# LED/resistor question



## Bwells (Mar 30, 2014)

I have an LED rated at 3.2 volts, 25 mA and want to feed it with a 12v supply. The resistor calculator says I need a 390Ω 1/2 watt resistor. Fine, the ohms seems to low but okay and the watts is going to dissipate the heat from such a drastic drop in voltage and it seems it may be too low. Does the resistor heat up dropping 9 volts? Just keeping you guys on your toes!!:dunno:


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## fcwilt (Sep 27, 2013)

Bwells said:


> I have an LED rated at 3.2 volts, 25 mA and want to feed it with a 12v supply. The resistor calculator says I need a 390Ω 1/2 watt resistor. Fine, the ohms seems to low but okay and the watts is going to dissipate the heat from such a drastic drop in voltage and it seems it may be too low. Does the resistor heat up dropping 9 volts? Just keeping you guys on your toes!!:dunno:


Yes the resistor will heat up since it is dissipating power (8.8 volts * 25 ma worth of heat).

Are you asking if it will be "hot" to the touch?

My experience is that LEDs are plenty bright (usually) at much less than rated current.

Get yourself a resistance substitution box and you can try values until you find just what makes the LED do what you want.

You can get them in a range of prices. 

There was a simple one on Amazon that had 24 common values for appx $20.

There was also a more versatile one with values from 1 ohm to 11,111,110 in 1 ohm increments for appx $34.

Frederick


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## mopac (Feb 24, 2011)

I am electronic challenged but I have done a few LEDs. 390 does sound too low. I would use something around 600 up to 1000. Yes, the resistor does get warm. I would not say
real hot.


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## wvgca (Jan 21, 2013)

Another alternative that you could consider would be to get around a 2k pot, and adjust the pot until you get a brightness that you are happy with. Then measure the pot value to obtain the desired resistor value.. I find that a reasonable brightness can be obtained at much less current [and a higher resistor value] than at full maximum rated current, plus they will quite likely last much longer as well .. 
On monday night I ran 1k resistors on white leds driven by 5V, and they were fine . it's surprising how large a resistor can be used and still get bright leds


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## rkenney (Aug 10, 2013)

W=VxA (watts = volts x amps remembered as the West Virginia (WVA) rule).

8.8 volts (drop across the resistor) x .025 (total circuit current) = .22 watts.


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## Bwells (Mar 30, 2014)

wvgca, the pot idea is a good one. With a volt meter across the LED, you would know exactly what you were putting into it. I played around with the resistance calculator and made a graph as follows:







Seems like 120Ω drops 3 volts until you get to 3.2 volts and then the stuff hits the fan. Something is happening here and the graph appears to level out. Why does a 470Ω do .5,1, and 2 volts? It appears that a resistor above 500Ω would not change anything.


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## Tom_C (Jan 8, 2016)

Bwell, I don't think that is giving you meaningful information because the voltage drop across the led isn't going to change that much. 

I'm assuming you used descrete resistors for these tests, so if you measure the voltage across the resistor then you can calculate the current through the LED, and calculate the power dropped by the resistor.

ADDED: But for your purpose all you care about is what resistor value to use for the desired brightness, right?


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## wvgca (Jan 21, 2013)

Bwells.. the resistor when used with a LED is used to limit the current passing through the LED ..not voltage
Most OnLine LED resistor calculators work with an input voltage, and a user determined current value through the LED ...
There is an good current calculator on Rob Paisley's site ..

http://home.cogeco.ca/~rpaisley4/LEDcalc.html

The ma value stated on most LED specs is the MAXIMUM current that a LED will withstand, and is for MAXIMUM brightness .. which for most model train uses is normally too bright in my opinion ..

Different colour LED's normally have different voltage drops across them, so there isn't really ONE resistor value that will work with all LED's ...

It's pretty hard to determine how bright a LED will be without the actual manufacturer's spec sheet, but it's pretty easy to do that physically by using a pot..

