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LED and Resistor Sizing

 

I'm trying to avoid extra "stuff" because of space reasons, so I'm looking at putting two LEDs in series. If I have two white LEDs* that operate on about 4V at 20 mA connected in series, using a 1 kOhm resistor at 12V, what rating does the resistor need to have?

Can I subtract the required voltage of the LEDS like this?
12V - 4V -4V = 4V
4V / 1000 ohm=.004A
.004A * 12V=.048W

Do I need to keep it in like this?
12V / 1000 ohm=.012A
.012A * 12V=.144W

*

Puckdropper
 

I have a program that I use and this is what it says to use. I hope this helps.?




On Tuesday, September 5, 2023 at 07:49:36 PM EDT, Puckdropper via groups.io <puckdropper@...> wrote:


I'm trying to avoid extra "stuff" because of space reasons, so I'm looking at putting two LEDs in series.? If I have two white LEDs* that operate on about 4V at 20 mA connected in series, using a 1 kOhm resistor at 12V, what rating does the resistor need to have?

Can I subtract the required voltage of the LEDS like this?
12V - 4V -4V = 4V
4V / 1000 ohm=.004A
.004A * 12V=.048W

Do I need to keep it in like this?
12V / 1000 ohm=.012A
.012A * 12V=.144W

*

Puckdropper
 



On Tuesday, September 5, 2023 at 08:00:47 PM EDT, PAUL HART via groups.io <psr436@...> wrote:


I have a program that I use and this is what it says to use. I hope this helps.?

Inline image


On Tuesday, September 5, 2023 at 07:49:36 PM EDT, Puckdropper via groups.io <puckdropper@...> wrote:


I'm trying to avoid extra "stuff" because of space reasons, so I'm looking at putting two LEDs in series.? If I have two white LEDs* that operate on about 4V at 20 mA connected in series, using a 1 kOhm resistor at 12V, what rating does the resistor need to have?

Can I subtract the required voltage of the LEDS like this?
12V - 4V -4V = 4V
4V / 1000 ohm=.004A
.004A * 12V=.048W

Do I need to keep it in like this?
12V / 1000 ohm=.012A
.012A * 12V=.144W

*

Puckdropper
 

Use about 14 V for your starting voltage, and white LEDs drop about 3V each. So

14-6 = 8 V on the resistor.
I = V/R = 8/1000 = .008 A = 8 mA.
P = IV = .064 Watts.... Well below the rating of most resistors. So you can use a 1/8 W resistor. However, that means each LED will be functioning with 8 mA. This is plenty, as far as I'm concerned. In fact, I use LEDs with 1-2 mA going through them, all the time in my installations.? You can also do the LEDs in parallel.

...Tom


On Tue, Sep 5, 2023 at 5:10?PM PAUL HART via <psr436=[email protected]> wrote:


On Tuesday, September 5, 2023 at 08:00:47 PM EDT, PAUL HART via <psr436=[email protected]> wrote:


I have a program that I use and this is what it says to use. I hope this helps.?

Inline image


On Tuesday, September 5, 2023 at 07:49:36 PM EDT, Puckdropper via <puckdropper=[email protected]> wrote:


I'm trying to avoid extra "stuff" because of space reasons, so I'm looking at putting two LEDs in series.? If I have two white LEDs* that operate on about 4V at 20 mA connected in series, using a 1 kOhm resistor at 12V, what rating does the resistor need to have?

Can I subtract the required voltage of the LEDS like this?
12V - 4V -4V = 4V
4V / 1000 ohm=.004A
.004A * 12V=.048W

Do I need to keep it in like this?
12V / 1000 ohm=.012A
.012A * 12V=.144W

*

Puckdropper
 

Puckdropper, using two LEDs in series is good. Use a 1/4 watt resistor as it has more surface area to keep its operating temperature down. Even though the LEDs are diodes, you should protect them from significant reverse voltage. use a bridge rectifier or a single diode in series (1/2 wave rectification from DCC track voltage) to further reduce resistor watts with almost no noticeable dimming of the LED.

