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about 1k, 1/2 watt

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If only 15 milliamps, then it is .3 watts and a 1/2 watt resistor will suffice as it is close to double.

Failure to size the resistor wattage will result in plastic melting, or resistor burning up and a repair.

If you can not easily find a 1/2 watt resistor, then use double the resistor value and place 2 1/4 watt in parallel. In this case that would be 2.2k as 2.0 k will be hard to find.

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20,880 Posts

I believe Radio Shack has 1 watt 1,000 ohm resistors off the shelf.

Conservatively yours,

Greg

Digitrax website has the db150 at 20 volt output. Got resistors before I say other replies 1k ohm 1/2 watt. The leds will be in gp40 headlight houseings and ditch lights direct contact with plastic. Will this be ok or will it still get hot.

Jason

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658 Posts

If your ditch tights are on at the same time, you could also hook them in series with the same results. If the ditch lights are lit individually, your 1K resistor, 1/2 wattresistor will still.

In each case, the power dissipated via the resistor will still be no worse than about 1/4 watt so a 1/2 watt resistor would be just fine.

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I calculate a safe way, which works well with a single LED, and will be off a little with 2 in series.

To have 13 ma through a 1k resistor means that the voltage drop across the resistor is 13 volts. So are you assuming that the voltage drop across the 2 LEDs in series is 7 volts (notice we are ignoring the 23 volt max by the way!).

So that would assume a 3.5 volt drop across each resistor. That's reasonable for many white LEDs, but you have to be sure, since this would be wrong for 2 red LEDs. Just wanting to understand your method.

Regards, Greg

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Assume worst case: 23 volts, drop a volt through the decoder, 22 volts. 2 LEDs in series would drop 6.6 volts...

Now you need a resistor for 22-6.6 volts = 15.4 volts. The led's run 25 milliamps. R= V/ I 15.4 / 0.025 = 616 ohms. If you want to use a 1k resistor, then the current would be 15.4 / 1000 = 15.4 milliamps, might turn out dim.

At 15.4 milliamps, you have .237 watts, so a half watt 1k will be fine (in my book) because the resistor is at least double the actual wattage.

At the 25 milliamp rating, you would have .385 watts, so getting warm, but just keep the 1/2 watt resistor away from things.

After all that, my opinion would be to buy the 680 ohm 1/2 watt resistor from Radio Shack, would give you 15.4 / 680 = 23 milliamps. Should be as bright as the 25 milliamp rating of the LED.

Regards, Greg

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I did a dimming headlight with an FL-4 function decoder. I found that the provided dimming function was not as pronounced as I wanted so I made the limiting resistor for the dimming lead much higher in value, about 5K as I remember.

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20,880 Posts

"Hey All,

This may have been asked but I couldn't find it. What resistor rating do I need to use 5mm led's with the qsi aristo pnp, and a tcs fl4 function decoder? Useing digitax dcc at

Thx Jason"

It clearly says 20-23 volts.

I clearly said "assume worst case", sound engineering principal.

He gives the radio shack part number, which gives the specs on the typical forward voltage (3.3) and the current (25 ma).

I do not think any of this data is in error, or can be in question.

I completely agree with your idea of series, smart, and reduces the voltage drop needed by the resistor, and thus the power dissipation requirements.

Does the FL-4 have a "led vs incandescent" setting?

Good info on having to reduce the current a lot to see a difference on this particular decoder. More and more modern decoders have the "led vs incandescent" setting so dimming LEDs works fine while setting them for full brightness.

Regards, Greg

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657 Posts

Just FYI, I use and stock 2W resistors for that purpose. Don't cost much more, and keep cool.

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20,880 Posts

Ohms law is always right.... V=IR and Power = I**2 times R ..... really nothing to argue with

Oh well, I tried... hope you did not burn your fingers...

again, for everyone else: 1/2 watt in a 1/2 watt resistor is HOT! 1/2 watt in a 1/4 watt resistor will normally burn it up! My recommendation is to size your resistor at least double the calculated wattage.

Greg

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381 Posts

The program handles circuits from one to fifty LEDs, dividing them into several series strings if necessary.

If you want to look in your junk drawer for the specified resistor, you can enter the value in the Resistor Calculator to find its color code: Resistor Calculator

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20,880 Posts

I see this states the resistor is 1100 ohms and 1/4 watt and the current is 15 milliamps.

Well this means the voltage drop across the resistor only is: (from Ohms law) the current times the resistance, or .015 * 1100 or 16.5 volts.

With a total "input" voltage of 23 volts, this means the two leds have a voltage drop of 23-16.5 or 6.5 volts, or 3.25 volts each, which makes sense.

Power is P = I**2 R (derived from Ohms law and the fact that P = IV by definition of power in watts).

So the power in the resistor is equal to the current squared times the resistance.

0.015 * 0.015 * 1100 is .2475 ... or just shy of 1/4 watt.

Your picture indicates that the total power is 105 milliwatt.... I cannot work backwards to see anything in the circuit at 105 milliwatt... at 15 milliamps, you need to be dropping 7 volts to dissipate 105 mW... nothing in the circuit is at 7 volts, the 2 leds are 6.5 volts, the resistor is 16.5 volts....

So I'm guessing that your calculator just indicates the resistor "next up" in size.

I think there may be an error here, but check my work.

The only thing I was hoping to contribute in this thread that any resistor running at it's full rated wattage is HOT! Too hot in my opinion, and could melt or soften plastic if in contact, and will definitely burn your fingers.

So maybe it would be good (after finding the discrepancy) to recommend a wattage rating of double the actual wattage dissipated in the resistor.

Regards, Greg

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381 Posts

To be conservative I go to the next highest standard resistor if the calculation doesn't match a standard value perfectly. For example, if the calculalted result were 1000 ohms, that's what the program would use. However, if the calculation was 1001 ohms, I would go to the next highest standard value, which is 1100.

For wattage, I derate the resistor 60%. Not quite the double that you suggested, but conservative nevertheless.

I'll check my work on the other computation you mentioned. BTW, I assumed a 3.5 volt drop.

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I'm still confused about the numbers, but from what you said, if you had the program assume 3.5 volts per LED (close to the actual), then the 105 milliwatts for the 2 LEDs works out.

Was just confused about the term "total power".... I guess it may mean "total power of the LEDs"

Regards, Greg

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