RDG1519 So for example. The new BLMA Dwarf Signals specifies no more than 2.0 to 2.2 volts and 25 to 35 mA resistor value = (supply voltage 5 volts - 2.1 volts) / .015 amps (conservative) = 193 ohms so chose maybe a 220 ohm resistor? If too dim decrease resistance (increases current flow) Let me know if I have this right. Thanks for your clarifications. Chris
So for example.
The new BLMA Dwarf Signals specifies no more than 2.0 to 2.2 volts and 25 to 35 mA
resistor value = (supply voltage 5 volts - 2.1 volts) / .015 amps (conservative)
= 193 ohms so chose maybe a 220 ohm resistor? If too dim decrease resistance (increases current flow)
Let me know if I have this right. Thanks for your clarifications.
Chris
That's a darn good starting point. :-)
Don - Specializing in layout DC->DCC conversions
Modeling C&O transition era and steel industries There's Nothing Like Big Steam!
rrinker Safest thing is to still use a resistor with the pot - that way if you accidently turn it to 0, you don;t fry the LED because you hvae a minimal fixed resistence still in the circuit. Does mean you need to do math - add the measured setting of the pot to the fixed resistor. Also all the pots and decade boxes don't mean a hill of beans if you aren't driving the thing with the same voltage that the LED will see in the final circuit. --Randy
Safest thing is to still use a resistor with the pot - that way if you accidently turn it to 0, you don;t fry the LED because you hvae a minimal fixed resistence still in the circuit. Does mean you need to do math - add the measured setting of the pot to the fixed resistor.
Also all the pots and decade boxes don't mean a hill of beans if you aren't driving the thing with the same voltage that the LED will see in the final circuit.
--Randy
Randy has a good point. Most questions revolve around DCC which is usually around 12 VDC. Until his message, I had forgot about that.
Some years ago I built a LED signal system using TTL chips which operate at 5VDC.
Having started using LED's in 1972 I knew I could use a 1k, 10 turn pot set at max resistance to find a standard value resistor. Ten turn pots give much better control. Majority of LED's are 20 ma though I do have 2 ma LED's for special battery operated projects.
Rich
If you ever fall over in public, pick yourself up and say “sorry it’s been a while since I inhabited a body.” And just walk away.
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
I have used the pot and ohm meter method also. Some make their own decade resistance box with rotary switch and a bunch of resistors.
If anyone works at the component level they really need a multimeter, set of clip leads and a bunch of resistors.
Easier method is probably using math and having various resistor.
Many are not aware you can find all kinds of calculators online.
I have been doing that for some years. You can even keep the links in your smartphone or iPad.
Always start high and work down, if you start too low you will blow the LED. Also I really don't like that all these articles start with recommending using 20ma in the calculation. Unless you want an intensely bright LED, and/or one on the verge of emulating a flashbulb, 20ma is too high - it's the upper limit for many white LEDs. The rating on the LED's data sheet is the MAXIMUM it can handle, not a nominal value, or how much it always draws - LEDs don;t work like that. When an incandescent bulb says it is 15ma, it draws 15ma, give or take. The fixed value on an LED is how much voltage it drops, when the LED says 3.5 volts, it will drop 3.5 volts across it. The current is what varies.
For example, most of the time for HO DCC I use a 1K resistor with golden white LEDs. This usually is not overly bright, because most of the time they shine through plastic light pipes and lenses which may look clear but actually attenuate a lot of the light - except Bowser, their plastic lenses seem exceptionally clear and I had to use higher value resistors. At standard HO voltages, with a typical 3.5V white LED, the use of a 1K resistor results in about 9ma through the LED. This is just under half the maximum of 20ma, and at least when viewed through the plastic lenses, results in a bright but not blinding light. The lower the current, the dimmer the LED, until you reach a point where there just isn't enough current to make it light up at all. So to calculate the resistor, pick a target current, but NOT 20ma, just because it "says that on the LED package".
I determine resistor value the same way Mel does. Start high and work down.
One suggestion I would make is to get a few resistors higher than the 3.3K he recommends. I just did a lighting project where I wanted the LEDs to glow like an early electrical bulb. I ended up using 30.1K resistors.
Dave
I'm just a dude with a bad back having a lot of fun with model trains, and finally building a layout!
PRRLionel Could someone, please, provide me with the reader's response?
Bob:
The reader recommends using a 20 mA current limiter circuit available from Hansen Hobbies so that regardless of the voltage the led will always see 20mA. I'm not familiar with these devices but with the variability in leds and the desire to control the brightness I don't see this as a universal answer. I still think that tailoring the resistor to the application is a better answer.
Joe
The way I used to do it, was to use a potentiometer wired in series with the LED and power supply. Turn the pot to it's highest resistance, turn on the power, then gradually turn the pot down until you reach the desired color of the light. Don't change the pot, remove it from the power and LED, and measure the ohm's of the pot. Then select the closest resistor value to that reading, and wire it into the circuit. This will also work with incandescent lamps.
The Oct. 2014 issue of MR, pp. 66-67, had an article entitled "Choosing headlight resistors." A subsequent issue had a reader's response that those extensive calculations to determine a suitable resistor were unnecessary because there was an alternate, easier way to provide the requisite resistance in the circuit, but I can not find that response. Could someone, please, provide me with the reader's response? TIA, Bob Blackson