If you are thinking about lighting passenger cars, you will need to add a Cary over capacitor to minimize blinking.

DonV
 

Tom's numbers look good to me, although I've not measured the voltage drop across white LEDs.?

If each LED gets 8 mA of current and the voltage across it is 3V, then 24 mW is used in each LED to produce the light. Two LEDs would then use 48 mW. (mA is milliAmps: 1/1000 of an Amp, mW is milliWatts: 1/1000 of a Watt, V is Volts.)

If the LEDs are in parallel, to get 8 mA in each requires 16 mA total. If the input voltage is 14 V, then the total circuit power will be 224 mW: nearly 1/4 Watt. And, since only 48 mW is actually used by the LEDs, the remaining 176 mW must be dissipated as heat by the resistor. The resistor would need to be about 680 Ohms, and rated for 1/4 W or more.

Putting the LEDs in series raises the circuit efficiency significantly. The total current then is only 8 mA, and total power only about 112 mW. The LEDs use the same 48 mW as before. That leaves about 64 mW to be dissipated in the series resistor: There will be about 6 V across the two LEDs. That leaves only about 8V across the resistor, which would need to be about 1k Ohm (1000 Ohms). A resistor rated at 1/8 W would be sufficient, and smaller than the 1/4 W or higher power resistor needed if the LEDs were in parallel.

Most small LEDs are rated for maximum currents of 20 mA. Most are bright enough for our uses at much lower currents. The "standard" 1k Ohm resistors used for many onboard LEDs produce more light than needed in many instances, leading to decoders needing brightness or dimmer adjustments. But, they are handy if using an IC that is 7 or 9 resistors of the same value, with a common connection. I use separate discrete resistors, and vary their values depending on the brightness I need.

As Tom said, you probably won't want your LEDs as bright as they will be if they are operating on 10 mA or more, perhaps on far less.

Also, resistor power ratings depend on heat having an exit path. Many power resistors rely on some of their heat being carried away through their leads, converting the circuit boards they are mounted on into heatsinks. Try to provide a way for air to flow past any resistors operating at half or more of their rated power level, or have them mounted where there is a circuit board or metal surface that can help carry away the heat.

Don W

--
Don Weigt
Connecticut
 

... just saying ...
w
? Although calculated resistor sizes are "correct" (you won't blow the LED) - most LEDs we use
are FAR too bright (to my eyes)?and so running them at reduced current/voltage is better
(in my book).? That means I use a resistor that is larger than the calculated value - because I
don't want full brightness.

? When deciding what resistor to use for an LED in an HO scale model running on DCC - I
start with a 3k, 1/4watt resistor and go UP from there to get the brightness that I want.? The
LED will often still be lighting up with a 20k resistor!? Dimming is not 'linear' so I experiment
using a resistor box until I get the desired brightness.? That sounds like a lot of "bother"?
and it is ... I usually just start with a 3k or 3.3k (can't tell the difference with the LEDs I
use which are 0406) and only do the resistor box stuff when that doesn't work out.

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?- Jim in the PNW
 

I would add only that the brightness you want and need will depend on your room lighting. If it's bright, your LEDs will need to be brighter (use lower resistances and higher currents) for the same visibility than if your room is relatively dim. I find I need more light than when I was younger to see to rerail equipment and do other precision work. So, again, YMMV!????? :--)

Don W

--
Don Weigt
Connecticut
 

Some decoders have dimming capabilities. That could, given people have the right software and programming capabilities, even be programmed to change with the time of day, for anyone who wants to go *really* over the top...

Wouter

On Fri, 8 Sept 2023, 14:27 Don Weigt, <dweigt47@...> wrote:
I would add only that the brightness you want and need will depend on your room lighting. If it's bright, your LEDs will need to be brighter (use lower resistances and higher currents) for the same visibility than if your room is relatively dim. I find I need more light than when I was younger to see to rerail equipment and do other precision work. So, again, YMMV!????? :--)

Don W

--
Don Weigt
Connecticut

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