[bull]

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 20-23 volts.
Thx Jason

 

[George Schreyer]

about 1k, 1/2 watt

 

[Dan Pierce]

For LED's, when dropping a large voltage, it is very important to use the larger wattage resistor like George stated. 20 volts dropped at 20 milliamps gives .4 watts. The resistor wattage should always be more (I like double) than what is needed.

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.

 

[Greg Elmassian]

And if it's at 23 volts as specified by Bull and 20 milliamps then it's over 1/2 watt. A half watt resistor is very hot, normally will burn your fingers at the full 1/2 watt. I normally double the power to keep from melting plastic if the resistor comes in contact with anything.

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

Conservatively yours,

Greg

 

[bull]

Thx Guy's
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

 

[Bill Swindell]

The GP-40 has dual headlights. If you hook them in series with your 1K resistor, the current will be limited to 13ma. In that case, a 1/2 watt resistor will be fine. The LED's will not get hot and if you are worried at all, I would keep the resistor away from the plastic shell.

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.

 

[Greg Elmassian]

Bill, are you saying to hook the LEDS in series? A good idea, but how did you calculate the current drop? Did you assume a nominal voltage drop on the LEDs, and if so what?

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

 

[bull]

Thx again,
What I am wanting is to have dimmable headlights (rule 17). Also ditchlights will be seperate I want them to pulse with the horn. Useing 5mm high brightness white leds radio shak item 276-0017.

 

[Greg Elmassian]

Well, that data helps.

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

 

[George Schreyer]

15 mA will be fine, the LEDs get a little less efficient at 20 mA so that the change in brightness from 15 to 20 mA is not all that great. I would not try to push 25 mA though

 

[Bill Swindell]

The average voltage drop of a white LED is 3.5 volts. The question regards a 20 volt supply, not 23. I put LEDs in series all of the time on my battery installs. They work fine that way since you only need 1 limiting resistor and a lower wattage works fine.

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.

 

[Greg Elmassian]

Here is the post:


"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 20-23 volts.
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

 

[bull]

the fl4 has led abitlity. Ok so headlights wired together in series ditch lights have to be seperate though so they'll flash. So I should beware of the resistor going to the ditch lights getting hot should keep away from plastic. Thanks Guys

 

[Axel Tillmann]

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

 

[bull]

Hey Axel thx will pick some up tommorow. Hooked up the other ones 1k 1/4w and they did get quite hot.
Jason

 

[Greg Elmassian]

Hmmm...

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

 

[stanman]

FYI - The latest version of Handy Converter (just released yesterday) includes an LED Calculator. You enter the source voltage, the forward voltage and current for the LED plus the number of LEDs and you quickly get a circuit like this: LED Circuit Design

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

 

[Greg Elmassian]

Stan, since one interesting thing in this thread is a discussion about correctly sizing the wattage of a resistor to a circuit, I looked at your page:

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

 

[stanman]

Hi Greg, thanks for your comments.
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.

 

[Greg Elmassian]

Thanks Stan, yeah, my rule of thumb of doubling is pretty conservative! 60% sounds good.

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