Hi all,
I need to find a power supply for some Tomar signals I bought recently. Their instruction insert refers to a '12.5-volt power supply' that can be used with their own resistors (different resistance values for each LED color). I bought the resistors but now I'm wondering: Does the power supply's current rating matter? I see 12-volt units whose outputs are rated starting at 300mA up to 10A. Somebody on another forum recommended this one, a 300mA unit, but I'd have to special-order it and pay for shipping. Otherwise I could drive to my local Radio Shack and buy a 12v supply rated at 1.5A. Would that one fry my signals' LED's?
-Ken in Maryland (B&O modeler, former CSX modeler)
Dr. Frankendiesel aka Scott Running BearSpace Mouse for president!15 year veteran fire fighterCollector of Apple //e'sRunning Bear EnterprisesHistory Channel Club life member.beatus homo qui invenit sapientiam
I'm sure some sharper minds will jump in here and give you a better explanation, but the short answer is no, a 1.5 amp power supply won't fry an LED if you're using the proper resistor.
An LED needs to be protected against high current, and most of then have an optimum rating, tpyically something on the order of 3.3 volts at 20 mA. This represents one point on the operating curve of the LED and is usually close to optimum, but there's a balance between brightness and life expectancy and you need to consider. Higher resistor values generally reduce current and therefore brightness, but will extend the life of the LED. Conversely, lower resistor values will yield brighter LEDs with shorter lives. My experience has been that dimmer is better; the light looks more realistic and lasts longer, but, depending on how far you go, can be difficult to see.
There is a good online resistor calculator at http://led.linear1.org/1led.wiz. It will give you some resistor values to get you started. If in doubt, you can't usually go too far wrong with 470-ohm resistors.
Chris
EDIT: I should add that the output rating of the transformer is not critical; it's simply a maximum value that the transformer can supply. A 1.5 amp power supply could drive 60 LEDs at 20 mA each, with a 20% buffer.
If the Tomar signals you have are LEDs, you don't need 12 Volts. The data sheet that comes with the signals indicates that each one draws only 2.2 to 2.4 Volts at 20 or 30 mA, depending on which LED is lit, with green being the highest. In order to determine the Amp rating you need for your power supply, multiply the number of signals by the mA rating for the green LED.
A 1 Amp power supply could theoretically power 33 green LEDs, but there will be some power loss in the wiring and resistors; and the power supply needs to provide only 2.4 Volts max; so a 5 Volt, 1 Amp DC power supply would be more than sufficient.
With a 12 Volt power supply, you're just burning up more power as heat in the resistors.
Jeff/Chris/Charles - Thanks for replying!
The Tomar data sheet confuses me a bit, because of how they worded this statement:
"If you have a 12.5volt DC power supply, you can use these 1/2 watt resistors ..."
...then they list the resistance rating for each color and have item numbers for each of those value resistors. I thought the chart at the bottom showed what I'd get if I connected their resistors and LEDs to a 12.5v power supply.
So I guess the mA rating refers to the capacity of the p/s (sort of like the amp ratings of various DCC command stations)?
The power supply needs to be equal to or preferably greater than the load placed on it. "Load" being whatever is connected - in your case the LEDs in the Tomar signals. If you have, for example, 10 LEDs that will be lit at one time, each drawing 20ma, you need at LEAST a 200ma power supply, although it would be loaded to the maximum. a 350 to 500ma power supply would be a better choice in this example. If you plan to add more signals later - you would need something even bigger. DO NOT EVER leave out the resistors with the LEDs - they will be destroyed in short order (like, seconds or less - we're not talking, well, they'll work for a couple of years without, or forever with). It's NOT optional. So long as each LEDs has it's proper resistor, and the power suppyl VOLTAGE is correct for the resistor and LED combo, the current rating of the power supply can be anything that's above the total of all the LED/resistor current draws - so a 50 amp power supply would work fine - the LED will use only as much as it needs, you won;t be pushing 50 amps through the LED.
This is probably one of the most misunderstood concept when it comes to electricity. You have something that draws 1 amp total, it WON'T take in 10 amps from a 10 amp power supply and be fried - if it draws at most 1 amp it will use at most 1 amp. And if each device draws a maximum of 1 amp, you can connect at most 10 of them to a 10 amp power supply
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
That pretty much confirms what I was thinking, Randy - thanks!
OK, I went and bought myself a 12v, 500mA power supply. Now I have another "noob" electrical question: Since both of the DC output wires are the same color, how do I figure out which one of them is '+' and which one is '-' without accidentally frying my LEDs? Can a voltmeter be used for this?
