# LED's , a lot of them



## tommyheadleycox (Oct 15, 2010)

Objective: Choose a power supply with enough current to light up a strip of 40 LED's. The LED specs are 3.3 volts 25 mA each. Leaving aside the dropping resistor calculations for the time being, am I correct in stating : 40 LEDS X .025 A each = 1 amp minimum is required from the power supply?

I foresee using these in passenger cars. BTW, I have read as many posts and threads here as I could find.

Thanks,
Tom


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## KeithRB (Sep 25, 2015)

Are they in series or parallel?


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## tommyheadleycox (Oct 15, 2010)

KeithRB said:


> Are they in series or parallel?


Actually, that's part of the reason I'm asking! Many articles recommend wiring LED's in series. But that uses more current than parallel wiring. I can use one of the online calculators to "draw a circuit" and it shows me how to do the whole thing, but I'm trying to UNDERSTAND the concepts, not just fill in the blanks. BTW, the online calculator, when I specify a 12v power supply and 40 LED's that are 3.3v / 25mA, replies as follows:

You will need 13 x 91 ohm 1/8 watt and 1 x 360 ohm 1/4 watt resistors.
The 91 ohm resistor is color coded: White, Brown, Black, Gold.
The 360 ohm resistor is color coded: Orange, Blue, Brown, Gold.
Each 91 ohm resistor consumes 56.875 milliwatt.
Each 360 ohm resistor consumes 225 milliwatt.
Total power consumed by the resistors is 964.375 milliwatt.
Total power consumed by the LEDs is 3300 milliwatt.
Total power consumed by the circuit is 4264.375 milliwatt.
Total current drawn by the circuit is 350 milliampere

Only 350 mA! But why?


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## Cmorais (Mar 11, 2013)

Hi Tom:

Since you specified a 12v power supply, you can link 3 LEDs in series: 3x3.3V=9.9V. You will need to drop the remainder 2.1V in a resistor placed in series with the three LEDs. Note that the current in this 3 LED string is 25 mA, the same current passes in succession through each LED. To drop 2.1 V with this 25mA current you need R=V/I = 2.1/.025=84ohm. Each 3 LED string and 84ohm resistor will draw 25 mA, and consume P=VxI=12 × 0.025 = 300 mW.

You can link in parallel up to 40 of these three LED and resistor strings to a 1A power supply (12W maximum power).

I don't know why the slight difference in resistor vales but they are irrelevant: the sets will workshop equaly well with 80, 90 or 100 ohms. 

You can easily calculate the effect of using different resistors. In terms of current, each LED appears as a R=V/I=3.3/0.025=132ohm resistor. So three LEDs will have a 132×3=396 ohm total resistance. If you add a 100 resistance you will have 496 ohm. The current through this string will be I=V/R=12/496=24 mA. A minimal reduction in LED current.

You can avoid using the resistors by linking four LEDS. In this case R=4×132=528ohm, so I=12/528=23mA, a quite acceptable and safe value. Try if it suits you.

Hope this helps

José Morais
Headmaster of the Lapa Furada RR


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## dbodnar (Jan 2, 2008)

Tom - you could likely put groups of 4 in series and have plenty of light from the LEDs.

You may want to look over this article on my web page as it deals with many of the things you are working with:

http://www.trainelectronics.com/LED_Articles_2007/LEDs_for_Coaches/index.htm

Please let me know if you have any questions

dave


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## BigRedOne (Dec 13, 2012)

Are you taking power from track power or battery? In either case, are you dealing with a certain voltage you'll have available?

Large scale model trains are generally 18 to 24 volts, though 12 volts could be a realistic track voltage running a variable voltage throttle at a moderately slow train speed. However, to protect yourself for future conversion to DCC, you might want to plan ahead for that with a voltage regulator. 

Having settled the voltage, I'd wire enough LED's in series to equal the voltage, then as many of those arrangements in parallel as is needed to provide the light output. For example, and assuming 20 volts at the track, six LED's in series equals the voltage available (3.3 x 6.) Then, seven arrangements of six are needed to give the output of 40 lights. Each strip draws 25 ma, so the total draw is 175 ma (25 x 7.) I think the resisters are unnecessary complexity.


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## Semper Vaporo (Jan 2, 2008)

tommyheadleycox said:


> Actually, that's part of the reason I'm asking! Many articles recommend wiring LED's in series. But that uses more current than parallel wiring. I can use one of the online calculators to "draw a circuit" and it shows me how to do the whole thing, but I'm trying to UNDERSTAND the concepts, not just fill in the blanks. BTW, the online calculator, when I specify a 12v power supply and 40 LED's that are 3.3v / 25mA, replies as follows:
> 
> You will need 13 x 91 ohm 1/8 watt and 1 x 360 ohm 1/4 watt resistors.
> The 91 ohm resistor is color coded: White, Brown, Black, Gold.
> ...



Wiring in Series does not take more Current. In Series you will have to increase the Voltage to cause the same amount of Current to flow. Wiring them in Parallel will take more Current from the power source, but the Voltage needed will remain the same. Either way, the Power (Wattage) will increase by the same amount.

If it takes 3.3Volts to push 25milliAmps through the LED to make it light, then;

If you connect 2 LEDS, in Series you will need to increase the Voltage to 6.6Volts to get that same 25milliAmps to flow. 