I'm not sure if this is the answer that you're looking for ..I'm assuming that you actually want to adjust the brightness, and not run the LED at maximum .. and to check with a multimeter you would set it on current, 50 ma range if the is available on your meter, and put the meter in series

edit .. Rob's calculator will also tell you how big of a resistor you need .


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## Bwells (Mar 30, 2014)

Tom, I don't know what a discrete resistor is, I assume one that has 0% tolerance. These were not tests that I did but from a calculator online and I seem to recall they said assuming 5% tolerance and rounded up to the next "standard" value resistor.


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## Bwells (Mar 30, 2014)

Shoot Warren, I thought it was to cut down the voltage!


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## Tom_C (Jan 8, 2016)

Bwells said:


> Tom, I don't know what a discrete resistor is, I assume one that has 0% tolerance.


A discrete resistor (or other component) just means it's something you can hold in your hand, as opposed to something built into another device. So a single resistor is a discrete resistor.


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## wvgca (Jan 21, 2013)

Bwells said:


> Shoot Warren, I thought it was to cut down the voltage!


lol .. not really ...
that's why it's normally referred to a _current_ limiting resistor ..

Do you have some kind of variable resistance pot laying around??
If not, check locally, or I can even mail you one??

have fun experimenting ..


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## Lemonhawk (Sep 24, 2013)

I'm with wvgca on this. I use 1k resistors when connecting up an LED to 12v. Its plenty bright and increases the lifetime to forever. The LED is a current device and there are IC's that will generate a constant current such as a CL2, which generates a constant 20 ma. And you want to measure the voltage drop across the resistor, not the LED. That way you can determine the current. You can also connect LED's in series which will reduce the resistor Ohms but do not eliminate the resistor entirely. Yes its a nice chance for a learning experience to use a multimeter and Ohm's Law (it's a hobby of many facets). A 12 v wall wart, some alligator clip leads, a 2k pot an your trusty multimeter will provide a nice leaning period. Remember the LED is a diode, works one way but not the other - connecting them the wrong way will not damage them.


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## rkenney (Aug 10, 2013)

Bwells said:


> Tom, I don't know what a discrete resistor is, I assume one that has 0% tolerance.


A discreet resistor is one who won't tell your wife! :laugh:

A discrete resistor is one that is not packaged with another component. ie a single resistor


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## Bwells (Mar 30, 2014)

rkenney said:


> A discreet resistor is one who won't tell your wife! :laugh:
> 
> Doesn't matter, she thinks I'm an idiot anyway!
> Warren, I do have a pot but a rheostat would be more my speed. The third leg is a bit confusing and I see no use for it.
> Good posts and thanks guys but I keep going back to the garden hose idea when it comes to pressure(amps) and volume(voltage) with a kink(resistor).:dunno:


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## highvoltage (Apr 6, 2014)

Bwells said:


> ...but I keep going back to the garden hose idea when it comes to pressure(amps) and volume(voltage) with a kink(resistor).:dunno:


Close, but pressure relates to volts, flow is current, and a restriction (or kink) is correct as a resistance.


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## Tom_C (Jan 8, 2016)

Pressure is voltage, volume is amps.


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## Bwells (Mar 30, 2014)

Ok, looks like my hose idea is hosed.
How about this, my skillsaw pulls around 13 amps at startup and on a 15 amp breaker will dim the lights till it picks up the RPM's. On a 20 amp breaker I assume it is still pulling 13 amps at startup but could care less what is available amp wise. It takes what it wants and is happy. Why doesn't the LED take 25mA and pass the rest down the road?


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## gunrunnerjohn (Nov 10, 2010)

The Skillsaw is designed totally different than a component LED. As others have explained, the LED is a current mode device. It goes from no light to maximum rated current over a very narrow voltage range. Here's an illustration. at 2.8 volts, the LED is drawing almost no current. If it's lit, it's very dim. However, at just over 3.3 volts it's at full rated 20ma current. At just .2 volts more at 3.5 volts, you're at over twice the rated 20ma current!

