Resistors for LEDs

[Bino]

How do I identify which resistors to buy for operating LEDs through 12 or 16v power supply?

[smokebox]

1000 ohm as a minimum, up to about 7000 ohm maximum depending on how bright you want them.

 

There are calculators for the basics https://www.digikey.co.uk/en/resources/conversion-calculators/conversion-calculator-led-series-resistor

[RAF96]

Here you go...

https://www.petervis.com/electronics/led/led-resistor-calculator.html

Rob

Edited November 26, 2018 by RAFHAAA96

[cliff park]

One problem with all of these calculators is that they don't take account of 'preferred values'. In other words if the calculation says '395 Ω' then the modeller tries to buy one he will come unstuck. Always go up to the next preferred value, and if in doubt start high, if it's too dim come down. Unless you are feeding a lot, or they are very high powered, a ¼ watt should always be adequate. Several companies sell a selection pack fairly cheap, they always come in handy.

[Junctionmad]

Most modern leds can operate on 2mA , I usually now pick 5mA as a good starting point

 

Hence R = V/I , where V = power supply dc voltage -led forward voltage drop

 

You can use 2 volts as a rule of thumb for most led fwd drops

 

Hence for 12 V it’s 10/.005 = 2000 ohms , pick the nearest higher value to whatever value comes out

 

I run my calculations for 3mm leds at 2mA these days

 

Note that 16 v in model railways is typically AC , which will need rectifying with a diode before driving the led , the actual peak voltage can vary a lot , often upto 20 v dc peak or more

Edited November 26, 2018 by Junctionmad

[DCB]

Here you go...

https://www.petervis.com/electronics/led/led-resistor-calculator.html

Rob

Not too sure about the Blue Green Pink Vf  mine gave been nearer 2.2 Volts as I power them with 2 X AA batteries.   The resistor chart gave me a minus figure for my set up which was reassuring. 

[AndyID]

gave me a minus figure for my set up which was reassuring. 

 

That's pretty scary. Sounds like you are violating the first law of Thermodynamics. That could easily rip a hole in space-time.

[kevinlms]

Not too sure about the Blue Green Pink Vf mine gave been nearer 2.2 Volts as I power them with 2 X AA batteries. The resistor chart gave me a minus figure for my set up which was reassuring.

Did you buy yourself some zero value resistors? They do exist! Used for machine soldering of PCBs.

[Junctionmad]

Not too sure about the Blue Green Pink Vf  mine gave been nearer 2.2 Volts as I power them with 2 X AA batteries.   The resistor chart gave me a minus figure for my set up which was reassuring.

 

Those charts are very dated , modern leds need about 2mA not 20mA

[melmerby]

Those charts are very dated , modern leds need about 2mA not 20mA

And the voltage drops vary somewhat from the values given.

 

e.g. Kingbright L-7104LGD green 3mm is typ 1.9v @ 2mA

 

Keith

Edited December 6, 2018 by melmerby

[meil]

I suggest you look here: http://led.linear1.org/led.wiz

[Brian]

Don't worry to much about the 2 milliamp or 20 milliamp ratings shown in some posts.  IMO they are actually confusing the question.

Based on your rated input voltage, a 1000 to 2200 Ohm (1K0 to 2K2) resistor in series with either lead of an LED will be fine. But where possible always use a DC power supply rather than an AC one, as AC may well stress a LED.   However, you can of course use AC ideally via a series diode. Or direct feed add an Inverse diode and then reduce the Ohm value of the series resistor by approx 50%.  Here is a typical LED chart with current around the 10MA range... https://www.rapidonline.com/kingbright-l-7104gd-3mm-green-led-20mcd-55-0105

Best for LED feeding is to use a DC regulated power source.  There are also AC rated LED available too. but possibly a little harder to find?

Edited December 9, 2018 by Brian

[Junctionmad]

Don't worry to much about the 2 milliamp or 20 milliamp ratings shown in some posts.  IMO they are actually confusing the question.

Based on your rated input voltage, a 1000 to 2200 Ohm (1K0 to 2K2) resistor in series with either lead of an LED will be fine. But where possible always use a DC power supply rather than an AC one, as AC may well stress a LED.   However, you can of course use AC ideally via a series diode. Or direct feed and then reduce the Ohm value of the series resistor by approx 50%.  Here is a typical LED chart with current around the 10MA range... https://www.rapidonline.com/kingbright-l-7104gd-3mm-green-led-20mcd-55-0105

Best for LED feeding is to use a DC regulated power source.  There are also AC rated LED available too. but possibly a little harder to find?