But if you wire them in Parallel, then the current will double to 50milliAmps (25milliAmps to each LED) being pushed by that same 3.3Volts.

Either way, you will consume twice the Power to light 2 LEDs... Power (Watts) equals Volts times Amps, so 

1 LED is 3.3V * 0.025A = 0.0825Watts

2 LEDs is either
in Series: 6.6V * 0.025A = 0.165Watts
or
in Parallel: 3.3V * 0-025A + 0.025A = 3.3V * 0.05A = 0.165Watts

The Resistors in your examples are there to consume power such that the LEDs are not subject to more power than they are designed to dissipate. i.e.: if you apply 6.6Volts across one LED then that will attempt to push 50milliAmps through the device (or cause it to handle .165Watts and that may cause it to lose all its magic smoke and quit working.

How are the 40 LEDs wired in your example... if Parallel then the Current would need to be constrained to just 25millAmps each, so a large resistor would be required to drop the 12Volts down to just 3.3Volts for all of them. If they are in Series then you would need more than 12Volts to push 25milliAmps through them (3.3Volts EACH; thus 3.3V*40 = 132Volts).

Either way All 40 LEDs will consume 3.3V*25milliAmps*40 = 3.3Watts.

The values of the resistors will depend on where they are in the circuit and how much voltage they have to drop at the current they are required to pass. I could probably figure out your wiring pattern given the info above, but I need to go someplace and don't have the time now. I'd bet you have 13 groups of 3 LEDS in Series with a Series resistor in each group and another one LED by itself with a single resistor for it, and maybe one other resistor in Series with the whole circuit.


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## Semper Vaporo (Jan 2, 2008)

Oh, and if you add up my guess at 13 strings of Series LEDs (0.025A each) and add one more single LED in parallel you get 13 * 0.025 +0.025 = 0.350Amps.


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## tommyheadleycox (Oct 15, 2010)

Thanks everybody. NOW I get it! You all explained things clearly. BTW - here are some things y'all said that helped me a lot and will help others.
"Having settled the voltage, I'd wire enough LED's in series to equal the voltage, then as many of those arrangements in parallel as is needed to provide the light output." (Thanks, Dave)

"Wiring in Series does not take more Current. In Series you will have to increase the Voltage to cause the same amount of Current to flow. Wiring them in Parallel will take more Current from the power source, but the Voltage needed will remain the same. Either way, the Power (Wattage) will increase by the same amount." (Thanks, Semper. BTW - the "design a circuit" application drew a circuit EXACTLY as you described)

And thanks to everybody else too! Now I can sit down with a pencil and paper and calculate any scenario.
Tom


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## Dan Pierce (Jan 2, 2008)

Where are you getting the 12 volts from???
If a regulated supply then the calculations above are great, but if an unregulated wall type supply then the voltage can be as high as 15 volts and goes close to 12 volts only at a full load!!!!!
SO, make the resistor larger, if 3 leds in series then the 84 ohm can be 120 or 1even 150 ohm. A little less current will hardly make a brightness difference.


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## Paul Burch (Jan 2, 2008)

I would set up a test circuit and play with the resistance some. You may find that you don't need to drive the leds at full current. Sometimes they are too bright at full power.


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## MyMiniatureWorlds (May 3, 2015)

Paul Burch said:


> I would set up a test circuit and play with the resistance some. You may find that you don't need to drive the leds at full current. Sometimes they are too bright at full power.


Experimenting is definitely the way to go. They will be too bright at full power, but it's also very easy to make them too dim (and lose the "effect").

See my post here: http://my-miniature-worlds.blogspot.com/2015/09/interior-lighting-tuning.html


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## CliffyJ (Apr 29, 2009)

You also might want to consider the pre-wired tape,
http://www.amazon.com/gp/product/B006079BCK

Every 3rd led, you can cut it; and there are solder pads there, ready to hook up to 12vdc supply. The resistors are baked in.

Cliff


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## Greg Elmassian (Jan 3, 2008)

Agreed, MUCH easier, and cheap, available in different whites, etc.

Greg


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## dpotp (May 25, 2011)

Those lights are what I've been looking for. Has anyone used these? How reliable are they and what color would be best for passenger cars. I have a hard time with colors since I'm so color blind.


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## Greg Elmassian (Jan 3, 2008)

they work well, are inexpensive, and you can get them in different temperatures.

2700k is like incandescent lighting... the higher the number the whiter/more blue the color.

cool white flourscents are about 4100k for reference, 6000k is like halogen.

Greg


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## dpotp (May 25, 2011)

Thanks Greg.


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## East Broad Top (Dec 29, 2007)

Here's a chart showing color temperatures:










Later,

K


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## CliffyJ (Apr 29, 2009)

The warm white I cited is 3000k, but here's a 2700k:

http://www.amazon.com/dp/B002Q907EW

You can also get amber color:

http://www.amazon.com/dp/B00T1DL9NE


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## Naptowneng (Jun 14, 2010)

I just used them for coach lighting. Pleased so far. Using a23 battery under car. Batt life TBD

Jerry


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## Greg Elmassian (Jan 3, 2008)

Normally warm white is defined as 2700k, but I do see 3000 and 3100 labelled warm white... I have tried a number of different temperatures in my house, and learned a bunch when I tried to keep incandescents, flourscents and halogens the same temperature...

Greg


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