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Model RR, LEDs on variable ac input voltage Answered

Led's for ac powered model railroading, to replace incandescent bulbs.  Voltage to track varies from 5 to 17 volts AC.  Leds will light up, but what is it doing to them. They do not pop like a flash bulb at top voltage. I've rectified some to dc voltage for lights in cars, the variable voltage seems to power them.

Ideally a circuit that put out a constant 3.2 DC volts, with a variable 5 to 17 volt AC input.  Or maybe would a constant 3.2 volt AC out would function?

I found that with HO DCC control, a white led with a 1k resistor will function, have not ran them for more than few hours though.

Prewired rectified leds go for around $3.00  each, 5 passenger cars, 2 bulbs each, gets pricey with 4 sets of cars.

Surly someone has gone through this.

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icengBest Answer (author)2014-03-12

A simple circuit can light one or two or three LEDs in series

keep C1 near the TO-92 LM317 current regulator.

The only drawback three LEDs will need over 6.7 VAC to light them up.

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31cssinc (author)iceng2014-03-13

Thank you,

Best bet would be to run 2 leds off one lm317, add another lm317 and 2 leds to get 4 in each car, they will turn at low voltage and they do not care about higher voltage with increased track voltage?

The cap and bridge will smooth out input fluxes from track splices and such?

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iceng (author)31cssinc2014-03-21

Thank you for the best answer.

You are intuitively correct, placing the cap next to the LM317 allows the regulator to function as though it was next to the source of power.

The 317 is the fastest regulator in response to input voltage variations of all the regulators on the market and as such can even be used on a PWM train without any undesirable output changes.

That bridge lines up the correct polarity independent of direction of travel or placement.

A

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Josehf Murchison (author)2014-03-12

A LED is a diode and it makes its own DC current, now at 50 to 60 Hz it lights up and goes out faster than you can see that is why moving film works.

As long as you have the right current passing through the LED it will light up.

The voltage drop across the diode will always be the same no matter the source as long as the current is right for the diode.

What that means is you can light up a 3.1 volt 20 ma LED with a 1000 volt source as long as the current doesn't excide 20 ma.

The reason you can do this is the voltage drop across the lead resistor is 996.9 volts and the voltage drop across the LED is 3.1 making 1000 volts.

All it needs to manage is its 3.1 volts.

So if you have a problem of your LEDs diming add a rectifier and a large capacitor to the power input to the LEDs supply and a current regulator to the output.

So if you are using a LM317 set for 20 ma current you need the ripple voltage to be above 2 volts above what you need for the LED.

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user

What they don't like is high reverse voltages, and when the diode is not conducting the reverse voltage = supply voltage= dead diode.

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user

That is why I said add a rectifier.

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steveastrouk (author)2014-03-12

An LED wants a constant CURRENT supply, not a constant voltage. If you want a circuit for a good way to get an adjustable current, I'll post one.

The simplest way to run an LED on AC is to put a resistor in series with it, and either another LED in parallel, backwards across it, or an ordinary diode.

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31cssinc (author)steveastrouk2014-03-13

Hello,
Thank you for help. I'm just having a hard time wrapping my brain around I= V/R.
The AC voltage to Model RR tracks is varied between 5-18 volts,
( high end depends on what each transformer is outputting).

As V goes up does the I not go down?

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steveastrouk (author)31cssinc2014-03-13

No, the I goes up !

I=V/R. Make V bigger, if R is the same , I gets bigger.

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31cssinc (author)steveastrouk2014-03-13

Just when I thought old age is the golden years, was thinking this all backwards, thanks.

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