Seriously , if you can’t apply ohms law , ie divide one number by another , I struggle to understand how to help people with this issue

[melmerby]

But where possible always use a DC power supply rather than an AC one, as AC may well stress a LED.   However, you can of course use AC ideally via a series diode. Or direct feed and then reduce the Ohm value of the series resistor by approx 50%. 

IMHO you should never use AC unless you have a) diode in series to block the reverse half cycles or b) shunt the LED with a diode back to back so that it passes the reverse half cycle.

Bear in mind LEDs have a very low reverse polarity breakdown voltage, typically 5V.

A 16v AC supply and a 1K series resistor will easily take the LED into a reverse breakdown situation.

 

Keith

Edited December 7, 2018 by melmerby

[Brian]

Yes correct.. I forgot to type and I've now corrected it, to add the words "Add an Inverse diode" in the comment.  Otherwise why reduce the resistor value?  But Hay Ho, we all make errors in typing at times!!!!

Don't forget AC rated LEDs are also available.

Edited December 9, 2018 by Brian

[Brian]

Seriously , if you can’t apply ohms law , ie divide one number by another , I struggle to understand how to help people with this issue

Bit harsh isnt it??  

Isn't it a case that we try and provide info. to help people who ask, not belittle them!! 

Edited December 9, 2018 by Brian

[Junctionmad]

Bit harsh isnt it??  

Isn't it a case that we try and provide info. to help people who ask, not belittle them!!

 

I was referring to the attempts to point people at various “ charts “ and precomouted solutions , when the reality is if you can divide two numbers you can derive the value especially in cases where the situations aren’t directly covered by charts etc

 

Surely we should ( as I did earlier in this thread ) teach people to fish , rather then point them at a chipper !!

[Brian]

Seems very old hat to me!  Like using a slide rule or pencil and paper to work out something, when a simple On line calculator is readily available!   

What has to be remembered more is with any calculation how ever reached, is that the figure calculated is the Minimum and this may well not be available, therefore the next highest value in the range should be chosen as a minimum.

[Junctionmad]

Seems very old hat to me!  Like using a slide rule or pencil and paper to work out something, when a simple On line calculator is readily available!   

What has to be remembered more is with any calculation how ever reached, is that the figure calculated is the Minimum and this may well not be available, therefore the next highest value in the range should be chosen as a minimum.

An online calculator that is largely simply dividing two numbers !

 

As for the statement the value is the minimum , this is only strictly true if the value chosen for current is approaching the max allowed , n many cases leds will operate across a wide ranges of currents , ie 2-20mA , hence if you are using low currents , then the next resistor size down may in fact be fine.

[sharris]

Another thing to remember is that different colours have different forward voltages, different efficiencies and perceived brightnesses.

 

If you're doing, say, a 4 aspect signal or traffic lights, once you've worked out what the minimum safe value of resistor is for each colour, balancing the light levels for each can be a bit trial-and-error, adding extra resistance into each. You may end up with very different resistors for each colour.

[Brian]

Actually the On Line calculators allow the current to be set, so if you want maximum permissible then use that value e.g. 20ma (0.02A) if you want to reduce it then enter that value instead e.g. 5ma. The answer given is for whatever current value you enter.

IMO it couldn't be simpler to obtain the minimum value required for a specified supply voltage, LED forward voltage, LED forward current and some will even allow you to add the number of LEDs in a series chain too if applicable.  Simples!! 

 

Altering any value after the calculation is down to the user, but the starting place (Resistors minimum value) is immediately given as too may also be shown the recommended resistor minimum wattage value .

[Tricky Dicky]

An online calculator that is largely simply dividing two numbers ! .

For many it is not as simple as dividing two numbers it is what numbers to use, whereas those in the know will consult manufacturers data sheets. Beginners would need to know where to first obtain that data let alone make sense of what appears gobblygook even if they could obtain the relevant data. This hobby is a broad church and not everyone wants to become an electronics expert to run a layout and why should they? I do not think it is unreasonable for people to take out a " black box" approach of saying " I have a certain colour LED and a certain power supply, what resistor do I use? After all we all have controllers and most have no idea what is going on in the box but it does not stop us using them or realising certain inputs will result in certain outputs. If online calculators get solutions for people to enjoy their hobby then I wholeheartedly condone their use! Rant over.

 

My own preference for a recommended calculator is Electronics Assistant available as Windows only FREE download from here;

 

https://www.electronics2000.co.uk/

 

The series resistor calculator allows you to get a solution with the minimum information of LED colour and supply voltage to putting whatever data you have to hand. It shows you the formula and what figures it is using and will not only provide the answer but suggest the nearest preferred value. Pressing the help button will educate you about the calculation and what all the abbreviations mean. Yes educating yourself on certain aspects of electronics will be useful to operate a layout but is not essential to enjoy the hobby nor is it everybody's desire.

 

Richard