Note that I'm looking at the white plot and this is a 20ma LED. The blue plot is actually for a blue LED, the ratings, as you can see, are similar.

You can always go higher in current limiting resistors, but you should never go lower.


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## Tom_C (Jan 8, 2016)

Your hose idea is accurate, except that pressure = voltage, and water flow = current.

The skill saw has built in resistance, so it will pull current equal to Voltage divided by it's resistance, I = E/R (I is current, E is voltage)

The LED doesn't have any resistance (or little) so it will pull enough current to burn itself out unless you give resistance to the circuit.

ADDED: A motor is a bit different than just a resistor, but that's not really important in this thread.


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## highvoltage (Apr 6, 2014)

Bwells said:


> Ok, looks like my hose idea is hosed.
> How about this, my skillsaw pulls around 13 amps at startup and on a 15 amp breaker will dim the lights till it picks up the RPM's. On a 20 amp breaker I assume it is still pulling 13 amps at startup but could care less what is available amp wise. It takes what it wants and is happy. Why doesn't the LED take 25mA and pass the rest down the road?


A motor requires more current to start than to run. The reason is the motor shaft is at rest and is trying to turn. Inertia keeps it from turning instantaneously, so the current consumption is high trying to get it started.


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## Bwells (Mar 30, 2014)

LEDs do seem to have a very narrow voltage range but wouldn't providing excessive mA be just as detrimental as over voltage, POOF! It's like shooting a bullseye from 300 yards while it's moving, you need to be good. A cannon would do it or just a big resistor and be happy.


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## Tom_C (Jan 8, 2016)

Bwells said:


> LEDs do seem to have a very narrow voltage range but wouldn't providing excessive mA be just as detrimental as over voltage, POOF!


Yes, it's not the voltage that damages the LED, it's the current. (keeping it simple, there could be reverse voltage damage, but that's not important for this thread) Look at that graph John posted; if you tweak the voltage just a little bit and current goes through the roof. 

Like drinking out of a firehose. You can do it, but I'd prefer if there was a large resistance to slow down the flow.


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## Bwells (Mar 30, 2014)

Okay Tom, I think things are starting to click. Here is my thoughts and see if I understand. You have an LED and a resistor with a 12 volt supply. The resistor is sized according to the rated spec of the LED mA-wise. Voltage is irrelevant as well as the supply current amps. We need to close the door on anything over 25mA but let in all the voltage as the LED can pass this but it can't tolerate too much current. Does this sum it up? I always thought that voltage is what blows em up.


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## Tom_C (Jan 8, 2016)

More or less.  Voltage does matter, but it's the current that's going to kill it.

The job of the resistor is to limit current and drop voltage.

Lets say you put 2 equal resistors in series and hooked them to your 12v supply. If you measured from ground to the center of the 2 resistors you'd get 6 volts, so each resistor drops 6v.

The current flowing through the resistors can be calculated as 12/2R. And, because the resistors are in series you can also calculate the current through either 1 resistor by 6/R.

Substitute your LED for one of the resistors and you need to adjust the value of R to get you down to 25ma. And, if you used the same LED as John posted in that graph then the voltage across the LED would match up to 25ma on that chart... the resistor would then have to be dropping the remaining voltage.


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## rkenney (Aug 10, 2013)

In a closed circuit the voltage is entirely used across its components.

In a series circuit (as with our LED and resistor) the voltage across (dropped by each device will be different. The current used by each device will be the same. (when we know the current through the resistor, we also know the current through the LED because the are the same).

In a parallel circuit voltage is the same across each component. The current through each component will vary.

The relationships between current, voltage and resistance is where the hydraulic analogy falls apart. In a hydraulic system a constriction (resistance) increases pressure while reducing volume (current).

All the plumbing theory does is confuse the whole issue.


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