Here is a closeup of the adapter, neither of the two holes on the end are labeled:
Do you have a meter? If so it will be easy to tell. Or, as long as you have the proper resistor to use the LED with 12 volts (1K is usually a safe bet), you can tell with an LED (with resistor). yeah I'll say it again - DO NOT hook up an LED without a resistor! Cut off that connector on the power supply unless you have the mating side, so you have wires to connect to. Put it like this:
<one side of power supply>---resistor---short leg of LED---(O)---long leg of LED----<other side of power supply>.
If the LED lights up, the side with the resistor is the negative lead of the power supply. If not, reverse the power supply wires and the LED should light up - the side with the resistor is the negative side of the power supply. If the LED leads are both the same length - the one that would be the 'short' lead should have a flat in the base next to it.
rrinker wrote: Do you have a meter? If so it will be easy to tell. ...
Do you have a meter? If so it will be easy to tell. ...
Yep, you were right - I found my old V-O-M (where the battery was, amazingly, still charged) and saw a distinct 'minus' sign when I placed the leads one way as opposed to the other. That +/- mystery is solved.
But one part of the Tomar instruction sheet is still confusing me: Their wiring diagram, shown in the embedded image, has a resistor attached to the ground (black) wire - leading me to interpret that there is supposed to be a resistor on either side of the load. Is this a misprint or what? Maybe it's to protect the diodes in case I accidentally wired it backwards; but they didn't even specify a value for it as they did for the others.
I realize this seems like a dumb-@$$ question, but given my limited exposure to electronics, I'd rather not risk frying a $40 B&O c-p light signal.
I don't know why Tomar shows a resistor on the black lead. I'm installing some now and am putting resistors only on the R/Y/G leads after computing the necessary values for a 5 Volt DC power supply. The black lead is the positive voltage input for the LEDs they use in their signals (note the + mark).
That wire on the ones I am using is white instead of black, but is for the same purpose. Mine are 3 position signals and don't have the lead marked "Marker" on their diagram, but they show a resistor on the white wire, which is unnecessary. None of the text explains why they show a resistor on the + line.
Those questions are not stupid or dumb. There are those of us constantly seeking out the Grail of Knowledge regarding wiring.
Get a set of resistors from a store. This set should have high ohms at the top and small ohms at the bottom. As you decrease (Work down the list) ohms You will find a resistor value that will burn the bulb just bright enough for your taste. That will be your resistor of choice for that light.
If you have one resistor at a ohms value that burns too dim of a light and the next ohms value too bight, then you will need a intermediate resistor with greater wattage.
Resistors are supposed to dump excess electricity that otherwise will burn out lights.
I have just about exhausted my knowledge here. Even then this is book learning from long ago during a failed electronics class. My instructor was beside himself because I could not do math then.
My DS 64 pulls off a 300 ma PS12 (12 volt) power supply. I can feed a DS64 with this safely, but cannot feed many of DS64's because the load provided by the power pack is spread too thin.
However, I dont dare put a 5 amp supply onto that poor ds64 because it will get fried.
On the relationship of Voltage and Current:
Voltage is the PUSH element. If you have a light bulb that takes 12 volts, using a higher voltage will damage it. You are pushing more voltage on the device than it is designed for.
Current on the other hand is the DRAW element. That is, if you have something that needs 20ma of current to work, and you have a power supply that has the correct voltage and a higher current output, you are OK. You CAN NOT PUSH CURRENT. If your 20ma device shorts out, that will try to use, or draw, all the current in the power supply and more. That is one reason most devices should have fuses, or at least the output line of the power supply should.
Resistors limit the current. But at the same time, when you use a resistor to limit the current, you also reduce the voltage to the device. That is why you can use a resistor to reduce the current for an LED, and also to reduce the voltage to a light bulb.
LED's are current dependant devices, and light bulbs are voltage dependant devices.
Elmer.
The above is my opinion, from an active and experienced Model Railroader in N scale and HO since 1961.
(Modeling Freelance, Eastern US, HO scale, in 1962, with NCE DCC for locomotive control and a stand alone LocoNet for block detection and signals.) http://waynes-trains.com/ at home, and N scale at the Club.
I have an NJ International 2-light SA-type signal that has three leads: one white lead from each LED and one green from the base. Each wire has a resistor soldered to it. All LEDs are bipolar.
I guess the theory is that, since the body of the signal acts as one of the leads, any other power supply that comes into contact with the mast could theoretically short out the LED if there was no resistor in the line to protect it. Consider it a safety feature.
If you want to do this to the Tomar signal (probably a good idea if the mast is used as a common return), use resistors with half of the value. Total resistance through the LED will be the same, since resistors in series are cumulative. This appears to be what NJ International did.