correct?

I agree that in parallel the voltage would be 1.5 volts with a current
draw of .0250 AMPS.

Is that correct?

Don


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## gunrunnerjohn (Nov 10, 2010)

Correct... and correct. 

Any current flowing in a series circuit has to go through every component and wire in the circuit. So, the current anywhere in the circuit will be exactly the same. 

In a parallel circuit, the current will be the sum of the individual devices current draw.

Note that in a series circuit, you have to insure that all devices operate at the same current, especially stuff like lamps. 

_Let's take an example_.

You put in series 8 1.5 volt bulbs, one is a .1 amp bulb, and the rest of them are much larger and consume an amp. The 7 larger bulbs will probably not light at all, even if you get to observe them in the brief moment before the .1 amp bulb gets toasted by the approximately 11 volts it would be dropping!


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## DonR (Oct 18, 2012)

Thanks again John

This has been a very interesting walk through
the electronic thicket. Hope some of the others
also feel more comfortable with their layout wiring
needs as a result.

Some of this ought to be attached to the electrical
knowledge sticky.

Don


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## Hutch (Dec 19, 2012)

I agree with you Don but this is a discussion forum and I like asking questions and getting various responses. Give an answer now and then too. We can have it both ways.


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## DonR (Oct 18, 2012)

Good point Hutch

Obviously, by the number of postings from me, I enjoy the
back and forth also. Get's the brain juices flowing when
someone comes up with a strange problem. 

Such as the guy with the 9 inch circle of N gauge track
and a lonely Amtrak loco that spins it's wheels instead
of moving. I'm thinking the radius is too tight and
binding the wheels. Everything else seems to check out.

Don


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